Mathematics Today - January 2016

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Vol. XXXIV

No. 1

January 2016

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contents 79

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Maths Musing Problem Set - 157

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10 Practice Paper - Jee Main

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21 Jee Work Outs 23 Ace Your Way CBSe XI

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mathematics today | JANUARY ‘16

7

M

aths Musing was started in January 2003 issue of Mathematics Today with the suggestion of Shri Mahabir Singh. The aim of Maths Musing is to augment the chances of bright students seeking admission into IITs with additional study material. During the last 10 years there have been several changes in JEE pattern. To suit these changes Maths Musing also adopted the new pattern by changing the style of problems. Some of the Maths Musing problems have been adapted in JEE benefitting thousand of our readers. It is heartening that we receive solutions of Maths Musing problems from all over India. Maths Musing has been receiving tremendous response from candidates preparing for JEE and teachers coaching them. We do hope that students will continue to use Maths Musing to boost up their ranks in JEE Main and Advanced.

Set 157 jee main 1. Let f be twice differentiable such that f ″(x) = –f (x)

and f ′(x) = g(x). If h(x) = {f(x)}2 + {g(x)}2, where h(5) = 11, then h(10) = (a) 11 (b) 10 (c) 9 (d) 8 2. The real values of a, the point (–2a, a + 1) will be an interior point of the smaller region bounded by the circle x2 + y2 = 4 and the parabola y2 = 4x is (a) [–1, 0] (b) (−1, − 5 + 2 6 ) (c) [0, 1] (d) [−1, − 5 − 2 6 ] 3. A plane whose equation is 2x – y + 3z + 5 = 0 is rotated through 90° about its line of intersection with the plane 5x – 4y – 2z + 1 = 0. Find the equation of the plane in the new position. (a) 27x – 24y – 26z = 13 (b) 27x + 24y + 26z = 13 (c) 27x – 24y + 26z = 13 (d) none of these 4. The coefficient of x50 in the series S=

101

∑ rx

r −1

r =1

101 − r

(1 + x )

is

100  101 102  103  (a)  50  (b)  50  (c)  50  (d)          50  5. A line through A(–5, –4) meets the lines x + 3y + 2 = 0, 2x + y + 4 = 0 and x – y – 5 = 0 at the points B, C, D respectively. 2

2

2

 15   10   6  + = If  , then the equation  AB   AC   AD  of the line is (a) 2x + 3y + 11 = 0 (b) 2x + 3y + 22 = 0 (c) x + 3y + 11 = 0 (d) 2x + y + 22 = 0 jee advanced 6. Two integers x and y are chosen (without random) at random from the set {x : 0 ≤ x ≤ 10, x is an integer} then the probability for|x – y| ≤ 5 is 80 91 85 90 (a) (b) (c) (d) 121 121 121 121 8


comprehension Let n ∈N. The A.M., G.M., H.M. and R.M.S. (root-mean square) of the n numbers n + 1, n + 2, n + 3, ...., n + n are An, Gn, Hn, Rn respectively. Then G 7. lim n = n→ ∞ n 1 2 3 4 (a) (b) (c) (d) e e e e Rn = 8. lim n→ ∞ n 5 7 (a) 3 (b) (c) (d) 3 3 3 integer match

9. If

n n

∑ ∑ nC j jCi = mn − 1, then the value of m2 is

i =0 j =1

matching list 10.

Column-I Column-II log e a log e b log e c 37 P. If 1. = = , then b−c c−a a −b 29 ab+c⋅ bc+a⋅ca+b = Q. In an equilateral triangle, 3 coins of 2. 2 radii 1 are kept so that they touch each other and also the sides of the triangle. The area of the triangle is R. The remainder when 82n – (62)2n+1 is 3. 1 divided by 9 is S. If ∫e–2x(3cos5x – 4sin5x)dx 4. 6 + 4 3 = e–2x(a cos5x + b sin5x) + c, then a+b= P Q R S (a) 3 4 1 2 (b) 3 4 2 1 (c) 4 3 1 2 (d) 4 1 2 3 See Solution set of Maths Musing 156 on page no. 78

1.

sin2 q

cos2 q

1 + 4 sin 4q

sin2 q

1 + cos2 q

4 sin 4q

1 + sin2 q

cos2 q

4 sin 4q

Let f(q) =

4 And the period of f(q) is T, then the value of × T is p (a) 2 (b) 3 (c) 4 (d) none of these 2. Let A = {x : x is a prime factor of 240} B = {x : x is the sum of any two prime factor of 240}. Then A ∪ B is (a) {2, 3, 5} (b) {2, 3, 5, 7, 8} (c) {2, 3, 5, 7} (d) none of these 3.

If x1 and x2 are the roots of the equation x +x 4x – 3(2x + 3) + 128 = 0 , then the value of 1 2 is | x1 − x2 | (a) 5 (b) 6 (c) 7 (d) 8 4. The number of values of ‘t’ for which there exist a non-zero 3 × 3 matrix B such that B′ = tB is (a) 0 (b) 1 (c) 2 (d) infinite 5. Let Tn be the number of possible quadrilaterals formed by joining vertices of an n sided regular polygon. If Tn + 1 – Tn = 20, then the number of triangles formed by joining the vertices of the polygon is (a) 15 (b) 16 (c) 20 (d) none of these 6.

Coefficient of x2 in the expansion of

E = (x

2 /3

(a) 20 (c) –18

+ 4x

1/3

 + 4)  +  1 / 3 2 / 3 1 / 3  x −1 x + x +1 5

1

1

(b) 24 (d) none of these

−9

is

7. If t1, t2, t3, …….., tn, …… are in A.P. such that t4 – t7 + t10 = p, then sum of the first 13 terms of the A.P. is (a) 10p (b) 12p (c) 13p (d) 15p 8. If (20)19 + 2(21) (20)18 + 3(21)2 (20)17 + ……. + 20(21)19 = m(20)19, then the value of m + 41 is (a) 20 (b) 21 (c) 22 (d) none of these 2

9.

If lim

(a) 1 (c) 4

x →0

e x − cos x x2

= b, then value of 2b is

 ex − e−x  10. If y = ln cos  tan −1  2   Statement I : y′(0) = 0 ex − e−x Statement II : y′(x) = 1 + x2 (a) Statement I is True, Statement II is True; Statement II is a correct explanation for Statement I. (b) Statement I is True, Statement II is True; Statement II is not a correct explanation for Statement I. (c) Statement I is True, Statement II is False. (d) Statement I is False, Statement II is True. 11. The length of the rectangle of the greatest area, that can be inscribed in the ellipse x2 + 2y2 = 8, are given by (a) 2√6 (b) 4√6 (c) 3√6 (d) none of these 12. If the primitive of


1

is f(x) – ln | g(x) | + C (e − 1)2 then f(x) and g(x) are respectively (a) (1 – ex) and 1 – e–x (b) (1 – ex)–1 and 1 – ex (c) (1 – ex)–1 and 1 – e–x (d) none of these

Contributed by : Sanjay Singh, Mathematics Classes, Chandigarh

10


x

∞

13. If ∫ 0

dx

(x 2 + 4)(x 2 + 9)

(a) ± 7 (c) ± 9

p

, then the value of k is = 2 k + 11 (b) ± 8 (d) none of these

dy 1 = is given by dx 2 x − y 2 x = cg(y) + ay2 + by + d where c, a, b, d are constants then a + b + d is 3 1 (a) (b) 4 4 5 (c) (d) none of these 4 15. If a, b, c are in A.P., a, x, b are in G.P. and b, y, c are in G.P., then the locus of the point (x, y) is (a) a straight line (b) a circle (c) an ellipse (d) a hyperbola 14. The solution of

16. If the circles x2 + y2 + 2ax + cy + a = 0 and x2 + y2 – 3ax + dy – 1 = 0 intersect at two distinct points P and Q, then the line 5x + by – a = 0 passes through P and Q for (a) infinitely many values of a (b) exactly two values of a (c) exactly one value of a (d) no value of a 17. AB is a chord of the parabola y2 = 4ax with the end A at the vertex of the given parabola. BC is drawn perpendicular to AB meeting the axis of the parabola at C. The projection of BC on the axis is 1 (a) (latus-rectum) (b) semi latus-rectum 4 (c) latus-rectum (d) twice latus-rectum 18. Consider the two curves C1 : y2 = 4x, C2 : x2 + y2 – 6x + 1 = 0. Statement I: C1 and C2 touch each other exactly at two points. Statement II: Equation of the tangent at (1, 2) to C 1 and C 2 both is x – y + 1 = 0 and at (1, –2) is x + y + 1 = 0. (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I. (b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I. (c) Statement I is true, Statement II is false. (d) Statement I is false, Statement II is true. 12


19. Let L1 be the line 2x – 2y + 3z – 2 = 0 = x – y + z + 1 and L2 be the line x + 2y – z – 3 = 0 = 3x – y + 2z – 1. The vector along the normal of the plane containing the lines L1 and L2 is (a) 7 i^ − 7 ^j + 8 k^ (b) 7 i^ + 7 ^j + 8 k^ (c) 7 i^ − 7 ^j + 7 k^ (d) none of these     20. If a, b, γ and δ are unit vectors, then             | a − b |2 + | b − γ |2 + | γ − δ |2 + | δ − a |2 + | γ − a |2 + | b − δ |2 does not exceed (a) 4 (c) 8

(b) 12 (d) 16

21. The distance between the line and the plane x + 5y + z = 5 is 3 10 (a) (b) 10 3 10 10 (c) (d) 9 3 3 22. If

18

18

i =1

i =1

x −2 y +2 z −3 = = 1 −1 4

∑ (xi − 8) = 9 and ∑ (xi − 8)2 = 45 , then the

standard deviation of x1, x2, ….., x18 is 4 9 (a) (b) 9 4 3 (c) (d) none of these 2 23. Let A = {x, y, z, u} and B = {a, b}. A function f : A → B is selected randomly. Probability that function is an onto function is 1 5 (a) (b) 8 8 3 (c) 7 (d) 8 8 24. If letters of the word “ASSASSIN” are written down at random in a row, the probability that no two S's occur together is 1 1 (a) (b) 17 14 1 1 (c) (d) 28 35 25. If cot a equals the integral solution of the inequality 4t2 – 16t + 15 < 0 and sinb equals to the slope of the bisector of the second quadrant, then sin (a + b) sin (a – b) is equal to

3 5 2

(a) − (c)

5

26. If sin a2

+ 1 is (a) 10

(b) −

Solutions

4 5 1. (a) : f (q) =

(d) 3 −1  a 

 5  + cosec

(b) 9

(d) 26

27. The value of sin–1 (sin 10) is (a) 10 (b) 3p – 10 (c) 10 – 3p (d) none of these 28. Statement I : ~(p ↔ ~ q) is equivalent to p ↔ q. Statement II : ~(p ↔ ~ q) is a tautology. (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I. (b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I. (c) Statement I is true, Statement II is false. (d) Statement I is false, Statement II is true.  tan −1 x  29. Statement I : lim   = 0, where [x] x  x →0  represents greatest integer ≤ x. −1 Statement II : tan x < 1 for all x ≠ 0. x (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I. (b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I. (c) Statement I is true, Statement II is false. (d) Statement I is false, Statement II is true. 30. The natural numbers are divided into rows as follows: 1 2 3 4 5 6 7 8 9 ……………………………….. Statement I : Sum of the numbers in the 10th row is a number which can be written as sum of two cubes in two different ways. Statement II : Sum of the numbers in the rth row is 1 3 (r + (r + 1)3) if r ≥ 2. 3 (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I. (b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I. (c) Statement I is true, Statement II is false. (d) Statement I is false, Statement II is true. 14


0 1

1 0

1 + sin2 q cos2 q 4 sin 4q

p  4  = 2 , then the value of

−1  5 

(c) 11

−1 −1

(R1 → R1 – R3; R2 → R2 – R3) 0 −1

=

0 1

1 0

1 + sin2 q + 4 sin 4q cos2 q 4 sin 4q

(C1 → C1 + C3 )

= – (cos2q + 1 + sin2q + 4 sin4q) = – 2(1 + 2 sin4q) 2p p = which is periodic function with period 4 2 4 4 p Hence the value of × T = . = 2 p p 2 2. (b) : 240 = 24 × 3 × 5 So, A = {2, 3, 5}, B = {5, 7, 8} Clearly, A ∪ B = {2, 3, 5, 7, 8} 3. (c) : Put 2x = y, given equation becomes y2 – 3(8y) + 128 = 0 ⇒ y2 – 24y + 128 = 0 ⇒ (y – 8) (y – 16) = 0 ⇒ y = 16, 8 ⇒ 2x = 16, 8 ⇒ x = 4, 3 x +x 4+3 Hence 1 2 = =7 | x1 − x2 | | 4 − 3 | 4. (c) : We have, B = (B′)′ = (tB)′ = tB′ = t2B As B ≠ 0, t2 = 1 ⇒ t = ± 1 5. (c) : Tn = Number of ways of choosing four vertices out of n = nC4 \ 20 = Tn + 1 – Tn = n + 1C4 – nC4 = n C4 + n C3 – n C4 = n C3 [

nC r–1

n(n − 1) (n − 2) 3. 2. 1 ⇒ n(n – 1) (n – 2) = 120 = 6.5.4

+ nCr =

n + 1C ] r

\ 20 =

⇒ n=6 \ Required number of triangles = nC3 = 6C3 = 20. 6. (c) : E =

= (2 + x

((x1/3

+

2)2)5

 x 2/3 + x1/3 + 1 + x1/3 − 1    x −1  

 x 2/3 + 2 x1/3  )   x −1  

1/3 10

−9

−9

(x − 1)9

= (2 + x1/3 )10

x1/3)x–3(x

1)9

= (2 + – (x1/3 )9 (x1/3 + 2)9 Coefficient of x2 = 2[coefficient of x5 in the expansion of x–3(x – 1)9] = 2(–1)9 [coefficient of x8 in (1 – x)9] = 2(–1) 9C8 (–1)8 = –18 7. (c) : Let d be the common difference of the A.P., then p = t4 – t7 + t10 = t4 + (t10 – t7) = t1 + 3d + 3d = t1 + 6d 13 Now, S13 = [2t1 + (13 – 1)d] = 13p 2 8. (b) : Dividing the given equation by (20)19, we get m = 1 + 2x + 3x2 + …… + 20x19, 21 where x = . 20 d  x(x 20 − 1)  d   m = [x + x 2 + ...... + x 20 ] = dx  x − 1  dx =

(x − 1)(21x

20

− 1) − x(x

20

− 1)

2

(x − 1)

 21  But 20x21 – 21x20 = 20    20  1 \ m= = 400 2  21  − 1  20 

21

=

20 x

21

− 21x

20

+1

 21  − 21    20 

=0

2

x →0

e x − 1 + 1 − cos x x2

 x2  e − 1 2 sin2 x / 2  = 1+ 1 = 3 = lim  + 2  x →0  x 2 2 2 ( / ) 4 x 2   Hence 2b = 2(3/2) = 3  ex − e−x  sin  tan −1  2   10. (c) : y ′(x ) = −  ex − e−x  cos  tan −1  2   ex + e−x 1 × × 2 2  ex − e−x  1+   2   x x −  e −e  2 x −x = − tan  tan −1  × x − x 2 × (e + e ) 2 e e ( + )   16


11. (a) : Any point on the ellipse (2 2 cosq, 2 sinq). A = Area of the inscribed rectangle

x2 y2 + = 1 is 8 4

= 4(2 2 cosq) (2 sinq) = 8 2 sin 2q dA = 16 2 cos 2q = 0 ⇒ q = p / 4 dq d2 A p Also = − 32 2 sin 2q < 0 for q = . 2 4 dq Hence, the inscribed rectangle is of largest area if the sides are p p 4 2 cos and 4 sin   i.e., 4 and 2 2 . 4 4 Hence the length of diagonal = 42 + (2 2 )2 = 16 + 8 = 24 = 2 6

(x − 1) 20

ex − e−x

ex + e−x y′(0) = 0

2

Hence, m + 41 = 441 = 21 9. (b) : b = lim

=−

12. (c) :

1

∫ (e x − 1)2

dx = ∫

x

e x dx x

2

e (e − 1)

=∫

dt t (t − 1)2

 1 1 1 1  1 1 = − + − dt   dt ∫ t − 1  t − 1 t   (t − 1)2 t − 1 t  1 = – ln | t – 1 | + ln | t | + C 1− t = (1 – ex)–1 – ln | 1 – e–x | + C x –1 Hence, f(x) = (1 – e ) and g(x) = 1 – e–x

=∫

13. (a) :

∞

dx

∫ (x 2 + 4)(x 2 + 9) 0

=

∞

1  1 1  dx − 2  ∫ 2 5  x + 4 x + 9  0 ∞

x 1 x 1 1 =  tan −1 − tan −1  5 2 2 3 3 0

1 p p p =  −  = , so k 2 + 11 = 60 5  4 6  60 ⇒ k2 = 49 ⇒ k = ± 7 dx = 2x – y2 dy This is a linear equation in x. 14. (c) : We have,

The integrating factor is e − ∫ 2dy = e −2 y d So (xe −2 y ) = − y 2e −2 y dy

Integrating, we have − y 2e −2 y − ∫ ye −2 y dy + Const −2 y 2 −2 y ye −2 y 1 −2 y = + − ∫ e dy + Const e 2 2 2 2 y −2 y y −2 y e −2 y = + e + +C e 2 2 4 y2 y 1 \ x = Ce 2 y + + + 2 2 4 On comparing with the above equation, we get xe −2 y =

1 1 1 a = ,b = ,d = 2 2 4

5 Hence a + b + d = . 4 15. (b) : We have 2b = a + c, x2 = ab, y2 = bc so that x2 + y2 = b(a + c) = 2b2 , which is a circle. 16. (d) : Equation of the line passing through the points P and Q is x2 + y2 + 2ax + cy + a – (x2 + y2 – 3ax + dy – 1) = 0 ⇒ 5ax + (c – d)y + a + 1 = 0 which coincides with 5x + by – a = 0 5a c − d a + 1 if = = −a 5 b or if a2 + a + 1 = 0 which does not give any real value of a. 17. (c) : Draw BD perpendicular to the axis of the parabola. Let the coordinates of B be (x, y) then slope y of AB is given by tanq = x Projection of BC on the axis of the parabola is DC = BD tanq  y  y 2 4ax = = 4a = y = x x x Y B(x, y)

A



 D

C

X

18. (a) : Solving for the points of intersection, we have x2 + 4x – 6x + 1 = 0 ⇒ (x – 1)2 = 0 ⇒ x=1 ⇒ y=±2 18


Thus the two curves meet at (1, 2) and (1, –2). Tangent at (1, 2) to y2 = 4x is y(2) = 2(x + 1) ⇒ x – y + 1 = 0 Tangent at (1, 2) to the circle C2 is x + 2y – 3(x + 1) + 1 = 0 or x – y + 1 = 0 same as the tangent to the curve C1. Similarly the tangent at the point (1, –2) to the two curves is x + y + 1 = 0 ⇒ Statement II is true and hence statement I is also true. 19. (a) : The plane through L1 is 2x – 2y + 3z – 2 + l(x – y + z + 1) = 0 and the plane through L2 is x + 2y – z – 3 + m(3x – y + 2z – 1) = 0 Two planes are same 2+ l 2+ l 3+ l 2− l So = = = 1 + 3m m − 2 2m − 1 3 + m 3 ⇒ m=– and l = 5 and the equation of the plane 2 is 7x – 7y + 8z + 3 = 0. ^

^

^

Hence the vector along the normal is 7 i − 7 j + 8 k . 20. (d) : We have         0 ≤ | a + b + γ + δ |2 = | a |2 + | b |2 + | γ |2 + | δ |2             + 2(a ⋅b + b ⋅ γ + γ ⋅ a + a ⋅ δ + γ ⋅ δ + b ⋅ δ)             = 4 + 2(a ⋅b + b ⋅ γ + γ ⋅ a + a ⋅ δ + γ ⋅ δ + b ⋅ δ)             So, a ⋅b + b ⋅ γ + γ ⋅ a + a ⋅ δ + γ ⋅ δ + b ⋅ δ ≥ −2         Now | a − b |2 + | b − γ |2 + | γ − δ |2 + | δ − a |2     + | γ − a |2 + | b − δ |2     = 3(| a |2 + | b |2 + | γ |2 + | δ |2 )             −2(a ⋅ b + γ ⋅ δ + δ ⋅ a + a ⋅ γ + b ⋅ δ + b ⋅ γ ) ≤ 3(4) − 2(−2) = 16 21. (d) : Direction ratios of the line are 1, –1, 4 and that of the normal to the plane are 1, 5, 1. Since 1 × 1 + (–1) (5) + 4 × 1 = 0 ... Normal to the plane is perpendicular to the line and thus the line is parallel to the plane and the distance

between them is the distance of any point, let (2, –2, 3) on the line from the plane and is equal to 2(1) + (−2)(5) + 3 × 1 − 5 1 + 25 + 1 22. (c) : Let di = xi – 8 but 1 1  s 2x = s 2d = ∑ di2 −  ∑ di   18  18 =

=

10 27

=

10 3 3

2

2

1 5 1 9 9 × 45 −   = − =  18  18 2 4 4

Therefore sx = 3/2

23. (c) : Total number of functions from A to B is 24 = 16. There are exactly two functions which are not onto viz. one in which all the elements of A are mapped to a and other in which all the elements of A are mapped to b. Thus, there are 14 onto functions. 14 7 Therefore probability of required event is = 16 8 24. (b) : The number of ways of permuting the letters of the word “ASSASSIN” is 8! 8 × 7 × 6 × 5 = = 840 2!4! 2 To find the number of ways is which two S are not together, We first arrange A, A, I, N. This can be done in 4! 2! = 12 ways.

