M ATH EM ATIC S
BANSAL CLASSES Tar g et IIT JE JEE 200 7 CLASS : XIII (XYZ)
D a i ly ly Pr Pra c t ic ic e Pr Pro b l e m s
DATE : 19-20//07/2006
TIME : 75 Min.
DPP. NO.-8
This is is the test paper of Class-XI (PQRS) held on 09-07-2007. Take exactly exact ly 75 minutes. Q. 1
[Sol.
Q. 2 [Sol.
Let y =
2
log 1 / 4 x 2
7
log 27 ( x 2 1) 3
3
4 log 49 x
2x
x 1
and
x 4 (x 2 2x 1) y= = x2 + x + 1; 2 x x 1 hence a = 2 and and b = 1 Ans. Ans. ]
dy = ax + b, find the value of a and b. dx
[4]
dy = 2x + 1 = ax + b dx
Show that cos2A + cos2(A + B) + 2 cosA co sA cos(180° + B) · cos(360° + A + B) is independent of A. A. Hence find its value when B = 810°. [Ans. 1] [4] 2 2 cos A + cos (A + B) B) – [2 cosA · cosB · cos (A + B)] cos2A + cos2(A + B) – [ {cos(A + B) + cos(A co s(A – B) } cos (A + B) ] 2 2 2 cos A + cos (A + B) – cos (A + B) – (cos2A – sin2B) = sin2B which is independent independe nt of A 2 now, sin (810°) = sin2(720° + 90°) = sin290° = 1 Ans. ]
Fin Find the the prod produc uctt of the the root rootss of of the the equ equat atiion, on, | x2 | + | x | – 6 = 0. [Ans. – 4] | x |2 + | x | – 6 = 0. 0. let |x|=t t 2 + t – 6 = 0 (t + 3)(t – 2) = 0 t = – 3 or t = 2 b but ut | x | = – 3 is not possible. possible. hence | x | = 2 x = 2 or – 2. Hen Hence prod product uct of roots = – 4 Ans. Alternate: case-I: x 0 x = – 3 , 2 x= 2 case-II: x < 0 x = 3, – 2 x=–2 ]
[4]
One root of mx2 – 10x + 3 = 0 is two third of the other root. roo t. Find the sum of the roo roots. ts. [Ans. 5/4]
[4]
Q.3 Q.3 [Sol.
Q. 4
[Hint: + also
2·
[Sol [Sol..
·
5 10 = 3 m
2 3 = 3 m
36 9 = m m2
Q.5
2 10 = 3 m
sum =
22 =
24 = 3 m
=
6 ; m
9 m
m= 8
10 10 5 = = Ans. ] m 8 4
Suppose Suppose x and y are real real num number berss such such that that tan x + tan tan y = 42 and and cot x + cot y = 49. Fi Find the the valu value of [Ans. 294] tan(x + y). [4] tan x + tan tan y = 42 and and cot x + cot y = 49
tan x tan y 1 tan x tan y cot x + cot y = 49
tan(x + y) = now,
1 1 = 49 tan x tan y
tan y tan x tan x · tan y = 49
tan x · tan y =
tan x tan y 42 6 = = 49 49 7
42 42 tan (x + y) = 1 (6 7) = 1 7 = 294 Ans. ] Q.6 [Sol.
Q.7
Find the solution set of k so that y = kx is secant to the curve y = x2 + k. [4] [Ans. k (– , 0) (4, )] put y = kx in y = x2 + k kx = x2 + k = 0 x2 – kx + k = 0 for line to be secant, D > 0 k 2 – 4k > 0 k(k – 4) > 0 hence k > 4 or k < 0 k (– , 0) (4, ) Ans. ] A quadratic polynomial p(x) has 1 + 5 and 1 –
5 as roots and it satisfies p(1) = 2. Find the 2 [Ans. – (x2 – 2x – 4)] [4] 5
quadratic polynomial. [Sol.
Q.8
sum of the roots = 2 product of the roots = – 4 ] let p(x) = a(x2 – 2x – 4) p(1) = 2 2 = a(12 – 2 · 1 – 4) a = – 2/5 2 p (x) = – (x2 – 2x – 4) Ans. ] 5 Solve the equation x 1 ( x 2 x ) log 2 (1 2 ) log x
[Sol.
zQ.9
0.5 log
x
(x 2 x )
3log9 4 . 2
log ( x x ) x 3 3 = 2 x x x2 – x = 2 x2 – x – 2 = 0 (x – 2)(x + 1) = 0 x = 2 or x = – 1 (rejected) x = 2 Ans. ]
log 2
Find the sum of the series, cos
3 5 + cos + cos + ........ upto n terms. 2n 1 2n 1 2n 1
Do not use any direct formula of summation. [Sol.
