Maths Previous Years Question Papers

MATHEMATICS Examination Papers 2008–2012

CONTENT n

CBSE Examination Paper–2008 (Delhi)

3

n

CBSE Examination Paper–2008 (All India)

32

n


67

n


93

n

CBSE Examination Paper–2009 (Foreign)

119

n


147

n


179

n


211

n


241

n


273

n


303

n


335

n


368

n


400

EXAMINATION PAPERS – 2008 MATHEMATICS CBSE (Delhi) CLASS – XII Time allowed: 3 hours

Maximum marks: 100

General Instructions: 1. All questions are compulsory. 2. The question paper consists of 29 questions divided into three sections-A, B and C. Section A comprises of 10 questions of one mark each, Section B comprises of 12 questions of four marks each and Section C comprises of 7 questions of six marks each. 3. All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question. 4. There is no overall choice. However, internal choice has been provided in 4 questions of four marks each and 2 questions of six marks each. You have to attempt only one of the alternatives in all such questions. 5. Use of calculators is not permitted.

Set–I SECTION–A 1. If f ( x) = x + 7 and g( x) = x - 7, x Î R, find ( fog) (7) 1 ù ép 2. Evaluate : sin - sin - 1 æç - ö÷ êë 3 è 2 ø úû é 1 3ù é y 0ù é5 6ù 3. Find the value of x and y if : 2 ê ú+ê ú=ê ú ë 0 x û ë 1 2û ë1 8û a + ib c + id 4. Evaluate: - c + id a - ib 2 -3

5

5. Find the cofactor of a 12 in the following: 6

0

4

1

5

-7

6. Evaluate: ò

x2 1 + x3

dx

1

dx

0

1 + x2

7. Evaluate: ò

®

8. Find a unit vector in the direction of a = 3i$ - 2j$ + 6k$

4

Xam idea Mathematics – XII ®

®

9. Find the angle between the vectors a = i$ - j$ + k$ and b = i$ + j$ - k$ ®

®

10. For what value of l are the vectors a = 2i$ + lj$ + k$ and b = i$ - 2j$ + 3k$ perpendicular to each other?

SECTION–B 11.

(i) Is the binary operation defined on set N, given by a * b =

a+b 2

for all a, b Î N, commutative?

(ii) Is the above binary operation associative? 12. Prove the following: 1 1 1 1 p tan - 1 + tan - 1 + tan - 1 + tan - 1 = 3 5 7 8 4 é 3 2 5ù 13. Let A = ê 4 1 3ú . ê ú êë 0 6 7 úû Express A as sum of two matrices such that one is symmetric and the other is skew symmetric. OR 1 2 2 If A = 2 1 2 , verify that A 2 - 4A - 5I = 0 2 2 1 14. For what value of k is the following function continuous at x = 2? ì 2x + 1 ; x < 2 ï f ( x) = í k ;x=2 ï 3x - 1 ; x > 2 î 15. Differentiate the following with respect to x : tan -

1

æ 1 + x - 1 - xö çç ÷÷ è 1 + x + 1 - xø

16. Find the equation of tangent to the curve x = sin 3t , y = cos 2t at t = p

x sin x

0

1 + cos 2 x

17. Evaluate: ò

dx

18. Solve the following differential equation: ( x 2 - y 2 ) dx + 2xy dy = 0 given that y = 1 when x = 1 OR Solve the following differential equation: dy x ( 2y - x) , if y = 1 when x = 1 = dx x( 2y + x)

p 4

5

Examination Papers – 2008

19. Solve the following differential equation : cos 2 x ®

®

dy + y = tanx dx

®

®

®

®

® ®

20. If a = i$ + j$ + k$ and b = j$ - k$, find a vector c such that a ´ c = b and a . c = 3. OR ®

®

®

®

®

®

®

®

If a + b + c = 0 and| a | = 3,| b | = 5 and| c | = 7, show that the angle between a and b is 60°. 21. Find the shortest distance between the following lines : x- 3 y -5 z -7 x+1 y +1 z +1 and = = = = 1 -2 1 7 -6 1 x+ 2

y+1

OR z- 3

at a distance 3 2 from the point (1, 2, 3). = = 3 2 2 22. A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability distribution of number of successes. Find the point on the line

SECTION–C 23. Using properties of determinants, prove the following : a b g a2

b2

b+g

b

2

b

b+g

= (a - b) (b - g ) ( g - a ) (a + b + g )

g +a a +b

a a

g2

2

g g

2

g +a a +b

a = (a + b + g ) a

2

1

b

g

2

g2

1

1

b

24. Show that the rectangle of maximum area that can be inscribed in a circle is a square. OR Show that the height of the cylinder of maximum volume that can be inscribed in a cone of 1 height h is h. 3 25. Using integration find the area of the region bounded by the parabola y 2 = 4x and the circle 4x 2 + 4y 2 = 9. a

a-x

-a

a+x

26. Evaluate: ò

dx

27. Find the equation of the plane passing through the point ( - 1, - 1, 2) and perpendicular to each of the following planes: 2x + 3y - 3z = 2 and 5x - 4y + z = 6 OR Find the equation of the plane passing through the points ( 3, 4, 1) and ( 0, 1, 0) and parallel to x+ 3 y- 3 z-2 the line = = 2 7 5

6

Xam idea Mathematics – XII

28. A factory owner purchases two types of machines, A and B for his factory. The requirements and the limitations for the machines are as follows : Machine

Area occupied 2

A

1000 m

B

1200 m 2

Labour force

Daily output (in units)

12 men

60

8 men

40

He has maximum area of 9000 m2 available, and 72 skilled labourers who can operate both the machines. How many machines of each type should he buy to maximise the daily output? 29. An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accident involving a scooter, a car and a truck are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver.

Set–II Only those questions, not included in Set I, are given p 20. Solve for x : tan - 1 ( 2x) + tan - 1( 3x) = . 4 x tan x p 21. Evaluate: ò dx. 0 sec x cosec x æ1 dy 1 ö 22. If y = x 2 + 1 - log ç + 1 + ÷, find . 2 ø dx èx x 23. Using properties of determinants, prove the following : 1 + a2 - b 2

2ab

2ab

2

1- a +b - 2a

2b p

x sin x

0

1 + cos 2 x

24. Evaluate: ò

- 2b 2

= (1 + a 2 + b 2 ) 3 .

2a 2

1- a -b

2

dx.

25. Using integration, find the area of the region enclosed between the circles x 2 + y 2 = 4 and ( x - 2) 2 + y 2 = 4.

Set–III Only those questions, not included in Set I and Set II, are given. æx-1ö -1 æ x +1 ö p 20. Solve for x : tan -1 ç ÷ + tan ç ÷= è x - 2ø è x + 2ø 4 é 1 + sin x + 1 - sin x ù dy ú, find 21. If y = cot -1 ê dx ê 1 + sin x - 1 - sin x ú ë û 1

22. Evaluate: ò cot -1 [1 - x + x 2 ] dx 0

7


23. Using properties of determinants, prove the following : a + b + 2c a b = 2 ( a + b + c) 3

c

b + c + 2a

b

c

a

c + a + 2b

24. Using integration, find the area lying above x-axis and included between the circle x 2 + y 2 = 8x and the parabola y 2 = 4x. x tan x p 25. Using properties of definite integrals, evaluate the following: ò dx 0 sec x + tan x

SOLUTIONS Set–I SECTION–A 1. Given f ( x) = x + 7 and g( x) = x - 7, x Î R fog( x) = f ( g( x)) = g ( x) + 7 = ( x - 7) + 7 = x Þ ( fog) (7) = 7. 1 ù p ép é p æ p öù 2. sin - sin - 1 æç - ö÷ = sin - = sin = 1 êë 3 êë 3 çè 6 ÷ø úû è 2 ø úû 2 1 3 y 0 5 6 é ù é ù é ù 3. 2ê ú + ê 1 2ú = ê1 8ú 0 x ë û ë û ë û Þ Þ

4.

é 2 6 ù é y 0ù é5 ê 0 2xú + ê 1 2ú = ê1 ë û ë û ë 6 ù é5 é2 + y = ê 1 2x + 2úû êë1 ë

6ù 8úû 6ù 8úû

Comparing both matrices 2 + y = 5 and 2x + 2 = 8 Þ y = 3 and 2x = 6 Þ x = 3, y = 3. a + ib c + id - c + id a - ib = ( a + ib) ( a - ib) - ( c + id) ( - c + id) = [a 2 - i 2 b 2 ] - [i 2 d 2 - c 2 ] = (a 2 + b 2 ) - (- d 2 - c 2 ) = a2 + b 2 + c 2 + d2

5. Minor of a 12 is M 12 =

6 4 = - 42 - 4 = - 46 1 -7

8


Cofactor C 12 = ( - 1) 1 + 2 M 12 = ( - 1) 3 ( - 46) = 46 6. Let I = ò

x2 1 + x3

dx

Putting 1 + x 3 = t Þ 3x 2 dx = dt or

dt 3 1 dt 1 I = ò = log|t| + C 3 t 3 1 = log|1 + x 3 | + C 3

x 2 dx =

\

7.

1

ò0

dx 1 + x2 = tan - 1 x

1 0

= tan - 1 (1) - tan - 1 ( 0)

p p = -0= . 4 4 8.

®

a = 3i$ - 2j$ + 6k$ ®

®

a

Unit vector in direction of a =

®

| a| =

9.

®

a = i$ - j$ + k$

®

b = i$ + j$ - k$ ® ®

3 2 + ( - 2) 2 + 6 2

=

1 $ ( 3i - 2j$ + 6k$) 7

®

| a | = 1 2 + ( - 1) 2 + 1 2 =

Þ

3

®

| b | = (1) 2 + (1) 2 + ( - 1) 2 =

Þ ®

3i$ - 2j$ + 6k$

3

®

a . b =| a || b |cos q

Þ

1-1-1=

3 . 3 cos q 1 cos q = 3

Þ 10.

®

Þ

- 1 = 3 cos q 1 q = cos - 1 æç - ö÷ è 3ø

Þ

®

a and b are perpendicular if ® ®

a.b =0

Þ ( 2i$ + lj$ + k$) . (i$ - 2j$ + 3k$) = 0 Þ 2 - 2l + 3 = 0

Þ

l=

5 . 2

9


SECTION–B 11. (i) Given N be the set a+b a *b = " a, b Î N 2 To find * is commutative or not. a+b b + a Now, a * b = = \ (addition is commulative on N) 2 2 = b*a So a *b = b * a \ * is commutative. (ii) To find a * (b * c) = ( a * b) * c or not æb + c ö a+ ç ÷ è 2 ø 2a + b + c æb + c ö Now a * (b * c) = a * ç = ÷= è 2 ø 2 4 a+b æa + bö ( a * b) * c = ç ÷ *c = è 2 ø =

2

...(i)

+c

2

a + b +2c

...(ii)

4

From (i) and (ii) ( a * b) * c ¹ a * (b * c) Hence the operation is not associative. 1 1 1 1 12. L.H.S. = tan - 1 + tan - 1 + tan - 1 + tan - 1 3 5 7 8 1 1 1 1 + + = tan - 1 3 5 + tan - 1 7 8 1 1 1 1 1- ´ 1- ´ 3 5 7 8 8 15 = tan - 1 + tan - 1 14 55 4 3 + -1 4 -1 3 -1 7 11 = tan + tan = tan 4 3 7 11 1- ´ 7 11 65 p -1 - 1 65 -1 = tan = tan = tan 1= 77 - 12 65 4

= R.H.S

13. We know that any matrix can be expressed as the sum of symmetric and skew symmetric. 1 1 So, A = ( A T + A) + ( A - A T ) 2 2

10


or A = P + Q where P is symmetric matrix and Q skew symmetric matrix. ì é 3 2 5 ù é 3 4 0ùü 1 1 ïê ï T P = ( A + A ) = í 4 1 3ú + ê 2 1 6úý ú ê ú 2 2 ïê ï î ëê 0 6 7 ûú ëê 5 3 7 ûúþ é3 3 5ù 6 6 5 2ú é ù ê ê 1ê 9ú ú = 6 2 9 = ê3 1 ú 2ê 2ú ú êë 5 9 14úû ê 5 9 7ú ê ë2 2 û 1 Q = (A - A T ) 2 0 ùü ìé 3 2 5 ù é - 3 - 4 1 ïê ï ú ê = í 4 1 3 + - 2 - 1 - 6úý ú ê ú 2 ïê ï î êë 0 6 7 úû êë - 5 - 3 - 7 úûþ 5 ùü é 0 -2 ï ê 2 0 - 3úý ê ú 3 0 úûïþ êë - 5 5ù é 0 -1 ê 2ú ê 3ú =ê 1 0 - ú 2 ê 5 ú 3 0ú ê2 ë 2 û ì 1ï = í 2ï î

OR 1 2 2 A= 2 1 2 2 2 1 \ \

A2 = A ´ A é1 ´ 1 + 2 ´ 2 + 2 ´ 2 1 ´ 2 + 2 ´ 1 + 2 ´ 2 1 ´ 2 + 2 ´ 2 + 2 ´ 1ù = ê 2 ´ 1 + 1 ´ 2 + 2 ´ 2 2 ´ 2 + 1 ´ 1 + 2 ´ 2 2 ´ 2 + 1 ´ 2 + 2 ´ 1ú ê ú êë 2 ´ 1 + 2 ´ 2 + 1 ´ 2 2 ´ 2 + 2 ´ 1 + 1 ´ 2 2 ´ 2 + 2 ´ 2 + 1 ´ 1úû é 9 8 8ù = ê 8 9 8ú ê ú êë 8 8 9úû 0 0 ù é5 0 0ù é 4 8 8ù é5 ´ 1 4A = ê 8 4 8ú and 5I = ê 0 5´1 0 ú = ê 0 5 0ú ê ú ê ú ê ú 0 5 ´ 1úû êë 0 0 5 úû êë 8 8 4úû êë 0

11


8-8 8 - 8 ù é 0 0 0ù é9 - 4 - 5 ê A - 4A - 5I = 8 - 8 9- 4-5 8 - 8 ú = ê 0 0 0ú. ê ú ê ú 8-8 9 - 4 - 5úû êë 0 0 0úû êë 8 - 8 2

14. For continuity of the function at x = 2 lim f ( 2 - h) = f ( 2) = lim f ( 2 + h) h®0

h®0

Now, f ( 2 - h) = 2 ( 2 - h) + 1 = 5 - 2h \ lim f ( 2 - h) = 5 h®0

Also, f ( 2 + h) = 3( 2 + h) - 1 = 5 + 3h lim f ( 2 + h) = 5 h®0

So, for continuity f(2) = 5. \ k = 5. æ 1 + x - 1 - xö ÷÷ = y 15. Let tan - 1 çç è 1 + x + 1 - xø

y = tan -1

æ ç1 ç ç ç1 + ç è

1-xö ÷ 1+x÷ ÷ 1-x ÷ 1 + x ÷ø

Þ y = tan - 1 1 - tan dy =0dx

1

æ 1 - xö çç ÷÷ è 1 + xø

1 2

.

d dx

æ 1- xö çç ÷÷ è 1+ xø

æ 1- xö ÷÷ 1 +çç è 1+ xø 1 ì -1 ü 1+x 1 - xï ï 2 1+x 1 + x ï2 1 - x ï =í ý 2 ï 1+x ï ïî ïþ ì 1+x ´ 1+x 1-x ´ 1-x ü + ï ï 1+x ´ 1-x ï 1+xï 1-x ´ 1+x = í ý 4 ï 1+x ï ï ï î þ 1 2 1 = . = 4 1 - x2 2 1 - x2

12


dy dx dy d(cos 2t) - 2 sin 2t dt = dt = = dx d(sin 3t) 3 cos 3t dt dt p - 2 ´ sin 2 = - 2´1 = 2 2 = 3p 3 æ 1 ö 3 ´ cos 3 ´ ç÷ 4 è ø 2

16. Slope of tangent =

\

æ dy ö ç ÷ è dx ø at t

Now

\

p 4

=

3p 1 x = sin æç ö÷ = è 4 ø 2 2 p y = cos æç ö÷ = 0 è 4 ø

Equation of tangent is dy æ æ 1 öö y-0= ÷÷ çx - ç è 2 øø dx è y=

2 2 æ 1 ö çx ÷ 3 è 2ø

2 2 2 x3 3 3y = 2 2 x - 2. x sin x dx 1 + cos 2 x y=

or 17. Let I = ò

p

0

a

a

0

0

Apply the property ò f ( x) dx = ò f ( a - x) dx p

( p - x) sin xdx

0

1 + cos 2 x

I=ò

I=pò I =p ò

dx

p

0

2

1 + cos x

p/ 2

1 + sec x

p/ 2

0

sec 2 x 2

0

I =p ò

-I

sec 2 x 2 + tan 2 x

Þ

dx

sec 2 xdx = dt

dx 1 + cos 2 x

é Using êë

dx

Putting tan x = t

p

0

2I = p ò

if x = 0, t = 0 if x =

p , t=¥ 2

2a

ò0

a

f ( x) dx = 2ò f ( x) dxù úû 0

13

Examination Papers – 2008 ¥

dt

0

( 2) 2 + t 2

I =p ò

1 tan 2 p æpö I= ç ÷ 2 è2ø I=p

I=

1

æ t ö ç ÷ è 2ø

¥ 0

p2 2 2

18. ( x 2 - y 2 ) dx + 2xy dy = 0 (x 2 - y 2 ) dy =dx 2xy It is homogeneous differential equation. xdu dy Putting y = ux Þ u+ = dx dx æ 1 - u2 ö (1 - u 2 ) du ÷ From (i) u + x = - x2 =-ç ç 2u ÷ dx 2x 2 u è ø Þ

é 1 - u2 ù xdu =-ê + uú dx êë 2u úû

Þ

é 1 + u2 ù xdu =-ê ú dx êë 2u úû

Þ

2u 1+u

2

du = -

dx x

Integrating both sides, we get 2udu dx Þ ò 1 + u2 = - ò x Þ Þ log

Þ Þ

log|1 + u 2 | = - log| x| + log C x2 + y2 x2

| x| = log C

x2 + y2

=C x x 2 + y 2 = Cx

Given that y = 1 when x = 1 Þ 1 + 1 = C Þ C = 2. \ Solution is x 2 + y 2 = 2x.

...(i)

14


OR dy x( 2y - x) = dx x( 2y + x)

...(i)

y = ux dy du =u+ x dx dx du æ 2u - 1 ö Þ u+ x. =ç ÷ dx è 2u + 1 ø du 2u - 1 x = -u dx 2u + 1 Let

x Þ Þ

[from(i)]

2 du 2u - 1 - 2u - u = dx 2u + 1 2u + 1 dx ò u - 1 - 2u2 du = ò x 2u + 1 dx ò 2u2 - u + 1 du = - ò x

1 3 , B= 2 2 3 4u - 1 2 ò 2u2 - u + 1 du + ò 2u2 - u + 1 du = - log x + k 3 du log ( 2u 2 - u + 1) + ò = - log x + k 4 æ 1ö2 7 çu - ÷ + è 4ø 16

Let 2u + 1 = A( 4u - 1) + B ; A =

Þ

1 2

Þ

1 2

log ( 2u 2 - u + 1) +

3 2

1 ù éæ çu - ö÷ ú ê è 1 4ø tan - 1 ê ú = - 2 log x + k ¢ 7 7 ú ê 4 êë 4 úû

y and then y = 1 and x = 1, we get x 6 3 k ¢ = log 2 + tan -1 7 7

Putting u =

æ 2y 2 - xy + x 2 ö 6 3 6 æ 4y - x ö ÷÷ + \ Solution is log çç tan -1 ç tan -1 ÷ + 2 log x = log 2 + 2 è ø 7 7 x 7 7 è ø x 19. cos 2 x

dy + y = tanx dx

dy + sec 2 x ´ y = sec 2 x tan x dx

15


It is a linear differential equation. Integrating factor = e ò

sec2 x dx

= e tan x General solution : y. IF = ò Q. IF dx y. e tan x = ò e tan x . tanx. sec 2 x dx Þ sec 2 x dx = dt

Putting tan x = t \

ye tan x

= ò e t . t . dt = e t . t -ò e t dt = e t . t - e t + k = e tan x (tan x - 1) + k

y. e tan x = e tan x (tan x - 1) + k

\

where k is some constant. ®

20. Given a = i$ + j$ + k$ and

®

b = j$ - k$

®

Let c = xi$ + yj$ + zk$ ®

i$

®

j$

k$

a ´ c = 1 1 1 = i$ (z - y) + j$ ( x - z) + k$ ( y - x) x y z ®

®

®

Given a ´ c = b

(z - y)i$ + ( x - z) j$ + ( y - x) k$ = j$ - k$. Comparing both sides z-y=0 \ x-z =1 \ y- x= -1 \

z=y x=1+z y= x-1

® ®

a.c =3 $ $ (i + j + k$) . ( xi$ + yj$ + zk$) = 3

Also,

x +y +z= 3 (1 + z) + z + z = 3 3z = 2 \

z= 2 / 3 y=2/ 3 2 5 x=1 + = 3 3 ® 1 c = (5i$ + 2j$ + 2k$) 3

16


OR ®

®

®

a + b + c =0 ®

®

®

( a + b ) 2 = ( - c) 2

Þ ®

®

®

®

® ®

Þ

( a + b ) .( a + b ) = c. c

Þ

| a |2 +| b |2 + 2 a . b =| c |2

®

®

® ®

®

® ®

Þ

9 + 25 + 2 a . b = 49 ® ®

Þ

2 a . b = 49 - 25 - 9 ®

Þ

®

2| a || b |cos q = 15

Þ

30 cos q = 15 1 Þ cos q = = cos 60° 2 Þ q = 60° x- 3 y -5 z -7 x+1 y +1 z +1 21. Let and = = =l = = =k 1 -2 1 7 -6 1 Now, let’s take a point on first line as A ( l + 3, -2l + 5, l + 7) and let B (7 k - 1, - 6k - 1, k - 1) be point on the second line The direction ratio of the line AB 7 k - l - 4, - 6k + 2l - 6, k - l - 8 Now as AB is the shortest distance between line 1 and line 2 so, (7 k - l - 4) ´ 1 + ( - 6k + 2l - 6) ´ ( -2) + ( k - l - 8) ´ 1 = 0 and (7 k - l - 4) ´ 7 + ( - 6k + 2l - 6) ´ ( -6) + ( k - l - 8) ´ 1 = 0 Solving equation (i) and (ii) we get l = 0 and k = 0 \ A º ( 3, 5, 7) and B º ( - 1, - 1, - 1)

line 1

B

line 2

...(i) ...(ii)

AB = ( 3 + 1) 2 + (5 + 1) 2 + (7 + 1) 2 = 16 + 36 + 64 = 116 units = 2 29 units

\

OR x+ 2

y+1

z- 3

= = =l 3 2 2 \ ( 3l - 2, 2l - 1, 2l + 3) is any general point on the line Now if the distance of the point from (1, 2, 3) is 3 2, then Let

A

( 3l - 2 - 1) 2 + ( 2l - 1 - 2) 2 + ( 2l + 3 - 3) 2 = ( 3 2 ) Þ

( 3l - 3) 2 + ( 2l - 3) 2 + 4l2 = 18

Þ

9l2 - 18l + 9 + 4l2 - 12l + 9 + 4l2 = 18

17


17 l2 - 30l = 0

Þ Þ

l(17 l - 30) = 0 30 Þ l = 0 or l= 17 56 43 77 ö \ Required point on the line is ( -2, -1, 3) or æç , , ÷ è 17 17 17 ø 22. Let X be the numbers of doublets. Then, X = 0, 1, 2, 3 or 4 (non doublet in each case) P (X = 0) = P 5 5 5 5 625 P (D 1D 2 D 3 D 4 ) = æç ´ ´ ´ ö÷ = è 6 6 6 6 ø 1296 1 5ù é n r r ê Alternatively use Cr p q where p = 6 , q = 6 úû ë

P (X = 1) = P (one doublet)

= P (D 1D 2 D 3 D 4 ) or P (D 1D 2 D 3 D 4 ) or P (D 1D 2 D 3 D 4 ) or P (D 1D 2 D 3 D 4 ) 1 5 5 5 5 1 5 5 5 5 1 5 5 5 5 1 = æç ´ ´ ´ ö÷ + æç ´ ´ ´ ö÷ + æç ´ ´ ´ ö÷ + æç ´ ´ ´ ö÷ è 6 6 6 6ø è 6 6 6 6ø è 6 6 6 6ø è 6 6 6 6ø 125 ö 125 = æç 4 ´ ÷= è 1296 ø 324 P(X = 2) = P (two doublets) = P (D 1D 2 D 3 D 4 ) or P (D 1D 2 D 3 D 4 ) or P (D 1D 2 D 3 D 4 ) or P (D 1D 2 D 3 D 4 ) or P (D 1D 2 D 3 D 4 ) or P (D 1D 2 D 3 D 4 ) 1 1 5 5 1 5 1 5 1 5 5 1 = æç ´ ´ ´ ö÷ + æç ´ ´ ´ ö÷ + æç ´ ´ ´ ö÷ è 6 6 6 6ø è 6 6 6 6ø è 6 6 6 6ø 5 1 1 5 5 1 5 1 5 5 1 1 + æç ´ ´ ´ ö÷ + æç ´ ´ ´ ö÷ + æç ´ ´ ´ ö÷ è 6 6 6 6ø è 6 6 6 6ø è 6 6 6 6ø 25 ö 25 = æç 6 ´ ÷= è 1296 ø 216 P (X = 3) = P (three doublets) = P (D 1D 2 D 3 D 4 ) or P (D 1D 2 D 3 D 4 ) or P (D 1D 2 D 3 D 4 ) or P (D 1D 2 D 3 D 4 ) 1 1 1 5 1 1 5 1 1 5 1 1 5 1 1 1 = æç ´ ´ ´ ö÷ + æç ´ ´ ´ ö÷ + æç ´ ´ ´ ö÷ + æç ´ ´ ´ ö÷ è 6 6 6 6ø è 6 6 6 6ø è 6 6 6 6ø è 6 6 6 6ø 5 ö 5 = æç 4 ´ ÷= è 1296 ø 324 P (X = 4) = P (four doublets) = P (D 1D 2 D 3 D 4 ) 1 1 1 1 1 = æç ´ ´ ´ ö÷ = è 6 6 6 6 ø 1296 Thus, we have X = xi Pi

0 625 1296

1 125 324

2 25 216

3 5 324

4 1 1296

18


SECTION–C a 23. L.H.S. = a

b

2

b

b+g

g

2

g2

g +a a +b

Applying R 3 ® R 3 + R 1 and taking common (a + b + g ) from R 3 . a b g 2 2 = (a + b + g ) a b g2 1 a = (a + b + g ) a 2 1

1

1

b -a

g -a

b2 - a 2 0

g2 -a2 0

(Applying C 2 ® C 2 - C 1 , C 3 ® C 3 - C 1 )

= (a + b + g )[( g 2 - a 2 ) (b - a ) - ( g - a ) (b 2 - a 2 )] (Expanding along R 3 ) = (a + b + g )( g - a ) (b - a ) [( g + a ) - (b + a )] = (a + b + g )( g - a ) (b - a ) ( g - b) = (a + b + g )(a - b) (b - g ) ( g - a ) 24. Let x and y be the length and breadth of rectangle and R be the radius of given circle, (i. e. R is constant). Now, in right D ABC, we have D

x 2 + y 2 = ( 2 R) 2 x 2 + y 2 = 4R 2

Þ y = 4R 2 - x 2

Þ

2R

....(i)

Now, area, of rectangle ABCD. A = xy

A

A = x 4R 2 - x 2

[from (i)]

For area to be maximum or minimum dA =0 dx 1 Þ x´ ´ - 2x + 4R 2 - x 2 ´ 1 = 0 2 2 2 4R - x Þ Þ

-x 2 4R 2 - x 2

+ 4R 2 - x 2 = 0

Þ

( 4R 2 - x 2 ) 2 - x 2 4R 2 - x 2

4R 2 - x 2 - x 2 = 0

Þ

4R 2 - 2x 2 = 0

x 2 - 2R 2 = 0

Þ

x = 2R

C

=0

O

x

y

B

19


d2A

Now,

dx 2 d2A

\

dx 2

=

= at x =

2 R

2x ( x 2 - 6R 2 ) ( 4R 2 - x 2 ) 3 / 2 -8 2 R3 ( 2R 2 ) 3 / 2

<0

So, area will be maximum at x = 2R Now, from (i), we have y = 4R 2 - x 2 = 4R 2 - 2R 2 = 2R 2 y = 2R Here x=y= 2 R So the area will be maximum when ABCD is a square. OR Let radius CD of inscribed cylinder be x and height OC be H and q be the semi-vertical angle of cone. B

Therefore, OC = OB - BC

q

Þ H = h - x cot q C

Now, volume of cylinder

D h

V = px 2 ( h - x cot q) Þ V = p ( x 2 h - x 3 cot q) For maximum or minimum value dV =0 Þ p( 2xh - 3x 2 cot q) = 0 dx Þ

px( 2h - 3x cot q) = 0

\

2h - 3x cot q = 0

Þ

x= Now, Þ Þ

(as x = 0 is not possible)

2h tan q 3 d 2V dx 2 d 2V dx 2

= p ( 2h - 6x cot q) = 2ph - 6px cot q

d 2V dx

2

2 h tan q at x = 3

= 2 ph - 6 p ´

2h tan q cot q 3

= 2 ph - 4 ph = - 2 ph < 0

O

A

20


Hence, volume will be maximum when x =

2h tan q. 3

Therefore, height of cylinder H = h - x cot q 2h 2h h =htan q cot q = h = . 3 3 3 1 Thus height of the cylinder is of height of cone. 3 9 25. x 2 + y 2 = ...(i) 4 y 2 = 4x

...(ii)

From (i) and (ii) 2

æ y2 ö 9 çç ÷÷ + y 2 = 4 è 4 ø

Y'

y2 = t

Let

2 X'

t2 9 +t = 16 4

(– 3 , 0) 2

– 2

t 2 + 16t = 36 t 2 + 18t - 2t - 36 = 0

Y

t(t + 18) - 2(t + 18) = 0 (t - 2) (t + 18) = 0 t = 2, - 18 y2 = 2 y=± 2 2

Required area = ò

-

2 2

=ò

-

= 2ò

2 2

0

éy = 2ê ë2 é 2 = 2ê ë 2

( x 2 - x 1 ) dy æ 9 y2 ö çç ÷ dy - y2 4 ÷ø è 4 2 æç 3 ö÷ - y 2 dy - 2 è2ø 4

ò0

2

y 2 dy 2

æy3 ö çç ÷÷ è 3 ø0

9 9 - y 2 + sin 4 8

1

y ù 3 / 2 úû 0

9 9 - 2 + sin 4 8

1

2 2ù 1 - 2 2 3 úû 6

-

1 2

2

y= 2 X ( 3 , 0) 2 y=– 2

21


1 9 + sin 2 4

= = 26. Let I = ò

1 3 2

+

1

9 sin 4

æ2 2ö 2 ç ÷3 è 3 ø 1

æ2 2ö ç ÷ sq. units è 3 ø

a-x dx a+x

a

-a

Put x = a cos 2q dx = a ( - sin 2q) 2dq If x = a, then cos 2q = 1 2q = 0 q=0 x = - a, cos 2 q = - 1 2q = p p q= 2 a - a cos 2q 0 \ I=ò a ( - sin 2q) 2dq p/ 2 a + a cos 2q p/ 2

2 sin 2 q

0

2 cos 2 q

=ò

= 2 aò

p/ 2

0

2a sin 2q dq

2 sin 2 q dq = 2aò

p/ 2

0

(1 - cos 2q) dq

sin 2q ù p / 2 sin 0 ö ù é æ p sin p ö æ é = 2a êq = 2a ê ç ÷ - ç0 ÷ ú è ø è 2 2 2 2 ø úû ë û0 ë é p ù = 2a æç - 0ö÷ = pa êë è 2 ø úû 27. Equation of the plane passing through ( - 1, - 1, 2) is a( x + 1) + b( y + 1) + c (z - 2) = 0 (i) is perpendicular to 2x + 3y - 3z = 2 \ 2a + 3b - 3c = 0 Also (i) is perpendicular to 5x - 4y + z = 6 \ 5a - 4b + c = 0 From (ii) and (iii) a b c = = =k 3 - 12 - 15 - 2 - 8 - 15 a b c = = =k - 9 - 17 - 23

Þ Þ

a = - 9k ,

b = - 17 k ,

c = - 23k

...(i) ...(ii) ...(iii)

22


Putting in equation (i) - 9k( x + 1) - 17 k ( y + 1) - 23k (z - 2) = 0 Þ 9( x + 1) + 17( y + 1) + 23(z - 2) = 0 Þ 9x + 17 y + 23z + 9 + 17 - 46 = 0 Þ 9x + 17 y + 23z - 20 = 0 Þ 9x + 17 y + 23z = 20. Which is the required equation of the plane. OR Equation of the plane passing through ( 3, 4, 1) is a( x - 3) + b( y - 4) + c (z - 1) = 0 Since this plane passes through (0, 1, 0) also \ a( 0 - 3) + b(1 - 4) + c( 0 - 1) = 0 or - 3a - 3b - c = 0 or 3a + 3b + c = 0 Since (i) is parallel to x+ 3 y- 3 z-2 = = 2 7 5 \ 2a + 7b + 5c = 0 From (ii) and (iii) a b c = = =k 15 - 7 2 - 15 21 - 6

...(i)

...(ii)

...(iii)

Þ a = 8k , b = - 13k , c = 15k Putting in (i), we have 8k( x - 3) - 13k( y - 4) + 15k (z - 1) = 0 Þ 8( x - 3) - 13( y - 4) + 15 (z - 1) = 0 Þ 8x - 13y + 15z + 13 = 0. Which is the required equation of the plane. 28. Let the owner buys x machines of type A and y machines of type B. Then ...(i) 1000x + 1200y £ 9000 ...(ii) 12x + 8y £ 72 (0, 9) 3x + 2y = 18 Objective function is to be maximize z = 60x + 40y 15 (0, ) From (i) 2 9 , 45 10x + 12y £ 90 4 8 or ...(iii) 5x + 6y £ 45 FR ...(iv) [from (ii)] 3x + 2y £ 18 We plot the graph of inequations shaded region in the feasible solutions (iii) and (iv) .

(

(0, 0)

) 5x + 6y = 45

(6, 0)

(9, 0)

23


The shaded region in the figure represents the feasible region which is bounded. Let us now evaluate Z at each corner point. at (0, 0) Z is 60 ´ 0 + 40 ´ 0 = 0 15 15 Z at æç 0, ö÷ is 60 ´ 0 + 40 ´ = 300 è 2ø 2 Z at ( 6, 0) is 60 ´ 6 + 40 ´ 0 = 360 9 45 9 45 Z at æç , ö÷ is 60 ´ + 40 ´ = 135 + 225 = 360. è4 8 ø 4 8 Þ max. Z = 360 Therefore there must be 9 45 but second case is not possible as x and y are whole ,y= 4 8 numbers. Hence there must be 6 machines of type A and no machine of type B is required for maximum daily output. 29. Let E1 be the event that insured person is scooter driver, E2 be the event that insured person is car driver, E 3 be the event that insured person is truck driver, and A be the event that insured person meets with an accident. æAö 2, 000 1 \ P(E1 ) = = , P ç ÷ = 0.01 12, 000 6 è E1 ø either x = 6, y = 0 or x =

æAö 4, 000 1 P(E2 ) = = , P ç ÷ = 0.03 12000 3 è E2 ø P(E 3 ) =

\

E1

E2

E3

æAö 6, 000 1 = , P ç ÷ = 0.15 12, 000 2 èE3 ø

A

æAö P(E1 ) . P ç ÷ è E1 ø

æE ö P ç 1÷ = èAø

æAö æAö æAö P(E1 ) . P ç ÷ + P(E2 ) . P ç ÷ + P(E 3 ) . P ç ÷ è E1 ø è E2 ø èE3 ø 1 ´ 0.01 1 1 6 = = = 1 1 1 ´ 0.01 + ´ 0.03 + ´ 0.15 1 + 6 + 45 52 6 3 2

Set–II 20. We have, tan - 1 ( 2x) + tan - 1 ( 3x) = Þ tan -

1

é 2x + 3x ù p ê ú= ë 1 - ( 2x) . ( 3x) û 4

p 4 [Using property tan - 1 x + tan - 1 y = tan -

1

x+y 1 - xy

]

24


tan -

Þ

1

5x 1 - 6x

2

5x

Þ

1 - 6x

2

=

p 4

=1

6x 2 + 5x - 1 = 0

1 6

which is the required solution.

2

Þ

6x + 6x - x - 1 = 0

Þ

6x( x + 1) - 1( x + 1) = 0

Þ

( x + 1) ( 6x - 1) = 0

Þ

x = - 1,

21. Let I = ò

p

0

p

x tan x dx sec x cosec x sin x x. cos x dx 1 1 . cos x sin x

Þ

I=ò

Þ

I = ò x sin 2 x dx

Þ

I = ò ( p - x) . sin 2 ( p - x) dx

Þ

Þ

0 p

...(i)

0 p 0 p

a

a

0

0

[Using property ò f ( x) dx = ò f ( a - x) dx]

2

I = ò ( p - x) sin x dx

...(ii)

0

Adding (i) and (ii) we have p

2I = ò p sin 2 x dx 0

p

Þ 2I = p ò sin 2 x dx = 0

Þ 2I =

Hence ò

p

0

p

ò0

(1 - cos 2x) dx

sin 2p ö æ sin éæ ê çè p - 2 ÷ø - çè 0 - 2 ë p2 . I= 4 x tan x p2 . dx = sec x . cosec x 4

p 2

\

p 2

0 öù ÷ú øû

æ1 1 ö 22. We have, y = x 2 + 1 - log ç + 1 + ÷ èx x2 ø æ1 + x 2 + 1 ö ÷ Þ y = x 2 + 1 - log ç ç ÷ x è ø Þ

y = x 2 + 1 - log æç1 + x 2 + 1 ö÷ + log x è ø

Þ Þ

sin 2x ù p é x êë 2 úû 0 p p2 2I = [p] = 2 2 2I =

p 2

25


On differentiating w.r.t. x, we have dy 1 1 1 1 = ´ 2x ´ ´ 2x + 2 2 dx 2 x 2 + 1 x ( x + 1 + 1) 2 x + 1 =

x x2 + 1 x

=

x

-

x 2 + 1 ( x 2 + 1 + 1) x

-

2

2

x +1

=

-

2

x +1 =

=

=

23. Let D =

´

2

x + 1 ( x + 1 + 1)

x

x æç x 2 + 1 - 1ö÷ è ø 2

2

+

( x + 1) ( x )

x

-

x2 + 1

( x 2 + 1 - 1) x x2 + 1

+

1 x ( x 2 + 1 - 1) 2

( x + 1 - 1)

1 x

1 x

+

x2 + 1 - x2+ 1 + x2+ 1 x x2+ 1 x2 + 1 x x2+ 1

x2+ 1

=

x

1 + a2 - b 2

2ab

2ab

2

- 2b

1- a +b

2

2a 1 - a2 - b 2

- 2a

2b

Applying C 1 ® C 1 - b. C 3 and C 2 ® C 2 + a. C 3 , we have 1 + a2 + b 2

0

0

2

D=

2

- 2b

1+ a +b 2

2

2

2a 2

b (1 + a + b ) - a (1 + a + b ) 1 - a 2 - b 2 Taking out (1 + a 2 + b 2 ) from C 1 and C 2 , we have 2

2 2

= (1 + a + b )

1

0

- 2b

0

1

2a

b

-a

1 - a2 - b 2

Expanding along first row, we have = (1 + a 2 + b 2 ) 2 [1 . (1 - a 2 - b 2 + 2a 2 ) - 2b ( - b)] = (1 + a 2 + b 2 ) 2 (1 + a 2 - b 2 + 2b 2 ) = (1 + a 2 + b 2 ) 2 (1 + a 2 + b 2 ) = (1 + a 2 + b 2 ) 3 .

+

1 x

26


24. Let I = ò

p

x sin x

p

dx 1 + cos 2 x ( p - x) sin ( p - x)

0

2

0

Þ I=ò Þ I=ò

1 + cos ( p - x) ( p - x) sin x dx 1 + ( - cos x) 2 ( p - x) sin x dx 1 + cos 2 x

p

0

Þ I=ò

p

0

...(i) a

a

0

0

[Using property ò f ( x) dx = ò f ( a - x) dx]

dx

...(ii)

Adding (i) and (ii), we have sin x p p sin x p 2I = ò dx = p ò dx 2 0 0 1 + cos x 1 + cos 2 x Let cos x = t Þ - sin x dx = dt Þ sin x dx = - dt As x = 0, t = 1 and x = p, t = - 1 Now, we have - 1 - dt 2I = ò 1 1 + t2 Þ 2I = ò

1

-1

dt 1+t

2

= [tan - 1 (t)] 1-

1

Þ 2I = tan - 1 (1) - tan - 1 ( - 1) =

p æ- pö p -ç ÷= 4 è 4 ø 2

p . 4 25. The equations of the given curves are Þ

and

I=

x2 + y2 = 4

...(i)

( x - 2) 2 + y 2 = 4

...(ii)

2

2

Clearly, x + y = 4 represents a circle with centre (0, 0) and radius 2. Also, ( x - 2) 2 + y 2 = 4 represents a circle with centre (2, 0) and radius 2. To find the point of intersection of the given curves, we solve (i) and (ii). Simultaneously, we find the two curves intersect at A (1, 3 ) and D(1, - 3 ). Since both the curves are symmetrical about x-axis, So, the required area = 2( Area OABCO) Now, we slice the area OABCO into vertical strips. We observe that the vertical strips change their character at A(1, 3 ). So, Area OABCO = Area OACO + Area CABC.

27

Examination Papers – 2008 y

When area OACO is sliced in the vertical strips, we find that each strip has its upper end on the circle ( x - 2) 2 + ( y - 0) 2 = 4 and the lower end on x-axis. So, the approximating rectangle shown in figure has length = y 1 width = Dx and area = y 1 Dx. As it can move from x = 0 to x = 1

x'

A (1, 3)

x B (2, 0) (4, 0)

O

1

\

Area OACO = ò y 1 dx

\

Area OACO = ò

0 1

C (1, 0) y' D (1, – 3)

4 - ( x - 2) 2 dx

0

Similarly, approximating rectangle in the region CABC has length = y 2 , width = Dx and area = y 2 Dx. As it can move from x = 1 to x = 2 \

2

2

1

1

Area CABC = ò y 2 dx = ò

4 - x 2 dx

Hence, required area A is given by 1

A = 2 éò êë 0

4 - ( x - 2) 2 dx + ò

2

1

4-x

é é ( x - 2) 4 Þ A = 2 êê . 4 - ( x - 2) 2 + sin 2 êë ë 2

2

1

dxù úû ( x - 2) ù 1 é x 4 + . 4 - x 2 + sin ú ê 2 û0 ë2 2

ì 3 1 3 Þ A = 2 í+ 2 sin - 1 æç - ö÷ - 2 sin - 1 ( - 1) + 2 sin - 1 (1) - 2 sin è ø 2 2 î 2 p p p p ù é = 2 - 3 - 2 æç ö÷ + 2 æç ö÷ + 2 æç ö÷ - 2æç ö÷ êë è6ø è2ø è2ø è 6 ø úû 2p = 2 æç - 3 + 2p ö÷ è ø 3 4p 8p = 2æç - 3 ö÷ = æç - 2 3 ö÷ sq. units. è 3 ø è 3 ø

Set–III 20. We have, tan -

Þ

1

æx-1ö ç ÷ + tan è x - 2ø

1

æx+1ö p ç ÷= è x + 2ø 4

ì x-1 x+1 ü + ï ï x-2 x+2 ï p -1ï tan í ý= ï 1 - æç x - 1 ö÷ æç x + 1 ö÷ ï 4 ï è x - 2 ø è x + 2 ø ïþ î

1

1

xù 2 ù ú 2 úû 1 ú û

1ü ý 2þ

28


Þ

ì ( x - 1) ( x + 2) + ( x - 2) ( x + 1) ü p tan - 1 í ý= î ( x - 2) ( x + 2) - ( x - 1) ( x + 1) þ 4

Þ

ìï x 2 + x - 2 + x 2 - x - 2 üï p tan - 1 í ý= x2 - 4 - x2 +1 ïî ïþ 4

Þ

tan -

Þ Þ

21. Given

æ 2x 2 - 4 ö p ç ÷= ç -3 ÷ 4 è ø

2x 2 - 4 -3

= tan

p

Þ

4

2x 2 - 4 -3

=1

2x 2 - 4 = - 3 Þ x2 =

2x 2 = 1

Þ Hence,

1

x=

1 2

Þ x=±

1 2

1 1 are the required values. ,2 2

é 1 + sin x + 1 - sin x ù y = cot -1 ê ú êë 1 + sin x - 1 - sin x úû é ( 1 + sin x + 1 - sin x ) ( 1 + sin x + 1 - sin x) ù = cot -1 ê ú êë ( 1 + sin x - 1 - sin x ) ( 1 + sin x + 1 - sin x ) úû é 1 + sin x + 1 - sin x + 2 1 - sin 2 x ù ú = cot -1 ê 1 + sin x - 1 + sin x ê ú ë û x 2 cos 2 -1 -1 é 2 (1 + cos x) ù 2 = cot ê ú = cot x x 2 sin x ë û 2 sin cos 2 2 x x = cot -1 æç cot ö÷ = è 2ø 2 dy 1 = = dx 2 1

22. Let

I = ò cot - 1 (1 - x + x 2 ) dx 0 1

= ò tan 0 1

= ò tan 0

1

1 1-x+x

1

2

dx

x + (1 - x) 1 - x (1 - x)

dx

éQ cot êë

1

x = tan -

1

1 ù x úû

[Q 1 can be written as x + 1 - x]

29

Examination Papers – 2008 1

= ò [tan 0 1

= ò tan -

é ì a+b ü x + tan - 1 (1 - x)] dx êQ tan - 1 í ý = tan 1 ab î þ ë

1

1

ù a + tan - 1 b ú û

1

x dx + ò tan - 1 (1 - x) dx

0 1

0 1

= ò tan - 1 xdx + ò tan - 1 [1 - (1 - x)] dx 0

0 1

1

= 2 ò tan - 1 xdx = 2 ò tan 0

1

é êQ êë

a

a ù f ( x ) = ò ò f ( a - x) dx úú 0 0 û

x. 1 dx, integrating by parts, we get

0

é = 2 ê {tan êë

1

1

ù × xdx ú 2 úû 01+ x

1

1

x. x}0 - ò 1

= 2 [tan - 1 1 - 0] - ò

2x

01

+x

2

dx = 2 ×

p - [log (1 + x 2 )] 10 4

p p - (log 2 - log 1) = - log 2 2 2 a + b + 2c a b

[Q log 1 = 0 ]

=

23. Let D =

1

c

b + c + 2a

b

c

a

c + a + 2b

Applying C 1 ® C 1 + C 2 + C 3 , we have 2( a + b + c) a

b

D = 2( a + b + c) b + c + 2a 2( a + b + c)

b c + a + 2b

a

Taking out 2( a + b + c) from C 1 , we have 1 a

b

D = 2( a + b + c) 1 b + c + 2a 1

a

b c + a + 2b

Interchanging row into column, we have 1 1 D = 2( a + b + c) a b + c + 2a b

b

1 a

c + a + 2b

Applying C 1 ® C 1 - C 2 and C 2 ® C 2 - C 3 , we have 0 0 D = 2( a + b + c) - ( a + b + c) 0

a+b + c

1 a

- ( a + b + c) c + a + 2b

30


Now expanding along first row, we have 2( a + b + c) [1 . ( a + b + c) 2 ] = 2( a + b + c) 3 = R.H.S. 24. We have, given equations x 2 + y 2 = 8x

...(i)

2

...(ii)

y = 4x

and

Equation (1) can be written as ( x - 4) 2 + y 2 = ( 4) 2 So equation (i) represents a circle with centre (4, 0) and radius 4. Again, clearly equation (ii) represents parabola with vertex (0, 0) and axis as x-axis. The curve (i) and (ii) are shown in figure and the required region is shaded. On solving equation (i) and (ii) we have points of y intersection 0(0, 0) and A ( 4, 4), C( 4, - 4) (4, 4) A Now, we have to find the area of region bounded B by (i) and (ii) & above x-axis. x' So required region is OBAO. O (4, 0) (0, 0) Now, area of OBAO is 4

A = ò ( 8x - x 2 - 4x ) dx 0 4

2

= ò ( ( 4) - ( x - 4) 0

2

C (4, – 4)

- 2 x ) dx

y' 4

é ( x - 4) ( x - 4) 16 2x 3 / 2 ù =ê ( 4) 2 - ( x - 4) 2 + sin - 1 -2´ ú 2 4 3 ë 2 û0 3ù é 4 = ê 8 sin - 1 0 - ( 4) 2 ú - [8 sin - 1 ( - 1) - 0] 3 ê ú ë û 4 p æ ö æ = ç 8 ´ 0 - ´ 8÷ - ç 8 ´ - ö÷ è ø è 3 2ø 32 32 + 4p = æç 4p - ö÷ sq.units è 3 3ø x tan x p 25. Let I = ò ...(i) dx 0 sec x + tan x ( p - x) tan ( p - x) p a a [Using property ò f ( x) dx = ò f ( a - x) dx] I=ò dx 0 sec ( p - x) + tan ( p - x) 0 0 p - ( p - x) tan x Þ I=ò dx 0 - sec x - tan x =-

x

31

Examination Papers – 2008 p

( p - x) tan x

0

sec x + tan x

Þ I=ò

dx

Adding (i) and (ii) we have p tan x p 2I = ò dx 0 sec x + tan x tan x p Þ 2I = p ò dx 0 sec x + tan x tan x (sec x - tan x) p Þ 2I = p ò ´ dx 0 (sec x + tan x) (sec x - tan x) p tan x (sec x - tan x) Þ 2I = p ò dx 0 sec 2 x - tan 2 x p

Þ 2I = p ò (tan x . sec x - tan 2 x) dx 0 p

Þ 2I = p ò [sec x tan x - (sec 2 x - 1)] dx 0

Þ 2I = p [sec x - tan x + x] p0 Þ 2I = p [(sec p - tan p + p) - (sec 0 - tan 0 + 0)] Þ 2I = p [( - 1 - 0 + p) - (1 - 0 )] Þ 2I = p ( p - 2) p \ I = ( p - 2) 2 x tan x p p Hence ò0 sec x + tan x = 2 ( p - 2)

...(ii)

EXAMINATION PAPERS – 2008 MATHEMATICS CBSE (All India) CLASS – XII Time allowed: 3 hours

Maximum marks: 100

General Instructions: As given in CBSE Examination paper (Delhi) – 2008.

Set–I SECTION–A 1. If f ( x) is an invertible function, find the inverse of f ( x) = 2. Solve for x : tan -

1

1-x 1+x

=

1 tan 2

1

3x - 2 5

.

x; x > 0

é x + 3y yù é 4 - 1ù 3. If ê , find the values of x and y. ú=ê 4 úû ë 7 - x 4û ë 0 4. Show that the points (1, 0), (6, 0), (0, 0) are collinear. x + cos 6x 5. Evaluate : ò dx 3x 2 + sin 6x 6. If ò ( e ax + bx) dx = 4e 4 x + ®

7. If| a | =

3x 2 , find the values of a and b. 2

®

®

®

® ®

3 ,| b | = 2 and angle between a and b is 60°, find a . b . ®

8. Find a vector in the direction of vector a = i$ - 2j$, whose magnitude is 7. x- 3 y+ 2 z-5 9. If the equation of a line AB is , find the direction ratios of a line parallel to AB. = = 1 -2 4 10. If

x+2 3 = 3, find the value of x. x+5 4

SECTION–B 11. Let T be the set of all triangles in a plane with R as relation in T given by R = {(T1 , T2 ) : T1 @ T2 }. Show that R is an equivalence relation. p 1 a p 1 a 2b 12. Prove that tan æç + cos - 1 ö÷ + tan æç - cos - 1 ö÷ = . è4 2 è4 2 bø bø a

33


OR 8 Solve tan ( x + 1) + tan ( x - 1) = tan 31 13. Using properties of determinants, prove that following: a + b + 2c a b -1

-1

-1

= 2( a + b + c) 3

c

b + c + 2a

b

c

a

c + a + 2b

14. Discuss the continuity of the following function at x = 0 : ì x 4 + 2x 3 + x 2 ï , x¹0 f ( x) = í tan - 1 x ï 0, x=0 îï OR Verify Lagrange’s mean value theorem for the following function: f ( x) = x 2 + 2x + 3, for [4, 6]. 15. If f ( x) =

sec x - 1

p , find f '( x). Also find f ¢ æç ö÷. è sec + 1 2ø OR

If x 1 + y + y 1 + x = 0, find 16. Show that ò

p/ 2

0

dy . dx

tan x + cot x = 2p

17. Prove that the curves x = y 2 and xy = k intersect at right angles if 8k 2 = 1. 18. Solve the following differential equation: dy x + y = x log x; x ¹ 0 dx 19. Form the differential equation representing the parabolas having vertex at the origin and axis along positive direction of x-axis. OR Solve the following differential equation: ( 3xy + y 2 ) dx + ( x 2 + xy) dy = 0 20. If i$ + j$ + k$ , 2i$ + 5j$ , 3i$ + 2j$ - 3k$ and i$ - 6j$ - k$ are the position vectors of the points A , B, C and ¾®

¾®

¾®

¾®

D, find the angle between AB and CD . Deduce that AB and CD are collinear. 21. Find the equation of the line passing through the point P(4, 6, 2) and the point of intersection x-1 y z +1 of the line and the plane x + y - z = 8. = = 3 2 7 22. A and B throw a pair of die turn by turn. The first to throw 9 is awarded a prize. If A starts the 9 game, show that the probability of A getting the prize is . 17

34


SECTION–C 23. Using matrices, solve the following system of linear equations: 2x - y + z = 3 -x + 2y - z = - 4 x - y + 2z = 1 OR Using elementary transformations, find the inverse of the following matrix: é 2 - 1 4ù ê4 0 2ú ê ú êë 3 - 2 7 úû 24. Find the maximum area of the isosceles triangle inscribed in the ellipse

x2 a2

+

y2 b2

= 1, with its

vertex at one end of major axis. OR Show that the semi–vertical angle of the right circular cone of given total surface area and 1 maximum volume is sin - 1 . 3 25. Find the area of that part of the circle x 2 + y 2 = 16 which is exterior to the parabola y 2 = 6x. x tan x p 26. Evaluate: ò dx 0 sec x + tan x x + 2 2y + 3 3z + 4 27. Find the distance of the point ( - 2, 3, - 4) from the line measured = = 3 4 5 parallel to the plane 4x + 12y - 3z + 1 = 0. 28. An aeroplane can carry a maximum of 200 passengers. A profit of Rs. 400 is made on each first class ticket and a profit of Rs. 300 is made on each second class ticket. The airline reserves at least 20 seats for first class. However, at least four times as many passengers prefer to travel by second class then by first class. Determine how many tickets of each type must be sold to maximise profit for the airline. Form an LPP and solve it graphically. 29. A man is known to speak truth 3 out of 4 times. He throws a die and report that it is a 6. Find the probability that it is actually 6.

Set–II Only those questions, not included in Set I, are given. 20. Using properties of determinants, prove the following: a a + b a + 2b a + 2b

a

a+b

a + 2b

a + b = 9b 2 ( a + b) a

35


21. Evaluate: ò

p/ 2

0

log sin x dx

22. Solve the following differential equation: dy (1 + x 2 ) + y = tan - 1 x dx 27. Using matrices, solve the following system of linear equations: 3x - 2y + 3z = 8 2x + y - z = 1 4x - 3y + 2z = 4 OR Using elementary transformations, find the inverse of the following matrix: é 2 5 3ù ê 3 4 1ú ê ú êë 1 6 2 úû 28. An insurance company insured 2000 scooter drivers, 3000 car drivers and 4000 truck drivers. The probabilities of their meeting with an accident respectively are 0.04, 0.06 and 0.15. One of the insured persons meets with an accident. Find the probability that he is a car driver. 29. Using integration, find the area bounded by the lines x + 2y = 2, y - x = 1 and 2x + y = 7.

Set–III Only those questions, not included in Set I and Set II are given. 20. If a, b and c are all positive and distinct, show that a b c D = b c a has a negative value. c a b 1

21. Evaluate: ò cot - 1 (1 - x + x 2 ) dx 0

22. Solve the following differential equation: dy x log x + y = 2 log x dx 27. Using matrices, solve the following system of linear equations: x+ y+z= 4 2x + y - 3z = - 9 2x - y + z = - 1 OR Using elementary transformations, find the inverse of the following matrix:

36


é 2 5 3ù ê 3 4 1ú ê ú êë 1 6 3úû 28. Find the area bounded by the curves ( x - 1) 2 + y 2 = 1 and x 2 + y 2 = 1. 29. An insurance company insured 3000 scooter drivers, 5000 car drivers and 7000 truck drivers. The probabilities of their meeting with an accident respectively are 0.04, 0.05 and 0.15 One of the insured persons meets with an accident. Find the probability that he is a car driver.

SOLUTIONS Set – I SECTION–A 3x - 2

1. Given f ( x) =

5 3x - 2

Let

y=

Þ

3x - 2 = 5y f - 1 ( x) =

Þ 2. tan Þ

Þ

5

1

Þ

tan -

1

1

tan -

1

Þ

tan -

1

Þ

tan -

1

Þ

1 - x2 2x

3

3 1

x

æ1 - x ö ç ÷ = tan è1 + x ø æ1 - x ö 2ç ÷ è1 + x ø

æ1 - x ö 1-ç ÷ è1 + x ø Þ

5y + 2

5x + 2

æ1 - x ö 1 ç ÷ = tan è1 + x ø 2 2 tan -

x=

2

1

x

= tan -

1

x

æ1 - x ö (1 + x) 2 2ç = tan ÷ è 1 + x ø (1 + x) 2 - (1 - x) 2 2 (1 + x) (1 - x)

= tan 4x æ1 - x 2 ö ç ÷ = tan - 1 x ç 2x ÷ è ø =x

Þ

1

1

x

x

1 - x 2 = 2x 2

37


3x 2 = 1

Þ

1 1 ,3 3 é x + 3y yù é 4 - 1ù 3. Given ê ú=ê 4 úû ë 7 - x 4û ë 0 Þ

x=

Hence

Þ

1 3 1 x= 3

x2 =

Þ Þ

(Q x > 0)

x + 3y = 4 y= -1 7 -x=0 x = 7, y = - 1 1 0 1

...(i) ...(ii) ...(iii)

4. Since 6 0 1 = 0 0 0 1 Hence (1, 0), (6, 0) and (0, 0) are collinear. x + cos 6x 5. Let I= ò dx 3x 2 + sin 6x 3x 2 + sin 6x = t

Let Þ

( 6x + 6 cos 6x) dx = dt dt ( x + cos 6x) dx = 6 dt 1 1 I= ò = log|t|+ C = log| 3x 2 + sin 6x|+C 6t 6 6

Þ \ 6.

3x 2 2 Differentiating both sides, we get ax 4x ò ( e + bx) dx = 4e +

( e ax + bx) = 16e 4 x + 3x On comparing, we get b = 3 But a cannot be found out. ®

7. | a | = ® ®

®

3 , | b |= 2 ®

®

a . b =| a |.| b |cos q = =

8.

3 . 2 . cos 60° 3

®

a = i$ - 2j$ ®

Unit vector in the direction of a =

i$ - 2j$

®

5

Hence a vector in the direction of a having magnitude 7 will be

7 $ 14 $ ij. 5 5

38


9. The direction ratios of line parallel to AB is 1, –2 and 4. x+2 3 10. =3 x+5 4 Þ Þ Þ

4x + 8 - 3x - 15 = 3 x-7 = 3 x = 10

SECTION–B 11.

(i) Reflexive R is reflexive if T R T "T1 1 1 Since T1 @ T1 \ R is reflexive. (ii) Symmetric R is symmetric if T R T Þ T R T 1 2 2 1 Since T1 @ T2 Þ T2 @ T1 \ R is symmetric. (iii) Transitive

R is transitive if T1 R T2 and T2 R T 3 Þ T1 R T 3 Since T1 @ T2 and T2 @ T3 Þ T1 @ T3 \ R is transitive From (i), (ii) and (iii), we get R is an equivalence relation. p 1 a p 1 a 12. L.H.S. = tan æç + cos - 1 ö÷ + tan æç - cos - 1 ö÷ è4 2 è4 2 bø bø p 1 a p 1 a tan + tan æç cos - 1 ö÷ tan - tan æç cos - 1 ö÷ è2 è2 4 bø 4 bø = + p 1 a p 1 a 1 - tan tan æç cos - 1 ö÷ 1 + tan tan æç cos - 1 ö÷ è ø è 4 2 b 4 2 bø 1 a 1 a 1 + tan æç cos - 1 ö÷ 1 - tan æç cos - 1 ö÷ è2 ø è b 2 bø = + 1 a 1 a 1 - tan æç cos - 1 ö÷ 1 + tan æç cos - 1 ö÷ è2 ø è b 2 bø

=

é ê1 + tan ë

2

é æ1 æ1 - 1 æ aööù - 1 æ aööù ç cos çè ÷ø ÷ ú + ê1 - tan ç cos çè ÷ø ÷ ú è2 ø è b û 2 b øû ë 1 a 1 - tan 2 æç cos - 1 ö÷ è2 bø

2

39


1 a 2 sec 2 æç cos - 1 ö÷ 2(1 + tan 2 q) è2 2 sec 2 q bø = = = 2 1 a 1 - tan 2 q 1 - tan 2 æç cos - 1 ö÷ 1 - tan q è2 bø 2 2 2 = = = a 1 cos 2q æ - 1 aö cos 2ç cos ÷ è2 ø b b 2b = R. H.S. = a OR 8 We have tan - 1 ( x + 1) + tan - 1 ( x - 1) = tan - 1 31 é ( x + 1) + ( x - 1) ù 8 Þ tan - 1 ê ú = tan - 1 2 31 êë 1 - ( x - 1) úû 2x 8 Þ = 2 31 2-x Þ

62x = 16 - 8x 2

Þ

8x 2 + 62x - 16 = 0

1 é êë Let 2 cos

1

æ aö = qù ç ÷ úû èb ø

4x 2 + 31x - 8 = 0 1 Þ x = and x = - 8 4 As x = - 8 does not satisfy the equation 1 Hence x = is only solution.. 4 a + b + 2c a b 13. Let D= c b + c + 2a b Þ

c

c + a + 2b

a

Applying C 1 ® C 1 + C 2 + C 3 , we get 2 ( a + b + c) a D = 2 ( a + b + c) b + c + 2a 2 ( a + b + c)

b c + a + 2b

a

Taking common 2( a + b + c) 1 a = 2( a + b + c) 1 b + c + 2a 1

b

0

b 0 c + a + 2b

[by R 2 ® R 2 - R 1 , R 3 ® R 3 - R 1 ]

40


1

a

b

= 2( a + b + c) 0 a + b + c 0

0 a+b + c

0

= 2 ( a + b + c) {( a + b + c) 2 - 0} expanding along C 1 . = 2( a + b + c) 3 = RHS 14. At x = 0 L.H.L.

= lim

( 0 - h) 4 + 2 ( 0 - h) 3 + ( 0 - h) 2 tan - 1 ( 0 - h)

h®0

h 4 - 2h 3 + h 2

= lim

- tan - 1 h

h 3 - 2h 2 + h

tan - 1 h h [On dividing numerator and denominator by h.] æ ö tan - 1 h ç as lim = 0÷÷ ç h®0 h è ø

h®0

=

= lim

0 -1

h®0

-

=0 R.H.L

= lim

( 0 + h) 4 + 2 ( 0 + h) 3 + ( 0 + h) 2

h®0

= lim

h®0

= lim

h®0

=

tan - 1 ( 0 + h) h 4 + 2h 3 + h 2 tan - 1 h h 3 + 2h 2 + h tan - 1 h h

0 1

(on dividing numerator and denominator by h)

æ ö tan - 1 h ç as lim ÷ = 1 ç h®0 ÷ h è ø

=0 and f ( 0) = 0 (given) so, L.H.L = R.H.L = f ( 0) Hence given function is continuous at x = 0 OR f ( x) = x 2 + 2x + 3 for [4, 6] (i) Given function is a polynomial hence it is continuous (ii) f ¢( x) = 2x + 2 which is differentiable f ( 4) = 16 + 8 + 3 = 27 f ( 6) = 36 + 12 + 3 = 51

41


Þ f ( 4) ¹ f ( 6) . All conditions of Mean value theorem are satisfied. \ these exist atleast one real value C Î( 4, 6) f ( 6) - f ( 4) 24 such that f ¢ ( c) = = = 12 6-4 2 Þ 2c + 2 = 12 or c = 5 Î(4,6) Hence, Lagrange's mean value theorem is verified sec x - 1 1 - cos x 1 - cos x 15. f ( x) = = ´ sec x + 1 1 + cos x 1 - cos x 1 - cosx

Þ

f ( x) =

Þ

f ¢( x) = - cosec x cot x + cosec 2 x

sin x

= cosec x - cot x

Þ

f ¢( p / 2) = - 1 ´ 0 + 1 2

Þ

f ¢( p / 2) = 1 OR

We have, x 1+y +y 1+x =0 Þ x 1+y =-y 1+x x+1 x Þ =y 1+y x2

Þ

y

2

=

x+1 y+1

Þ

x 2 y + x 2 = xy 2 + y 2

Þ

x 2 y - xy 2 + x 2 - y 2 = 0

Þ Þ

xy( x - y) + ( x - y) ( x + y) = 0 ( x - y) ( xy + x + y) = 0 but x ¹ y y(1 + x) = - x \

16.

p/ 2

ò0

p/ 2

ò0

é (1 + x).1 - x ´ 1 ù dy = -ê ú dx (1 + x) 2 êë úû

\

xy + x + y = 0 -x \ y= 1+x -1

=

(1 + x) 2

{ tan x + cot x} dx æ sin x ç ç cos x + è = 2

cos x ö ÷ dx sin x ÷ø p/ 2

ò0

(sin x + cos x) 2 sin x cos x

p/ 2

dx = 2 ò

0

(sin x + cos x) 1 - (sin x - cos x) 2

dx

42


Let sin x - cos x = t (cos x + sin x) dx = dt Now x = 0 Þ t = - 1, and x = \

p/ 2

ò0

p Þt = 1 2

{ tan x + cot x} dx = 2

1

ò- 1

dt 1 - t2

= 2

[sin t] -1

1

-1

= 2 [sin - 1 1 - sin - 1 ( - 1)] = 2 [2 sin - 1 1]. p = 2 2 æç ö÷ = 2 p = RHS è2ø 17. Given curves

x = y2

...(i)

xy = k

...(ii)

Solving (i) and (ii), y 3 = k \y = k 1/ 3 , x = k 2 / 3 Differentiating (i) w. r. t. x, we get dy 1 = 2y dx dy 1 Þ = dx 2y \

1 æ dy ö = = m1 ç ÷ è dx ø ( k 2 / 3 , k 1/ 3 ) 2k 1/ 3

And differentiating (ii) w.r.t. x we get dy x + y=0 dx dy y =dx x \

k 1/ 3 æ dy ö == - kç ÷ 2/ 3 è dx ø ( k 2 / 3 , k 1/ 3 ) k

1/ 3

= m2

\

m1 m2 = – 1 1 1 Þ = -1 Þ k 2/ 3 = 1 / 2 1/ 3 1/ 3 2k k dy 18. Given x + y = x log x dx dy y + = log x dx x This is linear differential equation

Þ

8k 2 = 1 ...(i)

43


Integrating factor I.F. = e

òx

dx

= e log e x = x

Multiplying both sides of (i) by

I.F. = x, we get dy x + y = x log x dx Integrating with respect to x, we get y. x = ò x. log x dx Þ

xy = log x.

Þ

xy =

x 2 log x

1 x2 +C 2 2 2 x 1 y = æç log x - ö÷ + C 2è 2ø

Þ 19.

x2 x2 1 -ò . dx 2 2 x -

Given y 2 = 4ax Þ

2y

Þ

y.

Þ

dy = 4a dx

dy dy y2 (from (i)) = 2a \ y = 2. dx dx 4x dy y which is the required differential equation = dx 2x OR

We have, ( 3xy - y 2 ) dx + ( x 2 + xy) dy = 0 ( 3xy - y 2 ) dx = - ( x 2 + xy) dy 2 dy y - 3xy = dx x 2 + xy

Let

...(i)

y = Vx dy æ dV ö = çV + x ÷ è dx dx ø

\

2 2 dV ö V x - 3x. V . x æ çV + x ÷= è dx ø x 2 + x. Vx

Þ

V+x

Þ

x

2 dV V - 3V = dx 1+V

2 dV V - 3V = -V dx 1+V

44


Þ

2 2 - 4V dV V - 3V - V - V = = dx (1 + V ) 1+V 1 +V dx ò V dV = - 4ò x 1 dx ò V dV + ò dV = - 4ò x log V + V = - 4 log x + C

Þ

log V + log x 4 + V = C

Þ

log (V . x 4 ) + V = C

Þ

æy ö y log ç x 4 ÷ + = C èx ø x

Þ

x

Þ Þ

or x log ( x 3 y) + y = Cx

20. Given ¾®

OA = i$ + j$ + k$ ¾®

OB = 2i$ + 5j$

¾®

OC = 3i$ + 2j$ - 3k$ ¾®

OD = i$ - 6j$ - k$ ¾®

¾®

¾®

¾®

¾®

¾®

AB = OB - OA = i$ + 4j$ - k$

CD = OD - OC = - 2i$ - 8j$ + 2k$ ¾®

CD = - 2(i$ + 4j$ - k$) ¾®

¾®

CD = - 2 AB ¾®

¾®

¾®

¾®

Therefore AB and CD are parallel vector so AB and CD are collinear and angle between them is zero. x-1 y z +1 21. Let =l ...(i) = = 3 2 7 Coordinates of any general point on line (i) is of the form º (1 + 3l , 2l , - 1 + 7 l) For point of intersection (1 + 3l) + 2l - (7 l - 1) = 8 1 + 3l + 2l - 7 l + 1 = 8 - 2l = 6 l=-3 Point of intersection º (– 8, – 6, – 22)

45


\

Required equation of line passing through P ( 4, 6, 2) and Q ( - 8, - 6, - 22) is: x-4 y-6 z-2 = = 4 + 8 6 + 6 2 + 22 x-4 y-6 z-2 z-2 . or x - 4 = y - 6 = \ = = 12 12 24 2 22. Let E be the event that sum of number on two die is 9. E = {( 3, 6), ( 4, 5), (5, 4), ( 6, 3)} 4 1 P(E) = = 36 9 8 P(E ¢) = 9 1 8 8 1 8 8 8 8 1 P (A getting the prize P( A) = + ´ ´ + ´ ´ ´ ´ +...... 9 9 9 9 9 9 9 9 9 ö 1æ 8 2 8 4 8 6 = çç1 + æç ö÷ + æç ö÷ + æç ö÷ + ....÷÷ è 9ø è 9ø è 9ø 9è ø =

1 1 1 92 9 . = . = 9 é æ 8 ö 2 ù 9 ( 9 2 - 8 2 ) 17 ê1 - ç ÷ ú êë è 9 ø úû

SECTION–C 23. Given System of linear equations 2x - y + z = 3 -x + 2y - z = - 4 x - y + 2z = 1 we can write these equations as 1 ù é xù é 3 ù é 2 -1 ê- 1 2 - 1ú ê yú = ê - 4ú ê úê ú ê ú 2 úû êë z úû êë 1úû êë 1 - 1 1ù é 2 -1 Þ A X = B, where, A = ê - 1 2 - 1ú ê ú 2 úû êë 1 - 1 é xù é 3ù ê ú X = y , B = ê - 4ú ê ú ê ú êë z úû êë 1 úû Þ

X = A - 1B

Now,| A| = 2( 4 - 1) - ( - 1) ( - 2 + 1) + 1 (1 - 2) = 6-1-1= 4

...(i)

46


Again Co-factors of elements of matrix A are given by é 2 - 1ù C 11 = ê = 4-1= 3 2 úû ë- 1 é - 1 - 1ù C 12 = - ê = - ( - 2 + 1) = 1 2 úû ë 1 2ù é- 1 C 13 = ê = (1 - 2) = - 1 1 1úû ë é- 1 1 ù C 21 = - ê ú = - ( - 2 + 1) = 1 ë - 1 2û é2 1ù C 22 = ê ú = ( 4 - 1) = 3 ë1 2û é2 - 1 ù C 23 = - ê ú = - ( - 2 + 1) = 1 ë1 - 1û é- 1 C 31 = ê ë 2

1ù = (1 - 2) = - 1 - 1úû

1ù é 2 C 32 = - ê ú = - ( - 2 + 1) = 1 ë - 1 - 1û é 2 - 1ù C 33 = ê = 4-1= 3 2 úû ë- 1 \

\

é 3 1 adj A = (C) = ê 1 3 ê êë - 1 1 é 3 adj. A 1 ê -1 A = = 1 | A| 4 ê êë - 1 T

- 1ù 1ú ú 3 úû 1 - 1ù 3 1ú ú 1 3 úû

\ From (i), we have é xù é 3 1 - 1ù ê yú = 1 ê 1 3 1ú ê ú 4ê ú 3 úû êë z úû êë - 1 1 é xù é 4ù é 1ù 1ê ú ê ú ê ú Þ y = -8 = -2 ê ú 4ê ú ê ú êë z úû êë - 4úû êë - 1 úû Þ

é 3ù ê - 4ú ê ú êë 1 úû

x = 1, y = - 2, z = - 1 OR

A = I 3. A

47


é 2 - 1 4ù é1 0 0ù ê4 0 2ú = ê 0 1 0ú A ê ú ê ú êë 3 - 2 7 úû êë 0 0 1 úû Applying R 2 ® R 2 - 2R 1 4 ù é 1 0 0ù é2 - 1 ê0 2 - 6ú = ê - 2 1 0ú A ê ú ê ú 7 úû êë 0 0 1 úû êë 3 - 2 Applying R 1 ® 1 / 2R 1 1 é1 - 1 2ù é 0 0ù ê ú ê ú 2 2 ê0 2 - 6ú = ê - 2 1 0ú A ê ú ê ú 7 ú ê 0 0 1ú ê3 - 2 ë û ë û Applying R 3 ® R 3 - 3R 1 ù 1 é ù é 1 2ú ê 0 0ú ê1 - 2 2 ê0 2 - 6ú = ê - 2 1 0ú A ú ê ú ê -3 1 ê0 1ú ê 0 1ú úû ë û êë 2 2 Applying R 2 ® R 2 / 2 é 1 1 é ù ê 1 2 ê ú ê 2 2 ê0 1 - 3ú = ê - 1 ê ú ê 1 ê0 1ú ê - 3 ë û ê 2 ë 2 Applying R 1 ® R 1 + R 2 + R 3 é é ù ê -2 0ú ê1 0 ê ê0 1 - 3ú = ê - 1 ê ú ê 1 ê0 1ú ê - 3 ë û ê 2 ë 2

ù 0ú ú 1 0ú A 2 ú 0 1ú úû 0

1 2 1 2

ù 0ú ú 0ú A ú 0 1ú úû

Applying R 3 ® R 3 + 1 / 2R 2 1 é ù é -2 1ù ê ú 0 ú 2 ê1 0 ê ú 1 ê0 1 - 3 ú = - 1 0ú A ê ê ú 2 -1 ú ê0 0 ú ê 1 ê ú 2 1 êë 2 úû ë 4 û Applying R 2 ® R 2 - 6R 3

48


1 ù é ù 1ú 0 ú ê -2 2 0 ú = ê 11 - 1 - 6ú A ú 1ú ê 1 - ú ê -2 1ú û 2û ë 4

é ê1 0 ê0 1 ê ê0 0 ë

Applying R 3 ® - 2R 3 1 é ù 1ú é1 0 0ù ê - 2 2 ê 0 1 0ú = ê 11 - 1 - 6ú A ê ú ê ú êë 0 0 1 úû ê 4 - 1 - 2ú ë û 2 1 é ù 1ú ê -2 2 -1 ê Hence A = 11 - 1 - 6ú ê ú 1 ê 4 - 2ú ë û 2 24. Let DABC be an isosceles triangle inscribed in the ellipse

x2

y2

= 1. a2 b 2 Then coordinates of points A and B are given by ( a cos q , b sin q) and(a cosq, - b sin q) 1 The area of the isosceles D ABC = ´ AB ´ CD 2 1 Þ A(q) = ´ ( 2b sin q) ´ ( a - a cos q) 2 Þ A(q) = ab sin q (1 - cos q) For Amax d( A(q)) =0 dq Þ

+

ab[cos q(1 - cos q) + sin 2 q] = 0 cos q - cos 2 q + sin 2 q = 0

Þ

cos q - cos 2q = 0 2p q= 3

Þ

A (a cos q, b sin q)

D

C

2

Now,

d ( A(q)) dq 2

= ab [- sin q + 2 sin 2q]

2 æ 2p d ( A(q)) 3 3ö , = ab ç -2´ ÷<0 2 3 2 ø è 2 dq 2p Hence for q = , Amax occurs 3

For

q=

B (a cos q, –b sin q)

49


\

Amax = ab sin

2p æ 2p ö ç1 - cos ÷ square units è 3 3ø

3æ 1ö 3 3 ab square units ç1 + ÷ = 2 è 2ø 4

= ab

OR Let r be the radius, l be the slant height and h be the vertical height of a cone of semi - vertical angle a. S = prl + pr 2

Surface area

l=

or

S - pr pr

The volume of the cone 1 1 V = pr 2 h = pr 2 3 3 2 2

2

=

=

...(i)

2

l2 - r 2 a

pr 3

( S - pr )

pr 2 3

( S - pr 2 ) 2 - p 2 r 4

2 2

p r

- r2

p 2r 2

S 2 - 2pSr 2 + p 2 r 4 - p 2 r 4 r pr 2 = ´ = 3 pr 3 2 r S V2 = S( S - 2pr 2 ) = ( Sr 2 - 2pr 4 ) 9 9 dV 2 S = ( 2Sr - 8pr 3 ) dr 9

\

d 2V 2 dr Now Þ

2

=

d 2V 2

dV 2 =0 dr S ( 2Sr - 8pr 3 ) = 0 9

=

r

S( S - 2pr 2 )

S ( 2S - 24pr 2 ) 9

...(ii)

or

Putting S = 4 pr 2 in (ii), 4 pr 2 [8pr 2 - 24pr 2 ] < 0 9

dr 2 Þ V is maximum when S = 4pr 2 Putting this value of S in (i) 4pr 2 = prl + pr 2 or

l h

3pr 2 = prl

S - 4 pr 2 = 0

Þ S = 4 pr 2

50


or \

r 1 = sin a = l 3 - 1 æ1ö a = sin ç ÷ è 3ø

1 Thus V is maximum, when semi vertical angle is sin -1 æç ö÷. è 3ø 25. First finding intersection point by solving the equation of two curves x 2 + y 2 = 16

...(i)

y 2 = 6x

...(ii)

and 2

Þ Þ

x + 6x = 16 2

x + 6x - 16 = 0

Þ x 2 + 8x - 2x - 16 = 0 Þ x( x + 8) - 2( x + 8) = 0 Þ ( x + 8) ( x - 2) = 0 x=-8

(not possible Q y 2 can not be – ve)

(only allowed value) x= 2 y=±2 3 2 3æ y2 ö ç 16 - y 2 ÷ dy Area of OABCO = ò ç 0 6 ÷ø è

or \

é 2 2 êò a - x êë

éy 16 = ê 16 - y 2 + sin 2 êë 2 x a2 xù = a2 - x2 + sin - 1 ú 2 2 a úû

4 3 16 4 3 40 + p + 8p = + p 3 3 3 3 4 = ( 3 + 10p ) sq. units 3

=

O

B (4, 0) 2, –2 3

y y3 ù ú 4 18 úû 0

æ2 3 8 ö 1 Required are = 2ç + p÷ + (p4 2 ) 3 ø 2 è 3

2, 2 3 C

2 3

1

é 3 24 3 ù = ê 3 . 16 - 12 + 8 sin - 1 ú 2 18 û ë p 4 ù 2 4 8 2 é = 3= ê 3.2+ 8 + p= ú 3 3 3û 3 3 ë \

A

3+

8 p 3

51

Examination Papers – 2008 p

x tan x

0

sec x + tan x

26. I = ò

...(i)

dx

a

a

0

0

Using property ò f ( x) dx = ò f ( a - x) dx, we have \

I=ò

( p - x) tan ( p - x)

p

dx sec ( p - x) + tan ( p - x) ( p - x) ( - tan x) dx - sec x - tan x p . tan x x . tan x p dx - ò dx 0 sec x + tan x sec x + tan x

0

I=ò

p

0

I=ò

p

0

...(ii)

Adding (i) and (ii) we have tan x p 2I = p ò dx 0 sec x + tan x p sin x Þ 2I = p ò dx 0 1 + sin x [ f ( x) = f ( 2a - x)] then ò

2a

0

Þ Þ

2I = p ´ 2 ´ ò

a

f ( x) dx = 2 . ò f ( x) dx 0

sin x

p/ 2

0

1 + sin x sin x + 1-1 p/ 2 I=pò dx 0 1 + sin x p/ 2

dx - p ò

p/ 2

1 dx 1 + sin x

Þ

I=pò

Þ

I=p

Þ

I=

Þ

I=

Þ

p/ 2 é tan x ù p p ê 2ú I= - .ê ú 2 2 ê 1 ú ë 2 û0

0

0

dx

p/ 2 p 1 -pò dx 0 2 1 + cos x

p/ 2 p2 1 -pò dx 0 x 2 2 cos 2 2

p 2 p p/ 2 x - .ò sec 2 . dx 0 2 2 2 2

I=

p2 p p - ´ 2 ´ é tan - tan 0ù úû 2 2 4 ëê

I=

p2 -p 2

é Using a f ( x) dx = a f ( a - x) dxù ò0 ò0 êë úû

52


x+2

=

2y + 3

=

3z + 4

=l 3 4 5 Any general point on the line is 4l - 3 5l - 4 , 3l - 2, 2 3 Now, direction ratio if a point on the line is joined to ( - 2, 3, - 4) are 4l - 9 5l + 8 Þ 3l , , 2 3 Now the distance is measured parallel to the plane 4x + 12y - 3z + 1 = 0 æ 4l - 9 ö æ 5l + 8 ö \ 4 ´ 3l + 12 ´ ç ÷- 3´ç ÷=0 è 2 ø è 3 ø

27. Let

Þ

12l + 24l - 54 - 5l - 8 = 0 31l - 62 = 0 Þ l=2 5 \ The point required is æç 4, , 2ö÷ . è 2 ø \

2 5 Distance = ( 4 + 2) 2 + æç - 3ö÷ + ( 2 + 4) 2 è2 ø

=

36 + 36 +

1 = 4

289 17 units = 4 2

28. Let there be x tickets of first class and y tickets of second class. Then the problem is to max z = 400x + 300y x = 20 Subject to x + y £ 200 y x ³ 20 x = 40 x + 4x £ 200 (0, 200) (20, 180) (40, 160) 5x £ 200 x £ 40 The shaded region in the graph represents the feasible x region which is proved. (200, 0) (20, 0) Le us evaluate the value of z at each corner point z at ( 20, 0), z = 400 ´ 20 + 300 ´ 0 = 8000 z at ( 40, 0) = 400 ´ 40 + 300 ´ 0 = 16000 z at ( 40, 160) = 400 ´ 40 + 300 ´ 160 = 16000 + 48000 = 64000 z at ( 20, 180) = 400 ´ 20 + 300 ´ 180 = 8000 + 54000 = 62000 max z = 64000 for x = 40, y = 160 \ 40 tickets of first class and 160 tickets of second class should be sold to earn maximum profit of Rs. 64,000.

53


29.

A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six. [CBSE 2005] Sol. Let E be the event that the man reports that six occurs in the throwing of the die and let S1 be the event that six occurs and S2 be the event that six does not occur. 1 Then P ( S1 ) = Probability that six occurs = 6 5 P ( S2 ) = Probability that six does not occur = 6 P (E S1 ) = Probability that the man reports that six occurs when six has actually occurred on the die 3 = Probability that the man speaks the truth = 4 P (E S2 ) = Probability that the man reports that six occurs when six has not actually occurred on the die 3 1 = Probability that the man does not speak the truth = 1 - = . 4 4 Thus, by Bayes’ theorem, we get P ( S1 E) = Probability that the report of the man that six has occurred is actually a six 1 3 ´ P ( S1 ) P (E S1 ) 3 6 4 = = = × 1 3 5 1 P ( S1 ) P (E S1 ) + P ( S2 ) P (E S2 ) 8 ´ + ´ 6 4 6 4

Set–II a 20. Let D = a + 2b a+b

a +b

a + 2b

a

a+b

a + 2b

a

Applying R 1 ® R 1 + R 2 + R 3 , we have 3( a + b) 3( a + b) 3( a + b) D=

a + 2b

a

a+b

a+b

a + 2b

a

Taking out 3( a + b) from 1st row, we have 1 1 1 D = 3( a + b) a + 2b a+b

a

a+b

a + 2b

a

Applying C 1 ® C 1 - C 2 and C 2 ® C 2 - C 3

54


0

0

D = 3( a + b) 2b

1

-b a+b

-b

2b

a

Expanding along first row, we have D = 3( a + b) [1. ( 4b 2 - b 2 )] = 3 ( a + b) ´ 3b 2 = 9b 2 ( a + b) 21. Let I = ò

p/ 2

0 p/ 2

Þ I=ò

0

Þ I=ò

p/ 2

0

...(i)

log sin x dx p log sin æç - xö÷ dx è2 ø

...(ii)

log cos x dx

Adding (i) and (i) we have, 2I = ò

p/ 2

0 p/ 2

Þ 2I = ò

0

Þ 2I = ò

p/ 2

0

Þ 2I = ò

p/ 2

0 p/ 2

Þ 2I = ò

0

Let 2x = t

(log sin x + log cos x) dx log sin x cos x dx log

2 sin x cos x

dx

2

(log sin 2 x - log 2) dx log sin 2x dx - ò

p/ 2

0

Þ

dx =

log 2dx

dt 2

p x = 0, , t = 0, p 2 1 p p \ 2I = ò log sin t dt - log 2. æç - 0ö÷ 0 è ø 2 2 p Þ 2I = I - log 2 2 p Þ 2I - I = - log 2 2 p Þ I = - log 2 2 22. We have dy (1 + x 2 ) + y = tan - 1 x dx When

Dividing each term by (1 + x 2 )

éQ a f ( x) dx = a f (t) dtxù ò0 ëê ò0 ûú

55


dy 1 tan - 1 x + .y= dx 1 + x 2 1 + x2 Clearly, it is linear differential equation of the form P=

So,

\

1

and Q =

1 + x2

dy + P. y = Q dx

tan - 1 x 1 + x2

Integrating factor, I. F. = e

ò

ò P dx

=e

1 1 + x2

dx

= e tan

- 1x

Therefore, solution of given differential equation is y ´ I . F. = ò Q ´ I . F. dx Þ y . e tan

- 1x

=ò

I=ò

Let

- 1x

tan - 1 x 1+x

2

. e tan

tan - 1 x e tan

e tan

Also

tan -1 x = log t I = ò log t dt

Þ

I = t log t - t + C I=e

tan - 1 x

. tan

dx e tan

Þ

=t

Þ

Þ

- 1x

1 + x2

Let

- 1 x dx

- 1x

1 + x2

dx = dt

[Integrating by parts] -1

x-e

tan - 1 x

+C

Hence required solution is y. e tan Þ

- 1x

= e tan

- 1x

(tan - 1 x - 1) + C

y = (tan - 1 x - 1) + C e -

tan - 1 x

27. The given system of linear equations. 3x - 2y + 3z = 8 2x + y - z = 1 4x - 3y + 2z = 4 We write the system of linear equation in matrix form é 3 - 2 3 ù é xù é 8ù ê 2 1 - 1ú ê yú = ê1 ú ê úê ú ê ú êë 4 - 3 2 úû êë z úû êë 4úû

56


é3 - 2 3 ù Þ A. X = B, where A = ê 2 1 - 1ú , X = ê ú êë 4 - 3 2 úû

é xù é 8ù ê yú and B = ê 1 ú ê ú ê ú êë z úû êë 4úû

X = A -1 B

Þ

Now, co-factors of matrix A are C 11 = ( - 1) 1 + 1 . ( 2 - 3) = ( - 1) 2 . ( - 1) = - 1 C 12 = ( - 1) 1 + 2 . ( 4 + 4) = ( - 1) 3 . 8 = - 8 C 13 = ( - 1) 1 + 3 . ( - 6 - 4) = ( - 1) 4 . ( - 10) = - 10 C 21 = ( - 1) 2 + 1 ( - 4 + 9) = ( - 1) 3 (5) = - 5 C 22 = ( - 1) 2 + 2 . ( 6 - 12) = ( - 1) 4 ( - 6) = - 6 C 23 = ( - 1) 2 + 3 ( - 9 + 8) = ( - 1) 5 ( - 1) = 1 C 31 = ( - 1) 3 + 1 ( 2 - 3) = ( - 1) 4 ( - 1) = - 1 C 32 = ( - 1) 3 + 2 ( - 3 - 6) = ( - 1) 5 . ( - 9) = 9 C 33 = ( - 1) 3 + \

adj A = c

and

T

3

( 3 + 4) = ( - 1) 6 7 = 7

é - 1 - 5 - 1ù =ê-8 -6 9 ú ê ú 7 úû êë - 10 1 3 -2 3

|A| = 2

1 - 1 = 3 ( 2 - 3) + 2 ( 4 + 4) + 3( - 6 - 4)

4 -3

\ A -1

Where c = matric of co-factors of elements.