If in the following figure X positions of show one such arrangement, then 4S’s can be arranged at four boxes out of the five boxes. × × × × Thus, corresponding to each arrangement of A, A, I, N, there are 5C4 = 5 ways of arranging S’s. \ The number of favourable ways = (12) × (5) = 60 60 1 = Thus, required probability = 840 14 25. (b) : We have, 4t2 – 16t + 15 < 0 3 5 ⇒
a 4 p ⇒ sin −1   + sin −1   = 5 5 2 16 p a ⇒ sin −1   + cos −1 1 − = 5 25 2 a 3 p ⇒ sin −1   + cos −1   = 5 5 2 p ⇒ a = 3 as sin–1 a + cos–1 a = "a 2 2 Hence a + 1 = 10 27. (b) : Let y = sin–1 (sin 10) ⇒ siny = sin 10 = sin (3p + (10 – 3p)) = – sin (10 – 3p) = sin (3p – 10) ⇒ y = 3p – 10 28. (c) : Table for basic logical connectives. p

q

~q

T

T

F

F

T

T

T

F

T

T

F

F

F

T

F

T

F

F

F

F

T

F

T

T

p ↔ ~ q ~ (p ↔ ~ q) (p ↔ q)

Note that ~ (p ↔ ~ q) is not a tautology. Statement II is false. And ~ (p ↔ ~ q) is equivalent to (p ↔ q) Thus, statement I is true. 29. (c) : Let f(x) = tan–1 x – x f ′( x ) =

1

−1 =

−x2

<0 1 + x2 1 + x2 Hence f is a decreasing function, so for x > 0, −1 tan–1x < x ⇒ tan x < 1 and for x < 0, f(x) > f(0) = 0 x tan −1 x ⇒ tan −1 x > x ⇒ < 1 (x < 0) x  tan −1 x   tan −1 x  = Thus lim  0 and lim =0   x  x  x →0+  x →0−  30. (c) : The rth row contains (2r – 1) numbers and its last term is r2. When read from right to left, the rth row forms an A.P. with first term r2 and common difference – 1. \ Sum of the terms in the rth row 2r − 1 = [2(r 2 ) + (2r − 1− 1)(−1)] 2 = (2r – 1) (r2 – r + 1) = r3 + (r – 1)3 Sum of the terms in the 10th row = 103 + 93 = 1729 = 123 + 13 nn mathematics today | JANUARY ‘16

19

PRESS RELEASE

VITEEE-2016

B.Tech Engineering Entrance Exam

Application forms sales begins

VIT University Chancellor Dr. G. Viswanathan is seen inaugurating the sale of the VITEEE-2016 application forms, to some of the aspiring students for the tests, in April 2016 at the Vellore Head Post Office on Friday. Seen others in the picture are Superintendent of Posts A. Natarajan, Deputy Superintendent R. Mahendran, VIT University’s VC Dr. Anand A Samuel, Pro-VC Dr. S. Narayan, VPs Sankar Viswanathan, Sekar Viswanathan and G.V. Selvam, Director UG Admissions, K. Manivannan, PRO- (Posts) Kathir Ahmed and Marketing Executive S. Selvakumar.

T

he sale of application forms for the VIT University Entrance Examinations (VITEEE-2016) to be held in April 2016, for B.Tech courses various streams, began in all the 92 Head Post Offices, with the VIT University Chancellor Dr. G. Viswanathan inaugurating it at the Head Post Office, here, on Friday. It is scheduled that the entrance examinations for the B.Tech offered in the VIT University, Vellore and Chennai Campuses, will be held from April 6th to April 17th 2016. This Computer Based Test (CBT) is held in 118 cities including Dubai, Kuwait and Muscat. The University offers courses in Vellore Campus - Bio-Medical Engineering, Biotechnology, Computer Science and Engineering (Specialisation in Bioinformatics), Civil Engineering, Chemical Engineering, Computer Science and Engineering, Electronics and Communication Engineering, Electrical and Electronics Engineering, Electronics and Instrumentation Engineering, Information Technology, Mechanical Engineering, Mechanical Engineering (Specialisation in Automotive Engineering), 20


Mechanical Engineering (Specialisation in Energy Engineering), Production and Industrial Engineering and in Chennai Campus – B.Tech in Civil Engineering, Computer Science and Engineering, Electronics and Communication Engineering, Electrical and Electronics Engineering and Mechanical Engineering. The entrance examination application forms from the Head Post Offices, it can be obtained by sending a Demand Draft for Rs.990/- drawn in favour of VIT University, payable at Vellore to the Director – UG Admissions or by cash payment at selected post offices across the country. Issuing of online and offline application has commenced from November 27th 2015. Candidates can also apply online at www.vit.ac.in (online applicants need to pay Rs.960/- only). The last date for applying is 29th February, 2016. VIT University has been consistently been ranked among the premier engineering institutions of the country by India Today. VIT has an impressive track record of placement. For the 8th year in a row, VIT has achieved the top slot in placements. Visit: www.vit.ac.in for further details.

nn

1. Let (x0, y0) be the solution of the following equations (2x)ln2 = (3y)ln3 and 3lnx = 2lny. Then x0 is (a) 1/6 (b) 1/3 (c) 1/2 (d) 6 2. Let S(K) = 1 + 3 + 5 + ... + (2K – 1) = 3 + K2. Then which of the following is true? (a) S(1) is correct (b) Principle of mathematical induction can be used to prove the formula (c) S(K) ⇒ / S(K + 1) (d) S(K) ⇒ S(K + 1) The real part of (1 – cosq + 2isinq)–1 is 1 1 (a) (b) 5 − 3 cos q 3 + 5 cos q 1 1 (c) (d) 3 − 5 cos q 5 + 3 cos q 3.

5.

If sr

equal to (a) 0 (c) (a + b + g)6

g r, then the value of

s0 s1 s2

Sita Sita Sita Sita

is is is is

not beautiful or Sita is not clever beautiful and she is clever not beautiful or Sita is clever beautiful and Sita is not clever

7. Ten persons, amongst whom are A, B and C to speak at a function. The number of ways in which it can be done if A wants to speak before B and B wants to speak before C is 10 ! (a) (b) 3! 7! 6 (c) 10P3 · 7! (d) none of these 8.

11

1   If the coefficient of x7 in  ax 2 +   bx 

is equal to

11

4. a, g, h are arithmetic mean, geometric mean and harmonic mean between two positive numbers x and y respectively. Then identify the correct statement among the following (a) h is the harmonic mean between a and g (b) No such relation exists between a, g and h (c) g is the geometric mean between a and h (d) A is the arithmetic mean between g and h = ar + br +

(a) (b) (c) (d)

s1 s2 s3

s2 s3 is s4

(b) (a – b)(b – g)(g – a) (d) (a – b)2(b – g)2(g – a)2

6. p : Sita is beautiful; q = Sita is clever. When ~ p ∨ q is written in verbal form it becomes

1   the coefficient of x–7 in  ax −  , then ab =  bx 2  (a) 1 (b) 1/2 (c) 2 (d) 3 9. If the probability density function of a random variable X is f(x) = x/2 in 0 ≤ x ≤ 2, then P(X > 1.5 | x > 1) is equal to (a) 7/16 (b) 3/4 (c) 7/12 (d) 21/64 10. The product of perpendiculars drawn from the origin to the lines represented by the equation, ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, will be bc ab (a) (b) 2 a − b 2 + 4h 2 a 2 − b 2 + 4h 2 ca c (c) (d) (a2 + b2 ) + 4h2 (a − b)2 + 4h2 11. If two distinct chords, drawn from the point (p, q) on the circle x2 + y2 = px + qy, (where pq ≠ 0) are bisected by the x-axis, then

By : Vidyalankar Institute, Pearl Centre, Senapati Bapat Marg, Dadar (W), Mumbai - 28. Tel.: (022) 24306367 mathematics today | JANUARY ‘16

21

(a) p2 = q2 (c) p2 < 8q2

(b) p2 = 8q2 (d) p2 > 8q2

(c)

12. A circle C1 of radius 2 touches both x-axis and y-axis. Another circle C2 whose radius is greater than 2 touches circle C1 and both the axes. Then the radius of circle C2 is (a) 6 − 4 2 (b) 6 + 4 2 (d) 6 + 4 3

(c) 6 − 4 3

sin a − cos a , then sina + cosa and 13. If tan q = sin a + cos a sina – cosa must be equal to (a) 2 cos q, 2 sin q (b) 2 sin q, 2 cos q (c)

2 sin q,

(d)

2 sin q

2 cos q,

2 cos q

14. A pair of straight lines drawn through the origin form with the line 2x + 3y = 6 an isosceles right angled triangle, then the area of the triangle thus formed is 36 12 (a) ∆ = (b) ∆ = 13 17 13 (c) ∆ = (d) none of these 5 15. Point D, E are taken on the side BC of a triangle ABC such that BD = DE = EC. If ∠BAD = x, ∠DAE = y, sin(x + y )sin( y + z ) ∠EAC = z, then the value of = sin x sin z (a) 1 (b) 2 (c) 4 (d) none of these tan–1x

16. If equal to (a) xyz

+

tan–1y

(b) 0

+

tan–1z

= p, then x + y + z is

(c) 1

(d) 2xyz

17. From the bottom of a pole of height h, the angle of elevation of top of a tower is a and the pole subtends angle b at the top of the tower. The height of the tower is h cot(a − b) h tan(a − b) (a) (b) cot(a − b) − cot a tan(a − b) − tan a (c)

cot(a − b) cot(a − b) − cot a

(d) none of these

18. If f is an even function defined on the interval (–5, 5), then four real values of x satisfying the equation  x +1  are f (x ) = f   x + 2  −3 − 5 −3 + 5 3 − 5 3 + 5 (a) , , , 2 2 2 2 (b) 22

−5 + 3 −3 + 5 3 + 5 3 − 5 , , , 2 2 2 2 mathematics today | JANUARY ‘16

3 − 5 3 + 5 −3 − 5 5 + 3 , , , 2 2 2 2

(d) −3 − 5 , − 3 + 5 , 3 − 5 , 3 + 5 x

19. The value of lim (a) ∞

x → p /2

(b) p/2

∫ tdt

p /2

sin(2 x − p) (c) p/4

is (d) p/8

20. If f ′(x) = g(x) and g′(x) = –f (x) for all x and f (2) = 4 = f ′(2) then f 2(4) + g2(4) is (a) 8 (b) 16 (c) 32 (d) 64 21. The value of p for which the function  (4 x − 1)3 , x ≠0   x2   x is continuous at f (x ) =  sin log e 1 +   p 3    x =0 12(log e 4)3 ,  x = 0, is (a) 1 (b) 2 (c) 3 (d) none of these 22. Maximize z = 3x + y subject to x + y ≤ 6, y ≤ 4, x ≥ 1, x ≥ 0 and y ≥ 0. (a) 20 (b) 18 (c) 10 (d) 15 x cos t

23. Let f (x ) = ∫ (a) (b) (c) (d)

dt , x > 0 then f (x) has t maxima when n = –2, –4, –6, .... maxima when n = –1, –3, –5, .... maxima when n = 1, 3, 5, ... none of these 0

24. Let P, Q, R and S be the points on the plane with ^

^

position vectors −2 i^ − ^j, 4 i^ , 3 i^ + 3 ^j and −3 i + 2 j respectively. The quadrilateral PQRS must be a (a) parallelogram, which is neither a rhombus nor a rectangle (b) square (c) rectangle, but not a square (d) rhombus, but not a square

25. Let A = {2, 3, 4, 5} and let R = {(2, 2), (3, 3), (4, 4), (5, 5), (2, 3), (3, 2), (3, 5), (5, 3)} be a relation in A. Then R is (a) reflexive and transitive (b) reflexive and symmetric (c) transitive and symmetric (d) equivalence

Contd. on Page No. 30

Series 8

YOUR WAY CBSE XI

XII

STATISTICS AND PROBABILITY YO

BSE

AYC

W UR

statistics

IntroductIon Statistics deals with data collected for specific purposes. Statistical analysis involves the process of collecting data, analyzing data and then summarizing it into numerical form. z Measures of central tendency are not sufficient to give complete information about a given data. Variability is another factor which is required to be z studied under statistics. Like ‘measures of central tendency’ we want to have a single number to describe variability. This single number is called a ‘measure of dispersion’. z

measures of dIspersIon It measures the degree of scatteredness of the observations in the given data around the central value. There are following measures of dispersion: (i) Range (ii) Quartile deviation (iii) Mean deviation (iv) Standard deviation range It is the difference between two extreme observations of the given distribution (arranged in ascending or descending order). i.e., Range = Maximum value – Minimum value mean devIatIon Mean deviation of a distribution is the arithmetic mean of the absolute deviations of the terms of the distribution from its statistical mean (A.M., median or mode). The sum of deviations from mean (x–) is zero.

Mean deviation for ungrouped data Let x1, x2, x3, ..., xn be the given n observations. Let x– be the mean and M be the median. Then, n

MD (x ) =

∑ | xi − x | i =1

n

n

MD (M ) =

∑ | xi − M | i =1

n Mean deviation for grouped data (1) For discrete frequency distribution : MD (x ) =

1 n f i | xi − x | N∑ i =1

MD (M ) =

n 1 n where N = f | x − M |, ∑ fi i i N∑ i =1 i =1

(2) For continuous frequency distribution : MD (x ) =

1 n f i | xi − x | N∑ i =1

MD (M ) =

1 n f i | xi − M | N∑ i =1

Here, xi’s are the mid-points of the corresponding classes Shortcut Method for finding mean deviation (1) About mean – For this, find the mean by stepdeviation method, i.e., n

x =a+

∑ f i di i =1

N assumed mean.

× h, where di =

xi − a and a is the h


23

And then applying the formula, MD (x ) =

n

1 f i | xi − x | N∑ i =1

(2) About median – For this, find the median by the formula,

For a discrete frequency distribution For continuous frequency distribution

N −C M =l+ 2 ×h f

1 n f i | xi − M | N∑ i =1

Variance and standard deViation The variance of a variate is the mean of the squares of all deviations of the values of the variate x from the arithmetic mean of the observations and is denoted by s2 or var (x). Thus, if x1, x2, ....., xn be n given values of the variate and x– be their mean, then 1 n Variance, s 2 = ∑ (x i − x )2 n i =1 The dispersion about mean of a set of observations is expressed as the positive square-root of the variance and is called the standard deviation which is denoted by s and S.D. is given by

24

s=

1 n f i (xi − x )2 or N∑ i =1

s=

1 n (x i − x )2 n i∑ =1


2

Shortcut method to find variance and standard deviation 2  n n   2 2 1 2 1 s =h z ∑ fi yi −  N ∑ fi yi    N i =1  i =1    z

Limitations of mean deviation (1) The mean deviation about median calculated for such series (where the degree of variability is very high) can not be fully relied. (2) The mean deviation about the mean is not very scientific since the sum of the deviations from the mean (if we ignored the minus signs) is more than the sum of the deviations from median. (3) The mean deviation is calculated on the basis of absolute values of the deviations and therefore, cannot be subjected to further algebraic treatment.

For ungrouped data

1 n f i (x i − x )2 N∑ i =1

n  n  1 N ∑ f i xi2 −  ∑ f i xi  s= N  i =1  i =1

where l, C, f and h are the lower limit, the cumulative frequency of the class just preceding the median class, the frequency and the width of the median class respectively. And then MD (M ) =

s=

s=

n  n  h N ∑ f i y i2 −  ∑ f i y i  N  i =1  i =1

2

where h and N are the width of the class-intervals and x −a , the sum of frequencies respectively and y i = i h a = assumed mean analysis of frequency distributions The measure of variability which is independent of units is called coefficient of variation and denoted as C.V. The coefficient of variation is defined as s C.V. = × 100, x ≠ 0, x Where s and x are the standard deviation and mean of the data respectively. For two series with equal means, the series with greater standard deviation (or variance) is called more variable or dispersed than the other. Also, the series with lesser value of standard deviation (or variance) is said to be more consistent than the other. The series having greater C.V. is said to be more variable than the other. The series having lesser C.V. is said to be more consistent than the other. Note: (1) The mean of two groups of observations taken together is n x +n x x= 1 1 2 2 n1 + n2 (2) The variance of two groups of observations taken together is 1  2 2  nn  n1s1 + n2s 22 +  1 2  ( x1 − x 2 )  s2 =  n1 + n2   n1 + n2   where, x1, x2 are the respective means of the two groups having n1 and n2 observations.

probability defInItIon Probability is a concept which numerically measures the degree of uncertainty and, therefore, of certainty of the occurrence of an event. experIment An operation which can produce some well-defined outcomes is known as an experiment. random experIment An experiment whose outcome cannot be predicted with certainty is called a random experiment. An experiment whose outcome can be foretold before hand is not a random experiment. outcomes and sample space Outcome : A possible result of a random experiment is called its outcome. z Sample space : The set of all possible outcomes of a random experiment is called the sample space for that experiment. It is usually denoted by S. Sample point : Each element of the sample space is z called a sample point. z

event A subset of the sample space S is called an event. z Occurrence of an event : For a random experiment, let E be an event and let E = {a, b, c}. If the outcome of the experiment is a or b or c, then we say that event E has occurred. Types of events (1) Impossible and Sure events : The empty set φ is also a subset of sample space S and it represents an impossible event. Sample space S is also a subset of S, it represents a sure event. (2) Simple event : An event is said to be a simple event if it is a singleton subset of the sample space S. (3) Compound event : An event which is not a simple event is called compound event. In other words, an event is said to be compound event if it has more than one sample point. algebra of events (1) Complementary event : Corresponding to every event A, we define an event “not A” which is said to occur when A does not occur. Thus, the – event “not A” denoted by A′ or A is called the complementary event of A.

(2) The event ‘A or B’ : Let A and B be two events associated with a sample space, then the event ‘A or B’ denoted by A ∪ B is the event ‘either A or B or both’ A ∪ B = { x : x : ∈ A or x ∈ B} (3) The event ‘A and B’ : If A and B are two events, then the set A ∩ B denotes the event ‘A and B’. A ∩ B = { x : x ∈ A and x ∈ B} (4) The event ‘A but not B’ : A – B is the set of all those elements which are in A but not in B. Therefore, the set A – B denotes the event ‘A but not B’. We know that A – B = A ∩ B′ mutually exclusIve events Two or more events are said to be mutually exclusive if one of them occurs, others cannot occur. Thus two or more events are said to be mutually exclusive if no two of them can occur together. Thus events A1, A2, ..., An are mutually exclusive if and only if Ai ∩ Aj = φ for i ≠ j exhaustIve events For a random experiment, a set of events (cases) is said to be exhaustive if atleast one of them must necessarily happen every time the experiment is performed. Note : If Ei ∩ Ej = φ for i ≠ j i.e., events Ei and Ej are n

pairwise disjoint and ∪ Ei = S, then events E1, E2,..., En i =1

are called mutually exclusive and exhaustive events. Remarks : (1) For any event E, P(E) ≥ 0. (2) P(S) = 1 (3) Let S be a sample space and E be an event, such that n(S) = l and n(E) = m. If each outcome is equally likely, then it follows that m Number of outcomes favourable to E P(E) = = l Total possible ou utcomes (4) P(A ∪ B) = P(A) + P(B) – P(A ∩ B) If A and B are disjoint sets, i.e., they are mutually exclusive events, then A ∩ B= φ. Therefore P(A ∩ B) = P(φ) = 0 Thus, for mutually exclusive events A and B, we have P(A ∪ B) = P(A) + P(B) (5) P(A′) = P(not A) = 1 – P(A) mathematics today | JANUARY ‘16

25

Very short answer type 1. If standard deviation of n numbers x1, x2, …, xn is a, then write the value of standard deviation of numbers 2x1, 2x2, ……, 2xn. 2. Two coins are tossed once. Find the probability of getting 1 head and 1 tail. 3. Find the mean deviation about the mean for the following data : 15, 17, 10, 13, 10 4. What is the probability that a number selected from the numbers 1, 2, 3,..., 25, is prime number when each of the given numbers is equally likely to be selected? 3 5. If is the probability that an event will happen, 10 what is the probability that it will not happen? short answer type 6. Find the mean deviation about the median of the following frequency distribution: Class Frequency

0-6 8

6-12 10

12-18 18-24 24-30 12 9 5

7. The probability that at least one of the events E1 and E2 occurs is 0.6. If the probability of the simultaneous – – occurrence of E1 and E2 is 0.2, find P(E 1) + P(E 2). 8. An urn contains 9 red, 7 white and 4 black balls. If two balls are drawn at random, find the probability that: (i) both the balls are red (ii) one ball is white 9. The sum and the sum of squares of length x (in cm) and weight y (in g) of 30 plant products are given below: 30

30

i =1

i =1

30

30

i =1

i =1

∑ xi = 112, ∑ x 2 = 502.4, ∑ yi = 194 and ∑ y 2 = 1275.4 i

i

Which is more variable, the length or weight? 10. If a leap year is selected at random, what is the chance that it will contain 53 Tuesdays? long answer type 11. Find the mean deviation about the mean for the following data: Marks obtained Number of students 26

1020

2030

3040

4050

5060

6070

7080

8

4

2

12

2

4

8


12. Find the probability that when a hand of 7 cards is dealt from a well-shuffled deck of 52 cards, it contains: (i) all 4 kings (ii) exactly 3 kings (iii) at least 3 kings 13. Calculate mean, variance and standard deviation for the following frequency distribution. Class

030

3060

6090

Frequency

2

4

9

90- 120- 150- 180120 150 180 210 10

9

4

2

14. A natural number is chosen at random from among the first 500 natural numbers. What is the probability that the number so chosen is divisible by 3 or 5? 15. Suppose that samples of polythene bags from two manufactures, A and B are tested by a prospective buyer for bursting pressure, with the following results: Bursting Number of bags manufactured pressure (in kg) by manufacturer A B 5-10 2 9 10-15 9 11 15-20 29 18 20-25 54 32 25-30 11 27 30-35 5 13 Which set of bag has the highest average bursting pressure? Which has more uniform pressure? solutions 1. If the standard deviation of n numbers x1, x2, …, xn is a. Then the standard deviation of numbers 2x1, 2x2, ……, 2xn is 2a. 2. When two coins are tossed once, the sample space is given by S = {HH, HT, TH, TT} and therefore, n(S) = 4. Let E = event of getting 1 head and 1 tail. Then, E = {HT, TH} and therefore, n(E) = 2. n(E) 2 1 \ P(E) = = = . n(S) 4 2 3. Let the mean of the given data be x–. Then, n

x=

∑ xi i =1

n

=

65 = 13 5

[ n = 5]

– – ⇒ {1 – P(E 1)} + {1 –P(E 2)} = 0.8 – – ⇒ P(E 1) + P(E 2) = (2 – 0.8) = 1.2 – – Hence, P(E 1) + P(E 2) = 1.2

The values of |xi – x–| are 2, 4, 3, 0, 3 n

MD (x ) =

∑| xi − x | i =1

=

n – Hence, MD (x ) = 2.4

12 = 2. 4 5

8. There are 20 balls in the bag out of which 2 balls can be drawn in 20C2 ways. So, total number of

4. Let S be the sample space associated with the given experiment and A be the event “selecting a prime number”. Then, S = {1, 2, 3,..., 25} and A = {2, 3, 5, 7, 11, 13, 17, 19, 23} \ Total number of elementary events = 25 Favourable number of elementary events = 9 9 Hence, required probability = . 25 5. Let E be the event. Then, 3 3 7  P(E) = ⇒ P(E) = {1 − P(E)} = 1 −  = . 10  10  10 7 . 10 6. Calculation of Mean Deviation about the Median Hence, P(not E) = Class

Mid Frequency Cumulative |xi – 14| fi|xi – 14| (fi) values Frequency (xi) (c.f.)