[4]
2n 1 S = cos + cos 3 + cos 5 + ........ cos (2n – 1) (2 sin ) S = 2 sin [cos + cos 3 + cos 5 + ........ cos (2n – 1) ] T1 = sin 2 – 0 T2 = sin 4 – sin 2 T3 = sin 6 – sin 4 Let
=
Tn = sin 2n – sin 2(n – 1) ——————————— (2 sin ) S = sin2n 2n 2n 1 = 1 Ans. ] S= 2 2 sin 2n 1 sin
[5]
zQ.10 Find the minimum and maximum value of f (x, y) = 7x2 + 4xy + 3y2 subjected to x2 + y2 = 1. [Sol.
Let x = cos and y = sin y = f () = 7 cos2 + 4 sin cos + 3 sin2 = 3 + 2 sin 2 + 2(1 + cos 2) = 5 + 2(sin 2 + cos 2) but – 2 (sin 2 + cos 2) 2
ymax = 5 + 2 2
and
[5]
[Ans. ymax = 5 + 2 2 and ymin = 5 – 2 2 ]
ymin = 5 – 2 2 ]
Find the minimum & maximum value of (sin x – cos x – 1) (sin x + cos x – 1) x R. [Ans. maximum = 4; minimum = – 1/2] 2 2 2 2 [Hint. y = (sin x – 1) – cos x = (sin x – 1) – (1 – sin x) = 2sin2x – 2 sin x Q.11
[5]
2 1 1 2 y = 2 (sin x – sin x) = 2 sin x 2 4
9 1 hence ymax = 2 = 4 when sin x = – 1 4 4 1 1 1 ymin = 2 0 = – when sin x = ] 2 2 4
Q.12
Given that log2a = s, log4 b =
[Sol.
(a, b, c > 0, c 1). Given log2a = s log2 b = 2s2 log8
c2 =
s3 1 2
s2 and
a 2 b 5 log 2 (8) = 3 . Write log2 4 as a function of 's' c s 1 c [Ans. 2s + 10s2 – 3(s3 + 1)] [5]
2
....(1) ....(2) ....(3)
2 log c s3 1 = 3 log 2 2 4 log2c = 3(s3 + 1) ....(4) to find 2 log2a + 5 log2 b – 4 log2c 2s + 10s2 – 3(s3 + 1) ] x 2 2x 8 Q.13 Find the range of the expression y = 2 , for all permissible value of x. x 4x 5 [Sol. x2y – 4xy – 5y = x2 – 2x – 8 [Ans. (– , )] (y – 1)x2 + 2x(1 – 2y) + 8 – 5y = 0 (y 1) x R hence D 0 4(1 – 2y)2 – 4(y – 1)(8 – 5y) 0 (4y2 – 4y + 1) – (13y – 8 – 5y2) 0 9y2 – 17y + 9 0 ....(1) 2 since coefficient of y > 0 and D = 289 – 324 < 0 Hence (1) is always true also if y = 1 x = 3/2 and hence range of y is (– , ) ]
[5]
zQ.14 Find whether a triangle ABC can exists with the tangents of its interior angle satisfying, tan A = x, tan B = x + 1 and tan C = 1 – x for some real value of x. Justify your assertion with adequate reasoning. [6]
[Sol.
tan A = tan A
In a triangle
(to be proved)
x + x + 1 + 1 – x = x(1 + x)(1 – x) 2 + x = x – x3 x3 = – 2 x = – 21/3 Hence tanA = x < 0 A is obtuse and tanB = x + 1 = 1 – 21/3 < 0 Hence A and B both are obtuse. Which is not possible in a triangle. Hence no such triangle can exist. ] zQ.15 Solve the equation, 5 sin x +
[Sol.
1 5 – 5 = 2 sin2x + if x (0, ). 2 sin x 2 sin 2 x
2 1 sin x 1 sin 2 x 1 sin x 1 – 5 = 2 5 2 = 2 [Ans. x 2 sin x 2 sin x 4 sin x
1 = t 2 sin x 5t – 5 = 2(t 2 – 1) 2t2 – 5t + 3 = 0 t = 1 or t = 3/2 t = 1, 2 sin2x – 2 sin x + 1 = 0 D < 0 no solution if t = 3/2, 2 sin2x – 3 sin x + 1 = 0 sin x = 1 or sin = 1/2 Let
Q.16
[Sol.