2

= 3 ´ - 1 + 2 ´ 8 + 3 ´ - 10 = - 3 + 16 - 30 = - 17 - 5 - 1ù é -1 adj A 1 ê = =-8 -6 9ú ú |A| 17 ê 1 7 úû êë - 10

Now, X = A -1 B Þ

Þ \

é xù ê yú = - 1 ê ú 17 êë z úû é xù ê yú = - 1 ê ú 17 êë z úû x = 1, y = 2, z = 3

é -1 ê-8 ê êë - 10 é-8 ê - 64 ê êë - 80

- 5 - 1ù é 8ù - 6 9 ú ê1 ú ú ê ú 1 7 úû êë 4úû -5 - 4 ù 1 - 6 + 36ú = ú 17 + 1 + 28 úû

é - 17 ù é 1 ù ê - 34ú = ê 2 ú ê ú ê ú êë - 51 úû êë 3úû

57


OR For elementary transformation we have, A = IA é 2 5 3ù é1 0 0ù ê 3 4 1 ú = ê 0 1 0ú A Þ ê ú ê ú êë 1 6 2 úû êë 0 0 1 úû Applying R 1 ® R 1 - R 3 é 1 -1 1 ù é 1 0 -1 ù ê 3 4 1ú = ê0 1 0 ú A Þ ê ú ê ú êë 1 6 2úû êë 0 0 1 úû Applying R 2 ® R 2 - 3R 1 , R 3 ® R 3 - R 1 é 1 - 1 1 ù é 1 0 -1 ù ê 0 7 -2 ú = ê -3 1 3 ú A Þ ê ú ê ú 1 úû êë -1 0 2 úû êë 0 7 4 Applying R 2 ® R 2 7 é 1 - 1 1 ù é 1 0 -1 ù ê -2 ú ê -3 1 3 ú Þ = A ê0 1 7 ú ê7 7 7ú ê0 7 ú ê ú 1 û ë -1 0 2 û ë Applying R 1 é1 ê ê ê0 ê0 ê ë

® R1 + R2 5 ù é 4 0 7 ú ê7 -2 ú ê -3 1 ú=ê 7 ú ê7 7 1 -1 ú ê û ë

Applying R 3 é1 ê ê ê0 ê0 ê ë

® R 3 + 7R2 5 ù é 4 0 7 ú ê7 -2 ú ê -3 1 ú=ê 7 ú ê7 0 3 2 ú ê û ë R3 ® 3 5 ù é 4 0 ú 7 ú ê7 -2 ú ê -3 1 =ê 7 ú ê7 ú ê -2 0 1 úû ë 3

Applying R 3 é ê1 ê ê0 ê ê ê0 ë

1 7 1 7 0

-4 ù 7 ú 3ú úA 7 ú 2 ú û

1 7 1 7 -1

-4 ù 7 ú 3ú úA 7 ú -1 ú û

1 7 1 7 -1 3

-4 ù 7 ú 3ú úA 7 ú -1 ú 3û

58


R1 ® R1

-5 R3, 7

é ù é 2 ê 1 0 0ú ê 21 ê ú ê -5 ê 0 1 0ú = ê 21 ê 0 0 1 ú ê -2 ê ú ê ë û ë 3

\

A -1

2 R3 7 8 -1 ù 21 3 ú 1 1 ú ú 21 3 ú -1 -1 ú 3 3û

R2 ® R2 +

8 -1 ù 21 3 ú 1 1 ú ú 21 3 -1 -1 ú ú 3 3û 8 -7 ù é 2 1 ê = -5 1 7ú ú 21 ê êë +14 -7 -7 úû

é 2 ê 21 ê -5 = ê ê 21 2 ê ë 3

28. Let S = Event of insurance of scooter driver C = Event of insurance of Car driver T = Event of insurance of Truck driver and A = Event of meeting with an accident Now, we have, P( S) = Probability of insurance of scooter driver 2000 2 Þ P( S) = = 9000 9 P (C) = Probability of insurance of car driver 3000 3 Þ P(C) = = 9000 9 P(T) = Probability of insurance of Truck driver 4000 4 Þ P(T) = = 9000 9 and, P( A / S) = Probability that scooter driver meet. with an accident Þ P ( A / S) = 0.04 P ( A / C) = Probability that car driver meet with an accident Þ P ( A / C) = 0.06 P ( A / T) = Probability that Truck driver meet with an accident Þ P ( A / T) = 0.15 By Baye’s theorem, we have the required probability

59


P(C). P( A / C)

P(C / A) =

P( S) . P( A / S) + P(C) .P( A / C) + P(T). P( A / T) 3 ´ 0.06 9 = 2 3 4 ´ 0.04 + ´ 0.06 + ´ 0.15 9 9 9 3 ´ 0.06 0.18 = = 2 ´ 0.04 + 3 ´ 0.06 + 4 ´ 0.15 0.08 + 0.18 + 0.60 0.18 18 9 = = = 0.86 86 43

29. Given,

x + 2y = 2 y–x=1 2x + y = 7 On plotting these lines, we have

...(i) ...(ii) ...(iii) Y

(0,7) Y

(2,3) B 2x+ 7 y=

(0,1) A X'

( 7 ,0) E2 C'

D (2,0)

1 (–1,0)O x= y–

(4,–1)

X

C x +2y =2

Y'

Area of required region 3

=

ò

-1

7-y dy 2 3

1

ò

-1

3

( 2 - 2y) dy -

ò

( y - 1) dy

1 3

é y2 ù y2 ù 1é 2 1 = ê7 y - yú ú - [2y - y ] -1 - ê 2 êë 2 úû êë 2 úû 1 -1

60


=

1æ 9 1ö 1 æ9 ö ç 21 - + 7 + ÷ - ( 2 - 1 + 2 + 1) - ç - 3 - + 1÷ è ø è ø 2 2 2 2 2

= 12 - 4 - 2 = 6 sq. units

Set–III 20. We have a b c D= b c a c a b Applying C 1 ® C 1 + C 2 + C 3 , we have ( a + b + c) b c D = ( a + b + c) c a ( a + b + c) a b taking out ( a + b + c) from Ist column, we have 1 b c D = ( a + b + c) 1 c a 1 a b Interchanging column into row, we have 1 1 1 D = ( a + b +c) b c a c a b Applying C 1 ® C 1 - C 2 and C 2 ® C 2 - C 3 , we have 0 0 1 D = ( a + b +c) b - c c - a a c - a a-b b Expanding along Ist row, we have D = ( a + b + c) [1 (b - c) ( a - b) - ( c - a) 2 ] = ( a + b + c) (ba - b 2 - ca + bc - c 2 - a 2 + 2ac) Þ

D = ( a + b + c) ( ab + bc + ca - a 2 - b 2 - c 2 )

D = - ( a + b + c) ( a 2 + b 2 + c 2 - ab - bc - ca) 1 Þ D = - ( a + b + c) {( a - b) 2 + (b - c) 2 + ( c - a) 2 } 2 Here, ( a + b + c) is positive as a, b , c are all positive and it is clear that ( a - b) 2 + (b - c) 2 + ( c - a) 2 is also positive 1 Hence D = - ( a + b + c) [( a - b) 2 + (b - c) 2 + ( c - a) 2 ] has negative value. 2 Þ

61


21. Let

I = ò cot - 1 (1 - x + x 2 ) dx 0 1

= ò tan 0 1

= ò tan 0 1

= ò [tan 0 1

= ò tan -

1

1

2

1-x+x

x + (1 - x)

1

x = tan -

é ì a+b ü x + tan - 1 (1 - x)] dx êQ tan - 1 í ý = tan 1 ab êë îï þ

1

1 ù x úû

1

ù a + tan - 1 b ú úû

1

0 1

= ò tan - 1 xdx + ò tan - 1 [1 - (1 - x)] dx 0

0 1

1

= 2 ò tan - 1 xdx = 2 ò tan 0

1

é êQ êë

a

a ù f ( x ) = ò ò f ( a - x) dx úú û 0 0

x. 1 dx, integrating by parts, we get

0

é = 2 ê {tan êë

1

1

ù × xdx ú 2 úû 01+ x

1

1

x. x}0 - ò 1

= 2 [tan - 1 1 - 0] - ò

01

2x +x

2

dx = 2 ×

p - [log (1 + x 2 )] 10 4

p p = - (log 2 - log 1) = - log 2 2 2 22. We have the differential equation dy x logx + y = 2 log x dx dy 1 2 Þ + .y = dx x log x x It is linear differential equation of the from P=

[Q log 1 = 0 ]

dy + Py = Q dx

1 2 and Q = x log x x 1

Now,

1

x dx + ò tan - 1 (1 - x) dx

0 1

So, Here

1

[Q 1 can be written as x + 1 - x]

dx

1 - x (1 - x)

1

éQ cot êë

dx

pdx I.F. = e ò =e

ò x log x dx

= e log |log

x|

= log x Hence, solution of given differential equation is y ´ I . F. = ò Q ´ I . F dx

62


2 . log x dx x

Þ

y log x = ò

Þ

y log x = 2 ò

Þ

y log x = (log x) 2 + C

(log x) 1 . log x dx = 2. x 2

2

+C

27. The given system of linear equations is x+ y+z = 4 2x + y - 3z = - 9 2x - y + z = - 1 We write the system of equation in Matrix form as 1 ù é xù é 4 ù é1 1 ê 2 1 - 3ú ê yú = ê - 9ú ê úê ú ê ú êë 2 - 1 1 úû êë z úû êë - 1 úû Þ AX = B, we have 1 ù é1 1 ê A = 2 1 - 3ú , X = ê ú êë 2 - 1 1 úû \

é xù é 4 ù ê yú and B = ê - 9ú ê ú ê ú êë z úû êë - 1 úû

X = A -1 B

Now, co-factors of A C 11 = ( - 1) 1 + C 13 = ( - 1)

1

1+ 3

C 22 = ( - 1) 2 +

C 12 = ( - 1) 1 +

(1 - 3) = - 2;

2

\

3

2+ 1

( 2 + 6) = - 8

( - 2 - 2) = - 4;

C 21 = ( - 1)

(1 - 2) = - 1;

C 23 = ( - 1) 2 + 3 ( - 1 - 2) = 3

C 31 = ( - 1) 3 + 1 ( - 3 - 1) = - 4; C 33 = ( - 1) 3 +

2

C 32 = ( - 1) 3 + 2 ( - 3 - 2) = 5

= (1 - 2) = - 1

é - 2 - 2 - 4ù adj A = (C) T = ê - 8 - 1 5 ú ê ú êë - 4 3 - 1 úû

Now,|A| = 1 ( - 2) - 1 ( 8) + 1 ( - 4) = - 2 - 8 - 4 = - 14 adj. A \ A- 1 = | A| é - 2 - 2 - 4ù ê- 8 - 1 5 ú ê ú êë - 4 3 - 1 úû 1 = = - 14 14

(1 + 1) = - 2

4 ù é2 2 ê 8 1 - 5ú ê ú êë 4 - 3 1 úû


Now, X = A -1 B Þ

Þ

Þ \

4 ùé 4 ù é xù é2 2 ê yú = . 1 ê 8 1 - 5ú ê - 9ú ê ú 14 ê úê ú êë z úû êë 4 - 3 1 úû êë - 1 úû é xù é 8 + ( - 18) + ( - 4) ù ê yú = 1 ê 32 + ( - 9) + 5 ú ê ú 14 ê ú êë z úû êë16 + 27 + ( - 1) úû é xù é - 14ù é - 1ù ê yú = 1 ê 28 ú = ê 2 ú ê ú 14 ê ú ê ú êë z úû êë 42 úû êë 3 úû x = - 1, y = 2 and z = 3 is the required solution. OR é 2 5 3ù Let A = ê 3 4 1 ú ê ú êë 1 6 3úû

Therefore, for elementary row transformation, we have A=I A é 2 5 3ù é1 0 0ù ê 3 4 1 ú = ê 0 1 0ú A Þ ê ú ê ú êë 1 6 3úû êë 0 0 1 úû Applying R 1 ® R 1 - R 3 é 1 -1 0 ù é 1 0 -1 ù ê 3 4 1 ú = ê0 1 0 ú A ê ú ê ú êë 1 6 3úû êë 0 0 1 úû Applying R 2 ® R 2 - 3R 1 é 1 -1 0 ù é 1 0 -1 ù ê 0 7 1 ú = ê -3 1 3 ú A ê ú ê ú êë 1 6 3úû êë 0 0 1 úû ® R 3 - R1 -1 0 ù é 1 0 -1 ù 7 1 ú = ê -3 1 3 ú A ú ê ú 7 3úû êë -1 0 2 úû 1 Applying R 1 ® R 1 + R 2 7 Applying R 3 é1 ê0 ê êë 0

63

64


é1 0 1 ù é 4 ê 7ú ê 7 ê 0 7 1 ú = ê -3 ê ú ê ê 0 7 3 ú ê -1 ë û ë R2 Applying R 2 ® 7 é1 0 1 ù é 4 ê 7ú ê 7 ê 1 ú ê -3 ê0 1 ú=ê 7 ê 0 7 3 ú ê -71 ê ú ê ë û ë

1 7 1 0

1 7 1 7 0

-4 ù 7 ú 3úA ú 2ú û

-4 ù 7 ú 3ú úA 7 ú 2 ú û

® R 3 - 7R2 1ù é 4 1 -4 ù 0 ú ê 7 7 7 7 ú 1 ú ê -3 1 3ú 1 ú=ê úA 7ú ê 7 7 7 ú 0 2 2 -1 -1 ú ê ú û ë û R3 Applying R 3 ® 2 1 -4 ù é1 0 1 ù é 4 ê ú ê 7 7 7 7 ú ê 1 ú ê -3 1 3ú ê0 1 ú=ê úA ê 0 0 71 ú ê 7 -71 -71 ú ê ú ê1 ú 2 2û ë û ë 1 1 Applying R 1 ® R 1 - R 3 , R 2 ® R 2 - R 3 7 7 3 -1 ù é 3 é 1 0 0ù ê 7 14 2 ú 1 ú ê 0 1 0 ú = ê -4 3 ê úA ê ú 7 14 2 ú ê êë 0 0 1 úû -1 -1 ê1 ú 2 2û ë 3 -1 ù é 3 ê 7 14 2 ú 3 -7 ù é6 3 1 ú 1 ê -1 ê -4 \ A =ê -8 3 7 ú ú= ê ú 2 14 ê 7 14 êë 14 -7 -7 úû -1 -1 ú ê1 ú 2 2û ë Applying R 3 é1 ê ê ê0 ê0 ê ë

28. The equations of the given curves are x2 + y2 = 1

...(i)

65


and,

( x - 1) 2 + ( y - 0) 2 = 1

...(ii)

Clearly, x 2 + y 2 = 1 represents a circle with centre at ( 0, 0) and radius unity. Also, ( x - 1) 2 + y 2 = 1 represents a circle with centre at (1, 0) and radius unity. To find the points of intersection of the given curves, we solve (1) and (2) simultaneously. Thus,

1 - ( x - 1) 2 = 1 - x 2

y 1 A( 1 , 3 ) 2 2 P (x, y ) 1 2 We find that the two curves intersect at Q (x, y2) A (1 / 2, 3 / 2) and D (1 / 2, - 3 / 2). y1 y2 x Since both the curves are symmetrical about x-axis. O Dx B (1, 0) x' So, Required area = 2 (Area OABCO) 1 Dx C( 2 , 0) Now, we slice the area OABCO into vertical strips. (x – 1)2 + y2 = 1 We observe that the vertical strips change their character at A(1 / 2, 3 / 2). So. y' Area OABCO = Area OACO + Area CABC. When area OACO is sliced into vertical strips, we find that each strip has its upper end on the circle ( x - 1) 2 + ( y - 0) 2 = 1 and the lower end on x-axis. So, the approximating rectangle

2x = 1

Þ

x=

x2

+ y2

=1

Þ

shown in Fig. has, Length = y 1 , Width = Dx and Area = y 1 Dx. As it can move from x = 0 to x = 1 / 2. \

Area OACO = ò

1/ 2

0

Þ Area OACO = ò

1/ 2

0

y 1 dx 1 - ( x - 1)

2

dx

éQ P ( x, y ) lies on ( x - 1) 2 + y 2 = 1 ù 1 ê ú ê\( x - 1) 2 + y 2 = 1 Þ y = 1 - ( x - 1) 2 ú 1 1 ë û

Similarly, approximating rectangle in the region CABC has, Length, = y 2 , Width Dx and Area 1 = y 2 Dx. As it can move form x = to x = 1. 2 \

Area CABC = ò

1

1/ 2

Þ Area CABC = ò

1

1/ 2

y 2 dx 1 - x 2 dx

éQQ ( x, y ) lies on x 2 + y 2 = 1ù 2 ê ú ê\ x 2 + y 2 = 1 Þ y = 1 - x 2 ú 2 2 ë û

Hence, required area A is given by 1/ 2 1 A = 2 éê ò 1 - ( x - 1) 2 dx + ò 1/ 2 ë0 Þ

1 - x 2 dxùú û

éé 1 1 A = 2 ê ê . ( x - 1) 1 - ( x - 1) 2 + sin 2 êë ë 2

1/ 2

1

æ x - 1 öù ç ÷ è 1 ø úû 0

1 é1 + ê x 1 - x 2 + sin 2 ë2

1

1 ù æ x öù ú ç ÷ú è 1 ø û 1/ 2 ú û

66


Þ

éì 3 A = êí + sin ëîï 4

Þ

A= -

1

æç è

ü ì 1ö 3 1 üù -1 -1 (1) - sin - 1 æç ö÷ ý ú ÷ - sin ( - 1)ý + í sin ø è 2 4 2 ø þû þ îï

3 p p p 3 p æ 2p 3ö - + + - =ç ÷ sq. units 4 6 2 2 4 6 è 3 2 ø

29. Let S = Event of insuring scooter driver C = Event of insuring Car driver T = Event of insuring Truck driver and A = Event of meeting with an accident. Now, we have 3000 3 = 15000 15 5000 5 P(C) = Probability of insuring car driver = = 15000 15 7000 7 P(T) = Probability of insuring Truck driver = = 15000 15 and, P( A / S) = Probability that scooter driver meet with an accident = 0.04 P ( A / C) = Probability that car driver meet with an accident = 0.05 P ( A / T) = Probability that Truck driver meet with an accident = 0.15 By Baye‘s theorem, we have P(C). P( A / C) Required probability = P(C / A) = P( S) . P( A / S) + P(C) .P( A / C) + P(T). P( A / T) 5 ´ 0.05 15 = 3 5 7 ´ 0.04 + ´ 0.05 + ´ 0.15 15 15 15 5 ´ 0.05 = 3 ´ 0.04 + 5 ´ 0.05 + 7 ´ 0.15 0.25 = 0.12 + 0.25 + 1.05 0.25 25 = = 1.42 142 P( S) = Probability of insuring scooter driver =


Maximum marks: 100

General Instructions: 1. All questions are compulsory. 2. The question paper consists of 29 questions divided into three Sections A, B and C. Section A comprises of 10 questions of one mark each, Section B comprises of 12 questions of four marks each and Section C comprises of 7 questions of six marks each. 3. All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question. 4. There is no overall choice. However, internal choice has been provided in 4 questions of four marks each and 2 questions of six marks each. You have to attempt only one of the alternatives in all such questions. 5. Use of calculators is not permitted.

Set–I SECTION–A ®

®

® ®

®

1. Find the projection of a on b if a . b = 8 and b = 2i$ + 6j$ + 3k$ . ®

2. Write a unit vector in the direction of a = 2i$ - 6j$ + 3k$ . ®

®

3. Write the value of p, for which a = 3i$ + 2j$ + 9k$ and b = i$ + pj$ + 3k$ are parallel vectors. 4. If matrix A = (1 2 3), write AA', where A' is the transpose of matrix A. 2 3 4 5. Write the value of the determinant 5 6 8 . 6x 9x 12x 6. Using principal value, evaluate the following: 3p ö sin -1 æç sin ÷ è 5 ø 7. Evaluate : ò 1

sec 2 x dx . 3 + tan x

8. If ò ( 3x 2 + 2x + k) dx = 0, find the value of k . 0

68


9. If the binary operation * on the set of integers Z, is defined by a * b = a + 3b 2 , then find the value of 2 * 4. 10. If A is an invertible matrix of order 3 and|A|= 5, then find|adj. A|.

SECTION–B ®

®

®

®

®

®

®

®

®

®

®

®

®

®

11. If a ´ b = c ´ d and a ´ c = b ´ d show that a - d is parallel to b - c , where a ¹ d and ®

®

b ¹ c.

4 5 16 p 12. Prove that: sin - 1 æç ö÷ + sin - 1 æç ö÷ + sin - 1 æç ö÷ = è5 ø è 13 ø è 65 ø 2 OR p Solve for x : tan -1 3x + tan -1 2x = 4 13. Find the value of l so that the lines 1 - x 7 y - 14 5z - 10 7 - 7x y - 5 6 - z and . = = = = 3 2l 11 3l 1 5 are perpendicular to each other. 14. Solve the following differential equation: dy + y = cos x - sin x dx 15. Find the particular solution, satisfying the given condition, for the following differential equation: dy y y - + cosec æç ö÷ = 0; y = 0 when x = 1 è dx x xø 16. By using properties of determinants, prove the following: x + 4 2x 2x 2x

x+4

2x

2x

2x

x+4

= (5x + 4)( 4 - x) 2

17. A die is thrown again and again until three sixes are obtained. Find the probability of obtaining the third six in the sixth throw of the die. 18. Differentiate the following function w.r.t. x : x sin x + (sin x) cos x . 19. Evaluate :

ò

ex 5 - 4e x - e 2 x

dx

OR Evaluate : ò

( x - 4) e x ( x - 2) 3

dx

20. Prove that the relation R on the set A = { 1, 2, 3, 4, 5} given by R = {( a, b) :|a - b|is even }, is an equivalence relation.

69


21. Find

dy if ( x 2 + y 2 ) 2 = xy. dx OR

If y = 3 cos (log x) + 4 sin(log x), then show that x 2 .

d2y 2

dx 22. Find the equation of the tangent to the curve y = 4x - 2y + 5 = 0 . OR

+x

dy + y= 0. dx

3x - 2 which is parallel to the line

Find the intervals in which the function f given by f ( x) = x 3 + (i) increasing

1 x3

, x ¹ 0 is

(ii) decreasing.

SECTION–C 23. Find the volume of the largest cylinder that can be inscribed in a sphere of radius r. OR A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m 3 . If building of tank costs Rs. 70 per sq. metre for the base and Rs. 45 per sq. metre for sides, what is the cost of least expensive tank? 24. A diet is to contain at least 80 units of Vitamin A and 100 units of minerals. Two foods F1 and F2 are available. Food F1 costs Rs. 4 per unit and F2 costs Rs. 6 per unit. One unit of food F1 contains 3 units of Vitamin A and 4 units of minerals. One unit of food F2 contains 6 units of Vitamin A and 3 units of minerals. Formulate this as a linear programming problem and find graphically the minimum cost for diet that consists of mixture of these two foods and also meets the minimal nutritional requirements. 25. Three bags contain balls as shown in the table below: Bag

Number of White balls

Number of Black Number of Red balls balls

I

1

2

3

II

2

1

1

III

4

3

2

A bag is chosen at random and two balls are drawn from it. They happen to be white and red. What is the probability that they came from the III bag? 26. Using matrices, solve the following system of equations: 2x - 3y + 5z = 11 3x + 2y - 4z = -5 x + y - 2z = -3 p

27. Evaluate:

e cos x

ò e cos x + e - cos x dx . 0

70


OR p/2

Evaluate:

ò ( 2 log sin x - log sin 2x) dx . 0

28. Using the method of integration, find the area of the region bounded by the lines 2x + y = 4, 3x - 2y = 6 and x - 3y + 5 = 0 . 29. Find the equation of the plane passing through the point (-1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0 .

Set–II Only those questions, not included in Set I, are given. 2. Evaluate: ò sec 2 (7 - x) dx ®

7. Write a unit vector in the direction of b = 2i$ + j$ + 2k$ . 11. Differentiate the following function w.r.t. x : y = (sin x) x + sin -1 x . 18. Find the value of l so that the lines

1-x y-2 z- 3 = = 3 2l 2

and

x-1 y-1 6-z are = = 3l 1 7

perpendicular to each other. 19. Solve the following differential equation : dy (1 + x 2 ) + y = tan -1 x . dx 21. Using the properties of determinants, prove the following: a b c a - b b - c c - a = a 3 + b 3 + c 3 - 3abc . b + c c + a a+b 23. Two groups are competing for the position on the Board of Directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3, if the second group wins. Find the probability that the new product was introduced by the second group. 26. Prove that the curves y 2 = 4x and x 2 = 4y divide the area of the square bounded by x = 0, x = 4, y = 4 and y = 0 into three equal parts.

Set–III Only those questions, not included in Set I and Set II, are given. (1 + log x) 2 4. Evaluate : ò dx x

71

Examination Papers – 2009 ®

®

9. Find the angle between two vectors a and b with magnitudes 1 and 2 respectively and when ®

®

| a ´ b |=

3.

15. Using properties of determinants, prove the following: 1 + a2 - b 2 2ab 2b

2ab 1 - a2 + b 2

-2b 2a

-2 a

2

1- a -b

= (1 + a2 + b2)3 2

17. Differentiate the following function w.r.t. x : ( x) cos x + (sin x) tan x 19. Solve the following differential equation: dy x log x + y = 2 log x . dx 20. Find the value of l so that the following lines are perpendicular to each other. 2 - y 1 - z x 2y + 1 1 - z x-5 . = = ; = = 5l + 2 5 -1 1 4l -3 24. Find the area of the region enclosed between the two circles x 2 + y 2 = 9 and ( x - 3) 2 + y 2 = 9. 27. There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up tail 25% of the times and the third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?

SOLUTIONS Set–I SECTION–A 1. Given

® ®

a.b =8 ®

b = 2i$ + 6j$ + 3k$ ®

® ®

®

We know projection of a on b =

a.b ®

|b| = 2. Given

8 8 = 4 + 36 + 9 7

®

a = 2i$ - 6j$ + 3k$ ®

Unit vector in the direction of a =

®

a

®

| a|

= a$

72


2i$ - 6j$ + 3k$ 4 + 36 + 9 2$ 6$ 3 $ a$ = i - j + k 7 7 7 a$ =

Þ Þ ®

®

®

®

3. Since a || b , therefore a = l b Þ 3i$ + 2j$ + 9k$ = l(i$ + pj$ + 3k$) Þ

l = 3 , 2 = lp , 9 = 3 l 2 or l = 3, p = 3 4. Given A = (1 2 3) æ1 ö ç ÷ A ¢ = ç 2÷ ç 3÷ è ø AA ¢ = (1 ´ 1 + 2 ´ 2 + 3 ´ 3) = (14) 2 3 4 5. Given determinant|A|= 5 6 8 6x 9x 12x 2 Þ

|A|= 3x 5 2

6. \

3 4 6 8 =0 3 4

3p 2p =p5 5 3p ö sin -1 æç sin ÷ è 5 ø 2p ö ù é = sin -1 ê sin æç p ÷ 5 ø úû ë è 2p ù 2p é p p ù = sin -1 é sin = Î - , 5 ûú 5 êë 2 2 ûú ëê

7.

ò

sec 2 x dx 3 + tan x Let 3 + tan x = t sec 2 x dx = dt

\

ò

sec 2 x dt dx = ò 3 + tan x t = log|t|+ c = log| 3 + tan x|+ c

(Q R 1 = R 3 )

73


ò ( 3x

8.

2

+ 2x + k) dx = 0

0 1

Þ

é 3x 3 2x 2 ù + + kxú = 0 ê 2 êë 3 úû 0

Þ

1+1+ k = 0

9. Given a * b = a + 3b \

Þ

2

k = -2

" a, b Î z

2

2 * 4 = 2 + 3 ´ 4 = 2 + 48 = 50 .

10. Given |A|= 5 We know |adj. A| = |A|2 |adj. A| = 5 2 = 25

\

SECTION–B 11.

®

®

®

®

®

®

®

®

®

a - d will be parallel to b - c , if ( a - d ) ´ ( b - c ) = 0

Now

®

®

®

®

®

®

®

®

®

®

®

®

®

®

®

®

®

®

®

®

( a - d) ´ ( b - c ) = a ´ b - a ´ c - d ´ b + d ´ c = a´ b - a´ c + b ´ d- c ´ d ®

®

\

®

®

®

®

®

®

®

®

( a - d ) || ( b - c )

12. We know sin -1 x + sin -1 y = sin -1 ( x 1 - y 2 + y 1 - x 2 ) \

4 5 16 sin -1 æç ö÷ + sin -1 æç ö÷ + sin -1 æç ö÷ è5 ø è 13 ø è 65 ø æ4 25 5 16 ö -1 æ 16 ö = sin -1 ç 1+ 1÷ + sin ç ÷ è 65 ø 169 13 25 ø è5 4 12 5 3 16 = sin -1 æç ´ + ´ ö÷ + sin -1 æç ö÷ è 5 13 13 5 ø è 65 ø 63 16 = sin -1 æç ö÷ + sin -1 æç ö÷ è 65 ø è 65 ø 63 Let sin -1 =q 65 Þ Þ

63 = sin q 65 cos 2 q = 1 -

63 2

Þ

65 63

2

65

2

2

65 - 63 2 65

2

... (i)

= sin 2 q

2

=

®

®

[Q given a ´ b = c ´ d and a ´ c = b ´ d ]

=0

=

( 65 + 63)( 65 - 63) 65 2

74


Þ

cos 2 q =

256

\

2

cos q =

16 65

65 \ Equation (i) becomes 63 16 63 16 sin -1 æç ö÷ + sin -1 æç ö÷ = cos -1 æç ö÷ + sin -1 æç ö÷ è 65 ø è 65 ø è 65 ø è 65 ø =

p 2

éQ sin -1 A + cos -1 A = p ù êë 2 úû OR

Given, Þ Þ Þ Þ

p tan -1 3x + tan -1 2x = 4 3x + 2x ö p tan -1 æç ÷= è 1 - 3x ´ 2x ø 4 5x =1 1 - 6x 2

é -1 -1 -1 x + y ù êQ tan x + tan y = tan ú 1 - xy û ë

5x = 1 - 6x 2 6x 2 + 5x - 1 = 0

6x 2 + 6x - x - 1 = 0 6x( x + 1) - 1( x + 1) = 0 ( 6x - 1)( x + 1) = 0 1 or x = -1. \ x= 6 13. The given lines 1 - x 7 y - 14 5z - 10 = = 3 2l 11 7 - 7x y - 5 6 - z and are rearranged to get = = 3l 1 5 x-1 y- 2 z- 2 = = 11 2l -3 5 7 x-1 y-5 z- 6 = = -3 l 1 -5 7 Direction ratios of lines are 2l 11 -3 l and -3, , , 1, - 5 7 5 7 As the lines are perpendicular 11 æ -3 l ö 2 l \ -3 ç ´ 1 + ( -5 ) = 0 ÷+ è 7 ø 7 5 9l 2l Þ + - 11 = 0 7 7 Þ Þ Þ

... (i) ... (ii)

75


11 l = 11 7 l =7

Þ

Þ 14. Given differential equation dy dy + y = cos x - sin x is a linear differential equation of the type + Py = Q . dx dx 1. dx Here I. F = e ò = ex Its solution is given by Þ

y e x = ò e x (cos x - sin x) dx

Þ

y e x = ò e x cos x dx - ò e x sin x dx Integrate by parts

15.

x

x

Þ

y e = e cos x - ò - sin x e x dx - ò e x sin dx

\

y e x = e x cos x + C

Þ

y = cos x + C e - x dy y y - + cosec æç ö÷ = 0 è xø dx x

... (i)

It is a homogeneous differential equation, y Let =v Þ y = vx x dy xdv =v+ dx dx (Substituting in equation (i)) dv Þ v+x = v - cosec v dx dv Þ x = - cosec v dx dv dx dx Þ =Þ sin v dv = cosec v x x Integrating both sides dx ò sin v dv = - ò x Þ cos v = log|x| + C y or cos = log|x|+ C x Given y = 0 , when x = 1 Þ cos 0 = log|1|+ C 1=C Þ

Þ

- cos v = - log|x|+ C

Hence, solution of given differential equation is cos

y = log|x|+ 1. x

76


16. Let|A|=

x+4

2x

2x

2x

x+4

2x

2x

2x

x+4

Apply C 1 ® C 1 + C 2 + C 3 5x + 4 2x 2x |A|= 5x + 4 x + 4

2x

5x + 4

x+4

2x

Take 5x + 4 common from C 1 1 2x 2x |A|= (5x + 4) 1 x + 4 1

2x

2x x+4

Apply R 2 ® R 2 - R 1 ; R 3 ® R 3 - R 1 1 2x 2x |A|= (5x + 4) 0 4 - x 0

0

0 4-x

Expanding along C 1 , we get

|A|= (5x + 4)( 4 - x) 2 = R.H.S.

17. If there is third 6 in 6th throw, then five earlier throws should result in two 6. 1 5 Hence taking n = 5 , p = , q = 6 6 \

P( 2 sixes) = P(5, 2) = 5 C 2 p 2 q 3 5 ! æ 1 ö 2 æ 5 ö 3 10 ´ 125 ç ÷ ç ÷ = 2! 3! è 6ø è 6ø 65 10 ´ 125 1 1250 625 P( 3 sixes in 6 throws) = ´ = = 5 6 6 6 6 3 ´ 65

Þ P( 2 sixes) = \

18. Let y = x sin x + (sin x) cos x Let u = x sin x and v = (sin x) cos x Then,

y=u+v dy du dv Þ = + dx dx dx Now, u = xsin x Taking log both sides, we get Þ log u = sin x log x Differentiating w.r.t. x 1 du sin x Þ = + log x . cos x u dx x

...(i)

77


du sin x = x sin x é + log x . cos xù ê úû dx ë x

Þ

Similarly taking log on v = (sin x) cos x log v = cos x log sin x Differentiating w. r. t. x 1 dv cos x = cos x . + log sin x .( - sin x) v dx sin x dv = (sin x) cos x [cos x. cot x - sin x . log sin x] dx Form (i), we have dy sin x = x sin x é + log x . cos xù + (sin x) cos x [cos x. cot x - sin x. log sin x] ê úû dx ë x 19. Let I = ò

ex 5 - 4e x - e 2 x

dx

Suppose e x = t Þ e x dx = dt dt dt Þ I=ò =ò 2 2 5 - 4t - t -(t + 4t - 5) Þ I=ò Þ I=ò

dt -(t 2 + 4t + 4 - 9 ) dt 2

3 - (t + 2)

2

= sin -1

t+2 +C 3

æ ex + 2ö ÷ +C Þ I = sin -1 ç è 3 ø OR Let I = ò

( x - 4) e

x

( x - 2) 3

dx

é ( x - 2) - 2 ù x = òê ú e dx êë ( x - 2) 3 úû =

= =

e x dx

e x dx

ò ( x - 2) 2 - 2ò ( x - 2) 3 ex ( x - 2) 2 ex ( x - 2) 2

+ 2ò +C

e x dx ( x - 2) 3

- 2ò

e x dx ( x - 2) 3

78


20. The relation given is R = {( a, b):|a - b|is even} where a, b Î A = {1, 2, 3, 4, 5} To check: Reflexivity Let a Î A Then aRa as|a - a| = 0 which is even. \ ( a, a) Î R. Hence R is reflexive. To check: Symmetry Let ( a, b) Î R Þ | a - b|is even Þ |b - a|is even (b – a) Î R. Þ Hence R is symmetric. To check: Transitivity Let ( a, b) Î R and (b , c) Î R Þ |a - b|is even and|b - c|is also even. Then, |a - c|=|( a - b) + (b - c)| £ |a -b|+|b - c| even

even

\ |a - c|= even So, ( a, c) Î R. It is transitive. As R is reflexive, symmetric as well as transitive, it is an equivalence relation. 21. Given equation is ( x 2 + y 2 ) 2 = xy Differentiating w.r.t. x dy ö dy æ Þ 2( x 2 + y 2 ) ç 2x + 2y ÷ = x +y è ø dx dx Þ 2( x 2 + y 2 ).2y Þ

dy dy -x = y - 4( x 2 + y 2 ) x dx dx

2 2 dy y - 4x( x + y ) = dx 4y( x 2 + y 2 ) - x

OR y = 3 cos(log x) + 4 sin(log x) Differentiating w.r.t. x dy -3 sin(log x) 4 cos(log x) Þ = + dx x x dy Þ x = -3 sin(log x) + 4 cos(log x) dx

79


Differentiating again w.r.t. x d 2 y dy -3 cos(log x) 4 sin(log x) Þ x + = x x dx 2 dx Þ x2

d2y dx

Þ x2

2

d2y 2

+x

dy = -y dx

+x

dy +y=0 dx

dx 22. Given curve is y = 3x - 2 dy 1´ 3 Þ = dx 2 3x - 2

...(i)

Since tangent is parallel to line 4x - 2y + 5 = 0 -4 3 Þ = slope of line = -2 2 3x - 2 9 Þ 4= 4( 3x - 2) 41 Þ 48x - 32 = 9 Þ x= 48 Substituting value of x in (i) 41 9 3 y= 3´ -2= = 48 16 4 41 3 Thus point of tangency is æç , ö÷ è 48 4 ø \

Equation of tangent is 3 41 y - = 2æç x - ö÷ è 4 48 ø 4y - 3 48x - 41 Þ = 4 24 Þ 24y - 18 = 48x - 41 Þ 48x - 24y - 23 = 0 is the equation of tangent. OR 1 3 Given f ( x) = x + x3 3 f ¢( x) = 3x 2 x4 =

3( x 6 - 1) x4

=

3( x 2 - 1)( x 4 + x 2 + 1) x4

But x 4 + x 2 + 1, x 4 are always > 0

80


\

f ¢( x) = 0 Þ x = ±1

\

Intervals

x–1

x+1

sign of f ¢( x)

x<–1

–ve

–ve

+ve

–1
–ve

+ve

–ve

x>1

+ve

+ve

+ve

Given function is increasing " x Î ( - ¥ , 1) È (1, ¥) and is decreasing " x Î (– 1, 1).

SECTION–C 23. Let a right circular cylinder of radius ‘R’ and height 'H' is inscribed in the sphere of given radius ‘r’. H2 = r2 4 Let V be the volume of the cylinder.

\

R2 +

Then, V = pR 2 H æ H2 ö ÷H ...(i) Þ V = p ççr 2 4 ÷ø è p Þ V = pr 2 H - H 3 4 Differentiating both sides w.r.t. H dV 3 pH 2 = pr 2 dH 4 For maximum volume Þ

3 pH 2 = pr 2 4

Þ

R r H

… (ii) dV =0 dH H2 =

4r 2 3

or

H=

2 r 3

Differentiating (ii) again w.r.t. H d 2V 6 pH d 2V ù -6 p 2 =Þ = ´ r<0 ú 2 2 4 4 3 2 dH dH úû H = r 3

\

Volume is maximum when height of the cylinder is 2 r in (i), we get 3 æ 4r 2 ö 2 p 2r 2 2r ÷. = p ççr 2 r = . 4 ´ 3 ÷ø 3 3 3 è

Substituting H = Vmax

=

4pr 3 cubic units. 3 3

2 r. 3

81


OR Let the length and breadth of the tank are L and B. 4 \ Volume = 8 = 2 LB Þ B = L The surface area of the tank, S = Area of Base + Area of 4 Walls = LB + 2( B + L) × 2 = LB + 4B + 4L The cost of constructing the tank is C = 70( LB) + 45( 4B + 4L) 4 4 = 70æç L × ö÷ + 180æç + L ö÷ è Lø èL ø 4 Þ C = 280 + 180æç + L ö÷ èL ø Differentiating both sides w.r.t. L dC 720 =+ 180 dL L2 dC For minimisation =0 dL 720 Þ = 180 L2 720 Þ L2 = =4 180 Þ L=2 Differentiating (iii) again w.r.t. L d 2C

=

… (ii)

… (iii)

1440

> 0 "L > 0 dL L3 \ Cost is minimum when L = 2 From (i), B=2 4 æ Minimum cost = 280 + 180 ç + 2ö÷ è2 ø 2

… (i)

(from (ii))

= 280 + 720 = Rs 1000 24. Let x units of food F1 and y units of food F2 are required to be mixed. Cost = Z = 4x + 6y is to be minimised subject to following constraints. 3x + 6y ³ 80 4x + 3y ³ 100 x ³ 0, y ³ 0 To solve the LPP graphically the graph is plotted as shown.

82

Xam idea Mathematics – XII Y 40 35 A 30 25 20 15 10 5 O

5

10

)24, 43 )

B

15

20 25

X 35 3 x+6 y=8 0

30 +3 4x 1 y=

00

The shaded regions in the graph is the feasible solution of the problem. The corner points are 100 ö æ 4ö æ 80 ö A æç 0, ÷ , Bç 24, ÷ and C ç , 0÷. The cost at these points will be è ø è ø è 3 ø 3 3 100 Z] A = 4 ´ 0 + 6 ´ = Rs 200 3 4 Z]B = 4 ´ 24 + 6 ´ = Rs 104 3 80 320 = Rs 106.67 Z]C = 4 ´ + 0 = Rs 3 3 Thus cost will be minimum if 24 units of F1 and 4/3 units of F2 are mixed. The minimum cost is Rs 104. 25. The distribution of balls in the three bags as per the question is shown below. Bag

Number of white balls

Number of black balls

Number of red balls

Total balls

I

1

2

3

6

II

2

1

1

4

III

4

3

2

9

As bags are randomly choosen 1 3 Let E be the event that one white and one red ball is drawn.

\

P( bag I) = P( bag II) = P( bag III) =

1

P(E/bag I) =

C 1 ´ 3C 1 6

C2

=

3´2 1 = 6´5 5

83


P(E/bag II) =

C 1 ´ 1C 1 4

4

P(E/bag III) =

=

C2

C 1 ´ 2C 1 9

2´2 1 = 4´ 3 3

=

C2

4´2´2 2 = 9´8 9

Now, required probability = P(bag III/E) =

P( bag III). P(E / bag III)

P( bag I). P(E / bag I) + P( bag II). P(E / bag II) + P ( bagIII). P (E / bagIII) 1 2 1 2 ´ ´ 3 9 3 9 = = 1 1 1 1 1 2 1 æ1 1 2 ö ´ + ´ + ´ ç + + ÷ 3 5 3 3 3 9 3 è5 3 9 ø 2 2 45 5 9 = = ´ = 9 + 15 + 10 9 34 17 45 26. Given system of equations is 2x - 3y + 5z = 11 3x + 2y - 4z = -5 x + y - 2z = -3 The equations can be expressed as matrix equation AX = B æ 2 -3 5 ö æ xö æ 11 ö ç ÷ç ÷ ç ÷ Þ ç 3 2 -4 ÷ ç y ÷ = ç -5 ÷ ç 1 1 -2 ÷ ç z ÷ ç -3 ÷ è øè ø è ø \ X = A–1 B Now, |A|= 2( -4 + 4) + 3( -6 + 4) + 5( 3 - 2) = -6 + 5 = -1 ¹ 0 Þ A -1 exists. The cofactors of elements of A are C 11 = 0 C 12 = 2 C 13 = 1 C 21 = -1 C 22 = -9 C 23 = -5 C 31 = 2

C 32 = 23 C 33 = 13

2 1ö æ0 ç ÷ Matrix of cofactors = ç -1 -9 -5÷ ç 2 23 13 ÷ è ø \

æ 0 -1 2 ö ç ÷ Adj A = ç 2 -9 23÷ ç 1 -5 13 ÷ è ø

84


Þ A

\

-1

æ 0 -1 2 ö ç ÷ = - ç 2 -9 23÷ ç 1 -5 13 ÷ è ø

æ ö 1 ( Adj A) ÷ çQ A -1 = |A| è ø

æ xö æ 0 -1 2 ö æ 11 ö æ 0 + 5 - 6 ö æ1 ö ç ÷ ç ÷ç ÷ ç ÷ ç ÷ X = ç y÷ = - ç 2 -9 23÷ ç -5÷ = - ç 22 + 45 - 69÷ = ç 2 ÷ çz÷ ç 1 -5 13 ÷ ç -3÷ ç11 + 25 - 39÷ ç 3÷ è ø è øè ø è ø è ø

Hence solution of given equations is x = 1, y = 2, z = 3. p

27. Let I =

e cos x

ò e cos x + e - cos x dx 0 p

Þ I=

e cos( p - x )

æQ a f ( x) dx = a f ( a - x) dxö ç ò ÷ ò0 è 0 ø

ò e cos( p - x ) + e - cos( p - x ) dx 0 p

=

...(i)

e - cos x

ò e - cos x + e cos x dx

...(ii)

0

Adding (i) and (ii), we get p

2I =

e cos x + e - cos x

ò e cos x + e - cos x

p

p

dx = ò dx = x]0 = p

0

0

Þ

I=

p 2

OR p 2

Let I = ò ( 2 log sin x - log sin 2x) dx 0 p 2

p p é ù Þ I = ò ê( 2 log sin æç - xö÷ - log sin 2æç - xö÷ ú dx è ø è ø 2 2 ë û 0

… (i)

æQ a f ( x) dx = a f ( a - x) dxö ç ò ÷ ò0 è 0 ø

p 2

Þ I = ò ( 2 log cos x - log sin 2x) dx 0

Adding (i) and (ii), we get p 2

2I = ò 2 log sin x + 2 log cos x - 2 log sin 2x 0 p 2

Þ 2I = ò 2 [log sin x + log cos x – log sin 2x]dx 0

… (ii)

85

Examination Papers – 2009 p 2

p 2

sin x cos x Þ I = ò log dx 2 sin x cos x

p

1 1 I = log ò dx = log × xù 2 2 2 úû 0

Þ

0

0

p 1 Þ I = log 2 2 28. The lines are plotted on the graph as shown. Y

3x –2

y= 6

5 4

B

3

(1,2) A

2

=0 y+5 x–3

(4,3)

1 C

X’

X

O

y=4 2x+

Y’

4

Area of DABC = ò 1

2

4

x+5 3x - 6 dx - ò ( 4 - 2x) dx - ò dx 3 2 1

2

4

2

4

öù æ öù 1 æ x2 2x 2 ö ù 1 æ 3x 2 ÷ú - ç = çç + 5x÷÷ ú - çç 4x - 6x÷÷ ú ÷ ç 3è 2 2 ø úû 2è 2 ø úû 1 è ø úû 2 1 1æ 1 1 ö ç 8 + 20 - - 5÷ - ( 8 - 4 - 4 + 1) - ( 24 - 24 - 6 + 12) ø 3è 2 2 1 45 1 = æç ö÷ - 1 - ( 6) 3è 2 ø 2 15 15 7 = -1- 3= - 4 = square units. 2 2 2 =

86


29. The equation of plane through ( -1, 3, 2) can be expressed as … (i) A( x + 1) + B( y - 3) + C(z - 2) = 0 As the required plane is perpendicular to x + 2y + 3z = 5 and 3x + 3y + z = 0, we get A + 2B + 3C = 0 3A + 3B + C = 0 A B C A B C Þ = = Þ = = 2- 9 9-1 3- 6 -7 8 -3 \ Direction ratios of normal to the required plane are -7 , 8, –3. Hence equation of the plane will be -7( x + 1) + 8( y - 3) - 3(z - 2) = 0 Þ -7 x - 7 + 8y - 24 - 3z + 6 = 0 or 7 x - 8y + 3z + 25 = 0

Set–II 2. Let I = ò sec 2 (7 - x) dx =

tan(7 - x)

+C -1 = - tan(7 - x) + C ®

7. Given b = 2i$ + j$ + 2k$ ®

®

b Unit vector in the direction of b = = b$ ® |b| $ 2i$ + j$ + 2k 2 1 2 = i$ + j$ + k$ \ b$ = 3 3 3 22 +12 + 22 11. Let y = (sin x) x + sin -1 x Suppose z = (sin x) x Taking log on both sides log z = x log sin x Differentiating both sides w.r.t. x 1 dz cos x = x. + log sin x z dx sin x dz Þ = (sin x) x ( x cot x + log sin x) dx dy 1 1 \ = (sin x) x [x cos x + log sin x] + dx 1-x 2 x 1 = (sin x) x ( x cos x + log sin x) + 2 x (1 - x)

87


18. The given lines can be expressed as x-1 y- 2 z- 3 and = = -3 2l 2 x-1 y-1 z- 6 = = 3l 1 -7 The direction ratios of these lines are respectively -3, 2l , 2 and 3l, 1, - 7. Since the lines are perpendicular, therefore -3( 3l) + 2l(1) + 2( -7) = 0 Þ -9l + 2l - 14 = 0 Þ -7 l = 14 Þ l = -2 19. Given differential equation is dy (1 + x 2 ) + y = tan -1 x dx The equation can be expressed as dy y tan -1 x + = dx 1 + x 2 1 + x2 This is a linear differential equation of the type ò

dx 1+ x 2

Here I . F = e = e tan Its solution is given by y e tan

-1 x

= ò e tan

Suppose I = ò e Let tan -1 x = t 1 1 + x2

-1 x

dy + Py = Q dx

-1 x

.

tan -1 x

tan -1 x 1 + x2

dx

tan -1 x 1 + x2

… (i)

dx

dx = dt

Þ I = ò e t . t dt Integrating by parts, we get I = t e t - ò e t dt Þ I = t e t - e t + C' Þ I = e tan

-1 x

(tan -1 x - 1) + C'

From (i) y e tan

-1 x

= e tan

-1 x

(tan -1 x - 1) + C

Þ y = tan -1 x - 1 + C e - tan

-1 x

which is the solution of given differential equation.

88


a

b

c

21. Let|A|= a - b b - c c - a Apply C 1 ® C 1 + C 2 + C 3 b + c c + a a+b a+b + c |A| =

b

c

b-c c-a

0

2( a + b + c) c + a a + b Taking (a + b + c) common from C 1 1

b

c

|A| = ( a + b + c) 0 b - c c - a 2 c + a a+b Apply R 3 ® R 3 - 2R 1 1

b

c

|A| = ( a + b + c) 0

b-c

c-a

0 c + a - 2b a + b - 2c Expand along C 1 to get |A| = ( a + b + c)[(b - c)( a + b - 2c) - ( c + a - 2b) ( c - a)] = ( a + b + c)[ab + b 2 - 2bc - ac - cb + 2c 2 - ( c 2 - ac + ac - a 2 - 2bc + 2ab)] = ( a + b + c)( a 2 + b 2 + c 2 - ab - bc - ca) = a 3 + b 3 + c 3 - 3abc = RHS 23. P(GI ) = 0.6 P(GII ) = 0.4 Let E is the event of introducing new product then P(E/GI ) = 0.7 P(E/GII ) = 0.3 To find P(GII /E) Using Baye’s theorem we get P(GII ). P(E / GII ) P(GII /E) = P(GI ). P(E / GI ) + P(GII ). P(E / GII ) 0.4 ´ 0. 3 0.6 ´ 0.7 + 0.4 ´ 0. 3 0.12 = 0.42 + 0.12 12 2 = = 54 9 =

26. We plot the curves y 2 = 4x and x 2 = 4y and also the various areas of the square. To show that area of regions I = II = III 4

4

Area of region I = ò 4dx - ò 2 xdx 0

0

89


x 3/ 2 ù = 4x - 2 ú 3 / 2 úû

Y

4

x2=4y

0

4 16 square units = 16 - ´ 8 = 3 3 4

4

Area of Region II = 2ò x dx - ò 0

0

=

2 3/ 2 x ù x ú 3 12 úû

y2=4x

y=4 I

x2 dx 4 3

= 2.

4

x=4

II III

4 O

4

X

0

4 64 128 - 64 64 16 square units ´8-0= = = 3 12 12 12 3 4

Area of Region III = ò 0

x2 dx 4 4

x3 ù 64 16 square units. = = ú = 12 úû 12 3 0

Thus, the curves y2 = 4x and x2 = 4y divide the area of given square into three equal parts.

Set–III 4. Let I = ò

(1 + log x) 2

x Let 1 + log x = t 1 dx = dt x

\

I = ò t 2 dt = =

t3 +C 3

(1 + log x) 3 3 ®

dx

+C

®

9. Given| a ´ b|=

3

Þ a b sin q = 3 Þ 1 ´ 2 sin q = 3 3 sin q = 2 p Þ q = radians. 3

(Q a = 1, b = 2)

90


1 + a2 - b 2 15. Let|A| = 2ab

2ab 1 - a2 + b 2

-2b 2a

- 2a

1 - a2 - b 2

2b Apply R 1 ® R 1 + bR 3 1 + a2 + b 2 |A| = 2ab

1 - a2 + b 2

-b - ba 2 - b 3 2a

-2 a

1 - a2 - b 2

0

2b

Taking 1 + a 2 + b 2 common from R 1 1 2

-b

0

2

2

|A| = (1 + a + b ) 2ab 1 - a + b

2

1 - a2 - b 2

-2 a

2b

2a

Apply R 2 ® R 2 - aR 3 1

0

|A| = (1 + a + b ) 0

2

2

2

1+ a +b

-b 2

3

a + a + ab 2 1 - a2 - b 2

-2 a

2b

Taking 1 + a 2 + b 2 common from R 2 2

2 2

|A| = (1 + a + b )

1

0

0

1

-b a 2

2b -2a 1 - a - b 2 Apply R 3 ® R 3 - 2bR 1 2

2 2

|A|= (1 + a + b )

1

0

-b

0

1

a

0 -2 a 1 - a 2 + b 2 Expanding along C 1 , we get

|A| = (1 + a 2 + b 2 ) 2 [1(1 - a 2 + b 2 + 2a 2 )] = (1 + a 2 + b 2 ) 3 = RHS

17. Let y = x cos x + (sin x) tan x Let u = x

cos x

, v = (sin x)

tan x

Taking log on either side log u = cos x. log x, log v = tan x log sin x Differentiating w.r.t. x 1 du 1 1 dv tan x. cos x = cos x. + log x( - sin x), = + log sin x. sec 2 x u dx x v dx sin x

… (i)

91


du cos x dv = x cos x æç - sin x log xö÷ , = (sin x) tan x (1 + sec 2 x log sin x) è ø dx x dx \ From (i) we get dy cos x = x cos x æç - sin x log xö÷ + (sin x) tan x [1 + sec 2 x log sin x] è x ø dx 19. Given differential equation is dy x log x + y = 2 log x dx This can be rearranged as dy y 2 + = dx x log x x It is a linear differential equation of the type

dy + Py = Q dx

1

Now, IF = e

ò x log x dx

= e log(log x ) = log x

Its solution is given by 2 y log x = ò log x dx x Þ y log x = 2

(log x) 2

Q ò f ( x). f ¢( x) dx = [ f ( x)] 2 + C

+C

2 C which is the solution of the given differential equation Þ y = log x + log x 20. The given lines on rearrangement are expressed as y- 2 z-1 x-5 x y+1/ 2 z-1 and = = = = 5l + 2 -5 1 1 2l 3 The direction ratios of the two lines are respectively 5l + 2, - 5, 1 and 1, 2l , 3 As the lines are perpendicular, \ (5l + 2) ´ 1 - 5( 2l) + 1( 3) = 0 Þ 5l + 2 - 10l + 3 = 0 Þ -5 l = -5 Þ l = 1 Hence l = 1 for lines to be perpendicular. 24. The two circles are re-arranged and expressed as … (i) y2 = 9 - x2 y 2 = 9 - ( x - 3) 2

… (ii)

To find the point of intersection of the circles we equate y 2 Þ 9 - x 2 = 9 - ( x - 3) 2 Þ 9 - x 2 = 9 - x 2 - 9 + 6x

92

Xam idea Mathematics – XII Y 3 x2+y2 = 9 2 The circles are shown in the figure and the shaded area is the required area. X’ 3/2 3 Now, area of shaded region O 3 é ù 3 ê2 ú = 2ê ò 9 - ( x - 3) 2 dx + ò 9 - x 2 dxú (x –3)2 + y2 = 9 ê0 ú 3 Y’ êë úû 2 x- 3 9 x - 3 ù 3/ 2 x 9 x 3 = 2é 9 - ( x - 3) 2 + sin -1 + 2é 9 - x 2 + sin -1 ù êë 2 êë 2 2 3 úû 0 2 3 úû 3

Þ

x=

2

é -3 ù é9 9 9 1 9 3 9 9 = 2ê 9 - + sin -1 æç - ö÷ - sin -1 ( -1) ú + 2ê sin -1 1 9 - - sin -1 è ø 4 4 2 2 2 2 4 4 2 ë û ë é -3 3 3 9 p 9 p ù é9 p 3 3 3 9 pù = 2ê . - . + . ú + 2ê . - . - . ú 2 2 6 2 2û 2 6û ë 4 ë2 2 4 2 é 9 3 3p 9p 9p 9 é -9 3 6p 18p ù 3p ù = 2ê + + 3+ ú = 2ê ú 8 4 4 4 8 4 û 4 4 û ë ë 4 é 9 3 12p ù 9 3 square units. = 2ê + ú = 6p 4 4 û 2 ë 27. The three coins C 1 , C 2 and C 3 are choosen randomly. 1 \ P(C 1 ) = P(C 2 ) = P(C 3 ) = 3 Let E be the event that coin shows head. Then , P(E/C 1 ) = 1 75 3 1 P(E/C 2 ) = P(E/C 3 ) = = 100 4 2 To find: P(C 1 /E) From Baye’s theorem, we have P(C 1 ). P(E / C 1 ) P(C 1 /E) = P(C 1 ). P(E / C 1 ) + P(C 2 ) P(E / C 2 ) + P(C 3 ). P(E / C 3 ) 1 1 ´1 3 3 = = 1 1 3 1 1 1æ 3 1ö ´1+ ´ + ´ ç1 + + ÷ 3 3 4 3 2 3è 4 2ø 1 4 4 = = = 3 1 4+ 3+2 9 1+ + 4 2 4 Thus, probability of getting head from the two headed coin is . 9

1ù ú 2û

X


Maximum marks: 100


Set–I SECTION–A æ 3x + y - yö æ 1 2 ö 1. Find the value of x, if ç ÷=ç ÷. è 2 y - x 3 ø è -5 3 ø 2. Let * be a binary operation on N given by a * b = HCF (a, b) a, b, ÎN. Write the value of 22 * 4. 1 2

3. Evaluate :

ò 0

4. Evaluate : ò

1 1 - x2

dx.

cos x dx. x

7p 5. Write the principal value of, cos -1 æç cos ö÷. è 6 ø a-b b - c c - a 6. Write the value of the following determinant : b - c c - a a - b c - a a-b b - c 7. Find the value of x, from the following:

x 4 =0 2 2x

® 8. Find the value of p, if ( 2i$ + 6j$ + 27 k$) ´ (i$ + 3j$ + pk$) = 0 . 9. Write the direction cosines of a line equally inclined to the three coordinate axes. ®

®

®

®

®

®

10. If p is a unit vector and ( x - p ). ( x + p ) = 80, then find| x|.

SECTION–B 11. The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8 cm and y = 6 cm, find the rate of change of (a) the perimeter, (b) the area of the rectangle.

94


OR Find the intervals in which the function f given by f ( x) = sin x + cos x, 0 £ x £ 2p , is strictly increasing or strictly decreasing. 12. If sin y = x sin( a + y), prove that

If (cos x)y = (sin y)x, find

2 dy sin ( a + y) . = dx sin a OR

dy . dx

ì n + 1 , if n is odd ï 13. Let f : N ® N be defined by f (n) = í 2 for all n Î N. n ï , if n is even î 2 Find whether the function f is bijective. dx 14. Evaluate : ò 5 - 4x - 2x 2 OR Evaluate : ò x sin 15. If y =

sin -1 x

-1

x dx

, show that (1 - x 2 )

d2y 2

- 3x

dy - y = 0. dx

dx 1 - x2 16. On a multiple choice examination with three possible answers (out of which only one is correct) for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing? 1 1+p 1+p+q 17. Using properties of determinants, prove the following : 2 3 + 2p 1 + 3p + 2q = 1 3 6 + 3p 1 + 6p + 3q dy y = y - x tan æç ö÷ è dx xø dy 19. Solve the following differential equation : cos 2 x + y = tan x. dx 20. Find the shortest distance between the following two lines : 18. Solve the following differential equation : x

®

r = (1 + l)i$ + ( 2 - l) j$ + ( l + 1) k$;

®

r = ( 2i$ - j$ - k$) + m( 2i$ + j$ + 2k$).

æ 1 + sin x + 1 - sin x ö x æ pö 21. Prove the following : cot -1 ç ÷ = , x Î ç 0, ÷ è 4ø è 1 + sin x - 1 - sin x ø 2 OR Solve for x : 2 tan

-1

(cos x) = tan

-1

( 2cosec x)

95


22. The scalar product of the vector i$ + j$ + k$ with the unit vector along the sum of vectors 2i$ + 4j$ - 5k$ and li$ + 2j$ + 3k$ is equal to one. Find the value of l.

SECTION–C 23. Find the equation of the plane determined by the points A( 3, - 1, 2), B(5, 2, 4) and C(–1, –1, 6). Also find the distance of the point P(6, 5, 9) from the plane. 24. Find the area of the region included between the parabola y 2 = x and the line x + y = 2. p

25. Evaluate : ò 0

xdx 2

2

a cos x + b 2 sin 2 x

26. Using matrices, solve the following system of equations : x+ y+z= 6 x + 2z = 7 3x + y + z = 12 OR é 3 0 -1 ù Obtain the inverse of the following matrix, using elementary operations : A = ê 2 3 0 ú. ê ú êë 0 4 1 úû 27. Coloured balls are distributed in three bags as shown in the following table : Bag

Colour of the ball Black

White

Red

I

1

2

3

II

2

4

1

III

4

5

3

A bag is selected at random and then two balls are randomly drawn from the selected bag. They happen to be black and red. What is the probability that they came from bag I? 28. A dealer wishes to purchase a number of fans and sewing machines. He has only Rs. 5,760 to invest and has a space for at most 20 items. A fan costs him Rs. 360 and a sewing machine Rs. 240. His expectation is that he can sell a fan at a profit of Rs. 22 and a sewing machine at a profit of Rs. 18. Assuming that he can sell all the items that he can buy, how should he invest his money in order to maximize the profit? Formulate this as a linear programming problem and solve it graphically. 29. If the sum of the hypotenuse and a side of a right angled triangle is given, show that the area p of the triangle is maximum when the angle between them is . 3 OR x ö A manufacturer can sell x items at a price of Rs. æç5 ÷ each. The cost price of x items is è 100 ø x Rs. æç + 500ö÷. Find the number of items he should sell to earn maximum profit. è5 ø

96


Set–II Only those questions, not included in Set I, are given sin x 2. Evaluate : ò dx x æ x - y 2ö æ 2 2ö 5. Find the value of y, if ç ÷=ç ÷. 5ø è 3 5ø è x 11. If y = 3e 2 x + 2e 3 x , prove that d2y 2

-5

dy + 6y = 0 dx

dx 18. Find the shortest distance between the following two lines: ®

r = (1 + 2l)i$ + (1 - l) j$ + lk$;

®

r = 2i$ + j$ - k$ + m( 3i$ - 5j$ + 2k$).

19. Form the differential equation of the family of circles touching the y axis at origin. 21. Using properties of determinants, prove the following: x+y x x 5x + 4y

4x 2x = x 3

10x + 8y 8x

3x

25. Find the area of the region included between the parabola 4y = 3x 2 and the line 3x - 2y + 12 = 0. 29. Coloured balls are distributed in three bags as shown in the following table: Bag I

Black 2

Colour of the ball White 1

II

4

2

1

III

5

4

3

Red 3

A bag is selected at random and then two balls are randomly drawn from the selected bag. They happen to be white and red. What is the probability that they came from bag II?

Set–III Only those questions, not included in Set I and Set II are given sec 2 x 7. Evaluate : ò dx x 10. Find the value of x from the following : æ 2x - y 5 ö æ 6 5 ö ç ÷=ç ÷. y ø è 3 -2 ø è 3

97


13. Find the shortest distance between the following two lines: ®

r = (i$ + 2j$ + 3k$) + l(i$ - 3j$ + 2k$);

®

r = ( 4 + 2m)i$ + (5 + 3m) j$ + ( 6 + m) k$.

14. Form the differential equation representing the family of curves given by ( x - a) 2 + 2y 2 = a 2 , where a is an arbitrary constant. 16. Using properties of determinants, prove the following: 1+x 1 1 1

1+y

1

1

1

1+z

= xyz + xy + yz + zx.

18. If y = e x (sin x + cos x), then show that d2y dx

2

-2

dy + 2y = 0 dx

23. Find the area of the region bounded by the curves y 2 = 4ax and x 2 = 4ay. 26. A man is known to speak the truth 3 out of 5 times. He throws a die and reports that it is a number greater than 4. Find the probability that it is actually a number greater than 4.

SOLUTIONS Set – I SECTION–A 1. Given, æ 3x + y - yö æ 1 2 ö ç ÷=ç ÷ è 2 y - x 3 ø è -5 3 ø Using equality of two matrices, we have 3x + y = 1, –y=2 Þy= -2 Substituting the values of y, we get x=1 3x + ( - 2) = 1 Þ 2. Given a * b = HCF (a, b), a, b Î N Þ 22 * 4 = HCF (22, 4) = 2 1 2

3.

ò 0

1 1 - x2

1/ 2

dx = sin

-1

ù xú úû 0

p æ 1 ö -1 = sin -1 ç ÷ - sin 0 = è 2ø 4

98


4. Let I = ò

cos x dx x

Let x = t 1 2 x

dx = dt

Þ I = ò cos t.2 dt Þ I = 2 sin t + C I = 2 sin x + C 7p -1 æ 5. cos ç cos ö÷ è 6 ø p ö æ = cos -1 ç cosæç p + ö÷ ÷ è è 6 øø p = cos -1 æç - cos ö÷ è 6ø æ p 3ö = cos -1 ç ÷=p6 è 2 ø 5p = 6 6. Given determinant is a-b b - c c - a A = b - c c - a a-b c - a a-b b - c Use the transformation C 1 ® C 1 + C 2 + C 3 0 b-c c-a A = 0 c - a a-b = 0 0 a-b b - c 7. We are given that x 4 =0 2 2x Þ 2x 2 - 8 = 0 Þ 2x 2 = 8 Þ x2 = 4 Þ x= ±2 ® 8. ( 2i$ + 6j$ + 27 k$) ´ (i$ + 3j$ + pk$) = 0 i$ Þ

j$

k$

®

2 6 27 = 0 1

3

p

99


Þ ( 6p - 81)i$ - ( 2p - 27) j$ + 0k$ = 0 Þ

6p = 81 81 27 Þ p= = . 6 2 9. Any line equally inclined to co-ordinate axes will have direction cosines l, l, l \

l2 + l2 + l2 = 1 1 3 1 1 1 1 1 1 Direction cosines are + or ,+ ,+ ,,3 3 3 3 3 3 3l 2 = 1

\

®

Þ

®

®

l=±

®

10. Given ( x - p ).( x + p ) = 80 ® ® Þ | x |2 -| p |2 = 80 ® Þ | x |2 -1 = 80 ® ® or Þ | x |2 = 81 x =9

SECTION–B dy dx = -5 cm/min = 4 cm/min dt dt where x = length of rectangle and y = breadth of rectangle. Perimeter of rectangle is given by P = 2( x + y) \ Rate of change of P is dy dP dx = 2. +2 dt dt dt dP Þ = 2( -5) + 2( 4) = - 2 dt dP Þ =-2 dt ( 8 , 6 )

11. Given

x = 8 cm = – 2 cm/min y = 6 cm. i.e., the perimeter is decreasing at the rate of 2 cm/min. Now, Area of rectangle is given by A = xy dy dA dx Þ =x +y dt dx dt = 4x – 5y

100


Þ

dP = 32 – 30 = 2 dt ( 8 , 6 )

i.e., the area is increasing at the rate of 2 cm2/min. OR Given function f ( x) = sin x + cos x 0 £ x £ 2p f ¢( x) = cos x - sin x For the critical points of the function over the interval v 0 £ x £ 2p is given by f ¢( x) = 0 Þ cos x - sin x = 0 Þ cos x = sin x p 5p Þ x= , 4 4 p p 5p 5p Possible intervals are æç 0, ö÷ , æç , ö÷ , æç , 2p ö÷ è 4ø è4 4 ø è 4 ø p If 0 < x < , f ¢( x) = cos x - sin x > 0 Q cos x > sin x 4 Þ f ¢( x) > 0 Þ f ( x) is strictly increasing. p 5p If < x < , f ¢( x) = cos x - sin x < 0 Q cos x < sin x 4 4 Þ f ( x) is strictly decreasing. 5p If < x < 2p Þ f ¢( x) = cos x - sin x > 0Q cos x > sin x 4 Þ f ( x) is again strictly increasing. p 5p \ Given function f ( x) = sin x + cos x [0, 2p] is strictly increasing "x Î æç 0, ö÷ and æç , 2p ö÷ è 4ø è 4 ø p 5p while it is strictly decreasing " x Î æç , ö÷ è4 4 ø 12. If sin y = x sin( a + y) sin y Þ =x sin( a + y) Differentiating both sides w.r.t. x dy dy sin( a + y). cos y - sin y cos( a + y). dx dx = 1 Þ 2 sin ( a + y) dy [sin( a + y) cos y - sin y. cos( a + y) ] dx Þ =1 sin 2 ( a + y) dy Þ [sin( a + y - y) ] = sin 2 ( a + y) dx 2 dy sin ( a + y) \ = dx sin a

101


OR Given (cos x) y = (sin y) x Taking log on both sides \

log (cos x) y = log(sin y) x

Þ y log(cos x) = x log(sin y) Differentiating both sides w.r.t. x, we get dy 1 d 1 d y . cos x + log(cos x). = x. . sin y + log sin y.1 cos x dx dx sin y dx dy cos y dy sin x Þ -y + log(cos x). =x + log sin y cos x dx sin y dx dy dy Þ - y tan x + log(cos x) = x cot y + log sin y dx dx dy dy Þ log(cos x). - x cot y = log sin y + y tan x dx dx dy Þ [log(cos x) - x cot y] = log sin y + y tan x dx dy log sin y + y tan x \ = dx log cos x - x cot y ì n + 1 , if n is odd ï 13. Given f : N ® N defined such that f (n) = í 2 n ï , if n is even î 2 Let x, y Î N and let they are odd then x+1 y+1 f ( x) = f ( y) Þ = Þx=y 2 2 If x, y Î N are both even then also x y f ( x) = f ( y) Þ = Þ x = y 2 2 If x, y Î N are such that x is even and y is odd then y x+1 and f ( y) = f ( x) = 2 2 Thus, x ¹ y for f ( x) = f ( y) Let x = 6 and y = 5 6 5+1 We get f ( 6) = = 3, f (5) = =3 2 2 \ f ( x) = f ( y) but x ¹ y So, f (x) is not one-one. Hence, f (x) is not bijective.

...(i)

102


dx

14. Let I = ò

5 - 4x - 2x 2 dx

Þ I=ò

-2æç x 2 + 2x è dx

Þ I=ò

7 -2é( x + 1) 2 - ù êë 2 úû 1 2

Þ I=

5ö ÷ 2ø

ò

dx

1 sin -1 2

=

ææ 7 ö 2 ö 2 çç ÷ - ( x + 1) ÷ çè 2 ø ÷ è ø

2 ( x + 1) 7

+C

OR Let I = ò x sin II

-1 I

I = sin -1 x. Þ I=

x dx

x2 x2 -ò dx 2 2 1 - x2

(using integration by parts)

x2 1 1 - x2 - 1 sin -1 x + ò dx 2 2 1 - x2

=

x2 1 sin -1 x + ò 2 2

=

x2 1 1 x 1 sin -1 x - sin -1 x + é 1 - x 2 + sin -1 xù + C ê úû 2 2 2 ë2 2

1 - x 2 dx -

1 sin -1 x 2

x2 1 1 sin -1 x - sin -1 x + x 1 - x 2 + C 2 4 4 1é 2 -1 = ( 2x - 1) sin x + x 1 - x 2 ù + C úû 4 êë =

15. If y =

sin -1 x 1 - x2 1 - x2 .

dy = dx dy 1 + xy Þ = dx 1 - x 2 Þ

2

Þ

d y dx 2

=

1 1 - x2

- sin -1 x.

-2 x 2 1 - x2

1 - x2 ...(i)

æ dy ö (1 - x 2 ) ç x + y÷ + 2x(1 + xy) è dx ø (1 - x 2 ) 2

103


Þ (1 - x 2 ) 2

d2y dx

Þ (1 - x 2 ) 2

d2y dx

Þ (1 - x 2 ) 2

2

= (1 - x 2 ) x.

dy + y(1 - x 2 ) + 2x(1 + xy) dx

= (1 - x 2 ) x.

dy dy + y(1 - x 2 ) + 2x.(1 - x 2 ) dx dx

= 3x(1 - x 2 )

2

d y dx

Þ (1 - x 2 )

2

d2y dx

Þ (1 - x 2 )

2

2

d2y dx

2

(using (i))

dy + y(1 - x 2 ) dx

= 3x

dy +y dx

- 3x

dy -y=0 dx

16. Let p = probability of correct answer =

1 3

Þ q = probability of incorrect answer =

2 3

Here total number of questions = 5 P(4 or more correct) = P(4 correct) + P(5 correct) = 5 C 4 p 4 q 1 + 5 C 5 p5 q 0 using P(r success) = n Cr pr qn-r 4

1 17. Let A = 2

1+p

1 = 5 ´ æç ö÷ è 3ø 1 =5 ´ ´ 81 11 = 243 1+p+q

5 æ2ö + 1 ´ æ1ö ç ÷ ç ÷ è 3ø è 3ø 2 1 + 3 243

3 + 2p 1 + 3p + 2q

3 6 + 3p 1 + 6p + 3q Using the transformation R 2 ® R 2 - 2R 1 , R 3 ® R 3 - 3R 1 1 1+p 1+p+q A= 0

1

-1 + p

0

3

-2 + 3 p

Using R 3 ® R 3 - 3R 2 1 1+p 1+p+q Þ A= 0

1

-1 + p

0

0

1

Expanding along column C 1 , we get A =1

104


18. Given differential equation is dy y x = y - x tan æç ö÷ è xø dx dy y y = - tan dx x x It is a homogeneous differential equation. Let y = xt dy dt Þ = x. +t dx dx dt \ x + t = t - tan t dx dt Þ x = - tan t dx dt dx Þ =tan t x dx Þ cot t. dt = x Integrating both sides dx \ ò cot t. dt = - ò x Þ log | sin t| = – log | x | + log C y Þ log sin æç ö÷ + log x = log C è xø Þ

y Þ log x. sin æç ö÷ = log C è xø y =C x 19. Given differential equation is dy cos 2 x. + y = tan x dx dy Þ + y sec 2 x = tan x. sec 2 x dx Hence x. sin

Given differential equation is a linear differential equation of the type 2

ò Pdx ò sec xdx I.F. = e =e = e tan x \ Solution is given by e tan x y = ò tan x. sec 2 x. e tan x dx Let I = ò tan x. sec 2 x. e tan x dx

dy + Py = Q dx

105


Let tan x = t , sec 2 xdx = dt Þ I = ò t e t dt Integrating by parts \

I = te t - ò e t dt = t e t - e t + C ,

Þ I = tan x e tan x - e tan x + C, Hence e tan x y = e tan x (tan x - 1) + C Þ y = tan x – 1 + C e–tan x 20. The given equation of the lines can be re-arranged as given below. ®

r = (i$ + 2j$ + k$) + l(i$ - j$ + k$) and

®

r = ( 2i$ - j$ - k$) + m( 2i$ + j$ + 2k$) ®

®

®

®

a 1 = i$ + 2j$ + k$ , b 1 = i$ - j$ + k$ ,

Thus

a 2 = 2i$ - j$ - k$, b 2 = 2i$ + j$ + 2k$ The given lines are not parallel ®

\

Shortest distance between lines =

®

®

®

®

®

( a 2 - a 1 ) .(b 1 ´ b 2 ) |b 1 ´ b 2|

®

¬

We have a 2 - a 1 = i$ - 3j$ - 2k$ ®

®

i$

j$

k$

b 1 ´ b 2 = 1 -1 1 = -3i$ + 0j$ + 3k$ 2 ®

1

2

®

|b 1 ´ b 2|= 9 + 9 = 3 2 \

Shortest distance =

(i$ - 3j$ - 2k$).( -3i$ + 3k$) 3 2

=

-3 - 6 3 2

3 3 2 units. = 2 2 é 1 + sin x + 1 - sin x ù p 21. cot -1 ê where x Î æç 0, ö÷ ú è 4ø ë 1 + sin x - 1 - sin x û é x x 2 x x 2ù ê æç cos + sin ö÷ + æç cos - sin ö÷ ú è 2 2ø 2 2ø ú ê è = cot -1 ê 2 2 ú ê æ cos x + sin x ö - æ cos x - sin x ö ú ÷ ç ÷ êë çè è 2 2ø 2 2 ø úû =

106


æ cos x + sin ç 2 = cot -1 ç ç cos x + sin è 2 x x = cot -1 é cot ù = êë ú 2û 2

x x + cos - sin 2 2 x x - cos + sin 2 2

xö ÷ 2÷ x÷ 2ø

OR Given 2 tan

-1

(cos x) = tan

-1

( 2cosec x)

æ 2 cos x ö -1 æ 2 ö Þ tan -1 ç ÷ ÷ = tan çè 2 è 1 - cos x ø sin x ø 2 cos x 2 Þ = 2 sin x sin x

æ 2A ö Q 2 tan -1 A = tan -1 ç ÷ è1 - A 2 ø

Þ cot x = 1 p \ x= 4 ®

22. Let sum of vectors 2i$ + 4j$ - 5k$ and li$ + 2j$ + 3k$ = a ®

a = ( 2 + l)i$ + 6j$ - 2k$

®

a$ =

a

®

=

( 2 + l)i$ + 6j$ - 2k$

| a|

( 2 + l) 2 + 36 + 4

( 2 + l)i$ + 6j$ - 2k$ Hence (i$ + j$ + k$) × a$ = (i$ + j$ + k$) × =1 ( 2 + l) 2 + 40 Þ ( 2 + l) + 6 - 2 = ( 2 + l) 2 + 40 Þ ( l + 6) 2 = ( 2 + l) 2 + 40 Þ l2 + 36 + 12l = 4 + l2 + 4l + 40 Þ 8l = 8 Þ l = 1.

SECTION–C 23. The equation of the plane through three non-collinear points A(3, –1, 2), B(5, 2, 4) and (–1, –1, 6) can be expressed as x- 3 y+1 z- 2 5- 3

2+1

4-2 =0

-1 - 3 -1 + 1 6 - 2 Þ

x- 3 y+1 z- 2 2 3 2 =0 -4

0

4

107


Þ 12( x - 3) - 16( y + 1) + 12(z - 2) = 0 Þ 12x - 16y + 12z - 76 = 0 Þ 3x - 4y + 3z - 19 = 0 is the required equation. Now distance of P(6, 5, 9) from the plane is given by 3 ´ 6 - 4(5) + 3( 9) - 19 6 6 units. = = = 9 + 16 + 9 34 34 24. Plot the two curves y 2 = x

… (i)

and x+y=2 Solving (i) and (ii), we have

… (ii) y

y2 + y = 2 Þ ( y + 2)( y - 1) = 0 Þ y = -2, 1 \ x = 4, 1 We have to determine the area.of the shaded region. 1

Required Area =

1

ò ( 2 - y) dy - ò y

-2

2

2 1

dy

3

p

Þ I=

(4,–2)

xdx

… (i) a

p-x

ò a 2 cos 2 ( p - x) + b 2 sin 2 ( p - x)

0 p

2 x+y=2

–2

ò a 2 cos 2 x + b 2 sin 2 x 0 p

Þ I=

1

1

1

y y ù = 2y ú 2 3 ûú -2 1 1 4 8 = æç 2 - - ö÷ - æç -4 - + ö÷ è ø è 2 3 2 3ø 9 = square units. 2 25. Let I =

x

O

-2

2

y2 = x

(1, 1)

a

[using ò f ( x) dx = ò f ( a - x) dx] 0

0

p-x

ò a 2 cos 2 x + b 2 sin 2 x

… (ii)

0

Adding (i) and (ii) p

2I =

p

ò a 2 cos 2 x + b 2 sin 2 x dx 0

p

I=

p dx 2 ò a 2 cos 2 x + b 2 sin 2 x 0

Divide numerator and denominator by cos 2 x p

I=

2

p 2

p sec xdx sec 2 xdx Þ I = p ò a 2 + b 2 tan 2 x 2 ò a 2 + b 2 tan 2 x 0 0

2a

[using

a

ò f ( x) dx = 2ò f ( x) dx] 0

0

108


Let b tan x = t b sec 2 x dx = dt x = 0, p x= 2

When

I=

p b

¥

dt

t=0 t=¥

ò a2 + t 2 0

=

p 1 . tan -1 b a

¥

tù a úû 0

p p p I = (tan -1 ¥ - tan -1 0) = . ab ab 2 p2 2ab 26. The given system of equation are x+ y+z= 6 x + 2z = 7 3x + y + z = 12 In matrix form the equation can be written as AX = B é 1 1 1 ù é xù é 6 ù ê 1 0 2ú ê yú = ê 7 ú Þ ê úê ú ê ú ëê 3 1 1 ûú ëê z ûú ëê12ûú I=

A = 1( 0 - 2) - 1(1 - 6) + 1(1 - 0) = 4 ¹ 0 Þ A -1 exists. To find Adj A we have C 11 = -2 C 12 = 5

\

\

Þ

C 13 = 1

C 21 = 0

C 22 = -2

C 23 = 2

C 31 = 2

C 32 = -1 C 33 = -1

1ù é -2 5 ê Matrix of co-factors of elements = 0 -2 2 ú ê ú êë 2 -1 -1úû 2ù é -2 0 Adj A = ê 5 -2 -1ú ê ú êë 1 2 -1úû 2ù é -2 0 1ê A = = 5 -2 -1 ú ú |A| 4ê êë 1 2 -1úû 2ùé6ù é -2 0 1ê -1 X=A B= 5 -2 -1 ú ê 7 ú úê ú 4ê êë 1 2 -1úû êë12úû -1

Adj A

109


é -12 + 24 ù 1ê = 30 - 14 - 12ú ú 4ê êë 6 + 14 - 12 úû é12ù é 3ù 1ê ú ê ú = 4 = 1 4ê ú ê ú ëê 8 ûú ëê 2 ûú \

Solution of the equations is x = 3, y = 1, z = 2 OR é 3 0 -1 ù Given matrix is A = ê 2 3 0 ú ê ú êë 0 4 1 úû

We know A = IA é 3 0 -1 ù é 1 0 0 ù \ ê 2 3 0 ú = ê 0 1 0ú A ê ú ê ú êë 0 4 1 úû êë 0 0 1 úû Apply R 1 ® R 1 - R 2 é 1 -3 -1 ù é 1 -1 0 ù Þ ê 2 3 0 ú = ê 0 1 0ú A ê ú ê ú êë 0 4 1 úû êë 0 0 1 úû Apply R 2 ® R 2 - 2R 1 é 1 -3 -1 ù é 1 -1 0 ù Þ ê0 9 2 ú = ê -2 3 0 ú A ê ú ê ú êë 0 4 1 úû êë 0 0 1 úû Apply R 2 ® R 2 - 2R 3 é 1 -3 -1 ù é 1 -1 0 ù Þ ê0 1 0 ú = ê -2 3 -2 ú A ê ú ê ú êë 0 4 1 úû êë 0 0 1 úû Apply R 1 ® R 1 + 3R 2 , R 3 ® R 3 - 4R 2 é 1 0 -1 ù é -5 8 -6 ù Þ ê 0 1 0 ú = ê -2 3 -2 ú A ê ú ê ú êë 0 0 1 úû êë 8 -12 9 úû Apply R 1 é1 0 Þ ê0 1 ê êë 0 0

® R1 + R 3 0 ù é 3 -4 3ù ú ê 0 = -2 3 -2 ú A ú ê ú 1 úû êë 8 -12 9 úû

110


Þ A

-1

3ù é 3 -4 ê = -2 3 -2 ú ê ú êë 8 -12 9 úû

27. Given distribution of the balls is shown in the table Bag


White

Red

I

1

2

3

II

2

4

1

III

4

5

3

1 = P( bag II ) = P( bag III ) 3 Let E be the event that 2 balls are 1 black and 1 red. As bags are selected at random P( bag I ) =

1

P(E/bag I) =

C1 ´ 3 C1 6 4

P(E/bag III) =

=

C2 3

C1 ´ C1 12

C2

2

1 5 =

P(E/bag II) =

C 1 ´ 1C 1 7

C2

2 11

We have to determine P(bag I). P(E / bag I) P(bag I/E) = III

å P(bag i) . P(E / bag i)

i=I

1 1 1 1 ´ ´ 3 5 3 5 = = 1 1 1 2 1 2 æ1 + 2 + 2 ö 1 ´ + ´ + ´ ç ÷ 3 5 3 21 3 11 è 5 21 11 ø 3 1 231 5 = = 1 2 2 551 + + 5 21 11 28. Let the no. of fans purchased by the dealer = x and number of sewing machines purchased = y then the L.P.P. is formulated as Z = 22x + 18y to be maximised subject to constrains … (i) [space only for 20 items] x + y £ 20 360x + 240y £ 5760 … (ii) Þ 3x + 2y £ 48 … (iii) x ³ 0, y ³ 0 We plot the graph of the constraints.