0-6

3

8

8

11

88

6-12

9

10

18

5

50

12-18

15

12

30

1

12

18-24

21

9

39

7

63

24-30

27

5

44

13

65

N= Σ fi

Σ fi|xi – 14| = 278

Here N = 44, so N = 22 and the cumulative 2 N frequency just greater than is 30. Thus 12-18 2 is the median class. N /2−C Now, median = l + × h, f where l = 12, h = 6, f = 12, C = 18 4×6 22 − 18 So, median = 12 + × 6 = 12 + = 14 12 12 \ Mean deviation about median 1 5 278 = ∑ f i | xi − 14 | = = 6.318 N i =1 44 7. Given, P(E1 ∪ E2) = 0.6 and P(E1 ∩ E2) = 0.2 \ P(E1 ∪ E2) = P(E1) + P(E2) – P(E1 ∩ E2) ⇒ P(E1) + P(E2) = P(E1 ∪ E2) + P(E1 ∩ E2) = (0.6 + 0.2) = 0.8 ⇒ P(E1) + P(E2) = 0.8

elementary events = 20C2 = 190 (i) There are 9 red balls out of which 2 balls can be drawn in 9C2 ways. \ Favourable number of elementary events = 9C2 = 36 36 18 So, required probability = = . 190 95 (ii) There are 7 white balls out of which one ball can be drawn in 7C1 ways. One ball from the remaining 13 balls can be drawn in 13C1 ways. Therefore, one white and one ball of other colour can be drawn in 7C1 × 13C1 ways. So, favourable number of elementary events = 7C1 × 13C1. 7 C × 13C1 91 Hence, required probability = 120 = . 190 C2

 1 30  1 30 9. We have, Var(x) = ⋅ ∑ x i2 −  ⋅ ∑ x i  30 i =1  30 i =1  2

 502.4   112  251.2  56  = − = −  15  30   30   15  =

2

251.2 3136 3768 − 3136 632 − = = = 2. 8 15 225 225 225

Var( y) =

=

2

2 2 1 30 2  1 30  1275.4  194   ⋅ ∑ y i −  ⋅ ∑ yi  =  −  30 i =1  30    30 i =1   30 

637.7  97  2 −  15  15

637.7 9409 9565.5 − 9409 156.5 − = = = 0.6956 15 225 225 225 \ Var(x) > Var(y). \ Length is more variable than weight. =

10. A leap year has 366 days i.e., 52 complete weeks and two days more. These two days will be the two consecutive days of a week. A leap year will have 53 Tuesday if out of the two consecutive days of a week selected at random, one is a Tuesday. Let S be the sample space and E be the event that out of the two consecutive days of a week, one is Tuesday. mathematics today | JANUARY ‘16

27

S = {(Mon, Tues), (Tues, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun), (Sun, Mon)} \ n(S) = 7 and E = {(Mon, Tues), (Tues, Wed)} \ n(E) = 2 n(E ) 2 = Required probability, P (E ) = n(S) 7

11. We will find the mean (x ), let us compute x by step-deviation method. The formula for the same is 1 n  x = a + h  ∑ f id i  N   i =1

x −a where di = i , a = assumed mean and h h = width of class Let us take assumed mean a = 45 and h = 10 and form the following table: xi − 45 10

fidi

|xi – x | = |xi – 45|

fi|xi – x |

15

–3

–24

30

240

4

25

–2

–8

20

80

30-40

2

35

–1

–2

10

20

40-50

12

45

0

0

0

0

50-60

2

55

1

2

10

20

60-70

4

65

2

8

20

80

70-80

8

75

3

24

30

240

Marks obtained

Number of students (fi)

Mid-points (xi)

10-20

8

20-30

di =

N = 40 1 n  \ x = a + h  ∑ f id i   N i =1  ⇒ x = 45 + 10 × 0 = 45 40 1 680 \ MD (x ) = ∑ f i | x i − x | = = 17 40 N 12. From a deck of 52 playing cards, 7 cards can be drawn in 52C7 ways. \ Total number of elementary events = 52C7 (i) There are 4 kings. Therefore, 4 kings out of 4 kings and 3 other cards from the remaining 48 cards can be chosen in 4C4 × 48C3 ways. \ Favourable number of elementary events = 4C4 × 48C3 4 C 4 × 48 C3 1 = Hence, required probability = 52 7735 C7 (ii) Three kings out of 4 kings and 4 other cards out of remaining 48 cards can be chosen in 4 C3 × 48C4 ways. \ Favourable number of elementary events = 4C3 × 48C4 4 C3 × 48 C 4 9 Hence, required probability = = 52 1547 C7 28


_

∑fidi = 0

∑fi |xi – x | = 680

(iii) When 7 cards are drawn from a deck of 52 playing cards, then getting at least 3 kings means: getting 3 kings and 4 other cards or getting 4 kings and 3 other cards. This can be done in 4 C3 × 48C4 + 4C4 × 48C3 ways. \ Favourable number of elementary events = 4C3 × 48C4 + 4C4 × 48C3 Hence, required probability C3 × 48 C 4 + 4C 4 × 48 C3 46 = 52 7735 C7 13. Here, h = 30. Let the assumed mean be a = 105. Now, we prepare the table given below. =

4

yi2

fi yi

fiyi2

–3

9

–6

18

45

–2

4

–8

16

75

–1

1

–9

9

Class

Frequency (fi)

Midpoint (xi)

0-30

2

15

30-60

4

60-90

9

yi =

(x i − 105) 30

90-120

10

105 = a

0

0

0

0

120-150

9

135

1

1

9

9

150-180

4

165

2

4

8

16

180-210

2

195

3

9

6

18

Total

N = 40

0

86

\ a = 105, h = 30, N = ∑fi = 40, ∑fiyi = 0 and

 498   500  \ n(E1) =  = 166, n(E2 ) =  = 100  3   5   495  and n(E1 ∩ E2 ) =  = 33  15 

∑fiyi2 = 86

∑ f i yi   \ x = a + × h N   0   ⇒ x = 105 + × 30  = 105 30   Thus, mean = 105. h2 Variance, s 2 = 2 ⋅ {N ⋅ ∑ f i yi2 − (∑ f i yi )2} N (30)2 = ⋅ {40 × 86 − (0)2} = 1935 (40)2 Standard deviation, s = 1935 = 43.98.

\ P(E1) =

P(E2 ) =

Bursting pressure

n(E2 ) 100 1 = = n(S) 500 5

n(E1 ∩ E2 ) 33 = n(S) 500 \ P(The chosen number is divisible by 3 or 5) = P(E1 or E2) = P(E1 ∪ E2) = P(E1) + P(E2) – P(E1 ∩ E2)  83 1 33  233 = + − =  250 5 500  500 233 Hence, the required probability is . 500 15. For determining the set of bags having higher average bursting pressure, we compute mean and for finding out set of bags having more uniform pressure we compute coefficient of variation. and P(E1 ∩ E2 ) =

14. Let S be the sample space. Then, clearly n(S) = 500 Let E1 = event of getting a number divisible by 3, and E2 = event of getting a number divisible by 5. Then, (E1 ∩ E2) = event of getting a number divisible by both 3 and 5 = event of getting a number divisible by 15. \ E1 = {3, 6, 9,...,495, 498}, E2 = {5, 10, 15, ..., 495, 500} and (E1 ∩ E2) = {15, 30, 45, ……, 495} Manufacture A:

n(E1) 166 83 , = = n(S) 500 250

Computation of mean and standard deviation

Mid-values (xi) Frequency (fi)

ui =

xi −17.5 5

ui2

fiui

fiui2

5-10

7.5

2

–2

4

–4

8

10-15

12.5

9

–1

1

–9

9

15-20

17.5 = a

29

0

0

0

0

20-25

22.5

54

1

1

54

54

25-30

27.5

11

2

4

22

44

30-35

32.5

5

3

9

15

45

N =110

∑ui = 3

∑fiui = 78

∑fiui2 = 160

 n   ∑ f iui   x A = a + h  i =1 N  ⇒ x A = 17.5 + 5 ×

78 110

[ h = 5, a = 17.5]

= 21.04 2  n n   2 2 1 2  1 sA = h ∑ fiui −  N ∑ fiui    N i =1 i =1  

160  78  2  ⇒ s 2A = 25  −   110  110    17600 − 6084  = 23.79 ⇒ s 2A = 25   110 × 110  ⇒ s A = 23.79 = 4.87 \ Coefficient of variation s 4.87 = A × 100 = × 100 = 23.14 xA 21.04 mathematics today | JANUARY ‘16

29

Manufacture B: Bursting pressure

Mid-values xi

frequency (fi)

5-10 10-15 15-20 20-25 25-30 30-35

7.5 12.5 17.5 = a 22.5 27.5 32.5

9 11 18 32 27 13 N = 110

ui =

∑ui = 3

 n   ∑ f iui   x B = a + h  i =1 N  96 = 17.5 + 4.36 = 21.86 ⇒ x B = 17.5 + 5 × 110 2 1 n 1 n   s 2B = h 2  (∑ f iui2) −  ∑ f iui    N i =1  N i =1     2  304  96  2 − ⇒ s B = 25     110  110  

26. The value of a so that the volume of parallelopiped ^

^

becomes minimum is (a) –3 27. ∫

(b) 3

(c)

dx (1 + x ) p + q2 (tan −1 x )2 2

2

^

^

^

^

^

j + a k and a i + k

i +a j +ak ,

by

1 3

(d)

3

=

1 log q tan −1 x + p2 + q2 (tan −1 x )2  + c q (b) log q tan −1 x + p2 + q2 (tan −1 x )2  + c 2 2 2 ( p + q tan −1 x )3/2 + c (c) 3q (d) none of these (a)

28. Mean of 100 items is 49. It was discovered that three items which should have been 60, 70, 80 were wrongly read as 40, 20, 50 respectively. The correct mean is 1 (a) 80 (b) 50 (c) 82 (d) 48 2 30


ui2 4 1 0 1 4 9

fiui

fiui2

–18 –11 0 32 54 39

36 11 0 32 108 117

∑fiui = 96

∑fiui2 = 304

 33440 − 9216  2 = 50.04 ⇒ s B = 25   110 × 110  ⇒ s B = 50.04 = 7.07 \ Coefficient of variation s 7.07 = B × 100 = × 100 = 32.34 xB 21.86 We observe that the average bursting pressure is higher for manufacture B. So, bags manufactured by B have higher bursting pressure. The coefficient of variation is less for manufacture A. So, bags manufactured by A have more uniform pressure. nn

Contd. from Page No. 22

formed

xi −17.5 5 –2 –1 0 1 2 3

29. A tangent to an ellipse

x2

y2

= 1 meets the a 2 b2 coordinate axes in P and Q. The least value of PQ is (a) a – b (b) a + b a +b (c) (d) 2a 2 30. The plane passing through the point (–2, –2, 2) and containing the line joining the points (1, 1, 1) and (1, –1, 2) makes intercepts on the co-ordinate axes, the sum whose length is (a) 3 (b) 4 (c) 6 (d) 12 1. 6. 11. 16. 21. 26.

(c) (c) (d) (a) (d) (c)

2. 7. 12. 17. 22. 27.

+

Answer keys (d) 3. (d) 4. (a) 8. (a) 9. (b) 13. (a) 14. (b) 18. (a) 19. (b) 23. (b) 24. (a) 28. (b) 29.

(c) (c) (d) (c) (a) (b)

5. (d) 10. (d) 15. (c) 20. (c) 25. (b) 30. (d)

For detailed solution to the Sample Paper, visit our website : www. vidyalankar.org.

nn

CIRCLES

*ALOK KUMAR, B.Tech, IIT Kanpur

This column is aimed at Class XI students so that they can prepare for competitive exams such as JEE Main/Advanced, etc. and be also in command of what is being covered in their school as part of NCERT syllabus. The problems here are a happy blend of the straight and the twisted, the simple and the difficult and the easy and the challenging.

CIRCLES A circle is the locus of a point which moves in such a way that its distance from a fixed point is constant. The fixed point is called the centre of the circle and the constant distance is called the radius of the circle. EquatIon of thE CIRCLE In VaRIouS foRmS z The simplest equation of the circle is x2 + y2 = r2, whose centre is (0, 0) and radius is r. z The equation (x – a)2 + (y – b)2 = r2 represents a circle with centre (a, b) and radius r. z The equation x2 + y2 + 2gx + 2fy + c = 0 is the general equation of a circle with centre (–g, –f) and radius g2 + f 2 − c . Diametric form z Equation of the circle with points P(x 1 , y 1 ) and Q(x 2 , y 2 ) as extremities of a diameter is (x – x1)(x – x2) + (y – y1)(y – y2) = 0. Equation of a Circle under different conditions Condition

Equation

Touches both the axes (x – with radius a

a)2

+ (y – a)2 = a2

Touches x-axis only (x – a)2 + (y – b)2 = a2 with centre (a, b) Touches y-axis only (x – a)2 + (y – b)2 = a2 with centre (a, b)

Parametric Equation of a Circle The equation x = a cosq, y = asinq are called z parametric equations of the circle x2 + y2 = a2 and q is called a parameter. The point (acosq, asinq) is also referred to as point q. The parametric coordinates of any point on the circle (x – h)2 + (y – k)2 = a2 are given by (h + a cosq, k + a sinq) with 0 ≤ q < 2p. Remarks z Since there are three independent constants g, f, c in the general equation of a circle x2 + y2 + 2gx + 2fy + c = 0, a circle can be found to satisfy three independent geometrical conditions and no more. For example, when three points on a circle or three tangents to a circle or two tangents to a circle and a point on it are given, the circle can be determined. z To find the condition for the general equation of the second degree ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 to represent the circle, viz., x2 + y2 + 2gx + 2fy + c = 0, we see that there is no term in xy and that the coefficient of x2 is same as that of y2 i.e., the coefficient of xy = 0 and coefficient of x2 = coefficient of y2. The equation of a straight line joining two z points a and b on the circle x 2 + y 2 = a 2 is a +b a +b a −b . x cos + y sin = a cos 2 2 2

* Alok Kumar is a winner of INDIAN NAtIoNAl MAtheMAtIcs olyMpIAD (INMo-91). he trains IIt and olympiad aspirants.


31

Intercepts made on axes Solving the circle x2 + y2 + 2gx + 2fy + c = 0 with y = 0 we get, x2 + 2gx + c = 0. If discriminant 4(g2 – c) is positive, i.e., if g2 > c, the circle will meet the x-axis at two distinct points, say (x1, 0) and (x2, 0) where x1 + x2 = –2g and x1x2 = c. The intercept made on x-axis by the circle = |x1 – x2| = 2 g 2 − c . In the similar manner if f 2 > c, intercept made on z

z z

y-axis = 2 f 2 − c . If g2 – c > 0 ⇒ circle cuts x-axis at two distinct points. If g2 = c ⇒ circle touches the x-axis If g2 < c ⇒ circle lies completely above or below the x-axis. Case is similar for intercepts on y-axis.

tangEntS & noRmaLS A tangent to a curve at a point is defined as the limiting positions of a secant obtained by joining the given point to another point in the vicinity on the curve as the second point tends to the first point along the curve. or As the limiting position of a secant obtained by joining two points on the curve in the vicinity of the given point as both the points tend to the given point. Two tangents, real or imaginary, can be drawn to a circle from a point in the plane. The tangents are real and distinct if the point is outside the circle, real and coincident if the point is on the circle, and imaginary if the point is inside the circle. Equation of tangents z If S = 0 be a curve than S 1 = 0 indicate the equation which is obtained by substituting x = x1 and y = y1 in the equation of the given curve and T = 0 is the equation which is obtained by substituting x2 = xx1, y2 = yy1, 2xy = xy1 + yx1, 2x = x + x1, 2y = y + y1 in the equation S = 0. If S ≡ x2 + y2 + 2gx + 2fy + c = 0, then z S1 ≡ x12 + y12 + 2gx1 + 2fy1 + c, and T ≡ xx1 + yy1 + g(x + x1) + f (y + y1) + c Equation of the tangent to x2 + y2 + 2gx + 2fy + c = 0 at A(x1, y1) is xx1 + yy1 + g(x + x1) + f (y + y1) + c = 0 In particular, equation of the tangent to the circle x2 + y2 = a2 at (x1, y1) is xx1 + yy1 = a2 32


normals The normal to a curve at a point is defined as the straight line passing through the point and perpendicular to the tangent at that point. In case of a circle, every normal passes through the centre of the circle. Equation of normal z The equation of the normal to the circle x2 + y2 + 2gx + 2fy + c = 0 at any point (x1, y1) x − x1 y − y1 lying on the circle is = . x1 + g y1 + f In particular, equation of the normal to the circle y x = . x2 + y2 = a2 at (x1, y1) is x1 y1 Equation of tangent in Slope form z The condition that the straight line y = mx + c is a tangent to the circle x2 + y2 = a2 is c2 = a2(1 + m2) and the point of contact is (–a 2m/c, a 2/c) i.e., y = mx ± a 1 + m2 is always a tangent to the circle x2 + y2 = a2 whatever be the value of m. Equation of tangent in Parametric form Since parametric co-ordinates of circle x2 + y2 = a2 is (acosq, asinq), then equation of tangent at (acosq, asinθ) is x · acosq + y · asinθ = a2 or xcosq + ysinq = a LEngth of a tangEnt anD PowER of a PoInt The length of a tangent from an external point (x1, y1) to the circle S ≡ x2 + y2 + 2gx + 2fy + c = 0 is

z

z

2 2 given by L = x1 + y1 + 2 gx1 + 2 fy1 + c = S1 . Square of length of the tangent from the point P is also called the power of a point w.r.t. a circle. note : Power of a point P is positive, negative or zero according as the point ‘P’ is outside, inside or on the circle respectively. Chord of contact : If two A t angents PA and PB are P drawn from an outside point (x , y ) 1 1 B P ( x 1, y 1) to t he c i rcl e 2 2 S ≡ x + y + 2gx + 2fy + c = 0, then the equation of the chord of contact AB is xx1 + yy1 + g(x + x1) + f(y + y1) + c = 0 Equation of Pair of tangents : A The joint equation of a pair P of tangents drawn from the (x , y ) 1 1 point P(x1, y1) to the circle, B x2 + y2 + 2gx + 2fy + c = 0 is T2 = SS1.

z

z

z

Director Circle: The locus of the point of intersection of perpendicular tangents is called director circle. If (x – a)2 + (y – b)2 = r2 is the equation of a circle then its director circle is (x – a)2 + (y – b)2 = 2r2. Equation of the Chord with a given middle Point : The equation of the chord of the circle S ≡ x2 + y2 + 2gx + 2fy + c = 0 with its mid point x +g M (x1 , y1 ) is y − y1 = − 1 (x − x1 ) . y1 + f This on simplification can be put in the form xx1 + yy1 + g(x + x1) + f (y + y1) + c = x12 + y12 + 2gx1 + 2fy1 + c which is designated by T = S1. The shortest chord of a circle passing through a point ‘M’ inside the circle, is one chord whose middle point is M. The chord passing through a point ‘M’ inside the circle and which is at a maximum distance from the centre is a chord with middle point M. Position of a Point with respect to a Circle : The point P(x1, y1) lies outside, on, or inside a circle S ≡ x2 + y2 + 2gx + 2fy + c = 0, according as S1 ≡ x12 + y12 + 2gx1 + 2fy1 + c > = or < 0. The greatest and the least C distance of a point P from P B a circle with centre C and (x1, y1) A radius r is PC + r and PC – r respectively.