[6]
, , 5 ] 6 2 6
sin x +
x=
2
or
5 6
,
6
(2t – 3)(t – 1) = 0
5 ] x , , 6 2 6
Find the value of x, y, z satisfying the equations log2x + log4y + log4z = 2 log9x + log3y + log9z = 2 and log16x + log16y + log4z = 2. from (1) log2(x2yz) = 4 x2yz = 24 ....(1) 2 4 |||ly y zx = 3 ....(2) 2 4 z xy = 4 ....(3) (1) × (2) × (3) x4y4z4 = (2 · 3 · 4)4 xyz = 24 from (1)
x=
16 2 = 24 3
from (2)
y=
81 27 = 24 8
from (3)
z=
256 32 = ] 24 3
[6]
M ATH EM ATIC S
BANSAL CLASSES Tar g et IIT JEE 200 7 CLASS : XIII (XYZ)
D a i ly Pra c t ic e Pro b l e m s
DATE : 21-22//07/2006
TIME : 50 Min.
DPP. NO.-9
Q.11 log Given that log (2) = 0.3010....., number of digits in the number 20002000 is (A) 6601 (B) 6602 (C*) 6603 (D) 6604 2000 [Sol. Let x = 2000 log x = 2000 · log10(2000) = 2000 (log102 + 3) = 2000 (3.3010) = 6602 number of digits = 6603 Ans. ] Q.212 p&c Using only the letter from the word WILDCATS with no repetitions allowed in a codeword, number of 4 letter codewords are possible that both start and end with a consonant, are (A) 360 (B*) 900 (C) 1680 (D) 2204 [Sol. consonants WLDCTS vowels IA (6 · 5)(6 · 5) = 900 ] 1
0
Q.311 def Find ( x lnx ) dx (A*) –
1 4
(B) –
1 2
(C) – 1
(D) 1
1
[Sol.
0
let ln x = t, x = et
I = ( x lnx ) dx 0
e2t t = e t dt = t · 2
0
0
1 1 1 0 1 e 2 t dt = 0 – e2 t – = – [1 – 0] = – Ans.] Ans.] 4 4 2 4
Q.414 QE If P(x) is a polynomial with rational coefficients and roots at 0, 1, 2 and 1 3 , then the degree of P(x) is at least (A) 4 (B) 5 (C*) 6 (D) 8 [Sol. Irrational roots always occurs in conjugate pair.
if one root is 1 –
|||ly
other roots for 2 will be – 2 In all there are six roots so P(x) has degree 6. ]
3 other roots will be 1 + 3
3 = e and f (0) = 1, then f (x) equals 2
Q.514 de Given f ''(x) = cos x, f ' [Sol.
(A) sin x – (e + 1)x (B) sin x + (e + 1)x f '' (x) = cos x f ' (x) = sin x + C e=–1+C C=e+1 f (x) = – cos x + (e + 1)x + C2 1 = – 1 + C2 C2 = 2 f (x) = (e + 1)x – cos x + 2 Ans. ]
(C) (e + 1)x + cos x
(D*) (e + 1)x – cos x + 2 [ST 12th pass (11-6-2006)]
Q.619 cir A circle of radius 2 has center at (2, 0). A circle of radius 1 has center at (5, 0). A line is tangent to the two circles at points in the first quadrant. Which of the following is the y-intercept of the line? (A) 3
2 4
(B)
(C)
8 3
(D*) 2 2
C1P 2 [Hint: C P = [ST 12th pass (11-6-2006)] 1 2 C2 is the midpoint of C1 and P P(8, 0) equation of line through P y – 0 = m(x – 8) mx – y – 8m = 0 perpendicular from (2, 0) = radius i.e. 2 2 m 8m 1 m2
= 2
9m2 = 1 + m2
m=–
1 2 2
or
1 2 2
(rejected)
1
y=–
(x – 8) 2 2 for y-intercept put x = 0 y=
8 2 2
= 2 2 Ans. ]
Q.7 p/c Four persons put their hats in a pile. When they pick up hats later, each one gets some one else's hat. Number of ways this can happen, is (A) 4 (B) 6 (C) 8 (D*) 9 Q.8
Lim x x x x is x
(A) equal to 0
Q.9s&p Evaluate:
(B) equal to 1
(D) equal to – 1/2
(C) – 1.75
(D) – 2
2
2 x 8 x 12 x 35
(A) – 1.25
[Sol.