=

2 21

111

Examination Papers – 2009 y 28 24 (0, 24) 20 A (0, 20) 16 B(8,12)

12 8 4 O

(16, 0) C 4

8

12

16

3x

(20, 0) x 20 +2 y= 4

x+ y= 20

8

As per the constraints the feasible solution is the shaded region. Possible points for maximising Z are A( 0, 20), B( 8, 12), C(16, 0) Z]A = 22 ´ 0 + 18 ´ 20 = 360 Z]B = 22 ´ 8 + 18 ´ 12 = 392 Z]C = 22 ´ 16 + 18 ´ 0 = 352 Hence profit is maximum of Rs 392 when the dealer purchases 8 fans and 12 sewing machines. 29. Let the hypotenuse and one side of the right triangle be h and x respectively. Then (given as constant) h+x=k Let the third side of the triangle be y y2 + x2 = h2 2

(using Pythagoras theorem)

2

Þ

y= h -x

Þ

A = Area of D =

1 1 yx = x h 2 - x 2 2 2

x ( k - x) 2 - x 2 2 x A= k 2 - 2kx 2 Squaring both sides

\

h

y

A=

q x

x2 2 ( k - 2kx) 4 dA For maxima we find dx A2 =

2A If

dA xk 2 3kx 2 = dx 2 2

dA xk 2 3kx 2 =0Þ = dx 2 2

...(i) Þ

k =x 3

112


Differentiating (i) again w.r.t. x we get 2

d2A k 2 æ dA ö 2ç ÷ + 2. A = - 3kx è dx ø 2 dx 2 Þ

2 ´ 0 + 2×A× d2A

Þ

=-

d2A dx 2

=

k k2 kù - 3k. ú at x = 3 2 3 úû

k2 1 . <0 2 2A

dx 2 \ Area is maximum x = k/3 h = 2k/3 Þ x k/ 3 1 p In the right triangle, cos q = = = Þ q= h 2 k/ 3 2 3 OR x ö Selling price of x items = SP = æç5 ÷x è 100 ø x Cost price of x items = CP = + 500 5 Let profit = P = 5x P=

x2 x - - 500 100 5

24x x 2 - 500 5 100

To find maximisation of profit function dP 24 x =0 = dx 5 50 24 x Þ =0 Þ 5 50 Þ x = 240 items. Differentiating (i) again w.r.t. x

dP =0 dx

Þ

d2P 2

\

=

...(i) 24 x = 5 50

-1 <0 50

dx Profit is maximum if manufacturer sells 240 items

Set–II 2. To find I = ò Let x = t

sin x dx x \

1 2 x

dx = dt

113


I = 2ò sin t dt

[Let x = t \

1 2 x

dx = dt]

= -2 cos t + c = -2 cos x + C 5. Using equality of two matrices, we have equating a 11 elements of two sides x-y=2 equating a 21 elements of two sides x= 3 Þ 3 - y = 2 Þ - y = -1 \ y = 1 11. Given y = 3e 2 x + 2e 3 x Differentiating w.r.t. x dy = 3.2e 2 x + 2. 3e 3 x = 6e 2 x + 6e 3 x dx 6( y - 3e 2 x ) dy (using (i)) Þ = 6e 2 x + dx 2 dy Þ = 6 e 2 x + 3 y - 9 e 2 x = -3 e 2 x + 3 y dx Differentiating again w.r.t. x d2y dy Þ = 3. - 6e 2 x 2 dx dx dy From (ii) - 3 y = -3 e 2 x dx dy - 3y dx Þ = e 2x -3 Substitute in (iii) æ dy ö - 3y ÷ ç d2y dy dx ÷ Þ = 3. - 6ç dx çç -3 ÷÷ dx 2 è ø d2y

Þ

2

dx d2y

Þ

2

=3 -

dx 18. Given lines are

dy dy +2 - 6y dx dx

5dy + 6y = 0 dx

®

r = (1 + 2l)i$ + (1 - l) j$ + lk$ = (i$ + j$) + l ( 2i$ - j$ + k$)

®

r = ( 2i$ + j$ - k$) + m( 3i$ - 5j$ + 2k$) ®

\

ü ® ® ï $ ý Þ a 2 - a 1 = i$ - k ® $ a 2 = 2i$ + j$ - kïþ a 1 = i$ + j$

… (i)

… (ii)

… (iii)

114

Xam idea Mathematics – XII ® ü b 1 = 2i$ - j$ + k$ ï ý Þ lines are not parallel ® b 2 = 3i$ - 5j$ + 2k$ïþ ®

\

Shortest distance =

®

®

®

®

®

( a 2 - a 1 ).(b 1 ´ b 2 ) |b 1 ´ b 2|

®

i$

®

k$

j$

b 1 ´ b 2 = 2 -1 1 = 3i$ - j$ - 7 k$ 3 -5 2 ®

®

Þ |b 1 ´ b 2| = 9 + 1 + 49 = 59 \

Shortest distance = =

(i$ - k$).( 3i$ - j$ - 7 k$) 59 10 units 59

19. As the circle touches y axis at origin, x axis is its diameter. Centre lies on x axis i.e., centre is (r, 0). Hence equations of circle will be ( x - r) 2 + ( y - 0) 2 = r 2

… (i)

Þ x 2 + y 2 - 2rx = 0 Differentiating w.r.t. ‘x’ we get dy dy 2x + 2y - 2r = 0 Þ r = x + y dx dx Putting value of r in (i) we get 2

dy ö dy ö æ æ 2 çx - x - y ÷ + y = çx + y ÷ è è dx ø dx ø 2

2

2

dy æ dy ö æ dy ö Þ y 2 ç ÷ + y 2 = x 2 + y 2 ç ÷ + 2xy è dx ø è dx ø dx dy + x 2 - y 2 = 0 which is the required differential equation. dx 21. Given determinant is x+y x x Þ 2xy

5x + 4y

4x 2x

10x + 8y 8x

3x

Taking x common from both C 2 and C 3 we get

115


x

2

x+y

1 1

5x + 4y

4 2

10x + 8y 8

3

Apply R 2 ® R 2 - 2R 1 , R 3 ® R 3 - 3R 1 we get x+y 1 1 x 2 3x + 2y 2 0 7 x + 5y 5 0 Expanding along C 3 we get

x 2 [(15x + 10y - 14x - 10y)] = x 3 = RHS

25. Given the equation of parabola 4y = 3x 2 Þ y =

3x 2 4

… (i)

and the line 3x - 2y + 12 = 0 3x + 12 Þ =y 2 The line intersect the parabola at (–2, 3) and (4, 12). Hence the required area will be the shaded region. 4

ò

Required Area =

-2

4

3x + 12 dx 2

ò

3

=

… (ii) y

(4, 12)

3x 2 dx 4

-2 4

4

3 2 x ù x + 6x ú 4 4 úû -2

(–2,3) –4

= (12 + 24 - 16) - ( 3 - 12 + 2) = 20 + 7 = 27 square units. 29. From the given distribution of balls in the bags. Bag

2 –2

O

2


White

Red

I

2

1

3

II

4

2

1

III

5

4

3

As bags are randomly selected P(bag I) = 1/3 = P(bag II) = P(bag III) Let E be the event that the two balls are 1 white + 1 Red 1

P(E/ bag I) =

C 1 ´ 3C 1 6

C2

=

1 P(E/ bag II ) = 5

2

C 1 ´ 1C 1 7

C2

=

2 21

4

x

116

Xam idea Mathematics – XII 4

P(E/bag III) = \

C 1 ´ 3C 1 12

P(bag II/E) =

=

C2

2 11

P( bag II) × P(E/ bag II) III

å P( bag i) × P(E/bag i)

i=I

1 2 1 2 ´ ´ 3 21 3 21 = = 1 1 1 2 1 2 1 æ1 2 2 ´ + ´ + ´ + ö÷ ç + è 3 5 3 21 3 11 3 5 21 11 ø 2 110 21 = = 1 2 2 551 + + 5 21 11

Set–III sec 2 x dx x 1 Let x = t Þ dx = 2dt x

7. Let I = ò

\

I = 2ò sec 2 t dt = 2 tan t + C

Þ I = 2 tan x + C 10. Using equality of two matrices é 2x - y 5 ù é 6 5 ù = ê 3 yúû êë 3 -2úû ë Þ

2x - y = 6 y = -2 \ x=2 13. The given lines are

equating a 11 equating a 22

®

r = (i$ + 2j$ + 3k$) + l(i$ - 3j$ + 2k$)

Þ a 1 = i$ + 2j$ + 3k$ ,

®

b 1 = i$ - 3j$ + 2k$

®

r = ( 4i$ + 5j$ + 6k$) + m( 2i$ + 3j$ + k$)

® a2

= 4i$ + 5j$ + 6k$

®

b 2 = 2i$ + 3j$ + k$

é i$ j$ k$ ù ê ú b 1 ´ b 2 = ê 1 -3 2 ú = -9i$ + 3j$ + 9k$ ê2 3 1ú ë û ®

®

… (i)

… (ii)

[by rearranging given equation]

117


®

|b 1 ´ b 2|= 81 + 9 + 81 = 171 = 3 19 ®

®

a 2 - a 1 = 3i$ + 3j$ + 3k$ As lines (i) and (ii) are not parallel, the shortest distance ®

=

®

®

®

( a 2 - a 1 ) × (b 1 ´ b 2 ) ®

®

=

( 3i$ + 3j$ + 3k$) × ( -9i$ + 3j$ + 9k$)

|b 1 ´ b 2| Shortest distance =

-27 + 9 + 27 3 units = 3 19 19

14. Equation of family of curves is ( x - a) 2 + 2y 2 = a 2 2

3 19

2

Þ x + 2y - 2ax = 0

… (i) … (ii)

Differentiating w.r.t. ‘x’ dy 2x + 4y - 2a = 0 dx Þ a = x + 2yy 1 Substituting value of ‘a’ in (ii) x 2 + 2y 2 - 2( x + 2yy 1 ). x = 0 Þ 2y 2 - x 2 - 4xyy 1 = 0 which is required differential equation. 16. Given determinant is 1+x 1 |A|=

1

1

1+y

1

1

1

1+z

Apply C 2 ® C 2 - C 3 1+x 0 1 |A| =

1

y

1

1

-z 1 + z

Apply C 1 ® C 1 - C 3 x 0 1 |A| = 0

y

1

-z -z 1 + z Apply C 1 ® C 1 - x C 3 0 0 |A| =

-x

y

1 1

-z - x - xz -z 1 + z Expand along R 1 |A| = 1( xz + yz + xy + xyz) = RHS

118


18. Given equation is y = e x (sin x + cos x) Differentiating w.r.t. ‘x’ we get dy = e x (cos x - sin x) + e x (sin x + cos x) dx dy Þ = e x (cos x - sin x) + y dx Differentiating again w.r.t. ‘x’ we get dy dy d2y dy Þ = e x ( - sin x - cos x) + e x (cos x - sin x) + = -y + -y+ 2 dx dx dx dx \

d2y dx

2

-2

dy + 2y = 0 dx 2

æ x2 ö 23. The curves y = 4ax and x = 4ay intersects at points where çç ÷÷ = 4ax è 4a ø 2

Þ

x4

2

= 4ax

16a 2 Þ x( x 3 - 64a 3 ) = 0

Þ Þ

x 4 = 64a 3 x

Y x2=4ay

x = 0 or x = 4a

We plot the curves on same system of axes to get the required region. 4 aæ x2 ö ÷ dx The enclosed area = ò çç 4ax 4a ÷ø è 0

y2=4ax

O

4a

4a

3 ù 2 2 x3 ú =2 a x 3 12a ú û0 3

( 4a) 3 4 32a 2 16a 2 16a 2 square units. a( 4a) 2 - 0= = 3 12a 3 3 3 26. Let E1 be event getting number > 4 E2 be event getting number £ 4 2 1 4 2 P(E1 ) = = P(E2 ) = = 6 3 6 3 Let E be the event that man reports getting number > 4. 3 2 P(E/E1 ) = P(E/E2 ) = 5 5 By Baye’s theorem 1 3 ´ P(E1 ) × P(E/E1 ) 3 3 3 5 P(E1 /E) = = = = 1 3 2 2 P(E1 ) × P(E/E1 ) + P(E2 ). P(E / E2 ) 3+4 7 ´ + ´ 3 5 3 5 =

X

EXAMINATION PAPERS – 2009 MATHEMATICS CBSE (Foreign) CLASS – XII Time allowed: 3 hours

Maximum marks: 100


Set–I SECTION–A 1. Evaluate:

1

ò x + x log x dx . 1

2. Evaluate:

ò 0

1 dx . 2x + 3

3. If the binary operation *, defined on Q, is defined as a * b = 2a + b - ab, for all a, b Î Q, find the value of 3 * 4 . æ y + 2x 5 ö æ 7 5 ö 4. If ç ÷=ç ÷ , find the value of y. 3 ø è -2 3 ø è -x ®

5. Find a unit vector in the direction of a = 2i$ - 3j$ + 6k$. 6. Find the direction cosines of the line passing through the following points: ( -2, 4, - 5), (1, 2, 3). æ 2 3 -5 ö æ 2 1 -1 ö ç ÷ ç ÷ 7. If A = ( aij ) = ç 1 4 9 ÷ and B = (bij ) = ç -3 4 4 ÷ , then find a 22 + b 21 . ç 0 7 -2 ÷ ç1 5 2÷ è ø è ø ®

8. If| a|=

®

® ®

3 , | b|= 2 and a . b =

®

®

3, find the angle between a and b .

æ1 2ö 9. If A = ç ÷ , then find the value of k if|2A|= k|A|. è 4 2ø 3p ù 10. Write the principal value of tan -1 éê tan . 4 úû ë

120


SECTION–B 11. Evaluate:

cos x

ò ( 2 + sin x)( 3 + 4 sin x) dx OR 2

Evaluate: ò x . cos

-1

x dx

12. Show that the relation R in the set of real numbers, defined as R = {( a, b) : a £ b 2 } is neither reflexive, nor symmetric, nor transitive. dy x + y æ yö 13. If log ( x 2 + y 2 ) = 2 tan -1 ç ÷ , then show that . = è xø dx x - y OR If x = a (cos t + t sin t) and y = a (sin t - t cos t), then find

d2y

. dx 2 14. Find the equation of the tangent to the curve y = 4x - 2 which is parallel to the line 4x - 2y + 5 = 0 . OR Using differentials, find the approximate value of f ( 2 × 01), where f ( x) = 4x 3 + 5x 2 + 2. 15. Prove the following: 1 2 1 3 tan -1 æç ö÷ + tan -1 æç ö÷ = cos -1 æç ö÷. è 4ø è 9ø 2 è5ø OR Solve the following for x : æ x2 - 1ö æ 2x ö 2p ÷ + tan -1 ç . cos -1 çç ÷= ÷ 2 è x2 - 1ø 3 è x + 1ø x + 1 3y + 5 3 - z 16. Find the angle between the line and the plane 10x + 2y - 11z = 3. = = 2 9 -6 17. Solve the following differential equation: ( x 3 + y 3 ) dy - x 2 ydx = 0 18. Find the particular solution of the differential equation. dy p + y cot x = 4x cosec x, ( x ¹ 0), given that y = 0 when x = . dx 2 19. Using properties of determinants, prove the following: a2 + 1 ab 2 ab b +1 ca

cb

ac bc c2 + 1

= 1 + a2 + b 2 + c 2 .

121


1 2 and the probability that B hits it is . If each one of A 3 5 and B shoots at the target, what is the probability that (i) the target is hit? (ii) exactly one-of-them-hits the target? dy 21. Find , if y x + x y = a b , where a, b are constants. dx 20. The probability that A hits a target is

® ® ®

® ®

® ®

®

®

®

® ®

22. If a , b , c are three vectors such that a . b = a . c and a ´ b = a ´ c , a ¹ 0, then show that ®

®

b = c.

SECTION–C 23. One kind of cake requires 200 g of flour and 25 g of fat, and another kind of cake requires 100 g of flour and 50 g of fat. Find the maximum number of cakes which can be made from 5 kg of flour and 1 kg of fat assuming that there is no shortage of the other integredients used in making the cakes. Formulate the above as a linear programming problem and solve graphically. 24. Using integration, find the area of the region: {( x, y):9x 2 + y 2 £ 36 and 3x + y ³ 6} x+ 3 y-1 z-5 x+1 y- 2 z-5 25. Show that the lines are coplanar. Also find the = = ; = = -3 1 5 -1 2 5 equation of the plane containing the lines. 26. Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of 2R radius R is . Also find the maximum volume. 3 OR Show that the total surface area of a closed cuboid with square base and given volume is minimum, when it is a cube. 27. Using matrices, solve the following system of linear equations: 3x - 2y + 3z = 8 2x + y - z = 1 4x - 3y + 2z = 4 28. Evaluate: ò

x 4 dx ( x - 1)( x 2 + 1)

OR 4

Evaluate: ò [|x - 1|+|x - 2|+|x - 4|]dx 1

29. Two cards are drawn simultaneously (or successively without replacement) from a well suffled pack of 52 cards. Find the mean and variance of the number of red cards.

122


Set–II Only those questions, not included in Set I, are given. 7. Evaluate: e 2 x - e -2 x

ò e 2 x + e -2 x dx 5ö æ 3 5ö æ 3x - 2y 10. If ç ÷=ç ÷ , find the value of y. -2 ø è -3 -2 ø è x x - 2 2y - 5 3 - z and the plane x + 2y + 2z - 5 = 0. = = 3 4 -6 15. Solve the following differential equation: dy 2 ( x 2 - 1) + 2xy = 2 dx x -1 13. Find the angle between the line

16. Using properties of determinants, prove the following: 1 x2

x 1

x

x2

x2 x = (1 - x 3 ) 2 . 1

18. If y = a cos(log x) + b sin(log x), then show that x2

d2y 2

+x

dy + y = 0. dx

dx 26. A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the mean and variance of the number of successes. 28. Using integration, find the area of the region: {( x, y) : 25x 2 + 9y 2 £ 225 and 5x + 3y ³ 15}

Set–III Only those questions, not included in Set I and Set II are given. 5 ö æ -21 5 ö æ 7y 1. If ç ÷=ç ÷ , find the value of x. è 2x - 3y -3ø è 11 -3ø 4. Evaluate: e ax - e - ax

ò e ax + e - ax dx 15. If 1 - x 2 + 1 - y 2 = a( x - y), show that

dy 1 - y2 = . dx 1 - x2

123


17. Using properties of determinants, prove the following: a + bx c + dx p + qx

a c

p

ax + b cx + d px + q = (1 - x ) b d

q

2

u

v

w

18. For the differential equation xy

u v w dy = ( x + 2)( y + 2), find the solution curve passing through dx

the point (1, –1). 20. Find the equation of the perpendicular drawn from the point (1, –2, 3) to the plane 2x - 3y + 4z + 9 = 0. Also find the co-ordinates of the foot of the perpendicular. 24. Using integration, find the area of the triangle ABC with vertices as A( -1, 0), B(1, 3) and C( 3, 2). 27. From a lot of 30 bulbs which includes 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the mean and variance of the number of defective bulbs.

124


SOLUTIONS Set – I SECTION–A 1. Let I = ò

1 dx dx = ò x + x log x x (1 + log x)

Let 1 + log x = t 1 dx = dt x dt \ I = ò = log|t|+ C t = log 1 + log x + C 1

2.

ò 0

1

1

1 dx = ò ( 2x + 3) 2 dx 2x + 3 0

=

1 3) 2

( 2x + 1 ´2 2 1

1

ù ú ú ú úû 0

1

=52 - 32 = 5 -

3

3. Given binary operation is a * b = 2a + b - ab \ 3* 4 = 2 ´ 3 + 4 - 3 ´ 4 Þ 3 * 4 = -2 4. Using equality of two matrices in æ y + 2x 5 ö æ 7 5 ö ç ÷=ç ÷ 3 ø è -2 3 ø è -x We get y + 2x = 7 - x = -2 \ y + 2( 2) = 7

Þ Þ

®

5. Given a = 2i$ - 3j$ + 6k$ ®

Þ | a |= 4 + 9 + 36 = 7 ®

\

a$ =

2i$ - 3j$ + 6k$ a = a 7

x=2 y= 3

125


Þ a$ = Unit vector in direction of a 2 3 6 = i$ - j$ + k$ 7 7 7 6. Direction ratios of the line passing through (–2, 4, –5) and (1, 2, 3) are 1 – (–2), 2 – 4, 3 – (–5) = 3, – 2, 8 3 -2 8 \ Direction cosines are = , , 9 + 4 + 64 9 + 4 + 64 9 + 4 + 64 3 -2 8 = , , 77 77 77 7. a 22 = 4, b 21 = -3 a 22 + b 21 = 4 - 3 = 1 ®

8. Given| a|=

®

® ®

| b|= 2, a × b =

3,

3

We know ® ®

®

®

a × b =| a|| b|cos q

Þ

3 = 3 ( 2) cos q 1 Þ = cos q 2 p Þ q= 3 æ1 2ö 9. Given A=ç ÷ è 4 2ø Þ

æ 2 4ö 2A = ç ÷ è 8 4ø

\

|2A| = 8 - 32 = -24 |A|= 2 - 8 = -6 Þ -24 = k( -6) 4=k 3 pö p öö æ æ -1 æ 10. tan -1 ç tan ÷ = tan ç tan ç p - ÷ ÷ è ø è è 4 4 øø p = tan -1 ( -1) = 4 3p ö -p . \ Principal value of tan -1 æç tan ÷= è 4 ø 4

SECTION–B 11. Let I = ò

cos xdx ( 2 + sin x)( 3 + 4 sin x)

Let sin x = t

126


cos x dx = dt dt \ I=ò ( 2 + t)( 3 + 4t) 1 A B Let = + ( 2 + t)( 3 + 4 t) 2 + t 3 + 4 t Þ 1 = A( 3 + 4t) + B( 2 + t) Þ 3A + 2B = 1 4A + B = 0 Þ B = -4A \ 3A - 8A = 1 1 4 A=Þ B= 5 5 dt -1 dt 4 dt Þ I=ò = + ( 2 + t)( 3 + 4t) 5 ò 2 + t 5 ò 3 + 4t -1 4 log| 3 + 4t| = log|2 + t| + +C 5 5 4 -1 1 = log|2 + sin x|+ log| 3 + 4 sin x|+ C 5 5 1 3 + 4 sin x = log +C 5 2 + sin x OR 2

Let I = ò x cos

-1

= cos -1 x .

=

x dx

-1 x3 x3 -ò ´ dx 3 3 1 - x2

[Integrating by parts]

x3 1 x 3 dx cos -1 x + ò 3 3 1 - x2

x3 1 cos -1 x + I 1 3 3 In I1, let 1 – x2 = t so that - 2xdx = dt 1 1 -t 1 æ1 ö \ I1 = - ò dt = - ò ç - t ÷ dt ø 2 2 è t t =

1æ 2 3/ 2 ö ç2 t - t ÷ +C¢ è ø 2 3 1 = - 1 - x 2 + (1 - x 2 ) 3 2 + C ¢ 3 =-

\

I=

x3 1 1 cos -1 x 1 - x 2 + (1 - x 2 ) 3 3 3 9

2

+C

127


12. Given relation is R = {( a, b): a £ b 2 } Reflexivity: Let a Î real numbers. aRa Þ a £ a 2 but if a < 1

Let a =

a
aRb Þ / bRa

For example, let a = 2, b = 5 then 2 £ 5 2 but 5 £ 2 2 is not true. Hence R is not symmetric. Transitivity Let a, b, c Î real numbers aRb Þ a £ b 2 and bRc Þ b £ c 2 Considering aRb and bRc Þ a £ c4 Þ / aRc Hence R is not transitive e.g., if a = 2, b = –3, c = 1 aRb Þ 2 £ 9 bRc Þ -3 £ 1 aRc Þ 2 £ 1 is not true. y 13. Given log ( x 2 + y 2 ) = 2 tan -1 æç ö÷ è xø Differentiating w.r.t.x dy dy 2x + 2y x -y 2 dx = . dx x2 + y2 y2 x2 1+ x2

1 2

Þ

a2 =

1 4

128


æ dy ö - y÷ çx dy 2 ç dx ÷ Þ 2x + 2y = 2x dx çç x 2 ÷÷ è ø dy dy =x -y dx dx dy dy Þ x+y=x -y dx dx dy x + y Þ = dx x - y Þ x+y

OR Given x = a(cos t + t sin t), y = a(sin t - t cos t) dy dx Þ = a( - sin t + t cos t + sin t) = at cos t , = a(cos t + t sin t - cos t) = at sin t dt dt dy dy / dt at sin t = = = tan t dx dx / dt at cos t Differentiating w.r.t.x again Þ

d2y dx

2

= sec 2 t.

dt dx

= sec 2 t. =

1 at cos t

sec 3 t at

14. Given curve is y = 4x - 2 Differentiating w.r.t.x dy 4 2 = = dx 2 4x - 2 4x - 2

… (i)

The tangent is parallel to the line 4x - 2y + 5 = 0. -4 The slope of this line is = =2 -2 2 \ Slope of tangent = =2 4x - 2 Þ 1 = 4x - 2 Þ 1 = 4x - 2

Þ

Put value of x in (i) 3 y= 4´ -2=1 4

x=

3 4

129


\

Equation of tangent will be 3 y - 1 = 2 æç x - ö÷ è 4ø

3 2 or 2y - 2 = 4x - 3 Hence equation of tangent is 4x - 2y - 1 = 0 OR Þ y - 1 = 2x -

Given f ( x) = 4x 3 + 5x 2 + 2 Þ f ¢( x) = 12x 2 + 10x We know for finding approximate values f ( x + Dx) = f ( x) + f ¢( x). Dx \

f ( 2 . 01) = f ( 2) + f ¢( 2)( 0 . 01) = [4( 2) 3 + 5( 2) 2 + 2] + [12( 2) 2 + 10( 2)]( 0 . 01) = [4 ´ 8 + 5 ´ 4 + 2] + [12 ´ 4 + 20]( 0.01) = 54 + ( 68)( 0 . 01) = 54 . 68 1 2 + tan -1 æç ö÷ è 4 9ø æ 1+2 ö ç ÷ = tan -1 ç 4 9 ÷ ç1 - 1 . 2 ÷ è 4 9ø æ 17 ö ç ÷ = tan -1 ç 36 ÷ ç 34 ÷ è 36 ø 1 1 = tan -1 = æç 2 tan -1 2 2è æ1 - 1 ö ç ÷ 1 4÷ = cos -1 ç 2 ç1 + 1 ÷ è 4ø

15. LHS of given equation = tan -1

=

1 3 cos -1 æç ö÷ = R.H.S. è5ø 2

1ö ÷ 2ø Using 2 tan -1 A = cos -1

1 - A2 1 + A2

130


OR 2 æ ö æ 2x ö 2p x -1 ÷ + tan -1 ç Given cos -1 çç ÷= ÷ 2 è x2 - 1ø 3 è x + 1ø æ -(1 - x 2 ) ö ÷ + tan -1 æç - 2x ö÷ = 2p Þ cos -1 ç ç 1 + x2 ÷ è 1 - x2 ø 3 è ø æ1 - x 2 ö æ 2x ö 2p ÷ - tan -1 ç Þ p - cos -1 çç ÷= 2÷ è1 - x 2 ø 3 è1 + x ø [Using cos -1 ( - A) = p - cos -1 A and tan -1 ( - A) = - tan -1 A] 2p Þ p - 2 tan -1 x - 2 tan -1 x = 3 2p -1 Þ p= 4 tan x 3 p p p p Þ = tan -1 x Þ x = tan = tan æç - ö÷ è4 6ø 12 12 1 p p 1- tan 3 4 6 \ x= = p p 1 1 + tan ´ tan 1+ 4 6 3 ( 3 - 1)( 3 - 1) 3 -1 Þ x= Þ x= 3 +1 ( 3 + 1)( 3 - 1) tan

3+1- 2 3 =2- 3 2 16. Given line can be rearranged to get x - ( -1) y - ( -5/ 3) z- 3 = = 2 3 6 Its direction ratios are 2, 3, 6. Direction ratios of normal to the plane 10x + 2y - 11z = 3 are 10, 2, – 11 Angle between the line and plane 2 ´ 10 + 3 ´ 2 + 6(–11) sin q = 4 + 9 + 36 100 + 4 + 121 20 + 6 – 66 -40 = = 7 ´ 15 105 -8 -8 or q = sin -1 æç ö÷ sin q = è 21 21 ø x=

17. ( x 3 + y 3 ) dy - x 2 ydx = 0 is rearranged as dy x2y = dx x 3 + y 3

131


It is a homogeneous differential equation. y Let = v Þ y = vx x dy dv =v+ x dx dx dv v \ v+x = dx 1 + v 3 Þ x Þ

-v4 dv v = -v= dx 1 + v 3 1 +v3

1 +v3 4

dv = -

dx x

v Integrating both sides, we get æ 1 1ö dx ò çè v 4 + v ÷ø dv = - ò x 1 Þ + log|v| = - log|x| + C 3v 3 Þ -

x3 3y

Þ

-x 3 3y 3

3

æ yö + log ç ÷ = - log|x|+ C è xø

+ log|y|= C is the solution of the given differential equation.

18. Given differential equation is

dy dy + y cot x = 4x cosec x and is of the type + Py = Q dx dx

cot xdx \ I . F. = e ò = e log|sin x| = sin x Its solution is given by Þ sin x . y = ò 4x cosec x. sin x dx

Þ y sin x = ò 4x dx =

4x 2 +C 2

Þ y sin x = 2x 2 + C Now y = 0 when x =

p 2

p2 p2 + C ÞC = 4 2 Hence the particular solution of given differential equation is

\

0=2´

y sin x = 2x 2 -

p2 2

132


a2 + 1 ab 19. Let|A|= ab b2 + 1 ca

ac bc c2 + 1

cb

Apply C 1 ® aC 1 , C 2 ® bC 2 , C 3 ® cC 3 1 Þ |A|= abc

a( a 2 + 1)

ab 2

c2a

a 2b

b(b 2 + 1)

c 2b

a2 c

b2c

c( c 2 + 1)

Take a, b, c common respectively from R 1 , R 2 and R 3 a2 + 1 b2 c2 abc |A|= a2 b2 + 1 c2 abc a2 b2 c2 + 1 Apply C 1 ® C 1 + C 2 + C 3 a2 + b 2 + c 2 + 1 b2 |A|= a 2 + b 2 + c 2 + 1 b 2 + 1 a2 + b 2 + c 2 + 1

b2

1 b2 = ( a 2 + b 2 + c 2 + 1) 1 b 2 + 1 1

b2

c2 c2 c2 + 1 c2 c2 c2 + 1

Apply R 2 ® R 2 - R 1 R 3 ® R 3 - R1 1 b2 \ |A|= ( a + b + c + 1) 0 1 0 0 2

2

2

c2 0 1

Expanding along C 1 |A| = a 2 + b 2 + c 2 + 1 20. Let P(A) = Probability that A hits the target =

1 3

P(B) = Probability that B hits the target = 2/5 (i) P(target is hit) = P(at least one of A, B hits) = 1 - P (none hits) 2 3 9 3 =1- ´ = = 3 5 15 5

133


(ii) P(exactly one of them hits) = P( A& B or A& B) = P( A) ´ P( B) + P( A). P( B) 1 3 2 2 7 = ´ + ´ = 3 5 3 5 15 21. y x + x y = a b Let v = y

...(i)

x

u = xy Taking log on either side of the two equation, we get log v = x log y, log u = y log x Differentiating w.r.t.x, we get dy 1 dv 1 dy 1 du y = x. + log y, = + log x. v dx y dx u dx x dx Þ

é x dy ù du dy ù dv éy = yx ê + log yú , = x y ê + log x. ú dx dx û ëx ë y dx û dx

From (i), we have u + v = ab du dv Þ + =0 dx dx é x dy ù dy ù éy Þ yx ê + log yú + x y ê + log x. ú = 0 dx û ëx ë y dx û Þ yx. Þ

dy y x dy + x y . log x = - y x log y - x y . y dx dx x

x y- 1 y dy - y log y - x = x 1 y dx y x + x . log x ® ®

® ®

22. Given a . b = a . c Þ Þ

® ®

® ®

® ®

®

a. b - a. c = 0 a.(b - c ) = 0 ®

®

®

®

®

®

®

®

®

® ®

®

®

®

Þ either b = c or a ^ b - c ®

Also given a ´ b = a ´ c Þ Þ

®

a´ b- a´ c = 0 ®

Þ

®

®

®

a´(b - c) = 0

®

a|| b - c or b = c

®

®

®

But a cannot be both parallel and perpendicular to ( b - c ). ®

®

Hence b = c .

134


SECTION–C 23. Let x = Number of cakes of Ist type while y = Number of cakes of IInd type The linear programming problem is to maximise z = x + y subject to. 200x + 100y £ 5000 Þ 2x + y £ 50 y 25x + 50y £ 1000 Þ x + 2y £ 40 and x ³ 0, y ³ 0 50 (0, 50) To solve the LPP we draw the graph of the in equations and get the feasible solution 40 shown (shaded) in the graph. A Corner points of the common shaded region 30 are A (25, 0), B (20, 10) and (0, 20). C(0, 20) 20 Value of Z at each corner points: Z] = 0 + 20 = 20 10

( 0 , 20 )

Z]

= 20 + 10 = 30

A(25, 0) O

10

30

20 +y 2x

( 20 , 10 )

= 25 + 0 = 25

0 =5

Z ]

B(20,10)

40

x 50 x+2 y=4 0

( 25 , 0 )

Hence 20 cakes of Ist kind and 10 cakes of IInd kind should be made to get maximum number of cakes. 24. Given region is {( x, y): 9x 2 + y 2 £ 36 and 3x + y ³ 6} y

We draw the curves corresponding to equations x2 y2 + = 1 and 3x + y = 6 4 9 The curves intersect at (2, 0) and (0, 6) \ Shaded area is the area enclosed by the two curves and is 9x 2 + y 2 = 36

(0, 6)

or

2 2 æ x2 ö ÷ dx - ò ( 6 - 3x) dx = ò 9çç1 4 ÷ø è 0 0

x’

(2, 0) O

2

éx 4 x x2 ù = 3ê 4 - x 2 + sin -1 - 2x + ú 2 2 2 úû êë 4 0 2 4 4 é ù -1 2 = 3ê 4 - 4 + sin - 4 + - 0ú 2 2 2 ë4 û p = 3éê 2 - 2ùú = 3( p - 2) square units ë 2 û

y’

x

135


25. Given lines are x+ 3 y-1 z-5 = = -3 1 5 x+1 y- 2 z-5 and = = -1 2 5 These lines will be coplanar if x 2 - x 1 y 2 - y 1 z2 - z1 a1

b1

c1

a2

b2

c2

… (i) … (ii)

=0

-1 + 3 2 - 1 5 - 5 \

-3

1

5

-1

2

5

= 2(5 - 10) - 1( -15 + 5) = 0

Hence lines are co-planar. The equation of the plane containing two lines is x+ 3 y-1 z-5 -3

1

5

-1

2

5

=0

Þ -5( x + 3) + 10( y - 1) - 5(z - 5) = 0 Þ -5x - 15 + 10y - 10 - 5z + 25 = 0 Þ -5x + 10y - 5z + 0 = 0 or Þ - x + 2y - z = 0 x - 2y + z = 0 26. Let r, h be the radius and height of the cylinder inscribed in the sphere of radius R. \ Using Pythagoras theorem 4r 2 + h 2 = 4R 2 Þ r2 =

4R 2 - h 2 4

… (i) r

Volume of cylinder = V = pr 2 h æ 4R 2 - h 2 ö p ÷ = pR 2 h - h 3 Þ V = p . hçç ÷ 4 4 è ø dV 3p 2 = pR 2 h dh 4 For finding maximum volume dV 3p 2 =0 Þ pR 2 = h dh 4 2 Þ h= R 3 Þ

R h

… (ii)

2r

136


Differentiating (ii) again d 2V dh

2

=-

6p h 4

d 2V dh 2 æç h = 2 R ö÷ è 3 ø

=-

3p æ 2 ö R ÷ = - 3Rp < 0 ç 2 è 3 ø

Hence volume is maximum when h = Maximum volume = V ] h=

2R

2 R. 3

æ 4R 2 - h 2 ö ÷ = phçç ÷ 4 è ø

3

Vmax

æ 4R 2 ç 4R 2 2R ç 3 =p´ 4 3ç ç è =

ö ÷ ÷ ÷ ÷ ø

2 pR 2 R 2 4 pR 3 cubic units. . = 3 3 3 3

OR The sides of the cuboid in the square base can be x, x and y Let total surface area = S = 2x 2 + 4xy

…(i)

2

As volume of cuboid is V = x y = constant V \ y= x2 V 4V \ S = 2x 2 + 4x. = 2x 2 + 2 x x To find condition for minimum S we find dS 4V = 4x dx x2

Þ If

dS =0 Þ dx

d2S dx

2

=4+

8V x3

[Substituting (ii) in (i)] dS dx … (iii)

4x 3 = 4V

Þ x3 =V Þx =V Differentiating (iii) again w.r.t. x

… (ii)

1 3

137


d2S dx 2 æç x = V 1 3 ö÷ è ø \

=4+

8V = 12 > 0 V

Surface area is minimum when x = V y=

Put value of x in (ii)

V V

\

2 3

x= y=V

=V

1 3

1 3

1 3

Hence cuboid of minimum surface area is a cube. 27. Given linear in equations are 3x - 2y + 3z = 8 2x + y - z = 1 4x - 3y + 2z = 4 The given equations can be expressed as AX = B æ 3 -2 3 ö æ x ö æ 8 ö ç ÷ç ÷ ç ÷ Þ ç 2 1 -1 ÷ ç y ÷ = ç 1 ÷ ç 4 -3 2 ÷ ç z ÷ ç 4 ÷ è øè ø è ø \

X = A–1B

To find A -1 we first find Adj. A Co-factors of elements of A are c 11 = -1, c 12 = -8, c 13 = -10 c 21 = -5,

c 22 = -6,

c 31 = -1,

c 32 = 9,

c 23 = 1 c 33 = 7

æ -1 -8 -10ö ç ÷ Matrix of co-factors = ç -5 -6 1 ÷ ç -1 9 7 ÷ø è æ -1 -5 -1 ö ç ÷ Adj A = ç -8 -6 9 ÷ ç -10 1 7 ÷ è ø

\

|A| = 3( 2 - 3) + 2( 4 + 4) + 3( -6 - 4) = -3 + 16 - 30 = -17 ¹ 0 æ -1 -5 -1 ö ÷ 1 ç -1 A =ç -8 -6 9 ÷ 17 ç ÷ è -10 1 7 ø

138


æ xö ç ÷ 1 Þ ç y÷ = 17 çz÷ è ø

æ -1 -5 -1 ö æ 8 ö ç ÷ç ÷ ç -8 -6 9 ÷ ç 1 ÷ ç -10 1 7 ÷ ç 4÷ è øè ø

1 =17

æ -8 - 5 - 4 ö ç ÷ ç -64 - 6 + 36÷ ç -80 + 1 + 28 ÷ è ø

1 =17

æ -17 ö ç ÷ ç -34 ÷ ç -51 ÷ è ø

Þ x = 1, y = 2, z = 3 is the required solution of the equations. 28. Let I = ò

x4 ( x - 1)( x 2 + 1)

Suppose

x4 ( x - 1)( x 2 + 1) = x+1+

dx

=

x4 -1 +1 ( x - 1) ( x 2 + 1) 1

( x - 1)( x 2 + 1) 1 A Bx + c Also let = + 2 x 1 ( x - 1)( x + 1) x2 + 1 Þ 1 = A( x 2 + 1) + ( Bx + C) ( x - 1) Equating coefficients of similar terms A+B=0 -B + C = 0 Þ B=C A-C =1 \ A- B=1 A+B=0 1 1 Þ 2A = 1 Þ A = Þ B= - =C 2 2 1 æ ö ç 1 x+1 ÷ 2 \ I = ò çx + 1 + ÷dx x - 1 2 x2 + 1÷ ç è ø =

x2 1 1 2x 1 dx + x + log|x - 1| - ò dx - ò dx 2 2 4 x2 + 1 2 x2 + 1

=

x2 1 1 1 + x + log|x - 1| - log|x 2 + 1| - tan -1 x + C 2 2 4 2

139


OR 4

Given I = ò [|x - 1| +|x - 2| +|x - 4|] dx 1 4

2

4

4

= ò ( x - 1) dx + ò -( x - 2) dx + ò ( x - 2) dx + ò -( x - 4) dx 1

1

2

4

=

1

2

4

4

ù é x2 ù ù æ x2 öù x2 x2 - xú + ê + 2xú + - 2xú + çç + 4x÷÷ ú 2 2 ø úû 1 úû 1 êë 2 úû 1 úû 2 è 2

16 1 1 16 16 1 = æç - 4 - + 1ö÷ + æç -2 + 4 + - 2ö÷ + æç - 8 - 2 + 4ö÷ + æç + 16 + - 4ö÷ è2 ø è ø è ø è ø 2 2 2 2 2 1 1 1 = æç5 - ö÷ + + 2 + 4 + è ø 2 2 2 1 23 = 11 + = 2 2 29. Total no. of cards in the deck = 52 Number of red cards = 26 No. of cards drawn = 2 simultaneously \ x = value of random variable = 0, 1, 2 P(x)

xi 0

26

C 0 ´ 26 C 2 52

1

26

C1 ´ 26 C1 52

2

26

C2

C2

C 0 ´ 26 C 2 52

C2

xi P( x)

x12 P( x)

=

25 102

0

0

=

52 102

52 102

52 102

=

25 102

50 102

100 102

Sxi P( x) = 1

Mean = m = Sxi P( x) = 1 Variance = s 2 = Sxi2 P( x) - m 2 152 50 25 -1= = 102 102 51 = 0.49

=

Sxi 2 P( x) =

152 102

140


Set–II 7. Let I = ò

e 2 x - e -2 x e 2 x + e -2 x

dx

Let e 2 x + e -2 x = t Þ 2( e 2 x - e -2 x ) dx = dt \

I=

1 dt 2ò t

1 log|t| + c 2 1 = log|e 2 x + e -2 x| + c 2 10. Using equality of two matrices, we have 3x - 2y = 3 x = -3 \ 3( -3) - 2y = 3 Þ -2y = 12 Þ y = -6 \ x = - 3, y = - 6 13. The given line is x - 2 2y - 5 3 - z = = 3 4 -6 It is rearranged as x- 2 y-5/ 2 z- 3 = = 3 2 6 DR’s of the line are = 3, 2, 6 The given equation of plane is x + 2y + 2z - 5 = 0 The DR’s of its normal are = 1, 2, 2 To find angle between line and plane 3(1) + 2( 2) + 6( 2) 19 sin q = = 9 + 4 + 36 1 + 4 + 4 21 19 Þ q = sin -1 æç ö÷ è 21 ø =

15. The differential equation given is dy 2 ( x 2 - 1) + 2xy = 2 dx x -1 dy 2x 2 Þ + y= 2 2 dx x - 1 ( x - 1) 2

141


dy + Py = Q dx

It is an equation of the type ò

2x x2 - 1

dx

\ I . F. = e = e log( x Its solution is given by ( x 2 - 1) y = ò ( x 2 - 1)

2 - 1)

= x2 - 1

2 2

( x - 1) 2

dx

1 x-1 Þ ( x 2 - 1) y = 2 . log +C 2 x+1 1 x-1 C is required solution. Þ y= log + 2 2 x+1 x -1 x -1 1 16. Let|A| = x 2

x 1

x2 x

x

x2

1

Apply C 1 ® C 1 + C 2 + C 3 1 + x + x2 |A| = 1 + x + x 2

x 1

x2 x

1 + x + x2

x2

1 x 1

x2 x

1 x2

1

1 Þ |A| = (1 + x + x ) 1 2

Apply R 2 ® R 2 - R 1 , R 3 ® R 3 - R 1 1 Þ |A|= (1 + x + x ) 0 2

x 1-x

x2 x - x2

0 x2 - x 1 - x2 Take (1 - x) common from R 2 and R 3 |A| = (1 + x + x 2 )(1 - x) 2

1 0

x 1

x2 x

0 -x 1 + x Expanding along C 1

|A| = (1 + x + x 2 )(1 - x) 2 (1 + x + x 2 ) = (1 - x 3 ) 2

18. Given y = a cos(log x) + b sin(log x) dy - a sin(log x) b cos(log x) Þ = + dx x x

[Q 1 - x 3 = (1 - x)(1 + x + x 2 )]

142


dy = - a sin(log x) + b cos(log x) dx Differentiating again w.r.t. x d 2 y dy - a cos(log x) b sin(log x) Þ x + = x x dx 2 dx Þ x

Þ x2

d2y dx

\

x2

2

d2y 2

+

xdy = -y dx

+

xdy +y=0 dx

dx 26. Here number of throws = 4 6 1 P(doublet) = p = = 36 6 30 5 P(not doublet) = q = = 36 6 Let X denotes number of successes, then 5 4 625 P(X = 0) = 4 C 0 p 0 q 4 = 1 ´ 1 ´ æç ö÷ = è 6ø 1296 P(X = 1) = 4 C 1

1 æ5 ö 3 125 500 ´ç ÷ =4´ = 6 è 6ø 1296 1296

1 2 5 2 25 150 P(X = 2) = 4 C 2 æç ö÷ ´ æç ö÷ = 6 ´ = è 6ø è 6ø 1296 1296 1 3 5 20 P(X = 3) + 4 C 3 æç ö÷ ´ = è 6ø 6 1296 1 4 1 P(X = 4) = 4 C 4 æç ö÷ = è 6ø 1296 Being a binomial distribution with 1 5 n = 4, p = and q = 6 6 1 2 m = mean = np = 4 ´ = 6 3 1 5 5 2 m = variance = npq = 4 ´ ´ = . 6 6 9 28. The region given is {( x, y):25x 2 + 9y 2 £ 225 and 5x + 3y ³ 15} Consider the equations 25x 2 + 9y 2 = 225 Þ

and

5x + 3y = 15

x2 y2 + = 1 which is an ellipse. 9 25

143

Examination Papers – 2009 y

The two curve intersect at points (0, 5) and (3, 0) obtained by equating values of y from two equations. Hence the sketch of the curves is as shown in the figure and required area is the shaded region. The required included area is 3

(0, 5)

3

= ò5 1 0

x2 15 - 5x dx - ò dx 9 3

x’

0

(3, 0) O

3

5 æx 9 x x 2 öù ÷ú = çç 9 - x 2 + sin -1 - 3x + 3 è2 2 3 2 ÷ø ûú 0 5 æ3 9 9 ö -1 3 = ç 9 - 9 + sin - 9 + - 0÷ ø 3 è2 2 3 2 5 9 p 9 15 æ p ö = æç ´ - ö÷ = ç - 1÷ square units. è ø è ø 3 2 2 2 2 2

y’

Set–III 1. Using equality of two matrices 7 y = -21 Þ y = -3 2x - 3y = 11 Þ 2x + 9 = 11 Þ x=1 \ x = 1, y = – 3 4. Let I = ò

e ax - e - ax e ax + e - ax

dx

=

ax - ax ) 1 a( e - e dx ò ax ax a e +e

=

1 log|e ax + e - ax|+ C a

é êQ ë

ò

15. Given 1 - x 2 + 1 - y 2 = a( x - y) Let x = sin A

Þ

A = sin -1 x

y = sin B

Þ

B = sin -1 y

1 - sin 2 A + 1 - sin 2 B = a(sin A - sin B) A+B A-B Þ cos A + cos B = a . 2 cos sin 2 2 A+B A-B A+B A-B Þ 2 cos . cos = 2a cos sin 2 2 2 2

\

ù dx = log| f ( x)| + C ú f ( x) û

f ¢( x)

x

144


1 A-B = tan a 2 A-B -1 1 Þ tan = a 2 1 Þ 2 tan -1 = sin -1 x - sin -1 y a Differentiating w.r.t. x, we get dy 1 1 0= 1 - x2 1 - y 2 dx Þ

Þ

1 - y2 dy 1 - y2 = = dx 1 - x2 1 - x2 a + bx c + dx p + qx

17. Let|A| = ax + b cx + d px + q u

v

w

Apply R 2 ® R 2 - xR 1 a + bx c + dx |A| = b - bx u

2

d - dx v

p + qx

2

q - qx 2 w

Taking 1 - x 2 common from R 2 a + bx c + dx p + qx 2

|A| = (1 - x )

b

d

q

u

v

w

Apply R 1 ® R 1 - xR 2 , we get a c p 2 |A| = (1 - x ) b d q = RHS u v w 18. Given differential equation is dy xy = ( x + 2)( y + 2) dx y x+2 Þ dy = dx y+2 x Integrating both sides y 2ö æ ò y + 2 dy = ò çè1 + x ÷ø dx Þ

æ

2 ö

æ

2ö

ò çè1 - y + 2 ÷ø dy = ò çè1 + x ÷ø dx

145


… (i) Þ y - 2 log|y + 2| = x + 2 log|x| + c The curve represented by (i) passes through (1, – 1). Hence -1 - 2 log 1 = 1 + 2 log|1 |+ C Þ C = -2 \ The required curve will be y - 2 log|y + 2| = x + 2 log|x| - 2 20. Let the foot of the perpendicular on the plane be A. P(1,–2,3) PA^ to the plane 2x - 3y + 4z + 9 = 0 \ DR’s of PA = 2, –3, 4 Equation of PA can be written as x-1 y+ 2 z- 3 A = = =l 2 -3 4 General points line PA = ( 2l + 1, - 3l - 2, 4l + 3) The point is on the plane hence 2( 2l + 1) - 3( -3l - 2) + 4( 4l + 3) + 9 = 0 Þ 29l + 29 = 0 or l = 1 \ Co-ordinates of foot of perpendicular are (–1, 1, –1). 24. We mark the points on the axes and get the triangle ABC as shown in the figure y 3

B(1,3)

C(3,2)

2

1

A (–1,0) –1

O

x 1 1

Required area of triangle =

2 3

3 3

ò AB + ò BC - ò AC

-1

3 Equation of line AB Þ y = ( x + 1) 2 x 7 Equation of line BC y = - + 2 2 x 1 Equation of line AC Þ y = + 2 2

1

-1

146


\

Area of DABC =

3

3ö æ3 æ x 7ö ò çè 2 x + 2 ÷ø dx + ò çè - 2 + 2 ÷ø dx -

-1

1

1

3

3

æx

1ö

ò çè 2 + 2 ÷ø dx

-1

3

æ -x 2 7 öù æ x 2 x öù 3x 2 3 ù = + xú + çç + x÷÷ ú - çç + ÷÷ ú 4 2 úû 4 2 ø úû 2 ø úû è 4 -1 è 1 -1 3 3 3 3 -9 21 1 7 ö æ 9 3 1 1 ö = æç + - + ö÷ + æç + + - ÷-ç + - + ÷ è4 2 4 2ø è 4 2 4 2ø è 4 2 4 2ø -9 + 42 + 1 - 14 æ 9 + 6 - 1 + 2 ö = 3+ -ç ÷ è ø 4 4 = 3 + 5 - 4 = 4 square units. 27. Total no. of bulbs = 30 Number of defective bulbs = 6 Number of good bulbs = 24 Number of bulbs drawn = 4 = n 6 1 p = probability of drawing defective bulb = = 30 5 4 q = probability of drawing good bulb = 5 The given probability distribution is a binomial distribution with 1 4 n = 4, p = , q = 5 5 1 r 4 4 -r Where P(r = 0, 1, 2, 3, 4 success) = 4 Cr æç ö÷ æç ö÷ è5 ø è 5 ø Hence mean = m = np 1 4 \ m=4´ = 5 5 Variance = s 2 = npq 1 4 16 \ s2 = 4 ´ ´ = 5 5 25


Maximum marks: 100

General Instructions: 1. All questions are compulsory. 2. The question paper consists of 29 questions divided into three Sections A, B and C. Section A comprises of 10 questions of one mark each, Section B comprises of 12 questions of four marks each and Section C comprises of 7 questions of six marks each. 3. All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question. 4. There is no overall choice. However, internal choice has been provided in 4 questions of four marks each and 2 questions of six marks each. You have to attempt only one of the alternatives in all such questions. 5. Use of calculator is not permitted.

Set–I SECTION–A Question numbers 1 to 10 carry 1 mark each. | x - 1| 1. What is the range of the function f ( x) = ? ( x - 1) æ 3ö 2. What is the principal value of sin -1 ç ÷? è 2 ø æ cos a - sin a ö 3. If A = ç ÷ , then for what value of a is A an identity matrix? cos a ø è sin a 0 2 0 4. What is the value of the determinant 2

3 4 ?

4 5 5. Evaluate : ò

log x

6

dx. x 6. What is the degree of the following differential equation? 2

d2y æ dy ö 5x ç ÷ - 6y = log x è dx ø dx 2

148


7. Write a vector of magnitude 15 units in the direction of vector i$ - 2j$ + 2k$. 8. Write the vector equation of the following line: x-5 y+ 4 6-z = = 3 7 2 1 2 3 1 7 11 æ öæ ö æ ö 9. If ç ÷ç ÷=ç ÷ , then write the value of k. è 3 4ø è 2 5ø è k 23ø 10. What is the cosine of the angle which the vector 2i$ + j$ + k$ makes with y-axis?

SECTION–B 11. On a multiple choice examination with three possible answers (out of which only one is correct) for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing? 12. Find the position vector of a point R which divides the line joining two points P and Q whose ®

®

®

®

position vectors are ( 2 a + b ) and ( a - 3 b ) respectively, externally in the ratio 1 : 2. Also, show that P is the mid-point of the line segment RQ. 13. Find the Cartesian equation of the plane passing through the points A ( 0, 0, 0) and x- 4 y + 3 z +1 B( 3, - 1, 2) and parallel to the line = = . 1 -4 7 14. Using elementary row operations, find the inverse of the following matrix : æ2 5 ö ç ÷ è1 3ø 15. Let Z be the set of all integers and R be the relation on Z defined as R = {( a, b) ; a, b Î Z , and ( a - b) is divisible by 5.} Prove that R is an equivalence relation. 16. Prove the following: æ1 - x ö 1 tan -1 x = cos -1 ç ÷ , x Î ( 0, 1) 2 è1 + x ø OR Prove the following : 12 3 56 cos -1 æç ö÷ + sin -1 æç ö÷ = sin -1 æç ö÷ è 13 ø è5ø è 65 ø 17. Show that the function f defined as follows, is continuous at x = 2, but not differentiable: ì 3x - 2, 0 < x £ 1 ï f ( x) = í 2x 2 - x, 1 < x £ 2 ï 5x - 4, x>2 î OR dy -1 Find , if y = sin [x 1 - x - x 1 - x 2 ]. dx

149


æ sin 4x - 4 ö 18. Evaluate : ò e x ç ÷ dx. è 1 - cos 4x ø OR Evaluate : ò 19. Evaluate : ò

1-x

2

dx. x (1 - 2x) p / 3 sin x + cos x

p/ 6

sin 2x

dx.

20. Find the points on the curve y = x 3 at which the slope of the tangent is equal to the y-coordinate of the point. 21. Find the general solution of the differential equation dy 2 x log x . + y = × log x dx x OR Find the particular solution of the differential equation satisfying the given conditions: dy = y tan x, given that y = 1 when x = 0. dx 22. Find the particular solution of the differential equation satisfying the given conditions: x 2 dy + ( xy + y 2 ) dx = 0 ; y = 1 when x = 1.

SECTION–C Question numbers 23 to 29 carry 6 marks each. 23. A small firm manufactures gold rings and chains. The total number of rings and chains manufactured per day is atmost 24. It takes 1 hour to make a ring and 30 minutes to make a chain. The maximum number of hours available per day is 16. If the profit on a ring is Rs. 300 and that on a chain is Rs 190, find the number of rings and chains that should be manufactured per day, so as to earn the maximum profit. Make it as an L.P.P. and solve it graphically. 24. A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn at random and are found to be both clubs. Find the probability of the lost card being of clubs. OR From a lot of 10 bulbs, which includes 3 defectives, a sample of 2 bulbs is drawn at random. Find the probability distribution of the number of defective bulbs. 25. The points A ( 4, 5, 10), B ( 2, 3, 4) and C (1, 2, - 1) are three vertices of a parallelogram ABCD. Find the vector equations of the sides AB and BC and also find the coordinates of point D. 26. Using integration, find the area of the region bounded by the curve x 2 = 4y and the line x = 4y - 2.

150


OR p

x tan x

0

sec x + tan x

Evaluate: ò

dx.

27. Show that the right circular cylinder, open at the top, and of given surface area and maximum volume is such that its height is equal to the radius of the base. 28. Find the values of x for which f ( x) = [x ( x - 2)] 2 is an increasing function. Also, find the points on the curve, where the tangent is parallel to x-axis. 29. Using properties of determinants, show the following: (b + c) 2 ab

ab ( a + c) 2

ca bc

ac

bc

( a + b) 2

= 2abc ( a + b + c) 3

Set-II Only those questions, not included in Set I, are given. æ 3ö 3. What is the principal value of cos -1 ç ÷? è 2 ø 7. Find the minor of the element of second row and third column ( a 23 ) in the following determinant: 2 -3 5 6 0 4 1

5

-7

11. Find all points of discontinuity of f, where f is defined as follows : x£- 3 ì| x| + 3, ï f ( x) = í -2x , -3 < x < 3 ï 6x + 2 , x³ 3 î OR dy Find , if y = (cos x) x + (sin x) 1/ x . dx 12. Prove the following: æ1 - x ö 1 tan -1 x = cos -1 ç ÷ , x Î( 0, 1) 2 è1 + x ø OR Prove the following: 12 cos -1 æç ö÷ + sin -1 è 13 ø

æ 3 ö = sin -1 ç ÷ è5 ø

æ 56 ö ç ÷ è 65 ø

151


14. Let * be a binary operation on Q defined by 3ab a *b = 5 Show that * is commutative as well as associative. Also find its identity element, if it exists. p x 18. Evaluate: ò dx. 0 1 + sin x 20. Find the equations of the normals to the curve y = x 3 + 2x + 6 which are parallel to the line x + 14y + 4 = 0. 3

23. Evaluate ò ( 3x 2 + 2x) dx as limit of sums. 1

OR Using integration, find the area of the following region: ìï x2 y2 x y üï + £1£ + ý í( x, y) ; 9 4 3 2 ïþ ïî 29. Write the vector equations of the following lines and hence determine the distance between them: x-1 y- 2 z+ 4 x- 3 y- 3 z+5 = = ; = = 2 3 6 4 6 12

Set-III Only those questions, not included in Set I and Set II, are given. 1 1 1. Find the principal value of sin -1 æç - ö÷ + cos -1 æç - ö÷ . è 2ø è 2ø 9. If A is a square matrix of order 3 and| 3A| = K| A|, then write the value of K. 11. There are two Bags, Bag I and Bag II. Bag I contains 4 white and 3 red balls while another Bag II contains 3 white and 7 red balls. One ball is drawn at random from one of the bags and it is found to be white. Find the probability that it was drawn from Bag I. 14. Prove that : tan -1 (1) + tan -1 ( 2) + tan -1 ( 3) = p.

If tan -1

æx - 1ö ç ÷ + tan -1 x 2 è ø

OR æx + 1ö p ç ÷ = , find the value of x. è x + 2ø 4

17. Show that the relation S in the set R of real numbers, defined as S = {( a, b) : a, b Î R and a £ b 3 } is neither reflexive, nor symmetric nor transitive. x-7 19. Find the equation of tangent to the curve y = , at the point, where it cuts the ( x - 2) ( x - 3) x-axis. 23. Find the intervals in which the function f given by f ( x) = sin x - cos x, 0 £ x £ 2p is strictly increasing or strictly decreasing.

152


24. Evaluate ò ( x 2 - x) dx as limit of sums. 1

OR Using integration find the area of the following region : {( x, y) :| x - 1| £ y £ 5 - x 2 }

SOLUTIONS Set–I SECTION–A 1. We have given | x - 1| f ( x) = ( x - 1) ì ( x - 1), if x - 1 > 0 or x > 1 | x - 1| = í î -( x - 1), if x - 1 < 0 or x < 1 ( x - 1) (i) For x > 1, f ( x) = =1 ( x - 1) - ( x - 1) (ii) For x < 1, f ( x) = = -1 ( x - 1) | x - 1| is {-1, 1}. \ Range of f ( x) = ( x - 1) æ 3ö 2. Let x = sin -1 ç ÷ è 2 ø Þ Þ

3 2 p x=3

sin x = -

Þ

p sin x = sin æç - ö÷ è 3ø

é êQ ë

3 pù = sin ú 2 3û

æ p 3ö The principal value of sin -1 ç ÷ is - × 3 è 2 ø 3. We have given é cos a - sin a ù A=ê cos a úû ë sin a For the identity matrix, the value of A11 and A12 should be 1 and value of A12 and A21 should be 0. i.e., cos a = 1 and sin a = 0 As we know cos 0° = 1 and sin 0° = 0 Þ a = 0°

153


0 2 0 4.

2

3 4 =0

4 5

6

3 5

=-2

4 2 -2 6 4 2 4

4 2 +0 6 4

3 (expanding the given determinant by R 1 ) 5

4 6

= - 2 (12 - 16) = 8 The value of determinant is 8. 5. We have given log x ò x dx 1 Let log x = t Þ dx = dt x Given integral = ò t dt =

(log x) t2 + c= 2 2

2

+c

2

d2y æ dy ö 6. 5x ç ÷ - 6y = log x è dx ø dx 2 Degree of differential equation is the highest power of the highest derivative. In above the highest order of derivative. Its degree = 1. \ 7. Let

®

A = i$ - 2j$ + 2k$ i$ - 2j$ + 2k$

®

Unit vector in the direction of A is A$ =

(1) 2 + ( -2) 2 + ( 2) 2

=

1 $ (i - 2j$ + 2k$) 3

® (i$ - 2j$ + 2k$) \ Vector of magnitude 15 units in the direction of A = 15A$ = 15 3 15 $ 30 $ 30 $ = ij+ k 3 3 3 = 5i$ - 10j$ + 10k$

8. We have given line as x-5

=

y+4

=

z-6

3 7 -2 By comparing with equation x - x 1 y - y 1 z - z1 , = = a b c

d2y dx 2

is

154


We get given line passes through the point ( x 1 , x 2 , x 3 ) i.e., (5, - 4, 6) and direction ratios are ( a, b , c) i.e., (3, 7, –2). Now, we can write vector equation of line as ®

A = (5i$ - 4j$ + 6k$) + l ( 3i$ + 7 j$ - 2k$)

é 1 2ù é 3 1ù é7 11 ù 9. ê úê ú=ê ú ë 3 4û ë 2 5û ë k 23û é 1 2ù é 3 1ù LHS =ê úê ú ë 3 4û ë 2 5û é (1) ( 3) + ( 2) ( 2) =ê ë( 3) ( 3) + ( 4) ( 2)

(1) (1) + ( 2) (5) ù é 7 11 ù = ( 3) (1) + ( 4) (5) úû êë17 23úû

Now comparing LHS to RHS, we get \ k = 17 10. We will consider ®

a = 2i$ + j$ + k$ 2i$ + j$ + k$

®

Unit vector in the direction of a is a$ =

= =

( 2 ) 2 + (1) 2 + (1) 2 2i$ + j$ + k$ 4

=

2i$ + j$ + k$ 2

2 $ 1 $ 1 $ 1 $ 1 $ 1 $ i + j + k= i+ j+ k 2 2 2 2 2 2

1 The cosine of the angle which the vector 2i$ + j$ + k$ makes with y-axis is æç ö÷ . è 2ø

SECTION–B 11. No. of questions = n = 5 Option given in each question = 3 p = probability of answering correct by guessing =

1 3

q = probability of answering wrong by guessing = 1 - p = 1 This problem can be solved by binomial distribution. 2 n - r æ 1 ör P(r) = nCr æç ö÷ ç ÷ è 3ø è 3ø where r = four or more correct answers = 4 or 5 2 1 4 (i) P( 4) = 5 C 4 æç ö÷ æç ö÷ è 3ø è 3ø

1 5 (ii) P (5) = 5 C5 æç ö÷ è 3ø

1 2 = 3 3

155


\ P = P ( 4) + P(5) 2 1 4 1 5 = 5 C 4 æç ö÷ æç ö÷ + 5 C5 æç ö÷ è 3ø è 3ø è 3ø 1 4 = æç ö÷ è 3ø

1 é 10 + 1 ù = êë 3 3 úû 3 ´ 3 ´ 3 ´ 3

é 11 ù = 11 = 0 × 045 êë 3 úû 243

12. The position vector of the point R dividing the join of P and Q externally in the ratio 1 : 2 is ®

®

®

OR =

1-2 ®

®

=

®

®

1 ( a - 3 b ) - 2 (2 a + b ) ®

®

®

a -3b -4 a -2b

-1 Mid-point of the line segment RQ is ®

®

®

®

(3 a + 5 b ) +( a - 3 b )

=

®

-3 a - 5 b

®

-1

®

®

= 3 a +5 b

®

=2 a + b

2 As it is same as position vector of point P, so P is the mid-point of the line segment RQ. 13. Equation of plane is given by a ( x - x 1 ) + b ( y - y 1 ) + c (z - z 1 ) = 0 Given plane passes through (0, 0, 0) …(i) \ a ( x - 0) + b ( y - 0) + c (z - 0) = 0 Plane (i) passes through (3, –1, 2) …(ii) \ 3a - b + 2c = 0 Also plane (i) is parallel to the line x- 4 y+ 3 z+1 = = 1 -4 7 …(iii) a - 4b + 7 c = 0 Eliminating a, b , c from equations (i), (ii) and (iii), we get x y z 3 -1 2 = 0 1 Þ x

-1 -4

2 3 -y 7 1

-4

2 3 +z 7 1

Þ Þ

7 -1 =0 -4

x ( -7 + 8) - y ( 21 - 2) + z ( -12 + 1) = 0 x - 19y - 11z = 0 , which is the required equation 2 5 é ù 14. Given, A=ê ú 1 3 ë û We can write,

A = IA

156


i.e.,

é 2 5ù é 1 0ù ê 1 3ú = ê 0 1ú A ë û ë û é1 2ù é 1 -1ù A ê1 3ú = ê 0 1úû ë û ë

[R 1 ® R 1 - R 2 ]

é 1 2ù é 1 ê 0 1ú = ê -1 ë û ë

-1ù A 2úû

[R 2 ® R 2 - R 1 ]

é 1 0ù é 3 ê 0 1 ú = ê -1 ë û ë é 3 A -1 = ê ë -1

-5 ù A 2úû

[R 1 ® R 1 - 2R 2 ]

-5 ù 2úû

15. We have provided R = {( a, b) : a, b Î Z , and ( a - b) is divisible by 5} (i) As (a - a) = 0 is divisible by 5. \ ( a, a) Î R " a Î R Hence, R is reflexive. (ii) Let ( a, b) Î R Þ ( a - b) is divisible by 5. Þ - (b - a) is divisible by 5. Þ (b - a) is divisible by 5. \ (b , a) Î R Hence, R is symmetric. (iii) Let ( a, b) Î R and (b , c) Î Z Then, ( a - b) is divisible by 5 and (b - c) is divisible by 5. ( a - b) + (b - c) is divisible by 5. ( a - c) is divisible by 5. \ ( a, c) Î R Þ R is transitive. Hence, R is an equivalence relation. 16. We have to prove æ1 - x ö 1 tan -1 x = cos -1 ç ÷ , x Î( 0, 1) 2 è1 + x ø 1 L.H.S. = tan -1 x = [2 tan -1 x ] 2 (1) 2 - ( x ) 2 ù 1é = ê cos -1 ú 2ê (1) 2 + ( x ) 2 úû ë æ1 - x ö 1 = cos -1 ç ÷ = R.H.S. Hence Proved. 2 è1 + x ø

157


OR 12 ö 56 æ -1 æ 3 ö cos ç ÷ + sin ç ÷ = sin -1 æç ö÷ è 13 ø è5ø è 65 ø 12 3 LHS = cos -1 æç ö÷ + sin -1 æç ö÷ è 13 ø è5ø -1

5 = sin -1 æç ö÷ + sin -1 è 13 ø

é -1 êëQ cos 2 2ù é5 ê ´ 1 - æç 3 ö÷ + 3 ´ 1 - æç 5 ö÷ ú è5ø è 13 ø ú 5 ê 13 ë û 5 4 3 12 56 é ´ + ´ ù = sin -1 = RHS êë 13 5 5 13 úû 65

= sin -1 = sin -1 LHS = RHS

æ 12 ö = sin -1 ç ÷ è 13 ø

æ 5 öù ç ÷ú è 13 ø û

Hence Proved ì 3x - 2, 0 < x £ 1 ï f ( x) = í 2x 2 - x, 1 < x £ 2 ï 5x - 4, x>2 î

17. We have given, At x = 2, (i) = lim

æ 3ö ç ÷ è5ø

x ® 2+

RHL f ( x)

LHL = lim f ( x) x ® 2-

= lim f ( 2 + h)

= lim f ( x)

= lim {5 ( 2 + h) - 4}

= lim {2 ( 2 - h) 2 - ( 2 - h)}

= 10 – 4 = 6

= lim {( 2 - h) ( 4 - 2h - 1)} = 2 × 3 = 6

x ® 2-

h®0 h®0

Also,

h®0

h®0

2

f ( 2) = 2( 2) - 2 = 8 - 2 = 6

LHL = RHL = f ( 2) Q \ f ( x) is continuous at x = 2 (ii) LHD f ( 2 - h) - f ( 2) = lim h®0 -h

RHD = lim

2

= lim

[2 ( 2 - h) - ( 2 - h)] - ( 8 - 2) -h

h®0

= lim

[8 + 2h - 8h - 2 + h) - 6 -h

h®0

h [5 ( 2 + h) - 4] - ( 8 - 2) h

h®0

2

= lim

f ( 2 + h) - f ( 2)

h®0

= lim

h®0

5h h

2

= lim

h®0

2h - 7 h -h

= lim ( -2h + 7) = 7 h®0

= lim (5) h®0

=5

158


LHD ¹ RHD Q \ f ( x) is not differentiable at x = 2 OR We have given y = sin -1 [x 1 - x - x 1 - x 2 ]. = sin -1 [x 1 - ( x ) 2 - x 1 - x 2 ] y = sin -1 x - sin -1

Þ

x

[using sin -1 x - sin -1 y = sin -1 [x 1 - y 2 - y 1 - x 2 ] Differentiating w.r.t. x, we get dy 1 1 d = ( x) dx 1 - x2 1 - ( x ) 2 dx 1

=

1-x 18.

òe

x

-

2

1 1 1 1 . = 2 1-x 2 x 2 x ( 1 - x) 1-x

æ sin 4x - 4 ö ç ÷ dx è 1 - cos 4x ø æ 2 sin 2x cos 2x - 4 ö ÷ dx = ò ex ç ç ÷ 2 sin 2 2x è ø

[sin 4x = 2 sin 2x cos 2x and 1 - cos 4x = 2 sin 2 2x]

= ò e x (cot 2x - 2 cosec 2 2x) dx = ò cot 2x . e x dx - 2 ò e x cosec 2 2x dx = [cot 2x . e x - ò ( -2 cosec 2 2x) . e x dx] - 2 ò e x cosec 2 2x dx = cot 2x . e x + 2 ò cosec 2 2x . e x dx - 2 ò cosec 2 2x . e x dx = e x cot 2x + c OR We have given 1 - x2

1 - x2

ò x (1 - 2x) dx = ò x - 2x 2 =ò

x2 - 1 2x 2 - x

dx

dx = ò

1 2

æ 2x 2 - 2 ö ç ÷ dx ç 2x 2 - x ÷ è ø

( 2x 2 - x) + ( x - 2)

=

1 2

ò

=

1 2

ò ççè1 + 2x 2 - x ÷÷ø dx

2x 2 - x æ

x-2 ö

dx

…(i)

159


By partial fraction x-2 x-2 A B = = + 2 x ( 2 x 1 ) x 2 x -1 2x - x x - 2 = A ( 2x - 1) + Bx Equating co-efficient of x and constant term, we get 2A + B = 1 and - A = -2 Þ A = 2 , B = -3 x-2 2 3 \ = + 2 2x - x x 1 - 2x

…(ii)

From equation (i) 1 - x2

1 æ2

1

3

ö

ò x (1 - 2x) dx = 2 ò 1 dx + 2 ò çè x + 1 - 2x ÷ø dx 1 3 x + log| x| - log|1 - 2x| + c 2 4 19. Given integral can be written as sin x + cos x sin x + cos x p/3 p/3 I=ò dx = ò dx p/ 6 p / 6 1 - (1 - sin 2x) 1 - ( sin x - cos x) 2 =

Put sin x - cos x = t (cos x + sin x) =

so that,

p p p , t = sin - cos = 6 6 6 p p p when x = , t = sin - cos = 3 3 3 when x =

Þ

3 1 2 2 1 3 2 2

I=ò

é = sin -1 ê ë é = sin -1 ê ë

dt 1 - t2

[

dt dx

1 3 2 2 3 1 2 2

= sin

-1

]

t

3 1 2 2 1 3 2 2

é1 3 1ù 3ù - ú - sin -1 ê ú 2 2û 2 û ë2 é 3 1ù 1 3 1ù - ú + sin -1 ê - ú = 2 sin -1 ( 3 - 1) 2 2û 2û 2 ë 2

20. Let P ( x 1 , y 1 ) be the required point. The given curve is y = x3

dy = 3x 2 dx æ dy ö ç ÷ è dx ø x

1 , y1

= 3x 12

…(i)

160


Q the slope of the tangent at ( x 1, y 1 ) = y 1 3x 12 = y 1

…(ii) y1 =

Also, ( x 1 , y 1 ) lies on (i) so

x 13

…(iii)

From (ii) and (iii), we have 3x 12 = x 13 Þ

x1 = 0

x 12 ( 3 - x 1 ) = 0

Þ

x1 = 3

or

When x 1 = 0, y 1 = ( 0)

3

=0

When x 1 = 3, y 1 = ( 3) 3 = 27 \ the required points are (0, 0) and (3, 27). dy 2 21. x log x + y = log x dx x dy 1 2 Þ + y= dx x log x x2

...(i)

This is a linear differential equation of the form dy + Py = Q dx 1 2 where P = and Q = x log x x2 1

ò Pdx I.F. = e ò \ = e x log 1 [Let log x = t \ dx = dt] x 1

=e

òt

dt

2

x

dx

= e log t = t = log x

\

y log x = ò

Þ

y log x = 2 ò log x . x -2 dx + c

x2

log x dx + C

I

Þ Þ Þ Þ

[ \ solution is y (I. F.) = ò Q (I. F.) dx + C]

II

é é x -1 ù 1 é x -1 ù ù y log x = 2 ê log x ê ú-ò ê ú dxú + C x êë -1 úû úû êë êë -1 úû é log x ù y log x = 2 ê + ò x -2 dxú + C x ë û é log x 1 ù y log x = 2 ê - ú +C x xû ë 2 y log x = - (1 + log x + C), which is the required solution x

161


OR dy = y tan x dx

Þ

dy = tan x dx y

By integrating both sides, we get dy ò y = ò tan x . dx log y = log|sec x| + C By putting x = 0 and y = 1 (as given), we get log 1 = log (sec 0) + C C=0 \ (i) Þ log y = log|sec x| Þ y = sec x

...(i)

22. x 2 dy + y ( x + y) dx = 0 x 2 dy = - y ( x + y) dx ( x + y) dy =-y dx x2 æ xy + y 2 ö dy ÷ =-ç ç x2 ÷ dx è ø

…(i)

dy dv in equation (i) =v+ x dx dx æ vx 2 + v 2 x 2 ö dv dv ÷ Þ v+x =-ç v+x = - (v + v 2 ) ç ÷ 2 dx dx x è ø

Putting y = vx and

x dv

Þ Þ

= - 2v - v 2

dx dv 2

=-

v + 2v 1

dx x

(by separating variable) 1

Þ

ò v 2 + 2v dv = - ò x dx

Þ

ò v 2 + 2v + 1 - 1 dv = - ò x dx

Þ

ò (v + 1) 2 - 1 2

1

1

(Integrating both sides) 1

dv = - ò

1 dx x

Þ

v+1-1 1 log = - log x + log C 2 v+1+1

Þ

1 v log = - log x + log C 2 v+2

162


Þ

log

v + 2 log x = 2 log C v+2

Þ

log

v + log x 2 = log k , v+2

Þ

log

vx 2 = log k v+2

Þ

y 2 ×x x =k y +2 x

Þ

x 2 y = k ( y + 2x)

Þ

where k = C 2 vx 2 =k v+2

éQ y = vù úû ëê x …(ii)

It is given that y = 1 and x = 1, putting in (ii), we get 1 1 = 3k Þ k = 3 1 Putting k = in (ii), we get 3 1 x 2 y = æç ö÷ ( y + 2x) è 3ø Þ

3x 2 y = ( y + 2x)

SECTION–C

2 (0,

23. Total no. of rings & chain manufactured per day = 24. Time taken in manufacturing ring = 1 hour Time taken in manufacturing chain = 30 minutes Y One time available per day = 16 32 Maximum profit on ring = Rs 300 28 Maximum profit on chain = Rs 190 Let gold rings manufactured per day = x 24 Chains manufactured per day = y 20 L.P.P. is maximize Z = 300x + 190y 16 Subject to x ³ 0, y ³ 0 12 x + y £ 24 1 x + y £ 16 8 2 Possible points for maximum Z are 4 (16, 0), (8, 16) and (0, 24). 0 4 At (16, 0), Z = 4800 + 0 = 4800 4)

(8,16) x + 1/2y=16

x+ y=

24

(16,0) X

8

12

16

20

24

163


At (8, 16), Z = 2400 + 3040 = 5440 ¬ Maximum At (0, 24), Z = 0 + 4560 = 4560 Z is maximum at (8, 16). \ 8 gold rings & 16 chains must be manufactured per day. 24. Let A1 , E1 and E2 be the events defined as follows: A : cards drawn are both club E1 : lost card is club E2 : lost card is not a club 13 1 39 3 Then, P(E1 ) = = , P(E2 ) = = 52 4 52 4 12 11 ´ 51 50 13 12 P( A / E2 ) = Probability of drawing both club cards when lost card is not club = ´ 51 50 To find : P(E1 / A) By Baye’s Theorem, P(E1 ) P( A / E1 ) P(E1 / A) = P(E1 ) P( A / E1 ) + P(E2 ) P( A / E2 ) P( A / E1 ) = Probability of drawing both club cards when lost card is club =

1 12 11 ´ ´ 12 ´ 11 11 11 4 51 50 = = = = 1 12 11 3 13 12 12 ´ 11 + 3 ´ 13 ´ 12 11 + 39 50 ´ ´ + ´ ´ 4 51 50 4 51 50 OR There are 3 defective bulbs & 7 non-defective bulbs. Let X denote the random variable of “the no. of defective bulb.” Then X can take values 0, 1, 2 since bulbs are replaced 3 3 7 and q = P (D ) = 1 p = P(D) = = 10 10 10 We have 7 C 2 ´ 3C 0 7´6 7 P (X = 0) = = = 10 10 ´ 9 15 C2 7

P (X = 1) =

10 7

P (X = 2) =

C 1 ´ 3C 1

=

7 ´ 3´2

C2

C 0 ´ 3C 2 10

C2

=

10 ´ 9 1´ 3´2 10 ´ 9

=

7 15

=

1 15

\ Required probability distribution is X

0

1

2

P ( x)

7/15

7/15

1/15

164


25. The points A ( 4, 5, 10), B ( 2, 3, 4) and C (1, 2, - 1) are three vertices of parallelogram ABCD. Let coordinates of D be ( x, y, z) Direction vector along AB is ® a = ( 2 - 4) i$ + ( 3 - 5) j$ + ( 4 - 10) k$ = - 2i$ - 2j$ - 6k$ \ Equation of line AB, is given by ®

b = ( 4i$ + 5j$ + 10k$) + l ( 2i$ + 2j$ + 6k$)

Direction vector along BC is ®

c = (1 - 2) i$ + ( 2 - 3) j$ + ( -1 - 4) k$ = - i$ - j$ - 5k$

\ Equation of a line BC, is given by . ®

d = ( 2i$ + 3j$ + 4k$) + m (i$ + j$ + 5k$)

Since ABCD is a parallelogram AC and BD bisect each other é 4 + 1 5 + 2 10 - 1 ù é 2 + x 3 + y 4 + z ù \ ê 2 , 2 , 2 ú=ê 2 , 2 , 2 ú ë û ë û Þ 2 + x = 5, 3 + y = 7, Þ x = 3, y = 4, z = 5 Co-ordinates of D are (3, 4, 5). 26. Given curve x 2 = 4y

4+z= 9

…(i)

Line equation …(ii) x = 4y - 2 Equation (i) represents a parabola with vertex at the origin and axis along (+)ve direction of y-axis. Equation (ii) represents a straight line which meets the coordinates axes at 1 (–2, 0) and æç 0, ö÷ respectively. è 2ø

Y

x=

2

x = 4y

x, P(

2

y 2)

Q(x, y1)

2

x=x -2

(-1, 1 ) 4

x 2 - x - 2 = 0 (by eliminating y)

Þ ( x - 2) ( x + 1) = 0 Þ x = - 1, 2 The point of intersection of given 1 parabola & line are (2, 1) and æç - 1, ö÷ . è 4ø

–

(2,1)

By solving two equations, we obtain Þ

4y

X' (-2,0)

(2,6)

(-1,0)

Y'

X

165


\ required area = ò

2

-1

( y 2 - y 1 ) dx.

...(iii)

Q P ( x, y 2 ) and Q ( x, y 1 ) lies on (ii) and (i) respectively x+2 x2 \ y2 = and y 1 = 4 4 2æ x + 2 x2 ö ÷ dx \ (iii) Þ Area = ò çç 4 4 ÷ø -1è 2

2

-1

-1

x 1 = ò dx + 4 2

4 2 8 1 =é + - ù-é êë 8 2 12 úû êë 8

2

2

é x2 1 x3 ù x dx = + x ê ú ò 2 12 úû êë 8 -1 -1 1 1ù 9 - + = sq. units. 2 12 úû 8

1 ò dx - 4

2

OR I=ò

x tan x

p

0

I=ò

p

0

I=ò

p

0

I=ò

p

0 p

\

2I = ò

Þ

2I = p ò

0

dx sec x + tan x sin x x cos x p dx = ò 0 sin x 1 + cos x cos x ( p - x) sin ( p - x) dx 1 + sin ( p - x) ( p - x) sin x dx 1 + sin x p sin x dx 1 + sin x sin x (1 - sin x) p

0

sin x - sin 2 x

0

1 - sin 2 x

=pò

p

0 p

1 + sin x éQ êë

a

ò0

...(i)

dx

a

f ( x) dx = ò f ( a - x) dxù úû 0 ...(ii)

[Using (i) and (ii)]

(1 + sin x) (1 - sin x)

p

=pò

x sin x

dx

sin 2 x ù p é sin x dx = p ò ê ú dx 2 2 0 ê ë cos x cos x úû

tan x sec x dx - p ò

p

0 p

tan 2 x dx

=pò

tan x sec x dx - p ò (sec 2 x - 1) dx

=pò

sec x tan x dx - p ò

0 p

0

0 p

0

p

p

sec 2 x dx + p ò

p

0

dx

= p [ sec x]0 - p [ tan x]0 + p [x]p0 + C = p [-1 - 1] - 0 + p [p - 0] = p ( p - 2) Þ

I=

p ( p - 2) 2

166


27. Let r be the radius and h be the height of the cylinder of given surface s. Then, s = pr 2 + 2phr h= Then

s - pr 2

…(i)

2 pr

é s - pr 2 ù v = pr 2 h = pr 2 ê ú êë 2pr úû v=

[From eqn. (i)]

sr - pr 3 2

2 dv s - 3pr = dr 2 For maximum or minimum value, we have dv =0 dr

Þ Þ

s - 3 pr 2 2

=0

…(ii)

Þ s = 3 pr 2

pr 2 + 2prh = 3pr 2

Þ r=h Differentiating equation (ii) w.r.t. r, we get d 2v

= - 3 pr < 0 dr 2 Hence, when r = h, i.e., when the height of the cylinder is equal to the radius of its base v is maximum. 28. We have given y = [x ( x - 2)] 2 = x 2 ( x 2 - 4x + 4) = x 4 - 4x 3 + 4x 2 dy = 4x 3 - 12x 2 + 8x dx For the increasing function, dy >0 dx Þ Þ

4x 3 - 12x 2 + 8x > 0 Þ 4x ( x 2 - 3x + 2) > 0

4x ( x - 1) ( x - 2) > 0 dy For 0 < x < 1, = ( +) ( -) ( -) = ( +) ve dx dy For x > 2, = ( +) ( +) ( +) = ( +) ve dx

…(i)

167


The function is increasing for 0 < x < 1 and x > 2 dy If tangent is parallel to x-axis, then =0 dx Þ 4x ( x - 1) ( x - 2) = 0 Þ x = 0, 1, 2 For x = 0, f ( 0) = 0 For x = 1, f (1) = [1 (1 - 2)] 2 = 1 For x = 2, f ( 2) = [2 ´ 0] 2 = 0 \ Required points are (0, 0), (1, 1), (2, 0). (b + c) 2 29. To prove: ab

ab ( a + c) 2

ca bc

bc

( a + b) 2

ac

Let

(b + c) 2 D= ab

= 2abc ( a + b + c) 3

ab ( a + c) 2

ca bc

bc

( a + b) 2

ac

[Multiplying R 1 , R 2 and R 3 by a, b , c respectively] 1 D= abc

a (b + c) 2 ab

2

ba 2 b ( a + c)

ac 2

1 = abc abc

a2 c 2

b2c ( a + b) 2 c

bc 2

(b + c) 2

a2

a2

b2

( c + a) 2

b2

c

2

c

2

( a + b) 2

[C 1 ® C 1 - C 3 and C 2 ® C 2 - C 3 ] (b + c) 2 - a 2 = 0

0 ( c + a) 2 - b 2

a2 b2

c 2 - ( a + b) 2

c 2 - ( a + b) 2

( a + b) 2

(b + c + a) (b + c - a) = 0

0 ( c + a + b)(b + c - a)

a2 b2

( c + a + b)( c - a - b)

( c + a + b)( c - a - b)

( a + b) 2

= ( a + b + c)

2

b + c -a 0

0 c + a -b

a2 b2

c - a-b

c - a-b

( a + b) 2

168


= ( a + b + c) 2

=

=

=

b+c-a 0

0 c + a-b

a2 b2

-2b

-2 a

2ab

( a + b + c) 2 ab

ab + ac - a 2 0

0 bc + ba - b 2

a2 b2

-2ab

- 2ab

2ab

ab + ac

( a + b + c) 2

b

ab

ab

2

a2 bc + ba

a2 b2

0

2ab

0

( a + b + c) 2

( R 3 ® R 3 - ( R 1 + R 2 ))

b+c

a

a

b

c+a

b

0

0

1

. ab . 2ab

= 2ab ( a + b + c) 2

[C 1 ® C 1 + C 3 , C 2 ® C 2 + C 3 ]

b+c

a

b

c+a

= 2ab ( a + b + c) 2 {(b + c) ( c + a) - ab} = 2abc ( a + b + c) 3 = RHS

Set–II 3. Let

æ 3ö x = cos -1 ç ÷ è 2 ø

3 2 5p p [as cos p / 6 = Þ cos x = cos é p - ù = cos êë ú 6û 6 5p Þ x= 6 æ 3 ö 5p The principal value of cos -1 ç . ÷ is 6 è 2 ø Þ

cos x = -

7. We have given 2 -3

5

6

0

4

1

5

-7

Minor of an element 2 a 23 = M 23 = 1

-3 = 10 + 3 = 13 5

3 / 2]

169


11. We have given ì| x| + 3, ï f ( x) = í -2x , ï 6x + 2 , î

x£- 3 -3 < x < 3 x³ 3

(i) For x = - 3 LHL

=

RHL

= lim

lim

x ® - 3-

x ® 3+

f ( x) = lim f ( 3 - h) = lim - 2 ( 3 - h) = - 6 h®0

h®0

f ( x) = lim f ( 3 + h) = lim 6 ( 3 + h) + 2 = 20 h®0

h®0

LHL ¹ RHL At x = 3, function is not continuous. OR Given,

y = (cos x) x + (sin x) 1/ x = e x log (cos x ) + e 1/ x log (sin x )

By differentiating w.r.t. x dy = e x log (cos x ) dx

ì ü x - (sin x)ý í log ( cos x) + cos x î þ 1 log (sin x ) x e

é cos x ù 1 ê - log (sin x) 2 + ú x sin x û x ë cot x ü ì 1 log sin x + = (cos x) x {log (cos x) - x tan x} + (sin x) 1/ x í ý 2 x þ î x +

14. For commutativity, condition that should be fulfilled is a *b = b * a 3ab 3ba Consider a * b = = =b *a 5 5 \ a *b = b * a Hence, * is commutative. For associativity, condition is ( a * b) * c = a * (b * c) 9abc 3 æ 3 ö 3 æ 3ab ö Consider ( a * b) * c = ç = a ç bc ÷ = a(b * c) = a * (b * c) ÷ *c= è 5 ø 25 5 è5 ø 5 Hence, ( a * b) * c = a * (b * c) \ * is associative. Let e Î Q be the identity element, Then a*e = e *a= a 3ae 3ea 5 Þ = =a Þ e= × 5 5 3

170


18. I = ò

p

0

x dx 1 + sin x

p

p-x

0

1 + sin ( p - x)

I=ò

p

p-x

0

1 + sin x

=ò

…(i) a é a ù êQ ò f ( x) dx = ò f ( a - x) dxú êë 0 úû 0

dx

…(ii)

dx

Adding equations (i) and (ii), we get p p 2I = ò dx 0 1 + sin x 1 - sin x p p 1 - sin x =pò dx = p ò dx 0 (1 + sin x) (1 - sin x) 0 cos 2 x p

= p ò (sec 2 x - sec x tan x) dx 0

p

= p [ tan x - sec x]0 = p[( 0 + 1) - ( 0 - 1)] = 2p Þ 2I = 2p or I = p 20. Given equation of curve y = x 3 + 2x + 6 Equation of line x + 14y + 4 = 0 Differentiating (i) w.r.t. x, we get dy dx 1 = 3x 2 + 2 Þ = 2 dx dy 3x + 2 -1 \ Slope of normal = . 3x 2 + 2

...(i) ...(ii)

and it is parallel to equation of line. -1 -1 \ = 2 3x + 2 14 Þ

3x 2 + 2 = 14 Þ 3x 2 = 12 x2 = 4

Þ x=±2

From equation of curve, if x = 2, y = 18 ; if x = - 2, y = - 6 \ Equation of normal at (2, 18) is 1 y - 18 = ( x - 2) or x + 14y - 254 = 0 14 and for (–2, –6) it is 1 y+6=( x + 2) or x + 14y + 86 = 0 14

171


23.