Position of the Line with respect to the Circle If p is perpendicular from the centre on the line, z then • p = r ⇔ the line touches the circle or the line is a tangent of the circle. • 0 < p < r ⇔ the line is a secant of the circle. • p = 0 ⇔ the line is a diameter of the circle. • Chord of contact exists only if the point ‘P’ is outside the circle. If R is radius of the circle and L is the length of the tangent from (x 1, y 1) on the circle S = 0, then 2LR • Length of chord of contact T1T2 = R2 + L2 • Area of the triangle formed by the pair of RL3 tangents and its chord of contact = R2 + L2

•

Tangent of the angle between the pair of

•

 2RL  tangents from (x1, y1) =   L2 − R2  Equation of the circle circumscribing the triangle PT1T2 is (x – x1)(x + g) + (y – y1)(y + f) = 0

RaDICaL axIS The radical axis of two circles is the locus of a point from which the tangent segments to the two circles are of equal length. Equation of the Radical axis z In general S – S′ = 0 represents the equation of the radical axis to the two circles. i.e., 2x(g – g′) + 2y(f – f ′) + c – c′ = 0 where S ≡ x2 + y2 + 2gx + 2fy + c = 0 and S′ ≡ x2 + y2 + 2g′x + 2f ′y + c′ = 0 If S = 0 and S′ = 0 intersect z (–g, –f ) (–g, –f ) in real and distinct points then S – S′ = 0 is the common chord equation of the common chord of the two circles. z If S = 0 and S′ = 0 touch each other, then S – S′ = 0 is the equation of the (–g, –f ) (–g, –f ) common tangent to the common two circles at the point of tagent contact. z radical axis is always perpendicular to the line joining the centres of the two circles. z radical axis will pass through the mid point of the line joining the centres of the two circles only if the two circles have equal radii. radical axis bisects a common tangent between the two circles. z A system of circles, every two of which have the same radical axis, is called a coaxial system. Pairs of circles which do not have radical axis are concentric. Radical Centre z The radical axes of three circles, taken in pairs, are concurrent. Let the equations of the three circles S1, S2 and S3 be S1 ≡ x2 + y2 + 2g1x + 2f1y + c1 = 0, S2 ≡ x2 + y2 + 2g2x + 2f2 y + c2 = 0 and S3 ≡ x2 + y2 + 2g3x + 2f3 y + c3 = 0 ...(1) mathematics today | JANUARY ‘16

33

S1 = 0

L

z

S2 = 0 O

M

Radical centre N

S3 = 0

Now, by the previous section, the radical axis of S1 and S2 is obtained by subtracting the equations of these circles ; hence it is S1 – S2 = 0 ...(2) Similarly, the radical axis of S2 and S3 is S2 – S3 = 0 ...(3) The lines (2) and (3) meet at a point whose coordinates say, (X, Y) satisfy S1 – S2 = 0 and S2 – S3 = 0; hence the coordinates (X, Y) satisfy (S1 – S2) + (S2 – S3) = 0; that is, S1 – S3 = 0 ...(4) But (4) is the radical axis of the circles S1 and S3 and hence the three radical axes are concurrent. The point of concurrency of the three radical axes is called the radical centre. famILy of CIRCLES z If S ≡ x2 + y2 + 2gx + 2fy + c = 0 and S′ ≡ x2 + y2 + 2g′x + 2f ′y + c′ = 0 are two intersecting circles, then S + lS′ = 0, l ≠ –1, is the equation of the family of circles passing through the points of intersection of S = 0 and S′ = 0. If S ≡ x 2 + y 2 + 2gx + 2fy + c = 0 is a circle z which is intersected by the straight line µ ≡ ax + by + c = 0 at two real and distinct points, then S + lµ = 0 is the equation of the family of circles passing through the points of intersection of S = 0 and µ = 0. z If µ = 0 touches S = 0 at P, then S + lµ = 0 is the equation of the family of circles, each touching µ = 0 at P. z The equation of a family of circles passing t h rou g h t w o g i v e n p oi nt s ( x 1 , y 1 ) a n d (x2, y2) can be written in the form. x y 1 (x − x1 )(x − x2 ) + ( y − y1 )( y − y2 ) + l x1 y1 1 = 0 x2 y2 1 where l is a parameter. 34

z


z

The equation of the family of circles which touch the line y – y1 = m(x – x1) at (x1, y1) for any value of m is (x – x1)2 + (y – y1)2 + l(y – y1 – m(x – x1)) = 0. Family of circles circumscribing a triangle whose sides are given by L1 = 0, L2 = 0 and L3 = 0 is given by L1L2 + lL2L3 + µL3L1 = 0 provided coefficient of xy = 0 and coefficient of x2 = coefficient of y2. Equation of circles circumscribing a quadrilateral whose sides in order are represented by the lines L 1 = 0, L 2 = 0, L 3 = 0 and L 4 = 0 are L 1 L 3 + lL 2 L 4 = 0 where value of l can be found out by using condition that coefficient of x2 = coefficient of y2 and coefficient of xy = 0.

thE ConDItIon that two CIRCLES ShouLD IntERSECt A necessary and sufficient condition for the two circles to intersect at two distinct points is r1 + r2 > C1C2 > |r1 – r2|, where C1, C2 be the centres and r1, r2 be the radii of the two circles. External and Internal Contacts of Circles If two circles with centres C1(x1, y1) and C2(x2, y2) and r1 A r2 radii r1 and r2 respectively, C2 C1 touch each other externally, C1C2 = r1 + r2. Coordinates of the point of contact are r x +r x r y +r y  A≡ 1 2 2 1, 1 2 2 1 . r r +r  C C2 2  r +r 1

2

1

2

The circles touch each other internally if C1C2 = r1– r2. Coordinates of the point of contact are r x −r x r y −r y  T ≡ 1 2 2 1, 1 2 2 1 r1 − r2   r1 − r2 •

1

r1

Common Tangents to Two Circles

r1

C1

r1

C1

r2 C2

r2 C2

number of Condition tangents 4 c o m m o n r 1 + r2 < C1C 2 tangents (2 direct and 2 transverse) 3 c o m m o n r 1 + r2 = C1C 2 tangents

T

C1

C2

C1

C1

The tangent at A to the circle S1 is perpendicular to the radius Q1A and the tangent at A to S2 is perpendicular to the radius Q2A. Hence, if the two tangents are at right angles, then the radii Q1A and Q2A must also be at right angles. Accordingly, the condition that S1 and S2 should be orthogonal is that ∠Q 1AQ 2 should be 90°; by Pythagoras’ theorem, this condition is equivalent to Q1Q22 = Q1A2 + Q2A2 = r12 + r22 ⇒ (g – G)2 + (f – F)2 = g2 + f 2 – c + G2 + F2 – C or, on simplification, 2(gG + fF) = c + C Since, AQ2 is perpendicular to the radius Q1A, the tangent at A to the circle S1 passes through the centre of the circle S2; similarly, the tangent at A to S2 passes through the centre of S1.

2 c o m m o n |r1 – r2| < C1C2 tangents < r1 + r2

C2

1 c o m m o n |r1 – r2| = C1C2 tangent

C2

No common C1C2 < |r1 + r2| tangent

Length of an External (or Direct) Common tangent & Internal (or transverse) Common tangent Length of an external (or direct) common tangent & internal (or transverse) common tangent to the two circles is given by :

z

z

z

Lext = d 2 − (r1 − r2 )2 and Lint = d 2 − (r1 + r2 )2 where d = distance between the centres of the two circles, r1 and r2 are the radii of the two circles. The length of internal common tangent is always less than the length of the external or direct common tangent. The direct common tangents to two circles meet on the line of centres and divide it externally in the ratio of the radii. The transverse common tangents also meet on the line of centres and divide it internally in the ratio of the radii. transverse tangent

C1

R transverse tangent

direct tangent

C2

F

direct tangent

oRthogonaL CIRCLES Two circles are said to be A orthogonal if the tangents r2 r1 S2= 0 to the circles at either S1= 0 Q 1 Q point of intersection are 2 (–G, –F) at right angles. In fig. Q1 (–g, –f ) B and Q2 are the centres of the circles S1 ≡ x2 + y2 + 2gx + 2fy + c = 0 and S2 ≡ x2 + y2 + 2Gx + 2Fy + C = 0 The circles, S1 and S2, intersect at A and B.

Problems section-i single correct Answer type

1. (a) (b) (c) (d)

Equation of circle touching the line |x – 2| + |y – 3| = 4 will be (x – 2)2 + (y – 3)2 = 12 (x – 2)2 + (y – 3)2 = 4 (x – 2)2 + (y – 3)2 = 10 (x – 2)2 + (y – 3)2 = 8

2. The equation of four circles are (x ± a)2 + (y ± a)2 = a2. The radius of a circle touching all the four circles is (a) ( 2 − 1)a or ( 2 + 1)a (b)

2a or 2 2a

(c) (2 − 2 )a or (2 + 2 )a (d) None of these 3. If two distinct chords, drawn from the point (p, q) on the circle x2 + y2 = px + qy (where pq ≠ 0) are bisected by the x-axis, then (a) p2 = q2 (b) p2 = 8q2 (c) p2 < 8q2 (d) p2 > 8q2 4. C1 and C2 are circles of unit radius with centres at (0, 0) and (1, 0) respectively. C3 is a circle of unit radius, passes through the centres of the circles C1 and C2 and have its centre above x-axis. Equation of the common tangent to C1 and C3 which does not pass through C2 is (a) x − 3 y + 2 = 0

(b)

(c)

(d) x + 3 y + 2 = 0

3x − y − 2 = 0

3x − y + 2 = 0


35

5. Circles are drawn through the point (2, 0) to cut intercept of length ‘5’ units on the x-axis. If their centres lie in the first quadrant then their equation is (a) x2 + y2 – 9x + 2ky + 14 = 0, k > 0 (b) 3x2 + 3y2 + 27x – 2ky + 42 = 0, k > 0 (c) x2 + y2 – 9x – 2ky + 14 = 0, k > 0 (d) x2 + y2 – 2ky – 9y + 14 = 0, k > 0 6. The equation of the circle passing through the point (2, –1) and having two diameters along the pair of lines 2x2 + 6y2 – x + y – 7xy – 1 = 0 is (a) x2 + y2 + 10x + 6y – 19 = 0 (b) x2 + y2 + 10x – 6y + 19 = 0 (c) x2 + y2 + 10x + 6y + 19 = 0 (d) x2 + y2 – 10x + 6y + 19 = 0 7. An equilateral triangle has two vertices (–2, 0) and (2, 0) and its third vertex lies below the x-axis. The equation of the circumcircle of the triangle is (a) 3(x 2 + y 2 ) − 4 y + 4 3 = 0 (b)

3(x 2 + y 2 ) − 4 y − 4 3 = 0

(c)

3(x 2 + y 2 ) + 4 y + 4 3 = 0

(d)

3(x 2 + y 2 ) + 4 y − 4 3 = 0

8. The coordinates of two points on the circle x2 + y2 – 12x – 16y + 75 = 0 one nearest to the origin and the other farthest from it, are (a) (3, 4), (9, 12) (b) (3, 2), (9, 12) (c) (–3, 4), (9, 12) (d) (3, 4), (9, –12) 9. The centre of a circle of radius 4 5 lies on the line y = x and satisfies the inequality 3x + 6y > 10. If the line x + 2y = 3 is a tangent to the circle, then the equation of the circle is 2

2

2

2

2

2

2

2

(a)  x − 23  +  y − 23  = 80  3  3 (b)  x + 17  +  y + 17  = 80  3  3 (c)  x + 23  +  y − 23  = 80  3  3

(a) x + y = 0 (c) 7x – y = 0

(b) x – y = 0 (d) x – 7y = 0

11. A variable circle passes through the fixed point A(p, q) and touches x-axis. The locus of the other end of the diameter through A is (a) (y – p)2 = 4qx (b) (x – q)2 = 4py (c) (x – p)2 = 4qy (d) (y – q)2 = 4px 12. The chord of contact of tangents from a point ‘P’ to a circle passes through Q. If l1 and l2 are the lengths of the tangents from P and Q to the circle, then PQ is equal to l −l l +l (a) 1 2 (b) 1 2 2 2 (c)

l12 + l22

(d)

l12 − l22

13. If the chord of contact of tangents from 3 points A, B, C to the circle x2 + y2 = a2 are concurrent, then A, B, C will (a) be concyclic. (b) be collinear. (c) form the vertices of triangle. (d) none of these. 14. The equation of circumcircle of a DABC is x2 + y2 + 3x + y – 6 = 0. If A = (1, –2), B = (–3, 2) and the vertex C varies then the locus of orthocentre of DABC is a (a) straight line (b) circle (c) parabola (d) ellipse 15. Point A lies on y = x and point B on y = mx so that length of AB = 4 units. Then value of ‘m’ for which locus of mid point of AB represents a circle is (a) m = 0 (b) m = –1 (c) m = 2 (d) m = –2 16. x2 + y2 + 6x + 8y = 0 and x2 + y2 – 4x – 6y – 12 = 0 are the equation of the two circles. Equation of one of their common tangent is (a) 7x – 5y – 1 – 5 74 = 0 (b) 7x – 5y – 1 + 5 74 = 0

(d)  x − 17  +  y − 17  = 80  3  3

(c) 7x – 5y + 1 – 5 74 = 0

10. Let L1 be a straight line passing through the origin and L2 be the straight line x + y = 1. If the intercepts made by the circles x2 + y2 – x + 3y = 0 on L1 and L2 are equal then which of the following equations can represent L1?

17. P and Q are any two points on the circle x2 + y2 = 4 such that PQ is a diameter. If a and b are the lengths of perpendicular from P and Q on x + y = 1 then the maximum value of ab is (a) 1/2 (b) 7/2 (c) 1 (d) 2

36


(d) 5x – 7y + 1 – 5 74 = 0

18. From a point P outside a circle with centre at C, tangents PA and PB are drawn such that 1 1 1 + = , then the length of chord AB is 2 2 16 (CA) (PA) (a) 8 (c) 16


19. If two circles (x – 1)2 + (y – 3)2 = r2 and x2 + y2 – 8x + 2y + 8 = 0 intersect in two distinct points, then (a) 2 < r < 8 (b) r < 2 (c) r = 2 (d) r > 2 20. The circle x2 + y2 = 1 cuts the x-axis at P and Q. Another circle with centre at Q and variable radius intersects to first circle at R above the x-axis and the line segment PQ at S. The maximum area of the triangle QSR is 2 4 3 2 5 2 (b) (c) (d) 9 9 13 7 21. The locus of the centre of a circle x2 + y2 – 6x – 6y + 14 = 0 which touches the circle externally and also the y-axis is given by (a) x2 – 6x – 7y + 14 = 0 (b) x2 – 10x – 6y + 14 = 0 (c) y2 – 6x – 10y + 14 = 0 (d) y2 – 10x – 6y + 14 = 0 (a)

section-ii multiple correct Answer type

22. Consider the circle x2 + y2 – 8x – 18y + 93 = 0 with centre ‘C ’ and point P(2, 5) outside it. From the point P, a pair of tangents PQ and PR are drawn to the circle with S as the midpoint of QR. The line joining P to C intersects the given circle at A and B. Which of the following hold(s) good ? (a) CP is the arithmetic mean of AP and BP (b) PR is the geometric mean of PS and PC (c) PS is the harmonic mean of PA and PB (d) The angle between the two tangents from P is 3 tan −1   4 23. Q is any point on the circle x2 + y2 = 9. QN is perpendicular from Q on the x-axis. Locus of the point of trisection of QN is (a) 4x2 + 9y2 = 36 (b) 9x2 + 4y2 = 36 2 2 (c) 9x + y = 9 (d) x2 + 9y2 = 9

24. (1, 2 2 ) is a point on circle x2 + y2 = 9. Then the points on the given circle which are at 2 unit distance from (1, 2 2 ) is are : (a) (−1, 2 2 )  23 10 2  (c)  , 9  9

(b) (2 2 , 1) 23 10 (d)  ,  9 9

25. Let C1 = x2 + y2 = r12, C2 = x2 + y2 = r22 (r1 < r2) be two circles. Let A be the fixed point (r1, 0) on C1 and B be a variable point on C2. Let the line BA meet the circle C2 again at E. Then, (a) the maximum value of BE is 2r2. (b) the minimum value of BE is 2 r22 − r12 . (c) if O is origin, then, the best possible lower bound for OA2 + OB2 + BE2 is 5r22 – 3r12 . (d) if O is origin, then, the best possible upper bound for OA2 + OB2 + BE2 is r12 + 5r22 . 26. If Q, S are two points on the circle x2 + y2 = 4 such that the tangents QP, SR are parallel. If PS, QR intersect 2

2

 QT   ST  at T then  + + PQ ⋅ RS ≠  PQ   RS  (a) 5

(b) 10

(c) 16

(d) 17

27. If A and B are two points on the circle x2 + y2 – 4x + 6y – 3 = 0, which are farthest and nearest respectively from the point (7, 2) then (a) A ≡ (2 − 2 2 , −3 − 2 2 ) (b) A ≡ (2 + 2 2 , −3 + 2 2 ) (c) B ≡ (2 + 2 2 , −3 + 2 2 ) (d) B ≡ (2 − 2 2 , −3 − 2 2 ) 28. If two circles x2 + y2 – 6x – 12y + 1 = 0 and x2 + y2 – 4x – 2y – 11 = 0 cut a third circle orthogonally then the radical axis of the two circles passes through (a) (1, 1) (b) (0, 6) (c) centre of the third circle. (d) mid-point of the line joining the centres of the given circles. 29. The equation of a circle C1 is x2 + y2 = 4. The locus of the intersection of orthogonal tangents to the circle is the curve C2 and the locus of the intersection of perpendicular tangents to the curve C2 is the curve C3 then mathematics today | JANUARY ‘16

37

(a) (b) (c) (d)

C3 is a circle. the area enclosed by the curve C3 is 8p. C2 and C3 are circles with the same centre. none of the above.

30. Consider the circles C1 ≡ x2 + y2 – 2x – 4y – 4 = 0 and C2 ≡ x2 + y2 + 2x + 4y + 4 = 0 and the line L ≡ x + 2y + 2 = 0, then (a) L is the radical axis of C1 and C2. (b) L is the common tangent of C1 and C2. (c) L is the common chord of C1 and C2. (d) L is perpendicular to the line joining centres of C1 and C2. 31. A circle touches the line x + y – 2 = 0 at (1, 1) and cuts the circle at x2 + y2 + 4x + 5y – 6 = 0 at P and Q. Then (a) PQ can never be parallel to the given line x + y – 2 = 0. (b) PQ can never be perpendicular to the given line x + y – 2 = 0. (c) PQ always passes through (6, –4). (d) PQ always passes through (–6, 4). 32. If al2 – bm2 + 2dl + 1 = 0, where a, b, d are fixed real numbers such that a + b = d2 then the line lx + my + 1 = 0 touches a fixed circle. Then the fixed circle (a) which cuts the x-axis orthogonally. (b) with radius equal to b. (c) on which the length of the tangent from the origin is d 2 − b . (d) with centre (d, 0). 33. Point M moved along the circle (x – 4)2 + (y – 8)2 = 20, then it broke away from it and moving along a tangent to the circle cuts the x-axis at the point (–2, 0). The co-ordinate of the point on the circle at which the moving point broke away can be  −3 46  (a)  ,   5 5 (c) (6, 4)

 −2 44  (b)  ,   5 5  (d) (3, 5) section-iii

comprehension type

Paragraph for question no. 34 to 36 Tangents PA and PB are drawn to t he circle (x – 4)2 + (y – 5)2 = 4 from the point P on the curve y = sinx, where A and B lie on the circle. Consider the function y = f (x) represented by the locus of the center of the circumcircle of triangle PAB. 38


34. range of y = f(x) is (a) [–2, 1] (b) [–1, 4] (c) [0, 2] (d) [2, 3] 35. Fundamental period of y = f(x) is (a) 2p (b) 3p (c) p (d) not defined 36. Which of the following is true? (a) f(x) = 4 has real roots (b) f(x) = 1 has real roots p   p (c) range of y = f–1(x) is  − + 2, + 2 4 4 (d) none of these Paragraph for question no. 37 to 39 Given equation of two intersecting circles S1 = 0 and S2 = 0 Equation of family of circles passing through the intersection point’s of S1 = 0 and S2 = 0 is S1 + lS2 = 0, (where l ≠ –1) Equation of common chord is S1 – S2 = 0 Equation of chord of contact for circle x2 + y2 + 2gx + 2fy + c = 0 with respect to external point (x1, y1) is xx1 + yy1 + g(x + x1) + f(y + y1) + c = 0. 37. The equation of the circle described on the common chord of the circles x2 + y2 + 2x = 0 and x2 + y2 + 2y = 0 as diameter is (a) x2 + y2 + x – y = 0 (b) x2 + y2 – x – y = 0 (c) x2 + y2 – x + y = 0 (d) x2 + y2 + x + y = 0 38. Let P be any moving point on the circle x2 + y2 – 2x – 1 = 0. AB be the chord of contact of this point P with respect to the circle x2 + y2 – 2x = 0. The locus of the circumcentre of the triangle PAB (C being centre of the circles) is (a) 2x2 + 2y2 – 4x + 1 = 0 (b) x2 + y2 – 4x + 2 = 0 (c) x2 + y2 – 4x + 1 = 0 (d) 2x2 + 2y2 – 4x + 3 = 0 39. The common chord of the circle x2 + y2 + 6x + 8y – 7 = 0 and a circle passing through the origin and touching the line y = x, always passes through the point  1 1 (a)  − ,  (b) (1, 1)  2 2 1 1 (c)  ,  2 2

(d) none of these

Paragraph for question no. 40 to 42 Given P, Q are two points on the curve y = log1/2 (x − 0.5) + log 2 4 x 2 − 4 x + 1, P also lies on the circle x2 + y2 = 10. If Q lies inside the given circle such that its abscissa is integer then   40. OP ⋅ OQ is equal to (a) 4 (b) 8 (c) 2 (d) 10  41. Max | PQ | is equal to (a) 1

{ }

(b) 2  42. Min | PQ | =

(c) 4

(d) 5

(a) 1

(c) 4

(d) 6

{ }

(b) 3

Paragraph for question no. 43 to 45 The line x + 2y + a = 0 intersects the circle x 2 + y 2 – 4 = 0 at two distinct points A and B. Another line 12x – 6y – 41 = 0 intersect the circle x2 + y2 – 4x – 2y + 1 = 0 at two distinct points C and D. 43. The value of a so that the line x + 2y + a = 0 intersect the circle x2 + y2 – 4 = 0 at two distinct points A and B is (a) −2 5 < a < 2 5 (b) 0 < a < 2 5 (c) − 5 < a < 5

(d) 0 < a < 5

44. The value of ‘a’ for which the four points A, B, C and D are concyclic is (a) 1 (b) 3 (c) 4 (d) 2 45. The equation of circle passing through the points A, B, C and D is (a) 5x2 + 5y2 – 8x – 16y – 36 = 0 (b) 5x2 + 5y2 + 8x – 16y – 36 = 0 (c) 5x2 + 5y2 + 8x + 16y – 36 = 0 (d) 5x2 + 5y2 – 8x – 16y + 36 = 0 section-iV matrix-match type

46. Match the following : Column I Column II A. The number of common tangents that (p) 2 can be drawn to the two circles C1 : x2 + y2 – 4x – 6y – 3 = 0 C2 : x2 + y2 + 2x + 2y + 1 = 0 is B. The common chord of x2 + y2 – 4x (q) 0 – 4y = 0 and x2 + y2 = 16 subtends at the origin an angle equal to p/k, then k = ?