(C*) equal to – 1
(B*) – 1.5
( x 7) (x 5) 1 1 1 3 = = – 1 – = – Ans. ] ( x 7) 2 2 x 8 ( x 7)( x 5) x 8 ( x 5)
Q.10 p&c Number of regular polygons that have integral interior angle measure, is (A) 20 (B) 21 (C*) 22 (D) 23 [Sol.
exterior angle =
2 n
2 360 = 180° – n n where n is the number of sides
interior angle = –
360 must be an integer < 180° n hence n 1, 2 (think !). We have to find the number of divisors of 360 other than 1 and 2. now
now 360 = 23 · 3 2 · 51 number of divisors = 4 · 3 · 2 = 24 required number of divisors = 24 – 2 = 22 Ans. ] SUBJECTIVE: 2
dx . 2 cos 2x
Q.11173/3 Evaluate : 2
[Sol.
0
[Ans.
dx dx = 2 2 cos 2x 0 2 cos 2x
0
[3]
1 (period of is ) 2 cos 2x
I=
2 ] 3
put 2x = t 2
I=
dt dt = 2 2 cos t 0 2 cos t
0
I=2
dt 2 cos t
0
(using King)
....(1)
....(2)
Adding (1) and (2)
2I = 2
0
2
I = 8
0
4 4 cos2 t dt = 8 4 cos2 t
2
0
sec 2 t dt = 2 3 4 tan 2 t
2
0
sec 2 t dt 2
3 tan 2 t 2
put tan t = y
= 2
0
2 2 = 2 · · = 2 3 2 3 3 2
dy
y 2
Ans. ]
1 0 Q.12263/5 Show that the matrix A = 2 1 can be decomposed as a sum of a unit and a nilpotent marix. Hence 1 0 [Ans. 4014 1 ]
2007 1 0 evaluate the matrix . 2 1
[Sol.
1 0 1 0 0 0 A = 2 1 = 0 1 + 2 0 = I + B (say)
....(1)
0 0 0 0 0 0 now B2 = 2 0 2 0 = 0 0 hence B is nilpotent with order of nilpotency 2. So Bk = O for k 2 Now from (1) using binomial expansion, we get A2007 = (I + B) 2007 = I + 2007 B (remaining terms would be null matrices) 1 0 0 0 1 0 = 0 1 + 4014 0 = 4014 1 Ans. ]
M ATH EM ATIC S
BANSAL CLASSES Tar g et IIT JEE 200 7 CLASS : XIII (XYZ)
D a i ly Pra c t ic e Pro b l e m s
DATE : 24-25//07/2006
TIME : 50 Min.
DPP. NO.-10
Q.117 p&c A florist has in stock several dozens of each of the following: roses, carnations, and lilies. How many different bouquets of half dozen flowers can be made? 8! (A*) 2!· 6!
9! (B) 3!· 6!
12! (C) 6!· 6!
(D) 56
[Hint: R, C, L are beggars distribute six identical flowers in 3 bouquets (A)] Q.25 inte If f ''(x) = 10 and f ' (1) = 6 and f (1) = 4 then f (–1) is equals (A) – 4 (B) 2 (C) 8 [Hint: f''(x) = 10 f '(x) = 10x + c and f (x) = 5x2 + cx + k
(D*) 12 ]
12
x 2 2 3 Q.312 bin The coefficient of x in the expansion of , is 4 x (A) 97
(B) 98
(C*) 99
(D) 100
12
[Sol.
2 (12 r ) x 2 2 2 r x general term T = 12C × r 4 x r+1 r 12 r x 4
24 – 3r = 3
r=7
7 12 1110 9 8 1 2 3 = 99 Ans. ] coefficient = 12C7 × 5 = 120 2 4
Q.4
Which one of the following quadrants has the most solutions to the inequality, x – y < 2? (A) I quadrant (B*) II quadrant (C) III quadrant (D) I and III quadrant have same [ST 12th pass (11-6-2006)]
Q.5
If T = 3 ln(x2 + x) with > 0 and x > 0, then 2x + is equal to (A*)
[Sol.