3

ò1 ( 3x

2

+ 2x) dx

We have to solve this by the help of limit of sum. So, a = 1, b = 3 3-1 h= Þ nh = 2 f ( x) = 3x 2 + 2x, n Q

3

ò1 ( 3x

2

+ 2x) dx = lim h [ f (1) + f (1 + h) + f (1 + 2h) + ... f (1 + (n - 1) h)] h® 0

f (1) = 3(1) 2 + 2(1) f (1 + h) = 3 (1 + h) 2 + 2 (1 + h) = 3h 2 + 8h + 5 f (1 + 2h) = 3 (1 + 2h) 2 + 2 (1 + 2h) = 12h 2 + 16h + 5 &: &: &:

&: &: &:

&: &: &:

&: &: &:

&: &: &:

&: &: &:

f (1 - (n - 1) h) = 3 (1 + (n - 1) h) 2 + 2 (1 + (n - 1) h) = 3 (n - 1) 2 h 2 + 8 (n - 1) h + 5 By putting all values in equation (i), we get 3

ò1 ( 3x

2

+ 2x) dx = lim h [(5) + ( 3h 2 + 8h + 5) + (12h 2 + 16h + 5) + ... h®0

+ [ 3 (n - 1) 2 h 2 + 8 (n - 1) h + 5]] = lim h [ 3h 2 {1 + 4 + K + (n - 1) 2 } + 8h {1 + 2 + K + (n - 1)} + 5n] h®0

(n - 1) ( 2n - 1) n 8h (n - 1) n é ù = lim h ê 3h 2 × + + 5nú h®0 ë 6 2 û ( n 1 ) ( 2 n 1 ) n (n - 1) n and {1 + 2 + K + (n - 1) = [Q{1 + 4 + .... + (n - 1) 2 = ] 6 2 é (nh - h) (nh) ( 2nh - h) ù = lim ê + 4 (nh - h) (nh) + 5nhú h®0ë 2 û é ( 2 - h) ( 2) ( 4 - h) ù = lim ê + 4 ( 2 - h) ( 2) + 10ú h®0ë 2 û 2 ´ 2 ´ 4 é ù =ê + 4 ´ 2 ´ 2 + 10ú [by applying limit] = 34 2 ë û OR We have given ìï x2 y2 x y üï + £1£ + ý í( x, y) ; 9 4 3 2 ïþ ïî

…(i)

172

Xam idea Mathematics – XII Y x+ y 3 2 =1

There are two equations (i) y 1 = equation of ellipse

x2 y2 + =1 9 4 2 Þ y1 = 9 - x2 3 and y 2 = equation of straight line x y i.e., + =1 3 2 2 Þ y 2 = ( 3 - x) 3 \ We have required area i.e.,

(0,2)

2

(–3,0) X'

(3,0)

–3

3

(0,–2) –2

x2 + y2 9 4 =1

3

= ò ( y 1 - y 2 ) dx 0 3

2 2 = ò ìí 9 - x 2 - ( 3 - x)üý dx 0 î3 3 þ 2 3ì 2 ü = ò í 9 - x - ( 3 - x)ý dx 3 0 î þ =

2 3

2 3 2 = 3 =

éx ê êë 2

Y'

3

9 - x2 +

9 x x2 ù sin -1 - 3x + ú 2 3 2 úû 0

9 9ö éæ 3 ù -1 êë çè 2 0 + 2 sin (1) - 9 + 2 ÷ø - ( 0 + 0 - 0 + 0) úû é 9 × p - 9 ù = 3 ( p - 2) sq. units. êë 2 2 2 úû 2

29. Let x-1

=

y-2

=

z+4

…(i) =m 2 3 6 From above, a point ( x, y, z) on line 1 will be ( 2m + 1, 3m + 2, 6m - 4) x- 3 y- 3 z+5 Line 2 : …(ii) = = =l 4 6 12 From above, a point ( x, y, z) on line 2 will be ( 4l + 3, 6l + 3, 12l - 5) Position vector from equation (i), we get Line 1 :

®

r = ( 2m + 1) i$ + ( 3m + 2) j$ + ( 6m - 4) k$ = (i$ + 2j$ - 4k$) + m ( 2i$ + 3j$ + 6k$)

®

®

a 1 = i$ + 2j$ - 4k$ , b 1 = 2i$ + 3j$ + 6k$ Position vector from equation (ii), we get ®

r = ( 4l + 3) i$ + ( 6l + 3) j$ + (12l - 5) k$ = ( 3i$ + 3j$ - 5k$) + l ( 4i$ + 6j$ + 12k$)

X

173


®

a 2 = 3i$ + 3j$ - 5k$,

b 2 = 4i$ + 6j$ + 12k$

®

®

From b 1 and b 2 we get b 2 = 2 b 1 ®

Shortest distance =

®

®

|( a 2 - a 1 ) ´ b | ®

|b| ®

®

( a 2 - a 1 ) = ( 3i$ + 3j$ - 5k$) - (i$ + 2j$ - 4k$) = 2i$ + j$ - k$ ® ( a2

-

® a1 )

®

i$

j$

´b = 2 2

1

®

3

k$ -1 = 9i$ - 14j$ + 4k$ 6

®

®

|( a 2 - a 1 ) ´ b | = ( 9) 2 + ( -14) 2 + ( 4) 2 = 81 + 196 + 16 = 293 ®

| b | = ( 2) 2 + ( 3) 2 + ( 6) 2 = 4 + 9 + 36 = 7 Shortest distance =

293 units 7

Set–III 1. We have given 1 1 sin -1 æç - ö÷ + cos -1 æç - ö÷ è 2ø è 2ø But, as we know sin -1 x + cos -1 x =

p × 2

p × 2 | 3A| = K| A|, where A is a square matrix of order 3.

\ principal value is 9. Given We know that

3

| 3A| = ( 3) | A| = 27| A|

…(i) …(ii)

By comparing equations (i) and (ii), we get K = 27 11. Let A, E1 , E2 be the events defined as follow: A : Ball drawn is white E1 : Bag I is chosen, E2 : Bag II is chosen Then we have to find P(E1 / A) Using Baye’s Theorem 1 4 4 ´ 40 2 5 7 P(E1 / A) = = = = 40 + 21 61 P(E1 ) P( A / E1 ) + P(E2 ) P( A / E2 ) 1 ´ 4 + 1 ´ 3 2 7 2 10 70 P(E1 ) P( A / E1 )

174


14. tan -1 (1) + tan -1 ( 2) + tan -1 ( 3) = p Consider L.H.S. tan -1 (1) + tan -1 ( 2) + tan -1 ( 3) Let

…(i)

Z = tan -1 (1) tan Z = 1 p Z= 4

And we know

…(ii) tan -1 x + tan -1 y = p + tan -1

x+y 1 - xy

Putting value of (ii) and (iii) in equation (i), we get x+y p 2+ 3 p LHS = + p + tan -1 = + p + tan -1 4 1 - xy 4 1-2´ 3 p p p = + p + tan -1 ( -1) = + p - = p = RHS 4 4 4 OR æ ö æx + 1ö p x 1 tan -1 ç ÷ + tan -1 ç ÷= è x - 2ø è x + 2ø 4 Consider above equation We know tan -1 x + tan -1 y = tan -1

Þ

Þ

x+y 1 - xy

ì x-1 x+1 ü + ï ï x-2 x+2 ï p -1 ï tan í ý= ï 1 - æç x - 1 ö÷ æç x + 1 ö÷ ï 4 ï è x - 2 ø è x + 2 ø ïþ î x2 + x - 2 + x2 - x - 2 2

= tan

2

x - 4- x +1 Þ

2x 2 - 4 -3

p 4

= 1 Þ 2x 2 - 4 = - 3

Þ

2x 2 = 1

i.e.,

x=

or

1 1 ,2 2

17. We have given S = {( a, b) : a, b Î R and a £ b 3 } 1 (i) Consider a = 2

x=±

1 2

…(iii)

175


Then

1 1 ( a, a) = æç , ö÷ Î R è 2 2ø

But

1 æ1 ö 3 £ ç ÷ is not true 2 è 2ø

\ ( a, a) Ï R , for all a Î R Hence, R is not reflexive. 1 (ii) Let a = , b = 1 2 1 1 Then, £ (1) 3 i.e., £ 1 2 2 Þ ( a, b) Î R 1 3 But 1 £/ æç ö÷ è 2ø

\ (b , a) Ï R

Hence, (a, b) Î R but (b, a) Ï R 3 4 (iii) Let a = 3, b = , c = 2 3 3

Then

3 3 £ æç ö÷ è2ø

\

( a, b) Î R

Also,

3 æ4ö 3 £ç ÷ 2 è 3ø

\

(b , c) Î R

But

3 £/ 1

æ4ö ç ÷ è 3ø

i.e., 3 £ 27

i.e.,

3 64 £ 2 27

3

i.e., 3 £/

64 27

\ ( a, c) Ï R Hence, ( a, b) Î R, (b , c) Î R but ( a, c) Ï R Þ R is not transitive. 19. We have given x-7 y= ( x - 2) ( x - 3) Let (i) cuts the x-axis at (x, 0) x-7 then = 0 Þ x=7 ( x - 2) ( x - 3) \ the required point is (7, 0). Differentiating equation (i) w.r.t. x, we get dy ( x - 2) ( x - 3) 1 - ( x - 7) [( x - 2) + ( x - 3)] = dx [( x - 2) ( x - 3)] 2

…(i)

176


=

x 2 - 5x + 6 - 2x 2 + 19x - 35 ( x 2 - 5x + 6) 2

=

- x 2 + 14x - 29 ( x 2 + 6 - 5x) 2

-49 + 98 - 29 dy ù 20 1 = = = ú 2 dx û ( 7 , 0 ) ( 49 - 35 + 6) 400 20 \ Equation of tangent is 1 y - y1 = (x - x2 ) 20 1 Þ y-0= ( x - 7) or 20 23. f ( x) = sin x - cos x, 0 £ x £ 2p Differentiating w.r.t. x, we get

x - 20y - 7 = 0

p f ¢( x) = cos x + sin x = 2 sin æç + xö÷ è4 ø dy =0 dx Þ cos x + sin x = 0 Þ sin x = - cos x Þ tan x = - 1 p Þ tan x = tan æç - ö÷ è 4ø p p 3p 7 p 11p Þ x = np - = - , , , +K 4 4 4 4 4 3p (i) For 0 < x < , 4 p p i.e., It lies in quadrant I, II. 0, è2 ø 3p 7p (ii) For 0, è4 ø For critical points,

177


Internal where function is strictly increasing is æ 0, 3p ö È æ 7 p , 2p ö ç ÷ ç ÷ è ø 4 ø è 4 Interval where function is strictly decreasing is æ 3p , 7 p ö ç ÷ è 4 4 ø 24.

4

ò1 ( x

2

- x) dx

We have to solve it by using limit of sums. b - a 4-1 Here, a = 1, b = 4, h = i.e., nh = 3 = n n 4

Limit of sum for ò ( x 2 - x) dx is 1

= lim h [ f (1) + f (1 + h) + f (1 + 2h) + .... + f {1 + (n - 1) h}] h®0

Now, f (1) = 1 - 1 = 0 f (1 + h) = (1 + h) 2 - (1 + h) = h 2 + h f (1 + 2h) = (1 + 2h) 2 - (1 + 2h) = 4h 2 + 2h …………………………………………… …………………………………………… …………………………………………… f [1 + (n - 1) h] = {1 + (n - 1) h} 2 - {1 + (n - 1) h} = (n - 1) 2 h 2 + (n - 1) h \

2

ò1 ( x

2

- x) dx = lim h [0 + h 2 + h + 4h 2 + 2h + ...: (n - 1) 2 h 2 + (n - 1) h] h®0

= lim h [h 2 {1 + 4 + .. + (n - 1) 2 } + h {1 + 2 + K + (n - 1)}] h®0

(n) (n - 1) ( 2n - 1) n (n - 1) ù é = lim h ê h 2 × +h h®0 6 2 úû ë n (n - 1) ( 2n - 1) n (n - 1) ù [Q 1 + 4 + K + (n - 1) 2 = 1 + 2 + K + (n - 1) = ú 6 2 û é nh (n h - h) ( 2nh - h) nh (nh - h) ù = lim ê + ú h®0 ë 6 2 û é ( 3 - h) ( 3) ( 6 - h) ( 3 - h) ( 3) ù = lim ê + ú h®0ë 6 2 û 3 ´ 3 ´ 6 3 ´ 3 9 27 æ ö æ ö =ç ÷+ç ÷=9+ = è ø è 2 ø 6 2 2

178


OR We have provided ( x, y) :| x - 1| £ y £ 5 - x 2 Equation of curve is y = 5 - x 2 or y 2 + x 2 = 5, which is a circle with centre at (0, 0) and 5 radius . 2 Y Equation of line is y =| x - 1| y = x - 1 and

Consider,

y=–x+1

y = 5 - x2

Eliminating y, we get x - 1 = 5 - x2 Þ

2x 2 - 2x - 4 = 0

Þ

x2 - x - 2 = 0

2

5x

x 2 + 1 - 2x = 5 - x 2

y=

Þ

y = x –1

(–1,0) X'

Þ ( x - 2) ( x + 1) = 0 Þ x = 2, - 1 The required area is =ò

2

-1

–3

–2

–1

0

1

2

Y'

5 - x 2 dx - ò

1

-1

2

( - x + 1) dx - ò ( x - 1) dx 1

1

2

2 é x2 ù é x2 ù 5 x ù éx =ê 5 - x 2 + sin -1 + x - xú ê ú ê ú 2 5 û -1 êë 2 ë2 úû -1 êë 2 úû 1

5 2 ö 5 æ -1 = ç1 + sin -1 ÷ + 1 - sin è 2 2 5ø =

1 1 æ 1 ö æ -1 + 1 + + 1ö÷ - æç 2 - 2 - + 1ö÷ ç÷-ç ø è ø è 2 2 5ø è 2

5æ 1 ö 1 -1 2 + sin -1 ç sin ÷+2-2è ø 2 2 5 5

é 2 5 1 1 4ù 1 sin -1 ê 1- + 1- ú2 5 5û 2 5 ë 5 5é 4 1 ù 1 = ê sin -1 æç + ö÷ ú è 5 5 øû 2 2ë 5 1 = sin -1 (1) 2 2 5 p 1 = æç - ö÷ sq. units è 4 2ø =

3

X


Maximum marks: 100


Set–I SECTION–A Question numbers 1 to 10 carry 1 mark each. 1. If f : R ® R be defined by f ( x) = ( 3 - x 3 ) 1/ 3 , then find fof ( x). 2. Write the principal value of sec -1 ( -2). 3. What positive value of x makes the following pair of determinants equal? 2x 3 16 3 , 5 x 5 2 4. Evaluate :

ò sec

2

(7 - 4x) dx

5. Write the adjoint of the following matrix : æ 2 -1 ö ç ÷ 3ø è4 6. Write the value of the following integral : p/2

ò-p / 2

sin 5 x dx

7. A is a square matrix of order 3 and| A| = 7. Write the value of| adj. A|. 8. Write the distance of the following plane from the origin : 2x - y + 2z + 1 = 0 9. Write a vector of magnitude 9 units in the direction of vector -2i$ + j$ + 2k$. ®

10. Find l if ( 2i$ + 6j$ + 14k$) ´ (i$ - lj$ + 7 k$) = 0 .

SECTION–B Question number 11 to 22 carry 4 marks each. 11. A family has 2 children. Find the probability that both are boys, if it is known that (i) at least one of the children is a boy (ii) the elder child is a boy.

180


12. Show that the relation S in the set A = {x Î Z : 0 £ x £ 12} given by S = {( a, b) : a, b Î Z ,| a - b| is divisible by 4} is an equivalence relation. Find the set of all elements related to 1. 13. Prove the following : æ 2x ö æ 3x - x 3 ö ÷ = tan -1 ç ÷ tan -1 x + tan -1 ç ç1 - x 2 ÷ ç 1 - 3x 2 ÷ è ø è ø OR Prove the following: cos [tan -1 {sin (cot -1 x)}] =

1 + x2 2 + x2

14. Express the following matrix as the sum of a symmetric and skew symmetric matrix, and verify you result: -2 -4 ö æ 3 ç ÷ 3 2 -5 ÷ ç ç -1 1 2÷ø è ®

®

®

15. If a = i$ + j$ + k$ , b = 4i$ - 2j$ + 3k$ and c = i$ - 2j$ + k$ , find a vector of magnitude 6 units which is ®

®

®

parallel to the vector 2 a - b + 3 c . OR ®

®

®

®

Let a = i$ + 4j$ + 2k$ , b = 3i$ - 2j$ + 7 k$ and c = 2i$ - j$ + 4k$. Find a vector d which is ®

®

® ®

perpendicular to both a and b and c . d = 18. x+ 2 y+1 z- 3 16. Find the points on the line at a distance of 5 units from the point = = 3 2 2 P (1, 3, 3). OR Find the distance of the point P ( 6, 5, 9) from the plane determined by the points A ( 3, - 1, 2), B(5, 2, 4) and C ( -1, - 1, 6). 17. Solve the following differential equation : dy 1 ( x 2 - 1) + 2xy = ; | x| ¹ 1 2 dx x -1 OR Solve the following differential equation : dy 1 + x 2 + y 2 + x 2 y 2 + xy =0 dx 18. Show that the differential equation ( x - y)

dy = x + 2y, is homogeneous and solve it. dx

181


19. Evaluate the following : x+2 ò ( x - 2) ( x - 3) dx 20. Evaluate the following : 5x 2

2

ò1

x 2 + 4x + 3

21. If y = e a sin

-1 x

(1 - x 2 ) 22. If y = cos -1

dx

, -1 £ x £ 1, then show that d2y 2

-x

dy - a2 y = 0 dx

dx æ 3x + 4 1 - x 2 ç ç 5 è

ö ÷ , find dy × ÷ dx ø

SECTION–C Question number 2 to 29 carry 6 marks each. 23. Using properties of determinants, prove the following: x

x2

y

y

2

z

z2

1 + px 3 1 + py 3 = (1 + pxyz) ( x - y) ( y - z) (z - x) 1 + pz 3

OR Find the inverse of the following matrix using elementary operations : 2 -2 ö æ 1 ç ÷ A = ç -1 3 0÷ ç 0 -2 1÷ø è 24. A bag contains 4 balls. Two balls are drawn at random, and are found to be white. What is the probability that all balls are white? 25. One kind of cake requires 300 g of flour and 15 g of fat, another kind of cake requires 150 g of flour and 30 g of fat. Find the maximum number of cakes which can be made from 7 × 5 kg of flour and 600 g of fat, assuming that there is no shortage of the other ingredients used in making the cakes. Make it as an L.P.P. and solve it graphically. 26. Find the coordinates of the foot of the perpendicular and the perpendicular distance of the point P ( 3, 2, 1) from the plane 2x - y + z + 1 = 0. Find also, the image of the point in the plane. 27. Find the area of the circle 4x 2 + 4y 2 = 9 which is interior to the parabola x 2 = 4y. OR Using integration, find the area of the triangle ABC, coordinates of whose vertices are A ( 4, 1), B ( 6, 6) and C ( 8, 4).

182


28. If the length of three sides of a trapezium other than the base is 10 cm each, find the area of the trapezium, when it is maximum. 29. Find the intervals in which the following function is : (a) strictly increasing (b) strictly decreasing

Set-II Only those questions, not included in Set I, are given 6. Write the principal value of cot -1 ( - 3 ). ®

®

® ®

®

®

®

10. If a and b are two vectors such that| a . b | =| a ´ b |, then what is the angle between a and ®

b? 11. Prove the following : 1 tan -1 æç ö÷ + tan -1 è 3ø

æ 1 ö + tan -1 ç ÷ è5 ø

æ 1 ö + tan -1 ç ÷ è7 ø

æ1 ö = p ç ÷ è 8ø 4

OR Solve for x : æx - 1ö tan -1 ç ÷ + tan -1 x 2 è ø æ2 ç 14. If A = ç 2 ç1 è

0 1 -1

æx + 1ö p ç ÷= è x + 2ø 4

1ö ÷ 3÷ , then find the value of A 2 - 3A + 2I . 0÷ø

18. Evaluate:

ò

5x + 3 x 2 + 4x + 10

dx

20. Show that the following differential equation is homogeneous, and then solve it : y y dx + x log æç ö÷ dy - 2x dy = 0 è xø 23. Find the equations of the tangent and the normal to the curve p x = 1 - cos q, y = q - sin q ; at q = × 4 24. Find the equation of the plane passing through the point P (1, 1, 1) and containing the line ® r = ( -3i$ + j$ + 5k$) + l ( 3i$ - j$ - 5k$). Also, show that the plane contains the line ®

r = ( -i$ + 2j$ + 5k$) + m (i$ - 2j$ - 5k$)

183


Set-III Only those questions, not included in Set I and Set II are given 4p 6. Find the value of sin -1 æç ö÷ . è 5 ø ®

®

®

7. Vectors a and b are such that| a | =

®

3 ,| b | =

®

®

® ® 2 and ( a ´ b ) is a unit vector. Write the angle 3

between a and b . 11. Show that the relation S defined on the set N ´ N by ( a, b) S ( c , d) Þ a + d = b + c 15. For the following matricesA and B, verify that ( AB) ¢ = B ¢ A ¢. æ 1ö ç ÷ A = ç -4÷ , B = ( -1, 2, 1) ç 3÷ è ø 17. Solve the following differential equation : dy ( x 2 + 1) + 2xy = x 2 + 4 dx OR Solve the following differential equation : dy ( x 3 + x 2 + x + 1) = 2x 2 + x dx 20. If y = cosec -1 x, x > 1. then show that x ( x 2 - 1)

d2y 2

+ ( 2x 2 - 1)

dy =0 dx

dx 23. Using matrices, solve the following system of equations : x + 2y - 3z = - 4 2x + 3y + 2z = 2 3x - 3y - 4z = 11 OR If a, b , c are positive and unequal, show that the following determinant is negative : a b c D= b

c

a

c

a

b

25. Show that the volume of the greatest cylinder that can be inscribed in a cone of height ‘h’ and 4 semi-vertical angle ‘a’ is ph 3 tan 2 a . 27

184


SOLUTIONS Set–I SECTION–A 1. If f : R ® R be defined by f ( x) = ( 3 - x 3 ) 1/ 3 then

( fof ) x = f ( f ( x)) = f [( 3 - x 3 ) 1/ 3 ] = [ 3 - {( 3 - x 3 ) 1/ 3 } 3 ] 1/ 3 = [ 3 - ( 3 - x 3 )] 1/ 3 = ( x 3 ) 1/ 3 = x

2. Let

x = sec -1 ( -2)

Þ

sec x = - 2

Þ

sec x = - sec

Þ

x=

p 2p p = sec æç p - ö÷ = sec è 3 3ø 3

2p × 3

3. We have given 2x 3 16 = 5 x 5

3 2

Þ

2x 2 - 15 = 32 - 15

Þ

2x 2 = 32

(solving the determinant)

Þ x 2 = 16 Þ x = ± 4

But we need only positive value \ x=4 4. Let I = ò sec 2 (7 - 4x) dx 7 - 4x = m , -4dx = dm -1 I= Þ sec 2 m dm 4 ò 1 1 = - tan m + c = - tan (7 - 4x) + c 4 4 5. We have given matrix : é 2 -1 ù ê4 3 ú ë û Let

C 11 = 3 C 21 = 1 \

é3 Adj. A = ê ë -4

C 12 = - 4 C 22 = 2 1ù 2úû

185


6.

p/2

ò- p / 2 Let

sin 5 x dx f ( x) = sin 5 x f ( - x) = [sin ( - x)]5 = ( - sin x) 5 = - sin 5 x

= - f ( x) Thus, f ( x) is an odd function. \

p/2

ò- p / 2

sin 5 x dx = 0

7. A is a square matrix of order 3 and| A| = 7 | adj. A| =| A|2 = (7) 2 = 49

then

8. We have given plane 2x - y + 2z + 1 = 0 Distance from origin =

9. Let

( 2 ´ 0) - (1 ´ 0) + ( 2 ´ 0) + 1 ( 2) 2 + ( -1) 2 + ( 2) 2

=

1 1 = 3 4+1+ 4

®

r = - 2i$ + j$ + 2k$ ®

®

r Unit vector in the direction of r = r$ = r |r| \ Vector of magnitude 9 = 9 r$ é ù ® -2i$ + j$ + 2k$ ú Units in the direction of r = 9 ê ê ( 2) 2 + (1) 2 + ( 2) 2 ú ë û $ é - 2i$ + j$ + 2k ù =9ê ú = - 6i$ + 3j$ + 6k$ 4 + 1 + 4 êë úû 10. We have given ®

( 2i$ + 6j$ + 14k$) ´ (i$ - lj$ + 7 k$) = 0 Þ

i$

j$

2

6

1

-l 6 -l

k$

®

14 = 0 7 14 2 - j$ 7 1

14 2 + k$ 7 1

® 6 = 0 -l

Þ

i$

Þ

i$ ( 42 + 14l) - 0 j$ + k$ ( -2l - 6) = 0

Þ

42 + 14l = 0 Þ

®

14l = - 42 Þ l = - 3

186


Also, \

-2 l - 6 = 0 l=-3

Þ

l=-3

SECTION–B 11. A family has 2 children, then Sample space = S = {BB, BG , GB, GG} where B = Boy, G = Girl (i) Let us define the following events: A : at least one of the children is boy : {BB, BG , GB } B : both are boys: { BB } \ A Ç B : {BB} P ( A Ç B) 1 / 4 1 Þ P( B/ A) = = = P ( A) 3 /4 3 (ii) Let

Þ

A : elder boy child : { BB, BG} B : both are boys: {BB } \ A Ç B : { BB} P ( A Ç B) 1 / 4 1 P( B/ A) = = = P ( A) 2 /4 2

12. We have given, A = {x Î Z : 0 £ x £ 12} and S = {( a, b) : a, b Î A ,| a - b|is divisible by 4} (i) for ( a, a) Î S, | a - a|= 0 is divisible by 4. \ It is reflexive. (ii) Let ( a, b) Î S Then | a - b|is divisible by 4 Þ | - (b - a)|is divisible by 4 Þ|b - a|is divisible by 4 \ ( a, b) Î S Þ (b , a) Î S \ It is symmetric. (iii) Let ( a, b) Î S and (b , c) Î S Þ | a - b|is divisible by 4 and |b - c|is divisible by 4 Þ ( a - b) is divisible by 4 and (b - c) is divisible by 4 Þ | a - c| =|( a - b) + (b - c)|is divisible by 4 \ ( a, c) Î S \ It is transitive. From above we can say that the relation S is reflexive, symmetric and transitive. \ Relation S is an equivalence relation. The set of elements related to 1 are {9, 5, 1}.

187


æ 2x ö ÷ = tan -1 13. We have to prove: tan -1 x + tan -1 ç ç1 - x 2 ÷ è ø

æ 3x - x 3 ö ç ÷ ç 1 - 3x 2 ÷ è ø

æ 2x ö ÷ = tan -1 x + tan -1 ç ç1 - x 2 ÷ è ø

LHS

é ù 2x ê x+ ú ê 1 - x2 ú ê ú æ 2x ö ú ê ç ÷ ê1 - x ç 2 ÷ú è 1 - x ø úû êë

= tan -1

é a+b ù [As we know tan -1 a + tan -1 b = tan -1 ê ú] ë 1 - ab û

é x - x 3 + 2x ù -1 = tan -1 ê ú = tan 2 2 êë 1 - x - 2x úû

é 3x - x 3 ù ê ú = RHS êë 1 - 3x 2 úû

OR cos [tan -1 {sin (cot -1 x)}] =

1 + x2 2 + x2

= cos [tan -1 {sin (cot -1 x)}]

LHS

Let x = cot q = cos [tan -1 (sin q)]

LHS

é = cos ê tan -1 ê ë

é æ öù 1 ç ÷ ú = cos ê tan -1 ç ê 2 ÷ú è 1 + cot q ø û ë

æ 1 q 1 = tan -1 ç ç 2 è 1+x

Let

1 + x2

cos q 1 =

2 + x2

ö ÷ ÷ ø

Þ

tan q 1 =

Þ

q 1 = cos -1

Now, put q 1 in equation (i), we get é 1 + x2 ù 1 + x2 ú= cos ê cos -1 ê 2 + x 2 úû 2 + x2 ë 14. Consider é 3 A=ê 3 ê êë -1

-2 -2 1

-4 ù -5 ú ú 2úû

æ 1 ç ç 2 è 1+x

öù ÷ú ÷ú øû

1 1 + x2 1 + x2 2 + x2

…(i)

188


A=

We can write where,

1 1 ( A + A ¢) + ( A - A ¢) 2 2

…(i)

1 ( A + A ¢) is a symmetric matrix 2

1 ( A - A ¢) is a skew symmetric matrix. 2 3 -1 ù é 3 ê A¢ = -2 -2 1ú ê ú -5 2úû êë -4 é æ 3 -2 -4 ö æ 3 3 ÷ ç 1 1 êç ( A + A ¢) = ê ç 3 -2 -5÷ + ç -2 -2 2 2 ç ÷ ç 1 2 ø è -4 -5 ëê è 1

and

Now,

é 6 1ê = 1 2ê êë -5

-5 ù é 3 ú ê -4 = 1 / 2 ú ê 4úû ë -5 / 2

1 -4 -4

é é 3 -2 1 1 êê ( A - A ¢) = ê 3 -2 2 2 ê êë êë -1 1 -5 é0 1ê = 5 0 2ê 6 êë 3

-4 ù é 3 -5 ú - ê -2 ú ê 2úû êë -4

-3 ù -6 ú ú 0úû -5 / 2

0 é ê = 5/2 ê êë 3 / 2

0 3

3 -2 -5

-1 ö ù ÷ú 1÷ ú 2÷ø ûú 1/ 2 -2 -2

®

®

®

®

…(ii)

-1 ù ù ú 1ú ú ú 2úû úû

-3 / 2 ù -3 ú ú 0úû

Putting value of equations (ii) and (iii) in equation (i), 3 1/2 -5 / 2 ù é 0 -5 / 2 é ê ú ê A= 1/2 -2 -2 + 5 / 2 0 ê ú ê -2 2úû êë 3 / 2 3 êë -5 / 2 -2 -4 ù é 3 ê = 3 -2 -5 ú ê ú 1 2úû êë -1 Hence Proved. ® ® ® 15. Given, a = i$ + j$ + k$, b = 4i$ - 2j$ + 3k$, c = i$ - 2j$ + k$ Consider,

-5 / 2ù -2ú ú 2û

…(iii)

-3 / 2 ù -3 ú ú 0úû

r =2 a - b +3 c = 2i$ + 2j$ + 2k$ - 4i$ + 2j$ - 3k$ + 3i$ - 6j$ + 3k$ = i$ - 2j$ + 2k$

189


Since the required vector has magnitude 6 units and parallel to r . \ Required vector = 6$r é ù $ù $ é$ i$ - 2j$ + 2k$ ú = 6 ê i - 2j + 2k ú = 2i$ - 4j$ + 4k$ =6ê ê (1) 2 + ( -2) 2 + ( 2) 2 ú êë 1 + 4 + 4 úû ë û OR Given, ®

®

®

a = i$ + 4j$ + 2k$, b = 3i$ - 2j$ + 7 k$, c = 2i$ - j$ + 4k$ ®

®

®

®

®

®

Vector d is perpendicular to both a and b i.e., d is parallel to vector a ´ b . i$

j$

k$

d = 1

4

2

3

-2

7

®

\

= i$

4 -2

2 1 - j$ 7 3

2 1 + k$ 7 3

4 = 32i$ - j$ - 14k$ -2

®

Now let d = m ( 32i$ - j$ - 14k$) Also,

® ®

c . d = 18

Þ

( 2i$ - j$ + 4k$) . m ( 32i$ - j$ - 14k$) = 18

Þ

m ( 64 + 1 - 56) = 18 Þ

\

9m = 18 or

m=2

®

d = 2 ( 32i$ - j$ - 14k$) = 64i$ - 2j$ - 28k$

16. Given cartesian form of line as: x+ 2 y+1 z- 3 = = =m 3 2 2 \ General point on line is ( 3m - 2, 2m - 1, 2m + 3) Since distance of points on line from P (1, 3, 3) is 5 units. \

( 3m - 2 - 1) 2 + ( 2m - 1 - 3) 2 + ( 2m + 3 - 3) 2 = 5

Þ

( 3m - 3) 2 + ( 2m - 4) 2 + ( 2m) 2 = 25

Þ

17m 2 - 34m = 0 Þ

17m (m - 2) = 0

Þ

m = 0, 2

\ Required point on line is (–2, –1, 3) for m = 0, or (4, 3, 7) for m = 2. OR Plane determined by the points A ( 3, - 1, 2), B (5, 2, 4) and C ( -1, - 1, 6) is x- 3 y+1 z-2 x- 3 y+1 z-2 5- 3

2+1

-1 - 3

-1 + 1

4-2 =0 6-2

Þ

2

3

2

-4

0

4

=0

190


Þ

3 0

( x - 3)

2 2 - ( y + 1) 4 -4

2 2 + (z - 2) 4 -4

3 =0 0

Þ 12x - 36 - 16y - 16 + 12z - 24 = 0 Þ 3x - 4y + 3z - 19 = 0 Distance of this plane from point P ( 6, 5, 9) is ( 3 ´ 6) - ( 4 ´ 5) + ( 3 ´ 9) - 19 2

2

( 3) + ( 4) + ( 3) 17. Given, ( x 2 - 1)

=

2

18 - 20 + 27 - 19 9 + 16 + 9

6 units 34

=

dy 1 + 2xy = ;| x| ¹ 1 2 dx x -1

By simplifying the equation, we get dy 2x 1 + y= 2 dx x 2 - 1 ( x - 1) 2 This is a linear differential equation of the form Here P =

2x 2

x -1 ò

I.F. = e

,Q =

2x x2 - 1

1 2

( x - 1) 2

dx

= e log | x

2 - 1|

\ Solution is ( x 2 - 1) y = ò ( x 2 - 1) ; Þ

( x 2 - 1) y =

dy + Py = Q dx

= x2 - 1 1 2

( x - 1)

2

dx = ò

1 2

x -1

dx

x-1 1 log +C 2 x+1 OR

1 + x 2 + y 2 + x 2 y 2 + xy

Given,

dy =0 dx

By simplifying the equation, we get dy xy = - 1 + x2 + y2 + x2y2 dx dy Þ xy = - (1 + x 2 ) (1 + y 2 ) = - (1 + x 2 ) (1 + y 2 ) dx y

Þ

1 + y2

dy = -

1 + x2 x

dx

Integrating both sides, we get

ò

y 1 + y2

dy = - ò

1 + x2 x

dx

...(i)

191


1 + y2 = t

Þ 2y dy = dt

(For LHS)

Þ 2x dx = 2m dm Þ x dx = m dm 1+ x =m 1 1 m \ (i) Þ dt = - ò . m dm 2 2ò t m -1

(For RHS)

Let Let

Þ

2

2

1 t 1/ 2 m2 +ò dm = 0 21/2 m2 - 1 æ 1 ö÷ t + ò ç1 + dm = 0 ç m 2 - 1 ÷ø è

Þ

m2 + 1 - 1

Þ

t+ò

Þ

t +m+

m2 - 1

dm = 0

m-1 1 log =0 2 m+1

Now substituting these value of t and m, we get 1 + y2 + 1 + x2 +

1 log 2

1 + x2 - 1 1 + x2 + 1

+C = 0

dy = x + 2y dx By simplifying the above equation we get dy x + 2y = dx x - y x + 2y Let F ( x, y) = x-y Ax + 2Ay A ( x + 2y) then F ( Ax, Ay) = = = F ( x, y) Ax - Ay A ( x - y) ( x - y)

18. Given,

…(i)

\ F( x, y) and hence the equation is homogeneous Now let y = vx dy dv Þ =v+ x dx dx Substituting these values in equation (i), we get dv x + 2vx v+x = dx x - vx Þ Þ

1 + 2v - v + v 2 1 + v + v 2 dv 1 + 2v = -v= = dx 1 - v 1-v 1-v 1-v dx dv = 2 x 1+v+v

x

By integrating both sides, we get 1-v dx ò v 2 + v + 1 dv = ò x

…(ii)

192


1-v

ò v 2 + v + 1 dv

LHS

1 - v = A ( 2v + 1) + B = 2Av + ( A + B) Comparing both sides, we get 2A = - 1, A + B = 1 1 3 or A=- , B= 2 2 1 3 - ( 2v + 1) + 1-v 2 2 \ ò v 2 + v + 1 dv = ò v 2 + v + 1 dv 2v + 1 1 3 dv =- ò dv + ò 2 2 2 v +v+1 2 v +v+1 1 ( 2v + 1) 3 dv =- ò dv + ò 2 2 v +v+1 2 æ 1ö2 3 çv + ÷ + è 2ø 4 Let

æ v+1 ö ç ÷ 1 3 2 2 ÷ = - log|v 2 + v + 1| + ´ tan -1 ç 2 2 3 ç 3 / 2÷ è ø Now substituting it in equation (ii), we get 1 æ 2v + 1 ö - log|v 2 + v + 1| + 3 tan -1 ç ÷ = log x + C è 3 ø 2 Þ

y2 y 1 - log + +1 + 2 x2 x

Þ

-

Þ

-

19. Given,

( x + 2) dx

=ò =

1 2

x 2 - 5x + 6

ò

æ 2y + 1 ö ç ÷ ç x ÷ = log x + C 3 ÷ ç è ø

1 1 log|x 2 + xy + y 2|+ log x 2 + 2 2

1 log| x 2 + xy + y 2 | + 2 ( x + 2) dx dx ( x - 2)( x - 3)

ò

3 tan

-1

dx

2x + 4 x 2 - 5x + 6

dx

æ 2y + x ö 3 tan -1 ç ÷ = log x + C è 3x ø

æ 2y + x ö 3 tan -1 ç ÷ =C è 3x ø

193


=

1 2

ò

=

1 2

ò

( 2x - 5) + 9 x 2 - 5x + 6 2x - 5 x 2 - 5x + 6

dx dx +

9 2

ò

dx x 2 - 5x + 6

I1

I2

For I1 Let Þ \

x 2 - 5x + 6 = m ( 2x - 5) dx = dm =

1 2

ò

1 dm m

1 ´ 2 m = m = x 2 - 5x + 6 2 9 1 9 dx I2 = ò dx = ò 2 2 2 x 2 - 5x + 6 æ x - 5 ö - 25 + 6 ç ÷ è 2ø 4 9 dx = ò 2 2 2 æx - 5 ö - æ 1 ö ç ÷ ç ÷ è è 2ø 2ø I1 =

= Thus,

( x + 2)

ò

( x - 2) ( x - 3) 2

5x 2

1

x 2 + 4x + 3

20. Given, ò

=5ò

2

9 log æç x è 2

x + 4x + 3

2

5ö ÷ + x 2 - 5x + 6 + C 2ø

2

2

1

1

dx = 5 ò dx - 5ò

4x + 3 2

x + 4x + 3

dx

é 2 2 ( 2x + 4) ù 2 dx dx = 5 - 5 ê ò dx - 5 ò ú 1 2 1 2 x + 4x + 3 x + 4x + 3 úû êë 1 x 2 + 4x + 3 2x + 4 2 dx dx + 25 ò 2 1 x + 4x + 3 ( x + 2) 2 - 1

= 5[x]21 - 5 ò

1

9 log æç x è 2

…(ii)

dx

2

= 5 - 10 ò

5ö ÷ + x 2 - 5x + 6 2ø

dx = I 1 + I 2 = x 2 - 5x + 6 +

( x 2 + 4x + 3) - ( 4x + 3)

1

…(i)

2

4x + 8 - 5

2

é x+1 ù 25 = 5 - ê10 log| x 2 + 4x + 3| log ú 2 x+ 3 û ë 1 25 3 25 = 5 - é10 log 15 log - 10 log 8 + log êë 2 5 2

8 25 6 1ù = 5 + 10 log + log ú 2û 15 2 5

194


21. We have given, y = e a sin

-1 x

, -1£ x£1

…(i)

dy - a2 y = 0 dx

…(ii)

and we have to prove (1 - x 2 )

d2y 2

-x

dx Now differentiating equation (i) w.r.t. x, we get -1 dy a = e a sin x . dx 1 - x2 1 - x2

Þ

dy = ay dx

2

æ dy ö Þ (1 - x 2 ) ç ÷ = a 2 y 2 è dx ø

(Squaring both sides)

Now again differentiating w.r.t. x, we get 2

dy d 2 y æ dy ö æ dy ö . - 2x ç ÷ = a 2 ç 2y ÷ è dx ø è dx ø dx dx 2 dy Dividing both sides by 2 , we get dx 2 (1 - x 2 )

(1 - x 2 )

d2y dx

(1 - x 2 )

Þ

22. Given, Let

2

d2y 2

-x

dy = a2 y dx

-x

dy - a2 y = 0 dx

dx æ 3x + 4 1 - x 2 y = cos -1 ç ç 5 è

Hence Proved.

ö ÷ ÷ ø

x = cos a so that a = cos -1 x

é 3 cos a + 4 1 - cos 2 a ù ú = cos -1 y = cos -1 ê ê ú 5 ë û 3 4 Let = cos q, then = sin q 5 5 Þ

é 3 cos a + 4 sin a ù êë 5 úû 5

y = cos -1 [cos a cos q + sin a sin q] = cos -1 [cos (a - q)] = a - q 3 Þ y = cos -1 x - cos -1 5 Differentiating w.r.t. x, we get dy -1 -1 = -0= 2 dx 1-x 1 - x2

\

…(i)

195


SECTION–C 23.

x

x2

1 + px 3

y z

y2 z2

1 + py 3 = (1 + pxyz) ( x - y) ( y - z) (z - x) 1 + pz 3

x

x2

1 + px 3

LHS = y

y2

1 + py 3

z

z2

1 + pz 3

By splitting into two parts, we get x = y

x2 y2

1 x 1 + y

x2 y2

px 3

z

z2

1

z2

pz 3

x = y

x2 y2

1 1 x x2 1 + pxyz 1 y y 2

z

z2

1

x = y

x2 y2

1 x 2 1 + ( -1) pxyz y

x2 y2

1 1

z

z2

1

z2

1

z

py 3

1 z

z2

z

[In second determinant, replacing c 1 and c 3 and then c 1 with c 2 ] x = (1 + pxyz) y

2

x y2

1 1

z

z2

1

By applying R 1 ® R 1 - R 2 , R 2 ® R 2 - R 3 , we get x - y ( x - y) ( x + y) 0 = (1 + pxyz) y - z ( y - z) ( y + z) 0 z

z2

1

1

x+y

0

= (1 + pxyz) ( x - y) ( y - z) 1 z

y+z

0 1

z

2

By expanding the determinant, we get Þ (1 + pxyz) ( x - y) ( y - z) [y + z - x - y] Þ (1 + pxyz) ( x - y) ( y - z) (z - x)

196


é 1 A = ê -1 ê êë 0 Let A = IA é 1 ê -1 Þ ê êë 0

2 3 -2 2 3

-2

OR -2 ù 0ú ú 1úû

-2 ù é 1 0ú = ê 0 ú ê 1úû êë 0

0 1 0

Applying R 2 ® R 2 + R 1 2 -2 ù é 1 é1 ê0 5 -2 ú = ê 1 ê ú ê 1úû êë 0 êë 0 -2

0 1 0

0ù 0ú A ú 1úû 0ù 0ú A ú 1úû

Applying R 1 ® R 1 + R 3 , R 2 ® R 2 + 2R 3 0 -1 ù é 1 0 1ù é1 ê0 ú ê 1 0 = 1 1 2ú A ê ú ê ú 1úû êë 0 0 1úû êë 0 -2 Applying R 3 ® R 3 + 2R 2 é 1 0 -1 ù é 1 0 1 ù ê 0 1 0 ú = ê1 1 2ú A ê ú ê ú êë 0 0 1 úû êë 2 2 5 úû Applying R 1 ® R 1 + R 3 0 0ù é 3 é1 ê0 1 0ú = ê 1 ê ú ê 0 1úû êë 2 êë 0 é3 -1 ê \ A = 1 ê êë 2

2 1 2 2 1 2

24. Let us define the following events. E: drawn balls are white A : 2 white balls in bag. B: 3 white balls in bag. C: 4 white balls in bag. 1 Then P (A) = P (B) = P (C) = 3

6ù 2ú A ú 5úû 6ù 2ú ú 5úû

197


E P æç ö÷ = è Aø E P æç ö÷ = è Bø

2

C2

4

C2

3

C2

4

C2

=

1 , 6

=

3 , 6

E P æç ö÷ = èC ø

4

C2

4

C2

=1

By applying Baye’s Theorem E P (C) . P æç ö÷ èC ø E E E P ( A) . P æç ö÷ + P ( B) . P æç ö÷ + P (C) P æç ö÷ è Aø è Bø èC ø 1 ´1 1 3 3 = = = æ 1 ´ 1 ö + æ 1 ´ 3 ö + æ 1 ´ 1ö 1 + 3 + 1 5 ç ÷ ç ÷ ç ÷ è 3 6ø è 3 6 ø è 3 ø 6 6

C P æç ö÷ = èEø

25. Let number of first kind and second kind of cakes that can be made be x and y respectively Then, the given problem is Y Maximize, z=x+y 2x+y=50 Subjected to x ³ 0, y ³ 0 50 – 300x + 150y £ 7500 Þ 2x + y £ 50 40 – 15x + 30y £ 600 Þ x + 2y £ 40 30 – From graph, three possible points are (0,20) x+2y=40 20 – (25, 0), (20, 10), (0, 20) (20,10) 10 – At (25, 0), z = x + y = 25 + 0 = 25 (25,0) | | | | At (20, 10), z = x + y = 20 + 10 X X' 10 20 30 40 = 30 ¬ Maximum Y' At (0, 20), z = 0 + 20 = 20 As Z is maximum at (20, 10), i.e., x = 20, y = 10. \ 20 cakes of type I and 10 cakes of type II can be made. 26. Let O (a , b , g ) be the image of the point P ( 3, 2, 1) in the plane 2x - y + z + 1 = 0 \ PO is perpendicular to the plane and S is the mid point of PO and the foot of the perpendicular. P(3,2,1) DR’s of PS are 2, –1, 1. x - 3 y - 2 z-1 \ Equation of PS are = = =m 2 -1 1 S \ General point on line is S ( 2m + 3, - m + 2, m + 1) If this point lies on plane, then 2x–y+z=–1 2 ( 2m + 3) - ( -m + 2) + 1 (m + 1) + 1 = 0 Þ 6m + 6 = 0 Þ m = - 1 0(a,b,g)

198


\ Coordinates of S are (1, 3, 0). As S is the mid point of PO, æ3+a 2+b 1+ gö \ , , ç ÷ = (1, 3, 0) è 2 2 2 ø By comparing both sides, we get 3+a =1 Þ a = -1 2 2+b =3 Þ b=4 2 1+g =0 Þ g = -1 2 \ Image of point P is (–1, 4, –1). 27. Equation of circle is 4x 2 + 4y 2 = 9

…(i)

and equation of parabola is x 2 = 4y

…(ii)

y = x2 / 4

…(iii)

By putting value of equation (iii) in equation (i), we get 2

æx 4x 2 + 4 çç è 4

y

x2=4y

2

ö ÷÷ = 9 ø

Þ

x 4 + 16x 2 - 36 = 0

Þ

( x 2 + 18) ( x 2 - 2) = 0

Þ

x 2 + 18 = 0, x 2 - 2 = 0

Þ Þ

x = - 18, x = ± 2 x=± 2

\ Required area = 2 ò

2

éx =2ê êë 2

– 2

2

0

y'

(Q x = - 18 is not possible) ( y 1 - y 2 ) dx

0

é = 2 êò êë 0

x'

2

ì 9 x 2 üù - x2 í ý ú dx 4 þ úû î 4

[As y 1 : x 2 + y 2 =

9 9 x x3 ù - x 2 + sin -1 ú 4 8 3 / 2 12 úû 0

9 , y 2 : x 2 = 4y] 4

2

é 2 9 2 2 2ù æ 2 9 2 2ö =2ê + sin -1 =ç + sin -1 ÷ sq. units. ú 8 3 6 û è 6 4 3 ø ë 4

x

199


OR Given triangle ABC, coordinates of whose vertices are A ( 4, 1), B ( 6, 6) and C ( 8, 4). Equation of AB is given by Y 6-1 5 7 – y-6= ( x - 6) or y = x - 9 B (6,6) 6-4 2 6 – 5 –

Equation of BC is given by 4-6 y-4= ( x - 8) or y = - x + 12 8-6

2 – 1 –

–

–

2

3

4

5

6 7

–

–

1

–

–

0

–

X'

A (4,1) –

Equation of AC is given by 4-1 3 y-4= ( x - 8) or y = x - 2 8-4 4

C (8,4)

4 – 3 –

X

8

Y'

\ Area of DABC =

6

ò4

8

( y AB - y AC ) dx + ò ( y BC - y AC ) dx 6

6 5 8 3 3 = ò æç x - 9 - x + 2ö÷ dx + ò æç - x + 12 - x + 2ö÷ dx 4 è2 6 è ø ø 4 4 6 æ7 8 7x = ò ç x - 7 ö÷ dx + ò æç + 14ö÷ dx 4 è4 6 è ø ø 4 6

8

é7 x 2 ù é 7x2 ù -63 é 63 ù é ù =ê - 7 xú + ê + 14xú = ê æç - 42ö÷ - (14 - 28) ú + ê( -56 + 112) - æç + 84ö÷ ú è ø è ø 2 2 8 8 ë û ë û êë úû 4 êë úû 6 63 63 = é - 42 - 14 + 28 - 56 + 112 + - 84ù = 63 - 56 = 7 sq. units. êë 2 úû 2 28. Given, the length of three sides of a trapezium other than the base is 10 cm, each i.e., AD = DC = BC = 10 cm. 10 D Let AO = NB = x cm.

C

DO = 100 - x 2 cm 1 ( AB + DC) . DO 2 1 = (10 + 2x + 10) 100 - x 2 2

10

Area ( A) =

\

A

A = ( x + 10) 100 - x 2

- x ( x + 10) + (100 - x 2 ) 100 - x 2

x 0

…(i)

Differentiating w.r.t. x, we get dA 1 = ( x + 10) . ( -2x) + 100 - x 2 . 1 dx 2 100 - x 2 =

10

=

-2x 2 - 10x + 100 100 - x 2

N

x

B

200


dA =0 dx

For maximum area,

2x 2 + 10x - 100 = 0 or x 2 + 5x - 50 = 0

Þ

Þ ( x + 10) ( x - 5) = 0 Þ x = 5, - 10 Þ x=5 Now again differentiating w.r.t. x, we get 100 - x 2 ( -4x - 10) - ( -2x 2 - 10x + 100) . d2A dx 2

=

( -2x) 2 100 - x 2

(100 - x 2 )

For x = 5 d2A dx

2

=

100 - 25 ( -20 - 10) - 0 (100 - 25)

=

75 ( -30) 75

<0

\ For x = 5, area is maximum Amax = (5 + 10) 100 - 25 cm 2

[Using equation (i)]

= 15 75 cm 2 = 75 3 cm 2 29. Question is incomplete.

Set–II x = cot -1 ( - 3 )

6. Let

p p = cot æç p - ö÷ è 6 6ø 5p Þ x= 6

Þ

cot x = -

3 = - cot

Þ

cot x = cot

5p 6

®

®

10. Given, a and b are two vectors such that ® ®

®

®

| a . b | =| a ´ b | Þ

®

®

®

®

| a || b |cos q =| a || b |sin q sin q Þ cos q = sin q Þ =1 cos q p Þ tan q = 1 Þ q= 4 11. We have to prove 1 1 1 1 p tan -1 æç ö÷ + tan -1 æç ö÷ + tan -1 æç ö÷ + tan -1 æç ö÷ = è 3ø è5 ø è7 ø è 8ø 4 1 1 1 1 LHS = tan -1 æç ö÷ + tan -1 æç ö÷ + tan -1 æç ö÷ + tan -1 æç ö÷ è 3ø è5 ø è7 ø è 8ø

201


= tan -1

= tan -1

= tan -1

Given,

Þ

Þ

é 1 +1 ù é 1+1 ù ê 3 5 ú ê ú é æ a + b öù + tan -1 ê 7 8 ú êQ tan -1 a + tan -1 b = tan -1 ç ÷ ê 1 1ú 1 1 ë è 1 - ab ø úû ê1 - ´ ú ê1 - ´ ú ë ë 3 5û 7 8û 4 3 æ ö + tan -1 æ ö ç ÷ ç ÷ è7 ø è 11 ø æ 4+ 3 ö ç ÷ p -1 æ 65 ö -1 ç 7 11 ÷ = tan ç ÷ = tan (1) = = RHS è ø 4 65 ç1 - 4 ´ 3 ÷ è 7 11 ø

tan -1

OR æx - 1ö ç ÷ + tan -1 x 2 è ø

tan -1

é x-1 x+1 ù + ê ú ê x-2 x+2 ú= p ê æx - 1ö æx + 1öú 4 ê1 - ç ÷ç ÷ú è x - 2 ø è x + 2 ø úû êë

x2 + x - 2 + x2 - x - 2 2

2

x - 4- x +1 Þ Þ Þ

æx + 1ö p ç ÷= è x + 2ø 4

2x 2 - 4 -3

p 4

=1

2x 2 - 4 = - 3 1 x2 = 2

14. We have given é2 A = ê2 ê êë 1

= tan

0 1 -1

Þ 2x 2 = 1 1 Þ x=± 2 1ù 3ú ú 0úû

For A 2 - 3A + 2I 0 1ù é 2 0 1ù é2 A 2 = ê2 1 3ú ê 2 1 3ú ê úê ú 0úû êë 1 -1 0úû êë 1 -1 0 1ù é 6 0 é2 ê ú ê 3A = 3 2 1 3 = 6 3 ê ú ê 0úû êë 3 -3 êë 1 -1

é5 = ê9 ê êë 0 3ù 9ú ú 0úû

-1 -2 -1

2ù 5ú ú -2úû

202


é2 2I = ê 0 ê êë 0

18.

ò

0

0ù 2 0ú ú 0 2úû é5 2 A - 3A + 2I = ê 9 ê êë 0 5x + 3 dx x 2 + 4x + 10

-1 -2

2ù é 6 5ú - ê 6 ú ê -2úû êë 3

-1

0 3

3ù é 2 9ú + ê 0 ú ê 0úû êë 0

-3

0 2 0

0ù é 1 0ú= ê 3 ú ê 2úû êë -3

Let 5x + 3 = A ( 2x + 4) + B = 2Ax + ( 4A + B) Comparing both sides, we get 5 2A = 5 Þ A = 2 4A + B = 3 Þ B = - 7 5 ( 2x + 4) - 7 5x + 3 \ dx = 2 dx

ò

ò

x 2 + 4x + 10

=

5 2

x 2 + 4x + 10

ò

2x + 4 2

dx - 7

x + 4x + 10

ò

I1

dx 2

x + 4x + 10 I2

For I1 Let Þ

I2 = 7

x 2 + 4x + 10 = m Þ ( 2x + 4) dx = dm 5 1 5 I1= ò dm = ´ 2 m = 5 m = 5 x 2 + 4x + 10 + C 1 2 2 m

ò

1 2

x + 4x + 10

dx = 7

ò

dx 2

( x + 2) - 4 + 10

= 7ò

dx ( x + 2) 2 + 6

= 7 log|( x + 2) + x 2 + 4x + 10|+C 2 Thus, ò

5x + 3 x 2 + 4x + 10

dx = I1 + I2

= 5 x 2 + 4x + 10 - 7 log|( x + 2) + x 2 + 4x + 10| + C, C = C 1 + C 2 y 20. y dx + x log æç ö÷ dy - 2x dy = 0 è xø Simplifying the above equation, we get é ù æ yö êë x log çè x ÷ø - 2xúû dy = - y dx

-1 -3 2

-1 ù -4 ú ú 0úû

203


dy = dx

Let

y

…(i) y 2x - x log æç ö÷ è xø y F ( x, y) = y 2x - x log æç ö÷ è xø my y F (mx, my) = = = F( x, y) æ my ö 2x - x log æ y ö ç ÷ 2mx + mx log ç ÷ è xø è mx ø

\ Function and hence the equation is homogeneous, Let y =vx dy dv Þ =v+ x dx dx Substituting in equation (i), we get dv vx v+x = dx 2x - x log v dv v dv v log v - v Þ x = -v Þ x = dx 2 - log v dx 2 - log v 2 - log v dx Þ dv = v log v - v x Integrating both sides, we get 2 - log v dx ò v log v - v dv = ò x 1 + (1 - log v) dx Þ ò v (log v - 1) dv = ò x Þ Let Þ Þ Þ Þ Þ

dv

ò v(log v - 1) - ò

dv dx =ò v x

1 dv = dm v 1 1 dx ò m dm - ò v dv = ò x log|m| - log|v| = log| x| + log| c| m log = log| cx| v m = cx Þ (log v - 1) = vcx v é ù æ yö êë log çè x ÷ø - 1úû = cy

log v - 1 = m Þ

which is the required solution.

204


23. We have given x = 1 - cos q ü p ý q= y = q - sin q þ 4 p At q = 4 p 1 , x = 1 - cos = 1 4 2 1 p 1 ö æ point is ç1 \ , ÷ è 2 4 2ø

…(i)

y=

p p p 1 - sin = 4 4 4 2

Now differentiating equation (i). w.r.t. q, we get dy dx and = sin q = 1 - cos q dq dq dy dy dq 1 - cos q \ = ´ = = cosec q - cot q dx dq dx sin q dy p p p = cosec - cot = 2 - 1 4 dx 4 4 which is slope of the tangent. \ Equation of the tangent is ì 1 ö 1 öü æp æ y-ç ÷ = ( 2 - 1) í x - ç1 ÷ý è4 ø è 2 2 øþ î At q =

= ( 2 - 1) x - ( 2 - 1) Þ

( 2 - 1) 2

1 ö 2+1- 2 2 æp y-ç ÷ = ( 2 - 1) x è4 2ø 2 3-2 2

p 1 =0 4 2 2 p 4-2 2 Þ ( 2 - 1) x - y + =0 4 2 p Þ ( 2 - 1) x - y + - 2 2 + 2 = 0 4 which is the equation of the tangent. -( 2 + 1) -1 -1 Slope of the normal = = = = - ( 2 + 1) dy dx 2 - 1 ( 2 - 1)( 2 + 1) Þ

( 2 - 1) x - y -

+

Equation of the normal is ì 1 ö 1 öü æp æ y-ç ÷ = - ( 2 + 1) í x - ç1 ÷ý è4 è 2ø 2 øþ î

205


( 2 - 1) p 1 + = - ( 2 + 1) x + ( 2 + 1) 4 2 2 p 1 2-1 Þ y- + = -( 2 + 1) x + 4 2 2 p 1 1 Þ ( 2 + 1) x + y - + =0 4 2 2 p Þ ( 2 + 1) x + y - = 0 4 which is the equation of the normal. 24. Plane through the point P (1, 1, 1) is Þ

y-

®

®

[ r - (i$ + j$ + k$)] . n = 0

...(i)

®

As plane contains the line r = ( -3i$ + j$ + 5k$) + l ( 3i$ - j$ - 5k$) ®

\

[-3i$ + j$ + 5k$ - i$ - j$ - k$]. n = 0

Þ

( -4i$ + 4k$) . n = 0

®

…(ii)

®

( 3i$ - j$ - 5k$) . n = 0

Also,

From (ii) and (iii), we get i$ j$ ®

n = -4 3

0 -1

…(iii) k$

4 = 4i$ - 8j$ + 4k$ -5

®

Substituting n in (i), we get ® [ r - (i$ + j$ + k$)] . ( 4i$ - 8j$ + 4k$) = 0 Þ Þ

®

r . ( 4i$ - 8j$ + 4k$) - ( 4 - 8 + 4) = 0

®

r . (i$ - 2j$ + k$) = 0

Which is the required equation of plane. ® r . (i$ - 2j$ + k$) = 0 contain the line ®

if

r = (i$ + 2j$ + 5k$) + m (i$ - 2j$ - 5k$) ( -i$ + 2j$ + 5k$) . (i$ - 2j$ + k$) = 0

i.e., -1 - 4 + 5 = 0, which is correct and (i$ - 2j$ + k$) . (i$ - 2j$ - 5k$) = 0 i.e.,

1 + 4 - 5 = 0, which is correct.

206


Set–III 4p ö 6. We are given sin -1 æç sin ÷ = sin -1 è 5 ø = sin -1 ®

®

7. Angle b/w a and b = sin q =

p öö æ æ ç sin çè p - ÷ø ÷ è 5 ø æ sin p ö = p ç ÷ è 5ø 5

®

®

®

®

| a ´ b| | a || b |

=

1´ 3 3´2

=

3 2

3 p = 2 3 11. ( a, b) S ( c , d) Þ a + d = b + c (i) For ( a, b) Î N ´ N a + b = b + a Þ ( a, b) S ( a, b) \ S is reflexive. (ii) Let ( a, b) S ( c , d) Þ a + d = b + c Þ d+ a= c +b Þ c +b = d+ a \ ( a, b) S ( c , d) Þ ( c , d) S ( a, b) i.e., S is symmetric. (iii) For ( a, b), ( c , d), ( e , f ) Î N ´ N Let ( a, b) S ( c , d) and ( c , d) S ( e , f ) Þ a + d = b + c and c + f = d + e Þ a+ d+ c + f =b + c + d+ e Þ a+ f =b + e Þ ( a, b) S ( e , f ) \ ( a, b) S ( c , d) and ( c , d) S ( e , f ) Þ ( a, b) S ( e , f ) \ S is transitive. \ Relation S is an equivalence relation. é 1ù 15. Given, A = ê -4 ú , B = ( -1 2 1) ê ú 3 êë úû 2 1ù é 1ù é -1 ê ú ê AB = -4 [-1 2 1] = 4 -8 -4ú ê ú ê ú 6 3úû êë 3úû êë -3 ¢ 2 1ù 4 -3 ù é -1 é -1 ê ú ê ( AB) ¢ = 4 -8 -4 = 2 -8 6ú ê ú ê ú 6 3úû 3úû êë -3 êë 1 -4 Þ

q = sin -1

207


¢ é 1ù é -1 ù 2 1) ¢ ê -4ú = ê 2ú [1 ê ú ê ú êë 3úû êë 1úû

B ¢ A ¢ = ( -1

é -1 3] = ê 2 ê êë 1

-4

4 -8 -4

\

( AB) ¢ = B ¢ A ¢. dy 17. ( x + 1) + 2xy = x 2 + 4 dx Simplifying the above equation, 2

x2 + 4 dy 2x + y= dx x 2 + 1 ( x 2 + 1) This is a linear differential equation of the form dy + Py = Q dx Here, P =

2x x2 + 1 ò

I.F. = e \ Þ

2x x2 + 1

( x 2 + 1)

dx

2

= e log ( x

( x + 1) y = ò ( x + 1) . ( x 2 + 1) . y =

x 2

2 + 1)

x2 + 4

2

( x 3 + x 2 + x + 1) Þ

x2 + 4

,Q =

2

= ( x 2 + 1) dx

= ò x 2 + 4 dx

( x + 1) 4 x 2 + 4 + log| x + x 2 + 4| + C 2 OR

dy = 2x 2 + x dx

2x 2 + x dy = dx x 3 + x 2 + x + 1

Þ dy =

2x 2 + x ( x 2 + 1) ( x + 1)

dx

Integrating both sides, we get

ò dy = ò

2x 2 + x ( x 2 + 1) ( x + 1)

By partial fraction, 2x 2 + x 2

( x + 1) ( x + 1)

=

...(i)

dx

Bx + C A + = A ( x 2 + 1) + ( Bx + C) ( x + 1) x + 1 x2 + 1

2x 2 + x = x 2 ( A + B) + x ( B + C) + ( A + C) Comparing both the sides, we get A + B = 2, B + C = 1 and

A+C = 0

-3 ù 6ú ú 3úû

208


Þ

\ (i) Þ

3 1 -1 , A= ,C = 2 2 2 3 1 é x- ù ê1 / 2 2 2 ú dx y = òê + ú 2 x +1 ú êx +1 ë û 1 1 3 x 1 = ò dx + ò dx 2 2 x+1 2 x +1 2

B=

y=

1

ò x 2 + 1 dx

1 3 1 log| x + 1| + log| x 2 + 1| - tan -1 x + C 2 4 2

20. Consider, y = cosec -1 x Differentiating both sides w.r.t. x dy -1 = Þ dx x x 2 - 1

x x2 - 1

dy = -1 dx

Again differentiating w.r.t. x, we get x x2 - 1 .

d2y dx

Þ

x ( x 2 - 1)

2

d2y dx

2

+ x2 - 1

+ ( 2x 2 - 1)

dy dy 2x +x =0 dx 2 x 2 - 1 dx dy =0 dx

23. We are given x + 2y - 3z = - 4 2x + 3y + 2z = 2 3x - 3y - 4z = 11 The matrix equation form of equations is 2 -3 ù é x ù é -4 ù é1 ê2 3 2ú ê yú = ê 2ú ê úê ú ê ú êë 3 -3 -4úû êë zúû êë 11úû i.e., where

A -1

AX = B Þ X = A -1 B 1 = Adj. A. |A| 1

2

| A| = 2

3

3

-3

-3 2 =1 -4

3 -3

2 2 -2 -4 3

2 2 -3 -4 3

3 -3

= ( -12 + 6) - 2 ( -8 - 6) - 3 ( -6 - 9) = - 6 + 28 + 45 = 67 ¹ 0

209


é- 6 Adj. A = ê 17 ê êë 13

14 5 -8

é -6 1 ê A = 14 67 ê êë -15 é -6 1 ê X= 14 67 ê êë -15

17

-1

5 9 17 5 9

Þ

é xù é 201ù é 3ù ê yú = 1 ê -134ú = ê -2ú ê ú 67 ê ú ê ú êë zúû êë 67 úû êë 1úû

\

x = 3, y = - 2, z = 1

¢ -15ù é -6 9ú = ê 14 ú ê -1úû êë -15 13ù -8 ú ú -1úû 13ù é -4ù -8 ú ê 2 ú úê ú -1úû êë 11úû

17 5 9

13ù -8 ú ú -1úû

OR a

b

c

D= b

c

a

c

a

b

=

a+b + c

a+b + c

b

c

a

c

a

b

1

1

= ( a + b + c) b

c

a

c

a

b

a+b + c

1

Applying C 1 ® C 1 - C 2 and C 2 ® C 2 - C 3 0 0 1 D = ( a + b + c) b - c

[by applying R 1 ® R 1 + R 2 + R 3 ]

c - a a = ( a + b + c)

c - a a-b b

b-c c-a c - a a-b

= ( a + b + c) [(b - c) ( a - b) - ( c - a) 2 ] = - ( a + b + c) [a 2 + b 2 + c 2 - ab - bc - ca] 1 = - ( a + b + c) [( a 2 + b 2 - 2ab) + (b 2 + c 2 - 2bc) + ( c 2 + a 2 - 2ac)] 2 1 Þ D = - ( a + b + c) [( a - b) 2 + (b - c) 2 + ( c - a) 2 ] 2 As a ¹ b ¹ c and all are positive. a + b + c > 0, ( a - b) 2 > 0, (b - c) 2 > 0 and ( c - a) 2 > 0 Hence, D is negative.

210


25. Let a cylinder of base radius r and height h1 is included in a cone of height h and semi-vertical angle a. Then AB = r , OA = ( h - h1 ), In right angled triangle OAB, AB r = tan a Þ = tan a OA h - h1 O or \

r = ( h - h1 ) tan a

V = p [( h - h1 ) tan a ] 2 . h1

(Q Volume of cylinder = pr 2 h)

V = p tan 2 a . h1 ( h - h1 ) 2

...(i)

A

Differentiating w.r.t. h1, we get dV = p tan 2 a [h1 . 2 ( h - h1 ) ( -1) + ( h - h1 ) 2 ´ 1] dh1 = p tan 2 a ( h - h1 ) [-2h1 + h - h1 ] = p tan 2 a ( h - h1 ) ( h - 3h1 ) For maximum volume V, Þ

h - h1 = 0

or

Þ

h = h1

or

dV =0 dh1 h - 3 h1 = 0 1 h1 = h 3

1 h 3 Again differentiating w.r.t. h1, we get Þ

h1 =

d 2V dh1

2

(Q h = h1 is not possible)

= p tan 2 a [( h - h1 ) ( -3) + ( h - 3h1 ) ( -1)]

1 At h1 = h 3 d 2V dh1

2

1 é ù = p tan 2 a ê æç h - hö÷ ( -3) + 0ú è ø 3 ë û = -2ph tan 2 a < 0

\ Volume is maximum for h1 = 1 Vmax = p tan 2 a . æç è3 4 = ph 3 tan 2 27

1 h 3

1 2 hö÷ æç h - hö÷ øè 3 ø a

[Using (i)]

r

B

h h1

r

EXAMINATION PAPERS – 2010 MATHEMATICS CBSE (Foreign) CLASS – XII Time allowed: 3 hours

Maximum marks: 100


Set–I SECTION–A Question number 1 to 10 carry 1 mark each. 1. Write a square matrix of order 2, which is both symmetric and skew symmetric. 3x - 4 2. If ‘f’ is an invertible function, defined as f ( x) = , write f -1 ( x). 5 3. What is the domain of the function sin -1 x ? 4. What is the value of the following determinant? 4 a b+c D= 4

b

c+a

4

c

a+b

®

®

®

®

®

®

5. Find| x |, if for a unit vector a , ( x - a ) . ( x + a ) = 15. 6. For what value of p, is (i$ + j$ + k$) p a unit vector? 7. If ò ( ax + b) 2 dx = f ( x) + c , find f ( x). 8. Evaluate:

1

ò0

1 1 + x2

dx.

9. Write the cartesian equation of the following line given in vector form : ®

r = 2i$ + j$ - 4k$ + l (i$ - j$ - k$)

10. From the following matrix equation, find the value of x : 4 ö æ 3 4ö æx + y ç ÷=ç ÷ 3 y ø è -5 6ø è -5

212


SECTION–B Question numbers 11 to 22 carry 3 marks each. 11. Consider f : R ® [-5, ¥ ) given by f ( x) = 9x 2 + 6x - 5. Show that f is invertible with æ y + 6 - 1ö ÷÷ . f -1 ( y) = çç 3 è ø OR Let A = N ´ N and * be a binary operation on A defined by ( a, b) * ( c , d) = ( a + c , b + d). Show that * is commutative and associative. Also, find the identity element for * on A, if any. a ù a ù 2b ép 1 ép 1 12. Prove the following: tan ê + cos -1 æç ö÷ ú + tan ê - cos -1 æç ö÷ ú = . è ø è b û b øû a ë4 2 ë4 2 13. Prove the following, using properties of determinants: a + b + 2c a b c

b + c + 2a

b

c

a

c + a + 2b

æ 3 Find the inverse of A = ç è -4

= 2 ( a + b + c) 3

OR -1 ö ÷ using elementary transformations. 1ø

dy d2y p p x 14. If y = log tan æç + ö÷ , show that at x = × = sec x. Also find the value of 2 è 4 2ø dx 4 dx æ 2x + 1 ö dy ÷ , find 15. If y = cos -1 çç . x÷ dx è1 + 4 ø 16. Evaluate: ò sin x . sin 2x . sin 3x dx. OR 2

Evaluate: ò 17. Evaluate: ò

x - 3x ( x - 1) ( x - 2) x tan x p

0

dx.

sec x + tan x

dx.

18. Form the differential equation representing the family of ellipses foci on x-axis and centre at the origin. 19. Find the particular solution of the following differential equation satisfying the given condition : dx ( 3x 2 + y) = x, x > 0, when x = 1, y = 1 dy OR y Solve the following differential equation: y dx + x log æç ö÷ dy = 2x dy è xø

213


®

®

®

®

20. Let a = i$ - j$ , b = 3j$ - k$ and c = 7i$ - k$. Find a vector d which is perpendicular to both a and ®

® ®

b , and c . d = 1.

21. Find the shortest distance between the following pair of lines and hence write whether the lines are intersecting or not : x-1 y+1 x+1 y- 2 = =z; = ; z=2 2 3 5 1 22. An experiment succeeds twice as often as it fails. Find the probability that in the next six trails there will be at least 4 successes.

SECTION–C Question numbers 2 to 29 carry 6 marks each. 23. A factory makes two types of items A and B, made of plywood. One piece of item A requires 5 minutes for cutting and 10 minutes for assembling. One piece of item B requires 8 minutes for cutting and 8 minutes for assembling. There are 3 hours and 20 minutes available for cutting and 4 hours for assembling. The profit on one piece of item A is Rs 5 and that on item B is Rs 6. How many pieces of each type should the factory make so as to maximise profit? Make it as an L.P.P. and solve it graphically. 24. An urn contains 4 white and 3 red balls. Let X be the number of red balls in a random draw of three balls. Find the mean and variance of X. OR In answering a question on a multiple choice test, a student either knows the answer or guesses. 3 2 Let be the probability that he knows the answer and be the probability that he guesses. 5 5 Assuming that a student who guesses at the answer will be correct with probability 1 , what is 3 the probability that the student knows the answer, given that he answered it correctly? 25. Find the coordinates of the point where the line through (3, –4, –5) and (2, –3, 1) crosses the plane determined by points A (1, 2, 3), B ( 2, 2, 1) and C ( -1, 3, 6). 5ö æ 2 -3 ç ÷ 26. If A = ç 3 2 -4÷ , find A -1 . Using A -1 solve the following system of equations : ç1 1 -2÷ø è 2x - 3y + 5z = 16 ; 3x + 2y - 4z = - 4 ; x + y - 2z = - 3 27. Using integration, find the area of the region bounded by the lines, 4x - y + 5 = 0; x + y - 5 = 0 and x - 4y + 5 = 0 OR Using integration, find the area of the following region : {( x, y) ;| x + 2| £ y £ 20 - x 2 } . 28. The lengths of the sides of an isosceles triangle are 9 + x 2 , 9 + x 2 and 18 - 2x 2 units. Calculate the area of the triangle in terms of x and find the value of x which makes the area maximum. 29. Evaluate the following : ò

3/ 2

0

| x cos px| dx.

214


Set–II Only those questions, not included in Set I, are given 2. If f : R ® R and g : R ® R are given by f ( x) = sin x and g( x) = 5x 2 , find gof ( x). 3. From the following matrix equation, find the value of x : æ1 3ö æ xö æ5 ö ç ÷ ç ÷=ç ÷ è 4 5 ø è 2ø è 6ø 11. Prove the following, using properties of determinants : b+c c+a a+b c+a

a+b

b + c = 2 ( 3abc - a 3 - b 3 - c 3 )

a+b

b+c

c+a OR

æ 3 2ö Find the inverse of the following matrix, using elementary transformations : A = ç ÷. è7 5 ø æ1 - x ö -1 æ x + 2 ö 14. Differentiate the following function with respect to x : f ( x) = tan -1 ç ÷ - tan ç ÷. è1 + x ø è 1 - 2x ø 5

17. Evaluate : ò

-5

| x + 2| dx.

21. Find the cartesian and vector equations of the plane passing through the points (0, 0, 0) and x- 4 y+ 3 z+1 (3, –1, 2) and parallel to the line = = . 1 -4 7 23. Using matrices, solve the following system of equations : 3x - 2y + 3z = - 1 ; 2x + y - z = 6 ; 4x - 3y + 2z = 5 24. Evaluate the following : ò

3/ 2

-1

| x sin px| dx.

Set–III Only those questions, not included in Set I and Set II are given 1. If f ( x) = 27 x 3 and g( x) = x 1/ 3 , find gof ( x). æ3 7. If ç è2

4ö æ xö æ19ö ÷ ç ÷ = ç ÷ , find the value of x. xø è 1 ø è15ø

13. Prove the following, using properties of determinants : a + bx 2 2

c + dx 2 2

ax + b

cx + d

u

v

p + qx 2

b

d

q

px + q = ( x - 1) a

c

p

v

w

2

4

w

u OR

æ6 Using elementary transformations, find the inverse of the following matrix : A = ç è5

5ö ÷. 4ø

215


d2y t 17. If x = a æç cos t + log tan ö÷ , y = a (1 + sin t) , find . è 2ø dx 2 1

19. Evaluate the following : ò x 2 (1 - x) n dx. 0

21. The scalar product of the vector i$ + 2j$ + 4k$ with a unit vector along the sum of vectors i$ + 2j$ + 3k$ and li$ + 4j$ - 5k$ is equal to one. Find the value of l. æ 2 ç 23. If A = ç 1 ç -2 è

1

3ö ÷ -1÷ , find A -1 . Using A -1 , solve the following system of equations : 1÷ø

3 1

2x + y + 3z = 9 ;

x + 3y - z = 2 ;

-2 x + y + z = 7

27. The sum of the perimeter of a circle and a square is K, where K is some constant. Prove that the sum of their areas is least when the side of the square is double the radius of the circle.

SOLUTIONS Set–I SECTION–A 1. Square matrix of order 2, which is both symmetric and skew symmetric is é 0 0ù ê 0 0ú ë û 2. We are given f ( x) =

5

which is invertible

3x - 4

Let

y=

Þ

5y = 3x - 4

\

3x - 4

f -1 ( y) =

5 Þ

5y + 3

and

3

x= f -1 ( x) =

5y + 4

3 5x + 4 3

3. -1 £ x £ 1 is the domain of the function sin -1 x. 4. We are given D=

4

a

b+c

4

b

c+a

4

c

a+b

Applying C 3 ® C 3 + C 2

216


D=

4

a

b+c+a

1

a

1

4

b

c + a + b = 4 ( a + b + c) 1

b

1

4

c

a+b + c

c

1

1

As we know if two columns are same in any determinant then its value is 0 \ D=0 ®

5. For a unit vector a , ®

®

®

®

®

®

( x - a ) . ( x + a ) = 15 ®

x 2 - a 2 = 15 ®

Þ

| x |2 - 1 = 15

Þ

| x |2 = 16

®

®

Þ | x |2 -| a |2 = 15 ®

[| a |2 = 1] ®

®

or | x |2 = ( 4) 2 or | x | = 4

®

a = p (i$ + j$ + k$)

6. Let,

®

®

Magnitude of a is| a | ®

| a | = ( p) 2 + ( p) 2 + ( p) 2 = ±

3p

®

As a is a unit vector, ®

\

| a |= 1 Þ

7. Given

( ax + b) 3

Þ 8.

1

ò0

ò ( ax + b) 3a

1 1 + x2

2

±

3p = 1 Þ

p=±

1 × 3

dx = f ( x) + C + C = f ( x) + C

Þ

f ( x) =

( ax + b) 3 3a

dx 1

[tan x] -1

0

= [tan -1 1 - tan -1 0] =

p 4

9. Vector form of a line is given as : ®

r = 2i$ + j$ - 4k$ + l (i$ - j$ - k$)

Direction ratios of above equation are (1, –1, –1) and point through which the line passes is (2, 1, –4). \Cartesian equation is x - x1 y - y1 z - z 1 = = a b c x- 2 y-1 z+ 4 i.e., or x - 2 = 1 - y = -z - 4 = = = 1 -1 -1

217


10. Given matrix equation 4ù é3 éx + y = ê -5 3yúû êë -5 ë

4ù 6úû

Comparing both sides we get, x + y = 3 and 3y = 6 i.e., y = 2 and x = 1 \ x = 1, y = 2.