C. Shortest distance from the point (r) 3 (2, –7) to the circle x2 + y2 – 14x – 10y – 151 = 0 is D. If real numbers x and y satisfy (s) 1 (x + 5)2 + (y – 12)2 = (14)2 , then the minimum value of

x 2 + y 2 is

47. Match the following : Column I Column II 2 2 A. If the circle x + y + 2x + (p) the set satisfying c = 0 and x2 + y2 + 2y + c = 0 c=1 touch each other, then the set of ‘c’ values is contained in or equal to B. If the circle x2 + y2 + 2x + 3y (q) + c = 0 and x2 + y2 – x + 2y + c = 0 intersect orthogonally, then the set of ‘c’ values is contained in or equal to C. If the circle x 2 + y 2 = 9 (r) contains the circle x2 + y2 – 2x + 1 – c2 = 0, then the set of ‘c’ values is contained in or equal to D. If the circle x 2 + y 2 = 9 (s) is contained in the circle x2 + y2 – 6x – 8y + 25 – c2 = 0, then the set of ‘c’ values is contained in or equal to (t)

the set satisfying |c| < 2

the set satisfying c =1/2

the set satisfying |c| > 8

the set satisfying 2 < |c| < 8

48. Match the following: Column I Column II A. The radical axis of two (p) s u b t e n d s a r i g h t circles angle at a point of intersection B. The common tangent (q) is perpendicular to to two intersecting the line joining the circles of equal radii centres C. The common chord (r) is parallel to the line of two intersecting joining the centres circles D. T h e l i n e j o i n i n g (s) is bisected by the line joining the centres. the centres of two circles intersecting orthogonally mathematics today | JANUARY ‘16

39

section-V integer Answer type

49. Let S1 ≡ x2 + y2 – 4x – 8y + 4 = 0 and S2 is its image in the line y = x. The radius of the circle touching y = x 3 , then l2 + 2 = at (1, 1) and orthogonal to S2 is l 50. The tangents drawn from the origin to the circle x2 + y2 – 2rx – 2hy + h2 = 0 are perpendicular then sum of all possible values of h/r is 51. The number of integral values of a for which the point (a – 1, a + 1) lies in the larger segment of the circle x2 + y2 – x – y – 6 = 0 made by the chord whose equation is x + y – 2 = 0 is 52. Let r be the radius of incircle of triangle formed by joining centres of (x – a)2 + (y – b)2 = 9, (x – a)2 + (y – b – 7)2 = 16 and circle touching above two circles and having radius 5 units. Find r2. solutions 1. (d) : Perpendicular distance y f rom c e nt re to t ange nt = x+y=9 radius C |2 + 3 − 9| 4 x r= = (2, 3) x 2 2 4 2 = =2 2 y 2 Equation of circle is (x – 2)2 + (y – 3)2 = 8 2. (a) : radius of smallest circle is r + a = a 2 ⇒ r= a 2–a \ The radius of another circle is a 2 + a 3. (d) : Let PQ be a chord of the given circle passing through P(p, q) P(p, q) and the coordinates of Q be (x, y). Since PQ is bisected by the x-axis, the mid-point of PQ lies on the R Q(x, y) x-axis which gives y = –q Now Q lies on the circle x2 + y2 – px – qy = 0 So x2 + q2 – px + q2 = 0 ⇒ x2 – px + 2q2 = 0 …(i) Which gives two values of x and hence the coordinates of two points Q and R (say), so that the chords PQ and PR are bisected by x-axis. If the chords PQ and PR are distinct, the roots of (i) are real and distinct. ⇒ the discriminant p2 – 8q2 > 0 ⇒ p2 > 8q2. 40


4. (b) : Equation of any circle through (0, 0) and (1, 0) is x (x − 0)(x − 1) + ( y − 0)( y − 0) + l 0 1 ⇒ x2 + y2 – x + ly = 0 If it represents C3, its radius =1

⇒ 1 = (1 / 4) + (l2 / 4)

y 1 0 1 =0 0 1

y

⇒ l=± 3 C3 As the centre of C3, lies above the x-axis , we x take l = − 3 and thus C1 C2 an equation of C 3 is (1, 1) (0, 0) x2 + y2 − x − 3 y = 0 . Since C1 and C3 intersect and are of unit radius. So their common tangents are parallel to the line joining their centres (0, 0) and (1 / 2, 3 / 2). So, let the equation of a common tangents be 3x − y + k = 0 . k = 1 ⇒ k = ±2 It will touch C1, if 3 +1 From the figure, we observe that the required tangent makes positive intercept on the y-axis and negative on the x-axis and hence its equation be 3x − y + 2 = 0 . 9  5. (c) : C =  , k  , k > 0 2  2

9  2 25 2  x −  + ( y − k ) = + k 2 4 or x2 + y2 – 9x – 2ky + 14 = 0, k > 0 6. (a) : 2x2 + 6y2 – x + y – 7xy – 1 = 0 x – 2y – 1 = 0 and 2x – 3y + 1 = 0 Centre is (–5, –3) \ The required equation of a circle be x2 + y2 + 10x + 6y – 19 = 0.

…(1) …(2)

7. (d) : Vertex A = (0, − 12 )  −2  4 4 Centroid G =  0,  , Circumradius = 4 + =  3 3 3 \ 3(x 2 + y 2 ) + 4 y − 4 3 = 0 8. (a) : C = (6, 8) and radius = 5 = AC (say) OC = 36 + 64 = 10 , OA = 5 Q OA : AC = 5 : 5 = 1 : 1 A is midpoint of OC i.e. (3, 4) Let Coordinate of B be (h, k) Q ‘C’ is the midpoint of AB \ h = 9, k = 12 \ B(9, 12)

9. (a) : We have, C = (a, a) radius = 4 5 = length of the ^ from (a, a) to the line | a + 2(a) − 3 |

23 −17 , 3 3 4 +1  23 23   −17 −17  \ Centre  ,  or  ,  3 3  3 3   23 23  only C =  , satisfies the given inequality  3 3  3x + 6y > 10 i.e.,

=±4 5 ⇒ a=

2

2

23   23   \  x −  +  y −  = 80  3  3  1 −3  10. (b) : C  , and let 4 : y = mx  2 2  Q Perpendicular distance from centre to two equal chords are equal. \ Perpendicular distance from L2 = Perpendicular distance from L1 m 3 1 3 − −1 + −1 2 2 ⇒ = 2 2 ⇒ m = 1, 2 7 1+1 1+ m \ Two chords are y = x and x + 7y = 0. 11. (c) : (x – p)(x – a) + (y – q)(y – b) = 0 or x2 + y2 – (p + a)x – (q + b)y + pa + qb = 0 …(1) Put y = 0, we get x2 – (p + a)x + pa + qb = 0 …(2) \ Locus of B (a, b) is (p – x)2 = 4qy or (x – p)2 = 4qy 12. (c) : We have, P = (x1, y1) and Q = (x2, y2) Chord of contact to the given circle is xx1 + yy1 = a2 Since it passes through Q(x2, y2) \ x 2x 1 + y 2y 1 = a2 …(i) Now, l1 = x12 + y12 − a2 , l2 = x22 + y22 − a2 and PQ = l12 + l22 13. (b) : Let the three points be A = (x1, y1), B = (x2, y2) and C = (x3, y3) Chord of contacts to the given circle are : xx1 + yy1 = a2, xx2 + yy2 = a2 and xx3 + yy3 = a2 Q These lines will be concurrent if x1

y1

−a 2

x2

y2

−a = 0 or

x3

y3

2

−a 2

x1 x2 x3

y1 y2 y3

−1 −1 = 0 −1

which is the condition of the collinearity of A, B, C 14. (b) : Equation of circumcircle of a DABC is 2

2

3  1  17   x +  +  y +  = 2 2 2

  −3 17 17 −1 cos q, sin q  C= + +   2 2 2 2  −3 −1  Circumcentre of DABC is  , .  2 2  Centroid can be obtained. In a triangle centroid, circumcentre and orthocentre are collinear. 15. (b) : Let co-ordinates of A(x1, x1) and B(x2, mx2). Clearly (x1 – x2)2 + (x1 – mx2)2 = 16 Let mid point of AB be P(h, k) ⇒ x1 + x2 = 2h and x1 + mx2 = 2k ⇒ (x1 – x2)2 + 4x1x2 = 4h2 and (x1 – mx2)2 + 4mx1x2 = 4k2 (x1 – x2)2 + (x1 – mx2)2 = 4h2 + 4k2 = 16, when m = –1. 16. (c) : Both the circles have radius = 5 and they intersect each other, therefore their common tangent is parallel to the line joining their centres. Equation of the line joining their centres is 7x – 5y + 1 = 0. \ Equation of the common tangent is 7x – 5y = c c +1 \ = 5 ⇒ c = ±5 74 − 1 74 \ Equation is 7x – 5y + 1 ± 5 74 = 0. 17. (b) : P(cosq, 2sinq), Q(–2cosq, –2sinq) | 2 cos q + 2 sin q || −2 cos q − 2 sin q − 1 | ab = 2 | 4(cos q + sin q)2 − 1 | 7 = ≤ 2 2 r 18. (a) : tan q = PA 1 1 1 cot 2 q + 1 1 Given + = ⇒ = 16 r 2 (PA)2 16 (PA)2 ⇒ (PA)sinq = 4 ⇒ Length of chord = 2(PA) sinq = 2 × 4 = 8. 19. (a) : Centres and radii of the given circles are C1 = (1, 3), r1 = r and C2 = (4, –1), r2 = 3 respectively, since circles intersect in two distinct points, then |r1 – r2| < C1C2 < r1 + r2 ⇒ |r – 3| < 5 < r + 3 …(i) From last two relations, r > 2 From first two relations, |r – 3| < 5 ⇒ –5 < r – 3 < 5 ⇒ –2 < r < 8 …(ii) From equations (i) and (ii), we get 2 < r < 8 20. (c) : Q is (–1, 0) The circle with centre at Q and variable radius r has the equation (x + 1)2 + y2 = r2 mathematics today | JANUARY ‘16

41

This circle meets the line segment QP at S where QS = r   r2 − 2 r It meets the circle x2 + y2 = 1 at R  , 4 − r2    2 2 found by solving the R equations of the two circles x2 + y 2 = 1 simultaneously. A = area of the triangle QSR Q T S P O 1 = QS × RT 2 1 r  = r  4 − r 2  since RT is the y coordinate of R   2 2 dA 1  r 2 (−r )  = 2r 4 − r 2 +  dr 4  4 − r 2  =

{2r (4 − r 2 ) − r 3 } 4 4 − r2

=

8r − 3r 3

4 4 − r2

dA 8 = 0, when r (8 − 3r2 ) = 0, giving r = dr 3 2 2 2 (−4r ) 4 4 − r (8 − 9r ) − (8r − 3r ) d2 A 4 − r2 = dr 2 16(4 − r 2 ) 2 8 d A , <0. where, r = 3 dr 2 8 Hence A is maximum when r = 3 and the maximum area 8 8 16 4 4 3 = 4− = = = 4×3 3 12 3 3 3 9 21. (d) : Let (x1, y1) be the centre. Since it touches y-axis its radius is |x1|. Also it touches the given circle externally. 2

2

\ (x1 − 3) + ( y1 − 3) = | x1 + 2 | Squaring both sides, we get x12 + y12 – 6x1 – 6y1 + 18 = x12 + 4x1 + 4 ⇒ y12 – 10x1 – 6y1 + 14 = 0 Thus, the locus of the centre of a circle at (x1, y1) is y2 – 10x – 6y + 14 = 0 22. (a, b, c) : We have, radius =

16 + 81 − 93 = 2

CP = 20 , AP = 20 − 2; BP = 20 + 2 AP + BP ⇒ CP = ⇒ (a) is correct 2 L P (2, 5)

42



=4

L=

Q

4

2

C(4, 9)

R


⇒ PR is the Geometric Mean of PS and PC ⇒ (b) is correct Now angle between the two tangents is 2q, then  1 2⋅ 1 1      2 tan −1   = tan −1  2   As tan q =   2 1 2  1 −  4 4 = tan −1   . 3 ⇒ (d) is incorrect. 23. (a, d) : Let Q = (3cosq, 3sinq), N = (3cosq, 0) Points of trisection are (3cosq, sinq) and (3cosq, 2sinq) Locus is

x2 x2 y2 + y 2 = 1; + =1 9 9 4

x − 1 y1 − 2 2 = =2 24. (a, c) : We have, 1 cos q sin q 2 2 Also, x1 + y1 = 9 on solving the above equations, we get 7 cos q = or − 1 9 Then the required points are  23 10 2  (−1, 2 2 ) and  ,  9 9  25. (a, b, c, d) : (BE)max = diameter of circle C2 = 2r2 (BE )min = 2 r22 − r12 (OA2 + OB2 + BE2)min is

2

A S

Now, let L = PR = (PC )2 − r 2 = 20 − 4 = 4 = PQ 2 1 and tan q = = 4 2 PS Also, cos q = PR  2  8 ⇒ PS = PR cos q = 4 ⋅  =  5  5 Harmonic mean between PA and PB 2( 20 − 2)( 20 + 2) 16 8 = = = = PS 2 20 2 5 5 ⇒ (c) is correct  8  ( 20 ) = 16 = (PR)2 Hence (PS)(PC ) =   5 

B

r12 + r22 + 4r22 − 4r12 = 5r22 − 3r12 (OA2 + OB2 + BE2)max is r12 + r22 + 4r22 = r12 + 5r22

26. (a, b, c) : QTS = PTQ = Q

QT ST = sin q, = cos q, PQ RS 4 4 = tan q and = cot q PQ RS 2

\

R

90° –  



2

q = 45° \ tanq = 1

Equation of PA is

x −7 1/ 2

−3 − 2 =1 2−7 =

y −2 1/ 2

S

O

T

P  QT   ST  + + PQ ⋅ RS  PQ   RS  = sin2q + cos2q + (4cotq)(4tanq)

27. (a, c) : Slope of PC = Q

p as QS is a diameter 2

Q

B

A 4 4 C (2, –3)

P 45° = r (7, 2)

 r r  \ 7 + ,2 +  lie on circle, then,  2 2 2

2

r   r  r  r     −3=0  + 6  2 +  − 4  7 +  + 2 +  7 + 2  2 2 2

⇒ r 2 + 10 2r + 34 = 0 \ r = −5 2 ± 4  −5 2 ± 4 −5 2 ± 4  ,2 + \ Point  7 +  2 2   ⇒ (2 ± 2 2 , −3 ± 2 2 ) taking –ve sign for A and +ve sign for B. 28. (a, c) : r adica l axis of t he g iven circles is x + 5y – 6 = 0, which passes through (1, 1) Let the given circles intersect the circle x2 + y2 + 2gx + 2fy + c = 0 orthogonally then 2g(–3) + 2f(–6) = c + 1 and 2g(–2) + 2f(–1) = c – 11 ⇒ 2g(–1) + 2f(–5) = 12 ⇒ –g – 5f – 6 = 0 ⇒ The radical axis passes through the centre (–g, –f) of the third circle. Also, it does not pass through the mid-point of the line joining the centres. 29. (a, c) : Q C2 is the director circle of C1. \ Equation of C2 is

x2 + y2 = 2(2)2 = 8 Again C3 is the director circle of C2. Hence the equation of C3 is x2 + y2 = 2(8) = 16 30. (a, c, d) : C1 ≡ x2 + y2 – 2x – 4y – 4 = 0 …(1) and C2 = x2 + y2 + 2x + 4y + 4 = 0 …(2) \ radical axis is C1 – C2 = 0 ⇒ –4x – 8y – 8 = 0 or x + 2y + 2 = 0, which is L = 0.

\ (a) option is correct. Now, centre and radius of C 1 = 0 are (1, 2) and 3 respectively. \ length of ^ from (1, 2) on L = 0 is 1+ 4 + 2 7 = ≠ radius 1+ 4 5 \ (b) option is wrong. Now, L is also the common chord of C1 and C2 \ (c) option is correct. Q Centres of C1 = 0 and C2 = 0 are (1, 2) and (–1, –2) respectively. \ Slope of line joining centres of circles C1 = 0 and −2 − 2 4 = = 2 = m1 (say) C2 = 0 is −1 − 1 2 1 And slope of L = 0 is − = m2 (say) 2 Hence, L is perpendicular to the line joining centres of C1 and C2. \ (d) option is correct. 31. (a, b, c) : (x – 1)2 + (y – 1)2 + l(x + y – 2) = 0 ⇒ x2 + y2 + (l – 2)x + (l – 2)y + 2 – 2l = 0 …(1) x2 + y2 + 4x + 5y – 6 = 0 …(2) Eq. of common chord PQ is S – S′ = 0 …(3) ⇒ (l – 6)x + (l – 7)y + 8 – 2l = 0 Let PQ|| to the given line x + y – 2 = 0 6−l ⇒ = −1 ⇒ 6 = 7, which is impossible. l−7 Let PQ ^ to the given line x + y – 2 = 0 6−l 13 ⇒ = 1 ⇒ l = , which is possible. l−7 2 13 But when l = , we can see that the circles (1) and 2 (2) are not intersecting each other and their radical axis is perpendicular to the given line x + y – 2 = 0. Now eq. (3) can be written as –6x – 7y + 8 + l(x + y – 2) = 0, which is in the form L1 + lL2 = 0. Solving L1 and L2, we get (6, –4). Q (c) option is true but not (d). 32. (a, c, d) : al2 + 2dl + 1 = bm2 bl2 + al2 + 2dl + 1 = bm2 + bl2 d2l2 + 2dl + 1 = b(l2 + m2) ⇒

dl + 1 l 2 + m2

= b

⇒ A fixed circle with centre = (d, 0) and radius = b. \ (x – d)2 + y2 = b is the required equation of a circle. mathematics today | JANUARY ‘16

43

33. (b, c) : x2 + y2 – 8x – 16y + 60 = 0 Equation of chord of contact from (–2, 0) is 3x + 4y – 34 = 0 Intersection of (1) and (2) is

…(1) …(2)

2

 34 − 3x   34 − 3x  x2 +  − 8 x − 16  + 60 = 0  4   4  ⇒ 5x2 – 28x – 12 = 0 ⇒ x = 6, –2/5  −2 44  \ (6, 4) and  ,   5 5  34 - 36 : 34. (d)

35. (c)

36. (c)

A C(4, 5) D

B

P

Centre of the given circle is C(4, 5). Points P, A, C, B are concyclic such that PC is a diameter of the circle. Hence, centre D of the circumcircle of DPAB is midpoint of PC, then we have t+4 sin t + 5 h= and k = 2 2 Eliminating t, we have k=

sin(2h − 4) + 5 sin(2 x − 4) + 5 or y = 2 2

sin −1 (2 x − 5) + 4 2 sin(2 x − 4) + 5 Thus range of y = is [2, 3] and period 2 is p. Also f (x) = 4 ⇒ sin(2x – 4) = 3, which has no real solutions. Also f (x) = 1 ⇒ sin(2x – 4) = –3, which has no real solutions. ⇒ f −1 (x ) =

But range of y = 37 - 39 :

p  sin −1 (2 x − 5) + 4  p is  − + 2, + 2  4 4  2

37. (d) : Equation of common chord is 2x – 2y = 0 Equation of family of circle is x2 + y2 + 2x + l(2x – 2y) = 0 Centre of circle is (–l –1, +l) 44


Centre lies on y = x l = –l – 1 ⇒ 2l = –1 ⇒ l = –1/2 Equation is x2 + y2 + x + y = 0 38. (a) 39. (c) : Let the second circle x2 + y2 + 2gx + 2fy = 0 Hence, x2 + y2 + 2gx + 2fy = 0 lies equal roots f + g = 0 Equation of common chord is 2(g – 3)x + 2(–g – 4)y + 7 = 0 (–6x – 8y + 7) + g(2x – 2y) = 0 Passes through the intersection point of –6x – 8y + 7 = 0 and 2x – 2y = 0 (h + 1)2 + (k − 1)2 1 1 ⇒  ,  ⇒ cos 60° = 2 2 2 40 - 42 : 40. (a)

41. (b)

42. (a) 2

y = log1/2 (x − 0.5) + log 2 4 x − 4 x + 1 = –log(2x – 1) + log22 + log(2x – 1) ⇒ y = 1, P(x, y) lies on the circle x2 + y2 = 10 ⇒ x = ±3, neglecting x = –3 ⇒ Q(1, 1), Q(2 , 1) lies inside the circle such that their abscissa is an integer . 43 - 45 : 43. (a) : If lines x + 2y + a = 0 intersect the circle x2 + y2 = 4, then 0+0+a <2 ⇒ −2 5

45

(C) C = (7, 5) r = 49 + 25 + 151 = 15 = OP OA = (7 − 2)2 + (5 − 7)2 AP = OP – OA AP = 15 – 13 = 2 (D) Let x + 5 = 14 cosq y – 12 = 14 sinq x2 + y2 = 365 + 28(12sinq – 5cosq)

P A

(2, –7) O (7, 5)

( x 2 + y 2 )min = 365 − 28 × 13 = 1 47. a – q, r; B – p, q; C – q; D – s (A) c = 1/2 (B) c = 1 (C) C1 = (0, 0), C2 = (1, 0) r1 = 3, r2 = |c| C1C2 < r1 – r2 ⇒ |c| < 2 (D) C1 = (0, 0), r1 = 3 C2 = (3, 4), r2 = |c| ⇒ |c| > 8 48. a - q; B - r; C - q, s; D - p Let the equations of the circles be x2 + y2 + 2g1x + 2f1y + c1 = 0 and x2 + y2 + 2g2x + 2f2y + c2 = 0 (a) Equation of the radical axis is 2(g1 – g2)x + 2(f1 – f2)y + c1 – c2 = 0 g −g Slope of the radical axis = − 1 2 f1 − f2 f −f Slope of the line joining the centres = 1 2 g1 − g 2 So the radical axis is perpendicular to the line joining the centre. (b) Common tangent to the intersecting circles of equal radii is at the same distance from the centres of the two circles and hence is parallel to the line joining the centres. (c) Since the line joining the centres of the circles to the mid-point of the common chord is perpendicular to the chord as well as it bisects the chord. (d) The line joining the centre of one to a point of intersection is tangent to the other circle. S o, by d e f i n it i on of or t h o gon a l it y, t h e y are perpendicular. 49. (6) : Centre of circle S1 = (2, 4) Centre of circle S2 = (4, 2) radius of circle S1 = radius of circle S2 = 4 \ Equation of circle S2 is (x – 4)2 + (y – 2)2 = 16 ⇒ x2 + y2 – 8x – 4y + 4 = 0 …(1) 46


Equation of circle touching y = x at (1, 1) can be taken as (x – 1)2 + (y – 1)2 + l(x – y) = 0 or x2 + y2 + x(l – 2) + y(–l – 2) + 2 = 0 …(2) As eqn. (2) is orthogonal to S2 l−2  −l − 2  \ 2 ⋅ (−4) + 2  ⋅ (−2) = 4 + 2  2   2  ⇒ –4l + 8 + 2l + 4 = 6 ⇒ –2l = –6 ⇒ l = 3 \ required equation of circle is x2 + y2 + x – 5y + 2 = 0. 1 25 26 − 8 18 3 + −2 = = = 2 radius = 4 4 4 4 2 50. (0) : Combined equation of the tangents drawn from (0, 0) to the circle is (x2 + y2 – 2rx – 2hy + h2)h2 = (–rx – hy + h2)2 Here coefficient of x2 + coefficient of y2 = 0 h = ±1 ⇒ (h2 – r2) = 0 ⇒ r 51. (1) : S(x, y) = x2 + y2 – x – y – 6 = 0 …(1) 1 1 has centre at C ≡  ,  2 2 According to the required conditions, the given point P(a – 1, a + 1) must lie inside the given circle. i.e. S(a – 1, a + 1) < 0 ⇒ (a – 1)2 + (a + 1)2 – (a – 1) – (a + 1) – 6 < 0 ⇒ a2 – a – 2 < 0, i.e., (a – 2)(a + 1) < 0 …(2) ⇒ –1 < a < 2 Also P and C must lie on the same side of the line P (see figure) C x+y–2=0 L(x, y) ≡ x + y – 2 = 0 …(3) i.e., L(1/2, 1/2) and L(a – 1, a + 1) must have the same sign. 1 1 1 1 Since L  ,  = + − 2 < 0 2 2 2 2 ⇒ L(a – 1, a + 1) = (a – 1) + (a + 1) – 2 < 0, i.e., a < 1 …(4) Inequalities (2) and (4) together give the permissible values of a as –1 < a < 1. 52. (5) : Q All three circles touch each other externally, \ C1C2 = 7, C1 C2C3 = 9 and C3C1 = 8 7+8+9 = 12 Also, s = C3 C2 2 D = s(s − a)(s − b)(s − c) = 12 × 5 × 3 × 4 r=

D = 5 s

nn

TRIGONOMETRY

*ALOK KUMAR, B.Tech, IIT Kanpur

This column is aimed at Class XII students so that they can prepare for competitive exams such as JEE Main/Advanced, etc. and be also in command of what is being covered in their school as part of NCERT syllabus. The problems here are a happy blend of the straight and the twisted, the simple and the difficult and the easy and the challenging.