2
(B)
4e T 3
x=
2
2
4e T 3
(C)
2
4e T 3
T = ln (x2 + x) 3
T = 3 ln (x2 + x)
(D)
2
4e T 3
eT/3 = x2 + x x2 + x – eT/3
4e T 3
2 Because x > 0 and > 0, then 2x + > 0, so only the positive roots is permissible.
Hence x +
2
=
2
4e T 3 2
2 x = 2
2
4e T 3 Ans. ]
Q.613 func The composite of two functions f and g is denoted by fog and defined by (fog)(x) = f g(x ) . When f (x) =
(A)
6x 5x and g (x) = which one of the following is equal to (fog)(x)? x 1 x2
4x x2
(B)
30 x 5x 2
(C)
x2 4x 2
(D*)
15x 2x 1
k 1 k = F(k) · Q.77 log The equation ln 1 ( k 1) (k 1)
1 1 ln 1 k 1 k ln k is true for all k wherever defined.
F(100) has the value equal to (A) 100 [Sol.
(B*)
1 101
(C) 5050
L.H.S. =
(k 1)ln k k ln (k 1) 1 1 ln(k + 1) = ln k – k (k 1) k 1 k
R.H.S. =
F(k ) [k ln k – k ln(k + 1) + ln k] k
(D)
1 100
F( k ) [(k + 1) ln k – k ln(k + 1)] k L.H.S. = R.H.S. =
(k 1)ln k k ln (k 1) F( k ) = [(k + 1) ln k – k ln(k + 1)] k (k 1) k
F (k) =
1 k 1
F(100) =
1 Ans. ] 101
Q.823 cir From a point P outside of a circle with centre at O, tangent segments PA and PB are drawn. If 1 1 1 , then the lenth of the chord AB is ( AO) 2 ( PA) 2 16 (A) 6 (B) 4 (C*) 8 [Sol.
Given
now
1 1 1 2 2 = 16 R L
(D) 9 [ST 12th pass (11-6-2006)]
1 R 2 L2 = 16 R 2L2 16d 2 = R 2L2 RL = 4d ....(1) l · d=R· L ....(2) l · d = 4d l = 4 AB = 2l = 8 Ans. ]
x 1 Q.9func If g x 1 = 3x then the value of g (3), is (A*) – [Sol.
put
15 4
(B)
x 1 = 3 x 1 9(x + 3) = x – 1
2 2
(C) 9
(D)
3 3
8x = – 10 x = – 5/4;
put x = – 5/4 in the equation
5 15 g(3) = 3 · = – Ans. ] 4 4
Q.10 s&p The value of this product of 98 numbers
1 2 1 2 1 2 2 2 2 ....... 1 1 1 , is 3 4 5 98 99 100 (A) [Sol.
1 10
(B)
Tn = 1 –
2 n2
98
P=
98 100
(C)
1 5050
(D*)
1 4950
n
=
n2 1· 2 · 3 · 4........... · 97 · 98
n
1
1 = 3 · 4 · 5 · 6........... · 98 · 99 ·100 = 99 ·100 = Ans. ] 4950 n 1 n 2
Subjective: 2 1 2
Q.11175/3 Evaluate :
0
2 1 2
[Sol.
I=
0
dx 2
(2 x 1) x x
.
[Ans.
2dx (2 x 1) ( 2x 1) 2 1 2 1 2
1
sec (2x 1) sec –1
4
0
2 – sec –1(1)
– 0 =
4
Ans.
1 1 ; dx = – 2 dt t 2t if x = 0 then t = 1
Alternatively : put 2x + 1 =
1 1 1 – then 2 1 1 = 2 2 t Hence integral becomes if x =
1
I= –
1
2
2t 1 t
2
·
t dt = – sin t 1 2 1 2t
t=
2
1 2
=
4
]
4
]
Q.12241/5 Let 1, 2 & 1, 2 be the roots of ax2 + bx + c = 0 & px2 + qx + r = 0, respectively. If the system of equations 1y + 2z = 0 & 1y + 2z = 0 has a nontrivial solution, then prove that b 2 ac . q 2 pr
[IIT ’87, 3]
[Sol.
b ; 1 2 a q 1 2 ; 1 2 r for non- trivial solution
1 2
c a r p
2 0 1 2 = 1 2 1 2 or
1 1 2 2
or
1 2 1 2 1 2 1 2
1 2 2 1 2 2 1 2 2 41 2 1 2 2 412 b 2 a2
q 2 r 2
b 2 b 2 4ac ac b 2 q 2 2 = ] 2 2 2 2 2 q q 4 pr pr b 4c b 4ac q 4 pr q 4r a a2 p 2 p