…(i)

SECTION–B 11. Given

f : R ® [-5, ¥) , given by f ( x) = 9x 2 + 6x - 5

(i) Let f ( x 1 ) = f ( x 2 ) Þ

9x 12 + 6x 1 - 5 = 9x 22 + 6x 2 - 5

Þ Þ Þ

9 (x1 - x2 ) (x1 + x2 ) + 6 (x1 - x2 ) = 0 ( x 1 - x 2 ) [9 ( x 1 + x 2 ) + 6] = 0 x 1 - x 2 = 0 or 9( x 1 + x 2 ) + 6 = 0

Þ

9x 2 + 6x - 5 = y

Þ

9x 2 + 6x - 5 - y = 0

Þ

9x 2 + 6x - (5 + y) = 0

Þ

x=

…(i) 6 Þ x 1 = x 2 or 9 ( x 1 + x 2 ) = - 6 i.e., ( x 1 + x 2 ) = - which is not possible. 9 \ x1 = x2 So, we can say, f ( x 1 ) = f ( x 2 ) Þ x 1 = x 2 \ f is one-one. (ii) Let y Î [-5, ¥] So that y = f ( x) for some x Î R +

Þ

x=

here x =

Þ x=

-6 ± 6 1 + 5 + y 18 -1 + y + 6 3 -1 + y + 6 3

,

=

-6 ±

36 + 4( 9)(5 + y) 2´9

-1 ± y + 6 3

-1 - y + 6 3

Î R+

\ f is onto. Since function is one-one and onto, so it is invertible. -1 + y + 6 x+ 6 -1 i.e., f -1 ( x) = f -1 ( y) = 3 3

218


OR Given A = N ´ N * is a binary operation on A defined by ( a, b) * ( c , d) = ( a + c , b + d) (i) Commutativity: Let (a, b), (c, d) Î N ´ N Then ( a, b) * ( c , d) = ( a + c , b + d) = ( c + a, d + b) (Qa, b , c , d Î N , a + c = c + a and b + d = d + c) = ( c , d) * b Hence, ( a, b) * ( c , d) = ( c , d) * ( a, b) \ * is commutative. (ii) Associativity: let (a, b), (b, c), (c, d) Then [( a, b) * ( c , d)] * ( e , f ) = ( a + c , b + d) * ( e , f ) = (( a + c) + e , (b + d) + f ) (Q set N is associative) = {a + ( c + e), b + ( d + f )] = ( a, b) * ( c + e , d + f ) = (a, b) * {(c, d) * (e, f)} Hence, [( a, b) * ( c , d)] * ( e , f ) = ( a, b) * {( c , d) * ( e , f )} \ * is associative. (iii) Let (x, y) be identity element for Ú on A, Then ( a, b) * ( x, y) = ( a, b) Þ ( a + x, b + y) = ( a, b) Þ a + x = a, b + y=b Þ x = 0, y=0 But (0, 0) Ï A \ For *, there is no identity element. a ù a ù 2b ép 1 ép 1 12. tan ê + cos -1 æç ö÷ ú + tan ê - cos -1 æç ö÷ ú = è ø è b û b øû a ë4 2 ë4 2 L.H.S. Let LHS

a ù a ù ép 1 ép 1 tan ê + cos -1 æç ö÷ ú + tan ê - cos -1 æç ö÷ ú è ø è øû 4 2 b 4 2 b ë û ë a a cos -1 = x Þ = cos x b b p 1 p 1 = tan é + xù + tan é - xù êë 4 2 úû êë 4 2 úû p x p x tan + tan tan - tan 4 2 + 4 2 = p x p x 1 - tan tan 1 + tan tan 4 2 4 2 tan a + tan b tan a - tan b ù é and tan ( a - b) = êQ tan ( a + b) = ú 1 - tan a tan b 1 - tan a tan b û ë

219


x 1 + tan æç ö÷ 1 - tan è 2ø = + x 1 - tan æç ö÷ 1 + tan è 2ø

æç x ö÷ è 2ø æç x ö÷ è 2ø

2

2

x é é æ x öù æ x öù 2 é1 + tan 2 ù êë1 + tan çè 2 ÷ø úû + êë1 - tan çè 2 ÷ø úû êë 2 úû = = x x 1 - tan 2 1 - tan 2 2 2 2 æ 2 1 + tan q ö çQ cos 2q = ÷ = cos x è 1 - tan 2 q ø =

2 2b = a/b a

LHS = RHS a + b + 2c 13. L.H.S. = c

Hence Proved. a b b + c + 2a

b

a

c + a + 2b

c

Applying C 1 ® C 1 + C 2 + C 3 2 ( a + b + c) = 2 ( a + b + c)

a

b

b + c + 2a

b

a

c + a + 2b

2 ( a + b + c) 1 = 2 ( a + b + c) 1

a

b

b + c + 2a

b

a

c + a + 2b

1

Applying R 1 ® R 1 - R 2 and R 2 ® R 2 - R 3 0 - ( a + b + c) = 2 ( a + b + c) 0

( a + b + c)

- ( a + b + c)

a

c + a + 2b

1 0 -1 1

-1

1

a

c + a + 2b

OR Given

0

= 2( a + b + c +) 3 0

é 3 A=ê ë -4

We know that é 3 Þ ê -4 ë

-1 ù 1úû

A = IA -1 ù é 1 = 1úû êë 0

0

0ù A 1úû

= 2( a + b + c) 3 [1(1 - 0)] = 2( a + b + c) 3 = RHS

220


1 R 2 2 -1 / 2 ù é 1 = 1úû êë 0

1 / 2ù A 1 úû

Applying R 2 ® R 2 + 4R 1 -1 / 2 ù é 1 é1 = ê0 -1 úû êë 4 ë

1 / 2ù A 3 úû

Applying R 2 ® - R 2 é 1 -1 / 2 ù é 1 = ê0 1 úû êë -4 ë

1 / 2ù A -3úû

Applying R 1 ® R 1 + é 1 ê -4 ë

1 R 2 2 -1 ù é 1 0 ù é -1 A ê 0 1 ú = ê -4 -3úû ë û ë 0 ù -1 é -1 -1 ù A =ê ú 1úû ë -4 -3 û

Applying R 1 ® R 1 +

Þ

é1 ê0 ë

é -1 -1 ù A -1 = ê ú ë -4 -3 û p x 14. Given y = log tan æç + ö÷ è 4 2ø Þ

By differentiating of w.r.t. x, we get dy 1 p x 1 = . sec 2 æç + ö÷ . è 4 2ø 2 p xö dx æ tan ç + ÷ è 4 2ø p x cos æç + ö÷ è 4 2ø 1 = = p x p x p x p x 2 sin æç + ö÷ cos 2 æç + ö÷ 2 sin æç + ö÷ cos æç + ö÷ è 4 2ø è 4 2ø è 4 2ø è 4 2ø 1 1 1 = = = = sec x p x p cos x sin 2 æç + ö÷ sin æç + xö÷ è 4 2ø è2 ø Now again differentiating w.r.t. x, d2y dx 2

= sec x tan x

p d2y p p At x = , = sec tan = 2 2 4 dx 4 4

221


æ 2x + 1 ö ÷ Þ y = cos -1 15. Given y = cos -1 çç x÷ è1 + 4 ø Let \

é 2 x .2 1 ù ê ú x êë 1 + 4 úû

2 x = tan a Þ a = tan -1 ( 2 x ) æ 2 tan a ö ÷ = cos -1 (sin 2a ) = cos -1 y = cos -1 ç ç 1 + tan 2 a ÷ è ø

é æp öù êë cos çè 2 - 2a ÷ø úû

p p - 2a = - 2 tan -1 ( 2 x ) 2 2 By differentiating w.r.t. x, we get Þ

y=

2 . 2 x log e 2 -2 x + 1 log e 2 dy d =-2 [tan -1 ( 2 x )] = = dx dx 1 + 4x 1 + 2 2x 16.

ò sin x . sin 2x . sin 3x dx Multiplying and dividing by 2 1 1 = ò 2 sin x sin 3x sin 2x dx = ò sin x [2 sin 3x sin 2x] dx 2 2 1 = ò sin x [cos x - cos 5x] dx [Q 2 sin a sin b = cos ( a - b) - cos ( a + b)] 2 1 1 = ò (sin x cos x - cos 5x sin x) dx = ò ( 2 sin x cos x - 2 cos 5x sin x) dx 2 4 1 1 é cos 2x cos 6x cos 4x ù = ò (sin 2x - sin 6x + sin 4x) dx = ê + +C 4 4ë 2 6 4 úû cos 2x cos 6x cos 4x =+ +C 8 24 16 OR 2 2 ( x - 3x) dx ( x - 3x) dx Givenò = ( x - 1) ( x - 2) ò x 2 - 3x + 2 ( x 2 - 3x + 2) - 2

é ù 2 dx = ò ê1 ú dx x 2 - 3x + 2 x 2 - 3x + 2 úû êë dx dx = ò dx - 2 ò =x-2ò 2 2 x - 3x + 2 æç x - 3 ö÷ - 1 è 2ø 4 3 1 ù é x- - ú ê dx 1 x-a é ù 2 2 +C = log + cú = x - 2 ê log êQ ò 2 2 3 1 ú 2 a x + a ë x -a û x- + ú ê 2 2 û ë =ò

= x - 2 log

x-2 x-1

+C

222

Xam idea Mathematics – XII p

x tan x

0

sec x + tan x

17. Let I = ò a

a

0

0

…(i)

dx

As ò f ( x) dx = ò f ( a - x) dx \

p

( p - x) tan ( p - x)

0

sec ( p - x) + tan ( p - x) ( p - x) tan x dx sec x + tan x

I=ò

=ò

p

0

dx …(ii)

By adding equations (i) and (ii), we get tan x p 2I = p ò dx 0 sec x + tan x Multiplying and dividing by (sec x - tan x) , we get p tan x (sec x - tan x) 2I = p ò dx 0 sec 2 x - tan 2 x p

= p ò (sec x tan x - tan 2 x) dx 0 p

=pò

0

sec x tan x dx - p ò

p

0

p

p

sec 2 x dx + ò dx 0

p

= p [ sec x]0 - p [ tan x]0 + p [x]p0 = p( -1 - 1) - 0 + p( p - 0) = p ( p - 2) p ( p - 2) 2 18. The family of ellipses having foci on x-axis and centre at the origin, is given by Þ

2I = p ( p - 2)

x2

Þ I=

+

y2

=1 a2 b 2 Differentiating w.r.t. x, we get 2x 2y æ dy ö + ç ÷=0 Þ a 2 b 2 è dx ø

2y dy 2x =2 dx b a2 æ dy ö yç ÷ è dx ø -b 2 = x a2

dy dx = - x Þ Þ b2 a2 Again by differentiating w.r.t. x, we get é d 2 y æ dy ö 2 ù æ dy ö x êy + ç ÷ ú - çy × ÷ 2 è dx ø ú è dx ø êë dx û =0 2 x The required equation is \ y

xy

d2y dx 2

2

dy æ dy ö +xç ÷ -y =0 è dx ø dx

223


19. We are given ( 3x 2 + y) Þ

Þ Þ

dx = x, x > 0 dy

dx x = 2 dy 3x + y 2 dy 3x + y y = = 3x + dx x x dy 1 - y = 3x dx x

This is a linear differential equation of the form Here P = I.F. = e

1 , Q = 3x x -

1

òx

dx

= e-

log x

= e log

x -1

y 1 = ò 3x dx = 3 ò dx x x y Þ = 3x + C Þ y = 3x 2 + Cx x But, it is given when x = 1, y = 1 Þ 1= 3+C Þ C=-2

\

\

y = 3x 2 - 2x

OR y Given y dx + x log æç ö÷ dy = 2x dy è xø é ù æ yö Þ êë x log çè x ÷ø - 2xúû dy = - y dx Þ

dy = dx

Þ

y 2x - x log æç ö÷ è xø

dy dv =v+ x dx dx dv vx v+x = vx dx 2x - x log æç ö÷ èxø dv vx x = -v dx x ( 2 - log v)

Let y = vx, \

y

Þ

=

1 x

dy + Py = Q dx

224


dv v - 2v + v log v v log v - v = = dx 2 - log v 2 - log v 2 - log v dx dv = v log v - v x 2 - log v dx ò v log v - v dv = ò x 1 + (1 - log v) dx ò v (log v - 1) dv = ò x

Þ

x

Þ Þ Þ

dx

ò v(log v - 1) - ò

Þ

...(i)

1 dv = dt v 1 1 dx ò t dt - ò v dv = ò x log|t| - log|v| = log| x| + log| c| t t log = log| cx| Þ = cx v v log v - 1 = cx v é ù æ yö êë log çè x ÷ø - 1úû = cx y x é ù æ yö êë log çè x ÷ø - 1úû = cy, which is the required solution.

log v - 1 = t

Let

dv dx =ò v x

\ (i) Þ Þ Þ Þ

Þ

Þ ®

20. Given a = i$ - j$,

Þ

®

b = 3j$ - k$,

®

c = 7i$ - k$

®

®

®

Q vector d is perpendicular to both a and b \

®

®

Also

\

®

®

i$

j$

k$

d = l ( a ´ b ) = l 1 -1 0 = l (i$ + j$ + 3k$) 0 3 -1

Þ

Þ

®

d is along vector a ´ b

® ®

c . d =1

l(7 + 0 - 3) = 1 ®

d =

1 $ $ (i + j + 3k$) 4

Þ Þ

(7i$ - k$) . l (i$ + j$ + 3k$) = 1 1 l= 4

225


21. Given, pair of lines x-1 y+1 = = z and 2 3 In vector form equations are

x+1 5

=

y-2 1

=

z-2 0

®

r = (i - j$) + m ( 2i$ + 3j$ + k$)

and

®

r = ( -i$ + 2j$ + 2k$) + l (5i$ + j$) ®

®

b 1 = 2i$ + 3j$ + k$

a 1 = i$ - j$ ,

®

®

a 2 = - i$ + 2j$ + 2k$ , ®

b 2 = 5i$ + j$

®

a 2 - a 1 = - 2i$ + 3j$ + 2k$ i$

j$

k$

b1 ´ b2 = 2

3

1 = - i$ + 5j$ - 13k$

5

1

0

®

®

®

\

®

®

®

^ ( a 2 - a 1 ) × (b 1 ´ b 2 ) = ( -2i$ + 3j$ + 2k) .( -i$ + 5j$ - 13k$)

= 2 + 15 - 26 = - 9 ®

As we know shortest distance =

®

®

®

®

®

( a 2 - a 1 ) . (b 1 ´ b 2 ) |b 1 ´ b 2 |

=

-9 2

2

( -1) + (5) + ( -13) =

2

=

-9 9 units = 195 195

Lines are not intersecting as the shortest distance is not zero. 22. An experiment succeeds twice as often as it fails. 2 \ p = P (success) = 3 1 and q = P (failure) = 3 no. of trials = n = 6 By the help of Binomial distribution, 2 r 1 6 -r P (r) = 6 Cr æç ö÷ æç ö÷ è 3ø è 3ø P (at least four success) = P ( 4) + P (5) + P ( 6)

-9 1 + 25 + 169

226


1 2 2 4 1 2 5 2 6 = 6 C 4 æç ö÷ æç ö÷ + 6 C5 æç ö÷ æç ö÷ + 6 C 6 æç ö÷ è 3ø è 3ø è 3ø è 3ø è 3ø 2 4 1 2 4 = æç ö÷ é 6 C 4 + 6 C5 + 6 C 6 ù úû è 3 ø êë 9 9 6 16 31 496 2 4 15 2 4 = æç ö÷ é + ´ 6 + ù = ´ = è 3 ø êë 9 9 9 ûú 81 9 729

SECTION–C 23. Let the factory makes x pieces of item A and B by pieces of item. Time required by item A (one piece) Y cutting = 5 minutes, assembling = 10 minutes 40 Time required by item B (one piece) cutting = 8 minutes, assembling 30 = 8 minutes (0,25) Total time (8,20) 20 cutting = 3 hours & 20 minutes, assembling = 4 hours 10 Profit on one piece (24,0) item A = Rs 5, item B = Rs 6 x' 0 10 20 30 40 Thus, our problem is maximized z = 5x + 6y y' 10x+8y=240 Subject to x ³ 0, y ³ 0 5x + 8y £ 200 10x + 8y £ 240 From figure, possible points for maximum value of z are at (24, 0), (8, 20), (0, 25). at (24, 0), z = 120 at (8, 20), z = 40 + 120 = 160 (maximum) at (0, 25), z = 150 \ 8 pieces of item A and 20 pieces of item B produce maximum profit of Rs 160. 24. Let X be the no. of red balls in a random draw of three balls. As there are 3 red balls, possible values of X are 0, 1, 2, 3. 3 C 0 ´ 4C 3 4 ´ 3 ´ 2 4 P ( 0) = = = 7 7 ´ 6 ´ 5 35 C 3

P (1) =

C1 ´ C2 7

3

P ( 2) =

3 4

C3

C 2 ´ 4C 1 7

=

C3

=

3´6´6 7 ´ 6´5 3´4´6 7 ´ 6´5

=

18 35

=

12 35

X

50 5x+8y=200

227


P ( 3) =

C 3 ´ 4C 0 7

=

C3

1´1´ 6 7 ´ 6´5

=

1 35

For calculation of Mean & Variance X

P(X)

XP (X)

X2 P (X)

0

4/35

0

0

1

18/35

18/35

18/35

2

12/35

24/35

48/35

3

1/35

3/35

9/35

Total

1

9/7

15/7

Mean = S XP(X) =

9 7

Variance = SX 2 . P(X) - ( SX. P(X)) 2 =

15 81 24 = 7 49 49

OR Let A, B and and E be the events defined as follows: A : Student knows the answer B : Student guesses the answer E : Student answers correctly 3 2 Then, P ( A) = , P ( B) = , P (E / A ) = 1 5 5 1 P(E / B) = 3 Using Baye’s theorem, we get 3 3´ 3 9 5 P ( A / E) = = = = P ( A) . P (E / A ) + P ( B) P (E / B) 3 + 2 ´ 1 9 + 2 11 5 5 3 25. The line through (3, –4, –5) and (2, –3, 1) is given by x- 3 y+ 4 z+5 = = 2 - 3 -3 + 4 1 + 5 x- 3 y+ 4 z+5 ...(i) Þ = = -1 1 6 The plane determined by points A (1, 2, 3), B ( 2, 2, 1) and C ( -1, 3, 6) x-1 y- 2 z- 3 P ( A) . P (E / A )

2-1

2-2 1- 3 =0

-1 - 1

3-2 6- 3

228


x-1 y- 2 z- 3 Þ

Þ

( x - 1)

1

0

-2

1

-2 =0 3

-2 1 - ( y - 2) 3 -2

0 1

-2 1 + (z - 3) 3 -2

0 =0 1

Þ ( x - 1) ( 2) - ( y - 2) ( -1) + (z - 3) (1) = 0 Þ 2x - 2 + y - 2 + z - 3 = 0 Þ 2x + y + z - 7 = 0 P ( -m + 3, m - 4, 6m - 5) is the general point for line (i). If this point lies on plane (ii), we get -2m + 6 + m - 4 + 6m - 5 - 7 = 0 Þ m = 2 \ P (1, - 2, 7) is the point of intersection. -3 5ù é2 ê 26. If A= 3 2 -4 ú ê ú 1 -2úû êë 1 1 A -1 = Adj. A | A| 2 -3

...(ii)

5

| A| = 3

2

- 4 = 2 ( - 4 + 4) + 3 ( -6 + 4) + 5 ( 3 - 2)

1

1

-2

= 2 (0) + 3 (– 2 ) + 5 (1) = – 1 ¹ 0 2 1 ù é 0 -1 2ù é 0 ê ú ê Adj. A = -1 -9 -5 = 2 -9 23ú ê ú ê ú 23 13úû êë 1 -5 13úû êë 2 -1 2ù é 0 1 -2 ù é0 1 ê -1 ú ê A = 2 -9 23 = -2 9 -23ú ú ê ú -1 ê -5 13úû êë -1 5 -13úû êë 1

…(i)

Given equations are 2x - 3y + 5z = 16 3x + 2y - 4z = - 4 x + y - 2z = - 3 Matrix form is -3 5ù é xù é 16ù é2 ê3 2 -4 ú ê y ú = ê -4 ú ê úê ú ê ú 1 -2úû êë z úû êë -3úû êë 1 AX = B Þ

X = A -1 B

...(ii)

229


From equations (i) and (ii), we get 1 -2ù é 16ù é xù é 0 ê y ú = ê -2 9 -23ú ê -4ú ê ú ê úê ú 5 -13úû êë -3úû êë z úû êë -1 é xù é 2 ù ê yú = ê 1 ú ê ú ê ú êë z úû êë 3úû Þ x = 2, y = 1, z = 3 27. We have given ...(i) 4x - y + 5 = 0 ...(ii) x+ y-5= 0 ...(iii) x - 4y + 5 = 0 By solving equations (i) and (iii), we get (–1, 1) and by solving (ii) and (iii), we get (3, 2) \ Area of region bounded by the lines is given by: 0 ì 3ì æ x + 5 öü æ x + 5 öü ò-1 íî( 4x + 5) - çè 4 ÷øýþ dx + ò0 íî(5 - x) - çè 4 ÷øýþ dx =ò

0

-1

é 15x + 15 ù dx + 3 ò0 êë 4 4 úû

Y 6

4

é 15 - 5x ù dx êë 4 4 úû

0

x – 4y + 5 = 0

3 2

3

é 15x 2 15x ù é 15x 5x 2 ù =ê + ú +ê ú 4 úû 4 8 úû êë 8 -1 êë 0 X' 15 15 45 45 = 0 - æç - ö÷ + æç - ö÷ - 0 è8 4ø è4 8ø 15 45 15 sq. unit. = + = 8 8 2 OR

4x – y + 5 = 0

5

(-1,1)

1

–6 –5 –4 –3 –2 –1 0 –1

1

2

3

4

5

6

X

x+y–5=0

–2 –3 –4 Y'

Given region is {( x, y) :| x + 2| £ y £ 20 - x 2 .} It consists of inequalities y ³| x + 2|

y

y £ 20 - x 2

5

Plotting these inequalities, we obtain the adjoining shaded region. Solving y=x+2 and

y 2 = 20 - x 2

Þ

( x + 2) 2 = 20 - x 2

Þ

2x 2 + 4x - 16 = 0

4

|

x'

|

–4 –3

|

–2

|

–1

0 y'

|

|

|

|

1

2

3

4

x

230


or ( x + 4) ( x - 2) = 0 Þ x = - 4, 2 The required area =ò

2

-4

éx =ê ë2

20 - x 2 dx - ò

-2

-4

20 - x 2 +

= 4 + 10 sin -1

- ( x + 2) dx - ò

2

-2

20 sin -1 2

( x + 2) dx -2

2

2 é x2 ù é x2 ù x ù + + 2 x + 2 x ê ú ê ú 20 úû -4 êë 2 úû -4 êë 2 úû -2

1 æ 2 ö + 4 + 10 sin -1 ç ÷ + [2 - 4 - 8 + 8] - [2 + 4 - 2 + 4] è 5ø 5

1 2 ö æ = 8 + 10ç sin -1 + sin -1 ÷-2-8 è 5 5ø 1 2 ö æ = - 2 + 10 ç sin -1 + sin -1 ÷ è 5 5ø é 1 4 2 1ù = - 2 + 10 sin -1 ê 1- + 1- ú 5 5û 5 ë 5 1 4 = - 2 + 10 sin -1 é + ù= - 2 + 10 sin -1 1 ëê 5 5 ûú p = - 2 + 10 = (5p - 2) sq. units. 2 28. As given, the lengths of the sides of an isosceles triangle are 9 + x 2 , 9 + x 2 and 18 - 2x 2 units. Using Heron's formula, we get 2s = 9 + x 2 + 9 + x 2 + 18 - 2x 2 = 36

Þ s = 18

A = 18 (18 - 9 - x 2 ) (18 - 9 - x 2 ) (18 - 18 + 2x 2 ) = 18( 9 - x 2 )( 9 - x 2 )( 2x 2 ) A = 6x ( 9 - x 2 ) A = 6 ( 9x - x 3 )

...(i)

Differentiating (i) w.r.t. x dA = 6( 9 - 3x 2 ) dx dA For maximum A, =0 dx Þ

9 - 3x 2 = 0

Þ

x=±

3

Now again differentiating w.r.t. x d2A dx 2 At x =

3,

d2A dx 2

= 6 ( -6x) = - 36x = - 36 3 < 0

231


d2A

= 36 3 > 0 dx 2 3 , area is maximum.

At x = - 3 , \ For x = 29.

3/ 2

ò0

| x cos px| dx

As we know that

\

p , nÎZ 2 1 3 x= , 2 2

cos x = 0

Þ x = ( 2n - 1)

cos px = 0

Þ

1 , x>0 2 cos px > 0 Þ x cos px > 0 1 3 0 2 2 cos px < 0 Þ x cos px < 0

For 0 < x <

For

\

3/ 2

ò0

| x cos px| dx

=ò

1/ 2

0

x cos p x dx + ò

3/ 2

( - x cos px) dx

1/ 2

I

II 1/ 2

é sin px ù = êx p úû 0 ë

-ò

1/ 2

0

1.

sin px

1/ 2

1 éx ù = ê sin px + cos pxú 2 p p ë û0

p

3/ 2 3 / 2 sin px é x sin px ù dx - ê -ò dx ú 1 /2 p p ë û 1/ 2 3/ 2

1 éx ù - ê sin px + cos pxú 2 p p ë û 1/ 2

1 ö æ 3 1 ö 5 1 æ 1 =ç +0÷= ÷ - çè 2 ø è 2p 2p 2p 2p p 2 p ø

Set–II f : R ® R and g : R ® R defined by

2. Given

f ( x) = sin x and g( x) = 5x 2 gof ( x) = g [ f ( x)] = g (sin x) = 5 (sin x) 2 = 5 sin 2 x

\ 3. Given :

Þ

…(i)

é1 3ù é xù é5 ù ê 4 5 ú ê 2ú = ê 6ú ë ûë û ë û ( 1 ) ( x ) + ( 3 ) ( 2) ù é 5 ù é ê 4 ( x) + (5) ( 2) ú = ê 6ú ë û ë û

Þ

é x + 6 ù é5 ù ê 4x + 10ú = ê 6ú ë û ë û

232


Comparing both sides, we get x+ 6=5 Þ x= -1 Also, 4x + 10 = 6 or Þ 4x = - 4 x= -1 \ x= -1 b+c c+a a+b 11. We have to prove c + a a+b b + c = 2 ( 3abc - a 3 - b 3 - c 3 ) a+b L.H.S

b+c

c+a

b+c

c+a

a+b

= c+a

a+b

b+c

a+b

b+c

c+a

é 2( a + b + c) = ê c+a ê êë a + b

2( a + b + c) a+b b+c

a+b + c =2

a+b + c

2( a + b + c) ù b + c ú [By applying R 1 ® R 1 + ( R 2 + R 3 )] ú c + a úû a+b + c

c+a

a+b

b+c

a+b

b+c

c+a

Applying R 2 ® R 2 - R 1 ,R 3 ® R 3 - R 1 1

1

1

1

1

1

2

= 2 ( a + b + c) -b

-c

- a = 2 ( -1) ( a + b + c) b

c

a

-c

-a

-b

a

b

c

Applying C 1 ® C 1 - C 2 , C 2 ® C 2 ® C 3 0 0 1 = 2 ( a + b + c) b - c

c -a

c-a

a-b

é b-c c-a ù a = 2 ( a + b + c) ê ú ë c - a a-b û b

= 2 ( a + b + c) [(b - c) ( a - b) - ( c - a) ( c - a)] = 2 ( a + b + c) ( - a 2 - b 2 - c 2 + ab + bc + ca) = 2 ( 3abc - a 3 - b 3 - c 3 ) = RHS OR We are given é 3 2ù A=ê ú ë7 5 û Þ

A = IA é 3 2ù é1 0ù ê7 5 ú = ê 0 1 ú A ë û ë û

Hence Proved.

233


é7 ê3 ë é1 ê3 ë

Þ Þ

5ù é0 1ù = A 2úû êë 1 0úû 1 ù é -2 1 ù = A 2úû êë 1 0úû

é 1 1 ù é -2 ê 0 -1 ú = ê 7 ë û ë 0ù é 5 é1 ê 0 -1ú = ê7 ë û ë é 1 0ù é 5 ê 0 1 ú = ê -7 ë û ë

Þ Þ Þ Hence,

é 5 A -1 = ê ë -7

[By applying R 1 « R 2 ] [By applying R 1 ® R 1 - 2R 2 ]

1ù A -3úû

[By applying R 2 ® R 2 - 3R 1 ]

-2 ù A -3úû

[By applying R 1 ® R 1 + R 2 ]

-2 ù A 3úû -2 ù 3úû

[By applying R 2 ® - R 2 ]

æ1 - x ö -1 æ x + 2 ö 14. f ( x) = tan -1 ç ÷ - tan ç ÷ è1 + x ø è 1 - 2x ø æ 1- x ö -1 æ x + 2 ö = tan -1 ç ÷ - tan ç ÷ è 1 + x.1 ø è 1 - 2. x ø = (tan -1 1 - tan -1 x) - (tan -1 x + tan -1 2)

æ ö -1 a - b = tan -1 a - tan -1 b ÷ çQ tan è ø 1 + ab

= tan -1 1 - tan -1 2 - 2 tan -1 x Differentiating w.r.t. x 2 f ¢( x) = 1 + x2 ì ( x + 2) if x + 2 > 0 i.e., x > -2 x+ 2 =í î -( x + 2) if x + 2 < 0 i.e., x < -2

17. \

5

-2

ò-5 x + 2 dx = ò-5

5

- ( x + 2) dx + ò

-2

-2

( x + 2) dx 5

é x2 ù é x2 ù = ê- 2xú + ê + 2xú êë 2 úû -5 êë 2 úû 2 4 25 25 4 = é - + 4ù - é + 10ù + é + 10ù - é - 4ù úû ëê 2 ûú ëê 2 ûú ëê 2 ûú ëê 2 5 45 =2+ + + 2 = 29 2 2 21. Plane passing through the point (0, 0, 0) is a ( x - 0) + b ( y - 0) + c (z - 0) = 0 Plane (i) passes through the point (3, –1, 2) \ 3a - b + 2c = 0

…(i) …(ii)

234


Also, Plane (i) is parallel to the line x- 4 y+ 3 z+1 = = 1 -4 7 \ a - 4b + 7 c = 0 From equations (i), (ii) and (iii) x

Þ

x

-1 -4

…(iii) y

z

3

-1

2 =0

1

-4

7

2 3 2 3 -y +z 7 1 7 1

-1 =0 -4

Þ x [-7 + 8] - y [21 - 2] + z [-12 + 1] = 0 Þ x - 19y - 11z = 0 and in vector form, equation is ®

r . (i$ - 19j$ - 11k$) = 0

SECTION–C 23. 3x - 2y + 3z = - 1 2x + y - z = 6 4x - 3y + 2z = 5 Now the matrix equation form of above three equations is 3 ù é x ù é -1 ù é 3 -2 ê2 1 -1 ú ê y ú = ê 6 ú ê úê ú ê ú 2úû êë zúû êë 5úû êë 4 -3 i.e.,

AX = B

we know that

Þ X = A -1 B 1 A -1 = Adj. A | A| -2

3

| A| = 2

1

-1

4

-3

2

3

=3

1 -3

-1 2 +2 2 4

-1 2 +3 2 4

= - 3 + 16 - 30 = - 17 ¹ 0 ¢ -8 -10ù é -1 é -1 ê ú Adj. A = -5 -6 1 = ê -8 ê ú ê 9 7 úû êë -1 êë -10

1 -3

-5 -6 1

-1 ù 9ú ú 7 úû

235


A

-1

é -1 1 ê = -8 -17 ê êë -10

-5 -6 1

é -1 1 ê -8 X = A B= 17 ê êë -10 é xù é -34ù é 2ù ê yú = - 1 ê 17 ú = ê -1ú ê ú ú ê ú 17 ê êë zúû êë 51úû êë -3úû

-1 ù 9ú ú 7 úû -5

-1

-6 1

-1 ù é -1 ù 9ú ê 6ú úê ú 7 úû êë 5úû

By comparing both sides, we get x = 2, y = - 1, z = - 3 24.

3/ 2

ò-1

| x sin p x| dx

As we know sin q = 0 Þ q = np , n Î Z \ sin px = 0 Þ x = 0, 1, 2, K For -1 < x < 0, x < 0, sin px < 0 Þ x sin px > 0 For 0 < x < 1, x > 0, sin px > 0 Þ x sin px > 0 3 For 1 < x < , 2 x > 0, sin px < 0 Þ x sin px < 0 \

3/ 2

ò-1

| x sin px| dx

=ò

1

-1

x sin px dx + ò

3/ 2

1

( - x sin px) dx

1

3/ 2

- cos px 1 é (cos px) ù é - cos px ù = êx × - ò 1. dx - ê x × ú úû p û -1 -1 p p ë ë 1 1

+ò

3/ 2

1

1.

cos px p

3/ 2

1 1 é x ù é x ù = ê - cos px + sin pxú - ê - cos px + sin pxú 2 2 p p ë p û -1 ë p û1

1 1ù 1 1 1 1ù 1 3 1 + 3p é 1 1 é . =ê + + + ú = é + 0 + - 0ù - ê 0 - ú= + = 2 2 2 ê ú pû ëp p pû p p û ë p p2 ë p p p

Set–III 1. Given f ( x) = 27 x 3 and g( x) = x 1/ 3 ( gof ) ( x) = g [ f ( x)] = g [27 x 3 ] = [27 x 3 ] 1/ 3 = 3x

dx

236


7. Given, é3 ê2 ë

Þ

é 3 ( x) + 4 (1) ù é19ù Þ ê ú=ê ú ë( 2) ( x) + ( x) (1) û ë15û

4ù é xù é19ù = xúû êë 1 úû êë15úû é 3x + 4ù é19ù ê 3x ú = ê15ú ë û ë û

Þ

Comparing both sides, we get and 3x + 4 = 19 3x = 15 Þ 3x = 19 - 4, 3x = 15 Þ 3x = 15, x=5 \ x=5 13. We have to prove a + bx 2

c + dx 2

2

2

ax + b

cx + d

u

v a + bx 2

L.H.S

p + qx 2

d

q

px + q = ( x - 1) a

c

p

v

w

4

w c + dx 2

2

=

b

2

u p + qx 2

2

ax + b

cx + d

px 2 + q

u

v

w

Multiplying R 1 by x 2 and dividing the determinant by x 2 =

1 x

2

ax 2 + bx 4

cx 2 + dx 4

px 2 + qx 4

ax 2 + b

cx 2 + d

px 2 + q

u

v

w

Applying R 1 ® R 1 - R 2 =

1 x

=

2

b ( x 4 - 1)

q ( x 4 - 1)

ax + b

cx + d

px 2 + q

u

v

w

x4 - 1 x2

2

b

2

2

ax + b

cx + d

px + q

u

v

w

x 2 ( x 4 - 1) x2

q

d

2

Applying R 2 ® R 2 - R 1 b x4 - 1 = ax 2 x2 u =

d ( x 4 - 1)

2

q

d cx

2

px 2

v

w

b

d

q

b

d

a

c

p = ( x 4 - 1) a

c

q p = RHS

u

v

w

v

w

u

237


é6 A=ê ë5

Given

OR 5ù 4úû

We can write A = IA 1 1 é ù é1 0ù ê5 4ú = ê 0 1 ú A ë û ë û [By applying R 1 ® R 1 - R 2 ] é1 1 ù é 1 -1ù Þ A ê5 4ú = ê 0 1úû ë û ë [By applying R 2 ® R 2 - 5R 1 ] 1 ù é 1 -1 ù é1 A ê 0 -1 ú = ê -5 6úû ë û ë [By applying R 1 ® R 1 + R 2 ] 0 ù é -4 5 ù é1 ê 0 -1 ú = ê -5 6 ú A ë û ë û [By applying R 2 é1 ê0 ë é1 ê0 ë \ 17. Given

® - R2 ] 0 ù é -4 = 1úû êë 5

0 ù -1 é -4 A =ê -1úû ë 5 é -4 A -1 = ê ë 5

5ù A -6úû 5ù -6úû 5ù -6úû

t x = a é cos t + log tan ù êë 2 úû

…(i)

…(ii) y = a (1 + sin t) Differentiating equation (i) w.r.t. t ì ü ì ü ï ï ï dx 1 1 2 t 1ï = a í - sin t + . sec × ý = a í - sin t + ý t t t dt 2 2ï ï ï tan 2 sin cos ï 2 2 2þ î þ î Þ

cos 2 t ì dx 1 ü ì - sin 2 t + 1 ü = a í - sin t + ý= a ý = aí sin t dt sin t þ î sin t î þ

Differentiating equation (ii), w.r.t. t dy = a ( 0 + cos t) = a cos t dt dy dy dt a cos t ´ sin t Now, = ´ = = tan t dx dt dx a cos 2 t

238


Now again differentiating w.r.t. x, we get d2y

dt d (tan t) = sec 2 t . dx dx dx sin t 1 = sec 2 t . = sec 4 t . sin t 2 a cos t a 2

=

1

19. Let I = ò x 2 (1 - x) n dx 0

1

I = ò (1 - x) 2 [1 - (1 - x)]n dx

Þ

0 1

1

0

0

a

a

0

0

(Q ò f ( x) dx = ò f ( a - x) dx)

= ò (1 - 2x + x 2 ) xn dx = ò ( xn - 2xn +

1

+ xn + 2 ) dx

1

é xn + 1 1 2 1 xn + 2 xn + 3 ù =ê - 2. + + ú = n + 1 n + 2 n + 3 n + 1 n + 2 n + 3 êë úû 0 (n + 2)(n + 3) - 2(n + 1)(n + 3) + (n + 1)(n + 2) = (n + 1)(n + 2)(n + 3) =

2 n 2 + 5n + 6 - 2n 2 - 8n - 6 + n 2 + 3n + 2 = (n + 1)(n + 2)(n + 3) (n + 1)(n + 2)(n + 3)

21. Sum of given vectors is ®

r = i$ + 2j$ + 3k$ + li$ + 4j$ - 5k$ = (1 + l) i$ + 6j$ - 2k$

We have given (i$ + 2j$ + 4k$) . r$ = 1 Þ

[(1 + l) i$ + 6j$ - 2k$] (i$ + 2j$ + 4k$) . =1 (1 + l) 2 + 36 + 4

Þ

(1 + l) + 12 - 8 = (1 + l) 2 + 40

Þ

l + 5 = (1 + l) 2 + 40

Squaring both sides, we get l2 + 10l + 25 = 1 + 2l + l2 + 40 Þ 23. Given and

8l = 16

Þ l=2 1 é 2 ê A= 1 3 ê 2 1 êë 2x + y + 3z = 9 x + 3y - z = 2 -2 x + y + z = 7

3ù -1 ú ú 1úû …(i) …(ii) …(iii)

239


A -1 =

As we know

| A| =

1 Adj. A | A|

2

1

1

3

-2

1

3 -1 = 2 1

3 1

-1 1 -1 1 -2

= 2 ( 4) - 1 ( -1) + 3 (7) = 30 ¹ 0 1 7 ù é4 2 é 4 ê ú ê Adj. A = 2 8 -4 = 1 8 ê ú ê 5 5úû êë7 -4 êë -10 A

-1

é4 1 ê = 1 30 ê 7 êë

2 8 -4

-1 1 +3 1 -2

3 1

-10ù 5ú ú 5úû

-10ù 5ú ú 5úû

Matrix equation form of equations (i), (ii), (iii), is given by 1 3ù é xù é 9ù é 2 ê 1 3 -1 ú ê y ú = ê 2 ú ê úê ú ê ú 1 1úû êë z úû êë7 úû êë -2 i.e., Þ

Þ

AX = B Þ X = A -1 B 2 é4 1 ê X= 1 8 30 ê 7 4 êë é xù é -30ù ê yú = 1 ê 60ú ê ú 30 ê ú êë z úû êë 90úû

-10ù 5ú ú 5úû Þ

é 9ù ê 2ú ê ú êë7 úû é x ù é -1 ù ê yú = ê 2ú ê ú ê ú êë z úû êë 3úû

By comparing both sides, we get x = - 1, y = 2, z = 3 27. Let side of square be a units and radius of a circle be r units. It is given, k - 4a \ 4a + 2pr = k where k is a constant Þ r = 2p Sum of areas, A = a 2 + pr 2 2

Þ

1 é k - 4a ù 2 2 A = a2 + p ê ú = a + 4p ( k - 4a) 2 p ë û

Differentiating w.r.t. x 2 ( k - 4a) dA 1 = 2a + × 2 ( k - 4a) . ( -4) = 2a da 4p p

…(i)

240


dA =0 da 2 ( k - 4a) Þ 2a =0 p 2 ( k - 4a) 2 ( 2pr) Þ 2a = Þ 2a = p p [As k = 4a + 2pr given] Þ a = 2r Now again differentiating equation (i) w.r.t. x For minimum area,

d2A da at a = 2p,

2

d2A 2

=2-

2 8 ( -4) = 2 + p p

=2+

8 >0 p

da \ For ax = 2r , sum of areas is least. Hence, sum of areas is least when side of the square is double the radius of the circle.

EXAMINATION PAPERS – 2011 CBSE (Delhi) Set–I

Time allowed: 3 hours

Maximum marks: 100


SECTION–A Question numbers 1 to 10 carry one mark each. 1. State the reason for the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} not to be transitive. 1 ù ép 2. Write the value of sin ê – sin –1 æç – ö÷ ú è 2 øû ë3 i 3. For a 2 × 2 matrix, A = [aij], whose elements are given by aij = , write the value of a12. j é5 – x x + 1ù 4. For what value of x, the matrix ê is singular? 4 úû ë 2 é2 5 ù 5. Write A–1 for A = ê ú ë1 3û 6. Write the value of ò sec x (sec x + tan x) dx 7. Write the value of ò

dx 2

x + 16

.

8. For what value of ‘a’ the vectors 2i$ – 3j$ + 4k$ and ai$ + 6j$ – 8k$ are collinear?

242

Xam idea Mathematics–XII

9. Write the direction cosines of the vector – 2i$ + j$ – 5k$. 10. Write the intercept cut off by the plane 2x + y – z = 5 on x–axis.

SECTION–B Question numbers 11 to 22 carry 4 marks each. 11. Consider the binary operation* on the set {1, 2, 3, 4, 5} defined by a * b = min. {a, b}. Write the operation table of the operation *. 12. Prove the following: é 1 + sin x + 1 – sin x ù x æ pö cot – 1 ê ú = , x Î ç 0, ÷ è 4ø 2 1 + sin x – 1 – sin x ëê ûú OR æ xö æ x – yö Find the value of tan –1 ç ÷ – tan – 1 ç ÷ è yø è x + yø 13. Using properties of determinants, prove that – a 2 ab ba –b 2 ca

cb

ac bc = 4a 2 b 2 c 2 –c2

14. Find the value of ‘a’ for which the function f defined as ì a sin p ( x + 1), x £ 0 ï f ( x) = í tan x2– sin x , x>0 ï x3 î is continuous at x = 0. x2 + 1 15. Differentiate x x cos x + w. r. t. x x2 – 1 OR d2y If x = a (q – sin q), y = a (1 + cos q), find dx 2 16. Sand is pouring from a pipe at the rate of 12 cm3/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm? OR Find the points on the curve x 2 + y 2 – 2x – 3 = 0 at which the tangents are parallel to x–axis. 17. Evaluate: ò

5x + 3 x 2 + 4x + 10

dx

243


OR Evaluate: ò

2x ( x 2 + 1) ( x 2 + 3)

dx

18. Solve the following differential equation: e x tan y dx + (1 – e x ) sec 2 y dy = 0 19. Solve the following differential equation: dy cos 2 x + y = tan x. dx ®

®

®

®

®

20. Find a unit vector perpendicular to each of the vectors a + b and a – b , where a ® = 3i$ + 2j$ + 2k$ and b = i$ + 2j$ - 2k$. 21. Find the angle between the following pair of lines: –x+2 y–1 z+ 3 x + 2 2y – 8 z – 5 and = = = = –2 7 –3 –1 4 4 and check whether the lines are parallel or perpendicular.

1 1 and 2 3 respectively. If both try to solve the problem independently, find the probability that (i) the problem is solved (ii) exactly one of them solves the problem.

22. Probabilities of solving a specific problem independently by A and B are

SECTION–C Question numbers 23 to 29 carry 6 marks each. 23. Using matrix method, solve the following system of equations: 2 3 10 4 6 5 6 9 20 + + = 4, – + = 1, + – = 2; x, y, z ¹ 0 x y z x y z x y z OR Using elementary transformations, find the inverse of the matrix æ 1 3 – 2ö ç ÷ ç– 3 0 – 1÷ ç 2 1 0 ÷ø è 24. Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area. 25. Using integration find the area of the triangular region whose sides have equations y = 2x + 1, y = 3x + 1 and x = 4. p/ 2

26. Evaluate:

ò 2 sin x cos x tan

–1

(sin x) dx

0

OR p/ 2

Evaluate:

ò 0

x sin x cos x sin 4 x + cos 4 x

dx

244


27. Find the equation of the plane which contains the line of intersection of the planes ® ® r . (i$ + 2j$ + 3k$) - 4 = 0, r . ( 2i$ + j$ - k$) + 5 = 0 and which is perpendicular to the plane ®

r . (5i$ + 3j$ - 6k$) + 8 = 0.

28. A factory makes tennis rackets and cricket bats. A tennis racket takes 1 × 5 hours of machine time and 3 hours of craftman’s time in its making while a cricket bat takes 3 hours of machine time and 1 hour of craftman’s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman’s time. If the profit on a racket and on a bat is ` 20 and ` 10 respectively, find the number of tennis rackets and crickets bats that the factory must manufacture to earn the maximum profit. Make it as an L.P.P. and solve graphically. 29. Suppose 5% of men and 0.25% of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume that there are equal number of males and females.

CBSE (Delhi) Set–II Only those questions, not included in Set–I, are given. 3p ù 9. Write the value of tan – 1 é tan . 4 ûú ëê 10. Write the value of ò

sec 2 x

dx. cosec 2 x 15. Form the differential equation of the family of parabolas having vertex at the origin and axis along positive y–axis. ®

16. Find a vector of magnitude 5 units, and parallel to the resultant of the vectors a = 2i$ + 3j$ – k$ ®

and b = i$ – 2j$ + k$. ì 3ax + b , if x > 1 ï 19. If the function f ( x) given by f ( x) = í 11, if x = 1 ï5ax – 2b , if x < 1 î is continuous at x = 1, find the values of a and b. 20. Using properties of determinants, prove the following: x y z x2

y2

z 2 = xyz ( x – y) ( y – z) (z – x)

x3

y3

z3

23. Bag I contains 3 red and 4 black balls and Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and is found to be red. Find the probability that it was drawn from Bag II. 29. Show that of all the rectangles with a given perimeter, the square has the largest area.

245


CBSE (Delhi) Set–III Only those questions, not included in Set I and Set II, are given. 7p 1. Write the value of cos–1 æç cos ö÷. è 6 ø 2. Write the value of ò

2 – 3 sin x cos 2 x

dx

11. Using properties of determinants, prove the following: x+4

2x

2x

2x

x+4

2x

2x

2x

x+4

= (5x + 4) ( 4 – x) 2

12. Find the value of a and b such that the following function f(x) is a continuous function: ì5; x £ 2 ï f ( x) = í ax + b; 2 < x < 10 ï 21; x ³ 10 î 13. Solve the following differential equation: (1 + y 2 ) (1 + log x) dx + xdy = 0 ®

®

®

®

® ®

14. If two vectors a and b are such that | a|= 2, | b|= 1 and a × b = 1, then find the value of ®

®

®

®

( 3 a – 5 b ) × ( 2 a + 7 b ). 23. A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six. 24. Show that of all the rectangles of given area, the square has the smallest perimeter.

246


Solutions CBSE (Delhi) Set–I SECTION – A 1. R is not transitive as (1, 2) Î R , ( 2, 1) Î R But (1, 1) Ï R [Note : A relation R in a set A is said to be transitive if ( a, b) Î R , (b , c) Î R Þ ( a, c) Î R " a, b , c Î R] 1 1 é é p p ùù 2. Let sin -1 æç - ö÷ = q êëQ - 2 Î [-1, 1] Þ q Î êë - 2 , 2 úû úû è 2ø 1 p Þ sin q = Þ sin q = sin æç - ö÷ è 2 6ø p p p 1 p Þ q = - Î é- , ù Þ sin -1 æç - ö÷ = ê ú è ø 6 ë 2 2û 2 6 1 p p ép ù é ù Now, sin ê - sin -1 æç - ö÷ ú = sin ê - æç - ö÷ ú è 2 øû ë3 ë 3 è 6 øû p p æ 2p + p ö = sin æç + ö÷ = sin ç ÷ è 3 6ø è 6 ø = sin 3. Q

aij =

i j

Þ

a 12 =

1 2

3p p = sin = 1 6 2 [Here i = 1 and j = 2]

é5 – x x + 1ù 4. If ê is singular matrix. 4 úû ë 2 5–x x+1 then =0 2 4

[QA square matrix A is called singular if| A| = 0]

Þ Þ

4 (5 - x) - 2 ( x + 1) = 0 20 - 4x - 2x - 2 = 0 Þ

Þ

6x = 18 Þ

18 - 6x = 0 18 x= =3 6

5. For elementary row operations we write A = IA é 2 5 ù é1 0ù Þ ê1 3ú = ê 0 1 ú . A ë û ë û Þ

é1 3ù é 0 1 ù ê 2 5 ú = ê1 0ú . A ë û ë û

Applying R 1 « R 2

247


é1 ê0 ë é1 ê0 ë é1 ê0 ë

Þ

3ù é 0 = -1 úû êë 1

1ù A -2úû

Applying R 2 ® R 2 - 2R 1

0 ù é 3 -5 ù = A -1úû êë 1 -2úû 0 ù é 3 -5 ù = A 1úû êë -1 2úû é 3 -5 ù I=ê A Þ 2úû ë -1

Þ Þ Þ

Applying R 1 ® R 1 + 3R 2 Applying R 2 ® ( -1) R 2 -5 ù 2úû

é 3 A -1 = ê ë -1

[Note : B is called inverse of A if AB = BA = 1] 6.

ò sec x (sec x + tan x) dx = ò sec 2 x dx + ò sec x . tan x dx

d é 2 êQ dx (tan x) = sec x ê ê and d (sec x) = sec x . tan êë dx

= tan x + sec x + C

7.

dx

ù ú ú xú úû

dx

ò x 2 + 16 = ò x 2 + 4 2 =

é êQ êë

1 x × tan -1 + C 4 4

dx

ò x2 + a2

=

ù 1 x tan -1 + C ú a a úû

8. If 2 i$ - 3 j$ + 4 k$ and ai$ + 6 j$ - 8 k$ are collinear 2 -3 4 = = a 6 -8

then Þ

Þ a=

2´6

a=

or

-3

2´-8 4

a=-4 ®

®

®

®

[Note : If a and b are collinear vectors then the respective components of a and b are proportional.] 9. Direction cosines of vector -2i$ + j$ - 5k$ are -2 2

2

( -2) + 1 + ( -5) -2 , 30

2

1

,

2

2

( -2) + 1 + ( -5)

2

,

-5 2

( -2) + 1 2 + ( -5) 2

1 -5 , 30 30

Note : If l, m, n are direction cosine of ai$ + bj$ + ck$ then l=

a 2

2

a +b + c

2

,

m=

b 2

2

a +b + c

2

,

n=

c 2

a + b2 + c2

248


10. The equation of given plane is 2x + y - z = 5 2x y z + - =1 5 5 5

y x z + + =1 5 / 2 5 -5 5 Hence, intercept cut off by the given plane on x-axis is × 2 Þ

Þ

[Note : If a plane makes intercepts a, b, c on x, y and z-axis respectively then its equation is x y z + + =1] a b c

SECTION – B 11. Required operation table of the operation * is given as

12. L.H.S.

*

1

2

3

4

5

1

1

1

1

1

1

2

1

2

2

2

2

3

1

2

3

3

3

4

1

2

3

4

4

5

1

2

3

4

5

é 1 + sin x + 1 – sin x = cot – 1 ê ê 1 + sin x – 1 – sin x ë

ù ú ú û

é 1 + sin x + 1 – sin x 1 + sin x + 1 - sin x ù = cot – 1 ê ´ ú 1 + sin x + 1 - sin x úû êë 1 + sin x – 1 – sin x é ( 1 + sin x + 1 - sin x ) 2 ù ú = cot -1 ê êë ( 1 + sin x ) 2 - ( 1 - sin x ) 2 úû é 1 + sin x + 1 - sin x + 2 (1 + sin x) (1 - sin x) ù = cot -1 ê ú 1 + sin x - 1 + sin x êë úû = cot

-1

= cot -1

= cot

-1

é 2 + 2 1 - sin 2 x ù ê ú ê ú 2 sin x ë û 1 + cos x æ ö ç ÷ è sin x ø æ 2 cos 2 x ö ç ÷ 2 ÷ ç ç 2 sin x . cos x ÷ è 2 2ø

é æ pö êQ x Î çè 0, 4 ÷ø ê êÞ 0 < x < p ê 4 ê x p êÞ 0 < < 2 8 ê ê x æ pö Î ç 0, ÷ Ì ( 0, p) êÞ 2 è 8ø ë

249


= cot -1 æç cot è =

xö ÷ 2ø

x = R.H.S. 2 OR

æ x - yö æ xö tan -1 ç ÷ - tan -1 ç ÷ = tan -1 è yø è x + yø

æ x x-y ö ç ÷ ç y x+y ÷ ç x x - y÷ ç1 + . ÷ y x + yø è

é ù x x-y > - 1ú ê Here . y x+y ë û

æ x 2 + xy - xy + y 2 ö y ( x + y) ÷ = tan -1 ç ´ ç y ( x + y) xy + y 2 + x 2 - xy ÷ø è æ x2 + y2 ö ÷ = tan -1 (1) = p = tan -1 ç ç x2 + y2 ÷ 4 è ø – a 2 ab 13. L.H.S. = ba –b 2 ca

= abc

ac bc –c2

cb -a a

b -b

c c

a

b

-c

-1

1

1

1

-1

1

1

1

-1

2 2 2

=a b c

2 2 2

=a b c

Taking out factor a, b, c from R 1 , R 2 and R 3 respectively

0

0

2

1

-1

1

1

1

-1

Taking out factor a, b, c from C 1 , C 2 and C 3 respectively.

Applying R 1 ® R 1 + R 2

2 2 2

= a b c [0 - 0 + 2 (1 + 1)] = 4a 2 b 2 c 2 = R.H.S. 14. Q f ( x) is continuous at x = 0. (L.H.L. of f ( x) at x = 0) = (R.H.L. of f ( x) at x = 0) = f ( 0) Þ Þ

Now,

lim f ( x) = lim

x ® 0-

x ® 0+

f ( x) = f ( 0)

lim f ( x) = lim a sin

x ® 0-

x®0

p ( x + 1) 2

…(i) éQ f ( x) = a sin p ( x + 1), if x £ 0ù êë úû 2

250


p p = lim a sin æç + è2 2 x®0 = lim a cos x®0

lim

x ® 0+

f ( x) = lim

p x = a . cos 0 = a 2

tan x - sin x

x®0

x sin x

= lim

cos x

tan x - sin x é ù if x > 0ú êQ f ( x) = 3 x ë û

3

- sin x

x3

x®0

= lim

xö÷ ø

sin x - sin x . cos x

x®0

cos x . x

3

= lim

x®0

x 2 sin 2 sin x 1 2 = lim . lim . 2 x ® 0 cos x x ® 0 x x ´4 4 æ sin ç 1 1 = .1 . lim ç 1 2 x®0 ç x è 2

sin x (1 - cos x) cos x . x 3 éQ 1 - cos x = 2 sin 2 x ù êë 2 úû

x ö2 ÷ 2÷ ÷ ø

x ö2 æ sin ÷ 1 ç 2÷ = 1 ´1= 1 = × ç lim 2 2 2 çx ® 0 x ÷ è2 2 ø p Also, f ( 0) = a sin ( 0 + 1) 2 p = a sin = a 2 1 Putting above values in (i) we get, a = 2 15. Let y = x x cos x +

Let y = u + v

x2 + 1 x2 - 1

where u = x x cos x , v =

\

dy du dv = + dx dx dx

Now,

u = x x cos x

x2 + 1 x2 - 1 …(i)

[Differentiating both sides w.r.t. x]

251


Taking log of both sides we get log u = log x x cos x

Þ

log u = x cos x . log x

Differentiating both sides w.r.t. x we get 1 du 1 . = 1 . cos x . log x + ( - sin x) . x log x + . x cos x u dx x 1 du . = cos x . log x - x . log x . sin x + cos x u dx du = x x cos x {cos x . log x - x log x sin x + cos x} dx v=

Again,

x2 + 1 x2 - 1

2 2 dv ( x - 1) . 2x - ( x + 1) . 2x = dx ( x 2 - 1) 2

\

3 3 dv 2x - 2x - 2x - 2x -4 x = = 2 2 2 dx ( x - 1) ( x - 1) 2

Putting the values of

du dv and in (i) we get dx dx

dy 4x = x x cos x {cos x . log x - x log x . sin x + cos x} 2 dx ( x - 1) 2 = x x cos x {cos x . (1 + log x) - x log x . sin x} -

4x 2

( x - 1) 2

OR Given, x = a (q - sin q) Differentiating w.r.t. (q) we get dx = a (1 - cos q) dq y = a (1 + cos q) Differentiating w.r.t. q we get dy = a ( - sin q) = - a sin q dq dy - a sin q dy dq Now, = = dx dx a (1 - cos q) dq q q - 2 sin . cos 2 2 = - cot q = q 2 2 sin 2 2

…(i)

…(ii)

[Putting values from (i) and (ii)]

252


16. Let V, r and h be the volume, radius and height of the sand-cone at time t respectively. dV Given, = 12 cm 3 /s dt r h= Þ r=6h 6 1 1 Now, V = pr 2 h Þ V = p 36h 3 = 12ph 3 3 3 Differentiating w.r.t. t we get dV dh = 12p. 3h 2 . dt dt dh 12 Þ = dt 36ph 2 Þ

é dh ù êë dt úû t

= =4

é dV ù 2 êëQ dt = 12 cm / súû

12 1 cm/s. = 36p ´ 16 48p

OR Let required point be ( x 1 , y 1 ) on given curve x 2 + y 2 - 2x - 3 = 0. Now, equation of curve is x 2 + y 2 - 2x - 3 = 0 Differentiating w.r.t. x we get dy dy -2x + 2 -2=0 Þ = dx dx 2y -2 x 1 + 2 - x 1 + 1 = = 2y 1 y1 ,y )

2x + 2y . Þ

æ dy ö ç ÷ è dx ø ( x

1

1

Since tangent at ( x 1 , y 1 ) is parallel to x-axis. \ Slope of tangent = 0 æ dy ö Þ =0 Þ ç ÷ è dx ø ( x ,y )

-x1 + 1

1 1

Þ

-x1 + 1 = 0 2

Þ

2

y1

=0

x1 = 1

Since ( x 1 , y 1 ) lies on given curve x + y - 2x - 3 = 0. \

x 12 + y 12 - 2x 1 - 3 = 0

Þ

1 2 + y 12 - 2 ´ 1 - 3 = 0

Þ

y 12 = 4

Þ

[Q x 1 = 1]

y1 = ± 2

Hence, required points are (1, 2) and (1, –2). æ dy ö [Note : Slope of tangent at a point ( x 1 , y 1 ) on curve y = f ( x) is ç ÷ è dx ø ( x

] 1 , y1 )

253


5x + 3 = A

17. Let, Þ

d 2 ( x + 4x + 10) + B dx

5x + 3 = A ( 2x + 4) + B

Þ

5x + 3 = 2Ax + ( 4A + B)

Equating coefficient of x and constant, we get 5 5 and 2A = 5 Þ A= 4A + B = 3 Þ B = 3 - 4 ´ = - 7 2 2 5 ( 2x + 4) - 7 5x + 3 Hence, dx = 2 dx

ò

=

= where I 1 = ò Now,

ò

x 2 + 4x + 10

5 2

2x + 4 dx

ò

2

x 2 + 4x + 10

-7

x + 4x + 10

dx

ò

2

x + 4x + 10

5 I1 - 7I2 2

( 2x + 4) dx 2

x + 4x + 10 I1 = ò

=ò

…(i)

and I 2 = ò

dx 2

x + 4x + 10

( 2x + 4) dx x 2 + 4x + 10 é Let x 2 + 4x + 10 = z ù ê ú ( 2x + 4) dx = dzúû êë

dz = z -1/ 2 dz z ò -

1

+ 1

z 2 = + C1 = 2 z + C1 -1 / 2 + 1 I 1 = 2 x 2 + 4x + 10 + C 1 Again

I2 = ò =ò

…(ii)

dx x 2 + 4x + 10 dx 2

x + 2×2×x + 4 + 6

=ò

dx ( x + 2) 2 + ( 6 ) 2

= log|( x + 2) + ( x + 2) 2 + ( 6 ) 2 | + C 2 I 2 = log|( x + 2) + x 2 + 4x + 10| + C 2 Putting the values of I 1 and I 2 in (i) we get

…(iii)

254


5x + 3

ò

x 2 + 4x + 10

dx =

5 ´ 2 x 2 + 4x + 10 - 7 log|( x + 2) + x 2 + 4x + 10| + C 2 é where C = 5 C - 7C ù 2ú 2 1 ëê û = 5 x 2 + 4x + 10 - 7 log|( x + 2) + x 2 + 4x + 10| + C OR

2

Let x = z Þ 2x dx = dz 2x dx

dz

ò ( x 2 + 1) ( x 2 + 3) = ò (z + 1) (z + 3)

\

1 A B = + (z + 1) (z + 3) z + 1 z + 3

Now,

…(i)

A (z + 3) + B (z + 1) 1 = (z + 1) (z + 3) (z + 1) (z + 3) Þ

1 = A (z + 3) + B (z + 1) Þ

1 = ( A + B) z + ( 3A + B)

Equating the coefficient of z and constant, we get A+B=0 and 3A + B = 1 Substracting (ii) from (iii) we get 1 2A = 1 Þ A = 2 1 \ B=2

…(ii) …(iii)

Putting the values of A and B in (i) we get 1 1 1 = (z + 1) (z + 3) 2 (z + 1) 2 (z + 3) \

2x dx

dz

ò ( x 2 + 1) ( x 2 + 3) = ò (z + 1) (z + 3) æ 1 ö 1 dz 1 dz 1 - ò = òç ÷ dz = ò 2 z+1 2 z+ 3 è 2 (z + 1) 2 (z + 3) ø =

1 1 1 1 log|z + 1| - log|z + 3| + C = log|x 2 + 1| - log| x 2 + 3| 2 2 2 2

=

x2+ 1 1 log +C 2 x2 + 3

= log

x2 + 1 x2 + 3

+C

é Note : log m + log n = log m. n ù ê and log m - log n = log m / nú ë û

255


18. e x tan y dx + (1 - e x ) sec 2 y dy = 0 (1 - e x ) sec 2 y dy = - e x tan y dx

Þ

Þ

Integrating both sides we get

ò

Þ

sec 2 y dy tan y

ò

Þ Þ

Þ

tan y

=

-e x 1 - ex

1 - ex

dz dt =ò z t Þ z=tC

x

tan y = (1 - e ) . C

[Putting the value of z and t]

dy + y = tan x dx tan x dy 1 + .y= 2 dx cos x cos 2 x

The above equation is in the form of,

Þ

dy + sec 2 xy = sec 2 x . tan x dx

dy + Py = Q dx

where P = sec 2 x, Q = sec 2 x . tan x I.F. = e ò

\

P dx

= eò

sec2 x dx

= e tan x

Hence, required solution is y ´ I. F. = ò Q ´ I. F. dx + C Þ

y . e tan x = ò sec 2 x . tan x . e tan x dx + C

Þ

y. e tan x = ò z. e z dz + C

Þ

y. e tan x = z. e z - ò e z dz + C

é Let tan x = z ù ê ú 2 êë Þ sec x dx = dzúû Þ

y. e tan x = tan x . e tan x - e tan x + C y = tan x - 1 + C. e -

Þ 20. Given

®

a = 3i$ + 2j$ + 2k$

®

b = i$ + 2j$ - 2k$

\

®

®

®

®

a + b = 4i$ + 4j$ a - b = 2i$ + 4k$

dx

tan y = z é Let ê Þ sec 2 y dy = dz ê x ê Also, 1 - e = t êÞ - e x dx = dt ë

- e x dx

log z = log t + log C

Þ 19. cos 2 x .

=ò

sec 2 y dy

tan x

y. e tan x = z. e z - e z + C

256

Xam idea Mathematics–XII ®

®

®

®

Now, vector perpendicular to ( a + b ) and ( a - b ) is ®

®

®

®

( a + b) ´( a - b) i$

j$

k$

= 4

4

0

2

0

4

= (16 - 0) i$ - (16 - 0) j$ + ( 0 - 8) k$ = 16i$ - 16j$ - 8k$ ®

®

®

®

\ Unit vector perpendicular to ( a + b ) and ( a - b ) is given by ±

®

®

®

®

®

®

®

®

( a + b) ´( a - b) |( a + b ) ´ ( a - b )|

=±

=± =±

16i$ - 16j$ - 8k$ 16 2 + ( -16) 2 + ( -8) 2 2i$ - 2j$ - k$ 9

=±

8 ( 2i$ - 2j$ - k$) 8 22 + 22 + 12

æ2 2 k$ ö = ± çç i$ - j$ - ÷÷ 3 3ø è3

2$ 2 $ 1 $ i + j+ k 3 3 3

21. The equation of given lines can be written in standard form as x - 2 y - 1 z - ( -3) = = 2 7 -3 x - ( -2) y - 4 z - 5 and = = -1 2 4 ®

…(i) …(ii)

®

If b 1 and b 2 are vectors parallel to lines (i) and (ii) respectively, then ®

®

b 1 = 2i$ + 7 j$ - 3k$ and b 2 = - i$ + 2j$ + 4k$ ®

®

Obviously, if q is the angle between lines (i) and (ii) then q is also the angle between b 1 and b 2 . ® ®

\

cos q =

b1 .b2 ®

®

|b 1 ||b 2 | =

( 2i$ + 7 j$ - 3k$) . ( -i$ + 2j$ + 4k$) 2 2 + 7 2 + ( -3) 2 . ( -1) 2 + 2 2 + 4 2

257


-2 + 14 - 12

= Þ

q=

62 . 21

=0

p 2

Angle between both lines is 90°. Hence, given lines are perpendicular to each other. 22. Let A and B be the events that the problem is solved independently by A and B respectively. 1 1 and P ( B) = \ P ( A) = 2 3 \ P ( A ¢) = Probability of event that the problem is not solved by A = 1 - P ( A) 1 1 =1- = 2 2 P ( B ¢) = Probability of event that the problem is not solved by B = 1 - P ( B) 1 2 =1- = 3 3 (i) P (event that the problem is not solved) = P (event that the problem is not solved by A and B) = P ( A ¢ Ç B ¢) = P ( A ¢) ´ P ( B ¢) 1 2 1 = ´ = 2 3 3

[Q A and B are independent events]

\ P (event that the problem is solved) = 1 - P (event that the problem is not solved) 1 2 =1- = 3 3 (ii) P (event that exactly one of them solves the problem) = P (solved by A and not solved by B or not solved by A and solved by B) = P ( A Ç B ¢) + P ( A ¢ Ç B) = P ( A) ´ P( B ¢) + P ( A ¢) ´ P ( B) 1 2 1 1 1 1 1 = ´ + ´ = + = 2 3 2 3 3 6 2 [Note : If A and B are independent events of same experiment then (i) A ¢ and B are independent (ii) A and B ¢ are independent (iii) A ¢ and B ¢ are independent]

258


SECTION – C 23. Let

1 1 1 = u, = v, = w x y z

Now the given system of linear equation may be written as 2u + 3v + 10w = 4,

4u - 6v + 5w = 1 and

6u + 9v - 20w = 2

Above system of equation can be written in matrix form as AX = B é2 where A = ê 4 ê êë 6

3 -6 9

2

3

| A| = 4

-6

6

9

Þ

X = A -1 B

…(i)

10ù é uù é 4ù ú ê ú 5 , X = v , B = ê 1ú ú ê ú ê ú êëw úû êë 2úû -20úû 10 5 = 2 (120 - 45) - 3 ( -80 - 30) + 10 ( 36 + 36) -20 = 150 + 330 + 720 = 1200 ¹ 0

For adj A : A11 = 120 - 45 = 75

A12 = - ( -80 - 30) = 110

A13 = 36 + 36 = 72

A21 = - ( -60 - 90) = 150,

A22 = - 40 - 60 = - 100

A23 = - (18 - 18) = 0

A 31 = 15 + 60 = 75

A 32 = - (10 - 40) = 30

A 33 = - 12 - 12 = - 24

\

\

é 75 adj. A = ê150 ê êë 75 A

-1

110 -100 30

¢ 72ù é 75 ú 0 = ê110 ú ê êë 72 -24úû

é 75 1 1 ê = . adj. A = 110 | A| 1200 ê êë 72

150 -100 0

150 -100 0

75ù 30ú ú -24úû

75ù 30ú ú -24úû

Putting the value of A -1 , X and B in (i), we get éuù êv ú = 1 ê ú 1200 êëw úû

é 75 ê110 ê êë 72

150

éuù êv ú = 1 ê ú 1200 êëw úû

é 300 + 150 + 150ù ê 440 - 100 + 60 ú ê ú êë 288 + 0 - 48 úû

-100 0

75ù é 4ù 30ú ê 1 ú úê ú -24úû êë 2úû

259


é 600ù ê 400ú ê ú êë 240úû

Þ

éuù êv ú = 1 ê ú 1200 êëw úû

Þ

é u ù é1 / 2 ù ê v ú = ê1 / 3ú ê ú ê ú êëw úû êë 1 / 5 úû

Equating the corresponding elements of matrix we get 1 1 1 u= ,v= ,w = Þ x = 2, y = 3, z = 5 2 3 5 OR é 1 Let A = ê -3 ê êë 2

3 0 1

-2 ù -1 ú ú 0úû

For finding the inverse, using elementary row operation we write

Þ

é 1 ê -3 ê êë 2

A = IA 3 -2 ù é 1 0 0 ù 0 -1 ú = ê 0 1 0 ú A ú ê ú 1 0úû êë 0 0 1 úû

Applying R 2 ® R 2 + 3R 1 and R 3 ® R 3 - 2R 1 , we get 3 -2 ù é 1 0 0ù é1 ê ú ê Þ 0 9 -7 = 3 1 0ú A ê ú ê ú êë 0 -5 4úû êë -2 0 1úû 1 Applying R 1 ® R 1 - R 2 3 0 1 / 3ù é 0 -1 / 3 0 ù é1 ê0 Þ 9 -7 ú = ê 3 1 0ú A ê ú ê ú êë 0 -5 4úû êë -2 0 1úû 1 Applying R 2 ® R 2 9 0 1 / 3ù é 0 -1 / 3 0ù é1 ê0 Þ 1 -7 / 9ú = ê1 / 3 1/9 0ú A ê ú ê ú êë 0 -5 4 úû êë -2 0 1úû Applying R 3 ® R 3 + 5R 2 1 / 3ù é 0 é1 0 ê 0 1 -7 / 9 ú = ê 1 / 3 Þ ê ú ê êë 0 0 1 / 9úû êë -1 / 3

-1 / 3 1/9 5/9

0ù 0ú A ú 1úû

260


Applying R 1 ® R 1 - 3R 3 , R 2 ® R 2 + 7 R 3 0 ù é 1 é1 0 ê ú ê Þ 0 1 0 = -2 ê ú ê êë 0 0 1 / 9úû êë -1 / 3 Applying R 3 ® 9R 3 é1 ê0 Þ ê êë 0

0

-2

0ù é 1 0 ú = ê -2 ú ê 1úû êë -3

1 0

Hence,

A

-1

5

-3 ù 7ú A ú 1úû

4 5/9 -3 ù 7ú A ú 9úû

4

é 1 = ê -2 ê êë -3

-2

-2 4 5

Þ

-2

é 1 I = ê -2 ê êë -3

-3 ù 7ú A ú 9úû

4 5

-3 ù 7ú ú 9úû

24. Let x and y be the length and breadth of a rectangle inscribed in a circle of radius r. If A be the area of rectangle then A = x. y A = x . 4r 2 - x 2 Þ

dA 1 = x. ´ ( -2x) + 4r 2 - x 2 2 2 dx 2 4r - x dA 2x 2 =+ 4r 2 - x 2 2 2 dx 2 4r - x

éQ DABC is right angled triangle ê 2 2 2 ê Þ 4r = x + y ê 2 2 2 ê Þ y = 4r - x ê …(i) y = 4r 2 - x 2 êë Þ

2 2 2 dA - x + 4r - x = dx 4r 2 - x 2

r

2 2 dA 4r - 2x = dx 4r 2 - x 2

O

dA =0 dx

4r 2 - 2x 2 2

4r - x

2

=0

A

Þ

2x 2 = 4r 2

Þ

4r 2 - x 2 .( -4x) - ( 4r 2 - 2x 2 ) . Now,

d2A dx 2

=

=

y

r

For maximum or minimum, Þ

C

D

x

x = 2r

1 ´ - 2x 2 4r 2 - x 2

( 4r 2 - x 2 ) 2 -4x ( 4r 2 - x 2 ) + x ( 4r 2 - 2x 2 ) ( 4r 2 - x 2 ) 3 / 2

=

x {-16r 2 + 4x 2 + 4r 2 - 2x 2 } ( 4r 2 - x 2 ) 3 / 2

B

261


= é d2Aù ê 2ú êë dx úû x =

x ( -12r 2 + 2x 2 ) ( 4r 2 - x 2 ) 3 / 2 2r ( -12r 2 + 2.2r 2 )

=

( 4r 2 - 2r 2 ) 3 / 2

2r

=

2r ´ - 8r 2 ( 2r 2 ) 3 / 2

=

-8 2r 3 2 2r 3

=-4<0

Hence, A is maximum when x = 2r. Putting x = 2r in (i) we get y = 4r 2 - 2r 2 = 2r i.e., x = y = 2r Therefore, Area of rectangle is maximum when x = y i.e., rectangle is square. 25. The given lines are

For intersection point of (i) and (iii) y= 2´ 4+1= 9 Coordinates of intersecting point of (i) and (iii) is (4, 9) For intersection point of (ii) and (iii) y = 3 ´ 4 + 1 = 13 i.e., Coordinates of intersection point of (ii) and (iii) is (4, 13)

O

+1

…(iii)

\

|

|

|

1

2

3

4

Þ x=0

y=1

Shaded region is required triangular region. Required Area = Area of trapezium OABD - Area of trapezium OACD 4

4

0

0

= ò ( 3x + 1) dx - ò 4

( 2x + 1) dx 4

é x2 ù é 2x 2 ù = ê3 + xú - ê + xú êë 2 úû 0 êë 2 úû 0 = [( 24 + 4) - 0] - [(16 + 4) - 0] = 28 - 20 = 8 sq. units

D

|

i.e., Coordinates of intersection point of (i) and (ii) is (0, 1). \

C (4, 9) x=4

For intersection point of (i) and (ii) 2x + 1 = 3x + 1

2x

x=4

y=

…(ii)

(4, 13)

B

+1

y = 3x + 1

14 – 13 – 12 – 11 – 10 – 9– 8– 7– 6– 5– 4– 3– 2– (0, 1)1 – A

3x

…(i)

y=

y = 2x + 1

|

|

5

6

262


26. Let I = 2 ò

p/ 2

sin x . cos x . tan -1 (sin x) dx

0

Let sin x = z, cos x dx = dz If x = 0, z = sin 0 = 0 If x = \

p p , z = sin = 1 2 2 1

I = 2 ò z tan -1 (z) dz 0

1

é 1 z2 ù 1 z2 = 2 ê tan -1 z . ú - 2 ò . dz 0 2 úû 1 + z2 2 êë 0 p 1 2 = 2 é . - 0ù êë 4 2 úû 2

1

ò0

z2 1 + z2

dz

=

2 1 1+z -1 1 1 p p dz -ò dz = - ò dz + ò 2 0 0 0 4 4 1+z 1 + z2

=

p p p p - [z]10 + [tan -1 z] 10 = - 1 + é - 0ù = - 1 êë 4 úû 2 4 4 OR p/ 2

Let

ò

I=

0 p/ 2

Þ

I=

ò 0

p/ 2

Þ

I=

ò

x sin x cos x sin 4 x + cos 4 x

æ p - xö . sin æ p - xö . cos æ p - xö ç ÷ ç ÷ ç ÷ è2 ø è2 ø è2 ø dx p p sin 4 æç - xö÷ + cos 4 æç - xö÷ è2 ø è2 ø æ p - xö cos x . sin x ç ÷ è2 ø cos 4 x + sin 4 x

0

Þ

Þ

I=

I=

p 2 p 2

dx

p/ 2

ò 0 p/ 2

ò 0

cos x . sin x sin 4 x + cos 4 x sin x . cos x dx sin 4 x + cos 4 x

é By Property ê a a ê ò f ( x) dx = ò f ( a - x) dx 0 ë0

dx p/ 2

dx -

ò 0

-I

x sin x . cos x sin 4 x + cos 4 x

dx

263


sin x . cos x Þ

p 2I = 2

p/ 2

sin x . cos x dx

p = 4 4 sin x + cos x 2

ò 0

p/ 2

ò 0

cos 4 x

dx

tan 4 x + 1

[Dividing numerator and denominator by cos 4 x] =

p 2´2

p/ 2

ò 0

2 tan x . sec 2 x dx 1 + (tan 2 x) 2

Let tan 2 x = z ; 2 tan x . sec 2 x dx = dz p If x = 0, z = 0 ; x = , z = ¥ 2 =

p 4

¥

ò 0

dz 1 + z2

p [tan -1 z] ¥ 0 4 p = (tan -1 ¥ - tan -1 0) 4 =

\

2I =

p æp ö ç - 0÷ ø 4 è2

Þ

I=

p2 16

27. The given two planes are ®

r (i$ + 2j$ + 3k$) - 4 = 0

®

r ( 2i$ + j$ - k$) + 5 = 0

and

…(i) …(ii)

The equation of a plane passing through line of intersection of the planes (i) and (ii) is given by ®

®

r . (i$ + 2j$ + 3k$) - 4 + l [ r . ( 2i$ + j$ - k$) + 5] = 0

®

r [(1 + 2l) i$ + ( 2 + l) j$ + ( 3 - l) k$] - 4 + 5l = 0

…(iii)

Since, the plane (iii) is perpendicular to the plane ®

r . (5i$ + 3j$ - 6k$) + 8 = 0

Þ Normal vector of (iii) is perpendicular to normal vector of (iv) Þ {(1 + 2l) i$ + ( 2 + l) j$ + ( 3 - l) k$} . {5i$ + 3j$ - 6k$} = 0 Þ

(1 + 2l) ´ 5 + ( 2 + l) ´ 3 + ( 3 - l) ´ ( -6) = 0

Þ

5 + 10l + 6 + 3l - 18 + 6l = 0

Þ

19l - 7 = 0

Þ

l=

7 19

…(iv)

264


Putting the value of l in (iii) we get equation of required plane ® é 7 7 7 7 ù r . ê æç1 + 2 ´ ö÷ i$ + æç 2 + ö÷ j$ + æç 3 - ö÷ k$ ú - 4 + 5 ´ =0 è ø è ø è ø 19 19 19 û 19 ë ®

33 45 $ 50 r . æç i$ + j+ è 19 19 19

®

41 k$ ö÷ =0 Þ ø 19

r . ( 33i$ + 45j$ + 50k$) - 41 = 0

Normal

Þ

Normal

[Note : Normals of two perpendicular planes are perpendicular to each other. 28. Let the number of tennis rackets and cricket bats manufactured by factory be x and y respectively. Here, profit is the objective function z. \

z = 20x + 10y

…(i)

We have to maximise z subject to the constraints 1 × 5x + 3y £ 42 …(ii)

[Constraint for machine hour]

3x + y £ 24

[Constraint for Craft man’s hour]

…(iii)

x³0 y³0

[Non-negative constraint]

Graph of x = 0 and y = 0 is the y-axis and x-axis respectively.

x=0

y-axis

\ Graph of x ³ 0, y ³ 0 is the Ist quadrant.

28 –

Graph of 1 × 5x + 3y = 42

24 – (0, 24)

Graph for 3x + y £ 24

16 – B 12 – 8– 4–

A 0

Graph of 3x + y = 24 x

0

8

y

24

0

C(4, 12)

4

8 (8, 0)

(28, 0) –

\ Graph for 1 × 5x + 3y £ 42 is the part of Ist quadrant which contains the origin.

(0, 14)

–

0

–

14

–

y

20 –

–

28

–

0

–

x

12 16

20

24

28

\ Graph of 3x + y £ 24 is the part of Ist quadrant in which origin lie Hence, shaded area OACB is the feasible region.

x-axis

265


For coordinate of C equation 1 × 5x + 3y = 42 and 3x + y = 24 are solved as 1 × 5x + 3y = 42

…(iv)

3x + y = 24 2 ´ (iv) – (v)

Þ

…(v)

3x + 6y = 84 _ 3x ± y = _ 24 5y = 60 Þ

y = 12

Þ

x=4

(Substituting y = 12 in (iv))

Now value of objective function z at each corner of feasible region is z = 20x + 10y

Corner Point O ( 0, 0)

0

A ( 8, 0)

20 ´ 8 + 10 ´ 0 = 160

B ( 0, 14)

20 ´ 0 + 10 ´ 14 = 140

C ( 4, 12)

20 ´ 4 + 10 ´ 12 = 200

Maximum

Therefore, maximum profit is ` 200, when factory makes 4 tennis rackets and 12 cricket bats. 29. Let E1 , E2 and A be event such that E1 = Selecting male person E2 = Selecting women (female person) A = Selecting grey haired person. 1 Then P (E1 ) = , 2 æAö 5 , Pç ÷= è E1 ø 100

P (E2 ) =

1 2

æ A ö 0 × 25 Pç ÷= è E2 ø 100

æE ö Here, required probability is P ç 1 ÷ . èAø æAö P (E1 ) . P ç ÷ è E1 ø

\


\

1 5 ´ E 5 500 20 æ ö 2 100 P ç 1÷ = = = = 5 1 0 × 25 5 + 0 × 25 525 21 èAø 1 ´ + ´ 2 100 2 100

æAö æAö P (E1 ) . P ç ÷ + P (E2 ) . P ç ÷ è E1 ø è E2 ø

266


CBSE (Delhi) Set–II 3p ù p ö æ 9. tan -1 é tan = tan -1 ç tan æç p - ö÷ ÷ êë ú è è 4 û 4 øø

é æ p p öù -1 êQ tan (tan x) = x if x Î çè - 2 , 2 ÷ø ú ê ú ê Here p Î æ - p , p ö ú ç ÷ êë úû 4 è 2 2ø

p = tan -1 æç tan ö÷ è 4ø p = 4 10.

sec 2 x

ò cosec 2 x

dx = ò

1 cos 2 x

´

sin 2 x 1

dx

= ò tan 2 x dx = ò (sec 2 x - 1) dx = ò sec 2 x dx - ò dx = tan x - x + c 15. The equation of parabola having vertex at origin and axis along +ve y-axis is x 2 = 4ay

…(i) 2x = 4a .

Differentiating w.r.t. x we get,

dy dx

i.e.,

x = 2ay ¢

Þ

a=

Putting a =

where a is parameters.

dy ù é ê where y¢ = dx ú ë û

x 2y ¢

x in (i) we get 2y ¢ x2 = 4 .

x .y 2y ¢

2y x

Þ

y¢ =

Þ

Þ

xy ¢ - 2y = 0

xy ¢ = 2y

It is required differential equation. 16. Given two vectors are ®

a = 2i$ + 3j$ - k$ and

®

®

®

If c is the resultant vector of a and b then ®

®

®

c = a +b

®

b = i$ - 2j$ + k$

267


= ( 2i$ + 3j$ - k$) + (i$ - 2j$ + k$) = 3i$ + j$ + 0. k$ ®

Now a vector having magnitude 5 and parallel to c is given by ®

5 c

=

®

5 ( 3i$ + j$ + 0k$)

=

2

2

3 +1 + 0

| c|

2

=

15 $ 5 $ i+ j 10 10

It is required vector. ®

®

[Note : A vector having magnitude l and parallel to a is given by l .

a

®

.]

| a| 19. Q f ( x) is continuous at x = 1. Þ

(L.H.L. of f ( x) at x = 1) = (R.H.L. of f ( x) at x = 1) = f (1)

Þ

lim f ( x) = lim f ( x) = f (1)

x ® 1-

lim f ( x) = lim 5ax - 2b

Now,

…(i)

x ® 1+

x ® 1-

x®1

[Q f ( x) = 5ax - 2 f if x < 1]

= 5a - 2b lim f (1) = lim 3ax + b

x ® 1+

x®1

[Q f ( x) = 3ax + b if x > 1]

= 3a + b f (1) = 11

Also,

Putting these values in (i) we get 5a - 2b = 3a + b = 11 Þ

5a - 2b = 11 3a + b = 11

…(ii) …(iii)

On solving (ii) and (iii), we get a = 3, b = 2 x

y

z

20. L.H.S. = x 2

y2

z2

x3

y3

z3

1

1

1

= xyz x

y

z

2

2

2

x

y

1 = xyz x x2

z

[Taking x, y, z common from C 1 , C 2 , C 3 respectively]

0

0

y-x

z-x

2

y -x

2

2

z -x

2

C 2 ® C 2 - C 1, C 3 ® C 3 - C 1

268


= xyz ( y - x) (z - x)

1

0

0

x

1

1

x2

y+x

z+x

[Taking common ( y - x) and (z - x) from C 2 and C 3 respectively]

= xyz ( y - x) (z - x) [1 (z + x - y - x)]

[Expanding along R 1 ]

= xyz ( y - x) (z - x) (z - y) = xyz ( x - y) ( y - z) (z - x) 23. Let E1 , E2 and A be event such that E1 = choosing the bag I E2 = choosing the bag II A = drawing red ball Then,

P (E1 ) =

1 , 2

P (E2 ) =

1 2

æAö 3 æAö 5 and P ç ÷ = , P ç ÷ = è E2 ø 11 è E1 ø 7

æE ö P ç 2 ÷ is required. èAø

By Baye’s theorem,

æE ö Pç 2÷= èAø

æAö P (E2 ) . P ç ÷ è E2 ø æAö æAö P (E1 ). P ç ÷ + P (E2 ) . P ç ÷ è E1 ø è E2 ø

1 5 5 ´ 2 11 11 = = 1 3 1 5 3 5 ´ + ´ + 2 7 2 11 7 11 5 77 35 = ´ = 11 68 68 29. Let the length and breadth of rectangle be x and y. If A and P are the area and perimeter of rectangle respectively then Þ

A = x.y P A = x æç - xö÷ è2 ø

and

P x - x2 Þ 2 For maximum and minimum of A. dA =0 dx P Þ - 2x = 0 Þ 2 Þ

A=

P = 2 ( x + y) æQ y = P - xö ç ÷ è ø 2 dA P = - 2x dx 2

x=

P 4

269


Again

d2A dx 2

=-2 æ d2Aö ç ÷ ç 2÷ è dx ø x =

Þ

=0 P 4

P 4 P P P y= - = 2 4 4

Hence, A is maximum for x = \

Therefore, for largest area of rectangle x = y =

P i.e., with given perimeter, rectangle having 4

largest area must be square.

CBSE (Delhi) Set–III 7pö -1 1. cos -1 æç cos ÷ = cos è ø 6

5p ö ö æ æ ç cos çè 2p - ÷ø ÷ è 6 ø

5p ö æ = cos -1 ç cos æç ö÷ ÷ è 6 øø è = 2. Let I = ò =ò

cos 2 x 2 2

cos x

[Q cos ( 2p - q) = cos q ] éQ cos -1 (cos x) = x if x Î [0, p]ù ê ú 5p ê ú Here Î [0, p] 6 ë û

5p 6

2 - 3 sin x

éQ 5p Î [0, p]ù êë úû 6

dx

dx - ò

3 sin x dx cos 2 x - dz

= 2 ò sec 2 x dx - 3 ò

z2

[Let cos x = z - sin x dx = dz]

z -2 + 1 +c -2 + 1 3 = 2 tan x +c cos x = 2 tan x + 3

11. L.H.S.

x+4

2x

2x

x+4

2x

2x

2x

x+4

2x

5x + 4 5x + 4 5x + 4 =

2x

x+4

2x

2x

2x

x+4

R1 ® R1 + R2 + R 3

270


1

1

1

x+4

2x

2x

2x

x+4

1

0

0

4-x 0

0 4-x

= (5x + 4) 2x

= (5x + 4) 2x 2x

[Taking (5x + 4) common from R 1 ]

C2 ® C2 - C1 C 3 ® C 3 - C1

= (5x + 4) [1 {( 4 - x) 2 - 0} + 0 + 0]


= (5x + 4) ( 4 - x) 2 = R.H.S. 12. Since f ( x) is continuous. Þ f ( x) is continuous at x = 2 and x = 10. Þ (L.H.L. of f ( x) at x = 2) = (R.H.L. of f ( x) at x = 2) = f ( x) Þ Similarly,

lim f ( x) = lim

x ® 2-

f ( x) =

lim

x ® 10 -

f ( x) = f ( 2)

x ® 2+

f ( x) = f (10)

lim

x ® 10 +

lim f ( x) = lim 5

x ® 2-

…(i)

x®2

…(ii) [Q f ( x) = 5 if x £ 2]

=5 lim

x ® 2+

f ( x) = lim ax + b x®2

[Q f ( x) = ax + b if x > 2]

= 2a + b f ( 2) = 5 Putting these values in (i) we get 2a + b = 5 Again

lim

x ® 10 -

f ( x) = lim

x ® 10

…(iii) ax + b

[Q f ( x) = ax + b if x < 10]

21

[Q f ( x) = 21 if x > 10]

= 10a + b lim

x ® 10 +

f ( x) = lim

x ® 10

= 21 f (10) = 21 Putting these values in (ii) we get 10a + b = 21 = 21 Þ

10a + b = 21

Substracting (iii) from (iv) we get

…(iv)

271


10a + b = 21 _ 2a ± b = _ 5 8a = 16 a=2 b =5- 2´ 2=1 a = 2, b = 1

\

13. (1 + y 2 ) (1 + log x) dx + xdy = 0 x dy = - (1 + y 2 ) (1 + log x) dx 1 + log x dy =dx 2 x 1+y

Þ

Integrating both sides we get dy

ò 1 + y2

=-ò

1 + log x x

Þ

tan -1 y = - ò z dz

Þ

tan -1 y = -

Þ

tan -1 ®

®

dx é Let 1 + log x = zù ê ú 1 ê dx = dz ú x ë û

z2 +c 2 1 y = - (1 + log x) 2 + c 2

® ®

14. Given | a | = 2,| b | = 1 and a . b = 1 Now,

®

®

®

®

®

®

®

®

®

®

®

®

( 3 a - 5 b ) .( 2 a + 7 b ) = 3 a . 2 a + 3 a . 7 b - 5 b . 2 a - 5 b . 7 b ® ®

® ®

® ®

® ®

= 6 a . a + 21 a . b - 10 b . a - 35 b . b ® ®

®

®

= 6| a |2 + 11 a . b - 35| b |2 = 6 ( 2) 2 + 11 ´ 1 - 35 (1) 2 = 24 + 11 - 35 = 0 ® ®

®

®

®

®

[Note : a . a =| a |.| a |cos 0° =| a |2 ´ 1 =| a |2 Also, scalar product of vectors is commutative \

® ®

23. Let E1 , E2 and A be event such that E1 = Occurring six on die. E2 = Not occurring six on die. A = Reporting six by man on die.