ELEMENTS OF A TriANgLE In a triangle ABC, the angles are denoted by capital letters A, B and C and the length of the sides opposite to these angles are denoted by small letters a, b and c. Semi perimeter of the a+b+c and its area is triangle is given by s = 2 denoted by D. A + B + C = 180° z z

1 D = (base)(height) 2

SiNE rULE ANd ArEA OF TriANgLE z In DBCD, C CD = a sinB = b sinA ⇒

a b = sin A sin B

b

A a b c = = sin A sin B sin C Hence in DABC, a b c = = = 2R (Sine rule) sin A sin B sin C (where R is circumradius)

Similarly

z

a D c

Area of DABC, 1 1 1 D = × AB × CD = × c × a sin B = ac sin B 2 2 2 1 1 1 Similarly D = bc sin A = ac sin B = ab sin C. 2 2 2

COSiNE rULE In DDBC BC2 = DC2 + DB2 b = DC2 + (AB – AD)2 = CD2 + AD2 + AB2 – 2AB⋅AD A = AC2 + AB2 – 2AB⋅AC⋅cosA ⇒ a2 = b2 + c2 – 2bc cosA z

cos A =

z

cos B =

z

cosC =

B

C a D c

B

b2 + c 2 − a 2 2bc c 2 + a 2 − b2 2ca a 2 + b2 − c 2

2ab If ∠A = 60°, then b2 + c2 – a2 = bc If ∠B = 60°, then a2 + c2 – b2 = ac If ∠C = 60°, then a2 + b2 – c2 = ab

PrOJECTiON FOrMULAE In DABC BC = BD + DC = AB cosB + AC cosC ⇒ a = c cosB + b cosC z

a = b cosC + c cosB

z

b = c cosA + a cosC

z

c = a cosB + b cosA

A

B

D

C

* Alok Kumar is a winner of INDIAN NAtIoNAl MAtheMAtIcs olyMpIAD (INMo-91). he trains IIt and olympiad aspirants.


47

NAPiEr’S ANALOgY (TANgENT rULE) From Sine formula, b sin B b − c sin B − sin C A B −C = ⇒ = = tan .tan c sin C b + c sin B + sin C 2 2 A  B −C  b−c \ tan  = cot  2  b + c 2 z

z

z

z

tan

B C − A c −a tan  = cot   2  c+a 2

B (s − c)(s − a) = 2 s(s − b)

z

tan

C (s − a)(s − b) = 2 s( s − c )

C  A− B  a −b tan  = cot  2  a + b 2 Since 2 sin2

\ sin

A b2 + c 2 − a 2 = 1 − cos A = 1 − 2 2bc 2 2 a − (b − c) (a + b − c)(a − b + c) = = 2bc 2bc (2s − 2c)(2s − 2b) = 2bc

A (s − b)(s − c) = 2 bc

z

sin

A (s − b)(s − c) = 2 bc

z

sin

B (s − c)(s − a) = 2 ca

C (s − a)(s − b) = 2 ab A 2 Since 2 cos = 1 + cos A 2 sin

b2 + c 2 − a2 (b + c)2 − a2 = 2bc 2bc (b + c + a)(b + c − a) 2s(2s − 2a) = = 2bc 2bc = 1+

\ cos

A s( s − a ) = 2 bc

z

cos

A s( s − a ) = 2 bc

z

cos

B s ( s − b) = 2 ca

z

cos

C s( s − c ) = 2 ab

48

z

sin

A  B −C  b−c tan  = cot   2  b+c 2

HALF ANgLE FOrMULAE

z

A 2 , we get A cos 2 A (s − b)(s − c) tan = 2 s (s − a )

A Using tan = 2


ArEA OF CYCLiC QUAdriLATErAL Let ABCD be a cyclic D quadrilateral whose sides c AB, BC, CD and DA are d respectively a, b, c and d. b Area of quadrilateral a A

C

B

= (s − a)(s − b)(s − c)(s − d )

Circumradius of a cyclic quadrilateral 1 R= (ac + bd )(ad + bc)(ab + cd ) 4D CENTrOid ANd MEdiANS OF A TriANgLE The line joining any vertex of a triangle to the mid point of the opposite side of the triangle is called median of the triangle. The three medians of a triangle are concurrent and the point of concurrency of the medians of any triangle is called the centroid of the triangle. The centroid divides the median in the ratio 2 : 1. The distance of centroid from vertex A of triangle 1 2b2 + 2c 2 − a2 and the distance ABC, GA = 3 of centroid from side BC of a triangle ABC, 2D Ga = . 3a APOLLONiUS THEOrEM : In a triangle ABC, if AD is median through A, then AB2 + AC2 = 2(AD2 + BD2) CirCUMCirCLE The circle which passes through the angular points of a DABC, is called its circumcircle. The centre of this circle i.e., the B point of concurrency of the

A E O A A a/2 D a/2

F C

perpendicular bisectors of the sides of the DABC, is called the circumcentre. Radius of the circumcircle is given by the following formulae a b c abc R= = = = 2 sin A 2 sin B 2 sin C 4 D Distance of circumcentre from vertex A of triangle ABC = R and distance of circumcentre from side BC of triangle ABC is R cos A. OrTHOCENTrE ANd PEdAL TriANgLE OF A TriANgLE. In a triangle, the altitudes drawn from the three vertices to the opposite sides are concurrent and the point of cuncurrency of the altitudes of the triangle is called the orthocentre of the triangle.

ac ab and y = b+c b+c Also, let d be the length of AD we have DABD + DACD = DABC ⇒ x=

1 A 1 A 1 cd sin + bd sin = bc sin A, 2 2 2 2 2 bc sin A i.e., d = A b+c sin 2  2bc A = cos B b+c 2

\

x

A

C

y

D

iNCirCLE The circle which can be inscribed within the triangle so as to touch each of the sides of the triangle is called its incircle. The centre of this circle i.e., the point of concurrency of angle bisectors of the triangle is called the incentre of the DABC. A A/2 E

PD = 2R cosB cosC PE = 2R cosA cosC PF = 2R cosA cosB PA = 2R cosA PB = 2R cosB PC = 2R cosC 1 2 2 1 AD = b + c + 2bc cos A = 2b2 + 2c 2 − a2 2 2 1 2 2 1 c + a + 2ac cos B = 2c 2 + 2a2 − b2 2 2 1 2 2 1 CF = a + b + 2ab cos C = 2a2 + 2b2 − c 2 2 2 The triangle formed by joining the feet of these perpendiculars is called the pedal triangle i.e., DDEF is the pedal triangle of DABC. Pedal triangle constructed on DABC w.r.t. orthocentre Its sides : R sin2A, R sin2B, R sin2C Its angles : P – 2A, P – 2B, P – 2C Its circumradius = R/2 and inradius = 2R cosA cosB cosC BE =

z z z

BiSECTOrS OF THE ANgLES If AD bisects the angle A and divide the base into portions x and y, we have, by Geometry, x AB c x y x+y a = = = \ = = y AC b c b b+c b+c

r I

90° – B/2 B

B/2

r D

r

F

C/2

C

Radius of the incircle is given by the following formulae : D A B r = = (s − a) tan = (s − b) tan s 2 2 C A B C = (s −c) tan = 4 R sin sin sin 2 2 2 2 r=

a sin (B / 2) sin (C / 2) b sin ( A / 2) sin (C / 2) = cos A / 2 cos B / 2 c sin (B / 2)) sin ( A / 2) = cos (C / 2)

iNSCriBEd & CirCUMSCriBEd POLYgONS (iMPOrTANT FOrMULAE) z Area of Polygon of n sides inscribed in a circle of 1 2p radius R = nR2 sin 2 n z Area of Polygon of n sides inscribing a circle of 1 p radius r = nr 2 tan 2 n mathematics today | JANUARY ‘16

49

z

Side of Inscribed polygon = 2R sin

p n

p n iNVErSE TrigONOMETriC FUNCTiON If f : A → B is one to one and onto function and g is a rule under which for every element

z

Side of Circumscribed polygon = 2r tan

y ∈ B, $ a unique element x ∈ A then g : B → A is called Inverse function of f : A → B. i.e., g = f –1 \ x = g(y) ⇒ x = f –1(y) So y = f(x) and x = g(y) such that f (g(y)) = y and x = g(f (x)) then f and g are said to be inverse function of each other.

dOMAiN, rANgE ANd grAPH OF iNVErSE TrigNOMETriC FUNCTiONS Function y = sin–1x

domain –1 ≤ x ≤ 1

range

graph

–p/2 ≤ y ≤ p/2

y

/2 –1

y = sin–1x

O

1

x

–/2

y = cos–1x

–1 ≤ x ≤ 1

y  y = cos�1 x

0≤y≤p

/2 �1

y = tan–1x

x∈R

–p/2 < y < p/2

O

1

x

y = tan–1x

/2

O –/2

y = cot–1x

x∈R

y

0
x

–1  y = cot x

/2 x

O

y = sec–1x

|x| ≥1

y

0 ≤ y ≤ p, y ≠ p/2

2 3/2 /2 x



–2 –1O

1 2

x

–/2

– y y = sec–1x

y = cosec–1x

|x| ≥1

y

–p/2 ≤ y ≤ p/2, y ≠ 0

2 3/2 x

50


/2 –2 –1O



1 2 –/2 – y y = cosec–1x

x

PrOPErTiES OF iNVErSE TrigONOMETriC FUNCTiONS Property i z sin–1(–x) = –sin–1x " x ∈ [–1, 1] z tan–1(–x) = –tan–1 x " x ∈ R z cosec–1(–x) = –cosec–1x " x ∈ (–∞, –1] ∪ [1, ∞) cos–1(–x) = p –cos–1x " x ∈ [–1, 1] z z cot–1(–x) = p –cot–1x " x ∈ R z sec–1(–x) = p – sec–1x " x ∈(–∞, –1] ∪ [1, ∞) Property ii 1 z sin −1   = cosec −1 x " x ∈(−∞, −1] ∪ [1, ∞) x −1  1 

cos

tan −1 x1 + tan −1 x2 + ..... + tan −1 xn  s − s + s ...  = tan −1  1 3 5  1 − s2 + s4 − s6 ...  Where s1 = x1 + x2 + ..... + xn = ∑ x1 s2 = x1x2 + x2 x3 + ..... + xn−1xn = ∑ x1x2 s3 = ∑ x1x2 x3 and so on. Property V

z

  sin −1 x 1 − y 2 + y 1 − x 2 ,   2 2 if − 1 ≤ x , y ≤ 1 & x + y ≤ 1 or if xy < 0 & x 2 + y 2 > 1     2   −1 −1 −1  x 1 − y sin x + sin y =  p − sin     + y 1 − x 2   if 0 < x , y ≤ 1 & x 2 + y 2 > 1    2   −1  x 1 − y p s in − −      + y 1 − x 2   if − 1 ≤ x , y < 0 & x 2 + y 2 > 1

z

 −1 2 2 sin x 1 − y − y 1 − x ,  2 2 if − 1 ≤ x , y ≤ 1 & x + y ≤ 1  2 2 or if xy > 0 & x + y > 1    2   −1  x 1 − y p sin −    2  − y 1 − x    sin −1 x − sin −1 y = if 0 < x ≤ 1, − 1 ≤ y < 0  & x2 + y2 > 1     2   −1  x 1 − y p sin − −    2  − 1 − y x    if − 1 ≤ x < 0 , 0 < y ≤ 1   & x2 + y2 > 1  

−1

  = sec x " x ∈(−∞, −1] ∪ [1, ∞) x −1 cot x , x > 0 z tan −1 (1 / x ) =  −1 − p + cot x , x < 0 Property iii p z sin −1x + cos −1x = , "x ∈[−1, 1] 2 p z sec −1x + cosec −1x = , "x ∈(−∞, −1] ∪ [1, ∞) 2 p − 1 − 1 z tan x + cot x = , "x ∈ R 2 Property iV  −1 x + y tan 1 − xy , xy < 1   p + tan −1 x + y , x > 0,  1 − xy  −1 −1 z tan x + tan y =  y > 0, xy > 1  x+y − p + tan −1 , x < 0, 1 − xy   y < 0, xy > 1    −1 x − y tan 1 + xy , xy > −1   p + tan −1 x − y , x > 0,  1 + xy  −1 −1 z tan x − tan y =  y < 0 & xy < −1  x−y − p + tan −1 , x < 0, 1 + xy   y > 0 & xy < −1   z

remark : If x1, x2, x3,.....xn ∈ R, then

{

}

{

}


51

Property Vi

z

 −1  2 2 cos  xy − 1 − x 1 − y    if − 1 ≤ x , y ≤ 1 & x + y ≥ 0  −1 −1 cos x + cos y =   xy − 1 − x 2  2 p − cos −1    1 − y2      if − 1 ≤ x , y ≤ 1& x + y ≤ 0

 −1  2 2 cos  xy + 1 − x 1 − y   if − 1 ≤ x , y ≤ 1, x ≤ y z cos −1 x − cos −1 y =  − cos −1  xy + 1 − x 2 1 − y 2      if 0 < x ≤ 1, − 1 ≤ y ≤ 0 & x ≥ y Property Vii  −1 1 −1 if ≤x≤ 2 sin x 2 2  1  z sin −1 (2 x 1 − x 2 )  p − 2 sin −1 x , if ≤ x ≤1 2  −1  −1 − p − 2 sin x if − 1 ≤ x ≤ 2  −1 2cos x if 0 ≤ x ≤ 1 z cos −1 (2 x 2 − 1) =  −1 2 p − 2cos x if − 1 ≤ x ≤ 0

z

2 tan −1 x if − 1 ≤ x ≤ 1  2x  sin −1 = p − 2 tan −1 x if x > 1 2  1+ x  −1 − p − 2 tan x if x < −1

z

cos −1

z

2 tan −1 x , if − 1 ≤ x ≤ 1  2x  tan −1 = p − 2 tan −1 x , if x < −1 2  1− x  −1 − p + 2 tan x if x > 1

52

z

 −1 −1  p + 3 tan x , x < 3 3   3x − x −1 1  tan −1  = 3 tan −1 x , 3 

Property iX : Conversion z

z


  1  −1  1   = sec −1   = cosec   x 2   1− x 

 1 − x2 cos −1 x = sin −1 1 − x 2 = tan −1   x

 = 

 1   x  −1  1  −1 cot −1    = sec   = cosec  x  1 − x2   1 − x2  z

−1 2 tan x if x ≥ 0 = 1 + x 2 −2 tan −1 x if x ≤ 0

1 −1  −1 3 sin x ; 2 ≤ x ≤ 2  1  sin −1 (3x − 4 x 3 ) =  p − 3 sin −1 x ; ≤ x ≤ 1 2  − p − 3 sin −1 x ; − 1 ≤ x ≤ −1  2

 x  sin −1 x = cos −1 1 − x 2 = tan −1    1 − x2   1 − x2 = cot −1   x

1 − x2

Property Viii

z

z

−1  −1 −2 p + 3cos x ; − 1 ≤ x ≤ 2  −1 1  ≤x≤ cos −1 (4 x 3 − 3x ) = 2 p − 3cos −1x ; 2 2  3cos −1x ; 1 ≤ x ≤ 1  2

 1   −1   = cos   1 + x2    2  −1  1  −1  2 −1  1 + x  = cot   = sec 1 + x = cosec    x  x  x tan −1 x = sin −1   1 + x2

Section-i Single correct Answer type

 p  If x = log cot  + q   , then sinh x =   4 (a) tan 2q (b) cot 2q (c) –tan 2q (d) –cot 2q 1.

2. AB is a vertical pole with B at the ground level and A at the top. A man finds that the angle of elevation of the point A from a certain point C on the ground is 60°. He moves away from the pole along the line BC to a point D such that CD = 7 m. From D the angle of

elevation of the point A is 45°. Then the height of the pole is

8.

If cos q =

a cos f + b , then tan q/2 is equal to a + b cos f

(a)

7 3 ( 3 + 1)m 2

(b)

7 3 ( 3 − 1)m 2

(a)

(c)

7 3 1 m 2 3 +1

(d)

7 3 1 m 2 3 −1

a −b  a + b  tan(f / 2) (b)

(c)

a −b  a + b  sin(f / 2) (d) none of these

3. The angle of elevation of a cloud from a point h m above the surface of a lake is q and the angle of depression of its reflection in the lake is f. The height of the cloud is hsin(f + q) hsin(f − q) (a) (b) sin(f − q) sin(f + q) hsin(q − f) sin(q + f)

(c)

hsin(q + f) sin(q − f)

4.

p y p x If tan  +  = tan3  +  , then 4 2 4 2

(d)

 3 + sin2 x  sin x   equals  1 + 3 sin2 x  (a) cos y

(b) sin y

(c) sin 2y

(d) 0

5. If x1, x2, x3, ...., xn are in A.P, whose common difference is a, then the value of sin a [sec x1 sec x2 + sec x2 sec x3 + ..... + sec xn−1 sec xn ] = (a)

sin na cos xn cos x1

(b)

sin(n −1)a cos xn cos x1

(c)

sin(n +1)a cos xn cos x1

(d)

cos(n −1)a cos xn cos x1

6.

If a sin2x + b cos2x = c, b sin2y + a cos2y = d and 2

a a tanx = b tany, then 2 = b (a + d )(c + a) (a − d )(c − a) (a) (b) (b + c)(d + b) (b − c)(d − b) (c)

(a − d )(b − a) (a − c)(c − b)

7.

If cos x sin 2 x = ∑ar sin(rx ), "x ∈ R then 3

1 (a) n = 5, a1 = 2 1 (c) n = 5, a2 = 8

(d)

(d − a)(c − a) (b − c)(d − b)

n

r =0

1 4 1 (d) n = 5, a2 = 4

(b) n = 5, a1 =

9.

The value of

(a)

−9 2

(b)

10

a +b  a − b  cos(f / 2)

∑ cos3

pr is equal to 3

−7 2

(c)

r=0

−9 8

(d)

−1 8

10. The value of sin3100 + sin3500 – sin3700 is equal to 3 3 3 3 (a) − (b) (c) − (d) − 2 4 8 4 sin 2b 11. If tan(a − b) = , then 3 − cos 2b (a) tana = 2tanb (b) tanb = 2tana (c) 2tana = 3tanb (d) 3tana = 2tanb 12. In a triangle ABC, if angle C is obtuse and angles A and B are given by roots of the equation tan2x + p tanx + q = 0, then the value of q is (a) greater than 1 (b) less than 1 (c) equal to 1 (d) 0 13. If 2sinx – cos2x = 1, then cos2x + cos4 x is equal to (a) 1 (b) – 1 (c) − 5 (d) 5 14. If ABCD is a cyclic quadrilateral such that 13 cosA + 12 = 0 and 3tanB – 4 = 0, then the quadratic equation whose roots are tanC and cosD is (a) 15x2 + 60x – 11 = 0 (b) 60x2 + 11x – 15 = 0 (c) 11x2 + 60x – 15 = 0 (d) none of these 15. If A, B, C are the angles of a triangle such that A C cot = 3 tan , then sinA, sinB, sinC are in 2 2 (a) A.P. (b) G.P . (c) H.P. (d) none of these 16. If

2 tan a 2 tan a / 2 is equal to = l, then 1 + sec a + tan a 1 + tan a / 2

1 (b) l (c) 1 – l (d) 1 + l l 17. If 0 ≤ A, B, C ≤ p and A + B + C = p, then the minimum value of sin 3A + sin 3B + sin 3C is 3 3 (a) –2 (b) − 2 (c) 0 (d) none of these (a)


53

18. The value of x which satisfies the equation 4x is 2 tan −1 2 x = sin −1 1 + 4x2 1  1  (a)  , ∞  (b)  −∞, −   2  2    1 1 (d)  − ,   2 2

(c) [–1, 1]

19. Number of integral solutions of the equation  1 − 3x 2  3 tan −1 x + cos −1   = 0 is  (1 + x 2 )3/2  (a) 1

(b) 2

(c) 0

(d) infinite

p 2p 4p 20. In a triangle ABC, with A = , B = ,C= , 7 7 7 then a2 + b2 + c2 is (R = circumradius of DABC) (a) 4R2

(b) 6R2

(c) 7R2

(d) 8R2

21. For what value of x, sin(cot–1(x + 1)) = cos(tan–1 x)? 1 1 (a) (b) 0 (c) 1 (d) − 2 2 22. If the equation x2 + 12 + 3 sin(a + bx) + 6x = 0 has atleast one real solution where a, b ∈ [0, 2p], then value of cosq, where q is least positive value of a + bx is p (a) p (b) 2p (c) 0 (d) 2 23. In any DABC, which is not right angled, ScosA cosecB cosecC is (a) constant (b) less than 1 (c) greater than 2 (d) none of these 24. If C = 90° in DABC, then  a   b  + tan −1  tan −1  is equal to  b +c   c + a  (a)

p 2

(b)

p 4

p 3 Section-ii (c)

(d) p

Multiple correct Answer type

25. In DPQR, if 3sinP + 4cosQ = 6 and 4sinQ + 3cosP = 1 then ∠R can be 3p p 5p p (a) (b) (c) (d) 4 6 6 4 2

 sin q  tan q 26. If  = = 3, then  sin f  tan f 54


(a) tan f =

1 3

(c) tan q = 3

(b) tan f = −

1 3

(d) tan q = − 3

27. Which of the following quantities are rational?  11p   5p  (a) sin  sin  12   12   9p   4 p  (b) cosec   sec    10   5  4 p 4 p (c) sin   + cos   8 8

2p   4p   8p   (d) 1 + cos  1 + cos  1 + cos       9 9 9  28. A solution (x, y) of the system of equations 1 1 x − y = and cos2 (px ) − sin2 (py ) = is 3 2 7 5 2 1 (a)  ,  (b)  ,  6 6 3 3  −5 −7  (c)  ,   6 6 