® ®

a .b = b . a

272


Here

P (E1 ) =

1 , 6

P (E2 ) =

5 6

æAö P ç ÷ = P (Speaking truth i.e., man reports six on die when six has occurred on the die) è E1 ø 3 = 4 æAö P ç ÷ = P (Not speaking truth i.e., man report six on die when six has not occurred on die) è E2 ø 3 1 =1- = 4 4 æE ö Required probability is P ç 1 ÷ . èAø æAö P (E1 ) . P ç ÷ è E1 ø

æE ö By Baye’s theorem, P ç 1 ÷ = èAø

æAö æAö P (E1 ) . P ç ÷ + P (E2 ) . P ç ÷ E è 1ø è E2 ø 1 3 ´ 3 24 3 6 4 = = ´ = 1 3 5 1 24 3 + 5 8 ´ + ´ 6 4 6 4 24. Let x, y be the length and breadth of rectangle whose area is A and perimeter is P. \ P = 2 ( x + y) éQ A = x. yù A ê Þ P = 2 æç x + ö÷ A ú è ê y= ú xø x û ë For maximum or minimum value of perimeter P æ dP Aö = 2 ç1 ÷=0 è dx x2 ø A Þ 1=0 Þ x2 = A x2 [Dimensions of rectangle is always positive] Þ x= A 2 æ d P -1 ö 2 A Now, = 2 ç0 - A ´ ÷= 2 è dx x3 ø x3 \

é d2P ù ê 2ú êë dx úû x =

= A

2a ( A) 3

>0

i.e., for x = A , P (perimeter of rectangle) is smallest. A A \ y= = = A x A Hence, for smallest perimeter, length and breadth of rectangle are equal ( x = y = A ) i. e. , rectangle is square.

EXAMINATION PAPERS –2011 CBSE (All India) Set–I

Time allowed: 3 hours

Maximum marks: 100

General Instructions: As given in Examination Paper (Delhi) – 2011.

SECTION–A Question numbers 1 to 10 carry 1 mark each. 1. Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2,5), (3, 6)} be a function from A to B. State whether f is one-one or not. 2p 2p ö 2. What is the principal value of cos – 1 æç cos ö÷ + sin – 1 æç sin ÷? è è 3ø 3ø 3. Evaluate: cos 15° sin 15° sin 75° cos 75° é2 4. If A = ê ë5

3ù , write A–1 in terms of A. – 2úû

5. If a matrix has 5 elements, write all possible orders it can have. 6. Evaluate: ò ( ax + b) 3 dx 7. Evaluate: ò

dx 1 – x2

8. Write the direction-cosines of the line joining the points (1, 0, 0) and (0, 1, 1). 9. Write the projection of the vector i$ – j$ on the vector i$ + j$. 10. Write the vector equation of the line given by

x–5 y+4 z– 6 . = = 3 7 2

SECTION–B Question numbers 11 to 22 carry 4 marks each. 11. Let f : R® R be defined as f ( x) = 10x + 7. Find the function g : R® R such that gof = fog = I R . OR A binary operation * on the set {0, 1, 2, 3, 4, 5} is defined as:

274


if a + b < 6 ì a + b, a *b =í î a + b – 6, if a + b ³ 6 Show that zero is the identity for this operation and each element ‘a’ of the set is invertible with 6–a, being the inverse of ‘a’. 12. Prove that: é 1 + x - 1 - xù p 1 1 –1 tan – 1 ê £ x £1 ú = – cos x, – 4 2 1 + x + 1 – x 2 êë úû 13. Using properties of determinants, solve the following for x: x – 2 2x – 3 3x – 4 x – 4 2x – 9

3x – 16 = 0

x – 8 2x – 27 3x – 64 14. Find the relationship between ‘a’ and ’b’ so that the function ‘f’ defined by: ì ax + 1, if x £ 3 is continuous at x = 3 . f ( x) = í îbx + 3, if x > 3 If x y

OR log x dy . = e x – y , show that = dx {log ( xe)} 2

15. Prove that y =

4 sin q

p – q is an increasing function in é 0, ù. ê ( 2 + cos q) ë 2 úû

OR If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, then find the approximate error in calculating its surface area. 1 16. If x = tan æç log yö÷, show that èa ø (1 + x 2 )

d2y dx

p/ 2

17. Evaluate:

ò 0

x + sin x 1 + cos x

2

+ ( 2x – a)

dy =0 dx

dx

18. Solve the following differential equation: x dy – y dx = x 2 + y 2 dx 19. Solve the following differential equation: dx ( y + 3x 2 ) =x. dy 20. Using vectors, find the area of the triangle with vertices A (1, 1, 2), B (2, 3, 5) and C (1, 5, 5).

275


21. Find the shortest distance between the following lines whose vector equations are: ®

®

r = (1 – t) i$ + (t – 2) j$ + ( 3 – 2t) k$ and r = ( s + 1) i$ + ( 2s – 1) j$ – ( 2s + 1) k$ .

22. A random variable X has the following probability distribution: X

0

1

2

3

4

5

6

7

P (X)

0

K

2K

2K

3K

K2

2K 2

7K 2 + K

Determine: (i) K

(ii) P (X < 3)

(iii) P (X > 6) (iv) P (0 < X < 3) OR Find the probability of throwing at most 2 sixes in 6 throws of a single die.

SECTION–C Question numbers 23 to 29 carry 6 marks each. 23. Using matrices, solve the following system of equations: 4x + 3y + 3z = 60, x + 2y + 3z = 45 and 6x + 2y + 3z = 70 24. Show that the right-circular cone of least curved surface and given volume has an altitude equal to 2 times the radius of the base. OR A window has the shape of a rectangle surmounted by an equilateral triangle. If the perimeter of the window is 12 m, find the dimensions of the rectangle that will produce the largest area of the window. p/ 3

25. Evaluate:

dx 1 + tan x p/ 6

ò

OR Evaluate: ò

6x + 7 ( x – 5) ( x – 4)

dx

26. Sketch the graph of y =|x + 3| and evaluate the area under the curve y =| x + 3| above x-axis and between x = – 6 to x = 0. 27. Find the distance of the point (–1, –5, –10), from the point of intersection of the line ® ® r = ( 2i$ – j$ + 2k) + l ( 3i$ + 4j$ + 2k$) and the plane r .(i$ – j$ + k$) = 5. 28. Given three identical boxes I, II and III each containing two coins. In box I, both coins are gold coins, in box II, both are silver coins and in box III, there is one gold and one silver coin. A person chooses a box at random and takes out a coin. If the coin is of gold, what is the probability that the other coin in the box is also of gold? 29. A merchant plans to sell two types of personal computers — a desktop model and a portable model that will cost ` 25,000 and ` 40,000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than ` 70 lakhs and his profit on the desktop model is ` 4,500 and on the portable model is ` 5,000. Make an L.P.P. and solve it graphically.

276


CBSE (All India) Set–II Only those questions, not included in Set-I, are given. 9. Evaluate:

ò

(log x) 2 x

dx. ®

$ 10. Write the unit vector in the direction of the vector a = 2i$ + j$ + 2k. 19. Prove the following: 1 1 31 2 tan –1 æç ö÷ + tan –1 æç ö÷ = tan –1 æç ö÷ è 2ø è7 ø è 17 ø 20. Using properties of determinants, solve the following for x: a+x a– x a-x a– x a+ x a– x =0 a– x a– x a+x 21. Evaluate: p/ 4

ò log (1 + tan x) dx 0

22. Solve the following differential equation: x dy – ( y + 2x 2 ) dx = 0 28. Using matrices, solve the following system of equations: x + 2y + z = 7, x + 3z = 11 and 2x – 3y = 1 29. Find the equation of the plane passing through the line of intersection of the planes ® ® r .(i$ + j$ + k$) = 1 and r .( 2i$ + 3j$ – k$) + 4 = 0 and parallel to x-axis.

CBSE (All India) Set–III Only those questions, not included in Set I and Set II, are given. –1

1. Evaluate:

e tan x

ò 1 + x2

dx ®

®

2. Write the angle between two vectors a and b with magnitudes 3 and 2 respectively having ® ®

a × b = 6.

1 1 1 p 11. Prove that : tan –1 æç ö÷ + tan –1 æç ö÷ + tan –1 æç ö÷ = è 2ø è5 ø è 8ø 4

277


12. Using properties of determinants, solve the following for x: x+a x x

1

13. Evaluate:

æ1

x

x+a

x

x

x

x+a

=0

ö

ò log çè x – 1÷ø dx 0

14. Solve the following differential equation: x dx + ( y – x 3 ) dx = 0 23. Using matrices, solve the following system of equations: x + 2y – 3z = – 4 , 2x + 3y + 2z = 2 and 3x – 3y – 4z = 11 24. Find the equation of the plane passing through the line of intersection of the planes x–1 y– 3 z–5 . 2x + y – z = 3 and 5x – 3y + 4z + 9 = 0 and parallel to the line = = 2 4 5

Solutions

CBSE (All India) Set–I f

SECTION – A 1. f is one-one because

1 2 3

f (1) = 4 ; f ( 2) = 5 ; f ( 3) = 6 No two elements of A have same f image.

4 5 6 7

2p ö 2p ö 2p ö p ö ö é 2p æ p p ö ù æ -1 æ -1 æ -1 æ 2. cos -1 æç cos Ï ç- , ÷ ÷ + sin ç sin ÷ = cos ç cos ÷ + sin ç sin ç p - ÷ ÷ êQ è è è è è 3ø 3ø 3ø 3 ø ø ë 3 è 2 2 ø úû 2p ö pö -1 æ = cos -1 æç cos ÷ + sin ç sin ÷ è è 3ø 3ø =

2p p + 3 3

=

3p =p 3

é Note : By Property of inverse functions ù ê ú p p ùú ê é -1 sin (sin x) = x if x Î - , ú ê êë 2 2 úû ê ú êë and úû cos -1 (cos x) = x if x Î [0, p]

3. Expanding the determinant, we get cos 15° . cos 75° - sin 15° . sin 75° = cos (15° + 75° ) = cos 90° = 0 [Note : cos ( A + B) = cos A . cos B - sin . sin B]

278


é2 4. A = ê ë5

3ù -2úû

\

| A| =

2 5

3 = - 4 - 15 = - 19 ¹ 0 -2

Þ A is invertible matrix. Here, C 11 = - 2, C 12 = - 5, C 21 = - 3, C 22 = 2 -5 ù T é -2 -3 ù é -2 adj A = ê =ê ú 2úû ë -3 ë -5 2 û 1 A -1 = . adj A | A|

\ \

é -2 ê -5 ë

=

1 -19

=

1 A 19

-3 ù 1 = 2úû 19

é2 ê5 ë

3ù -2úû

[Note : Cij is cofactor aij of A = [aij ] ]

5. Possible orders are 1 ´ 5 and 5 ´ 1. 6.

ò ( ax + b)

3

dx ax + b = z

Let

adx = dz

Þ

dx =

ò ( ax + b)

\

3

dz a

dx = ò z 3 . =

7.

ò

dx 1-x

2

= sin -1 x + c. Because

dz a

1 1 z4 +c= ( ax + b) 4 + c a 4 4a d 1 (sin -1 x) = . dx 1 - x2

8. Direction ratios of line joining (1, 0, 0) and (0, 1, 1) are 0 –1,

1– 0,

1– 0

–1,

1,

1

i.e.,

\ Direction cosines of line joining (1, 0, 0) and (0, 1, 1) are -1 1 1 , , ( -1) 2 + (1) 2 + (1) 2 ( -1) 2 + (1) 2 + (1) 2 ( -1) 2 + (1) 2 + (1) 2 -

1 1 1 , , 3 3 3

279


®

9. Let a = i$ - j$ , b = i$ + j$ ®

® ®

®

a.b

Now, projection of a on b =

®

|b| =

(i$ - ^j ) . (i$ + j$) = |i$ + j$|

1-1 12 + 12

=0

10. The given equation of line may written as x - 5 y - ( -4) z - 6 = = 3 7 2 ®

®

Here, a = 5i$ - 4j$ + 6k$ and b = 3i$ + 7 j$ + 2k$ Hence, required vector equation is ®

®

®

r = a +l b

®

r = (5i$ - 4j$ + 6k$) + l ( 3i$ + 7 j$ + 2k$)

i.e.,

SECTION – B 11. Q

gof = fog = I R

Þ

fog = I R

Þ

fog ( x) = I ( x)

Þ

f ( g( x)) = x

[Q I ( x) = x being identity function]

Þ

10 ( g( x)) + 7 = x x-7 g( x) = 10

[Q f ( x) = 10x + 7]

Þ

i.e., g : R ® R is a function defined as g( x) =

x-7 10

.

OR For Identity Element : Let a be an arbitrary element of set {0, 1, 2, 3, 4, 5} Now, Eq. (i) and (ii) Þ

a*0= a+ 0= a

…(i)

0*a= 0+ a= a

…(ii)

[Q a + 0 = 0 + a < 6 " a Î {0, 1, 2, 3, 4, 5}]

a * 0 = 0 * a = a "a Î {0, 1, 2, 3, 4, 5}

Hence, 0 is identity for binary operation * . For Inverse : Let a be an arbitrary element of set {0, 1, 2, 3, 4, 5}. Now,

a * ( 6 - a) = a + ( 6 - a) - 6

[Q a + ( 6 - a) ³ 6]

280


=a+6-a-6 = 0 (identity) ( 6 - a) * a = ( 6 - a) + a - 6

Also,

…(i) [Q a + ( 6 - a) ³ 6]

=6-a+a-6 = 0 (identity)

…(ii)

Eq. (i) and (ii) Þ a * ( 6 - a) = ( 6 - a) * a = 0 (identity) " a Î {0, 1, 2, 3, 4, 5} Hence, each element ‘a’ of given set is invertible with inverse 6 - a. 12. Let x = sin q Þ q = sin Now,

-1

x

é 1 + x - 1 - xù tan -1 ê ú êë 1 + x + 1 - x úû

1 é êQ - 2 £ x £ 1 ê p p ê Þ sin æç - ö÷ £ sin q £ sin è ø 4 2 ê p pù ê é ê Þ q Î êë - 4 , 2 úû ë

é 1+x - 1-x 1 + x - 1 - xù = tan -1 ê ´ ú 1 + x - 1 - x úû êë 1 + x + 1 - x é ( 1 + x - 1 - x) 2 ù ú = tan -1 ê êë ( 1 + x ) 2 - ( 1 - x ) 2 úû é ( 1 + x) 2 + ( 1 - x) 2 - 2 . 1 + x . 1 - x ù ú = tan -1 ê 1+ x-1+ x êë úû é1 + x + 1 - x - 2 1 - x 2 = tan -1 ê ê 2x ë

ù ú = tan -1 ú û

é1 - 1 - x 2 ê ê x ë

ù ú ú û

é 1 - 1 - sin 2 q ù ú = tan -1 éê 1 - cos q ùú = tan -1 ê ê ú sin q ë sin q û ë û æ 2 sin 2 q ö ç ÷ -1 2 ÷ = tan -1 æç tan q ö÷ = tan ç è 2ø ç 2 sin q . cos q ÷ è ø 2 2 q 1 -1 = = sin x 2 2 1 = 2

æ p - cos -1 xö ç ÷ è2 ø

é Q sin -1 x + cos -1 x = p ù ê 2 ú ê ú 1 ù é ê and x Î ê - , 1ú Ì [-1, 1]ú ë 2 û ë û

ù ú ú ú ú ú ú û

281


13. Given,

x – 2 2x – 3

3x – 4

x – 4 2x – 9

3x – 16 = 0

x – 8 2x – 27 3x – 64 Þ

Þ

x– 2

1

x– 4 x– 8

-1 -11

x– 2

1

-2

-2

–6

-12

2

C 2 ® C 2 - 2C 1

-4 = 0 -40

C 3 ® C 3 - 3C 1

2

R2 ® R2 - R1

-6 = 0

R 3 ® R 3 - R1

-42

expanding along R 1 we get Þ

( x - 2) ( 84 - 72) - 1 ( 84 - 36) + 2 ( 24 - 12) = 0

Þ

12x - 24 - 48 + 24 = 0

Þ

x=4

Þ

12x = 48

14. Since, f ( x) is continuous at x = 3. Þ

lim

f ( x) =

lim

f ( x) = lim f ( 3 - h)

x ® 3-

Now,

x ® 3-

lim

x ® 3+

f ( x) = f ( 3)

…(i) é Let x = 3 - h êx ® 3 - Þ h ® 0 ë

h®0

[Q f ( x) = ax + 1 " x £ 3]

= lim a ( 3 - h) + 1 h®0

= lim 3a - ah + 1 = 3a + 1 h®0

lim

x ® 3+

f ( x) = lim f ( 3 + h) h®0

= lim b ( 3 + h) + 3 h®0

é Let x = 3 + h ù ê ú + êë x ® 3 Þ h ® 0úû [Q f ( x) = bx + 3 " x > 3]

= 3b + 3 3a + 1 = 3b + 3

From (i) ,

3a - 3b = 2 2 a-b = 3

or

3a - 3b = 2

OR Given,

y

x =e

x-y

Taking log of both sides Þ

log x y = log e x -

y

which is the required relation.

282


Þ

y . log x = ( x - y) log e

Þ

y . log x = ( x - y) x y= 1 + log x

Þ

Þ

[Q log e = 1] y log x + y = x

1 (1 + log x) . 1 - x . æç 0 + ö÷ è xø

Þ

dy = dx

Þ

log x dy 1 + log x - 1 = = 2 dx (1 + log x) (log e + log x) 2

Þ

log x dy = dx (log ex) 2 y=

15. Given,

(1 + log x) 2

4 sin q 2 + cos q

Þ

[Q1 = log e]

log x dy . = dx {log ( ex)} 2

-q

dy ( 2 + cos q) . 4 cos q - 4 sin q . ( 0 - sin q) = -1 dx ( 2 + cos q) 2

\

=

=

8 cos q + 4 cos 2 q + 4 sin 2 q - ( 2 + cos q) 2 ( 2 + cos q) 2 8 cos q + 4 - 4 - cos 2 q - 4 cos q ( 2 + cos q) 2

Þ

2 dy 4 cos q - cos q cos q ( 4 - cos q) = = dx ( 2 + cos q) 2 ( 2 + cos q) 2

Þ

dy +ve ´ + ve = dx +ve

éQ q Î [0, p / 2] Þ cos q > 0 ù ê 4 - cos q is + ve as - 1 £ cos q £ 1ú ë û

dy >0 dx 4 sin q p y= - q is increasing function in é 0, ù . ê 2 + cos q ë 2 úû

Þ i.e., OR

Here, radius of the sphere r = 9 cm. Error in calculating radius, dr = 0 × 03 cm. Let ds be approximate error in calculating surface area.

283


If S be the surface area of sphere, then S = 4 pr 2 ds = 4p . 2r = 8pr dr

Þ Now by definition, approximately

é ds ds ù = lim êQ ú d r ® 0 dr dr û ë

ds ds = dr dr Þ

æ ds ö ds = ç ÷ . dr è dr ø

Þ

ds = 8 p r . dr = 8p ´ 9 ´ 0 × 03 cm 2 = 2 × 16 p cm

[Q r = 9 cm]

2

1 x = tan æç log yö÷ èa ø 1 tan -1 x = log y a

16. Given Þ

a tan -1 x = log y

Þ

Differentiating w.r.t. x, we get a 1 dy Þ = . 2 y dx 1+x dy ay = dx 1 + x 2

Þ

dy = ay dx Differentiating w.r.t. x, we get (1 + x 2 )

Þ

(1 + x 2 )

d2y dx

(1 + x 2 )

Þ

d2y dx

p/ 2

17. I =

ò 0 p/ 2

=

ò 0

x + sin x 1 + cos x

2

+ 2x .

dy dy = a. dx dx

+ ( 2x - a)

dy =0 dx

dx

x dx + 1 + cos x

I = I 1+ I 2

2

p/ 2

ò 0

sin x 1 + cos x

dx …(i)

284


where I 1 = ò

x dx

p/ 2

0

p/ 2

0

=ò

p/ 2

1 2

1 + cos x

dx

x dx 1 + cos x

0

=

sin x

p/ 2

0

1 + cos x

I1 = ò

Now,

and I 2 = ò

x dx 2 cos 2

x 2

1 2

=

p/ 2

ò0

x . sec 2

x dx 2

éì p/ 2 xü p / 2 x ù -2ò tan dxú êí 2x . tan ý 0 2þ 0 2 ûú êëî

[Q ò sec 2 x dx = tan x + c]

p/ 2

p x ü ì = ìí . 1 - 0üý - í 2 log sec ý 2 þ0 î2 þ î

[Q

p p ì ü - 2 í log sec - log sec 0ý 2 4 î þ p = - log ( 2 ) 2 2 p I 1 = - log 2 2 p / 2 sin x dx I2 = ò 0 1 + cos x

[Q log 1 = 0]

=

Again,

Let 1 + cos x = z Also, if x = - sin x dx = dz Þ

ò tan x dx = log sec x + c]

p p , z = 1 + cos = 1 + 0 = 1 2 2

if x = 0, z = 1 + 1 = 2

sin x dx = - dz

\

I2 = ò

1

2

=ò

2

1

- dz z dz z

éQ b f ( x) dx = - a f ( x) dx ù òb êë òa úû

2

= [ log z]1 = log 2 - log 1 = log 2 Puting the values of I 1 and I 2 in (i), we get p / 2 x + sin x p p ò0 1 + cos x dx = 2 - log 2 + log 2 = 2 18. Given Þ

x dy - y dx = x 2 + y 2 dx x dy = ( y + x 2 + y 2 ) dx

Þ

2 2 dy y + x + y = dx x

285


F ( x, y) =

Let \

F ( lx, ly) =

y + x2 + y2 x l y + l2 x 2 + l2 y 2 lx l {y + x 2 + y 2 }

=

lx

= l° . F ( x, y)

Þ F( x, y) is a homogeneous function of degree zero. 2 2 dy y + x + y = dx x

Now,

y = vx dy dv =v + x. dx dx

Let Þ

Putting above value, we have 2 2 2 dv vx + x + v x = dx x dv v + x. = v + 1 + v2 Þ dx dx dv = x 1 + v2

v + x. Þ Þ

x.

dv = 1 + v2 dx

Integrating both sides, we get dx dv ò x =ò 1 + v2 é êQ ê ë

log x + log c = log|v + 1 + v 2 |

Þ

Þ

cx = v + 1 + v 2

Þ

y cx = + x

x2 + y2 x

ò

dx x2 + a2

y y2 + 1+ x x2

Þ

cx =

Þ

cx 2 = y + x 2 +`y 2

Þ

dy y = + 3x dx x

dx 19. ( y + 3x ) =x dy 2

Þ

2 dy y + 3x = dx x

ù = log| x + x 2 + a 2 | + c ú ú û

286


dy æ 1 ö + ç - ÷ . y = 3x dx è x ø dy It is in the form of + Py = Q dx 1 Here P = - and Q = 3x x Þ

ò P dx I.F. = e ò =e

\

=e

- log x

-

=e

1 dx x log

1 x

=

1 x

[Q e log z = z]

Hence, general solution is y.

1 1 = ò 3x . dx + c [General solution y ´ 1. F = ò Q ´ I . F . dx + C] x x

Þ

y = 3x + c x

Þ

y = 3x 2 + cx

20. Given, A º (1, 1, 2); B º ( 2, 3, 5) ; C º (1, 5, 5) \

®

AB = ( 2 - 1) i$ + ( 3 - 1) j$ + (5 - 2) k$ ®

AB = i$ + 2j$ + 3k$ ®

AC = (1 - 1) i$ + (5 - 1) j$ + (5 - 2) k$ = 0. i$ + 4j$ + 3k$ ® 1 ® \ The area of required triangle = | AB ´ AC| 2 $i j$ k$ ®

®

AB ´ AC = 1

2

3

0

4

3

= {( 6 - 12) i$ - ( 3 - 0) j$ + ( 4 - 0) k$} = - 6i$ - 3j$ + 4k$ \

®

®

| AB ´ AC| = ( -6) 2 + ( -3) 2 + ( 4) 2 = 61

1 61 sq. units. 61 = 2 2 21. The given equation of lines may be written as \ Required area =

®

r = (i$ - 2j$ + 3k$) + t ( -i$ + j$ - 2k$)

®

r = (i$ - j$ - k$) + s (i$ + 2j$ - 2k$)

…(i) …(ii)

287


®

®

®

®

®

Comparing given equation (i) and (ii) with r = a 1 + l b 1 and r = a 2 + l b 2 , we get ®

a 1 = i$ - 2j$ + 3k$ ,

®

b 1 = - i$ + j$ - 2k$

®

b 2 = i$ + 2j$ - 2k$

®

a 2 = i$ - j$ - k$ , ®

®

®

®

a 2 - a 1 = j$ - 4k$ j$

k$

1

-2

2

-2

i$

b 1 ´ b 2 = -1 1

= ( -2 + 4) i$ - ( 2 + 2) j$ + ( -2 - 1) k$ = 2i$ - 4j$ - 3k$ ®

®

|b 1 ´ b 2 | = 2 2 + ( -4) 2 + ( -3) 2 = 29

\

®

\ Required shortest distance =

®

( a 2 - a 1 ) .(b 1 ´ b 2 ) ®

®

|b 1 ´ b 2 | = =

( j$ - 4k$) . ( 2i$ - 4j$ - 3k$) 29

=

-4 + 12 29

8 units. 29

n

22. Q

å

Pi = 1

j=1

\

0 + k + 2k + 2k + 3k + k 2 + 2k 2 + 7 k 2 + k = 1

Þ

10k 2 + 9k - 1 = 0

Þ

10k 2 + 10k - k - 1 = 0

Þ

10k ( k + 1) - 1 ( k + 1) = 0 1 Þ ( k + 1) (10k - 1) = 0 Þ k = - 1 and k = 10 But k can never be negative as probability is never negative. 1 \ k= 10 Now, 1 (i) k = 10 (ii) P (X < 3) = P (X = 0) + P (X = 1) + P (X = 2) 3 . = 0 + K + 2K = 3K = 10

288


(iii) P (X > 6) = P (X = 7) = 7 K 2 + K =7 ´

1 1 17 + = 100 10 100

(iv) P ( 0 < X < 3) = P (X = 1) + P (X = 2) 3 . = K + 2K = 3K = 10 OR The repeated throws of a die are Bernoulli trials. Let X denotes the number of sixes in 6 throws of die. Obviously, X has the binomial distribution with n = 6 1 1 5 and p= , q=1- = 6 6 6 where p is probability of getting a six and q is probability of not getting a six Now, Probability of getting at most 2 sixes in 6 throws = P (X = 0) + P (X = 1) + P (X = 2) = 6 C 0 . p 0 . q 6 + 6 C 1 p 1 q5 + 6 C 2 p 2 q 4 6 ! 1 æ 5 ö5 6 ! æ1 ö2 5 6 = æç ö÷ + . .ç ÷ + .ç ÷ è 6ø 1 !5 ! 6 è 6ø 2 ! 4 ! è 6ø

5 . æç ö÷ è 6ø

4

5 6 1 5 5 6 ´ 5 æ1 ö2 æ5 ö4 = æç ö÷ + 6 . æç ö÷ + ´ ç ÷ .ç ÷ è 6ø è 6ø è 6ø 6 è 6ø 2 5 4 = æç ö÷ è 6ø

é 25 + 5 + 5 ù êë 36 6 12 úû

5 4 25 + 30 + 15 æ 5 ö 4 70 = æç ö÷ ´ =ç ÷ ´ è 6ø è 6ø 36 36 =

21875 23328

SECTION – C 23. The system can be written as AX = B é4 where A = ê 1 ê êë 6

3 2 2

Þ

2ù é xù é 60ù 3ú , X = ê yú and B = ê 45ú ú ê ú ê ú êë z úû êë 70 úû 3úû

| A| = 4 ( 6 - 6) - 3 ( 3 - 18) + 2 ( 2 - 12) = 0 + 45 - 20 = 25 ¹ 0

X = A -1 B

…(i)

289


For adj A A11 = 6 - 6 = 0

A21 = - ( 9 - 4) = - 5

A 31 = ( 9 - 4) = 5

A12 = - ( 3 - 18) = 15

A22 = (12 - 12) = 0

A 32 = - (12 - 2) = - 10

A13 = ( 2 - 12) = - 10

A23 = - ( 8 - 18) = 10

é 0 adj A = ê -5 ê êë 5

\

\

A

-10ù é 0 10ú = ê 15 ú ê êë -10 5úû

15 0 -10

-5

é 0 1 ê = 15 25 ê êë -10

-1

0 10 -1

é 0 5 ê = 3 25 ê êë -2 Now putting values in (i), we get é xù ê yú = 1 \ ê ú 5 êë z úû

A 33 = ( 8 - 3) = 5

T

0 2 é 0 ê 3 ê êë -2

-5 0 10

5ù -10ú ú 5úû

5ù -10ú ú 5úû 1ù 1 -2 ú = ú 5 1úû -1 0 2

é 0 ê 3 ê êë -2

-1 0 2

1ù -2 ú ú 1úû

1ù é 60ù -2ú ê 45ú úê ú 1úû êë 70 úû

Þ

é xù ê yú = 1 ê ú 5 êë z úû

é 0 - 45 + 70 ù ê 180 + 0 - 140 ú ê ú êë -120 + 90 + 70úû

Þ


é 25ù é 5 ù ê 40ú = ê 8ú ê ú ê ú êë 40úû êë 8úû

Hence, x = 5, y = 8, z = 8. 24. Let ABC be right-circular cone having radius ‘r’ and height ‘h’. If V and S are its volume and surface area (curved) respectively, then A

S = prl S = pr h 2 + r 2

…(i)

Putting the value of h in (i), we get S = pr

Þ

9V 2 p 2r 4

+ r2

æ 9V 2 + p 2 r 6 ö ÷ S2 = p 2r 2 ç ç ÷ p 2r 4 è ø

l h

éQ V = 1 pr 2 h r C B D ê 3 ê ê h = 3V êë pr 2 [Maxima or Minima is same for S or S 2 ]

290


S2 =

Þ

9V 2 r

2

-18V 2

Þ

(S2 ) ¢ =

Now,

(S2 ) ¢ = 0

Þ

- 18

V2 3

+ p 2r 4

r3

+ 4p 2 r 3

…(ii)

+ 4p 2 r 3 = 0

r 4p 2 r 6 = 18V 2 1 4p 2 r 6 = 18 ´ p 2 r 4 h 2 9

Þ Þ

2r 2 = h 2

Þ

[Differentiating w.r.t. ‘r’]

Þ

[Putting value of V] r=

h 2

Differentiating (ii) w.r.t. ‘r’, again 54V 2 ( S 2 ) ¢¢ = + 12p 2 r 2 r4 Þ ( S 2 ) ¢¢] h >0 r =

(for any value of r)

2

h or h = 2 r. 2 i.e., For least curved surface, altitude is equal to 2 times the radius of the base. OR Let x and y be the dimensions of rectangular part of window and x be side of equilateral part. If A be the total area of window, then 3 2 A = x. y + x 4 Also x + 2y + 2x = 12 x x Þ 3x + 2y = 12 12 - 3x Þ y= 2 (12 - 3x) 3 2 y y \ A = x. + x 2 4 3x 2 3 2 Þ A = 6x + x x 2 4 3 [Differentiating w.r.t. x] Þ A¢ = 6 - 3x + x 2 Now, for maxima or minima A¢ = 0 3 6 - 3x + x=0 2 2

Hence, S i.e., S is minimum for r =

291


Þ

x=

12 6- 3 3 < 0 (for any value of x) 2

A ¢¢ = - 3 +

Again

A ¢¢]

x=

12 6- 3

12 and y = 6- 3

i.e., A is maximum if x =

<0 æ 12 ö ÷ 12 - 3 ç è6 - 3ø 2

.

i.e., For largest area of window, dimensions of rectangle are x= p/ 3

25. Let

I=

12 6- 3

dx ò 1 + tan x = p/ 6 p/ 3

I=

p/ 6

1+

ò

dx sin x cos x …(i)

sin x dx

ò

…(ii)

sin x + cos x

p/ 6

p/ 3

sin x + cos x

ò

sin x + cos x

p/ 6

dx

p/ 3

ò

dx = [x]pp // 63

p/ 6

\

1 é p p ù 1 é 2p - p ù = 2 êë 3 6 úû 2 êë 6 úû p I= 12 I=

.

p p p p cos æç + - xö÷ + sin æç + - xö÷ è3 6 ø è3 6 ø

p/ 3

2I =

3

p p cos æç + - xö÷ dx è3 6 ø

p/ 6

Adding (i) and (ii), 2I =

ò

6-

cos x + sin x

p/ 3

I=

p/ 3

18 - 6 3

cos x dx

ò

p/ 6

=

and y =

292


OR 6x + 7

Let

I=ò

Now, Let

6x + 7 = A .

( x - 5) ( x - 4)

dx = ò

6x + 7 x 2 - 9x + 20

dx

d 2 ( x - 9x + 20) + B dx

6x + 7 = A ( 2x - 9) + B Þ

6x + 7 = 2Ax - 9A + B

Comparing the coefficient of x, we get 2A = 6 and -9A + B = 7 A = 3 and B = 34 3 ( 2x - 9) + 34 I=ò dx x 2 - 9x + 20

\

( 2x - 9) dx

= 3ò

where

Now,

2

x - 9x + 20

I = 3I 1 + 34I 2 ( 2x - 9) dx I1 = ò x 2 - 9x + 20 I1 = ò

+ 34 ò

dx 2

x - 9x + 20 …(i)

and I 2 = ò

dx 2

x - 9x + 20

( 2x - 9) dx x 2 - 9x + 20

Let x 2 - 9x + 20 = z 2 ( 2x - 9) dx = 2z dz \

I1 = 2 ò

z dz z

= 2z + c 1

I 1 = 2 x 2 - 9x + 20 + c 1 I2 = ò

=ò

I2 = ò

dx 2

x - 9x + 20

dx

=ò

dx 2 æx - 9 ö - 1 ç ÷ è 2ø 4

dx 2 2 æx - 9 ö - æ 1 ö ç ÷ ç ÷ è è 2ø 2ø

x2 - 2 .

9 9 2 81 x + æç ö÷ + 20 è 2ø 2 4

293


9 9 2 1 2 = log æç x - ö÷ + æç x - ö÷ - æç ö÷ + C 2 è è è 2ø 2ø 2ø é êQ ê ë

dx

ò

2

x -a

2

= log| x + x 2 - a 2| + x

9 = log æç x - ö÷ + x 2 - 9x + 20 + C 2 è 2ø Putting the value of I 1 and I 2 in (i) ì ü 9 I = 6 x 2 - 9x + 20 + 3c 1 + 34 í log æç x - ö÷ + x 2 - 9x + 20 ý + 34C 2 è ø 2 î þ 9 = 6 x 2 - 9x + 20 + 34 log æç x - ö÷ + x 2 - 9x + 20 + C è 2ø

\

where C = 3c 1 + 34c 2 . 26. For graph of y =| x + 3| 0

–3

–6

–2

–4

y

3

0

3

1

1

+

3| +

|x

y

=

3|

|x

=

x=0

y

x = –6

y-axis

x

(0, 3)

3 |

(–6, 3)

2 1

–6

|

|

|

|

|

–5 –4 –3 –2 (–3, 0)

|

–1 0

|

1

x-axis

Shaded region is the required region. Hence, Required area

=ò

0

-6

=ò

-3

-6

=ò

-3

-6

| x + 3| dx | x + 3| dx + ò

0

| x + 3| dx [By Property of definite integral]

-3

- ( x + 3) dx + ò

0

-3

-3

é x + 3 ³ 0 if - 3 £ x £ 0 ( x + 3) dx ê ë x + 3 £ 0 if - 6 £ x £ - 3 0

é x2 ù é x2 ù =-ê + 3xú + ê + 3xú êë 2 úû -6 êë 2 úû -3

294


36 9 é 9 ù é ù = - ê æç - 9ö÷ - æç - 18ö÷ ú + ê 0 - æç - 9ö÷ ú è ø è ø è øû 2 2 ë 2 û ë 9 9 = + = 9 sq. units. 2 2 27. Given line and plane are ®

r = ( 2i$ - j$ + 2k$) + l ( 3i$ + 4j$ + 2k$)

…(i)

®

r . (i$ - j$ + k$) = 5

…(ii) ®

For intersection point, we solve equations (i) and (ii) by putting the value of r from (i) in (ii). [( 2i$ - j$ + 2k$) + l ( 3i$ + 4j$ + 2k$)] .(i$ - j$ + k$) = 5 Þ

( 2 + 1 + 2) + l ( 3 - 4 + 2) = 5 Þ 5 + l = 5 Þ l = 0 Hence, position vector of intersecting point is 2i$ - j$ + 2k$. i.e., coordinates of intersection of line and plane is (2, –1, 2). Hence, Required distance = ( 2 + 1) 2 + ( -1 + 5) 2 + ( 2 + 10) 2 = 9 + 16 + 144 = 169 = 13 28. Let E1 , E2 , E 3 be events such that E1 º Selection of Box I ; E2 º Selection of Box II ;

E 3 º Selection of Box III

Let A be event such that A º the coin drawn is of gold Now, P (E1 ) =

æAö 2 P ç ÷ = P (a gold coin from box I) = = 1 2 è E1 ø

1 1 1 , P (E2 ) = , P (E 3 ) = , 3 3 3

æAö P ç ÷ = P (a gold coin from box II) = 0, è E2 ø

æAö 1 P ç ÷ = P (a gold coin from box III) = 2 èE3 ø

æE ö the probability that the other coin in the box is also of gold = P ç 1 ÷ èAø

\


æAö P (E1 ) . P ç ÷ è E1 ø æAö æAö æAö P (E1 ) . P ç ÷ + P (E2 ) . P ç ÷ + P (E 3 ) . P ç ÷ è E1 ø è E2 ø èE3 ø

1 ´1 2 3 = = × 1 1 1 1 3 ´1+ ´ 0+ ´ 3 3 3 2

295


29. Let the number of desktop and portable computers to be sold be x and y respectively. Here, Profit is the objective function z. \

z = 4500x + 5000y

…(i)

we have to maximise z subject to the constraints x + y £ 250

…(ii) (Demand Constraint)

25000x + 40000y £ 70, 00, 000 Þ

…(iii) (Investment constraint)

5x + 8y £ 1400 x ³ 0, y ³ 0

…(iv) (Non-negative constraint)

Graph of x = 0 and y = 0 is the y-axis and x-axis respectively. \ Graph of x ³ 0, y ³ 0 is the Ist quadrant. Graph of x + y £ 250 :

Graph of 5x + 8y £ 1400 : Graph of 5x + 8y = 1400 x

0

280

y

175

0

\ Graph of 5x + 8y £ 1400 is the part of Ist quadrant where origin lies.

300 – 250 – ( 200 – 0, 1 75 ) 150 –

100 – C(200, 50) 50 – 0

(280, 0) –

\ Graph of x + y £ 250 is the part of Ist quadrant where origin lies.

–

0

–

250

–

y

–

250

–

0

–

x

x=0

y-axis

Graph of x + y = 250

50 100 150 200 250 300 350 y = 0 A x+ y= 25 0

For cooridnates of C, equation x + y = 250 and 5x + 8y = 1400 are solved and we get x = 200, y = 50 Now, we evaluate objective function Z at each corner Corner Point

z = 4500x + 5000y

O ( 0, 0)

0

A ( 250, 0)

1125000

C ( 200, 50)

1150000

B ( 0, 175)

875000

maximum

Maximum profit is ` 11,50,000/- when he plan to sell 200 unit desktop and 50 portable computers.

296


CBSE (All India) Set–II 9. Let Þ

log x = z 1 dx = z x

(differentiating both sides)

ò

Now,

(log x) 2 x

dx = ò z 2 dz =

1 z3 + c = (log x) 3 + c 3 3

®

10. Required unit vector in the direction of a ®

a

=

®

| a| 19. L.H.S.

D= a-x a-x 3a - x

Now,

2

2

2 +1 + 2

2

=±

1 1 = 2 tan -1 æç ö÷ + tan -1 æç ö÷ è 2ø è7 ø 1 2´ -1 2 + tan -1 æç 1 ö÷ = tan è7 ø 1 2 1 - æç ö÷ è 2ø 4 1 = tan -1 + tan -1 æç ö÷ è 3 7ø 4 1 + = tan -1 3 7 4 1 1- ´ 3 7 31 = tan -1 æç ö÷ = R.H.S. è 17 ø a+x

20. Given,

2i$ + j$ + 2k$

=

1 $ $ ( 2i + j + 2k$) 3

[By Property -1 £

[Q

4 1 ´ < 1] 3 7

a-x a-x a+ x a- x =0 a-x a+x 3a - x 3a - x

D= a-x

x+a

a - x R1 ® R1 + R2 + R 3

a-x

a-x

a+x

1

1

1

= ( 3a - x) a - x

a+x

a-x

a-x

a-x

a+x

1 < 1] 2

297


= ( 3a - x)

0

0

1

0

2x

a-x

-2 x

-2 x

a+x

C 1 ®C 1 - C 3 C2 ® C2 - C 3

= ( 3a - x) [1 ( 0 + 4x 2 )]


= 4x 2 ( 3a - x) 4x 2 ( 3a - x) = 0

\ Þ

x = 0 or

x = 3a

p/ 4

21. Let

I=

ò log (1 + tan x) dx 0 p/ 4

=

ò 0 p/ 4

=

ò 0 p/ 4

=

ò 0 p/ 4

=

ò 0

é log ê1 + tan ë

ò

éQ êë

æ ö 2 log ç ÷ dx è 1 + tan x ø p/ 4

log 2dx -

ò

log (1 + tan x) dx

0

I = log 2 [x] p0 / 4 - I

Þ

p log 2 4 p I = log 2 8

2I =

22. x dy - ( y + 2x 2 ) dx = 0 The given differential equation can be written as dy dy 1 x - y = 2x 2 or - . y = 2x dx dx x I.F. = e

-

a

f ( x) dx = ò f ( a - x) dxù úû 0

p/ 4 æ æ 1 + tan x + 1 - tan x ö 1 - tan x ö log ç1 + ÷ dx = ò log ç ÷ dx 1 + tan x ø 1 + tan x è è ø 0

0

Þ

a

ò0

p æ tan - tan x ö÷ ç 4 log ç1 + ÷ dx p ç 1 + tan tan x ÷ è ø 4

p/ 4

=

æ p - xö ù dx ç ÷ú è4 øû

1

òx

dx

= e-

log x

= e log

x -1

=

1 x

298


1 1 = ò 2x . dx x x 1 y . = 2x + C or x

\ Solution is

y.

Þ

y = 2x 2 + Cx

28. The given system can be written as AX = B é1 where A = ê 1 ê êë 2

2

Þ

X = A -1 B

…(i)

-3

1ù é xù é 7ù ú ê ú 3 , X = y and B = ê11ú ú ê ú ê ú êë zúû êë 1úû 0úû

1

2

1

| A| = 1

0

3 = 1 ( 0 + 9) - 2 ( 0 - 6) + 1 ( -3 - 0) = 18 ¹ 0

2

-3

0

0

For adj A

\

\

A11 = 0 + 9 = 9

A12 = - ( 0 - 6) = 6

A13 = - 3 - 0 = - 3

A21 = - ( 0 + 3) = - 3

A22 = 0 - 2 = - 2

A23 = - ( -3 - 4) = 7

A 31 = 6 - 0 = 6

A 32 = - ( 3 - 1) = - 2

A 33 = 0 - 2 = - 2

-3 ù 7ú ú -2úû

6 é 9 ê adj. A = -3 -2 ê êë 6 -2 1 A -1 = . adj. A | A| -3

é 9 1 ê = 6 18 ê êë -3

-2 7

T

é 9 =ê 6 ê êë -3

-3 -2 7

6ù -2 ú ú -2úû

6ù -2 ú ú -2úû

Now putting above values in (i), we get -3 6ù é 7 ù é xù é 9 ê yú = 1 ê 6 -2 -2ú ê11ú ê ú 18 ê úê ú êë z úû êë -3 7 -2úû êë 1 úû é 63 - 33 + 6 ù ê 42 - 22 - 2 ú ê ú êë -21 + 77 - 2úû

Þ


Þ

é xù ê yú = ê ú êë z úû

Þ

x = 2, y = 1, z = 3

é 36ù 1 ê ú Þ = 18 18 ê ú êë 54 úû

é 2ù ê1 ú ê ú êë 3úû [From equality of matrices]

299


29. Two given planes are ®

r . (i$ + j$ + k$) - 1 = 0

®

r . ( 2i$ + 3j$ - k$) + 4 = 0

It’s cartesian forms are and

x+ y+z-1= 0

…(i)

2x + 3y - z + 4 = 0

…(ii)

Now, equation of plane passing through line of intersection of plane (i) & (ii) is given by ( x + y + z - 1) + l ( 2x + 3y - z + 4) = 0 (1 + 2l) x + (1 + 3l) y + (1 - l) z - 1 + 4l = 0

…(iii)

Since (iii) is parallel to x-axis Þ Normal of plane (iii) is perpendicular to x-axis. Þ Þ

(1 + 2l) . 1 + (1 + 3l) . 0 + (1 - l) . 0 = 0 [Q Direction ratios of x-axis are (1, 0, 0)] 1 1 + 2l = 0 Þ l=2

Hence, required equation of plane is 3 1 1 0x + æç1 - ö÷ y + æç1 + ö÷ z - 1 + 4 ´ - = 0 è ø è ø 2 2 2 1 3 Þ - y+ z-1- 2= 0 2 2 ®

y - 3z + 6 = 0 or r . ( j$ - 3k$) + 6 = 0

Þ

CBSE (All India) Set–III 1. Let tan -1 x = z 1 1 + x2 \

ò

e tan

-1

1+x

2

dx = dz x

[Differentiating we get]

dx = ò e z . dz = e z + c = e tan ®

-1 x

®

2. If q be the angle between a and b , then ® ®

®

®

a . b =| a |.| b |cos q

+c

300


6= \

cos q =

2´

3

1 = tan -1 æç ö÷ + tan -1 è 2ø = tan

-1

3. 2

=

2 3

=

2 1 = 2 2

= tan

-1

æ 1 ö + tan -1 ç ÷ è5 ø

p × 4

x

x

x+a

x

x

x

x+a

3x + a =

x

3x + a 3x + a

x

x+a

x

x

x

x+a

0

0

3x + a

= 0

a

x

-a

x+a

-a

éQ 1 ´ 1 = 1 < 1ù êë 2 5 10 úû

æ 7 +1 ö æ 7 ö + tan -1 æ 1 ö = tan -1 ç 9 8 ÷ = tan -1 æ 65 ´ 72 ö ç ÷ ç ÷ ç ÷ ç ÷ è 9ø è 8ø è 72 65 ø ç1 - 7 ´ 1 ÷ è 9 8ø

x+a D=

æ1 ö ç ÷ è 8ø

1 1 + 2 5 + tan -1 æç 1 ö÷ è 8ø 1 1 1- ´ 2 5

= tan -1 (1) =

12. Let

6

æ 1 ö p q = cos -1 ç ÷= è 2ø 4

\ 11. L.H.S.

3 . 2 cos q

R1 ® R1 + R2 + R 3

C1 ® C1 - C 3 C2 ® C2 - C 3

= ( 3x + a) ( 0 + a)


= a ( 3x + a) = 3ax + a Given \

2

D=0 3ax + a 2 = 0 x=-

a a2 =3a 3

301


1 log æç - 1ö÷ dx èx ø 1 xö 1 æ = ò log ç ÷ dx 0 è x ø

13. Let I = ò

1

0

I=ò

æ 1 - (1 - x) ö log ç ÷ dx è 1-x ø

1

0

I=ò

…(i) éQ êë

a

ò0

a

f ( x) dx = ò f ( a - x) dxù úû 0

æ x ö log ç ÷ dx è1 - x ø

1

0

…(ii)

Adding (i) and (ii), we get æ x ö 1 1 æ1 - x ö 2I = ò log ç ÷ dx ÷ dx + ò0 log ç 0 è x ø è1 - x ø =ò

æ1 - x x ö log ç . ÷ dx è x 1 - xø

1

0

=ò

1

log 1 dx

0

2I = 0 \ 14. x dy + ( y - x 3 ) dx = 0

I=0 x dy = - ( y - x 3 ) dx

Þ

dy - y = + x2 dx x

Þ It is in the form of where P =

Þ Þ

3 dy - y + x = dx x dy æ 1 ö + ç ÷ .y = x2 dx è x ø

dy + Py = Q dx

1 and Q = x 2 x

1

\ I.F. = e

[Q log A + log B = log ( A ´ B)]

òx

dx

= e log

x

=x

Hence, solution is y . x = ò x . x 2 dx + C x4 x3 C +C Þ y= + 4 4 x 23. The given system of equation can be written as AX = B Þ X = A -1 B xy =

é1 where A = ê 2 ê êë 3

2 -3

1

2

3

Now, | A| = 2

3

3

-3

-3 ù é xù é -4 ù ú ê ú 2 , X = y , B = ê 2ú ú ê ú ê ú êë zúû êë 11úû -4úû -3 2 = 1 ( -12 + 6) - 2 ( -8 - 6) - 3 ( -6 - 9) = 67 ¹ 0 -4

…(i)

302


For adj A : A11 = - 6 A12 = 14 A13 = - 15

A21 = 17 A 31 = 13 A22 = 5 A 32 = - 8 A23 = 9 A 33 = - 1 T 6 14 15 17 13ù é ù é -6 ê ú ê \ adj. A = 17 5 9 = 14 5 -8 ú ê ú ê ú êë 13 êë -15 -8 -1úû 9 -1úû 1 \ A -1 = . adj. A | A| 17 13ù é -6 1 ê = 14 5 -8 ú ú 67 ê êë -15 9 -1úû Putting the value of X , A -1 and B in (i), we get 17 13ù é -4ù é xù é -6 é xù é 24 + 34 + 143ù é 201ù 1 ê 1 ê ê yú = 1 ê 14 ú ê ú ê ú ú 5 -8 2 Þ y = -56 + 10 - 88 = -134ú ê ú 67 ê úê ú ê ú 67 ê ú 67 ê ú êë z úû êë -15 êë z úû êë 67 úû 9 -1úû êë 11úû ëê 60 + 18 - 11 úû é xù é 3ù ê y ú = ê -2 ú Þ ê ú ê ú êë z úû êë 1úû Þ x = 3, y = - 2, z = 1 24. The given planes are …(i) 2x + y - z - 3 = 0 and …(ii) 5x - 3y + 4z + 9 = 0 The equation of the plane passing through the line of intersection of (i) and (ii) is given by ( 2x + y - z - 3) + l (5x - 3y + 4z + 9) = 0 …(iii) Þ ( 2 + 5l) x + (1 - 3l) y + ( 4l - 1) z + ( 9l - 3) = 0 x-1 y- 3 z-5 It is given that plane (iii) is parallel to = = . 2 4 5 Þ Normal of (iii) is perpendicular to given line. \ ( 2 + 5l) . 2 + (1 - 3l) . 4 + ( 4l - 1) . 5 = 0 Þ 18l + 3 = 0 1 Þ l=6 Putting the value of l in (iii), we get the required plane. 1 ( 2x + y - z - 3) - (5x - 3y + 4z + 9) = 0 6 Þ 12x + 6y - 6z - 18 - 5x + 3y - 4z - 9 = 0 Þ 7 x + 9y - 10z - 27 = 0

EXAMINATION PAPERS –2011 CBSE (Foreign) Set–I Time allowed: 3 hours

Maximum marks: 100

General Instructions : As given in Examination Paper (Delhi) – 2011.

SECTION – A Question numbers 1 to 10 carry one mark each. 1. If f : R ® R is defined by f ( x) = 3x + 2, define f [ f ( x)]. 2. Write the principal value of tan -1 ( -1). 3. Write the values of x - y + z from the following equation : é x + y + zù é 9 ù ê x + z ú = ê5 ú ê ú ê ú êë y + z úû êë7 úû 4. Write the order of the product matrix : é1 ù ê 2 ú [2 3 4] ê ú êë 3úû 5. If

x x 3 4 = , write the positive value of x. 1 x 1 2

6. Evaluate :

ò

(1 + log x) 2 x

dx.

7. Evaluate :

ò1

3

dx 1 + x2

.

8. Write the position vector of the mid-point of the vector joining the points P ( 2, 3, 4) and Q ( 4, 1, -2). ® ®

® ®

®

9. If a . a = 0 and a . b = 0, then what can be concluded about the vector b ? 10. What are the direction cosines of a line, which makes equal angles with the co-ordinates axes?

304


SECTION – B Question numbers 11 to 22 carry 4 marks each. 11. Consider f : R + ® [4, ¥] given by f ( x) = x 2 + 4. Show that f is invertible with the inverse ( f -1 ) of f given by f -1 ( y) = y - 4, where R + is the set of all non-negative real numbers. 12. Prove the following : æ2 2ö 9p 9 1 9 - sin -1 æç ö÷ = sin -1 ç ÷ è 3ø 4 8 4 è 3 ø OR Solve the following equation for x : æ1 - x ö 1 tan -1 ç ÷ = tan -1 ( x), x > 0 è1 + x ø 2 13. Prove, using properties of determinants : y+k y y y

y+k

y

y

y

y+k

= k 2 ( 3 y + k)

14. Find the value of k so that the function f defined by ì k cos x p ïï p - 2x , if x ¹ 2 f ( x) = í ï 3 , if x = p ïî 2 p is continuous at x = . 2 15. Find the intervals in which the function f given by f ( x) = sin x + cos x, 0 £ x £ 2p is strictly increasing or strictly decreasing. OR Find the points on the curve y = x 3 at which the slope of the tangent is equal to the y-coordinate of the point. 16. Prove that : d éx a2 x ù a2 - x2 + sin -1 æç ö÷ ú = a 2 - x 2 ê è a øú dx ëê 2 2 û OR d2y dy If y = log [x + x 2 + 1 ] , prove that ( x 2 + 1) +x = 0. 2 dx dx 17. Evaluate :

òe

2x

sin x dx OR

Evaluate :

ò

3x + 5 x 2 - 8x + 7

dx

305


18. Find the particular solution of the differential equation : (1 + e 2 x ) dy + (1 + y 2 ) e x dx = 0, given that y = 1, when x = 0. 19. Solve the following differential equation : dy p + y cot x = 4x cosec x, given that y = 0 when x = . dx 2 ® ® ® ® ® 20. If vectors a = 2i$ + 2j$ + 3k$, b = - i$ + 2j$ + k$ and c = 3i$ + j$ are such that a + l b is ®

perpendicular to c , then find the value of l. 21. Find the shortest distance between the lines : ® r = 6 i$ + 2j$ + 2k$ + l (i$ - 2j$ + 2k$) and ®

r = - 4 i$ - k$ + m ( 3i$ - 2j$ - 2k$)

22. Find the mean number of heads in three tosses of a fair coin.

SECTION – C Question numbers 23 to 29 carry 6 marks each. 2 ù é -2 0 1ù é 1 -1 ê ú ê 23. Use product 0 2 -3 9 2 -3ú to solve the system of equations : ê úê ú êë 3 -2 4úû êë 6 1 -2úû x - y + 2z = 1 2y - 3z = 1 3x - 2y + 4z = 2 OR Using elementary transformations, find the inverse of the matrix : æ 2 0 -1 ö ç ÷ 0÷ ç5 1 ç0 1 3÷ø è 24. A window is in the form of a rectangle surmounted by a semi-circular opening. The total perimeter of the window is 10 metres. Find the dimensions of the rectangle so as to admit maximum light through the whole opening. 25. Using the method of integration, find the area of the region bounded by the lines : 2x + y = 4 3x - 2y = 6 x - 3y + 5 = 0 4

26. Evaluate ò ( x 2 - x) dx as a limit of sums. 1

OR Evaluate : p/ 4

ò0

sin x + cos x 9 + 16 sin 2x

dx

27. Find the equation of the plane passing through the point (–1, 3, 2) and perpendicular to each of the planes :

306


x + 2y + 3z = 5 and 3x + 3y + z = 0 28. A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of grinding/cutting machine and a sprayer. It takes 2 hours on the grinding/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp. It takes one hour on the grinding/cutting machine and 2 hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at the most 20 hours and the grinding/cutting machine for at the most 12 hours. The profit from the sale of a lamp is ` 5 and that from a shade is ` 3. Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximise his profit? Make an L.P.P. and solve it graphically. 29. A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Futher, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B?

CBSE (Foreign) Set–II 9. Write fog, if f : R® R and g : R ® R are given by f ( x) =| x| and g( x) =|5x - 2|. 10. Evaluate : e 2 x - e -2 x

ò e 2 x + e -2 x

dx

19. Prove, using properties of determinants : a-b - c 2a

2a

2b

b-c-a

2b

2c

2c

c - a-b

= ( a + b + c) 3

20. Find the value of k so that the function f, defined by ì kx + 1, if x £ p f ( x) = í î cos x, if x > p is continuous at x = p. 21. Solve the following differential equation: dy p + 2y tan x = sin x. given that y = 0, when x = . dx 3 22. Find the shortest distance between the lines : ®

r = (i$ + 2j$ + 3k$) + l (i$ - 3j$ + 2k$) and

®

r = ( 4i$ + 5j$ + 6k$) + m ( 2i$ + 3j$ + k$)

307


28. Find the vector equation of the plane, passing through the points A ( 2, 2, - 1), B ( 3, 4, 2) and C (7 , 0, 6). Also, find the cartesian equation of the plane. 29.

Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II at random. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.

CBSE (Foreign) Set–III 1. Write fog, if f : R ® R and g : R ® R are given by f ( x) = 8x 3 and g( x) = x 1/ 3 . 2. Evaluate :

ò

cos x x

dx

11. Prove, using properties of determinants : x + y + 2z

x

y

z

y + z + 2x

y

z

x

z + x + 2y

= 2 ( x + y + z) 3

12. For what value of l is the function 2 ïì l ( x - 2x) , if x £ 0 f ( x) = í ïî 4x + 1 , if x > 0

continuous at x = 0 ? 13. Solve the following differential equation : dy 1 (1 + x 2 ) + 2xy = , given y = 0 when x = 1. dx 1 + x2 14. Find the shortest distance betwen the lines : ®

r = (i$ + 2j$ + k$) + l (i$ - j$ + k$)

and

®

r = ( 2i$ - j$ - k$) + m ( 2i$ + j$ + 2k$)

23. Find the equation of the palne passing through the point (1, 1, –1) and perpendicular to the planes x + 2y + 3z - 7 = 0 and 2x - 3y + 4z = 0. 24. There are three coins. One is a two headed coin (having heads on both faces), another is a biased coin that comes up heads 75% of the times and the third is an unbiased coin. One of the three coins is chosen at random and tossed, and it shows heads. What is the probability that it was the two-headed coin?

308


Solutions CBSE (Foreign) Set–I Section – A f ( f ( x)) = f (3x + 2)

1.

= 3. (3x + 2) + 2 = 9x + 6 + 2 = 9x + 8 2.

Let

tan -1 ( -1) = q

Þ

tan q = -1

Þ

tan q = - tan

p 4

p æ pö tan q = tan ç - ÷ Þ q = 4 è 4ø p Þ tan -1( -1) = 4 p \ Principal value of tan -1 ( -1) is - . 4 3. We have éx + y + z ù é9ù ê x + z ú = ê5 ú ê y + z ú ê7 ú ë û ë û Þ

p æ p pö \- Î ç - , ÷ range of the 4 è 2 2ø principal value branch of tan -1 æ pö function and tan ç - ÷ = -1 è 4ø

By definition of equality of matrices, we have x+ y + z =9 x + z =5 y + z =7 (i) – (ii) Þ x + y + z - x - z = 9 -5 Þ y =4 (ii) – (iv) Þ x - y + z =5 -4 Þ x - y + z =1 4. Order is 3 ´ 3 because it is product of two matrices having order 3 ´ 1 and 1 ´ 3. 5. We have x x 3 4 Q = 1 x 1 2 Þ

x2 -x = 6 -4

Þ x2 -x -2 = 0

... (i) .... (ii) ... (iii) ... (iv)

309


x 2 - 2x + x - 2 = 0 Þ x( x - 2) + 1 ( x - 2) = 0

Þ Þ Þ Þ

( x - 2) ( x + 1) = 0 x = 2 or x = -1 (Not accepted) x=2 (1 + log x) 2 I =ò dx x 1 + log x = z 1 dx = dz Þ I = ò z 2 dz x

6. Let

= 7.

I =ò

1

3

1 z3 + C = (1 + log x) 3 + C 3 3 dx

1 + x2 é d 1 ù -1 êQ dx (tan x) = ú ë 1+ x2 û

= [tan -1 x]1 3 = tan -1 ( 3 ) - tan -1 (1) p p p = - = 3 4 12 ® ®

8. Let a . b be position vector of points P( 2, 3, 4) and Q( 4, 1, - 2) respectively. ®

a = 2$i + 3$j + 4k$

\

®

b = 4$i + $j - 2k$ ®

®

a + b 6$i + 4$j + 2k$ \ Position vector of mid point of P and Q = = 2 2 = 3$i + 2$j + k$ ® ®

9. \ a . a = 0 ®

Þ

®

Þ

®

a . a . cos q = 0 ®

® 2

Þ

[\cos 0 = 1]

a . a =0

a

®

=0

Þ

a =0

® ® é ® ® ® ® ù Þ b may be any vector êas a . b = a . b . cos q = 0 b . cos q = 0ú ë û 10. Let a be the angle made by line with coordinate axes. Þ Direction cosines of line are cos a , cos a , cos a

310


cos 2 a + cos 2 a + cos 2 a = 1

Þ

3 cos 2 a = 1 Þ cos 2 a =

Þ Þ

cos a = ±

1 3

1

3 Hence, the direction cosines, of the line equally inclined to the coordinate axes are 1 1 1 ± ,± ,± 3 3 3 [Note : If l, m, n are direction cosines of line, then l 2 + m 2 + n 2 = 1]

Section – B 11. For one-one Let

x1 , x 2 Î R (Domain) f ( x1 ) = f ( x 2 ) Þ x12 + 4 = x 22 + 4 x12 = x 22

Þ Þ f is one-one function. For onto Let y Î [ 4, ¥) s.t.

[\x1 , x 2 are +ve real number]

x1 = x 2

y = f ( x) " x Î R t y = x2 + 4

(set of non-negative reals)

Þ Þ

x = y -4

[\x is + ve real number]

Obviously, " y Î [ 4, a ] , x is real number Î R (domain) i.e., all elements of codomain have pre image in domain. Þ f is onto. Hence f is invertible being one-one onto. For inverse function : If f -1 is inverse of f, then (Identity function) fof -1 = I fof -1 ( y) = y

Þ Þ

f (f

-1

-1

" y Î [ 4, ¥)

( y)) = y

( y)) 2 + 4 = y

Þ

(f

Þ

f -1 ( y ) = y - 4

[ Q f ( x) = x 2 + 4]

Therefore, required inverse function is f -1 [ 4, ¥ ] ® R defined by f -1 ( y) = y - 4 " y Î [ 4, a). 12. L.H.S.

9p 9 1 - sin -1 8 4 3 9æp -1 1 ö = ç - sin ÷ 4è2 3ø =

311


=

9 æ1ö cos -1 ç ÷ 4 è3ø

é 1 ù êëQ 3 Î [ -1, 1]úû

æ1ö Let cos -1 ç ÷ = q è3ø Þ \ Þ Þ

\

Þ

Þ

cos q =

1 3

[ q Î [ 0, p] ]

æ1ö sin q = + 1 - ç ÷ è3ø

é Q q Î [ 0, p] ù ëêÞ sin q is + ve ûú

8 2 2 = 9 3 æ2 2ö æ2 2ö 1 ÷ Þ cos -1 æç ö÷ = sin -1 ç ÷ q = sin -1 çç ÷ ç 3 ÷ è3ø è 3 ø è ø Putting the value of q] æ ö 9 2 2 ÷ = R.H.S. L.H.S = sin -1 çç ÷ 4 è 3 ø sin q =

OR æ ö 1-x 1 x 1 tan -1 ç ÷ = tan -1 x Þ 2 tan -1 1+ x è1 + x ø 2 æ1-x ö 2ç ÷ è1 + x ø tan -1 = tan -1 x 2 æ1-x ö 1-ç ÷ è1 + x ø æ ö 2 (1 - x 2 ) ÷ = tan -1 x tan -1 ç ç (1 + x) 2 - (1 - x) 2 ÷ è ø

1-x2 =x 2x 1 Þ x= 3 y+k y 13. L.H.S. = y y+k y y y Þ

2

= tan -1 x éBy property ù ê ú 1– x < 1 as x > 0ú êHere – 1 < ë û 1+ x

Þ 3x 2 = 1 [\ x > 0] y y +k

3y + k y y = 3y + k y + k y 3y + k y y+k 1 y y = (3y + k) 1 y + k y 1 y y+k

[Applying C1 ® C1 + C 2 + C 3

[Taking common (3y + k) from C1

312


éApplying êR 2 ® R 2 - R1 êR ® R - R 3 1 ë 3

1 y y = (3y + k) 0 k 0 0 0 k Expanding along C1 we get = (3y + k) {1 ( k 2 - 0) - 0 + 0} = (3y + k). k 2 = k 2 (3y + k)

p é êLet x = 2 - h ê pê x® Þ h=0 ë 2

æp ö 14. lim f ( x) = lim f ç - h ÷ h ®0 è2 ø p x® 2

æp ö k cos ç - h ÷ è2 ø = lim p h ®0 æ ö p -2 ç -h ÷ è2 ø k sin h = lim h ®0 2h sin h k k = lim = 2 h ®0 h 2

k cos x pù é êë\f ( x) = p - 2x if x =/ 2 úû …(i) p é ù êLet x = 2 + h ú ê ú + p ê x® Þ h = 0ú ë û 2

æp ö lim f ( x) = lim f ç + h ÷ + h ®0 è2 ø p

x®

2

æp ö k cos ç + h ÷ é k cos x pù è2 ø = lim \ f ( x) = if x =/ ú ê h ®0 p - 2x 2 û æp ö ë p-2ç + h÷ è2 ø -k sin h = lim h ® 0 p - p - 2h k sin h k sin h k = lim = lim = h ®0 2 h 2 h ®0 h 2 æ pö f ç ÷ =3 è2ø

Also Since f ( x) is continuous at x = \

pù é êë\f ( x) = 3 if x = 2 úû

p 2

æ pö lim f ( x) = lim f ( x) = f ç ÷ + è2ø p p

x®

2

x®

2

k k = =3 2 2

Þ

k = 6.

… (ii)

313


15. f ( x) = sin x + cos x Differentiating w.r.t. x, we get f ¢( x) = cos x - sin x For critical points f ¢( x) = 0 Þ cos x - sin x = 0 Þ cos x = sin x æp ö Þ cos x = cos ç - x ÷ è2 ø æp ö where n = 0, ± 1, ± 2, K Þ x = 2n p ± ç - x ÷ è2 ø p p Þ x = 2n p + - x or x = 2n p - + x 2 2 p (Not exist) Þ 2x = 2n p + 2 p Þ x=n p+ 4 p 5p x= , [\0 £ x £ 2p] 4 4 p 5p The critical value of f ( x) are , . 4 4 é p ö æ p 5p ö æ 5p ù Therefore, required intervals are ê0, ÷ , ç , ÷ and ç , 2pú ë 4ø è4 4 ø è 4 û p ö æ 5p é ù Obviously, f ¢( x) > 0 if x Î ê0, ÷ È ç , 2pú ë 4ø è 4 û p 5 p æ ö and f ¢ ( x) < 0 if x Î ç , ÷ è4 4 ø é p ö æ 5p ù i.e., f ( x) is strictly increasing in ê0, ÷ È ç , 2xú ë 4ø è 4 û æ p 5p ö and strictly decreasing in ç , ÷ è4 4 ø OR Let ( x1 , y 1 ) be the required point on the curve y = x 3 , Now \

y = x3 dy æ dy ö = 3x 2 Þ ç ÷ = 3x12 dx è dx ø( x 1 , y 1 )

æ dy ö Þ Slope of tangent at point ( x1 , y 1 ) on curve ( y = x 3 ) is ç ÷ è dx ø( x 1 y 1 )

314


From question 3x12 = y 1

… (i)

Also since ( x1 , y 1 ) lies on curve y = x 3 y 1 = x13

\

… (ii)

From (i) and (ii) 3x12 = x13 Þ 3x12 - x13 = 0 x12 (3 - x1 ) = 0 Þ x1 = 0, x1 = 3

Þ

If x1 = 0, y 1 = 0 If x1 = 3, y 1 = 27 Hence, required points are (0, 0) and (3, 27). 16. Prove that d éx a2 æ x öù a 2 -x2 + sin -1 ç ÷ú = a 2 - x 2 ê dx ë 2 2 è a øû 2 d æx ö d æç a æ x öö L.H.S. = a 2 -x2 ÷ + sin -1ç ÷ ÷ ç ç dx è 2 ø dx è 2 è a ø ÷ø üï a 2 1 ìï 1 = íx . ´ - 2x + a 2 - x 2 ý + . 2 ïî 2 a 2 - x 2 2 ïþ

=

-x 2

+

2 a 2 -x2 =

1 1-

´ x2

1 a

a2

a 2 -x2 a2 + 2 2 a 2-x2

- x2 + a 2 - x2 + a 2 2 a 2 - x2

=

a 2 - x2 2

a -x

= a 2 - x 2 = R.H.S.

2

OR Given

Þ

y = log éx + x 2 + 1 ù êë úû é ù dy 1 2x ú = ´ ê1 + dx x + x 2 + 1 ê 2 x 2 + 1 ú ë û =

2 ( x + x 2 + 1) ( x + x 2 + 1) ´ 2 x 2 + 1

dy 1 = dx x2 +1

[Differentiating]

315


Differentiating again, we get \

Þ

d2y dx 2

d2y 1 -x -x = - ( x 2 + 1) -3 / 2 . 2x = Þ ( x 2 + 1) = 2 3 / 2 2 2 ( x + 1) dx x2 +1

( x 2 + 1)

d2y dx

17. Let

2

+x

dy =0 dx

I = ò e 2 x sin x dx = - e 2 x cos x - ò 2e 2 x ( - cos x) dx = - e 2 x cos x + 2 ò e 2 x cos x dx = - e 2 x cos x + 2 [ e 2 x sin x - ò 2e 2 x sin x dx] = - e 2 x cos x + 2e 2 x sin x - 4 ò e 2 x sin x dx + C¢ = e 2 x ( 2 sin x - cos x) - 4I + C¢

Þ

I=

e 2x [ 2 sin x - cos x] + C 5

[where C =

C¢ ù 5 úû

OR d 2 Now 3x + 5 = A. ( x - 8x + 7) + B dx Þ 3x + 5 = A ( 2x - 8) + B Þ 3x + 5 = 2Ax - 8A + B Equating the coefficient of x and constant, we get 2A = 3 and -8A + B = 5 3 3 A = and -8 ´ + B = 5 2 2 Þ B = 5 + 12 = 17 3 ( 2x - 8) + 17 3x + 5 2 Hence dx = dx ò 2 ò 2 x - 8x + 7 x - 8x + 7 = = Where Now Let

3 2

ò

( 2x - 8) x 2 - 8x + 7

3 I 1 , + 17 I 2 2 2x - 8

I1 = ò I1 = ò

2

x - 8x + 7 2x - 8

dx + 17 ò

dx x 2 - 8x + 7 …(i)

dx , I 2 = ò

dx 2

x - 8x + 7

dx

x 2 - 8x + 7

x 2 - 8x + 7 = z 2 Þ

( 2x - 8) dx = 2zdz

316


\

2zdz z = 2ò dz = 2z + C1

I1 = ò

I 1 = 2 x 2 - 8 x + 7 + C1 I2 = ò =ò

…(ii)

dx 2

x - 8x + 7 dx 2

x - 2. x.4 + 16 - 16 + 7

=ò

dx ( x - 4) 2 - 3 2

= log ( x - 4) + ( x - 4) 2 - 3 2 + C 2 I 2 = log ( x - 4) + x 2 - 8x + 7

… (iii)

+ C2

Putting the value of I 1 and I 2 in (i) 3x + 5 dx 3 = . 2 x 2 - 8x + 7 + 17 log | ( x - 4) + x 2 - 8x + 7 | + ( C1 + C 2 ) ò 2 x - 8x + 7 2 = 3 x 2 - 8x + 7 + 17 log | ( x - 4) x 2 - 8x + 7 | + C. é ù dx = log x + x 2 - a 2 + Cú êNote : ò êë x2 -a 2 ûú 18. Given equation is (1 + e 2 x ) dy + (1 + y 2 ) e x dx = 0 Þ

(1 + e 2 x ) dy = - (1 + y 2 ) e x dx Þ

dy 1+ y2

=-

e x dx 1 + e 2x

Integrating both sides, we get tan -1 y = - ò Þ Þ

tan -1 y = - ò tan

-1

e x dx 1 + (e x ) 2 dz 1+ z

Let e x = z, e x dx = dz

2

y = - tan -1 z + C

Þ

tan -1 y + tan -1 e x = c

For particular solution : Putting y =1 and x = 0, we get tan -1 (1) + tan -1 e 0 = C p p Þ + =C 4 4 Therefore, required particular solution is

tan -1 (1) + tan -1 (1) = c p C= 2

Þ Þ

tan -1 y + tan -1 e x =

p 2

317


19. Given differential equation is dy + y cot x = 4x cosec x dx dy Þ + cot x . y = 4x cosec x dx dy Comparing the given equation with + Py = Q, we get dx P = cot x, Q = 4x cosec x cot x dx I.F. = e ò \ = e log (sin x ) = sin x Hence the General solution is y. sin x = ò 4x. cosec x. sin x dx + C Þ

y sin x = ò 4x dx + C

[ cosec x . sin x =1]

y sin x = 2x 2 + C p Putting y = 0 and x = , we get 2

Þ

0=2

p2 p2 + C Þ C=4 2

Therefore, required solution is y sin x = 2x 2 -

p2 2

Note: When the given differential equation is in the form of

dy + Py = Q, where P, Q are dx

constant or function of x only, then general solution is y ´ (I. F. ) = ò ( Q ´ I. F. ) dx + C I. F. = e ò

where

Pdx

20. Here ®

®

®

a = 2$i + 2$j + 3k$ , b = - $i + 2$j + k$ , c = 3$i + $j

®

®

a + l b = ( 2$i + 2$j + 3k$ ) + l ( -$i + 2$j + k$ ) = ( 2 - l) $i + ( 2 + 2l) $j + (3 + l) k$

®

®

®

Since ( a + l b ) is perpendicular to c ®

®

®

Þ

( a + l b) . c = 0

Þ

6 - 3l + 2 + 2l = 0 Þ l = 8 ®

Þ ( 2 - l) . 3 + ( 2 + 2l). 1 + (3 + l) . 0 = 0 ®

® ®

®

®

[Note : If a is perpendicular to b , then a . b = | a | . | b | . cos 90° = 0]

318


21. Given equation of lines are ®

r = 6$i + 2$j + 2k$ + l ( $i - 2$j + 2k$ )

… (i)

®

r = - 4$i - k$ + m (3$i - 2$j - 2k$ ) ®

Comparing (i) and (ii) with r

® = a1

+

® l b1

® =a2

and r

®

a 1 = 6$i + 2$j + 2k$

®

®

®

b1 = $i - 2$j + 2k$ ®

… (ii) ®

a

2

+

® l b2,

we get

= - 4$i - k$

b 2 = 3$i - 2$j - 2k$

®

= ( 6$i + 2$j + 2k$ ) - ( -4$i - k$ ) = 10 $i + 2$j + 3k$ $i $j k$ ® ® b 1 ´ b 2 = 1 -2 2 3 -2 -2

a1 - a

2

= ( 4 + 4) $i - ( -2- 6) $j + ( -2 + 6) k$ = 8$i + 8$j + 4k$ ®

\

®

| b 1 ´ b 2 | = 8 2 + 8 2 + 4 2 = 144 = 12

Therefore, required shortest distance ® ö æ ® ® ö æ® çç a 1 - a 2 ÷÷ . çç b 1 ´ b 2 ÷÷ è ø è ø = ® b1

=

®

´ b2

(10 $i + 2$j + 3k$ ) . (8$i + 8$j + 4k$ ) 12

80 + 16 + 12 12 108 =9 = 12 =

®

®

®

®

®

®

Note : Shortest distance (S.D) between two skew lines r = a 1 + l b1 and r = a 2 + l b 2 is given by

S.D. =

æ® ®ö æ® ®ö çç a 1 - a 2 ÷÷ . çç b1 ´ b 2 ÷÷ è ø è ø ®

®

b2 ´ b2 22. The sample space of given experiment is S = {( HHH), ( HHT), ( HTT), ( TTT), ( TTH), ( THH), ( HTH), ( THT)} Let X denotes the no. of heads in three tosses of a fair coin Here, X is random which may have values 0, 1, 2, 3.

319


1 3 , P ( X = 1) = 8 8 3 1 , P ( X = 2) = P ( X = 3) = 8 8 Therefore, Probability distibution is Now,

P ( X = 0) =

X

0

1

2

3

P(X)

1/8

3/8

3/8

1/8

1 3 3 1 +1´ + 2 ´ + 3 ´ 8 8 8 8 3 6 3 12 3 =0+ + + = = 8 8 8 8 2

Mean number ( E( x)) = 0 ´

\

Section – C 23. Given system of equation is x - y + 2z = 1, 2y - 3z = 1, 3x - 2y + 4z = 2 Above system of equation can be written in matrix form as A X = B Þ X = A -1 B 1 A= 0 3

whrere

-1 2 -2

2 éx ù é1 ù -3 , X = êy ú , B = ê1 ú êz ú ê2ú 4 ë û ë û 0 1ù 2 -3 ú 1 -2úû -1 2ù é-2 0 1ù 2 -3 ú ê 9 2 -3 ú -2 4úû êë 6 1 -2úû

é-2 C=ê 9 ê 6 ë é1 AC = ê0 ê3 ë é -2 - 9 + 12 = ê 0 + 18 -18 ê-6 -18 + 24 ë é1 0 0ù =ê 0 1 0ú ê0 0 1ú ë û

Let

Now

Þ Þ

A

Þ

(A -1 A ) C = A -1

-1

AC = I (AC) = A -1 I -1

0 -2 + 2 0 + 4 -3 0 -4 + 4

1+3 -4 0 -6 + 6 3 + 6 -8

[Pre multiplication by A -1 ] [By Associativity] Þ

-1

Þ

I C =A

Þ

é -2 0 1 ù A -1 = ê 9 2 -3 ú ê 6 1 -2 ú ë û

A

=C

ù ú ú û

… (i)

320


Putting X , A -1 and B in (i) we get éx êy êz ë Þ

0 1ùé1 ù ù é -2 éx ú =ê 9 2 -3 ú ê 1 ú Þ ê y ú ê 6 êz 1 -2 úû êë 2 úû û ë ë é x ù é0ù êy ú = ê5 ú Þ x = 0, y = 5 and z = 3 ê z ú ê3 ú ë û ë û

ù é -2 + 0 + 2 ú =ê 9+ 2-6 ú ê 6 + 1-4 û ë

OR é 2 0 -1 ù Let A =ê5 1 0 ú ê0 1 3ú ë û For elementary row operation, we write A =I A é2 0 -1ù é 1 0 0 ù ê5 1 0ú = ê 0 1 0 ú A ê0 1 3ú ê 0 0 1 ú ë û ë û Applying R 2 « R1 1 0ù é0 1 0ù é5 ê2 0 -1ú = ê 1 0 0ú A ê0 1 3úû êë0 0 1úû ë Applying R1 ® R1 - 2R 2 2ù é-2 1 0ù é1 1 ê2 0 -1ú = ê 1 0 0ú A ê0 1 3úû êë 0 0 1úû ë R3 « R2 2ù é-2 1 0ù é1 1 ê0 1 3ú = ê 0 0 1ú A ê2 0 -1ú ê 1 0 0ú ë û ë û R1 ® R1 - R 2 é1 ê0 ê2 ë R 3 ® R 3 - 2 R1 é1 ê0 ê0 ë R1 ® R1 + R 3 , R 2 é1 ê0 ê0 ë

0 1 0 0 1 0

-1ù é-2 3ú = ê 0 -1úû êë 1

-1ù é-2 3ú = ê 0 1úû êë 5 ® R 2 - 3R 3 0 0ù é 3 1 0ú = ê-15 0 1úû êë 5

1 0 0 1 0 -2 -1 6 -2

-1ù 1ú A 0úû -1ù 1ú A 2úû 1ù -5 ú A 2úû

ù ú ú û

321


é 3 -1 1 ù I = ê -15 6 -5 ú A ê 5 -2 2 ú ë û é 3 -1 1 ù Þ A -1 = ê -15 6 -5 ú ê 5 -2 2 ú ë û 24. Let x and y be the length and width of rectangle part of window respectively. Let A be the opening area of window which admit Light. x/2 x/2 Obviously, for admitting the maximum light through the opening, A must be maximum. y Now A = Area of rectangle + Area of semi-circle 2 1 x A = xy + p. 2 4 x px 2 Þ A = xy + 8 x( p + 2) ü px 2 ì é Þ A =x í 5ý+ êFrom question 4 8 î þ ê ê\ x + 2y + p × x = 10 ( p + 2) × x 2 px 2 Þ A = 5x + ê 2 4 8 ê æp ö æ p + 2 pö 2 êÞ x ç 2 + 1÷ + 2y = 10 Þ A = 5x - ç - ÷x è ø ê 8ø è 4 êÞ 2y = 10 - x æç p + 2 ö÷ p+ 4 2 dA æp+ 4ö è 2 ø Þ A = 5x x Þ =5 -ç ÷ 2x ê 8 dx ê è 8 ø x( p + 2) K( i) êÞ y = 5 For maximum or minimum value of A, ë 4 dA =0 dx p+ 4 æp+ 4ö Þ 5 -ç . 2x = 5 ÷ .2x = 0 Þ 8 è 8 ø 20 Þ x= p+ 4 Now i.e.,

d 2A

p+ 4 p+ 4 ´ 2 =8 4 dx 2 ù d A <0 ú dx 2 û x = 20 2

=-

p+ 4

20 Hence for x = , A is maximum p+ 4 ù p+ 2 é 20 20 and thus y =5 ´ Putting x = in ( i) ú ê p+ 4 4 p+ 4 ë û

322


5 ( p + 2) p+ 4 5p + 20 - 5p -10 10 = = p+ 4 p+ 4 Therefore, for maximum A i.e., for admitting the maximum light 20 Length of rectangle = x = . p+ 4 10 Breadth of rectangle = y = p+ 4 25. Given lines are 2x + y = 4 3x - 2y = 6 x - 3y + 5 = 0 For intersection point of (i) and (ii) Multiplying (i) by 2 and adding with (ii), we get 4x + 2y = 8 3x - 2y = 6 7x = 14 Þ x = 2 \ y =0 Here, intersection point of (i) and (ii) is (2, 0). 4– For intersection point of (i) and (iii) Multiplying (i) by 3 and adding with (iii), we get 5 3– y=– 6x + 3y = 12 (1, 2) x – 3 A 2– x - 3y = - 5 7x = 7 Þ x =1 1 – \ y =2 C 0 Hence, intersection point of (i) and (iii) is (1, 2). 1D 2 3 (2, 0) For intersection point of (ii) and (iii) Multiplying (iii) by 3 and subtracting from (ii), we get 3x - 2y = 6 _ 3x m 9y = m 15 7y = 21 Þ y =3 \ x=4 Hence intersection point of (ii) and (iii) is (4, 3). With the help of intersecting points, required region DABC in ploted. Shaded region is required region. \ Required Area = Area of DABC = Area of trap ABED – Area of DADC - Area of DCBE 4 x+5 2 4 3x - 6 =ò dx - ò ( 4 - 2x) dx - ò dx 1 1 2 3 2 =5 -

]

1

4

-

ù 1 é 3x 2 - 6xú ê 2ë 2 û2

6

y=

–

–

–

–

–

=4

[

2

3x

+y

ù 1 é x2 + 5x ú - 4x - x 2 ê 3ë 2 û1

B (4, 3)

–2

2x

4

=

… (i) … (ii) … (iii)

4E 5

323


1 ì æ 16 1 ì æ 3 ´ 16 ö æ3´ 4 öü ö æ1 öü - 24 ÷ - ç - 12 ÷ý í ç + 20 ÷ - ç + 5 ÷ý - {(8 - 4) - ( 4 -1) } - í ç 3 îè 2 2 îè 2 2 ø è2 øþ ø è øþ 1ì 11 ü 1 = í 28 - ý - { 4 - 3} - { 0 + 6} 3î 2 þ 2 1 45 = ´ -1 - 3 3 2 7 = sq. unit. 2 =

26. Comparing

4

ò1

( x 2 - x) dx with f ( x) = x 2 - x

b

òa

f ( x) dx, we get

and

a = 1, b = 4

By definition b

òa

f ( x) dx = lim h [ f ( a) + f ( a + h) + f ( a + 2h) + .... + f ( a + ( n -1) h) ] h ®0

4 -1 3 = n n nh = 3

Here Þ 4

\

b -a n Also n ® a Û h ® 0 where h =

h=

ò1

( x 2 - x) dx = lim h [ f (1) + f (1 + h) + f (1 + 2h) + .... + f (1 + ( n -1) h)] h ®0 2

= lim h [ 0 + {(1 + h) - (1 + h)} + {(1 + 2h) 2 - (1 + 2h)} + .... + {(1 + ( n -1) h) 2 - (1 + ( n -1) h)}] h ®0

= lim h [ 0 + {1 + h 2 + 2h -1 - h} + {1 + 4h 2 + 4h -1 - 2h} h ®0

+ ... + {1 + ( n -1) 2 h 2 + 2( n -1) h -1 - ( n -1) h}] = lim h [ 0 + ( h 2 + h) + ( 4h 2 + 2h) + ... {( n -1) 2 h 2 + ( n -1) h}] h ®0

= lim h [ h 2 {1 + 2 2 + ... + ( n -1) 2} + h {1 + 2 + ... + ( n -1)}] h ®0

( n -1) n ( 2n -1) ( n -1) n ù é = lim h ê h 2 . +h úû h ®0 ë 6 2 1ö 1ö é 3 3 æ 1 öæ 2 2 æ ê h . n çè1 - n ÷ø çè 2 - n ÷ø h . n çè1 - n ÷ø = lim ê + h ®0 ê 6 2 êë 1ö é æ 1 öæ æ 1 öù ê 27 çè1 - n ÷ø çè 2 - n ÷ø 9 çè1 - n ÷ø ú = lim ê + ú n ®¥ ê 6 2 ú ëê ûú =

ù ú ú ú úû

27 ´ (1 - 0) ( 2 - 0) 9(1 - 0) 54 9 9 27 + = + =9+ = 6 2 6 2 2 2

3 é ù êQ h = n ú êë\ h ® 0 Þ n ® ¥ úû

324


Let

OR sin x - cos x = z (cos x + sin x) dx = dz

If x = 0, z = -1 p If x = , z = 0 4

Also, Q Þ

sin x - cos x = z (sin x - cos x) 2 = z 2 Þ sin 2 x + cos 2 x - 2 sin x .cos x = z 2

Þ

1 - sin 2x = z 2 Þ sin 2x = 1 - z 2 p / 4 sin x + cos x 0 dz ò0 9 + 16 sin 2x dx = ò-1 9 + 16 (1 - z 2 ) 0 0 dz dz =ò =ò 2 -1 -1 9 + 16 -16 z 25 -16 z 2

Now

=

1 16

0

ò-1

dz 2

5 é +z 1 1 ê 4 = log ê 5 16 5 ê 2. -z 4 ë 4

0

ù ú ú ú û -1

æ5ö 2 ç ÷ -z è4ø 1 é 1ù 1 = log 1 - log ú = [log1 - log1 + log 9] ê 40 ë 9 û 40 1 = log 9 40 27. Let equation of plane passing through (–1, 3, 2) be … (i) a( x + 1) + b( y - 3) + c( z - 2) = 0 Since (i) is perpendicular to plane x + 2y + 3z = 5 Þ a .1 + b . 2 + c . 3 = 0 … (ii) Þ a + 2b + 3c = 0 Again plane (i) is perpendicular to plane 3x + 3y + z = 0 Þ a . 3 + b . 3 + c .1 = 0 … (iii) Þ 3a + 3 b + c = 0 From (ii) and (iii) a b c = = 2 - 9 9 -1 3 - 6 a b c (say) = = =l -7 8 -3 Þ a = -7 l, b = 8l, c = -3l Putting the value of a, b, c in (i), we get -7 l ( x + 1) + 8 l ( y - 3) - 3l ( z - 2) = 0 Þ -7x - 7 + 8y - 24 - 3z + 6 = 0 Þ -7x + 8y - 3z - 25 = 0 Þ 7x - 8y + 3z + 25 = 0 It is required plane. 28. Let the number of padestal lamps and wooden shades manufactured by cottage industry be x and y respectively. Here profit is the objective function z. … (i) \ z = 5x + 3y

325


We have to maximise z subject to the constrains 2x + y £ 12 3x + 2y £ 20 x ³ 0ü y ³ 0ýþ Q Graph of x = 0, y = 0 is the y-axis and x-axis respectively. \ Graph of x ³ 0, y ³ 0 is the Ist quadrant. Graph for 2x + y £ 12 Graph of 2x + y = 12 x

0

6

y

12

0

… (ii) … (iii) … (iv)

Since (0, 0) satisfy 2x + y £ 12 Þ Graph of 2x + y £ 12 is that half plne in which origin lies. Graph of 3x + 2y = 20 Graph for3x + 2y £ 20 x

0

20/3

y

10

0

y-axis 12 –

10 – C 8–

6– B

4–

–

A 6

–

4

–

2

–

–

0

–

2–

8

10

12

x-axis

3x + 2y = 20 2x + y = 12

Since (0, 0) Satisfy 3x + 2y £ 20 Þ Graph of 3x + 2y £ 20 is that half plane in which origin lies. The shaded area OABC is the feasible region whose corner points are O, A , B and C.

326


For coordinate B. Equation 2x + y = 12 and 3x + 2y = 20 are solved as 3x + 2 (12 - 2x) = 20 Þ 3x + 24 - 4x = 20 Þ x = 4 \ Þ y = 12 - 8 = 4 Coordinate of B = ( 4, 4) Now we evaluate objective function Z at each corner. Corner points

z = 5x + 3y

0 (0, 0)

0

A (6, 0)

30

B (4, 4)

32

C (0, 10)

30

Hence maximum profit is ` 32 when manufacturer produces 4 lamps and 4 shades. 29. Let E1 , E 2 and A be event such that E1 = Production of items by machine A E 2 = Production of items by machine B A = Selection of defective items. 60 3 40 2 P( E1 ) = = , P (E 2 ) = = 100 5 100 5 æA ö 2 1 æ A ö 1 ÷÷ = ÷÷ = Pçç = , Pçç è E1 ø 100 50 è E 2 ø 100 æE ö Pç 2 ÷ is required è A ø By Baye's theorem æ A ö ÷÷ P( E 2 ) . P çç æ E2 ö è E2 ø Pç ÷= æA ö æ A è A ø ÷÷ + P (E 2 ) . P çç P( E1 ) . P çç E è 1ø è E2 æE Pç 2 è A

2 1 ´ ö 5 100 ÷= ø 3´ 1 + 2´ 1 5 50 5 100 2 2 500 1 500 = = ´ = 3 2 500 6 + 2 4 + 250 500

ö ÷÷ ø

maximum

327


CBSE (Foreign) Set-II 9.

fog ( x) = f ( g( x)) = f (| 5x - 2 | ) = 5x - 2 = 5x - 2 I =ò

10. Let

(e e

2x

2x

-e

-2 x

+e

)

-2 x

dx

e 2 x + e -2 x = z ( 2e 2 x - 2e -2 x ) dx = dz

dz 2 1 dz \ I= ò 2 z 1 = log| z |+ C 2 1 = log e 2 x + e -2 x + C 2 a -b -c 2a 2a 19. L.H.S. = 2b b -c-a 2b 2c 2c c-a -b ( e 2 x - e -2 x ) dx =

Applying R1 ® R1 + R 2 + R 3 , we get a + b+c a + b+ c a + b + c = 2b b -c-a 2b 2c 2c c-a -b 1 = ( a + b + c) 2b 2c

1 b -c-a 2c

1 2b c-a -b

Applying C1 ® C1 - C 3 and C 2 ® C 2 - C 3 , we get 0 0 = ( a + b + c) 0 -b - c - a c + a + b c+a +b

1 2b c-a -b

Expanding along R1 , we get = ( a + b + c) [ 0 - 0 + 1 {0 - ( -b - c - a). ( c + a + b)}] = ( a + b + c). ( a + b + c) 2 = ( a + b + c) 3 = RHS

328 20.


éLet x = p - h lim f ( x) = lim f ( p - h) ê h ®0 x ® pëx ® p Þ h ® 0

[Q f ( x) = kx + 1 for x £ p

= lim K ( p - h) + 1 h ®0

= K p +1 lim h ®p

+

éLet x = p + h ù f ( x) = lim f ( p + h) ê + ú h ®0 ëx ® p Þ h ® 0û = lim cos ( p + h) [Q f ( x) = cos x for x > p] h ®0

= lim - cos h

= -1

h ®0

Also f ( p) = k p + 1 Since f ( x) is continuous at x = p Þ

lim f ( x) = lim f ( x) = f ( p)

x ® p-

x ® p+

Þ Þ

k p + 1 = -1 = k p + 1 k p = -2 2 Þ k =p 21. Given differential equation is dy + 2 tan x. y = sin x dx dy Comparing it with + Py = Q, we get dx P = 2 tan x, Q = sin x 2 tan xdx \ I. F. = e ò = e 2 log sec x = e log

sec2 x

= sec 2 x Hence general solution is y. sec 2 x = ò sin x. sec 2 x dx + C y .sec 2 x = ò sec x . tan x dx + C Þ y. sec 2 x = sec x + C y = cos x + C cos 2 x

Þ Putting y = 0 and x =

p , we get 3

p p + C . cos 2 3 3 1 C 0= + Þ C = -2 2 4 \ Required solution is y = cos x - 2 cos 2 x 0 = cos

[Q e log z = z ]

329


22. Given equation of lines are ®

r = ( $i + 2$j + 3k$ ) + l ( $i - 3$j + 2k$ )

… (i)

®

r = ( 4$i + 5$j + 6k$ ) + m ( 2$i + 3$j +k$ ) ®

Comparing (i) and (ii) with r ® a1 ® b1 ®

® = a1

®

®

+ l b and r ® a2 ® b2

= $i + 2$j + 3k$ = $i - 3$j + 2k$

® =a2

… (ii) ®

+ l b respectively we get.

= 4$i + 5$j + 6k$ = 2$i + 3$j + k$

®

Now a 2 - a 1 = 3$i + 3$j + 3k$ ® b1

´

® b2

´

® b1

$i $j = 1 -3 2 3

k$ 2 1

= ( -3 - 6) $i - (1 - 4) $j + (3 + 6) k$ = - 9$i + 3$j + 9k$ \

® b1

= ( -9) 2 + 3 2 + 9 2 = 3 19 ®

\

S.D. =

®

®

®

( a 2 - a 1 ) . ( b1 ´ b 2 ) ®

®

b1 ´ b 2 =

(3$i + 3$j + 3k$ ) . ( -9$i + 3$j + 9k$ )

=

3 19 28. Let equation of plane passing through A ( 2, 2, -1) be a ( x - 2) + b ( y - 2) + c ( z + 1) = 0 Since, B (3, 4, 2) lies on plane (i) Þ a (3 - 2) + b ( 4 - 2) + c ( 2 + 1) = 0 Þ a + 2b + 3c = 0 Again C (7, 0, 6) lie on plane (i) Þ a (7 - 2) + b ( 0 - 2) + c ( 6 + 1) = 0 Þ 5a -2b +7c = 0 From (ii) and (iii) a b c = = 14 + 6 15 - 7 -2 -10 a b c = = = l (say) 20 8 -12 a = 20l, b = 8l, c = -12l Putting the value of a, b, c in (i) 20l ( x - 2) + 8l( y - 2) - 12l( z + 1) = 0 Þ 20x - 40 + 8y -16 -12z -12 = 0 Þ 20x + 8y -12z - 68 = 0 Þ 5x + 2y - 3z -17 = 0

-27 + 9 + 27 3 19

=

3

.

19 … (i)

… (ii)

… (iii)

330


Þ 5x + 2y - 3z = 17 which is required cartesian equation of plane. Its vector form is ( xi$ + yj$ + zk$ ). (5$i + 2$j - 3k$ ) = 17 ®

Þ r . (5$i + 2$j - 3k$ ) = 17 29. Let E1 , E 2 and A be event such that E1 = red ball is transferred from Bag I to Bag II E 2 = black ball is transferred from Bag I to Bag II A = drawing red ball from Bag II 3 4 Now P( E1 ) = P( E 2 ) = 7 7 æA ö 5 æ A ö 4 æE ö ÷÷ = ÷÷ = , P çç , P ç 2 ÷ is required. P çç E 10 E 10 è A ø è 1ø è 2ø From Baye's theorem. æ A ö ÷÷ P(E 2 ) . Pçç æ E2 ö è E2 ø Pç ÷= æA ö æ A ö è A ø ÷÷ + P( E 2 ) . Pçç ÷÷ P( E1 ) . Pçç è E1 ø è E2 ø 4 4 ´ 16 16 7 10 = = = 3 5 4 4 15 + 16 31 ´ + ´ 7 10 7 10

CBSE (Foreign) Set-III 1.

fog ( x) = f ( g( x)) = f ( x1/ 3 ) = 8 ( x1/ 3 ) 3 = 8x

2.

I =ò Let

cos x

dx

x

x =t 1

dx = dt

Þ

2 x \

1 x

I = 2 ò cos t dt = 2 sin t + C = 2 sin x + C

dx = 2dt

331


11. L.H.S =

x + y + 2z z z

x y + z + 2x x

y y z + x + 2y

Applying C1 ® C1 ® C 2 + C 3 we get 2( x + y + z) x = 2( x + y + z) y + z + 2x 2( x + y + z) x

y y z + x + 2y

1 x y = 2 ( x + y + z) 1 y + z + 2 x y 1 x z + x + 2y

[Taking common from C1 ]

Applying R 2 ® R 2 - R1 and R 3 ® R 3 - R1 , we get 1 x y = 2( x + y + z) 0 x + y +z 0 0 0 x+y +z Expanding along C1 , we get = 2 ( x + y + z) [1{( x + y + z) 2 - 0} - 0 + 0] = 2 ( x + y + z) 3 = RHS 12.

lim f ( x) = lim l ( x 2 - 2x) x ®0

-

[\f ( x) = l ( x 2 - 2x) for x £ 0]

x ®0

= l ( 0 - 0) = 0 [\f ( x) = 4x + 1 for x > o]

lim f ( x) = lim 4x + 1

x ® 0+

x ®0

= 4 ´ 0 +1 = 1 Since lim f ( x) =/ lim f ( x) for any value of l. Hence for no value of l, f is continuous at x = 0 x ® o-

x ® 0+

13. Given differential equation is dy dy 1 2x 1 (1 + x 2 ) + 2xy = Þ + .y = 2 2 dx dx 1 + x 1+ x (1 + x 2 ) 2 dy Comparing this equation with + Py = Q we get dx 2x 1 P= , Q= 2 1+ x (1 + x 2 ) 2 \

I.F. = e ò I.F. = e

ò

=e

Pdx 2x 1+ x 2

ò

dx

dt t

= e log t = t =1 + x 2

éLet ê ë

t =1 + x 2 dt = 2xdx

332


Hence general solution is 1

y. (1 + x 2 ) = ò

. (1 + x 2 ) dx + C (1 + x 2 ) 2 dx y . (1 + x 2 ) = ò +C 1 + x2

Þ

Þ y. (1 + x 2 ) = tan -1 x + C Putting y = 0 and x =1 we get 0 = tan -1 (1) + C p C=4 Hence required solution is y . (1 + x 2 ) = tan -1 x -

p 4

14. Given lines are ®

r = ( $i + 2$j + k$ ) + l ( $i - $j + k$ )

… (i)

®

r = ( 2$i - $j - k$ ) + m ( 2$i + $j + 2k$ ) ®

Comparing the equation (i) and (ii) with r

® = a1

+

® l b1

… (ii) ®

and r

® =a2

+

® l b2.

We get a 1 = $i + 2$j + k$

®

a 2 = 2$i - $j - k$

®

®

b 2 = 2$i + $j + 2k$

®

b1 = $i - $j + k$ ®

Now

®

a 1 - a 1 = ( 2$i - $j - k$ ) - ( $i + 2$j + k$ ) = $i - 3$j - 2k$ ® b1

´

® b2

$i $j = 1 -1 2 1

k$ 1 2

= ( -2 -1) $i - ( 2 - 2) $j + (1 + 2) k$ = - 3$i + 3k$ \

® | b1

´

® b2

| = ( -3) 2 + (3) 2 = 3 2 ®

\

Shortest distance =

®

®

®

( a 2 - a 1 ) . ( b1 ´ b 2 ) ®

®

b1 ´ b 2

=

( $i - 3$j - 2k$ ) . ( -3$i + 0$j + 3k$ ) ®

®

b1 ´ b 2

333


=

-3 - 0 - 6 3 2

=

9 3 2

=

´

2 2

9 2 3 2 = 3´ 2 2

23. Let the equation of plane passing through point (1, 1, –1) be a( x - 1) + b( y - 1) + c( z + 1) = 0 Since (i) is perpendicular to the plane x + 2y + 3z - 7 = 0 \ 1. a + 2 . b + 3 . c = 0 a + 2b + 3c = 0 Again plane (i) is perpendicular to the plane 2x - 3y + 4z = 0 \ 2 . a -3 . b + 4 . c = 0 2a - 3b + 4c = 0 From (ii) and (iii), we get a b c = = 8 + 9 6 - 4 -3 - 4 a b c = = =l 17 2 -7 Þ a = 17 l, b = 2l, c = - 7l Puttting the value of a, b, c in (i) we get 17 l ( x -1) + 2l ( y -1) - 7l ( z + 1) = 0 Þ 17 ( x -1) + 2( y -1) - 7( z + 1) = 0 Þ 17 x + 2y - 7z -17 - 2 - 7 = 0 Þ 17x + 2y - 7z - 26 = 0 It is required equation. [Note: The equation of plane pasing through ( x1 , y 1 , z 1 ) is given by Þ

a ( x - x1 ) + b ( y - y 1 ) + c ( z - z 1 ) = 0 where a, b, c are direction ratios of normal of plane.] 24. Let E1 , E 2 , E 3 and A be events such that E1 = event of selecting two headed coin. E 2 = event of selecting biased coin. E 3 = event of selecting unbiased coin. A = event of getting head. 1 P( E1 ) = P( E 2 ) = P( E 3 ) = 3

… (i)

… (ii)

… (iii)

334


æA ö æ A ö 75 3 æ A ö 1 ÷÷ = 1, Pçç ÷÷ = ÷÷ = P çç = , Pçç è E1 ø è E 2 ø 100 4 è E 3 ø 2 æE ö Pç 1 ÷ is required. èA ø By Baye’s Theorem, æA ö ÷÷ P( E1 ) . Pçç æ E1 ö è E1 ø Pç ÷= æA ö æ A ö æ A èA ø ÷÷ + P(E 2 ) . Pçç ÷÷ + P( E 3 ) . Pçç P( E1 ) . Pçç è E1 ø è E2 ø è E3 1 ´1 3 = 1 1 3 1 1 ´1 + ´ + ´ 3 3 4 3 2 1 æE ö 3 Pç 1 ÷ = èA ø 1+1+1 3 4 6 1 12 4 = ´ = 3 9 9

ö ÷÷ ø

CBSE Examination Paper (Delhi 2012) Time allowed: 3 hours

Maximum marks: 100


Set–I SECTION–A Question numbers 1 to 10 carry 1 mark each. 1. If a line has direction ratios 2, –1, –2, then what are its direction cosines? ®

®

2. Find ‘l’ when the projection of a = li$ + j$ + 4k$ on b = 2i$ + 6j$ + 3k$ is 4 units. ®

®

®

3. Find the sum of the vectors a = i$ - 2j$ + k$, b = -2i$ + 4j$ + 5k$ and c = i$ - 6j$ - 7 k$. 3

4. Evaluate:

1

ò x dx. 2

5. Evaluate: ò (1 - x) x dx. 5 3 8 6. If D = 2 0 1 , write the minor of the element a 23 . 1

2

3

é 2 3 ù é 1 -3 ù é -4 6 ù 7. If ê úê ú=ê ú , write the value of x. ë 5 7 û ë -2 4 û ë -9 x û

336


é cos q 8. Simplify: cos q ê ë - sin q

sin q ù é sin q + sin q ê ú cos q û ë cos q

- cos q ù . sin q úû

9. Write the principal value of 1 1 cos -1 æç ö÷ - 2 sin -1 æç - ö÷. è 2ø è 2ø 10. Let * be a ‘binary’ operation on N given by a * b = LCM (a, b) for all a, b Î N. Find 5 * 7.

SECTION–B Question numbers 11 to 22 carry 4 marks each. dy 11. If (cos x) y = (cos y) x , find . dx OR 2 dy sin ( a + y) . = dx sin a 12. How many times must a man toss a fair coin, so that the probability of having at least one head is more than 80%? 13. Find the Vector and Cartesian equations of the line passing through the point (1, 2, –4) and x - 8 y + 19 z - 10 x - 15 y - 29 z - 5 perpendicular to the two lines and . = = = = 3 -16 7 3 8 -5

If sin y = x sin( a + y), prove that

® ® ®

®

®

®

®

®

®

®

14. If a , b , c are three vectors such that| a |= 5, | b |= 12 and| c |= 13, and a + b + c = O, find the ® ®

® ®

® ®

value of a . b + b . c + c . a . 15. Solve the following differential equation: dy 2x 2 - 2xy + y 2 = 0. dx 16. Find the particular solution of the following differential equation: dy = 1 + x 2 + y 2 + x 2 y 2 , given that y = 1 when x = 0. dx 17. Evaluate: ò sin x sin 2x sin 3x dx OR Evaluate:

2

ò (1 - x)(1 + x 2 ) dx

18. Find the point on the curve y = x 3 - 11x + 5 at which the equation of tangent is y = x - 11. OR Using differentials, find the approximate value of 49.5. 19. If y = (tan -1 x) 2 , show that ( x 2 + 1) 2

d2y dx

2

+ 2x( x 2 + 1)

dy = 2. dx

337


20. Using properties of determinants, prove that b + c q+r y+z a p x c+a r+p z+x =2 b a+b p+ q x+ y

q y

c r

z

cos x ö p x p p 21. Prove that tan -1 æç ÷ = - , x Î æç - , ö÷ . è 1 + sin x ø 4 2 è 2 2ø OR 8 3 36 Prove that sin -1 æç ö÷ + sin -1 æç ö÷ = cos -1 æç ö÷ . è 17 ø è5ø è 85 ø x - 2ö 22. Let A = R - { 3} and B = R - {1}. Consider the function f : A ® B defined by f ( x) = æç ÷. Show è x - 3ø that f is one-one and onto and hence find f -1 .

SECTION–C Question numbers 23 to 29 carry 6 marks each. 23. Find the equation of the plane determined by the points A(3, –1, 2), B(5, 2, 4) and C(–1, –1, 6) and hence find the distance between the plane and the point P(6, 5, 9). 24. Of the students in a college, it is known that 60% reside in hostel and 40% day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain ‘A’ grade and 20% of day scholars attain ‘A’ grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an ‘A’ grade, what is the probability that the student is a hosteler? 25. A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of `17.50 per package on nuts and `7 per package of bolts. How many packages of each should be produced each day so as to maximize his profits if he operates his machines for at the most 12 hours a day? Form the linear programming problem and solve it graphically. p 4

26. Prove that: ò ( tan x + cot x ) dx = 2 . 0

p 2

OR 3

Evaluate: ò ( 2x 2 + 5x) dx as a limit of a sum. 1

27. Using the method of integration, find the area of the region bounded by the lines 3x - 2y + 1 = 0, 2x + 3y - 21 = 0 and x - 5y + 9 = 0. 28. Show that the height of a closed right circular cylinder of given surface and maximum volume, is equal to the diameter of its base.

338


29. Using matrices, solve the following system of linear equations: x - y + 2z = 7 3x + 4y - 5z = -5 2x - y + 3z = 12 OR Using elementary operations, find the inverse of the following matrix: æ -1 1 2 ö ç ÷ ç 1 2 3÷ ç ÷ è 3 1 1ø

Set–II Only those questions, not included in Set I, are given. 9. Find the sum of the following vectors: ®

®

®

a = i$ - 2j$, b = 2i$ - 3j$, c = 2i$ + 3k$. 5

10.

3 8

If D = 2 0 1 , write the cofactor of the element a 32 . 1 2 3

19. Using properties of determinants, prove the following: 1 1 1 c = ( a - b)(b - c)( c - a)( a + b + c)

a

b

3

3

a

b

c3

20. If y = 3 cos(log x) + 4 sin(log x), show that x2

d2y 2

+x

dy +y=0 dx

dx 21. Find the equation of the line passing through the point (–1, 3, –2) and perpendicular to the lines x y z x+ 2 y-1 z+1 = = and = = 1 2 3 -3 2 5 22. Find the particular solution of the following differential equation: dy ( x + 1) = 2e - y - 1 ; y = 0 when x = 0. dx 28. A girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin two times and notes the number of heads obtained. If she obtained exactly two heads, what is the probability that she threw 1, 2, 3 or 4 with the die? 29. Using the method of integration, find the area of the region bounded by the following lines: 3x - y - 3 = 0 2x + y - 12 = 0 x - 2y - 1 = 0

339


Set–III Only those questions, not included in Set I and Set II, are given. 9. Find the sum of the following vectors: ®

®

®

a = i$ - 3k$, b = 2j$ - k$, c = 2i$ - 3j$ + 2k$. 1

2

3

10. If D = 2 0 1 , write the minor of element a 22 . 5

3 8

19. Using properties of determinants, prove the following: 1+a 1 1 1

1+b

1

1

1

1+c

= ab + bc + ca + abc

20. If y = sin -1 x, show that (1 - x 2 )

d2y 2

-x

dy = 0. dx

dx 21. Find the particular solution of the following differential equation: dy xy = ( x + 2)( y + 2); y = -1 when x = 1 dx 22. Find the equation of a line passing through the point P( 2, - 1, 3) and perpendicular to the lines ®

®

r = (i$ + j$ - k$) + l( 2i$ - 2j$ + k$) and r = ( 2i$ - j$ - 3k$) + m(i$ + 2j$ + 2k$).

28. Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. Two balls are transferred at random from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred balls were both black. 29. Using the method of integration, find the area of the region bounded by the following lines: 5x - 2y - 10 = 0 x+y-9=0 2x - 5y - 4 = 0

340


Solutions Set–I SECTION–A 1. Here direction ratios of line are 2, –1, –2 \ Direction cosines of line are

2 2

2

2 + ( -1) + ( -2)

2

,

-1 2

2

2 + ( -1) + ( -2)

2

,

-2 2

2 + ( -1) 2 + ( -2) 2

2 -1 -2 , , 3 3 3

i.e.,

[Note: If a, b, c are the direction ratios of a line, the direction cosines are

a 2

2

a +b + c b a2 + b 2 + c 2

c

,

a2 + b 2 + c 2

]

®

®

® ®

a. b

2. We know that projection of a on b =

®

| b| ® ®

Þ

a. b

4=

...(i)

®

| b| ® ®

a . b = 2l + 6 + 12 = 2l + 18

Now,

®

| b| = 2 2 + 6 2 + 3 2 = 4 + 36 + 9 = 7

Also

Putting in (i) we get 2l + 18 4= 7 Þ 3.

®

®

Þ

l=

®

1

3

ò x dx = [log x]2

= log 3 - log 2

2

5.

10 =5 2

a + b + c = (1 - 2 + 1)i$ + ( -2 + 4 - 6) j$ + (1 + 5 - 7) k$ = - 4j$ - k$

3

4.

2l = 28 - 18

ò (1 - x) x dx = ò xdx - ò x =ò

1 2 x dx

-ò

1+

1 2 dx

3 2 x dx

2

,

341


6. Minor of a 23 7. Given Þ Þ

+1

3

+1

3

5

2 2 x2 x2 = + c = x2 - x2 + c 1 3 3 5 +1 +1 2 2 5 3 = = 10 - 3 = 7. 1 2

é 2 3 ù é 1 -3 ù é -4 6 ù ê 5 7 ú . ê -2 4 ú = ê -9 x ú ë û ë û ë û é 2 ´ 1 + 3 ´ ( -2) 2 ´ ( -3) + 3 ´ 4ù é -4 6ù ê 5 ´ 1 + 7 ´ ( -2) 5 ´ ( -3) + 7 ´ 4 ú = ê -9 xú ë û ë û é -4 6 ù é -4 6 ù ê -9 13ú = ê -9 xú ë û ë û

Equating the corresponding elements, we get x = 13 é cos q sin q ù é sin q - cos q ù 8. cos q ê + sin q ê ú ú ë - sin q cos q û ë cos q sin q û é cos 2 q =ê ë - sin q. cos q

sin q. cos q ù é sin 2 q ú+ê cos 2 q û ë sin q. cos q

- sin q. cos q ù ú sin 2 q û

é sin 2 q + cos 2 q ù é1 0ù 0 =ê = 2 2 ú ê0 1ú 0 sin q + cos q û ë û ë 1 p 9. We have, cos -1 æç ö÷ = cos -1 æç cos ö÷ è 2ø è 3ø p éQ p Î [0, p]ù = êë 3 úû 3 1 p Also sin -1 æç - ö÷ = sin -1 æç - sin ö÷ è 2ø è 6ø p ö æ = sin -1 ç sin æç - ö÷ ÷ è è 6 øø p p p p ù é =Q - Îé - , ù ú ê ê 6 6 ë 2 2 úû û ë 1 1 p p \ cos -1 æç ö÷ - 2 sin -1 æç - ö÷ = - 2æç - ö÷ è 2ø è 2ø 3 è 6ø p p 2p = + = 3 3 3 p p [Note: Principal value branches of sin x and cos x are é - , ù and [0, p] respectively.] êë 2 2 úû 10. 5 * 7 = LCM of 5 and 7 = 35

342


SECTION–B 11. Given, (cos x) y = (cos y) x Taking logrithm of both sides, we get log (cos x) y = log (cos y) x Þ

[Q log mn = n log m]

y . log (cos x) = x . log (cos y)

Differentiating both sides we get dy dy 1 1 Þ y. ( - sin x) + log (cos x) . = x. . ( - sin y) . + log (cos y) cos x dx cos y dx y sin x x sin y dy dy Þ + log (cos x) . =. + log (cos y) cos x dx cos y dx y sin x dy x sin y dy Þ log (cos x) . + . = log (cos y) + dx cos y dx cos x x sin y y sin x ù dy é Þ log (cos x) + = log (cos y) + ê ú dx ë cos y û cos x y sin x log (cos y) + dy cos x = log (cos y) + y tan x Þ = x sin y log (cos x) + x tan y dx log (cos x) + cos y Here Þ

Þ Þ Þ Þ

OR sin y = x sin ( a + y) sin y =x sin ( a + y) dy dy sin ( a + y) . cos y . - sin y . cos ( a + y). dx dx = 1 sin 2 ( a + y) dy {sin ( a + y) . cos y - sin y . cos ( a + y)} = sin 2 ( a + y) dx sin 2 ( a + y) dy = dx sin ( a + y - y) 2 dy sin ( a + y) = dx sin a

12. Let no. of times of tossing a coin be n. Here, Probability of getting a head in a chance = p =

1 2

Probability of getting no head in a chance = q = 1 -

1 1 = 2 2

343


Now, P (having at least one head) = P (X ³ 1) = 1 - P (X = 0) = 1 - nC 0 p 0 . qn- 0 1 n 1 n = 1 - 1 . 1 . æç ö÷ = 1 - æç ö÷ è 2ø è 2ø From question 1 n 80 1 - æç ö÷ > è 2ø 100 8 1 1 n 8 1 - æç ö÷ > Þ 1> n è 2ø 10 2 10 1 1 n Þ > Þ 2 >5 5 2n Þ n³ 3 A man must have to toss a fair coin 3 times. 13. Let the cartesian equation of line passing through (1, 2, -4) be x-1 y- 2 z+ 4 ...(i) = = a b c Given lines are x - 8 y + 19 z - 10 ...(ii) = = 3 -16 7 x - 15 y - 29 z - 5 ...(iii) = = 3 8 -5 Þ

®

®

®

Obviously parallel vectors b 1 , b 2 and b 3 of (i), (ii) and (iii) respectively are given as ®

b 1 = ai$ + bj$ + ck$ ®

b 2 = 3i$ - 16j$ + 7 k$ ®

b 3 = 3i$ + 8j$ - 5k$ From question ®

®

®

®

(i) ^ (ii)

Þ b1 ^ b2

(i) ^ (iii)

Þ b1 ^ b 3

®

®

®

®

Þ b1 . b2 = 0 Þ b1 . b 3 = 0

Hence, 3a - 16b + 7 c = 0 and 3a + 8b - 5c = 0 From equation (iv) and (v) a b c = = 80 - 56 21 + 15 24 + 48 a b c Þ = = 24 36 72

...(iv) ...(v)

344


Þ

a b c = = = l (say) 2 3 6

Þ

a = 2l , b = 3l , c = 6l

Putting the value of a, b, c in (i) we get required cartesian equation of line as x-1 y- 2 z+ 4 = = 2l 3l 6l x-1 y- 2 z+ 4 Þ = = 2 3 6 Hence vector equation is ®

r = (i$ + 2j$ - 4k$) + l( 2i$ + 3j$ + 6k$)

14. Q Þ Þ Þ Þ

®

®

® ®

®

®

®

a+ b+ c =O ®

...(i) ® ®

a .( a + b + c ) = a . O

® ®

® ®

® ®

® ®

® ®

®

® ®

® ®

a. a + a. b + a. c = 0 é ® ® ® 2ù êQ a . a =| a | ú ë û

a . b + a . c = -| a |2 a . b + c . a = - 25

é ® ® ® ®ù êQ a . c = c . a ú ë û

...(ii)

®

®

Similarly taking dot product of both sides of (i) by b and c respectively we get ® ®

® ®

®

® ®

® ®

®

a . b + b . c = -| b |2 = -144

and

...(iii)

c . a + b . c = -| c |2 = -169

...(iv)

Adding (ii), (iii) and (iv) we get ® ®

® ®

® ®

® ®

® ®

® ®

a . b + c . a + a . b + b . c + c . a + b . c = - 25 - 144 - 169

Þ Þ 15. Given Þ

® ®

® ®

® ®

2( a . b + b . c + c . a ) = - 338 ® ®

® ®

® ®

a. b + b . c + c . a = -

338 = - 169 2

dy - 2xy + y 2 = 0 dx dy 2x 2 = 2xy - y 2 dx 2x 2

dy 2xy - y 2 = dx 2x 2 It is homogeneous differential equation. dy dv Let y = vx Þ =v+ x dx dx Þ

...(i)

345


Equation (i) becomes v+x

Þ

v+x

Þ

x

dv 2x. vx - v 2 x 2 = dx 2x 2 dv = dx

æ v2 ö ÷ 2x 2 çv 2 ø è 2x 2

dv v2 =dx 2

dv v2 =v-v dx 2

Þ

x

Þ

dx 2dv =x v2

Þ

log|x|+ c = 2 .

Integrating both sides we get dx dv Þ ò x = -2ò v 2 Þ

log|x|+ c = -2

v -2 + 1 -2 + 1

1 v

y Putting v = , we get x log|x|+ c = 16. Given: Þ Þ

2x y

dy = 1 + x2 + y2 + x2y2 dx dy = (1 + x 2 ) + y 2 (1 + x 2 ) dx dy (1 + x 2 ) dx = (1 + y 2 )

Þ

dy = (1 + x 2 ) (1 + y 2 ) dx

Þ

x+

Integrating both sides we get dy 2 ò (1 + x ) dx = ò (1 + y 2 ) Þ

2 ò dx + ò x dx = ò

dy (1 + y 2 )

x3 +c 3 Putting y = 1 and x = 0, we get tan–1 (1) = 0 + 0 + c Þ p Þ c = tan -1 (1) = 4 Therefore required particular solution is Þ

tan -1 y = x +

tan -1 y = x +

x3 p + 3 4

x3 + c = tan -1 y 3

346


17. Let I = ò sin x . sin 2x . sin 3x dx. 1 2 sin x . sin 2x . sin 3x dx 2ò 1 = ò sin x . ( 2 sin 2x . sin 3x) dx 2 1 = ò sin x . (cos x - cos 5x) dx [Q 2 sin A sin B = cos ( A - B) - cos ( A + B)] 2 1 1 = 2 sin x . cos x dx 2 sin x . cos 5x dx ò 2´2 2´2 ò =

1 1 sin 2x dx - ò (sin 6x - sin 4x) dx 4ò 4 cos 2x cos 6x cos 4x =+ +C 8 24 16 =

OR Here Now,

Þ

2

ò (1 - x)(1 + x 2 ) dx 2 2

=

(1 - x)(1 + x ) 2 (1 - x)(1 + x 2 )

=

A Bx + C + 1 - x 1 + x2 A(1 + x 2 ) + ( Bx + C)(1 - x) (1 - x)(1 + x 2 )

Þ

2 = A(1 + x 2 ) + ( Bx + C)(1 - x)

Þ

2 = A + Ax 2 + Bx - Bx 2 + C - Cx

Þ

2 = ( A + C) + ( A - B) x 2 + ( B - C) x

Equating co-efficient both sides, we get A+C = 2 A-B=0 B-C = 0 From (ii) and (iii) A = B = C Putting C = A in (i), we get A+A=2 Þ 2A = 2 Þ A = 1 i.e., A=B=C=1 2 1 x+1 \ = + 2 (1 - x)(1 + x ) 1 - x 1 + x 2 \

2

1

...(i) ...(ii) ...(iii)

x+1

ò (1 - x)(1 + x 2 ) = ò 1 - x dx + ò 1 + x 2 dx

347


= - log|1 - x|+ ò

x 1+x

2

dx + ò

1 1 + x2

dx

1 log|1 + x 2|+ tan -1 x + c 2 18. Let the required point of contact be ( x 1 , y 1 ). Given curve is = - log|1 - x|+

\ Þ

y = x 3 - 11x + 5 dy = 3x 2 - 11 dx é dy ù = 3x12 - 11 êë dx úû ( x1 , y1 )

...(i)

i.e., Slope of tangent at ( x 1 , y 1 ) to give curve (i) = 3x12 - 11 From question 3x12 - 11 = Slope of line y = x - 11, which is also tangent 3x12 - 11 = 1 Þ

x12 = 4

Þ x 1 = ±2

Since ( x 1 , y 1 ) lie on curve (i) \

y 1 = x13 - 11x 1 + 5

When

x 1 = 2, y 1 = 2 3 - 11 ´ 2 + 5 = - 9 x 1 = -2, y 1 = ( -2) 3 - 11 ´ ( -2) + 5 = 19

But ( -2, 19) does not satisfy the line y = x - 11 Therefore ( 2, - 9) is required point of curve at which tangent is y = x - 11 OR Let where x = 49 f ( x) = x, let dx = 0.5 \ f ( x + dx) = x + dx = 49.5 Now by definition, approximately we can write f ( x + dx) - f ( x) f '( x) = dx Here f ( x) = x = 49 = 7 dx = 0.5 1 1 1 f '( x) = = = 2 x 2 49 14 Putting these values in (i), we get 49.5 - 7 1 = 14 0.5

...(i)

348


Þ

49.5 = =

19. We have

0.5

+7 14 0.5 + 98 14

=

98.5 14

= 7.036

y = (tan -1 x) 2

... (i)

Differentiating w.r.t. x, we get dy 1 = 2 tan -1 x × dx 1 + x2 or

... (ii)

(1 + x 2 ) y 1 = 2 tan -1 x

Again differentiating w.r.t. x, we get dy d 1 (1 + x 2 ). 1 + y 1 (1 + x 2 ) = 2 . dx dx 1 + x2 2 Þ (1 + x 2 ). y 2 + y 1 . 2x = 1 + x2 or

(1 + x 2 ) 2 y 2 + 2x(1 + x 2 ) y 1 = 2 b+c

20. LHS

q+r

y+z

D= c+a r+p z+x a+b p+ q x+ y

Applying, R 1 « R 3 and R 3 « R 2 , we get a+b p+ q x+ y = b + c q+r y+z c+a r+p z+x Applying, R 1 ® R 1 + R 2 + R 3 , we get 2 ( a + b + c) 2 ( p + q + r) 2 ( x + y + z) D=

b+c

q+r

y+z

c+a

r+p

z+x

a+b + c p+ q+r x+ y+z =2

b+c

q+r

y+z

c+a

r+p

z+x

a =2 b+c

p q+r

x y + z [Applying R 1 ® R 1 - R 2 ]

c+a r+p z+x a

p

x

= 2 b + c q+r y+z c

r

z

[Applying R 3 ® R 3 - R 1 ]

349


Again applying R 2 ® R 2 - R 3 , we get a p x D = 2 b q y = RHS c r 21. Now,

tan

z

x x æ ö cos 2 - sin 2 ç ÷ cos x ö -1 2 2 ç ÷ = tan ç ÷ è 1 + sin x ø ç cos 2 x + sin 2 x + 2 cos x . sin x ÷ è 2 2 2 2ø

-1 æ

éæ x x öæ x xö ù ê çè cos - sin ÷ø çè cos + sin ÷ø ú 2 2 2 2 ú = tan -1 ê 2 ê ú æç cos x + sin x ö÷ ê ú è ø 2 2 ë û x x æ cos sin ö÷ ç 2 2 ç ÷ x x÷ æ cos x - sin x ö ç cos cos ç 2 2 ÷ = tan -1 ç 2 2÷ = tan -1 ç ÷ x x x x ç ÷ ç cos + sin ÷ ç cos 2 sin 2 ÷ è 2 2ø + ç x x÷ ç cos cos ÷ è 2 2ø æ 1 - tan x ö æ tan p - tan x ö ç ÷ -1 1 2 = tan ç 4 2 ÷ = tan ç ÷ ç ÷ ç 1 + tan x ÷ ç 1 + tan p tan x ÷ è è 2ø 4 2ø p x ù p p é = tan -1 ê tan æç - ö÷ ú Q x Î æç - , ö÷ è ø è 4 2 û 2 2ø ë p x p p = Þ - - >4 2 4 p p p x p p Þ + > - >- + 4 4 4 2 4 4 p p x Þ > - >0 2 4 2 p x p p p Þ æç - ö÷ Î æç 0, ö÷ Ì æç - , ö÷ è 4 2ø è 2 ø è 2 2 ø

350


Let sin Þ Þ Þ Þ Þ Þ Now, Þ Þ Þ Þ

OR 8ö -1 æ 3 ö ç ÷ = a and sin ç ÷ = b è 17 ø è5ø 8 3 and sin b = sin a = 17 5

-1 æ

cos a = 1 - sin 2 a

cos b = 1 - sin 2 b

64 9 and cos b = 1 289 25 289 - 64 25 - 9 and cos b = cos a = 289 25 225 16 and cos b = cos a = 289 25 15 4 and cos b = cos a = 17 5 cos(a + b) = cos a . cos b - sin a . sin b 15 4 8 3 cos(a + b) = ´ ´ 17 5 17 5 60 24 36 cos(a + b) = Þ cos(a + b) = 85 85 85 -1 æ 36 ö a + b = cos ç ÷ è 85 ø cos a = 1 -

sin -1

8 3 36 + sin -1 = cos -1 æç ö÷ è 85 ø 17 5

22. Let x 1 , x 2 Î A. Now,

and

f (x1) = f (x2 )

Þ

Þ Þ Þ Þ Hence f is one-one function. For Onto x-2 Let y= x- 3 Þ xy - 3y = x - 2 Þ Þ x( y - 1) = 3y - 2 3y - 2 Þ x= y-1

x1 - 2 x1 - 3

[Putting the value of a , b]

=

x2 - 2 x2 - 3

( x 1 - 2)( x 2 - 3) = ( x 1 - 3)( x 2 - 2) x 1 x 2 - 3x 1 - 2x 2 + 6 = x 1 x 2 - 2x 1 - 3x 2 + 6 -3 x 1 - 2 x 2 = -2 x 1 - 3 x 2 -x1 = -x2 Þ x1 = x2

xy - x = 3y - 2 ...(i)

From above it is obvious that " y except 1, i.e., " y Î B = R - { 1} $ x Î A

351


Hence f is onto function. Thus f is one-one onto function. It f -1 is inverse function of f then 3y - 2 [from (i)] f -1 ( y) = y-1

SECTION–C 23. The equation of the plane through three non-collinear points A(3, –1, 2), B(5, 2, 4) and C (–1, –1, 6) can be expressed as x- 3 y+1 z- 2 5- 3

2+1

4-2 =0

-1 - 3 -1 + 1 6 - 2 x- 3 y+1 z- 2 Þ

2

3

2

-4

0

4

=0

Þ 12( x - 3) - 16( y + 1) + 12(z - 2) = 0 Þ 12x - 16y + 12z - 76 = 0 Þ 3x - 4y + 3z - 19 = 0 is the required equation. Now, distance of P(6, 5, 9) from the plane is given by 3 ´ 6 - 4(5) + 3( 9) - 19 6 6 units. = = = 9 + 16 + 9 34 34 24. Let E1 , E2 and A be events such that E1 = student is a hosteler E2 = student is a day scholar A = getting A grade. Now from question 60 6 40 4 P(E1 ) = = , P(E2 ) = = 100 10 100 10 30 3 20 2 æ ö æ ö A A Pç E ÷ = = , Pç E ÷ = = è 1 ø 100 10 è 2 ø 100 10 æE ö We have to find P ç 1 ÷ . èAø Now

E P æç 1 A ö÷ = è ø

P(E1 ). P æç A E ö÷ è 1ø P(E1 ). P çæ A E ÷ö + P(E2 ). P çæ A E ÷ö è 1ø è 2ø

6 3 18 . 18 100 18 9 10 10 100 = = = = = ´ 6 3 4 2 18 8 100 26 26 13 . + . + 10 10 10 10 100 100

352


25. Let x package nuts and y package bolts are produced Let z be the profit function, which we have to maximize. Here z = 17.50x + 7y ... (i) is objective function. And constraints are ...(ii) x + 3y £ 12 ...(iii) 3x + y £ 12 ...(iv) x³0 ...(v) y³0 On plotting graph of above constraints or inequalities (ii), (iii), (iv) and (v) we get shaded region as feasible region having corner points A, O, B and C. Y 13 12 11 10 9 8 7 6

5 x+

3y

=1

A(0, 4)

2

4 C(3, 3)

3 2 1

B(4, 0) X’ –2

–1

–1

1

2

3

5

4

6

3x +

O (0, 0)

y= 12

Y’

For coordinate of ‘C’ two equations ...(vi) x + 3y = 12 ...(vii) are solved 3x + y = 12

7

8

9

10

11

12

X

353


Applying (vi) × 3 – (vii), we get 3x + 9y - 3x - y = 36 - 12 and Þ 8y = 24 Þ y = 3 x= 3 Hence coordinate of C are (3, 3). Now the value of z is evaluated at corner point as Corner point

z = 17. 5 x + 7y

(0, 4)

28

(0, 0)

0

(4, 0)

70

(3, 3)

73.5

¬

Maximum

Therefore maximum profit is `73.5 when 3 package nuts and 3 package bolt are produced. p 4

= ò ( tan x + cot x ) dx

26. LHS

0 p 4æ

cos x ö ÷ dx = sin x ø

sin x = òç + è cos x 0 p 4

= 2ò 0

p 4

sin x + cos x dx sin x . cos x 0

ò

p 4

(sin x + cos x) sin x + cos x dx = 2 ò dx 2 2 sin x. cos x 0 1 - (sin x - cos x)

Let sin x - cos x = z Þ

(cos x + sin x) dx = dz

Also if x = 0, z = -1 p 1 1 and x = , z = =0 4 2 2 0

\ LHS

= 2ò

-1

dz 1 - z2

= 2 [sin -1 z]

0 -1

= 2 [sin -1 0 - sin -1 ( -1)]

p p ù é = 2 ê 0 - æç - ö÷ ú = 2 . = RHS è ø 2 û 2 ë OR Let

2

f ( x) = 2x + 5x

Here a = 1, b = 3

b - a 3-1 2 = = n n n Þ nh = 2 Also, n ® ¥ Û h ® 0.

\ h=

354

Xam idea Mathematics–XII b

Q

h[ f ( a) + f ( a + h) + ... + f {a + (n - 1) h}] ò f ( x) dx = hlim ®0 a 3

\

ò ( 2x

2

1

+ 5x) dx = lim h [ f (1) + f (1 + h) + ... + f {1 + (n - 1) h}] h ®0

= lim h[{2 ´ 1 2 + 5 ´ 1} + {2(1 + h) 2 + 5(1 + h)} + ... + {2(1 - (n - 1) h) 2 + 5((1 + (n - 1) h}] h ®0

= lim h[( 2 + 5) + {2 + 4h + 2h 2 + 5 + 5h} + ... + {2 + 4(n - 1) h + 2(n - 1) 2 h 2 + 5 + 5(n - 1) h}] h ®0

= lim h[7 + {7 + 9h + 2h 2 } + ... + {7 + 9(n - 1) h + 2(n - 1) 2 h 2 }] h ®0

= lim h[7n + 9h{1 + 2 + ... + (n - 1)} + 2h 2 {1 2 + 2 2 + ... + (n - 1) 2 }] h ®0

(n - 1). n (n - 1). n( 2n - 1) ù é = lim ê7nh + 9h 2 + 2h 3 úû h ®0 ë 2 6 1 1 1 ù é 9(nh) 2 . æç1 - ö÷ 2(nh) 3 . æç1 - ö÷ . æç 2 - ö÷ ú ê è è nø nø è nø = lim ê7(nh) + + ú h ®0 ê 2 6 ú êë úû 1 1 1 ù é 36æç1 - ö÷ 16æç1 - ö÷ . æç 2 - ö÷ ú ê è è nø nø è nø [Q nh = 2] = lim ê14 + + ú n®¥ ê 2 6 ú êë úû 1 8 1 1 ù é = lim ê14 + 18æç1 - ö÷ + æç1 - ö÷ . æç 2 - ö÷ ú è n®¥ ë nø 3 è nø è nø û 8 = 14 + 18 + ´ 1 ´ 2 3 16 96 + 16 112 = 32 + = = 3 3 3 27. Given lines are ...(i) 3x - 2y + 1 = 0 ...(ii) 2x + 3y - 21 = 0 ...(iii) x - 5y + 9 = 0 For intersection of (i) and (ii) Applying (i) × 3 + (ii) × 2, we get 9x - 6y + 3 + 4x + 6y - 42 = 0 Þ 13x - 39 = 0 Þ x= 3 Putting it in (i), we get 9 - 2y + 1 = 0

355


3x

–2

y+

1=

0

Y Þ 2y = 10 Þ y=5 8 Intersection point of (i) and (ii) is (3, 5) For intersection of (ii) and (iii) 2x +3 7 y– Applying (ii) – (iii) × 2, we get 21 =0 6 2x + 3y - 21 - 2x + 10y - 18 = 0 Þ 13y - 39 = 0 (3, 5) 5 Þ y= 3 Putting y = 3 in (ii), we get 4 (6, 3) 2x + 9 - 21 = 0 3 Þ 2x - 12 = 0 +9=0 x – 5y 2 Þ x=6 (1, 2) Intersection point of (ii) and (iii) is (6, 3) 1 For intersection of (i) and (iii) Applying (i) – (iii) × 3, we get O X’ –1 1 2 5 6 3 4 3x - 2y + 1 - 3x + 15y - 27 = 0 Y’ Þ 13y - 26 = 0 Þ y = 2 Putting y = 2 in (i), we get 3x - 4 + 1 = 0 Þ x=1 Intersection point of (i) and (iii) is (1, 2) With the help of point of intersection we draw the graph of lines (i), (ii) and (iii) Shaded region is required region.

3

\ Area of Required region = ò 1

6

8

X

6

3x + 1 -2x + 21 x+9 dx + ò dx - ò dx 2 3 5 3

3

=

7

3

6

1

3

1 6

6

6

3

1

1 6

3 1 2 1 9 x dx + ò dx - ò x dx + 7 ò dx - ò x dx - ò dx ò 2 2 3 5 5 1

6

1 3 2 é x2 ù 1 é x2 ù 9 [x] 1 - ê ú + 7[x] 63 - ê ú - [x] 61 2 3ë 2 û 5ë 2 û 5 1 3 1 3 1 2 1 9 = ( 9 - 1) + ( 3 - 1) - ( 36 - 9) + 7( 6 - 3) - ( 36 - 1) - ( 6 - 1) 4 2 6 10 5 7 = 6 + 1 – 9 + 21 – - 9 2 7 20 - 7 13 = 10 – = = 2 2 2 28. Let r and h be radius and height of given cylinder of surface area S. If V be the volume of cylinder then =

V = pr 2 h

3 é x2 ù ê ú 2ë 2 û

3

+

356


V=

pr 2 .( S - 2pr 2 ) 2 pr

[Q S = 2pr 2 + 2prh Þ

Sr - 2pr 3 2 dV 1 Þ = ( S - 6 pr 2 ) dr 2 For maximum or minimum value of V dV =0 dr 1 Þ ( S - 6 pr 2 ) = 0 Þ S - 6 pr 2 = 0 2 S S Þ r2 = Þ r= 6p 6p d 2V 1 Now = - ´ 12pr 2 dr 2 Þ

Þ Þ

S - 2 pr 2 = h] 2 pr

V=

h

d 2V

= -6 pr dr 2 é d 2V ù = –ve ê 2 ú ë dr û r = S 6p

S . Volume V is maximum. 6p S S - 2p. 3S - S 6p 6p Þ h= h= ´ 3 ´ 2p S S 2p 6p 2S 6p S h= . =2 6p 6p S é S ù h = 2r (diameter) êQ r = 6p ú ë û

Hence for r =

Þ

Þ Þ

Therefore, for maximum volume height of cylinder in equal to diameter of its base. 29. The given system of equation can be written in matric form as AX = B é 1 -1 2 ù é xù é7 ù ê ú ê ú A = 3 4 -5 , X = y , B = ê -5 ú ê ú ê ú ê ú êë 2 -1 3 úû êë z úû êë 12 úû 1 -1 2 Now,|A|= 3 4 -5 = 1(12 - 5) + 1( 9 + 10) + 2( -3 - 8) 2 -1

3 = 7 + 19 - 22 = 4 ¹ 0

r

357


Hence A -1 exist and system have unique solution. 4 -5 = 12 – 5 = 7 C 11 = ( -1) 1+ 1 -1 3 C 12 = ( -1) 1+ 2

3 -5 = -( 9 + 10) = -19 2 3

C 13 = ( -1) 1+ 3

3 4 = +(– 3 – 8) = –11 2 -1

C 21 = ( -1) 2 + 1

-1 2 = –(–3+2) = 1 -1 3

C 22 = ( -1) 2 + 2

1 2 = +(3 – 4) = –1 2 3

C 23 = ( -1) 2 + 3

1 -1 = – (–1 + 2) = –1 2 -1

C 31 = ( -1) 3 + 1

-1 2 = + (5 – 8) = –3 4 -5

C 32 = ( -1) 3 + 2

1 2 = – (–5 – 6) = 11 3 -5

C 33 = ( -1) 3 + 3

1 -1 = +(4 + 3) = 7 3 4 T

\

Þ Q Þ Þ

Þ

1 -3 ù é 7 -19 -11ù é 7 adjA = ê 1 -1 -1 ú = ê -19 -1 11 ú ê ú ê ú 7 úû êë -3 11 êë -11 -1 7 úû 1 -3 ù é 7 1 1ê -1 A = adj A = -19 -1 11 ú ú |A| 4ê êë -11 -1 7 úû AX = B X = A -1 B 1 -3 ù é 7 ù é xù é 7 ê yú = 1 ê -19 -1 11 ú ê -5ú ê ú 4ê úê ú êë z úû êë -11 -1 7 úû êë 12 úû é xù é 49 - 5 - 36 ù ê yú = 1 ê -133 + 5 + 132ú ê ú 4ê ú êë z úû êë -77 + 5 + 84 úû

358


Þ

é xù é8ù ê yú = 1 ê 4 ú ê ú 4ê ú êë z úû êë12úû

é xù é 2 ù Þ ê yú = ê 1 ú ê ú ê ú êë z úû êë 3úû

Equating the corresponding elements, we get x = 2, y = 1, z = 3

Let

OR é -1 1 2 ù A = ê 1 2 3ú ê ú êë 3 1 1 úû

For applying elementary row operation we write, A = IA é -1 1 2 ù é 1 0 0 ù ê 1 2 3ú = ê 0 1 0ú A ê ú ê ú êë 3 1 1 úû êë 0 0 1 úû Applying R 1 « R 2 , we get é 1 2 3ù é 0 1 0ù ê -1 1 2 ú = ê 1 0 0 ú A ê ú ê ú êë 3 1 1 úû êë 0 0 1 úû Applying R 2 ® R 2 + R 1 and R 3 ® R 3 - 3R 1 , we get 3 ù é 0 1 0ù é1 2 ê 0 3 5 ú = ê1 1 0ú A ê ú ê ú êë 0 -5 -8úû êë 0 -3 1 úû 2 Applying R 1 ® R 1 - R 2 , we get 3 1 é1 0 - 1 ù é - 2 ù 3ú ê 3 3 0ú ê 5 ú=ê 1 1 0ú A ê0 3 êë 0 -5 -8 úû êë 0 -3 1 úû 1 Applying R 2 ® R 2 , we get 3 1 é1 0 - 1 ù é - 2 ù 3ú ê 3 3 0ú ê 5 1 1 ê0 1 3 ú=ê 3 3 0ú A êë 0 -5 -8 úû êë 0 -3 1 úû Applying R 3 é1 0 ê ê0 1 êë 0 0

® R 3 + 5R 2 , we get - 1 3ù é - 2 3 1 3 0ù ú 5 ú=ê 1 1 3 ú ê 3 3 0ú A 1 ú ê 5 4 3 û ë 3 - 3 1 úû

359


Applying R 1 é1 0 ê ê0 1 êë 0 0

® R 1 + R 3 and R 2 ® R 2 - 5R 3 0ù é1 -1 1ù ú ê ú 0 ú = ê -8 7 -5 ú A 1 ú ê5 4 3û ë 3 - 3 1 úû

Applying R 3 ® 3R 3 , we get é 1 0 0 ù é 1 -1 1 ù ê 0 1 0 ú = ê -8 7 -5 ú A ê ú ê ú êë 0 0 1 úû êë 5 -4 3 úû Hence

A

é 1 -1 1 ù = ê -8 7 -5 ú ê ú êë 5 -4 3 úû

-1

Set–II 9.

®

®

®

a + b + c = (i$ - 2j$) + ( 2i$ - 3j$) + ( 2i$ + 3k$) = 5i$ - 5j$ + 3k$

10. Co-factor of a 32 = ( -1) 3 + 2 1

1

1

19. LHS = a

b

c

3

3

c3

a

b

5 8 = -(5 - 16) = 11 2 1

Applying C 2 ® C 2 - C 1 , C 3 ® C 3 - C 1 , we get 1 0 0 = a b-a c-a a3 b 3 - a3 c3 - a3 Taking out (b - a), ( c - a) common from C 2 and C 3 respectively, we get 1 0 0 = (b - a)( c - a) a 1 1 a3

b 2 + ab + a 2

c 2 + ac + a 2

Expanding along R 1 , we get

= -( a - b)( c - a)[1( c 2 + ac + a 2 - b 2 - ab - a 2 ) - 0 + 0] = -( a - b)( c - a)( c 2 + ac - b 2 - ab) = -( a - b)( c - a){-(b 2 - c 2 ) - a(b - c)} = -( a - b)( c - a){(b - c)( -b - c - a)} = ( a - b)(b - c)( c - a)( a + b + c)

360


20. Given, y = 3 cos (log x) + 4 sin (log x) Differentiating w.r.t. x, we have 3 sin (log x) 4 cos (log x) dy =+ dx x x 1 Þ y 1 = [ - 3 sin (log x) + 4 cos (log x) ] x Again differentiating w.r.t. x, we have é - 3 cos (log x) 4 sin (log x) ù xê 2 úû - [– 3 sin (log x) + 4 cos (log x)] d y x x ë = dx 2 x2 – 3 cos (log x) - 4 sin (log x) + 3 sin (log x) + 4 cos (log x) = x2 2 d y - sin (log x) - 7 cos (log x) = dx 2 x2 - sin (log x) - 7 cos (log x) Þ y2 = x2 Now, L.H.S. = x 2 y 2 + xy 1 + y æ - sin (log x) - 7 cos (log x) ö 1 = x2 ç ÷ + x ´ [- 3 sin (log x) + 4 cos (log x)] 2 è ø x x + 3 cos (log x) + 4 sin (log x) = - sin (log x) - 7 cos (log x) - 3 sin (log x) + 4 cos (log x) + 3 cos (log x) + 4 sin (log x) = 0 = RHS 21. Let the direction ratios of the required line be a, b, c. Since the required line is perpendicular to the given lines, therefore, ...(i) a + 2b + 3c = 0 and ...(ii) -3a + 2b + 5c = 0 Solving (i) and (ii), by cross multiplication, we get a b c = = = k (let) 10 - 6 -9 - 5 2 + 6 Þ a = 4k , b = -14k , c = 8k Thus, the required line passing through P( -1, 3, - 2) and having the direction ratios x+1 y- 3 z+ 2 . a = 4k , b = -14k , c = 8k is = = 4k -14k 8k x+1 y- 3 z+ 2 x+1 y- 3 z+ 2 Removing k, we get or which is the required = = = = 4 -14 8 2 -7 4 equation of the line.

361


22. Given Þ

dy = 2e - y - 1 dx dy dx = -y 2e - 1 x + 1

( x + 1)

Integrating both sides we get dy dx ò 2e - y - 1 = ò x + 1 e y dy

Þ

ò 2 - ey

Þ

-ò

Þ

- log z = log|x + 1|+ c

Þ

- log|2 - e y|= log|x + 1|+ c

Þ

log|x + 1|+ log|2 - e y|= log k

Þ

log|( x + 1).( 2 - e y )|= log k

Þ

( x + 1)( 2 - e y ) = k

= log|x + 1|+ c

dz = log|x + 1|+ c z

[Let 2 - e y = z Þ - e y dy = dz Þ e y dy = - dz]

Putting x = 0, y = 0, we get 1.( 2 - e 0 ) = k

Þ

k =1

Therefore, required particular solution is ( x + 1)( 2 - e y ) = 1 28. Let E1 , E2 , A be events such that E1 = getting 5 or 6 in a single throw of die E2 = getting 1, 2, 3 or 4 in a single throw of a die A = getting exactly two heads æE ö P ç 2 ÷ is required. èAø Now, P(E1 ) =

2 1 4 2 = and P(E2 ) = = 6 3 6 3

æAö 3 Pç ÷ = è E1 ø 8

[Q {HHH, HHT, HTT, TTT, TTH, THH, THT, HTH}]

æAö 1 Pç ÷ = è E2 ø 4

[{HH, HT, TH, TT}]

362


\

æE ö Pç 2 ÷ = èAø

æAö P(E2 ). P ç ÷ è E2 ø æAö æAö P(E1 ). P ç ÷ + P(E2 ). P ç ÷ è E1 ø è E2 ø

2 1 1 1 ´ 1 24 4 3 4 = = 6 = 6 = ´ = 1 3 2 1 1 1 3+4 6 7 7 ´ + ´ + 3 8 3 4 8 6 24 29. Given lines are ...(i) 3x - y - 3 = 0 ...(ii) 2x + y - 12 = 0 ...(iii) x - 2y - 1 = 0 For intersecting point of (i) and (ii) (i) + (ii) Þ 3x - y - 3 + 2x + y - 12 = 0 Þ 5x - 15 = 0 Þ x= 3 Putting x = 3 in (i), we get 9-y- 3=0 Þ y=6 Intersecting point of (i) and (ii) is (3, 6) For intersecting point of (ii) and (iii) (ii) – 2 × (iii) Þ 2x + y - 12 - 2x + 4y + 2 = 0 Þ 5y - 10 = 0 Þ y=2 Putting y = 2 in (ii) we get 2x + 2 - 12 = 0 Þ x=5 Intersecting point of (ii) and (iii) is (5, 2). For Intersecting point of (i) and (iii) (i) – 3 × (iii) Þ 3x - y - 3 - 3x + 6y + 3 = 0 Þ 5y = 0 Þ y=0 Putting y = 0 in (i), we get 3x - 3 = 0 Þ x=1 Intersecting point (i) and (iii) is (1, 0).

363


–1

y–3

+y

=0

2x

Y

0

3x –

2=

7 6

(3, 6)

5 4 3 2

(5, 2)

1 x

y –2

–1

=0

(1, 0) O

1

2

3

5

4

6

X

7

Shaded region is required region. 3

5

5

\ Required Area = ò ( 3x - 3) dx + ò ( -2x + 12) dx - ò 1

3 3

3

1 5

x-1 dx 2

5

= 3 ò x dx - 3 ò dx - 2 ò x dx + 12 ò dx 1

1

3

3

3 2 ù5

5

5

1

1 5 2ù

1 1 x dx + ò dx ò 2 2

é x2 ù éx 1 éx 1 3 = 3ê ú - 3 [x] - 2ê ú + 12[x]5 - ê ú + [x]5 1 3 1 2 2 2 2 ë û1 ë û3 ë û1 2 3 1 1 = ( 9 - 1) - 3( 3 - 1) - ( 25 - 9) + 12(5 - 3) - ( 25 - 1) + (5 - 1) 2 4 2 = 12 – 6 – 16 + 24 – 6 + 2 = 10 sq. unit

Set–III 9.

®

®

®

a + b + c = i$ - 3k$ + 2j$ - k$ + 2i$ - 3j$ + 2k$ = 3i$ - j$ - 2k$

10. Minor of a 22 =

1 3 = 8 - 15 = -7 5 8

364


19. LHS = D =

1+a

1

1

1

1+b

1

1

1

1+c

Taking out a, b, c common from I, II, and III row respectively, we get 1 1 1 +1 a a a 1 1 1 D = abc +1 b b b 1 1 1 +1 c c c Applying R 1 ® R 1 + R 2 + R 3 1 1 1 1 1 1 1 1 1 + + +1 + + +1 + + +1 a b c a b c a b c 1 1 1 D = abc +1 b b b 1 1 1 +1 c c c 1 1 1 1 1 æ ö = abc ç + + + 1÷ èa b c ø b 1 c

1 1 1 1 +1 b b 1 1 +1 c c

Applying C 2 ® C 2 - C 1 , C 3 ® C 3 - C 1 , we get 1 1 1 D = abc æç + + + 1ö÷ èa b c ø

1 0 0 1 1 0 b 1 0 1 c

1 1 1 = abc æç + + + 1ö÷ ´ (1 ´ 1 ´ 1) èa b c ø

(Qthe determinant of a triangular matrix

is the product of its diagonal elements.) æ bc + ac + ab + abc ö 1 1 1 = abc æç + + + 1ö÷ = abc ç ÷ = ab + bc + ca + abc = R.H.S. èa b c ø è ø abc 20. Q Þ

y = sin -1 x dy 1 = dx 1 - x2

Þ

1 - x2

dy =1 dx

Again differentiating w.r.t. x, we get d 2 y dy 1 ´ (– 2x) =0 1 - x2 + . dx 2 dx 2 1 - x 2

365


Þ

(1 - x 2 )

d2y 2

–

xdy =0 dx

dx 21. Given differential equation is dy xy = ( x + 2)( y + 2) dx y x+2 Þ dy = dx y+2 x Integrating both sides y æ ò y + 2 dy = ò çè1 + æ

2 ö

2ö ÷ dx xø æ

2ö

ò çè1 - y + 2 ÷ø dy = ò çè1 + x ÷ø dx

Þ

… (i) Þ y - 2 log|y + 2| = x + 2 log|x| + c Given that y = – 1 when x = 1 \ -1 - 2 log 1 = 1 + 2 log|1 |+ C Þ C = -2 \ The required particular solution is y - 2 log|y + 2| = x + 2 log|x| - 2 22. Let the equation of line passing through the point (2, –1, 3) be x- 2 y+1 z- 3 ...(i) = = a b c Given lines are ®

r = (i$ + j$ - k$) + l( 2i$ - 2j$ + k$)...(ii)

®

r = ( 2i$ - j$ - 3k$) + m(i$ + 2j$ + 2k$)

Since (i), (ii) and (i), (iii) are perpendicular to each other Þ 2a - 2b + c = 0 a + 2b + 2c = 0 a b c Þ = = -4 - 2 1 - 4 4 + 2 a b c Þ = = = l (say) -6 -3 6 Þ a = -6 l, b = -3 l, c = 6 l Putting it in (i) we get required equation of line as x- 2 y+1 z- 3 = = -6l -3 l 6l x-2 z- 3 Þ = y+1= 2 –2

...(iii)

366


28. Let E1 , E2 , E 3 and A be events such that E1 = Both transfered ball from Bag I to Bag II are red. E2 = Both transfered ball from Bag I to Bag II are black. E 3 = Out of two transfered ball one is red and other is black. A = Drawing a red ball from Bag II. æE ö Here, P ç 2 ÷ is required. èAø 3

Now, P(E1 ) = P(E2 ) =

C2

7

C2

4

C2

7

C2

3

P(E 3 ) =

=

2!´ 5! 1 3! ´ = 2! 1! 7! 7

=

2!´ 5! 2 4! ´ = 2! 2! 7! 7

C1 ´ 4C1 7

C2

=

3´ 4 7!

æAö 6 æAö 4 Pç ÷ = , Pç ÷ = , è E1 ø 11 è E2 ø 11

\

æE ö Pç 2 ÷ = èAø

´

2 !5 ! 4 = 1 7

æAö 5 Pç ÷ = è E 3 ø 11

æAö P(E2 ). P ç ÷ è E2 ø æAö æAö æAö P(E1 ). P ç ÷ + P(E2 ). P ç ÷ + P(E 3 ). P ç ÷ è E1 ø è E2 ø èE3 ø

2 4 8 ´ 8 77 4 7 11 77 = = = ´ = 1 6 2 4 4 5 6 8 20 77 34 17 ´ + ´ + ´ + + 7 11 7 11 7 11 77 77 77 29. Given lines are ...(i) 5x - 2y - 10 = 0 ...(ii) x+y-9=0 ...(iii) 2x - 5y - 4 = 0 For intersecting point of (i) and (ii) (i) + 2 × (ii) Þ 5x - 2y - 10 + 2x + 2y - 18 = 0 Þ 7 x - 28 = 0 Þ x = 4 Putting x = 4 in(i), we get 20 - 2y - 10 = 0 Þ y=5 Intersecting point of (i) and (ii) is (4, 5). For intersecting point of (i) and (iii) (i) × 5 – (iii) × 2 Þ 25x - 10y - 50 - 4x + 10y + 8 = 0 Þ 21x - 42 = 0 Þ x = 2

367


Putting x = 2 in (i) we get 10 - 2y - 10 = 0 Þ y=0 i.e., Intersecting points of (i) and (iii) is (2, 0) For intersecting point of (ii) and (iii) 2 × (ii) × (iii) Þ 2x + 2y - 18 - 2x + 5y + 4 = 0 Þ 7 y – 14 = 0 Þ y = 2 Putting y = 2 in (ii) we get x+ 2- 9= 0 Þ x=7 Intersecting point of (ii) and (iii) is (7, 2). Y 6 (4, 5)

5

x + –

0

y

4

=

10 =

9

5x –

2y –

0

3 2

(7, 2) 1

2x –

(2, 0) 1

O

2

3

5y –

0 4=

5

4

6

8 X

7

Shaded region is required region. 4

\

Required Area = ò æç è 2

7

4

4

=

7

( 2x - 4) 5x - 10 ö dx ÷ dx + ò ( - x + 9) dx - ò 2 ø 5 2

4

7

2

4

7

7

7

4

2

2 7 2ù

5 2 4 x dx - 5ò dx - ò x dx + 9ò dx - ò x dx + ò dx 2ò 5 5 2

4

2 ù7

éx 5 é x2 ù 2 éx 4 7 4 7 ê ú - 5 [x] 2 - ê ú + 9 [x] 4 - ê ú + [x] 2 2ë 2 û 2 5 2 5 ë û4 ë û2 2 5 1 1 4 = (16 - 4) - 5( 4 - 2) - ( 49 - 16) + 9 (7 – 4) - ( 49 - 4) + (7 – 2) 4 2 5 5 33 33 54 - 33 21 = 15 - 10 + 27 - 9 + 4 = 27 = = sq. unit 2 2 2 2 =

CBSE Examination Paper (All India 2012) Time allowed: 3 hours

Maximum marks: 100


Set–I SECTION–A Question numbers 1 to 10 carry 1 mark each. 1. The binary operation * : R × R ® R is defined as a * b = 2a + b. Find (2 * 3) * 4 2. Find the principal value of tan–1 3 - sec–1 (-2). 3. Find the value of x + y from the following equation: 5 ù é 3 -4 ù é 7 6 ù éx 2ê ú+ê ú=ê ú ë7 y - 3û ë 1 2 û ë15 14û é 3 4. If A = ê -1 ê êë 0 T

4ù é -1 2ú and B = ê ú ë 1 1 úû

2 2

1ù , then find AT – BT. 3 úû

5. Let A be a square matric of order 3 × 3. Write the value of 2A, where A = 4. 2

6. Evaluate: 7. Given

òe

ò 0 x

4 – x 2 dx (tan x + 1) sec x dx = e x f ( x) + c.

Write f(x) satisfying the above. 8. Write the value of (i$ ´ j$ ) . k$ + i$ . j$

.

369


9. Find the scalar components of the vector AB with initial point A (2,1) and terminal point B (–5, 7). 10. Find the distance of the plane 3x – 4y + 12z = 3 from the origin.

SECTION–B Question numbers 11 to 22 carry 4 marks each. 3 3 6 11. Prove the following: cos æç sin –1 + cot –1 ö÷ = è ø 5 2 5 13 12. Using properties of determinants, show that b+c a a b

c+a

b

c

c

a+b

= 4abc

13. Show that f : N ® N, given by ì x + 1, if x is odd f (x) = í î x – 1, if x is even is both one-one and onto. OR Consider the binary operations * : R × R ® R and o : R × R ® R defined as a * b =|a -b|and aob = a for all a, b Î R. Show that ‘*’ is commutative but not associative, ‘o’ is associative but not commutative. –1 –1 dy y 14. If x = a sin t , y = a cos t , show that =– dx x OR é 1 + x2 - 1ù ú with respect to x. Differentiate tan–1 ê ê ú x ë û d2y p d2x d2y 15. If x = a (cos t + t sin t) and y = a (sin t – t cos t), 0 < t < , find , and . 2 dt 2 dt 2 dx 2 16. A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall? 2

17. Evaluate:

ò |x

3

- x|dx

–1 p

Evaluate:

ò

OR x sin x

dx 2 1 + cos x 0 18. Form the differential equation of the family of circles in the second quadrant and touching the coordinate axes. OR Find the particular solution of the differential equation dy y = 0 when x = 2 x ( x 2 - 1) = 1; dx

370


19. Solve the following differential equation: (1 + x2) dy + 2xy dx = cot x dx; x ¹ 0 ®

® ® 20. Let a = i$ + 4j$ + 2k$, b = 3i$ – 2j$ + 7 k$ and c = 2i$ – j$ + 4k$. ®

®

®

® ®

Find a vector p which is perpendicular to both a and b and p . c = 18. 21. Find the coordinates of the point where the line through the points A (3, 4, 1) and B (5, 1, 6) crosses the XY-plane. 22. Two cards are drawn simultaneously (without replacement) from a well-shuffled pack of 52 cards. Find the mean and variance of the number of red cards.

SECTION–C Question numbers 23 to 29 carry 6 marks each. 23. Using matrices, solve the following system of equations: 2x + 3y + 3z = 5, x – 2y + z = – 4, 3x – y – 2z = 3 24. Prove that the radius of the right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone. OR An open box with a square base is to be made out of a given quantity of cardboard of area c3 c2 square units. Show that the maximum volume of the box is cubic units. 6 3 25. Evaluate:

ò

x sin –1 x 1 – x2

dx OR

2

Evaluate:

x +1

ò ( x - 1) 2 ( x + 3) dx

26. Find the area of the region {(x, y) : x2 + y2 £ 4, x + y ³ 2}. x– 1 y– 2 z– 3 x– 1 y- 2 z - 3 27. If the lines = = and are perpendicular, find the value of k = – –3 –2k 2 k 1 5 and hence find the equation of plane containing these lines. 28. Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin 3 times and notes the number of heads. If she gets 1,2,3 or 4 she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1,2,3, or 4 with the die? 29. A dietician wishes to mix two types of foods in such a way that the vitamin contents of the mixture contains at least 8 units of vitamin A and 10 units of vitamin C. Food I contains 2 units/kg of vitamin A and 1 units/kg of vitamin C while Food II contains 1 unit/kg of vitamin A and 2 units/kg of vitamin C. It costs `5 per kg to purchase Food I and `7 per kg to purchase Food II. Determine the minimum cost of such a mixture. Formulate the above as a LPP and solve it graphically.

371


Set–II Only those questions, not included in Set I, are given 10. Write the value of ( k$ ´ j$) . i$ + j$ . k$ 4 12 33 19. Prove that: cos -1 æç ö÷ + cos -1 æç ö÷ = cos -1 æç ö÷ è5 ø è 13 ø è 65 ø 20. If y = (tan–1 x)2 , show that ( x 2 + 1) 2 21.

22. 28. 29.

d2y 2

+ 2x ( x 2 + 1)

dy = 2. dx

dx Find the particular solution of the differential equation dy p + y cot x = 4x cos ec x, (x ¹ 0) given that y = 0 when x = . dx 2 Find the coordinates of the point where the line through the points (3,–4, –5) and (2,–3, 1) crosses the plane 2x + y + z = 7. Using matrices, solve the following system of equations: x + y – z = 3; 2x + 3y + z = 10; 3x – y – 7z = 1 Find the length and the foot of the perpendicular from the point P (7, 14, 5) to the plane 2x + 4y - z = 2. Also find the image of point P in the plane.

Set–III Only those questions, not included in Set I and Set II are given 10. Find the value of x + y from the following equation: é 1 3ù é y 0ù é5 6ù 2ê ú+ê ú=ê ú ë 0 x û ë 1 2û ë1 8û d2y d2y t 19. If x = a æç cos t + log tan ö÷, y = a sin t, find and . è 2ø dt 2 dx 2 20. Find the co-ordinates of the point where the line through the points (3, –4, –5) and (2, –3, 1) crosses the plane 3x + 2y + z + 14 = 0. 21. Find the particular solution of the following differential equation. dy æ yö x – y + x sin ç ÷ = 0, given that when x = 2, y = p è xø dx 12 3 56 22. Prove that: cos –1 æç ö÷ + sin –1 æç ö÷ = sin –1 æç ö÷ è 13 ø è5ø è 65 ø 28. Find the coordinates of the foot of perpendicular and the length of the perpendicular drawn ® from the point P (5, 4, 2) to the line r = - i$ + 3j$ + k$ + l( 2i$ + 3j$ - k$). Also find the image of P in this line. 29. Using matrices, solve the following system of equations. 3x + 4y + 7z = 4 2x – y + 3z = –3 x + 2y – 3z = 8

372


Solutions Set–I SECTION–A 1. (2 * 3) * 4 = (2 × 2 +3) * 4 = 7*4 = 2 × 7 + 4 = 18 2. tan - 1 ( 3 ) - sec - 1 ( - 2) p p = tan - 1 æç tan ö÷ - sec - 1 æç - sec ö÷ è è 3ø 3ø p p ù p 2p ö é = - sec - 1 ê sec æç p - ö÷ ú = - sec - 1 æç sec ÷ è ø è 3 3 û 3 3ø ë p 2p p = =- . 3 3 3 5 ù é 3 -4 ù é 7 6 ù éx 3. Given: 2ê ú+ê ú=ê ú ë7 y - 3û ë 1 2 û ë15 14û Þ Þ

10 ù é 3 -4ù é 7 6 ù é 2x ê 14 2y - 6ú + ê 1 2 ú = ê15 14ú ë û ë û ë û 6 ù é7 6ù é 2x + 3 = ê 15 2 y - 4úû êë15 14úû ë

Equating the corresponding element we get 2x + 3 = 7 and 2y - 4 = 14 7-3 14 + 4 and y = Þ x= 2 2 and y = 9 Þ x=2 x + y = 2 + 9 = 11 \ é -1 2 1 ù 4. Given: B=ê ú ë 1 2 3û é –1 1 ù BT = ê 2 2ú ê ú êë 1 3úû 4ù é –1 é3 T T ê Now A - B = –1 2ú – ê 2 ê ú ê 1 úû êë 1 êë 0

\

1ù é4 2 ú = ê –3 ú ê 3úû êë –1

3ù 0ú ú – 2úû

é ù æ p pö -1 ê tan (tan x) = x if x Î çè - 2 , 2 ÷ø ú ê ú ê ú p æ p pö ê Here 3 Î çè - 2 , 2 ÷ø ú ê ú pú ê -1 also, sec (sec x ) = x if x Î [ 0 , p ] ê 2ú ê ú 2p p ê Here ú Î [0, p ] êë úû 3 2

373


2A = 2n A Where n is order of matrix A. A = 4 and n = 3 2A = 23 ×4 = 32

Q Here \

5.

2 éx 22 6. Let I = ò 4 - x 2 dx = ê 4 - x2 + sin -1 2 2 êë 0

é êQ êë

= (0 + 2 sin–1 1) – (0 + 0) = 2´

xù ú 2 úû

x

Þ

f(x) = sec x

Þ

ò

a 2 - x 2 dx =

ù x 2 a2 x a - x2 + sin -1 + c ú 2 2 a úû

x

Þ

Þ

0

p =p 2

ò e (tan x + 1) sec x dx = e f ( x) + c x x ò e (tan x sec x + sec x) dx = e f ( x) + c x x ò e (sec x + tan x sec x) dx = e f ( x) + c x x ò e sec x + c = e f ( x) + c

7. Given

2

[Note: ò e x [ f ( x) + f ¢( x)] dx = e x . f ( x) + c, Here f(x) = sec x] 8. (i$ ´ j$). k$ + i$. j$ = k$ . k$ + 0 =1+0=1 [Note: i$. j$ = j$ . k$ = k$ . i$ = 0, i$ . i$ = j$ . j$ = k$ . k$ = 1, i$ ´ j$ = k$, j$ ´ k$ = i$ and k$ ´ i$ = j$] 9. Let AB = ( -5 - 2)i$ + (7 - 1) j$ = -7i$ + 6j$ Hence scalar components are –7, 6 ® [Note: If r = x i$ + yj$ + zk$ then x, y, z are called scalar component and xi$ , yj$ , zk$ are called

vector component.] 10. Given plane is 3x – 4y + 12z – 3 = 0 \

Distance from origin =

= =

3 ´ 0 + (– 4) ´ 0 + 12 ´ 0 – 3 3 2 + (– 4) 2 + (12) 2 –3 9 + 16 + 144 –3

169 3 = units 13

374


SECTION–B 11. Here 3 3 LHS = cos æç sin –1 + cot -1 ö÷ è 5 2ø 3 3 Let sin –1 = q and cot –1 = f 5 2 3 3 Þ sin q = and cot f = 5 2 4 2 3 , cos f = Þ cos q = and sin f = 5 13 13 \

LHS = cos (q + f) = cos q. cos f – sin q ´ sin f 4 3 3 2 12 6 6 = . – . = = – 5 13 5 13 5 13 5 13 5 13 b+c a a 12. LHS = b c+a b c

c

a +b

Applying R1 ® R1 + R2 + R3 we get 2(b + c) 2( c + a) 2( a + b) = b c+a b c

c

a+b

Taking 2 common from R1 we get (b + c) ( c + a) ( a + b) =2 b c+a b c

c

a+b

Applying R 2 ® R 2 - R 1 and R 3 ® R 3 - R 1 we get (b + c) ( c + a) ( a + b) = 2 –c 0 -a –b

-a

0

Applying R 1 ® R 1 + R 2 + R 3 we get 0 c b = 2 – c 0 -a -b - a

0

Expanding along R1 we get = 2 [0 – c (0 – ab) + b (ac – 0)] = 2 [abc + abc] = 4 abc


375

13. For one-one Case I : When x1, x2 are odd natural number. f(x1) = f(x2) Þ x1+1 = x2 +1 \ " x1 , x2 Î N Þ x1 = x2 i.e., f is one-one. Case II : When x1, x2 are even natural number f(x1) = f(x2) Þ x 1 - 1 = x 2 - 1 \ Þ x1 = x2 i.e., f is one-one. Case III : When x1 is odd and x2 is even natural number f(x1) = f(x2) Þ x1+1 = x2 -1 Þ x2 - x1 = 2 which is never possible as the difference of odd and even number is always odd number. Hence in this case f (x1) ¹ f(x2) i.e., f is one-one. Case IV: When x1 is even and x2 is odd natural number Similar as case III, We can prove f is one-one For onto: f(x) = x +1 if x is odd \ = x -1 if x is even For every even number ‘y’ of codomain $ odd number y - 1 in domain and for every Þ odd number y of codomain $ even number y +1 in Domain. i.e. f is onto function. Hence f is one-one onto function. OR For operation ‘*’ ‘*’ : R ´ R ® R s.t. a*b = a - b "a , b Î R Commutativity a*b = a - b = b - a = b*a i.e., ‘*’ is commutative Associativity " a, b, c Î R (a * b) * c = a - b * c = a-b - c a * (b * c) = a * b - c = a- b-c But a -b - c ¹ a- b - c Þ ( a * b)* c ¹ a * ( b * c) " a, b, c Î R

376


Þ * is not associative. Hence, ‘*’ is commutative but not associative. For Operation ‘o’ o : R × R ® R s.t. aob = a Commutativity " a, b Î R aob = a and boa = b Q a ¹ b Þ aob ¹ boa ‘o’ is not commutative. Þ Associativity: " a, b, c Î R (aob) oc = aoc = a ao(boc) = aob = a (aob) oc = ao (boc) Þ ‘o’ is associative Þ Hence ‘o’ is not commutative but associative. 14. Given

x = a sin

- 1t

Taking log on both sides, we have log x = log ( a sin

- 1 t 1/ 2

)

-1 1 1 log ( a sin t ) = ´ sin - 1 t . log a 2 2 1 -1 log x = sin t . log a 2 Differentiating both sides w.r.t. t, we have 1 dx 1 1 = log a ´ x dt 2 1 - t2

=

\

Again,

æ dx 1 1 = x çç log a ´ dt 2 1 - t2 è y = a cos

ö ÷ ÷ ø

- 1t

Taking log on both sides, we have -1 1 log y = log a cos t 2 1 Þ log y = ´ cos - 1 t log a 2 Differentiating both sides w.r.t. t, we have -1 1 dy 1 = log a ´ y dt 2 1 - t2

377


-1 dy 1 = y ´ log a ´ dt 2 1 - t2 1 1 log a ´ 2 1 - t2 dy dy / dt = = 1 1 dx dx / dt x ´ log a ´ 2 1 - t2 y´

\

OR 1 + x2 - 1ù ú x ú û

é Let y = tan -1 ê ê ë Let x = tan q Now,

dy y =dx x

Þ

Þ q = tan -1 x

æ 1 + tan 2 q - 1 ö ÷ y = tan –1 ç ç ÷ tan q è ø = tan

æ 1 - 1ö ç ÷ - 1ö –1 cos q ç ÷ = tan ç sin q ÷ è tan q ø ç ÷ è cos q ø

–1 æ sec q

æ 2 sin 2 q ö ç ÷ - cos q ö –1 2 ÷ = tan ç ÷ = tan ç è sin q ø ç 2 sin q . cos q ÷ è 2 2ø q = tan -1 æç tan ö÷ è 2ø q = 2 1 y = tan -1 x 2 dy 1 = dx 2(1 + x 2 ) –1 æ 1

Þ Þ

15. Given x = a (cos t + t sin t) Differentiating both sides w.r.t. x we get dx = a ( - sin t + t cos t + sin t) dt dx = a t cos t Þ dt Differentiating again w.r.t. t we get d2x dt 2

= a (–t sin t + cos t) = a (cos t - t sin t).

...(i)

é êQ ê êÞ ê ê êÞ ê ê êÞ ê ê êÞ ë

ù ú ú æ pö æ p öú tan ç - ÷ < tan q < tan ç ÷ è 2ø è 2 øú ú p p ú - < q< 2 2 ú ú p q p - < < ú 4 2 4 ú q æ p pö æ p pö ú Îç - , ÷ Ì ç - , ÷ ú 2 è 4 4ø è 2 2ø û

– µ< x <µ

378


Again y = a (sin t - t cos t) Differentiating w.r.t. t we get dy = a (cos t + t sin t - cos t) dt dy = at sin t Þ dt Differentiating again w.r.t. t we get d2y

= a (t cos t + sin t)

dt 2 dy = dx

Now,

...(ii)

dy dt dx

[from (i) and (ii)]

dt

dy at sin t = dx at cos t

Þ

dy = tan t dx Differentiating w.r.t. x we get Þ

d2y dx

2

= sec2t .

dt dx

= sec 2 t .