(d)  13 , 11  6 6

2 y2 − y + 1 ≤ 2 is 29. If 0 ≤ x ≤ 2p then 2cosec x 2 (a) satisfied by exactly one value of y (b) satisfied by exactly two values of x (c) satisfied by x for which cos x = 0 (d) satisfied by x for which sin x = 0

30. If x cosa + y sina = x cosb + y sinb = 2a and a  b 2 sin   sin   = 1 then  2  2 (a) y2 = 4a(a – x) (b) cos a + cos b = cos a cos b 4a 2 + y 2 (c) cos a. cos b = 2 x + y2 4ax (d) cos a + cos b = 2 x + y2 tan 3 A = k (k ≠ 1), then 31. If tan A cos A k − 1 = cos 3 A 2 1 (c) k < 3 (a)

(b)

sin 3 A 2k = sin A k − 1

(d) k > 3

Section-iV

q 32. Let fn (q) = tan (1 + sec q)(1 + sec 2q) 2

Matrix-Match type

n

(1 + sec 4q)....(1 + sec 2 q), then (a)

p f2   = 1  16 

(b)

n

Column i

p f3   = 1  32 

(A)

 p  =1 (d) f5   128 

p (c) f 4   = 1  64  33. Sum of series

(B)

2r + 1

  2   is 2  r (r + 1) r + 2r + r − 1     

∑ sin−1 

r =1

p  1  − sin −1   n + 1  2  1  (c) cos −1   n + 2  (a)

(C) (D)

 1  (b) cos −1   n + 1 

Paragraph for Question No. 35 to 37 p 3p 5p Let cos , cos , cos are the roots of equation 7 7 7 3 2 8x – 4x – 4x + 1 = 0 p  3p   5p  35. The value of sec   + sec   + sec   is 7  7   7  (a) 2 (b) 4 (c) 8 (d) none of these

(b)

1 8

p 3p 5p sin sin is 14 14 14 (c)

7 4

(d)

7 8

p  3p   5p  37. The value of cos   cos   cos   is  14   14   14  (a)

1 2

(b)

1 8

(c)

7 2

(d)

cot160 cot 440 + cot440 cot760 (r) – cot 760 cot 160 = 9 (s)  rp  ∑ sin2  18  =

3 5

Column i

Column ii

(A)

cos200 (p)

x3 – 3x2 – 3x + 1 = 0

(B)

sin100

32x5 – 40x3 + 10x – 1 = 0

(C)

tan150 (r)

8x3 – 6x – 1 = 0

(D)

sin60

8x3 – 6x + 1 = 0

(q) (s)

40. Match the following : Column i

comprehension type

1 4

If tanq is the G.M. between (p) 1 sinq and cosq, then 2 – 4 sin2q + 3sin4q – sin6q = (q) 0 3 cot 200 − 4 cos 200 =

39. Match the following trigonometric ratios with the equations whose one of the roots is given :

(d) none of these

Section-iii

(a)

Column ii

r =1

34. Minimum positive values of x and y such that p x + y = and sec x + sec y = 2 2 are 2 p p (a) x = (b) y = 4 4 p p (c) x = − (d) y = − 4 4

36. The value of sin

38. Match the following :

7 8

Column ii (A) If sinq = 3sin(q + 2a), then (p) 0 the value of tan(q + a) + 2 tana is (B) If p sinq + q cosq = a and (q) 1 p cosq – q sinq = b, then p+a q −b + +1is equal to q +b p−a (C) Then value of the expression (r) sec q p 2p 10 p cos cos cos 7 7 7 p 3p 5p − sin sin sin is 14 14 14 (D) If secq + tanq = 1, then one (s) 1 − root of the equation 4 (a – 2b + c)x2 + (b – 2c + a) x + (c – 2a + b) = 0 is (t) 1 − 2 mathematics today | JANUARY ‘16

55

Section-V

3. (a) : tan q =

integer Answer type

41. Let f(x) = 0 be an equation of degree six, having p integer coefficients and whose one root is 2 cos . 18 Then, the sum of all the roots of f ′(x) = 0, is 42. If cosq + cos2q + cos3q = 1 and sin6q = a + b sin2q + c sin4q, then a + b + c = sin q sin 3q sin 9q 1 + + = [tan Bq − tan Cq], cos 3q cos 9q cos 27q A then (27A – B – 27C) =

p , then the value of 14

(tana tan2a + tan2a tan 4a + tan 4a tan a) is 47. In a triangle ABC, if r1, r2, r3 are the ex-radius and bc ac ab abc  s s s  + + =k + + − 3 , then k is equal to 2D  a b c  r1 r2 r3 48. If a + b + g = p and a +b − g   g + a −b g +b − a tan  tan     = 1, 4 4 4       then the value of 1 + cosa + cosb + cosg is K – 1, where K is

tan 

SolutionS 1. (c) : x = log [cot(p/4 + q)]

h

⇒ h − h/ 3 = 7

56

7 3 3 −1

m

D


60° 45° 7m C x B

x B



h A

3



y

F h C x+h E

y  x 1 + tan  2 = 2 4. (b) :  y x  1 − tan  1 − tan 2 2  1 + sin y (1 + sin x )3 Squaring both sides, we get = 1 − sin y (1 − sin x )3 Using componendo and dividendo 2 sin y (3 + sin2 x ) = sin x 2 1 + 3 sin2 x 5. (b) : sina (secx1secx2 + secx2secx3 + .... + secxn–1secxn) sin(xn − xn−1 ) sin( x2 − x1 ) sin( x3 − x2 ) + + .... + cos xn−1 cos xn cos x1 cos x2 cos x2 cos x3 = tanx2 – tanx1 + tanx3 – tanx2 +....+ tanxn – tanxn–1 sin( xn − x1 ) sin(n − 1)a = tan xn − tan x1 = = cos xn cos x1 cos xn cos x1

=

6. (a) : a tan2 x + b = c(1 + tan2 x) c −b d −a ⇒ tan2 x =  , tan2 y =   a−c   b − d  a2

cos q − sin q  cos q − sin q  = log  ⇒ ex =  cos q + sin q  cos q + sin q  x −x 1  (cos q − sin q)2 − (cos q + sin q)2  e −e sinh x = =   2 2  (cos q + sin q)(cos q − sin q)  1  −4 cos q sin q  − sin 2q = = − tan 2q =  2  cos2 q − sin2 q  cos 2q A

⇒ h=

2h ⇒ x= cot q ⋅ tan f − 1

1 + tan

44. If tan x = tan y = tan z and x + y + z = p, 2 3 5 38 tan2x + tan2y + tan2z = , then K = K 45. If sinq + sin2q + sin3q = 1, then the value of cos6q – 4cos4q + 8cos2q must be

2. (a) : x = h cot 60° = h/ 3 x + 7 = hcot45°

2h + x y

h sin(f + q) CD = h + x = sin(f − q)

43. If

46. If a =

tan f =

D

x y

2

=

tan2 y 2

=

(a − d )(c − a) (b − c)(d − b)

tan x b 7. (b) : cos3x sin2x = cos2 x cosx sin2x  1 + cos 2 x   2 sin 2 x cos x  =     2 2 1 = (1 + cos 2 x )(sin 3x + sin x ) 4 1 1 1 = [sin 3x + sin x + (2 sin 3x cos 2 x ) + (2 cos 2 x sin x )] 4 2 2 1 1 1 = [sin 3x + sin x + (sin 5x + sin x ) + (sin 3x − sin x )] 4 2 2 1 3 1 = [sin x + sin 3x + sin 5x] 4 2 2 1 By comparing, we get n = 5, a1 = 4

12. (b) : We have A + B = p – C ⇒ tan(A + B) = – tanC

 1 − cos q  8. (a) : tan q / 2 =   1 + cos q 

⇒

 a cos f + b  1−   a + b cos f  (a − b)(1 − cos f) = = (a + b)(1 + cos f)  a cos f + b  1+    a + b cos f  =

=

= tanA⋅tanB < 1 ⇒ q<1 13. (a) : Given, 2 sinx + 2sin2x – 1 = 1 or sin2 x + sinx – 1 = 0

a −b tan(f / 2) a +b

9. (d) : Let I = 10

10

1

∑ 4  cos 3

r =0

1

∑ 4  cos pr + 3 cos

r =0

\ I1 =

\ sinx =

pr pr  + 3 cos  3 3 

−1 + 5 2 3− 5 5 −1 ⇒ sin2 x = ⇒ cos2 x = 2 2

10

r =0

11 p 10 p    cos 2 ⋅ 3 sin 6  pr I2 = 3 ∑ cos = 3 1 +  p 3 r =0   sin   6 3 1 3 1    3 = 3 1 −  = − ⇒ I = 1 −  = −  2 2 4  2 8 10

10. (d) : We have, sin3100 + sin3500 – sin3700 1 = [(3 sin100 − sin 300 ) + (3 sin 500 − sin1500 ) 4 − (3 sin 700 − sin 2100 )]

=

1 3 3 3(sin 100 − 2 cos 600 ⋅ sin 100 ) −  = −  4 2 8

sin 2b 2 sin b ⋅ cos b = 11. (a) : We have 3 − cos 2b 2 − 2 cos 2b + 1 + cos 2b =

2 sin b ⋅ cos b 2

2

4 sin b + 2 cos b

 tan(a − b) =

=

tan b 2

1 + 2 tan b

=

2 tan b − tan b 1 + 2 tan2 b

tan a − tan b 1 + tan a ⋅ tan b

5 −1 5 +1 × =1 2 2 14. (b) : In a cyclic quadrilateral, no angle is greater than 180° 12 p Here cos A = − ⇒ < A < p and 0 < C < p/2 13 2 (since A + C = 180°) 5 5 \ tan A = − ⇒ tan C = 12 12 4 p p Also tan B = ⇒ 0 < B < and < D < p 3 2 2 \ cos2 x(1 + cos2 x ) =

∑ cos pr = 1 − 1 + 1 − 1 + ..... − 1 + 1 = 1

1 3 3(sin 100 + sin 500 − sin 700 ) −   4 2

…(i)

(since B + D = 180°) 3 3 \ cos B = ⇒ cos D = − 5 5 Now, the required equation is  5 3  5   3 x2 −  −  x +    −  = 0  12 5   12   5  ⇒ 60x2 + 11x – 15 = 0 A C ⋅ cot = 3 2 2 A C A−C cos ⋅ cos cos 2 =2 2 2 =3 ⇒ A C A+C sin ⋅ sin cos 2 2 2 (using componendo and dividendo)

15. (a) : Given cot

⇒

…(ii)

\ By (i) and (ii), we get tan a − tan b 2 tan b − tan b ⇒ tana = 2tanb = 1 + tan a ⋅ tan b 1 + 2 tan2 b

−1 ± 5 2

Consider, sinx =

pr  1  = (I1 + I2 ) 3  4

=

tan A + tan B > 0  tan A > 0, tan B > 0, tan C < 0 1 − tan A ⋅ tan B

⇒

A+C A−C cos 2 2 = 2 ⇒ 2sinB = sinA + sinC A+C A+C 2 sin ⋅ cos 2 2 2 sin


57

16. (b) : We have =2

2 tan a 2 sin a = 1 + sec a + tan a 1 + cos a + sin a 2 tan a / 2

2

(1 + tan a / 2) + (1 − tan2 a / 2) + 2 tan a / 2 2 tan a / 2 = 1 + tan a / 2 17. (a) : Since A + B + C = p ⇒ All of sin3A, sin3B, sin3C can’t be negative. Let us take sin3A = –1 ⇒ A = p/2 ⇒ sin3A = –1, sin3B = –1 and sin 3C = 0 So minimum value is – 2.

  1 and cos(tan −1 x ) = cos  cos −1    1 + x2 By (i) and (ii), we get

cos A

∑ cos A cosec B cosec C = ∑ sin B sin C

23. (a) : Since,

So, number of integral solutions is 2. 20. (c) : a2 + b2 + c2 = 4R2 (sin2 A + sin2B + sin2C) = 2R2(1 – cos2A + 1 – cos2B + 1 – cos2C)

But in right angled DABC

  2p 4p 8p   = 2R2 3 −  cos + cos + cos   7 7 7      2p 4p 6p   = 2R2 3 −  cos + cos + cos    7 7 7      2p 4p  p p 2 sin ⋅ cos + 2 sin ⋅ cos     1 7 7 7 7 = 2 R 2 3 −    p 6 p p   +2 sin ⋅ cos  2 sin     7  7 7  5p 3p   p  3p sin − sin + sin − sin   1 7 7 7 7   = 2 R 2 3 −  7p 5p    2 sin p   + sin − sin  7 7 7    1 = 2 R 2 3 +  = 7 R 2  2    1 −1 −1 21. (d) : sin(cot (x + 1)) = sin  sin     x 2 + 2 x + 2   1 ⇒ sin(cot −1 (x + 1)) = …(i) x 2 + 2x + 2 58


1

= x + 2x + 2 1 + x 2 22. (c) : (x + 3)2 + 3 + 3sin (a + bx) = 0 x = –3, sin(a + bx) = – 1 ⇒ sin(a – 3b) = –1 p a − 3b = (4n − 1) , n ∈ I 2 If n = 1 ⇒ a – 3b = 3p/2 ⇒ cos(a – 3b) = 0 2

1 1 p p ≤ 2 tan −1 2 x ≤ ⇒ − ≤ x ≤ 2 2 2 2  p p 19. (b) : Let tan −1 x = q, q ∈  − ,   2 2 3q + cos–1 (cos 3q) = 0 cos–1 (cos 3q) = – 3q ⇒ –p ≤ 3q ≤ 0 p ⇒ − ≤q≤0 3 ⇒ x ∈ − 3 , 0 18. (d) : −

 1  =   1 + x2 …(ii) 1

=

− ∑ cos( B + C ) sin B sin C

= ∑ (1 − cot B cot C ) = 3 − ∑ cot A cot B = 2 b   a + ab   24. (b) : tan −1  b + c c + a  as <1 a b (b + c)(c + a)  1− ⋅  b+c c+a  −1 c2 = a2 + b2 \ tan (1) =

p 4

25. (a) : Given equations ⇒ 16 + 9 + 24 sin(P + Q) = 37 5p p ⇒ P +Q = or 6 6 p 5p If P + Q = then R = 6 6 p 3 3 If P < , 3 sin P < , then 3 sin P + 4 cos Q < + 4 < 6 6 2 2 p p \ P + Q = is not possible \ R = 6 6 sin q sin q sin q cos f = sin f sin f sin f cos q

26. (a, b, c, d) : ⇒ ⇒ ⇒

sin q cos f = ⇒ sin 2q = sin 2f sin f cos q 2 tan q 2

1 + tan q

=

6 tan f

2

1 + 9 tan f

2 tan f 1 + tan2 f

=

2 tan f

1 ⇒ tan2 f = ⇒ tan q = ± 3 3 1 + tan f 2

27. (a, b, c, d) : (a) sin

11p 5p sin 12 12

p  p  1 p 1 = sin   cos   = sin   = ∈Q  12   12  2  6  4  9p   4 p   p  p (b) cosec   sec   = − cosec   sec   = −4 ∈Q  10   5   10   5  2 p

2 p

1 3 (c) 1 − 2 sin   cos   = 1 − = ∈Q 8 8 4 4

1 and 3

cos(2r q) n

= 2n+1 ⋅ tan(q / 2) ∏

cos2 (2r −1 q) cos(2r q)

= 2n+1 ⋅ tan(q / 2) ⋅ cos2 (q / 2) ∏

r =0

cos(2r −1 q) cos(2n q)

sin(2n q) = tan(2nq) = 2n ⋅ sin q ⋅ n 2 .sin q.cos(2n q) p p \ Alternate (a) : f2   = tan   = 1  16  4 p p Alternate (b) : f3   = tan   = 1  32  4

2 ( y − 1)2 + 1 29. (a, b, c) : 2cosec x ≤ 2 2

p p Alternate (c) : f 4   = tan   = 1  64  4

2 ⇒ 2cosec x ( y − 1)2 + 1 ≤ 2

 p  p = tan   = 1 Alternate (d) : f5    128  4

⇒ cosec2 x = 1 and y = 1 p 3p ⇒ x = , , y =1 2 2 30. (a, b, d) : a and b satisfy x cosq + y sinq = 2a ⇒ (x2 + y2) cos2q – 4ax cosq + (4a2 – y2) = 0 2

2

4ax 4a − y cos a + cos b = 2 , cos a ⋅ cos b = 2 2 x +y x + y2

a  b a b 2 sin   sin   = 1 ⇒ 4 sin2   sin2   = 1  2  2 2 2 ⇒ cosa + cosb = cosa⋅cosb 4a 2 − y 2 ⇒ y 2 = 4a(a − x ) = x2 + y2 x2 + y2 4ax

k +1 = 2 cos 2 A k −1 sin 3 A 2k = 1 + 2 cos 2 A = sin A k −1

31. (a, b, c, d) :

Also k =

r =0

2 cos2 (2r −1 q)

n

1 cos {p(x + y)} cos {p(x – y)} = 2 ⇒ x + y = 2n, n ∈ Z

⇒

n

= tan(q / 2) ∏

r =0

2p   4p  1 p  (d)  2 cos2   2 cos2 2 cos2 = ∈Q    9 9  9  8 28. (a, c, d) : x – y =

n 1 + cos(2r q)    = tan(q / 2) ∏   r r =0   cos(2 q) 

3 − tan2 A

1 ⇒ k < ,k > 3 3 1 − 3 tan A 2

n

32. (a, b, c, d) : fn (q) = tan(q / 2) ∏ (1 + sec 2r q) r =0

 r 2 + 2r − r 2 − 1  33. (a, b) : Tr = sin −1   r (r + 1)     1 1 1 1 Tr = sin −1  1 − − 1−   r r 2  (r + 1)2 r + 1 1  1  Tr = sin −1   − sin −1  r   r + 1  ⇒ Sn =

p  1   1  − sin −1  = cos −1   n + 1   n + 1  2

 x + y  sec x + sec y 34. (a, b) : sec  ≤  2  2 p 3p 5p 35. (b) : sec , sec , sec are the roots of 7 7 7 x3 – 4x2 – 4x + 8 = 0  3p   5p  p sec   + sec   + sec   = 4 7  7   7  p 3p   8 x 3 − 4 x 2 − 4 x + 1 = 8  x − cos   x − cos     7 7  5p    x − cos 7  mathematics today | JANUARY ‘16

59

36. (b) : Put x = 1  p   3p   5p  1 ⇒ sin   sin   sin   =  14   14   14  8 37. (d) : Put x = –1 7 p  3p   5p  ⇒ cos   cos   cos   =  14   14   14  8 38. A - p; B - p; C - r; d - s (A) tan2q = sinqcosq ⇒ sinq = cos3q \ (1 – sin2q) + (1 – 3sin2q) + 3sin4q – sin6q = cos2q + (1 – sin2q)3 = cos2q + cos6q = cos2q + sin2q = 1 (B) sin40° = sin(60° – 20°) 2 sin 200 cos 200 =

3 1 cos 200 − sin 200 2 2

4 cos 200 = 3 cot 200 − 1 (C)

= =

3 + cot 760 cot 160

=

cot 760 + cot 160

3 sin 760 sin160 + cos 760 cos160 siin(760 + 160 )

sin(760 + 160 ) 0

sin 92

=

1 − cos 920 0

sin 92

= tan 460

= cot 44° p  2p  p (D) sin2   + sin2   + .... + sin2   = 5  18   18  2 39. A - r; B - s; C - p; d - q (A) Let A = 200 ⇒ 3 A = 600 ⇒ cos 3 A = ⇒ 4 cos3 A − 3 cos A = where x = cos 200

1 2

1 ⇒ 8x3 – 6x – 1 = 0, 2

1 (B) Let A = 100 ⇒ sin 3 A = ⇒ 8 x 3 − 6 x + 1 = 0, 2 where x = sin100 (C) Let A = 15° ⇒ 3A = 45° ⇒ tan 3A = 1 ⇒

3 tan A − tan3 A 2

1 − 3 tan A

=1

= x3 – 3x2 – 3x + 1 = 0, where x = tan15° 1 (D) Let A = 60 ⇒ sin 5 A = ⇒ 32 x 5 − 40 x 3 + 10 x − 1 = 0, 2 0 where x = sin6 60

p 3p 5p sin sin 14 14 14 p p   p 3p   p 5p  = cos  −  cos  −  cos  −   2 14   2 14   2 14 

(C) Consider, sin

2 sin 760 sin 160 + cos(760 − 160 ) cos 600 − cos 920 + cos 600

40. A - p; B - q; C - s; d - q, r (A) Given, sinq = 3sin(q + 2a) ⇒ sin(q + a – a) = 3sin(q + a + a) ⇒ sin(q + a) cosa – cos(q + a)sina = 3sin(q + a) cosa + 3cos(q + a)sina ⇒ –2sin(q + a) cosa = 4cos(q + a) sina sin(q + a) 2 sin a ⇒ − = ⇒ –tan(q + a) = 2 tana cos(q + a) cos a ⇒ tan(q + a) + 2 tana = 0 (B) We have, psinq + qcosq = a ...(1) and pcosq – qsinq = b ...(2) Squaring (1) and (2) and then adding, we get (psinq + qcosq)2 + (pcosq – qsinq)2 = a2 + b2 ⇒ p2(1) + q2(1) – a2 – b2 = 0 ⇒ (p2 – a2) + (q2 – b2) = 0 ⇒ (p + a)(p – a) + (q + b)(q – b) = 0 p+a q −b ⇒ + =0 q +b p−a


 3p   2p  p = cos   cos   cos    7   7  7 p  2p   4p  = − cos   cos   cos   7  7   7  10 p 4p = cos 7 7 p 2p 10 p p 3p 5p So, cos cos cos − sin sin sin 7 7 7 14 14 14 1 p  2p   4p  = 2 cos   cos   cos   = − 7  7   7  4 Also, cos

(D) Clearly, secq – tanq = 1 Also, 1 satisfy the given equation. So, the roots of the given equation are 1 & secq. p p 1 ⇒ 6q = ⇒ cos 6q = 18 3 2 1 ⇒ 4 cos3 2q − 3 cos 2q = 2 2 3 ⇒ 8(2 cos q − 1) − 6(2 cos2 q − 1) = 1. 41. (0) : Let q =

Let 2 cosq = x


61

3

 x2   x2  ⇒ 8  2. − 1  − 6  2. − 1  = 1  4   4 

46. (1) : a + 2a + 4a = 7a =

p = tan(a + 2a + 4a) 2 1 tan a + tan 2a + tan 4a − tan a ⋅ tan 2a ⋅ tan 4a ⇒ = 0 1 − tan a ⋅ tan 2a − tan a ⋅ tan 4a − tan 2a ⋅ tan 4a ⇒ tan