1

= d2x

Hence

sec 3 t at

dx

dt

=

sec 2 t

[from (i)]

at cos t

.

= a ( cos t - t sin t ),

d2y

= a (t cos t + sin t ) and

d2y

=

sec 3 t . at

dt 2 dt 2 dx 2 16. Let x, y be the distance of the bottom and top of the ladder respectively from the edge of the wall. dx Here, = 2 cm/s dt x 2 + y 2 = 25 When x = 4 m, ( 4) 2 + y 2 = 25

Þ

y 2 = 25 - 16 = 9

La

y= 3m 2

Now, x + y 2 = 25 Differentiating w.r.t. t, we have dy dx 2x + 2y =0 Þ dt dt

5m

dd

xm

x

dy dx +y =0 dt dt

er

ym

379


Þ

4´2+ 3´

dy =0 dt dy 8 =dt 3

Hence, the rate of decrease of its height = 17. If Þ

8 cm/s 3

x3 - x = 0 x( x 2 - 1) = 0

Þ x = 0 or x 2 = 1 Þ x = 0 or x = ±1 Þ x = 0 , -1 , 1 Hence [-1, 2] divided into three sub intervals [–1, 0], [0, 1] and [1, 2] such that

and Now

x3 - x ³ 0

on

[-1, 0]

x3 - x £ 0

on

[0, 1]

on

[1, 2]

x

3

- x³ 0

2

0

1

2

-1

–1 0

0 1

1 2

3 3 3 3 ò|x - x|dx = ò|x - x|dx + ò|x - x|dx + ò|x - x|dx

= ò ( x 3 - x)dx + ò -( x 3 - x) dx + ò ( x 3 - x) dx -1

0

1

0

1

2

é x4 x2 ù é x4 x2 ù é x4 x2 ù =ê ú –ê ú +ê ú 2 úû 4 2 úû 4 2 úû êë 4 -1 êë 0 êë 1

Let

or

1 1 ü ì 1 1 1 1 ü ì ü ì = í 0 - æç - ö÷ý – í æç - ö÷ - 0ý + í( 4 - 2) - æç - ö÷ý è 4 2 øþ î è 4 2 ø è 4 2 øþ î þ î 1 1 1 1 1 1 =- + - + +2- + 4 2 4 2 4 2 3 3 11 = - + 2= 2 4 4 OR x sin x dx. 1 + cos 2 x

p

I=ò

0

I=ò

p

( p - x) sin ( p - x) dx

0

2

1 + cos ( p - x)

2I = p ò

p

0

sin x dx 2

1 + cos x

or

=ò

p

( p - x) sin x dx 2

0

I=

1 + cos x p 2

p

ò0

=pò

p

sin x dx

0

1 + cos 2 x

-I

sin x dx 1 + cos 2 x

Put cos x = t so that - sin x dx = dt. When x = 0, t = 1 and when x = p, t = - 1. Therefore, we get

380


I=

-p 2

dt

-1

ò1

1+t

2

=pò

1 0

dt 1+t

2

éQ – a f ( x) dx = - a f ( x) dx and 2 a f ( x) dx = 2 a f ( x) dxù ò- a ò0 ò0 êë ò a úû

p p2 = p [tan -1 t] 10 = p [tan -1 1 - tan -1 0] = p é - 0ù = êë 4 úû 4 18. Let C denotes the family of circles in the second quadrant and touching the coordinate axes. Let ( - a, a) be the coordinate of the centre of any member of this family (see figure). Y Equation representing the family C is or

( x + a) 2 + ( y - a) 2 = a 2

...(i)

x 2 + y 2 + 2ax - 2ay + a 2 = 0

...(ii)

Differentiating equation (ii) w.r.t. x, we get dy dy 2x + 2y + 2a - 2a =0 dx dx dy æ dy ö or x+y = a ç - 1÷ è dx ø dx x + yy ¢ or a= y¢ - 1

(–a, a) X'

O

X

Y'

dy ö æ çy ¢ = ÷ è dx ø

Substituting the value of a in equation (i), we get 2

2

é é é x + yy ¢ ù x + yy ¢ ù x + yy' ù êx + ú + êy ú =ê ú y¢ - 1 û y¢ - 1 û ë ë ë y¢ - 1 û

2

or

[xy ¢ - x + x + yy ¢] 2 + [yy ¢ - y - x - yy ¢] 2 = [x + yy ¢] 2

or

( x + y) 2 y ¢ 2 + ( x + y) 2 = ( x + yy ¢) 2

or

( x + y) 2 [( y ¢) 2 + 1] = [x + yy ¢] 2 ,

is the required differential equation representing the given family of circles.

OR Given differential equation is dy = 1, x ( x 2 - 1) dx dx dy = x( x 2 - 1) Þ

dy =

dx x( x - 1)( x + 1)

Integrating both sides we get dx ò dy = ò x( x - 1)( x + 1) Þ

y =ò

dx x( x - 1)( x + 1)

...(i)

381


Þ

1 A B C = + + x( x - 1)( x + 1) x x - 1 x + 1 A( x - 1)( x + 1) + B x ( x + 1) + C x ( x - 1) 1 = x( x - 1)( x + 1) x( x - 1)( x + 1)

Þ

1 = A( x - 1)( x + 1) + Bx ( x + 1) + Cx ( x - 1)

Putting

x = 1 we get 1= 0 + B. 1. 2 + 0

Putting

x = -1 we get 1 = 0 + 0 + C .(-1).(-2) Þ C =

Putting

x = 0 we get 1= A (-1).1 Þ A = -1 1 -1 1 1 = + + x( x - 1)( x + 1) x 2( x - 1) 2( x + 1)

Hence

Þ B=

1 2 1 2

From (i) æ 1 1 1 ö y = ò ç- + + ÷ dx è x 2( x - 1) 2( x + 1) ø Þ Þ Þ Þ

dx 1 dx 1 dx + ò + ò x 2 x-1 2 x+1 1 1 y = – log x + log x - 1 + log x + 1 + log c 2 2 1 2y = 2 log + log x 2 - 1 + 2 log c x y = -ò

2y = log

x2 - 1 x

2

+ log c 2

x = 2, y = 0 4-1 0 = log Þ + log c 2 4 3 log c2 = – log Þ 4 3 2 Putting log c = - log in (ii) we get 4 When

2y = log

x2 - 1 x

Þ

y=

2

– log

3 4

1 x2 - 1 1 3 log – log 2 2 2 4 x

19. Given differential equation is

Þ

(1 + x 2 ) dy + 2xy dx = cot x. dx cot x dy 2x + .y = dx 1 + x 2 1 + x2

...(ii)

382


dy + Py = Q. Where dx 2x cot x P= ,Q= 2 1+x 1 + x2 Pdx I. F. = e ò

It is in the form of

\

=e

ò

2x 1+ x 2

dx

dz

ò = e z [Let 1 + x2 = z Þ 2x dx = dz] 2 = e log z = e log (1 + x ) = 1 + x2 [ Q e log z = z] Hence the solution is y ´ I . F = ò Q ´ I . F dx + c cot x

.(1 + x 2 ) dx + c

Þ

y(1 + x 2 ) = ò

Þ

y(1 + x 2 ) = ò cot x dx + c

Þ

y(1 + x 2 ) = ò

Þ

y(1 + x 2 ) = log sin x + c

Þ

y=

1 + x2

cos x dx sin x

log sin x 1+x

2

+c

+

c 1 + x2

20. Given, ®

®

®

a = i$ + 4j$ + 2k$, b = 3i$ - 2j$ + 7 k$, c = 2i$ - j$ + 4k$ ®

®

®

®

®

®

Vector p is perpendicular to both a and b i.e., p is parallel to vector a ´ b . \

i$ a´b = 1

j$ 4

3

-2

®

®

®

k$ 2 = i$ 7 ®

4 -2

2 1 - j$ 7 3

2 1 + k$ 7 3

®

Since p is parallel to a ´ b \ Also,

®

p = m ( 32i$ - j$ - 14k$)

® ®

p . c = 18

Þ

m ( 32i$ - j$ - 14k$) .( 2i$ - j$ + 4k$) = 18

Þ

m ( 64 + 1 - 56) = 18 Þ

\

®

9m = 18 or

p = 2 ( 32i$ - j$ - 14k$) = 64i$ - 2j$ - 28k$

m=2

4 = 32i$ - j$ - 14k$ -2

383


21. Let P (a, b, g) be the point at which the given line crosses the xy plane. Now the equation of given line is x- 3 y- 4 z-1 = = …(i) 2 -3 5 Since P (a, b, g) lie on line (i) a - 3 b - 4 g -1 = = = l (say) \ 2 -3 5 Þ a = 2l + 3; b = -3l + 4 and g = 5l + 1 Also P (a, b, g) lie on given xy plane, i.e., z = 0 \ 0. a + 0. b + g = 0 Þ 5l + 1 = 0 Þ l = - 15 . Hence the coordinates of required points are 1 13 a = 2 ´ æç - ö÷ + 3 = è 5ø 5 1 23 b = –3 ´ æç - ö÷ + 4 = è 5ø 5 1 g = 5 ´ æç - ö÷ + 1 = 0 è 5ø 13 23 ö i.e., required point in æç , , 0÷. è5 5 ø

A (3,4,1)

P(a,b,g) z=0

B (5,1,6)

22. Total no. of cards in the deck = 52 Number of red cards = 26 No. of cards drawn = 2 simultaneously \ X = value of random variable = 0, 1, 2 P(X)

X or xi 0

26

C 0 ´ 26 C 2 52

1

26

C1 ´ 26 C1 52

2

26

C2 C2

C 0 ´ 26 C 2 52

C2

xi P( X )

x12 P( X )

=

25 102

0

0

=

52 102

52 102

52 102

=

25 102

50 102

100 102

Sxi P(X) = 1

Mean = m = Sxi P(X) = 1 Variance = s 2 = Sxi2 P(X) - m 2 =

152 50 25 = 0.49 -1= = 102 102 51

Sxi 2 P(X) =

152 102

384


SECTION–C 23. The given system of equation can be represented in matrix form as AX = B, where é2 3 3 ù éx ù é 5ù ê ú ê ú A = 1 -2 1 , X = y , B = ê -4 ú ê ú ê ú ê ú êë 3 -1 -2úû êë z úû êë 3 úû 2 3 3 Now A = 1 -2 1 = 2 (4 + 1) –3 (–2–3) + 3 (–1 + 6) 3

-1

C11 = (-1) 1+1

-2

= 10 + 15 + 15 = 40 ¹ 0 -2 1 =4+1=5 -1 - 2

C12 = (-1) 1+2

1 1 = (-2 - 3) = 5 3 -2

C13 = (-1) 1+3

1 -2 = (-1 + 6) = 5 3 -1

C21 = (-1) 2+1

3 3 = -(-6 + 3) = 3 -1 - 2

C22 = (-1) 2+2

2 3 = (-4 - 9) = - 13 3 -2

C23 = (-1) 2+3

2 3 = -(-2 - 9) = 11 3 -1

C31 = (-1) 3+1

3 -2

3 = (3 + 6) = 9 1

C32 = (-1) 3+2

2 1

3 = -(2 - 3) = 1 1

C33 = (-1) 3+3

2 1

3 = -4 - 3 = - 7 -2 T

5 5 ù 3 9ù é5 é5 ê ú ê Adj A = 3 - 13 11 = 5 - 13 1 ú ê ú ê ú 1 - 7 úû êë 9 êë5 11 - 7 úû 3 9ù é5 1 1 ê –1 A = adj A = 5 - 13 1 ú ú A 40 ê êë5 11 - 7 úû

385


\ \

AX = B Þ X = A –1 B 9 ù é 5ù é xù é5 3 ê yú = 1 ê5 - 13 1 ú ê – 4ú ê ú 40 ê ú ê ú êëz úû ëê5 11 - 7 úû êë 3úû é 25 - 12 + 27 ù 1 ê = 25 + 52 + 3 ú ú 40 ê êë 25 - 44 - 21 úû é 40ù 1 ê = 80ú ê ú 40 êë – 40 úû é xù é 1 ù ê yú = ê 2 ú ê ú ê ú êëz úû êë -1úû

Equating the corresponding elements we get x = 1, y = 2, z = -1 24. Let r and h be the radius and height of right circular cylinder inscribed in a given cone of radius R and height H. If S be the curved surface area of cylinder then S = 2prh Q D AOC ~ D FEC (R - r) OC AO S = 2pr . Þ .H Þ = R EC FE 2pH R H S= Þ (rR - r 2 ) Þ = R R -r h Differentiating both sides w.r.t. r, ( R - r). H Þh= we get R dS 2pH = ( R - 2r) dr R For maxima and minima dS =0 dr 2pH Þ ( R - 2r) = 0 R R Þ R - 2r = 0 Þ r= 2 d 2 S 2pH Now, = ( 0 - 2) R dr 2 Þ

é d 2 Sù 4pH == -ve ê ú R êë dr úû r = R 2

A

r

G

F

H

h

B

D

O

E R

C

386


R S is maximum. 2 i.e., radius of cylinder in half of that of cone. OR Let the length, breadth and height of open box with square base be x, x and h unit respectively. If V be the volume of box then V = x.x. h Hence for r =

V = x2h

Þ

....(i)

2

2

c = x + 4xh

Also

c2 - x2 4x Putting it in (i) we get Þ

h=

x 2 (c 2 - x 2 )

V=

4x Differentiating w.r.t. x we get

Þ

V=

c2x x 3 4 4

dV c 2 3x 2 = dx 4 4 Now for maxima or minima dV =0 dx Þ

c 2 3x 2 =0 4 4

Þ

3x 2 c 2 = 4 4

Þ

x2 =

c2 3

Þ

x=

Now,

d 2V dx

\

2

=-

c 3

6x 3x =4 2

é d 2V ù ê 2ú êë dx úû x =

=c

3c = -ve. 2 3

3

c Hence, for x = volume of box is maximum. 3 \

h=

c2 - x2 4x

c2 2 3 = 2c ´ 3 = c = c 3 4c 2 3 4 3 c2 -

x

h x

387


Therefore maximum volume = x 2 . h =

c2 c c3 = . 3 2 3 6 3

25. Let sin -1 x = z Þ x = sin z 1 \ Þ dx = dz 1 - x2 \

ò

x sin -1 x 1 - x2

dx = ò z. sin z dz = -z cos z + ò cos z dz = -z cos z + sin z + c = - sin -1 x. 1 - x 2 + x + c = x - 1 - x 2 sin –1 x + c

[\ cos z = 1 - sin 2 z = 1 - x 2 ]

OR 2

Now let

Þ Þ

x +1 2

( x - 1) ( x + 3) x2 + 1 ( x - 1) 2 ( x + 3)

=

=

A B C + + x - 1 ( x - 1) 2 x + 3 A( x - 1)( x + 3) + B( x + 3) + C( x - 1) 2 ( x - 1) 2 ( x + 3)

x 2 + 1 = A( x - 1)( x + 3) + B( x + 3) + C( x - 1) 2

Putting x = 1 in (i) we get 1 2 = 4B Þ B = Þ 2 Putting x = -3 in (i) we get 10 = 16C 10 5 C= Þ = 16 8 1 5 Putting x = 0, B = , C = in (i) we get 2 8 1 5 1 = A( -1).( 3) + ´ 3 + ( -1) 2 2 8 3 5 1 = - 3A + + 2 8 12 + 5 17 9 3A = Þ -1= -1= 8 8 8 3 Þ A= 8

…(i)

388


x2 + 1

\

2

( x - 1) ( x + 3)

=

3 1 5 + + 2 8( x - 1) 2( x - 1) 8( x + 3)

æ ö 3 1 5 ç = dx + + ò ( x - 1) 2 ( x + 3) ò çè 8( x - 1) 2( x - 1) 2 8( x + 3) ÷÷ø dx x2 + 1

\

3 dx 1 5 dx + ò ( x - 1) -2 dx + ò ò 8 x-1 2 8 x+ 3 3 1 5 = log x - 1 + log x + 3 + c 8 2( x - 1) 8

=

26. Let R = {( x, y): x 2 + y 2 £ 4, x + y ³ 2} Þ

R = {( x, y): x 2 + y 2 £ 4} Ç {( x, y): x + y ³ 2}

i.e.,

R = R1 Ç R2 where R 1 = {( x, y): x 2 + y 2 £ 4} and R 2 = {( x, y): x + y ³ 2}

For region R1 Obviously x 2 + y 2 = 4 is a circle having centre at (0,0) and radius 2. Since (0,0) satisfy x 2 + y 2 £ 4. Therefore region R1 is the region lying interior of circle x2 + y2 = 4 For region R2 x

0

2

y

2

0

x + y = 2 is a straight line passing through (0, 2) and (2, 0). Since (0, 0) does not satisfy x + y ³ 2 therefore R2 is that region which does not contain origin (0, 0) i.e., above the line x + y = 2 Hence, shaded region is required region R. Now area of required region 2

2

2

–1

O

1

2

–1

= ò 4 - x dx - ò ( 2 - x) dx 0

–2

x+y=2

0 2

2 é x2 ù 1 x ù é1 = ê x 4 - x 2 + 4 sin -1 æç ö÷ ú - 2 [x] 20 + ê ú è 2 øû 0 2 ë2 êë 2 úû 0 4 = [2 sin -1 1 - 0] - 2 [2 - 0] + é - 0ù êë 2 úû p = 2´ - 4 + 2= p - 2 2

389


27. Given lines are x-1 y- 2 z- 3 = = -3 -2 k 2 x-1 y- 2 z- 3 = = k 1 5 ®

...(i) ...(ii) ®

Obviously, parallel vectors b 1 and b 2 of line (i) and (ii) respectively are: ® b 1 = -3j$ - 2kj$ + 2k$ ® b 2 = kj$ + j$ + 5k$

Þ

®

®

Lines (i) ^ (ii) Þ b 1 ^ b 2 ®

®

Þ

b1 . b2 = 0

Þ

-3k - 2k + 10 = 0

Þ

-5k + 10 = 0

Þ

k=

-10 =2 -5

Putting k = 2 in (i) and (ii) we get x-1 y- 2 z- 3 = = -3 -4 2 x-1 y- 2 z- 3 = = 2 1 5 Now the equation of plane containing above two lines is y-2 z - 3ù éx - 1 ê-3 -4 2 ú=0 ê ú 1 5 úû êë 2 Þ Þ Þ Þ Þ

(x - 1) (-20 -2) - (y - 2) (-15 - 4) + (z - 3) (-3 + 8) = 0 - 22 (x - 1) + 19 (y - 2) + 5 (z - 3) = 0 - 22x + 22 + 19y - 38 + 5z - 15 = 0 - 22x + 19y + 5z - 31 = 0 22x - 19y - 5z + 31= 0 x - x 1 y - y 1 z - z1 Note: Equation of plane containing lines and = = a1 b1 c1 x - x2 a2

=

y - y2 b2

=

z - z2 c2

x - x1 is

y - y1

z - z1

a1

b1

c1

a2

b2

c

=0

2

28. Consider the following events: E1 = Getting 5 or 6 in a single throw of a die. E2 = Getting 1, 2, 3, or 4 in a single throw of a die. A = Getting exactly one head.

390


2 1 4 2 = , P(E2 ) = = 6 3 6 3 P( A / E1 ) = Probability of getting exactly one head when a coin is tossed three times We have, P(E1 ) =

1 1 1 2 3 = 3 C 1 æç ö÷ æç ö÷ = è 2ø è 2ø 8

P( A / E2 ) = Probability of getting exactly one head when a coin is tossed once only =

1 2

Now, Required probability = P(E2 / A)

2 1 ´ 3 2 = = P(E1 ) P( A / E1 ) + P(E2 ) P( A / E2 ) 1 ´ 3 + 1 ´ 2 3 8 2 3 1 1 24 8 = 3 = ´ = 1 1 3 11 11 + 8 3 29. Let the mixture contain x kg of Food I and y kg of Food II. According to question we have following constraints: ...(i) 2x + y ³ 8 ...(ii) x + 2y ³ 10 ...(iii) x³0 ...(iv) y³0 It z be the total cost of purchasing x kg of Food I and y kg of Food II then ...(v) z = 5x + 7 y Here we have to minimise z subject to the constraints (i) to (iv) On plotting inequalities (i) to (iv) we get shaded region having corner points A, B, C which is required feasible region. Now we evaluate z at the corner points A (0, 8), B (2, 4) and C (10, 0) P(E2 ) P( A / E2 )

Corner Point

z = 5x + 7y

A (0, 8)

56

B (2, 4)

38

C (10, 0)

50

¬ Minimum

Since feasible region is unbounded. Therefore we have to draw the graph of the inequality. ...(vi) 5x + 7 y < 38 Since the graph of inequality (vi) is that open half plane which does not have any point common with the feasible region.

391

Examination Papers – 2012 Y 9 8

A(0, 8)

7

x+

6 2y

=1

0

5 B(2, 4)

4 3 2 1

C(10, 0) X’ –2

–1

O

1

2

3

6

7

+ 2x y=

–1

5

4

8 5x +

9

7y

8

=3

10

X

8

–2 Y’

So the minimum value of z is 38 at (2, 4). i.e., the minimum cost of food mixture is `38 when 2kg of Food I and 4 kg of Food II are mixed.

Set–II 10. ( k$ ´ j$) . i$ + j$. k$ = - i$ . i$ + 0 = -1 + 0 = -1 4 12 19. Let cos -1 = x , cos –1 =y 5 13 4 12 Þ cos x = , cos y = 5 13 \ Þ Now

[x, y Î[o, p]]

4 2 12 2 sin x = 1 - æç ö÷ , sin y = 1 - æç ö÷ è5 ø è 13 ø 3 5 sin x = , sin y = 5 13 cos ( x + y) = cos x . cos y - sin x . sin y 4 12 3 5 = ´ - ´ 5 13 5 13

[Q x, y Î [0, p] Þ sin x and sin y are +ve]

392


Þ Þ

33 65 -1 æ 33 ö x + y = cos ç ÷ è 65 ø cos ( x + y) =

éQ 33 Î [-1, 1]ù ëê 65 ûú

Putting the value of x and y we get 4 12 33 cos -1 + cos -1 = cos –1 æç ö÷ Proved. è 65 ø 5 13 20. Refer to CBSE Delhi Set-I Q.No. 19. dy dy 21. Given differential equation is + y cot x = 4x cosec x and is of the type + Py = Q where dx dx P = cot x, Q = 4x cosec x Pdx I.F. = e ò \ cot xdx I.F. = e ò \ = e log|sin x| = sin x Its solution is given by sin x . y = ò 4x cosec x. sin x dx Þ

Þ

y sin x = ò 4x dx =

Now y = 0 when x =

4x 2 +C 2

Þ

y sin x = 2x 2 + C

p 2

p2 p2 + C ÞC = 4 2 Hence, the particular solution of given differential equation is

\

0=2´

p2 2 22. The equation of line passing through the point (3, - 4, -5) and (2, -3, 1) is x- 3 y+ 4 z+5 = = 2 - 3 -3 + 4 1 + 5 x- 3 y+ 4 z+5 ...(i) Þ = = -1 1 6 A (3, –4, –5) Let the line (i) crosses the plane 2x + y + z = 7 ...(ii) at point P (a , b , g ) Q P lies on line (i), therefore (a , b , g ) satisfy equation (i) a - 3 b + 4 g +5 \ = = = l (say) P (a,b,g) -1 1 6 Þ a = -l + 3 2x + y + z = 7 b=l-4 g = 6l - 5 B (2, –3,1) Also P (a , b , g ) lie on plane (ii) \ 2a + b + g = 7 y sin x = 2x 2 -


393

Þ 2 ( - l + 3) + ( l - 4) + ( 6l - 5) = 7 Þ -2 l + 6 + l - 4 + 6 l - 5 = 7 Þ 5l = 10 Þ l=2 Hence the co-ordinate of required point P is (-2 + 3, 2 - 4, 6 × 2 -5) i.e., (1, -2, 7) 28. The given system of linear equations may be written in matrix form as: AX = B é 1 1 – 1 ù é xù é 3 ù ê2 3 i.e., 1 ú ê yú = ê10ú ê úê ú ê ú êë 3 – 1 – 7 úû êë z úû êë 1 úû 1 1 –1 Now, |A| = 2 3 1 3 –1 –7

\

Þ Now, Þ

= 1 (– 21 + 1) – 1 (– 14 – 3) – 1(– 2– 9) = – 20 + 17 + 11 = 8 ¹ 0 C11 = – 20 C12 = 17 C13 = – 11 C21 = + 8 C22 = – 4 C23 = 4 C31 = 4 C32 = – 3 C33 = 1 ¢ 4 ù é - 20 17 - 11ù é - 20 8 ê ú ê Adj A = + 8 - 4 4 = 17 - 4 – 3ú ê ú ê ú 1 úû êë - 11 4 1 úû êë 4 – 3 A

–1

8 4ù é - 20 1 1ê = Adj A = + 17 - 4 –3ú ú |A| 8ê 4 1 úû êë - 11

AX = B Þ X = A–1B 8 4ù é 3 ù é xù é - 20 é – 60 + 80 + 4ù é 24ù é 3ù ê yú = 1 ê + 17 - 4 – 3ú ê10ú = 1 ê 51 – 40 – 3 ú = 1 ê 8 ú = 1 ê 1 ú ê ú 8ê úê ú 8ê ú 8ê ú 8ê ú 4 1 úû êë 1 úû êë z úû êë - 11 êë – 33 + 40 + 1úû êë 8 úû êë 1 úû

On equating, we get x = 3, y = 1, z = 1 29. Let Q (a, b, g) be the foot of perpendicular from P to the given plane ...(i) 2x + 4y - z = 2 Let P ¢ ( x 1 , y 1 , z 1 ) be the image of P in the plane (i) Now

®

P (7, 14, 5)

Q (a, b, g)

PQ = (a - 7 )i$ + (b - 14) j$ + ( g - 5) k$

Also, Normal vector of plane (i) is

2x + 4y – z = 2

®

N = 2i$ + 4j$ - k$

P' (x1, y1, z1)

394


®

Since PQ || N a - 7 b - 14 g - 5 = = = l (say) \ 2 4 -1 Þ a = 2l + 7 b = 4l + 14 g = –l + 5 Again Q Q (a , b , g ) lie on plane (i) \ 2a + 4b - g = 2 2 ( 2l + 7) + 4( 4l + 14) - ( - l + 5) = 2 Þ 4l + 14 + 16l + 56 + l - 5 - 2 = 0 Þ 21l + 63 = 0 Þ 21l = -63 Þ l = –3 Þ the coordinates of Q are (2×(-3)+7, 4×(-3)+14, - (-3) + 5) i.e., (1, 2, 8) \ Length of perpendicular = (7 - 1) 2 + (14 - 2) 2 + (5 - 8) 2 = 36 + 144 + 9 = 189 = 3 21 Also Q (1, 2, 8) in mid point of PP’ 7 + x1 1= \ Þ x 1 = -5 2 14 + y 1 2= Þ y 1 = -10 2 5 + z1 8= Þ z1 = 11 2 Hence the required image is (-5, -10, 11).

Set–III 10. Given:

Þ Þ

é 1 3ù é y 0ù é5 2ê ú+ê ú=ê ë 0 x û ë 1 2û ë1 é 2 6 ù é y 0ù é5 ê 0 2xú + ê 1 2ú = ê1 ë û ë û ë 6 ù é5 é2 + y = ê 1 2x + 2úû êë1 ë

6ù 8úû 6ù 8úû 6ù 8úû

Equating the corresponding elements we get and 2+ y=5 2x + 2 = 8 and Þ y= 3 x= 3 \ x + y = 3 + 3 = 6.

395


19. Q

t x = a æç cos t + log tan ö÷ è 2ø

Differentiating w.r.t. t, we get æ ö ç ÷ dx 1 2 t 1 = a ç - sin t + . sec . ÷ t dt 2 2÷ ç tan è ø 2 ì ü ì ï ï 1 1 ü = a í - sin t + = a í - sin t + ý t tý sin t þ î ï 2 sin . cos ï î 2 2þ 2 ö 2 æ 1 - sin t cos t dx ÷=a =aç ç ÷ dt sin t è sin t ø Q y = a sin t Differentiating w.r.t t, we get dy d2y = a . cos t Þ = – a sin t dt dt 2 dy dy / dt a cos t . sin t \ = = = tan t dx dx / dt a cos 2 t \ Hence,

d2y dx 2 d2y

= sec 2 t .

1 ´ sin t 1 dt = sec 2 t . = sec 4 t . sin t 2 dx a cos t a

= – a sin t and

d2y

=

sec 4 t sin t

a dt 2 dx 2 20. The equation of the line passing through the point (3, -4, -5) and (2, -3, 1) is x- 3 y+ 4 z+5 = = 2 - 3 -3 + 4 1 + 5 x- 3 y+ 4 z+5 ...(i) Þ = = -1 1 6 Let the line (i) crosses the plane 3x + 2y + z + 14 = 0 ...(ii) at point P(a , b , g ). P lie on line (i) therefore (a , b , g ) satisfy equation (i) Q A (–3, – 4, –5) a - 3 b +4 g +5 = = = l (say) \ -1 1 6 Þ a = - l + 3; b = l - 4 and g = 6l - 5 Also P (a , b , g ) lie on plane (ii) P (a,b,g) \ 3a + 2b + g + 14 = 0 Þ 3( - l + 3) + 2( l - 4) + ( 6l - 5) + 14 = 0 –3l + 9 + 2l – 8 + 6l – 5 + 14 = 0 Þ 5l + 10 = 0 Þ l = – 2 Þ B (2, –3,1)

396


Hence the coordinate of required point P is given as ( 2 + 3, - 2 - 4, 6 ´ - 2 - 5) º (5, - 6, – 17) 21. Given differential equation is dy y x - y + x sin æç ö÷ = 0 è dx xø dy y æ yö ...(i) Þ - + sin ç ÷ = 0 è xø dx x It is homogeneous differential equation. y Let = v Þ y = vx x dy dv =v + x Þ dx dx Putting these values in (i) we get dv v+x - v + sin v = 0 dx dv Þ x + sin v = 0 Þ dx dv - dx Þ = sin v x dx Þ cosec v dv = x Integrating both sides we get dx Þ ò cosec v dv = - ò x Þ log cos ec v - cot v = - log|x|+ c y y Þ log cos ec - cot + log x = c x x Putting x = 2, y = p we get Þ log cos ec p 2 - cot p 2 + log 2 = c log 1 + log 2 = c Þ c = log 2 Þ Hence particular solution is y y log cos ec - cot + log x = log 2 x x Þ

log x.(cos ec y x - cot y x ) = log 2

Þ

y yö æ xç cos ec - cot ÷ = 2 è x xø

x

dv = - sin v dx

[Q log 1 = 0]

397


= cos -1

22. LHS

12 3 + sin -1 13 5

12 = sin -1 1 - æç ö÷ è 13 ø

2

+ sin -1

3 5

144 3 + sin -1 169 5 5 3 = sin -1 + sin -1 13 5 é5 3 2 3 5 2ù = sin -1 ê 1 - æç ö÷ + 1 - æç ö÷ ú è5ø è 13 ø ú 5 ê 13 ë û = sin -1 1 -

éæ 5 ö 2 æ 3 ö 2 ù ê ç ÷ + ç ÷ £ 1ú è5ø êë è 13 ø úû

é5 9 3 25 ù = sin -1 ê 1+ 1ú 13 25 5 169 ë û 5 4 3 12 20 36 ù = sin -1 é ´ + ´ ù = sin -1 é + êë 13 5 5 13 úû êë 65 65 úû 56 = sin -1 é ù = RHS êë 65 úû 28. Given line is ®

r = -i$ + 3j$ + k$ + l( 2i$ + 3j$ - k$)

It can be written in cartesian form as x+1 y- 3 z-1 ...(i) = = 2 3 -1 Let Q (a , b , g ) be the foot of perpendicular drawn from P(5, 4, 2) to the line (i) and P’ ( x 1 , y 1 , z 1 ) be the image of P on the line (i) P (5, 4, 2) Q (a , b , g ) lie on line (i) Q a +1 b - 3 g -1 = l (say) \ = = 2 3 -1 Þ a = 2l - 1; b = 3l + 3 and g = - l + 1 ®

Now PQ = (a - 5)i$ + (b - 4) j$ + ( g - 2) k$ ®

Parallel vector of line (i) b = 2i$ + 3j$ - k$. ®

Q (a, b, g) Ù Ù Ù ® Ù Ù Ù r = –i + 3j + k + l (2i + 3j – k)

®

Obviously PQ ^ b ® ®

\

PQ . b = 0

Þ Þ Þ

2(a - 5) + 3(b - 4) + ( -1)( g - 2) = 0 2a - 10 + 3b - 12 - g + 2 = 0 2a + 3b - g - 20 = 0 2( 2l - 1) + 3( 3l + 3) - ( - l + 1) - 20 = 0

P' (x1, y1, z1)

[Putting a , b , g ]

398


Þ 4l - 2 + 9l + 9 + l - 1 - 20 = 0 Þ 14l - 14 = 0 Þ l =1 Hence the coordinates of foot of perpendicular Q are (2 × 1 – 1, 3 × 1 + 3, – 1 + 1), i.e., (1, 6, 0) \ Length of perpendicular = (5 - 1) 2 + ( 4 - 6) 2 + ( 2 - 0) 2 = 16 + 4 + 4 = 24 = 2 6 unit. Also since Q is mid-point of PP’ x +5 \ 1= 1 Þ x 1 = -3 2 y +4 6= 1 Þ y1 = 8 2 z +2 0= 1 Þ z 1 = -2 2 Therefore required image is (-3, 8, -2). 29. The given system of linear equations may be written in matrix form as AX = B Where é3 4 7 ù é xù é4ù ê ú ê ú A = 2 -1 3 , X = y and B = ê -3ú ê ú ê ú ê ú êë 1 2 -3úû êë z úû êë 8 úû 3 4 7 Now, A = 2 - 1 3 1

2

-3

= 3 ( 3 - 6) - 4( -6 - 3) + 7( 4 + 1) = - 9 + 36 + 35 = 62 ¹ 0 -1 3 C11 = (-1)1+1 = 3 - 6 = -3 2 -3 C12 = (-1)1+2

2 3 = -{-6 - 3} = 9 1 -3

C13 = (-1)1+3

2 -1 =4+1=5 1 2

C21 = (-1)2+1

4 7 = - (-12 -14)= 26 2 -3

C22 = (-1)2+2

3 7 = - 9 - 7= -16 1 -3

C23 = (-1)2+3

3 4 = - (6 - 4)= –2 1 2

399


C31 = (-1)3+1

4 7 = 12 + 7= 19 –1 3

C32 = (-1)3+2

3 7 = - (9 -14) = 5 2 3

C33 = (–1)3 +3

3 4 = (-3 -8) = -11 2 -1

\

5 ù é -3 9 ê Adj. A = 26 -16 -2 ú ê ú 5 -11úû êë 19

¢

é -3 26 19 ù = ê 9 -16 5 ú ê ú -2 -11úû êë 5 \

A -1 =

1 Adj A A

é -3 26 19 ù 1 ê = 9 -16 5 ú ú 62 ê -2 -11úû êë 5 \

AX = B

Þ

X = A -1 B é xù é -3 26 19 ù é 4 ù ê yú = 1 ê 9 -16 5 ú ê -3ú ê ú 62 ê úê ú -2 -11úû êë 8 úû êëz úû êë 5 é -12 - 78 + 152ù 1 ê = 36 + 48 + 40 ú ú 62 ê êë 20 + 6 - 88 úû é xù é 62 ù ê yú = 1 ê 124 ú ê ú 62 ê ú êë z úû êë - 62úû é xù é 1 ù ê yú = ê 2 ú ê ú ê ú êëz úû êë -1úû

Þ

Þ

Þ

Equating the corresponding elements we get x = 1 , y = 2 , z = -1

CBSE Examination Paper (Foreign 2012) Time allowed: 3 hours

Maximum marks: 100


Set–I SECTION–A Question number 1 to 10 carry 1 mark each. 1. If the binary operation * on the set Z of integers is defined by a * b = a + b - 5, then write the identity element for the operation * in Z. 2. Write the value of cot (tan -1 a + cot -1 a). 3. If A is a square matrix such that A2 = A, then write the value of (I + A)2 - 3A. é 2ù é -1ù é10ù 4. If x ê ú + y ê ú = ê ú, write the value of x. 3 ë û ë1 û ë5û 5. Write the value of the following determinant: 102 18 36 1

3

4

17

3

6

x-1 6. If ò æç 2 ö÷ e x dx = f ( x) e x + c, then write the value of f(x). è x ø a

7. If ò 3x 2 dx = 8, write the value of ‘a’. 0

8. Write the value of (i$ ´ j$) . k$ + ( j$ ´ k$) . i$ $ 9. Write the value of the area of the parallelogram determined by the vectors 2i$ and 3j.

401


10. Write the direction cosines of a line parallel to z-axis.

SECTION–B Question numbers 11 to 22 carry 4 marks each. 4x + 3 2 2 11. If f(x) = , x ¹ , show that fof ( x) = x for all x ¹ . What is the inverse of f ? 6x - 4 3 3 -1 æ 63 ö -1 æ 5 ö -1 æ 3 ö 12. Prove that: sin ç ÷ = sin ç ÷ + cos ç ÷ è 65 ø è 13 ø è5ø OR Solve for x: p 2 13. Using properties of determinants, prove that a a+b a+b + c 2 tan -1 (sin x) = tan -1 ( 2 sec x), x ¹

2a

3a + 2b

4a + 3b + 2c = a 3

3a

6a + 3b

10a + 6b + 3c

14. If x m y n = ( x + y) m+ n , prove that 15. If y = e a cos

-1

x

dy y = . dx x

, - 1 £ x £ 1, show that

(1 - x 2 )

d2y dx

2

-x

dy - a 2 y = 0. dx OR

If x 1 + y + y 1 + x = 0, –1 < x < 1, x ¹ y, then prove that

dy 1 . =dx (1 + x) 2

2x , x > -1 is an increasing function of x throughout its domain. 2+x OR Find the equation of the normal at the point (am2, am3) for the curve ay2=x3. 17. Evaluate: ò x 2 tan -1 x dx 16. Show that y = log (1 + x) -

OR 3x - 1 Evaluate: ò dx ( x + 2) 2 18. Solve the following differential equation: é e -2 x y ù dx = 1, x ¹ 0 ê ú x úû dy êë x 19. Solve the following differential equation: p 3e x tan y dx + ( 2 - e x ) sec 2 y dy = 0, given that when x = 0 , y = . 4

402


®

®

®

®

®

®

$ then express b in the form b = b + b , where b is 20. If a = 3i$ + 4j$ + 5k$ and b = 2i$ + j$ – 4k, 1 2 1 ®

®

®

parallel to a and b 2 is perpendicular to a . 21. Find the vector and cartesian equations of the line passing through the point P (1, 2, 3) and ® ® parallel to the planes r .(i$ - j$ + 2k$) = 5 and r .( 3i$ + j$ + k$) = 6. 22. A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability distribution of the number of successes and hence find its mean.

SECTION–C Question numbers 23 to 29 carry 6 marks each. 23. Using matrices, solve the following system of equations: x - y + z = 4; 2x + y - 3z = 0; x + y + z = 2 OR é 3 -1 1 ù é 1 2 -2 ù –1 ê ú If A = -15 6 -5 and B = ê -1 3 0 ú, find (AB)–1. ê ú ê ú êë 5 -2 2 úû êë 0 -2 1 úû 24. Show that the altitude of the right circular cone of maximum volume that can be inscribed in 4R a sphere of radius R is . 3 25. Find the area of the region in the first quadrant enclosed by x-axis, the line x = 3 y and the circle x 2 + y 2 = 4. 3

26. Evaluate: ò ( x 2 + x) dx 1

OR p4

Evaluate:

ò 0

2

cos x dx cos x + 4 sin 2 x 2

27. Find the vector equation of the plane passing through the points (2, 1, –1) and (–1, 3, 4) and perpendicular to the plane x - 2y + 4z = 10. Also show that the plane thus obtained contains ® the line r = - i$ + 3j$ + 4k$ + l( 3i$ - 2j$ - 5k$). 28. A company produces soft drinks that has a contract which requires that a minimum of 80 units of the chemical A and 60 units of the chemical B go into each bottle of the drink. The chemicals are available in prepared mix packets from two different suppliers. Supplier S had a packet of mix of 4 units of A and 2 units of B that costs `10. The supplier T has a packet of mix of 1 unit of A and 1 unit of B costs `4. How many packets of mixed from S and T should the company purchase to honour the contract requirement and yet minimize cost? Make a LPP and solve graphically. 29. In a certain college, 4% of boys and 1% of girls are taller than 1.75 metres. Furthermore, 60% of the students in the college are girls. A student is selected at random from the college and is found to be taller than 1.75 metres. Find the probability that the selected student is a girl.

403


Set–II Only those questions, not included in Set I, are given 9. If the binary operation * on set R of real numbers is defined as a * b =

3ab , write the identity 7

element in R for *. 2 10. Evaluate: ò dx 1 + cos 2x dy y = . dx x 20. Find the particular solution of the following differential equation: y e x 1 - y 2 dx + dy = 0, x = 0, y = 1 x ® ® ® ® ® ® ® ® ® $ then express b in the form b = b + b , where b is parallel 21. If a = 3 i - j and b = 2i$ + j$ - 3k, 1 1 2 19. If x13 y 7 = ( x + y) 20 , prove that

®

®

®

to a and b 2 is perpendicular to a . 22. Find the vector and cartesian equations of the line passing through the point P (3, 0, 1) and ® ® parallel to the planes r .(i$ + 2j$) = 0 and r .( 3j$ - k$) = 0. 28. Find the area of the region in the first quadrant enclosed by x - axis, the line y = 3x and the circle x 2 + y 2 = 16. 29. Find the vector equation of the plane passing through the points (3, 4, 2) and (7, 0, 6) and perpendicular to the plane 2x - 5y - 15 = 0. Also show that the plane thus obtained contains ® the line r = i$ + 3j$ - 2k$ + l (i$ - j$ + k$).

Set–III Only those questions, not included in Set I and Set II are given 9. If the binary operation * on the set Z of integers is defined by a * b = a + b + 2, then write the identity element for the operation * in Z. dy 2y 19. If x16 y 9 = ( x 2 + y) 17 , prove that = . dx x 20. Find the particular solution of the following differential equation: ( x 2 - yx 2 ) dy + ( y 2 + x 2 y 2 ) dx = 0; y = 1, x = 1 21. Find the distance between the point P (6, 5, 9) and the plane determined by the points A ( 3, - 1, 2), B (5, 2, 4) and C (–1, –1, 6). 22. The two adjacent sides of a parallelogram are 2i$ - 4j$ + 5k$ and i$ - 2j$ - 3k$. Find the unit vector parallel to one of its diagonals. Also, find its area. 28. Using the method of integration, find the area of the DABC, coordinates of whose vertices are A (2, 0), B(4, 5) and C(6, 3). 29. Find the equation of the plane passing through the points (2, 2, 1) and (9, 3, 6) and perpendicular to the plane 2x + 6y + 6z = 1. Also, show that the plane thus obtained contains the line ®

r = 4i$ + 3j$ + 3k$ + l (7i$ + j$ + 5k$).

404


Solutions Set–I SECTION–A 1. Let e Î Z be required identity \ a* e = a " a ÎZ Þ a+ e -5= a Þ e = a- a+5 Þ e =5 p 2. cot (tan -1 a + cot -1 a) = cot æç - cot -1 a + cot -1 è2 p = cot = 0 2 -1 -1 p [Note: tan x + cot x = 2 " x Î R] 3. (I + A)2 – 3A = I 2 + A 2 + 2A - 3A = I 2 + A2 - A = I2 + A - A [Q 2 =I =I.I=I é 2ù é -1ù é10ù 4. Given xê ú+ yê ú= ê ú ë 3û ë1 û ë5û

aö÷ ø

A2 = A]

é 2x ù é - yù é10ù ê 3xú + ê y ú = ê 5 ú ë û ë û ë û é 2x - y ù é10ù ê 3x + yú = ê 5 ú ë û ë û

Þ

Equating the corresponding elements we get. 2x - y = 10 3x + y = 5 (i) and (ii) Þ 2x - y + 3x + y = 10 + 5 Þ 5x = 15 Þ x = 3. 102 18 36 5. Let D = 1 3 4 17

3

...(i) ...(ii)

6

Applying R1 ® R1 - 6R 3 0 0 0 D= 1 17

3 4 =0 3 6

[Q R1 is zero]

405


6. Given Þ

æ x - 1 öe x dx = f ( x). e x + c ÷ x2 ø æ1 1 ö x x ò çè x - x 2 ÷øe dx = f ( x). e + c

ò çè

1 x . e + c = f ( x). e x + c x Equating we get 1 f ( x) = x [Note: ò [ f ( x) + f '( x)] e x = f ( x) e x + c] Þ

a

7. Given

ò 3x

2

dx = 8

0

a

Þ

éx3 ù 3ê ú = 8 ë 3 û0

Þ a=2 a3 = 8 Þ $ $ $ $ $ $ $ $ $ 8. (i ´ j ) . k + ( j ´ k) . i = k . k + i . i$ =1+1=2 ® ®

®

®

®

®

®

[Note a . b =| a|.| b| cos q. Also| i |=| j|=| k|= 1] 9. Required area of parallelogram = 2i$ ´ 3j$ = 6|i$ ´ j$|= 6|k$| = 6 square unit. ®

®

®

®

[Note: Area of parallelogram whose sides are represented by a and b is| a ´ b|] 10. The angle made by a line parallel to z axis with x, y and z axis are 90°, 90° and 0° respectively. \ The direction cosines of the line are cos 90° , cos 90° , cos 0° i.e, 0, 0, 1.

SECTION–B 11. Given f ( x) = \

4x + 3 6x - 4

,x¹

2 3

æ 4x + 3 ö fof ( x) = f ( f ( x)) = f ç ÷ è 6x - 4 ø

=

æ 4x + 3 ö 4ç ÷+3 è 6x - 4 ø æ 4x + 3 ö 6ç ÷-4 è 6x - 4 ø

=

16x + 12 + 18x - 12 24x + 18 - 24x + 16

=

34x =x 34

406


Now for inverse of f, 4x + 3 Let y= 6x - 4 \

\

6xy - 4y = 4x + 3

Þ

x( 6y - 4) = 3 + 4y

Þ

6xy - 4x = 3 + 4y 4y + 3 x= 6y - 4

Inverse of f is given by 4x + 3 f -1 ( x) = 6x - 4

5 3 12. Let sin -1 æç ö÷ = a , cos -1 æç ö÷ = b è 13 ø è5ø Þ sin a =

5 3 , cos b = 13 5

5 2 3 2 Þ cos a = 1 - æç ö÷ , sin b = 1 - æç ö÷ è 13 ø è5ø 12 4 , sin b = 13 5 sin (a + b) = sin a . cos b + cos a . sin b 5 3 12 4 = . + . 13 5 13 5 15 48 63 = = + 65 65 65 63 a + b = sin -1 æç ö÷ è 65 ø

Þ cos a = Now

Þ

Putting the value of a and b we get 5 3 63 + cos -1 = sin -1 æç ö÷ sin -1 è 65 ø 13 5 Given, Þ Þ Þ Þ Þ

OR 2 tan (sin x) = tan -1 ( 2 sec x) 2 sin x ö –1 tan -1 æç ÷ = tan (2 sec x) è 1 - sin 2 x ø 2 sin x éQ x ¹ p Þ 1 - sin 2 x ¹ 0ù = 2 sec x 2 êë úû 2 1 - sin x 2 sin x = 2 sec x Þ sin x = sec x . cos 2 x 2 cos x 1 sin x = . cos 2 x Þ sin x = cos x cos x p tan x = 1 Þ x= 4 -1

407


a+b

a+b + c

= 2a

3a + 2b

4a + 3b + 2c

3a

6a + 3b

10a + 6b + 3c

a 13. L.H.S.

a

a+b + c

a

b

a+b + c

4a + 3b + 2c + 2a

2b

4a + 3b + 2c

a

= 2a

3a

3a 6a 10a + 6b + 3c 1

1

=a 2

3

2

a+b + c

1

1

a+b + c

4a + 3b + 2c + ab 2

2

4a + 3b + 2c

3 6 10a + 6b + 3c 1

1

=a 2

3

2

3b 10a + 6b + 3c

3a

3 10a + 6b + 3c

3

a+b + c

[Qc1 = c 2 in second det.]

4a + 3b + 2c + ab .0

3 6 10a + 6b + 3c a+b + c

1 1 = a2 2 3

4a + 3b + 2c

3 6 10a + 6b + 3c 1 1 = a2 2 3

1

1

b

4a + a 2 2

3

3b + a 2 . c 2

a

3 6 10a 1

1

2

= a .a 2

3 6 1

4 + a .b 2

3

3 6 10 =a

3

6b

1 2

1

1 3

3 6

1

1

1

2

3

4 + a 2b. 0 + a 2 c.0

1

1 2

3 + a .c 2 6

=a

1

1

1

2

3

4

3 6 10 Applying C 2 ® C 2 - C1 and C 3 ® C 3 - C1 we get 1 0 0 a3 2 1 3

2

3 7

Expanding along R1 we get = a 3 . 1 (7 - 6) - 0 + 0 = a 3.

c

3 2c

3 6

3 6 10 3

1

3c 1

1

3 2

3 6

3 éQc 2 @ c 3 in second det.ù ê c @ c in 3rd det. ú 3 ë 1 û

408


14. Given x m . y n = ( x + y) m+ n Taking logarithm of both sides we get log ( x m . y n ) = log ( x + y) m+ n Þ log x m + log y n = (m + n) . log ( x + y) Þ m log x + n log y = (m + n) . log ( x + y) Differentiating both sides w.r.t. x we get dy ö m n dy m + n æ + . = . ç1 + ÷ è x y dx x + y dx ø Þ Þ Þ Þ

m m + n æ m + n n ö dy =ç - ÷ x x + y è x + y y ø dx mx + my - mx - nx x( x + y) my - nx

æ my + ny - nx - ny ö dy =ç ÷. y( x + y) è ø dx

my - nx dy . x ( x + y) y ( x + y) dx dy my - nx y( x + y) y = = . dx x( x + y) my - nx x =

15. We have, y = e a cos

-1

x

Taking log on both sides log y = a log cos - 1 x Differentiating w.r.t. x, we have -1 1 dy =a´ y dx 1 - x2 - ay dy = dx 1 - x2

…(i)

Again differentiating w.r.t. x, we have é ù dy 1 ê 1 - x2 -y´ ´ - 2xú dx ê ú 2 1 - x2 d2y ë û = a 2 2 dx (1 - x ) Þ

(1 - x 2 )

Þ

(1 - x 2 )

\

(1 - x 2 )

é - ay xy ê 1 - x2 ´ = a + 2 2 dx ê 1-x 1 - x2 ë d2y

d2y dx

2

d2y dx

2

= a2y = a2y + x

axy 1 - x2 dy [From (i)] dx

ù ú ú û

409


We have, (1 - x 2 )

d2y dx

2

-x

dy - a2y = 0 dx

OR Given, x 1+y +y 1+x=0 Þ x 1+y =-y 1+x Squaring both sides, we have x 2 (1 + y) = y 2 (1 + x) Þ x 2 + x 2 y = y 2 + xy 2 Þ x 2 - y 2 = xy ( y - x) Þ x + y + xy = 0 [Q x ¹ y] x Þ y=1+x Differentiating w.r.t. x, we get -1 dy (1 + x) ( - 1) + x = = 2 dx (1 + x) (1 + x) 2 16. Here Þ

f ( x) = log (1 + x) – f ¢( x) =

2x 2+x

[Where y = f(x)]

é ( 2 + x) . 1 - x ù 1 -2ê ú 2 1+x ë ( 2 + x) û

=

2( 2 + x - x) 1 1 4 = – 1+x 1 + x ( 2 + x) 2 ( 2 + x) 2

=

4 + x 2 + 4x - 4 - 4x x2 = 2 ( x + 1)( x + 2) ( x + 1)( x + 2) 2

For f(x) being increasing function f ¢( x) > 0 x2 Þ >0 Þ ( x + 1)( x + 2) 2

1 x2 . >0 x + 1 ( x + 2) 2

Þ

1 >0 x+1

é x2 ù > 0ú ê 2 ë ( x + 2) û

Þ Þ

x+1> 0 x+1> 0

[Q 1 > 0] or

x>–1 2x i.e., f ( x) = y = log(1 + x) – is increasing function in its domain x > -1 i.e. ( -1, µ). 2+x

410


Given, curve We have, Þ

OR ay 2 = x 3 dy 2ay = 3x 2 dx dy 3x 2 = dx 2ay

3 ´ a 2m 4 3m dy at ( am 2 , am 3 ) = = dx 2 2a ´ am 3 1 1 2 \ Slope of normal = ==3 m Slope of tangent 3m 2 Equation of normal at the point ( am 2 , am 3 ) is given by y - am 3 2 =2 3m x - am Þ

3my – 3am4 = – 2x + 2am2 Þ Þ 2x + 3my - am 2 ( 2 + 3m 2 ) = 0 Hence, equation of normal is 2x + 3my - am 2 ( 2 + 3m 2 ) = 0 x3 1 x3 17. ò x 2 tan -1 x dx = tan -1 x . -ò . dx 3 1 + x2 3

=

x 3 tan -1 x 1 æ x ö - ò çx - 2 ÷ dx 3 3 è x + 1ø

=

x 3 tan -1 x 1 é x xdx - ò 2 dxù ò ê 3 3ë x + 1 úû

=

x 3 tan -1 x 1 x 2 1 dz + ò 3 3 2 3 2z

x 3 tan -1 x x 2 1 + log|z|+ c 3 6 6 x 3 tan -1 x x 2 1 = + log x 2 + 1 + c 3 6 6 =

é ù ê ú x ê ú ê1 + x 2 x 3 ú ê ú 3 ê –x ± xú ê - x úû ë

é ê Let ê êÞ ê êÞ ë

ù x2 + 1 = z ú ú 2xdx = dz ú dz ú xdx = ú 2û

411


Let Þ

OR 3x - 1 A B = + ( x + 2) 2 x + 2 ( x + 2) 2 A( x + 2) + B 3x - 1 = 2 ( x + 2) ( x + 2) 2

Þ 3x - 1 = A( x + 2) + B Þ 3x - 1 = Ax + ( 2A + B) Equating we get A = 3, 2A + B = – 1 2×3+B=–1 Þ B=–7 Þ 3x - 1 3 7 \ = ( x + 2) 2 x + 2 ( x + 2) 2 3x - 1 3 7 Þ ò ( x + 2) 2 dx = ò x + 2 dx - ò ( x + 2) 2 dx = 3 log x + 2 - 7 = 3 log x + 2 + 18. Given

dy e -2 x y = dx x x

Þ

dy 1 e -2 x + .y = dx x x dy + py = Q. dx

It is linear equation of form

\

1 e -2 x , Q= x x Pdx I. F. = e ò P=

=e =e

1 dx x

ò

òx 1

=

-1 2

dx

+1

x2 1 - +1 e 2 =

e2

x

-1

7 +c ( x + 2)

æ e -2 x y ö dx ç ÷ = 1, x ¹ 0 x ø dy è x

Þ

Where

( x + 2) -1

+c

412


Therefore General solution is y. e 2

x

= ò Q ´ I . F dx + c

Þ

y.e 2

x

=ò

e -2 x 2 .e x

Þ

y.e 2

x

=ò

dx +c x

Þ

y.e 2

x

=2 x +c

x

dx + c 1

y.e 2

Þ

x

=

x - 2+1 +c - 12 + 1

19. Given 3e x tan y dx + ( 2 - e x ) sec 2 y dy = 0 ( 2 - e x ) sec 2 y dy = -3e x tan y dx

Þ

sec 2 y -3 e x dy = dx tan y 2 - ex

Þ

Þ x

Þ

log tan y = 3 log 2 - e

+ log c

Þ

log tan y = log c ( 2 - e x )

3

tan y = c ( 2 - e x ) 3 p Putting x = 0, y = we get 4 p Þ tan = c ( 2 - e° ) 3 4 Þ

Þ

1 = 8c

c=

1 8

Therefore particular solution is (2 - e x ) 3 . tan y = 8 ®

20. Q

®

b 1 is parallel to a ®

®

Þ

b 1 = l a where l is any scalar quantity.

Þ

b 1 = 3li$ + 4lj$ + 5lk$

®

®

®

®

Also If, b = b 1 + b 2 ®

Þ

2i$ + j$ - 4k$ = ( 3li$ + 4lj$ + 5lk$) + b 2

Þ

b 2 = ( 2 - 3l) i$ + (1 - 4l) j$ - ( 4 + 5l) k$

®

®

®

It is given b 2 ^ a

Þ ( 2 - 3l) . 3 + (1 - 4l) . 4 - ( 4 + 5l) . 5 = 0 Þ 6 - 9l + 4 - 16l - 20 - 25l = 0

ò

sec 2 y dy tan y

= 3ò

- e x dx 2 - ex

413


Þ – 10 – 50l = 0

Þ

l=

-1 5

®

3$ 4$ $ i+ j -k 5 5 ® 3 4 b 2 = æç 2 + ö÷i$ + æç1 + ö÷ j$ - ( 4 - 1) k$ è 5ø è 5ø 13 9 = i$ + j$ - 3k$ 5 5 Therefore required expression is 3 4 13 9 ( 2i$ + j$ - 4k$) = æç - i$ - j$ - k$ ö÷ + æç i$ + j$ - 3k$ ö÷ è 5 ø è5 ø 5 5 Therefore b 1 = –

21. Let required cartesian equation of line be x-1 y- 2 z- 3 = = a b c Given planes are ® r .(i$ - j$ + 2k$) = 5 ®

r .( 3i$ + j$ + k$) = 6

...(i)

...(ii) ...(iii)

Since line (i) is parallel to plane (ii) and normal vector of plane (ii) is i$ - j$ + 2k$ Þ

a - b + 2c = 0

...(iv)

Similarly line (i) is parallel to plane (iii) and normal vector of plane (iii) is 3i$ + j$ + k$ ...(v) Þ 3a + b + c = 0 From (iv) and (v) a b c = = -1 - 2 6 - 1 1 + 3 a b c = = =l -3 5 4 Þ a = -3 l , b = 5 l , c = 4 l Putting value of a, b and c in (i) we get required cartesian equation of line x-1 y- 2 z- 3 x-1 y- 2 z- 3 = = Þ = = -3 l 5l 4l -3 5 4 Its vector equation is ® r = (i$ + 2j$ + 3k$) + l( -3i$ + 5j$ + 4k$) 22. Here, number of throws = 4 6 1 P(doublet) = p = = 36 6 30 5 P(not doublet) = q = = 36 6

414


Let X denotes number of successes, then 5 4 625 P(X = 0) = 4C 0 p 0 q 4 = 1 ´ 1 ´ æç ö÷ = è 6ø 1296 P(X = 1) = 4C1

1 æ5 ö 3 125 500 ´ç ÷ =4´ = 6 è 6ø 1296 1296

1 2 5 2 25 150 P(X = 2) = 4C 2 æç ö÷ ´ æç ö÷ = 6 ´ = è 6ø è 6ø 1296 1296 1 3 5 20 P(X = 3) = 4C 3 æç ö÷ ´ = è 6ø 6 1296 1 4 1 P(X = 4) = 4C 4 æç ö÷ = è 6ø 1296 Therefore the probability distribution of X is X or x1

0 625 1296

P(X) or pi \

1 500 1296

2 150 1296

Mean (M) = å x i p i 625 500 150 20 1 = 0´ +1 ´ +2´ + 3´ + 4´ 1296 1296 1296 1296 1296 500 300 60 4 864 2 = = = + + + 1296 1296 1296 1296 1296 3

SECTION–C 23. Given equations x -y+z = 4 2x + y - 3z = 0 x+ y+z= 2 We can write this system of equations as 1ù é xù é 4ù é1 - 1 ê2 1 - 3ú ê yú = ê 0ú ê ú ê ú ê ú 1 1úû êë z úû êë 2úû êë 1 Let AX = B 1ù é1 - 1 é xù ê ú where A= 2 1 - 3 , X = ê yú and ê ú ê ú 1 1úû êë 1 êë z úû 1 -1 \

| A| = 2 1

é 4ù B = ê 0ú ê ú êë 2úû

1

1 -3 1

1

= 1 (1 + 3) - ( - 1) ( 2 + 3) + 1 ( 2 - 1) = 4 + 5 + 1 = 10

3

4

20 1296

1 1296

415


Now X = A -1 B -1 For A , we have é4 Cofactors matrix of A = ê 2 ê êë 2 2 é 4 ê \ adj A = -5 0 ê êë 1 - 2

-5

1ù 0 - 2ú ú 5 3úû

2ù 5ú ú 3úû 2 2ù é 4 adj A 1 ê -1 \ A = = -5 0 5ú ú | A| 10 ê êë 1 - 2 3úû 2 2ù é 4 ù é 4 é 16 + 0 + 4 ù 1 ê 1 ê -1 ú ê ú Thus, X = A . B = -5 0 5 0 = - 20 + 0 + 10ú ú ê ú 10 ê ú 10 ê êë 1 - 2 3úû êë 2 úû ëê 4 - 0 + 6 úû é xù é 20ù é 2 ù ê yú = 1 ê - 10ú = ê - 1 ú ê ú ú ê ú 10 ê êë z úû êë 10úû êë 1 úû The required solution is \ x = 2, y = - 1, z = 1 OR –1

For B

2

-2

B = -1

3

0

0

-2

1

1

= 1(3 - 0) -2(-1 - 0) -2(2 - 0) =3+2-4=1¹0 i.e., B is invertible matrix B–1 exist. Þ 3 0 Now C11 = (–1)1+1 = 3-0= 3 -2 1 C12 = ( -1) 1+ 2

-1 0 = -( -1 - 0) = 1 0 1

C13 = ( -1) 1+ 3

-1 3 =2-0=2 0 -2

C 21 = ( -1) 2+1

2 -2 = -( 2 - 4) = 2 -2 1

416


\

Þ

C 22 = ( -1) 2+ 2

1 -2 =1- 0=1 0 1

C 23 = ( -1) 2+ 3

1 2 = -( -2 - 0) = 2 0 -2

C 31 = ( -1) 3+1

2 -2 =0+6=6 3 0

C 32 = ( -1) 3+ 2

1 -2 = -( 0 - 2) = 2 -1 0

C 33 = ( -1) 3+ 3

1 2 = ( 3 + 2) = 5 -1 3

é 3 1 2ù ¢ é 3 2 6ù Adj B = ê 2 1 2ú = ê 1 1 2ú ê ú ê ú êë 6 2 5 úû êë 2 2 5 úû 1 B–1 = ( adj B) B é 3 2 6ù é 3 2 6ù 1ê = 1 1 2ú = ê 1 1 2ú ú ê ú 1ê êë 2 2 5 úû êë 2 2 5 úû

Now (AB)–1 = B–1. A–1 é 3 2 6 ù é 3 -1 1 ù = ê 1 1 2ú . ê -15 6 -5ú ê ú ê ú -2 2 úû êë 2 2 5 úû êë 5 = é 9 - 30 + 30 -3 + 12 - 12 3 - 10 + 12ù ê 3 - 15 + 10 -1 + 6 - 4 1-5+ 4 ú ê ú êë 6 - 30 + 25 -2 + 12 - 10 2 - 10 + 10 úû é 9 -3 5 ù = ê -2 1 0 ú ê ú 0 2úû êë 1 24. Let h be the altitude of cone inscribed in a sphere of radius R. Also let r be radius of base of cone. If V be volume of cone then B 1 2 V = pr h 3 1 [In DOBD BD2 = OB2 - OD2 V = p( 2hR - h 2 ). h 3

A

R O R r

r D

C

417


V=

Þ

p ( 2h 2 R - h 3 ) 3

Þ r 2 = R 2 - ( h - R) 2

dV p = ( 4hR - 3h 2 ) dh 3

Þ r 2 = R 2 - h 2 - R 2 + 2hR Þ r 2 = 2hR - h 2

For maximum or minimum value dV =0 dh p Þ ( 4hR - 3h 2 ) = 0 3 Þ 4hR - 3h 2 = 0 Þ h( 4R - 3h) = 0 4R . Þ h = 0, h = 3 d 2V p Now = ( 4R - 6h) 3 dh 2 d 2V ù d 2V ù = +ve and = -ve ú 2 ú dh û h= 0 dh 2 û h= 4R 3

4R Hence for h = , volume of cone is maximum. 3 25. Obviously x 2 + y 2 = 4 is a circle having centre at ( 0, 0) and radius 2 units. For graph of line x = 3y x

0

1

y

0

0.58

For intersecting point of given circle and line Putting x = 3 y in x 2 + y 2 = 4 we get ( 3y) 2 + y 2 = 4 Þ 3y 2 + y 2 = 4 Þ Þ y = ±1 4y 2 = 4 \ x=± 3 Intersecting points are ( 3 , 1), ( - 3 , -1). Shaded region is required region. 3

Now required area =

ò 0

x dx + 3

2

ò

3

4-x

2

2

3y x= ( 3,1)

1 –2

O

–1 –1 –2

1

3 2

418


=

=

3 é x 4 - x2 4 1 é x2 ù xù + sin -1 ú ê 2 ú +ê 2 2 2ú 3 ë û0 êë û

2

3

é æ 3 3 öù ( 3 - 0) + ê 2 sin -1 1 - ç + 2 sin -1 ÷ú 2 øû è 2 2 3 ë 1

3 é p 3 2p ù + ê2 2 2 3 úû ë 2 3 3 2p = + p2 2 3 2p p =p= sq. unit. 3 3 b - a 3-1 2 26. Here a = 1, b = 3, h = = = n n n Þ nh = 2 Also f ( x) = x 2 + x =

b

By definition

h{ f ( a) + f ( a + h) +.............+ f ( a + (n - 1) h} ò f ( x) dx = lim h ®0 a

3

h { f (1) + f (1 + h) +...........+ f (1 + (n - 1) h} ò f ( x) dx = lim h ®0 1

Now f(1) = 12 + 1 = 2 f (1 + h) = (1 + h) 2 + (1 + h) = 1 2 + h 2 + 2h + 1 + h = 2 + 3h + h 2 f (1 + 2h) = (1 + 2h) 2 + (1 + 2h) = 1 2 + 2 2 h 2 + 4h + 1 + 2h = 2 + 6h + 2 2 h 2 f (1 + (n - 1) h = {1 + (n - 1) h} 2 + {1 + (n - 1)h} = 2 + 3 (n - 1) h + (n - 1) 2 h 2 Hence 3

ò (x 1

2

+ x) dx = lim h{2 + ( 2 + 3h + h 2 ) + ( 2 + 6h + 2 2 h 2 ) + ....... + ( 2 + 3(n - 1) h + (n - 1) 2 . h 2 )} h ®0

= lim h.[{2n + 3h {1 + 2+...(n - 1)} + h 2 {1 2 + 2 2 +......+(n - 1) 2}] h ®0 (n - 1). n (n - 1). n( 2n - 1) ü ì = lim h í 2n + 3h . + h2 ý 2 6 þ h®0 î 1 1 1 ì n 2 . h 2 æç1 - ö÷ n 3 h 3 æç1 - ö÷ æç 2 - ö÷ ü ï è è nø nø è nø ï = lim í 2n. h + 3 + ý h ®0 2 6 ï ï î þ 12 æ 1 8 1 1 ü = lim ìí 4 + ç1 - ö÷ + æç1 - ö÷ æç 2 - ö÷ý è ø è ø è n®¥ î 2 n 6 n n øþ

éQ n h = 2 ù ê h ® 0 Þ n ® ¥ú ë û

419


4 (1 - 0) (2 - 0) 3 4 8 38 = 4 + 6 + ´ 2 = 10 + = 3 3 3 OR = 4 + 6 (1 - 0) +

p

Let

I

=

cos 2 x ò cos 2 x + 4 sin 2 xdx 0

p

=

cos 2 x ò cos 2 x + 4(1 - cos 2 x) dx 0

p

=

2

2

cos 2 x 1 ò 4 - 3 cos 2 x dx = - 3 0 2

p

1 2 4 = - ò æç1 è 30 4 - 3 cos 2 =

p

-1 p 2 4 [x]0 + 3 3

1 p 4 =- . + 3 2 3

p

2

p

4 - 3 cos 2 x - 4 dx 4 - 3 cos 2 x 0 2

ò

p

2 ö dx = - 1 dx + 4 ÷ 3 ò0 3 xø

dx

ò 4 - 3 cos 2 x 0

sec x dx

ò 4 sec 2 x - 3 0

sec 2 x dx

2

ò 4 (1 + tan 2 x) - 3 0

¥

p 4 dz + ò 6 3 0 4 + 4z 2 - 3

=-

p 4 + 6 3´4

¥

ò 0

dz 1 2 z 2 + æç ö÷ è 2ø ¥

é p 1 zù + 2 . tan -1 ú 1 6 3 êë 2û 0

p 2 ¥ + [tan -1 2z] 0 6 3 p 2 = - + [tan -1 ¥ - tan -1 0] 6 3 p 2 p = - + é - 0ù úû 6 3 êë 2 p p p =- + = . 6 3 6 =-

2

2

=-

=-

p

é Let tan x = z Þ sec 2 x dx = dz ù ê ú ê Also, x = p Þ z = ¥; x = 0 Þ z = 0ú 2 ë û

420


27. Let the equation of plane through (2, 1, –1) be a ( x - 2) + b ( y - 1) + c (z + 1) = 0 (i) passes through (–1, 3, 4) Q a (–1 –2) + b (3 – 1) + c (4 + 1) = 0 \ Þ -3a + 2b + 5c = 0 Also since plane (i) is perpendicular to plane x - 2y + 4z = 10 \ a - 2b + 4c = 0 From (ii) and (iii) we get a b c + = 8 + 10 5 + 12 6 - 2 a b c Þ = = = l (say) 18 17 4 Þ a = 18 l , b = 17 l, c = 4l, Putting the value of a, b , c in (i) we get 18 l( x - 2) + 17 l( y - 1) + 4l(z + 1) = 0 Þ 18x - 36 + 17 y - 17 + 4z + 4 = 0 Þ 18x + 17 y + 4z = 49 Required vector equation of plane is \ ®

r .(18i$ + 17 j$ + 4k$) = 49

...(i)

...(ii) ...(iii)

...(iv)

Obviously plane (iv) contains the line ®

...(v) r = ( -i$ + 3j$ + 4k$) + l( 3i$ - 2j$ - 5k$) if point ( -i$ + 3j$ + 4k$) satisfy equation (iv) and vector (18i$ + 17 j$ + 4k$) is perpendicular to ( 3i$ - 2j$ + 5k$). Here, ( -i$ + 3j$ + 4k$).(18i$ + 17 j$ + 4k$) = -18 + 51 + 16 = 49 Also, (18 i$ + 17 j$ + 4k$) .( 3i$ - 2j$ - 5k$) = 54 - 34 - 20 = 0 Therefore (iv) contains line (v). 28. Let x and y units of packet of mixes are purchased from S and T respectively. If z is total cost then ...(i) z = 10x + 4y is objective function which we have to minimize Here constraints are. ...(ii) 4x + y ³ 80 ...(iii) 2x + y ³ 60 Also, ...(iv) x³0 ...(v) y³0 On plotting graph of above constraints or inequalities (ii), (iii) , (iv) and (v) we get shaded region having corner point A, P, B as feasible region.

421


For coordinate of P Y

80

B(0, 80)

70

60 50 P(10, 40) 40 30 20 10 A (30, 0) X’ –1

O

10

40

50

60

70

80 X

+ 2x

10x

60 y = 260 y=

+4

= 80

Y’

30 y 4x +

–1

20

Point of intersection of ...(vi) 2x + y = 60 and ...(vii) 4x + y = 80 (vi) – (vii) Þ 2x + y - 4x - y = 60 - 80 Þ -2x = -20 Þ x = 10 Þ y = 40 co-ordinate of P º (10, 40) Q Now the value of z is evaluated at corner point in the following table Corner point

z = 10x + 4y

A (30, 0)

300

P (10, 40)

260

B (0, 80)

320

¬

Minimum

Since feasible region is unbounded. Therefore we have to draw the graph of the inequality. ...(viii) 10x + 4y < 260 Since the graph of inequality (viii) does not have any point common. So the minimum value of z is 260 at (10, 40).

422


i.e., minimum cost of each bottle is ` 260 if the company purchases 10 packets of mixes from S and 40 packets of mixes from supplier T. 29. Let E1, E2, A be events such that E1 = student selected is girl E2 = student selected is Boy A = student selected is taller than 1.75 metres. Here P( E1 A) is required. 60 3 40 2 Now = , = P(E1 ) = P(E2 ) = 100 5 100 5 æAö æAö 1 4 , Pç ÷ = Pç ÷= è E1 ø 100 è E2 ø 100

\

E P æç 1 ö÷ = èAø

æAö P(E1 ). P ç ÷ è E1 ø æAö æAö P(E1 ). P ç ÷ + P(E2 ). P ç ÷ è E1 ø è E2 ø

3 1 3 ´ 3 500 3 5 100 500 = = = = ´ 3 1 2 4 3 8 500 11 11 ´ + ´ + 5 100 5 100 500 500

Set–II 9. Let e Î R be identity element. a*e = a \ " a ÎR 3ae 7a =a e= Þ Þ 7 3a 7 e= Þ 3 2 2 10. ò dx = ò dx 1 + cos 2x 2 cos 2 x = ò sec 2 x dx = tan x + c 19. Given x13 y 7 = ( x + y) 20 Taking logarithm of both sides, we get log ( x13 y 7 ) = log ( x + y) 20 Þ log x13 + log y 7 = 20 log( x + y) Þ 13 log x + 7 log y = 20 log ( x + y) Differentiating both sides w.r.t. x we get dy ö 13 7 dy 20 æ + . = . ç1 + ÷ x y dx x + y è dx ø

423


Þ

æ 20 13 20 7 ö dy =ç - ÷ x x + y è x + y y ø dx

Þ

13x + 13y - 20x æ 20y - 7 x - 7 y ö dy =ç ÷ x ( x + y) è ( x + y). y ø dx

Þ

13y - 7 x æ 13y - 7 x ö dy =ç ÷. x ( x + y) è x ( x + y) ø dx

dy 13y - 7 x y( x + y) = ´ dx x( x + y) 13y - 7 x y 20. Given e x 1 - y 2 dx + dy = 0 x y x dy = - e 1 - y 2 dx x Þ

Þ

Þ

dy y = dx x

y 1-y

2

dy = - x e x dx

Integrating both sides we get y x ò 1 - y 2 dy = – ò x e dx -z dz

Þ

= - [x . e x - ò e x dx] + c z -z = - x e x + e x + c

Þ

- 1 - y 2 = -x e x + e x + c

Þ

ò

[Let 1 - y 2 = z 2 Þ - 2ydy = 2zdz Þ ydy = -zdz]

Þ x e x - e x - 1 - y2 = c

Putting x = 0, y = 1 we get Þ -1 - 1 - 1 = c Þ c = -1 Hence particular solution is Þ

x e x - e x - 1 - y 2 = -1

Þ

e x ( x - 1) - 1 - y 2 + 1 = 0

21. Q Þ

®

®

b 1 is parallel to a ®

®

Þ

b1 = l a ®

®

®

b 1 = 3li$ - lj$

®

Also b = b 1 + b 2 Þ Þ

®

2i$ + j$ - 3k$ = ( 3li$ - lj$) + b 2 ®

b 2 = ( 2i$ + j$ - 3k$) - ( 3li$ - lj$) = ( 2 - 3l)i$ + (1 + l) j$ - 3k$ ®

®

It is given b 2 is perpendicular to a \ Þ Þ

( 2 - 3l) 3 + (1 + l) .( -1) + ( -3) . 0 = 0 6 - 9l - 1 - l = 0 5 1 5 - 10l = 0 Þ l= = 10 2

424


1$ 1$ 3$ 1$ i- j= i- j 2 2 2 2 ® 1 3 1 1 b 2 = æç 2 - 3 ´ ö÷i$ + æç1 + ö÷ j$ - 3k$ = i$ + j$ - 3k$ è ø è ø 2 2 2 2

\

b1 = 3 ´

Therefore required expression is 3 1 1 3 2i$ + j$ - 3k$ = æç i$ - j$ ö÷ + æç i$ + j$ - 3k$ ö÷ è2 ø 2 ø è2 2 22. Let the cartesian equation of the line passing through the point P (3, 0, 1) be x- 3 y- 0 z-1 ...(i) = = a b c Given planes are ®

r .(i$ + 2j$) = 0

...(ii)

®

r .( 3i$ - k$) = 0

...(iii)

Since line (i) is parallel to plane (ii) and (iii) Þ ( ai$ + bj$ + ck$) .(i$ + 2j$) = 0 Þ a + 2b + 0. c = 0

...(iv)

( ai$ + bj$ + ck$) .( 3i$ - k$) = 0 Þ 3a + 0. b - c = 0

...(v)

and

From (iv) and (v) a b c = = -2 - 0 0 + 1 0 - 6 a b c Þ = = = l (say) -2 1 -6 Þ a = -2l, b = l, c = -6l Putting the value of a = -2l , b = l , c = -6l in (i) we get required cartesian equation of line x- 3 y z-1 x- 3 y z-1 = = Þ = = -2 l l -6 l -2 1 -6 Therefore required vector equation is ®

r = ( -3i$ + k$) + l( -2i$ + j$ - 6k$)

28. Obviously x 2 + y 2 = 16 is a circle having centre at (0, 0) and radius 4 units. For graph of line y = 3x 5 x

0

4

1

(2, 2Ö3) 3

3 = 1732 .

(1, 1 3)

1 O1 –3 –2 –1 –1

2

3 4

=

16

–2

+

y2

–3 –4 –5

x2

For intersecting point of given circle and line –5 –4 Putting y = 3x in x 2 + y 2 = 16 we get 2 2 x + ( 3x) = 16 Þ Þ x =±2 4x 2 = 16 \ y = ±2 3. Therefore, intersecting point of circle and line is ( ± 2, ± 2 3 )

2

3x

0

y=

y

5

425


Now shaded region is required region 2

4

\ Required Area = ò 3x dx + ò 16 - x 2 dx. 0

2

=

2

éx ù x 16 x 4 3ê ú +é 16 - x 2 + sin -1 ù ê 2 4 úû 2 ë 2 û0 ë2 2

3 x x 4 ´ 4+é 16 - x 2 + 8 sin -1 ù 2 4 úû 2 ëê 2 8p æ 8p ö ù 4p ù = 2 3 + é0 + - ç 12 + ÷ ú = 2 3 + é 4 p - 12 êë ê è ø 2 6 û 3 úû ë 4p 4p 8p = 4p = . sq. unit. = 2 3 + 4p - 2 3 3 3 3 29. Let the equation of plane through (3, 4, 2) be ...(i) a ( x - 3) + b ( y - 4) + c (z - 2) = 0 (i) passes through (7, 0, 6) Q \ a (7 - 3) + b ( 0 - 4) + c ( 6 - 2) = 0 Þ 4a - 4b + 4c = 0 ...(ii) Þ a-b + c = 0 Also, since plane (i) is perpendicular to plane 2x - 5y - 15 = 0 ...(iii) 2a - 5b + 0c = 0 From (ii) and (iii) we get a b c = = = l (say) Þ a = 5l , b = 2l, c = -3l. 5 2 -3 Putting the value of a, b, c in (i) we get 5l ( x - 3) + 2l( y - 4) - 3l(z - 2) = 0 Þ 5x - 15 + 2y - 8 - 3z + 6 = 0 Þ 5x + 2y - 3z = 17 ® ...(iv) \ Required vector equation of plane is r .(5i$ + 2j$ - 3k$) = 17 =

Obviously plane (iv) contains the line ®

...(v) r = (i$ + 3j$ - 2k$) + l (i$ – j$ + k$) if point (i$ + 3j$ - 2k$) satisfy the equation (iv) and vector (5i$ + 2j$ - 3k$) is perpendicular to (i$ - j$ + k$). Here (i$ + 3j$ - 2k$).(5i$ + 2j$ - 3k$) = 5 + 6 + 6 = 17 Also (5i$ + 2j$ - 3k$) .(i$ - j$ + k$) = 5 - 2 - 3 = 0 Therefore (iv) contains line (v).

426


Set–III 9. Let e be the identity for * in Z. \ a* e = a " aÎ Z Þ a+e+2=a Þ e =a-a-2 Þ e = -2 19. Given x16 y 9 = ( x 2 + y) 17 Taking logarithm of both sides, we get log( x16 y 9 ) = log( x 2 + y) 17 Þ log x16 + log y 9 = 17 log( x 2 + y) Þ 16 log x + 9 log y = 17 log( x 2 + y) Differentiating both sides w.r.t. x, we get dy ö 16 9 dy 17 æ Þ + . = ç 2x + ÷ x y dx x 2 + y è dx ø dy 16 9 dy 34x 17 Þ + . = 2 + 2 . x y dx x + y x + y dx Þ

æ9 17 ö dy 34x 16 = 2 ç - 2 ÷ y dx è x + yø x +y x

Þ

æ 9x 2 + 9y - 17 y ö dy 34x 2 - 16x 2 - 16y = ç ÷. 2 x( x 2 + y) è y( x + y) ø dx

Þ

2 2 dy 18x 2 - 16y y( x + y) 2( 9x - 8y). y 2y = = ´ = dx x x( x 2 + y) 9x 2 - 8y x( 9x 2 - 8y)

20. Given Þ Þ

( x 2 - yx 2 ) dy + ( y 2 + x 2 y 2 ) dx = 0 x 2 (1 - y) dy + y 2 (1 + x 2 ) dx = 0 (1 - y) . dy æ 1 + x 2 ö = ç 2 ÷ dx è x ø y2

Integrating both sides we get 1-y 1 + x2 Þ dy = ò y2 ò x 2 dx y 1 1 Þ ò y 2 dy - ò y 2 dy = ò x 2 dx + ò dx 1 -2 -2 Þ ò y dy - ò y dy = ò x dx + ò dx y -2+1 x -2+1 Þ - log y = +x+c -2 + 1 -2 + 1

427


1 1 - log y = - + x + c y x Putting x = 1, y = 1 we get 1 1 Þ - - log 1 = - + 1 + c 1 1 Þ -1-0=-1+1+c Þ c=-1 Putting c = -1 in (i) we get particular solution 1 1 - - log y = - + x - 1 y x Þ

-

Þ

log y =

1 1 - x+1x y

Þ

log y =

...(i)

y - x 2 y + xy - x xy

21. Plane determined by the points A ( 3, - 1, 2), B (5, 2, 4) and C ( -1, - 1, 6) is x- 3 y+1 z-2 x- 3 y+1 z-2

Þ

5- 3

2+1

-1 - 3

-1 + 1 3 0

( x - 3)

4-2 =0

Þ

6-2

2 2 - ( y + 1) 4 -4

2 2 + (z - 2) 4 -4

2

3

2

-4

0

4

3 =0 0

Þ 12x - 36 - 16y - 16 + 12z - 24 = 0 Þ 3x - 4y + 3z - 19 = 0 Distance of this plane from point P ( 6, 5, 9) is ( 3 ´ 6) - ( 4 ´ 5) + ( 3 ´ 9) - 19

=

( 3) 2 + ( 4) 2 + ( 3) 2

18 - 20 + 27 - 19 9 + 16 + 9

=

6 units. 34

22. Let two adjacent sides of a parallelogram be ®

®

®

a = 2i$ - 4j$ + 5k$ and b = i$ - 2j$ - 3k$ i$ j$ k$

®

Now a ´ b = 2 -4 5 = 22i$ + 11j$ 1 -2 -3 ®

Þ

®

Area of given parallelogram =| a ´ b| = ( 22) 2 + (11) 2 = 484 + 121 = 605 = 11 5 square unit. ®

®

®

®

Let a and b be represented by AB and AD respectively. ®

®

\

BC = b

Þ

AC = AB + BC

Þ

AC = a + b = ( 2i$ - 4j$ + 5k$) + (i$ - 2j$ - 3k$) = 3i$ - 6j$ + 2k$

® ®

®

®

®

®

=0

428


Also|AC|=

3 2 + ( -6) 2 + 2 2

= 9 + 36 + 4 = 49 = 7 \ Required unit vector parallel to one diagonal is 1 = ( 3i$ - 6j$ + 2k$) 7

SECTION C 28. Vertices of DABC are A (2, 0), B (4, 5), C (6, 3). Equation of line AB is y-0 5-0 y 5 = Þ = x-2 4-2 x-2 2 5 Þ y = ( x - 2) 2

...(i)

Y 6 B(4, 5)

5

y= –x +

4

y= 5 2 (x – 2

)

9

3

C(6, 3)

2

3 (x y= 4

1

–2

)

A(2, 0) O

1

2

3

4

5

6

7

8 X

Equation of line BC is y-5 3-5 -2 = Þ y-5= ( x - 4) x-4 6-4 2 Þ y = -x + 4 + 5 ...(ii) Þ y = -x + 9 Equation of line AC y-0 3-0 y 3 = = Þ x-2 6-2 x-2 4 3 ...(iii) Þ y = ( x - 2) 4 Now Area of DABC = Area of region bounded by line (i), (ii) and (iii)

429


6

6

5 3 = ò ( x - 2) dx + ò ( - x + 9) dx - ò ( x - 2) dx 2 4 2 4 2 4

6

6

2 2 é ( x - 9) 2 ù 5 é ( x - 2) ù 3 é ( x - 2) ù = ê ú -ê ú - ê ú 2ë 2 û 2 û 4ë 2 û ë 2 4 2 5 1 3 = ( 4 - 0) - ( 9 - 25) - (16 - 0) 4 2 8 = 5 + 8 – 6 = 7 sq. unit 29. Let the equation of plane through (2, 2, 1) be a ( x - 2) + b ( y - 2) + c (z - 1) = 0 (i) passes through (9, 3, 6) Q \ a ( 9 - 2) + b ( 3 - 2) + c ( 6 - 1) = 0 Þ 7 a + b + 5c = 0 Also since plane (i) is perpendicular to plane 2x + 6y + 6z = 1 2a + 6b + 6c = 0 From (ii) and (iii) a b c = = 6 - 30 10 - 42 42 - 2 a b c Þ = = -24 -32 40 a b c Þ = = = m (say) -3 -4 5 Þ a = -3m, b = - 4m, c = 5m Putting the value of a, b, c in (i) we get -3m( x - 2) - 4m( y - 2) + 5m(z - 1) = 0 Þ -3x + 6 - 4y + 8 + 5z - 5 = 0 Þ -3x - 4y + 5z = -9 It is required equation of plane. Its vector form is ® ...(iv) r .( -3i$ - 4j$ + 5k$) = -9

Obviously, plane (iv) contains the line ® r = ( 4i$ + 3j$ + 3k$) + l(7i$ + j$ + 5k$)

...(i)

...(ii) ...(iii)

...(v) $ $ $ if point ( 4i + 3j + 3k) satisfy equation (iv) and vector (7i$ + j$ + 5k$) is perpendicular to $ -3i$ - 4j$ + 5k. Here ( 4i$ + 3j$ + 3k$).( -3i$ - 4j$ + 5k$) = -12 - 12 + 15 = -9 Also (7i$ + j$ + 5k$) . ( -3i$ - 4j$ + 5k$) = –21 – 4 + 25 = 0 Therefore plane (iv) contains line (v).

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