⇒ (x2 – 2)3 – 3(x2 – 2) = 1 ⇒ x6 – 6x4 + 9x2 – 3 = 0 f ′(x) = 6x (x4 – 4x2 + 3) f ′(x ) = 0 ⇒ x = 0, ± 1, ± 3

⇒ 1 – tana ⋅ tan2a – tana ⋅ tan4a – tan2a ⋅ tan4a = 0 ⇒ tana ⋅ tan2a + tan2a ⋅ tan4a + tan4a ⋅ tana = 1 D D D , r2 = , r3 = 47. (2) : r1 = , Substitute this s−a s −b s−c

42. (0) : cosq (1 + cos2q) = sin2q (1 – sin2q) (2 – sin2q)2 = sin4q sin6q = 4 – 8sin2q + 4sin4q On comparing with the given equation, we get a = 4, b = –8, c = 4 \ a+b+c=0 43. (0) :

value and take abc common, we get abc  s − a s − b s − c  abc  s  L.H.S. = + + = − 3 ∑    D  a b c  D  a 

sin q 2 sin q cos q = cos 3q 2 cos 3q cos q

⇒ k=2 48. (1) : Let A =

sin 2q sin(3q − q) = = 2 cos 3q cos q 2 cos 3q cos q sin q 1 \ = [tan 3q − tan q] cos 3q 2 sin 3q 1 = [tan 9q − tan 3q] cos 9q 2 sin 9q 1 = [tan 27q − tan 9q] cos 27q 2 \ (1) + (2) + (3) ⇒ sin q sin 3q sin 9q 1 + + = [tan 27q − tan q] cos 3q cos 9q cos 27q 2 Now, on comparing, we get A = 2, B = 27, C = 1 \ 27A – B – 27 C = 0 44. (3) : Let tanx = 2t, tany = 3t, tanz = 5t 1 ∑ tan x = tan x ⋅ tan y ⋅ tan z ⇒ t 2 = 3 Now, tan2x + tan2y + tan2z = t2 (4 + 9 + 25) = 38t2 , K = 3 45. (4) : We have, sinq(1 + sin2q) = 1 – sin2q ⇒ sinq(2 – cos2q) = cos2q Squaring both sides, we get sin2q(2 – cos2q)2 = cos4q ⇒ (1 – cos2q)(4 – 4cos2q + cos4q) = cos4q ⇒ –cos6q + 5cos4q – 8cos2q + 4 = cos4q \ cos6q – 4cos4q + 8cos2q = 4 62


p 2

C= …(1) …(2) …(3)

b+g−a g +a −b ,B = and 4 4

a +b− g 4

⇒ tanA tanB tanC = 1 sin A sin B 1 = cos A cos B tan C sin A sin B − cos A cos B 1 − tan C = ⇒ sin A sin B + cos A cos B 1 + tan C [By Componendo and Dividendo] ⇒

p  sin  − C    − cos( A + B) 4 ⇒ = cos( A − B)  p cos  − C  4  p  p  ⇒ 2 sin  − C  cos( A − B) + 2 cos  − C  cos( A + B) = 0 4  4 

p  p  ⇒ sin  − C + A − B  + sin  − C − A + B  4  4  p  p  + cos  − C + A + B  + cos  − C − A − B  = 0 ...(1) 4  4  A− B −C =

p p − a, B − A − C = − b, 4 4 p p C − A − B = − g and A + B + C = 4 4

\ (1) ⇒ cosa + cosb + cosg + 1 = 0 Thus, K = 1.

nn

8

LINEAR PROGRAMMING AND PROBABILITY

Linear Programming OptimiSatiOn prOblem A problem in which a linear function is to be optimised (maximise or minimise) satisfying certain linear inequalities is called optimisation problem. linear prOGramminG prOblem (l.p.p.) Linear programming (L.P.) is an optimisation technique in which a linear function is optimised (i.e., minimised or maximised) subject to certain constraints which are in the form of linear inequalities. mathematical fOrmulatiOn Of l.p.p. The procedure for mathematical formulation of a linear programming problem consists of the following steps : In every L.P.P. certain decisions are to be made. Step I These decisions are represented by decision variables. These decision variables are those quantities whose values are to be determined. Identify the variables and denoted them by x1, x2, x3, ..... Step II Identify the objective function and express it as a linear function of the variables introduced in step I. Step III In a L.P.P., the objective function may be in the form of maximising profits or minimising costs. So, after expressing the objective functions as a linear function of the decision variables, we must find the type of optimisation i.e., maximisation or minimisation. Identify the type of the objective function. Step IV Identify the set of the constraints, stated in terms of decision variables and express them as linear inequations or equations as the case may be.

uSeful termS in l.p.p. Objective function

The linear function Z = c1x1 + c2x2 + ...... + cnxn which is to be optimised (maximised or minimised) is called the objective function. Decision The variables x1, x2, x3, ....., xn whose values are to be decided, are called variables decision variables. Linear The linear inequations (inequalities) or constraints restrictions on the decision variables of a linear programming problem are called linear constraints. The conditions x ≥ 0, y ≥ 0 are called non-negative constraints. Convex A set S of points in a plane is said to be region convex, if the line segment joining any two points in it, lies in it completely. Feasible The set of points, whose co-ordinates region satisfy the constraints of a linear programming problem, is said to be the feasible region. Bounded A feasible region of a system of linear inequations is said to be bounded if it and unbounded can be enclosed within a circle. But if the feasible region extends indefinitely feasible in any direction, then the feasible region region is called unbounded. Points in the region which is the Corner intersection of two boundary lines are points of a feasible called corner points. region mathematics today | JANUARY ‘16

63

Feasible & infeasible solution

Optimal (feasible) solution

Points within and on the boundary of the feasible region of a L.P.P. represent feasible solutions. Any point outside the feasible region is called infeasible solution. The optimal solution of the feasible region is the optimal value (maximum or minimum) of the function at any point.

theOremS •

•

Let R be the feasible region (convex polygon) for a linear programming problem and let Z= ax + by be the objective function. When Z has an optimal value (maximum or minimum), where the variables x and y are subject to constraints described by linear inequalities, this optimal value must occur at the corner point (vertex) of the feasible region. Let R be the feasible region for a linear programming problem, and let Z = ax + by be the objective function. If R is bounded, then the objective function Z has both a maximum and minimum value on R and each of these occurs at a corner point (vertex) of R. Note : If R is unbounded, then a maximum or minimum value of the objective function may not exist. However, if it exists, it must occur at a corner point of R.

Graphical methOdS Of SOlvinG l.p.p. The graphical methods for solving L.P.P. is applicable to those problems which involve only two variables. Corner Point Method To solve a linear programming problem by corner point method, we follow the following steps : (i) Formulate the given L.P.P. in mathematical form. (ii) Convert all inequations into equations and draw their graphs.

(iii) Determine the region represented by each inequation. (iv) Obtain the region in xy-plane containing all points that simultaneously satisfy all constraints including non-negativity restrictions. The polygonal region so obtained is the feasible region and is known as the convex polygon of the set of all feasible solutions of the L.P.P. (v) Determine the coordinates of the vertices (corner points) of the convex polygon obtained in step (ii). These vertices are known as the extreme points of the set of all feasible solutions of the L.P.P. (vi) Obtain the values of the objective function at each of the vertices of the convex polygon. The point where the objective function attains its optimum (maximum or minimum) value is the optimal solution of the given L.P.P. different typeS Of l.p.p. A few important linear programming problems are listed below : • Manufacturing problems : In this problem, we determine the number of units of different products which should be produced and sold by a firm when each product requires a fixed manpower, machine hours, labour hour per unit of product, warehouse space per unit of the output etc., in order to make maximum profit. • Diet problems : In these problems, we determine the amount of different kinds of constituents/ nutrients which should be included in a diet so as to minimise the cost of the desired diet such that it contains a certain minimum amount of each constituent/nutrients. •

Transportation problems : In these problems, we determine a transportation schedule in order to find the cheapest way of transporting a product from plants/factories situated at different locations to different markets.

ProbabiLity cOnditiOnal prObability If two events A and B are associated with the same random experiment, the conditional probability P(A|B) of the occurrence of A knowing that B has occurred is given by P ( A | B) =

64

P ( A ∩ B) , P ( B) ≠ 0 P ( B)


Similarly, P (B | A) =

P ( A ∩ B) , P ( A) ≠ 0 . P ( A)

Properties of Conditional Probability • If E1 and E2 are any two events associated with an experiment and P(E2) ≠ 0, then 0 ≤ P(E1 | E2) ≤ 1. • If E is any event associated with an experiment, then P(E|E) = P(S|E) = 1, S being the sample space.

•

•

If E, E 1 and E 2 are events associated with an experiment and P(E) ≠ 0, then P((E1 ∪ E2)|E) = P(E1| E) + P(E2 |E) – P((E1 ∩ E2)|E) In particular, if E1 and E2 are mutually exclusive, then P((E1 ∪ E2)|E) = P(E1|E) + P(E2 |E). If E1 and E2 are any two events associated with an experiment, then P(E1c |E2) = 1 – P(E1 |E2).

multiplicatiOn theOrem On prObability If A and B are two events associated with a random experiment, then P(A ∩ B) = P(A) · P(B|A) ; P(A) ≠ 0 or P(A ∩ B) = P(B) · P(A|B) ; P(B) ≠ 0 If A1, A2, A3, ...., An are n events related to a random experiment, then P(A1 ∩ A2 ∩ .... ∩ An) = P(A1) × P(A2|A1) × P(A3|(A1 ∩ A2)) × .... × P(An|(A1 ∩ A2 ∩ ... ∩ An– 1)), where P(An|(A1 ∩ A2 ∩ ..... ∩ An–1)) represents the conditional probability of the event An, given that the events A1, A2, ....., An–1 have happened. independent eventS Two random experiments are said to be independent iff the probability of occurrence or non-occurrence of any event E2 associated with the second experiment is independent of the outcome of the first experiment. This means that if E1 is any event associated with the first experiment, then P(E1 ∩ E2) = P(E1) × P(E2) Note : • Two events E and F are said to be dependent if they are not independent, i.e., if P(E ∩ F) ≠ P(E) . P(F) • Three events A, B and C are said to be mutually independent, if P(A ∩ B) = P(A) . P(B) P(A ∩ C) = P(A) . P(C) P(B ∩ C) = P(B) . P(C) and P(A ∩ B ∩ C) = P(A) . P(B) . P(C) If at least one of the above is not true for three given events, we say that the events are not independent. theOrem Of tOtal prObability Partition of a sample space : A set of events E1, E2, ....., En is said to represent a partition of the sample space S if (i) Ei ∩ Ej = f, i ≠ j, i, j = 1, 2, 3 ....., n (ii) E1 ∪ E2 ∪ ..... ∪ En = S and (iii) P(Ei) > 0 for all i = 1, 2, ....., n.

where the events, E1, E2, ....., En represent a partition of the sample space S if they are pairwise disjoint, exhaustive and have non zero probabilities. For example any non empty event E and its complement E′ form a partition of the sample space S since they satisfy E ∩ E′ = f and E ∪ E′ = S. Law of total probability : Let S be the sample space and let E1, E2, ..., En be n mutually exclusive and exhaustive events associated with a random experiment. If A is any event of non-zero probability which occurs with E1 or E2 or ..... or En, then P(A) = P(E1)P(A|E1) + P(E2)P(A|E2) + ..... + P(En)P(A|En) n

= ∑ P (Ei )P ( A | Ei ) i =1

bayeS' theOrem Let S be the sample space and let E1, E2 , ....., En be n mutually exclusive and exhaustive events associated with a random experiment. If A is any event of non- zero probability which occurs with E1 or E2 or ..... or En, then P (E )P ( A | Ei ) P (Ei | A) = n i , i = 1, 2, ....., n

∑ P(Ei )P(A | Ei ) i =1

randOm variable and itS prObability diStributiOn • Random variable : Random variable is simply a variable whose values are determined by the outcomes of a random experiment; generally it is denoted by capital letters such as X, Y, Z etc. and their values are denoted by the corresponding small letters x, y, z etc. • Probability distribution : The system consisting of a random variable X along with P(X), is called the probability distribution of X. The probability distribution of a random variable X is the system of numbers X : x1 x2 ..... xn P(X) : p1 p2 ..... pn n

where •

pi > 0, ∑ pi = 1, i = 1, 2, ......, n i =1

Mean of a Random Variable The mean of a random variable X is also called the expectation of X. i.e., denoted by E(X). So, E( X ) = µ = p1x1 + p2 x2 + ..... + pn xn or mathematics today | JANUARY ‘16

n

∑ pi xi i =1

65

•

Variance of a Random Variable n

Var( X ) = ∑ (xi − µ)2 pi i =1

OR

2

n  n  Var( X ) = ∑ pi xi 2 −  ∑ pi xi  = E( X 2 ) − [E( X )]2 i =1  i =1

σ x = Var( X ) =

n

∑ (xi − µ)2 pi i =1

is called the Standard Deviation of the random variable X. bernOulli trialS and binOmial diStributiOn • Bernoulli trials : A sequence of independent trials which can result in one of the two mutually exclusive possibilities success or failure such that the probability of success or failure in each trial is constant, then such repeated independent trials are called Bernoullian trials. • Binomial distribution : A random variable X taking values 0, 1, 2, ...., n is said to have a binomial distribution with parameters n and p, if its probability distribution is given by P(X = r) = nCr prqn – r where p represents probability of success and q represents probability of non success, or failure and n is the number of trials. Note : While using above probability density functions of the binomial distribution in solving any problem we should first of all examine whether all the conditions, given below are satisfied : (i) There should be a finite number of trials. (ii) The trials are independent. (iii) Each trial has exactly two outcomes : success or failure. (iv) The probability of an outcome remains the same in each trial. • Mean, variance and standard deviation : (i) Mean = np (ii) Variance = npq

a chair at a profit of ` 15. Assume that he can sell all the items that he buys. Formulate this problem as an L.P.P., so that he can maximise the profit. 2. Two cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the probability of drawing two aces. 3. Six coins are tossed simultaneously. Find the probability of getting 3 heads. 4. If A and B are two independent events such that P(A ∪ B) = 0.7 and P(A) = 0.4. Find P(B). 5. A pair of dice is thrown 200 times. If getting a sum of 9 is considered a success, find the variance of the number of successes. short answer type 6. Bag I contains 3 red and 4 black balls while another Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from Bag I. 7. An instructor has a test bank consisting of 200 easy true/false questions, 100 difficult true/false questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from test bank, what is the probability that it will be a difficult question given that it is a multiple choice question? 8. Solve the following L.P.P. graphically : Minimize Z = 3x + 5y subject to –2x + y ≤ 4, x + y ≥ 3, x – 2y ≤ 2, x, y ≥ 0 9. A speaks truth in 60% of the cases and B in 90% of the cases. In what percentage of cases are they likely to contradict each other in stating the same fact. 10. A dice is thrown 6 times. If 'getting an even number' is a success, what is the probability of (i) 5 successes (ii) at most 5 successes Long answer type

Very short answer type

11. Find the probability distribution of the number of green balls drawn when 3 balls are drawn, one by one, without replacement from a bag containing 3 green and 5 white balls.

1. A furniture dealer deals in only two items viz., tables and chairs. He has ` 10, 000 to invest and a space to store at most 60 pieces. A table costs him ` 500 and chair ` 200. He can sell a table at a profit of ` 50 and

12. Anil wants to invest at most `12,000 in Bonds A and B. According to the rules, he has to invest at least `2,000 in Bond A and at least `4,000 in Bond B. If the rate of interest on Bond A is 8% per annum

(iii) Standard deviation = npq

66


and on Bond B is 10% per annum, how should he invest his money for maximum interest? Solve the problem graphically. 13. A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both spades. Find the probability of the lost card being a spade. 14. There are 5 cards numbered 1 to 5, one number on one card. Two cards are drawn at random without replacement. Let X denote the sum of the numbers on the two cards drawn. Find the mean and variance of X. 15. A brick manufacturer has two depots, A and B, with stocks of 30,000 and 20,000 bricks respectively. He receives orders from three buildings, P, Q and R, for 15,000, 20,000 and 15,000 bricks respectively. The cost of transporting 1000 bricks to the builders from the depots (in rupees) are given below : From

To A B

P

Q

R

40 20

20 60

30 40

How should the manufacturer fulfil the orders so as to keep the cost of transportation minimum? soLUtions 1. Let x and y be the number of tables and chairs respectively. We have, 500x + 200y ≤ 10000 Thus, the mathematical formulation of the L.P.P. is Maximize Z = 50x + 15y subject to the constraints 5x + 2y ≤ 100, x + y ≤ 60, x ≥ 0, y ≥ 0 2. p = 4 = 1 52 13 \

Required probability =

1 1 1 × = 13 13 169

3. Let p be the probability of getting a head in the toss of a coin. 1 1 Then, p = ⇒ q = 1 − p = 2 2 Let X = No. of successes in the experiment. Then P(X = r) = nCr pr. qn–r \

3

1 1 P ( X = 3) = 6C3   ⋅   2 2

6−3

=

20 20 5 = = 26 64 16 [... n = 6]

4. We know, P(A ∪ B) = P(A) + P(B) – P(A ∩ B) ⇒ P(A ∪ B) = P(A) + P(B) – P(A)P(B) [... A and B are independent events] ⇒ 0.7 = 0.4 + P(B) – 0.4 P(B) 0. 3 1 ⇒ P ( B) = = = 0. 5 0. 6 2 4 1 5. p = getting a sum of 9 on a pair of dice = = 36 9 1 8 ⇒ q = 1 − = . Here n = 200 9 9 Variance = npq = 200 × 1 × 8 = 1600 9 9 81 6. Let E1 be the event of choosing the bag I, E2 the event of choosing the bag II and A be the event of drawing a red ball. Then P (E1 ) = P (E2 ) = 1 2 3 Also P(A|E1) = P (drawing a red ball from bag I) = 7 and P(A|E2) = P(drawing a red ball from bag II) = 5 11 Now, the probability of drawing a ball from bag I, being given that it is red, is P(E1|A). By using Bayes' theorem, we have P (E1 )P ( A | E1 ) P (E1 | A) = P (E1 )P ( A | E1 ) + P (E2 )P ( A | E2 ) 1 3 × 33 2 7 = = 1 3 1 5 68 × + × 2 7 2 11 7. The given data may be tabulated as Easy Difficult Total True/False 200 100 300 Multiple choice 500 400 900 Total 700 500 1200 Let us denote E = Easy questions, D = Difficult questions, T = True/False questions and M = Multiple choice questions Number of difficult multiple choice questions = 400 Total number of questions = 1200 P(D ∩ M) = Probability of selecting a difficult and multiple choice question 400 = 1200 Total number of multiple choice questions = 500 + 400 = 900 mathematics today | JANUARY ‘16

67

P(M) = Probability of selecting one multiple choice question 900 = 1200 \

P(D | M ) =

\ A and B are likely to contradict each other in 42% cases.

P(D ∩ M ) 400 900 4 = ÷ = P(M ) 1200 1200 9

8. First we draw the lines – 2x + y = 4, x + y = 3 and x – 2y = 2 The feasible region has been shown shaded which is unbounded. The vertices of the feasible region are : 8 1 P  ,  , D (0, 3), B (0, 4) 3 3 Given, Z = 3x + 5y 8 1 29 8 1 at P  ,  , Z = 3 × + 5 × = = m (minimum) 3 3 3 3 3 at D(0, 3), Z = 0 + 5 × 3 = 15 at B(0, 4), Z = 0 + 5 × 4 = 20 29 8 1 at P , i.e., when Minimum value of Z = 3 3 3 8 1 x = and y = 3 3 Y

10. We have, the probability of success = the probability of even number on the die 3 1 = = 6 2 (i) The probability of getting 5 successes 5 6 6 3 1 1 1 P(5) = 6C5   ⋅   = 6 ⋅   = = 2 2 2 64 32 (ii) Probability of at most 5 successes = 1 – P(6) 6 1 63 1 = 1−   = 1− = 2 64 64 11. Let X denote the total number of green balls drawn in three draws without replacement. Clearly, there may be all green, 2 green, 1 green or no green at all. Thus, X can assume values 0, 1, 2 and 3. Let Gi denote the event of getting a green ball in ith draw. Now, P(X = 0) = Probability of getting no green ball in three draws ⇒ P(X = 0) = P(G1 ∩ G2 ∩ G3 ) ⇒ P(X = 0) = P(G1)P(G2 | G1)P(G3 |(G1 ∩ G2 ))

x+

5 4 3 5 ⇒ P(X = 0) = × × = 8 7 6 28 P(X = 1) = Probability of getting one green ball in three draws ⇒ P(X = 1) = P ((G1 ∩ G2 ∩ G3 ) ∪ (G1 ∩ G2 ∩ G3 )

y=

3

4 B(0, 4)

3 D(0, 3) 2 X

1 O

1

P 2 3 E C (2, 0)(3, 0)

X

Y

9. Here the random experiment is "The stating of the fact by A and B". Let E = the event of A speaking the truth and F = the event of B speaking the truth 60 3 90 9 Then, P(E) = = and P(F ) = = 100 5 100 10 Required probability, P (A and B contradicting each other) = P(EF or EF ) = P(EF ) + P(EF ) = P(E) ⋅ P(F ) + P(E) ⋅ P(F ) = P(E) · [1 – P(F)] + [1 – P(E)] · P(F) 3 9 3 9 21 = 1 −  + 1 −  ⋅ = 5  10   5  10 50 68


∪ (G1 ∩ G2 ∩ G3 ))

⇒ P(X = 1) = P(G1 ∩ G2 ∩ G3 ) + P (G1 ∩ G2 ∩ G3 ) + P(G1 ∩ G2 ∩ G3 ) ⇒ P(X = 1) = P(G1)P(G2 | G1)P(G3 |(G1 ∩ G2 )) + P (G1)P(G2 | G1)P(G3 |(G1 ∩ G2 )) + P(G1)P(G2 | G1)P(G3 |(G1 ∩ G2 )) 3 5 4 5 3 4 5 4 3 15 ⇒ P(X = 1) = × × + × × + × × = 8 7 6 8 7 6 8 7 6 28 P(X = 2) = Probability of getting two green balls in the three draws. P(X = 2) = P ((G1 ∩ G2 ∩ G3 ) ∩ (G1 ∩ G2 ∩ G3 ) ∪(G1 ∩ G2 ∩ G3 ))

⇒ P(X = 2) = P(G1)P(G2 | G1)P(G3 |(G1 ∩ G2 ))

+ P (G1)P(G2 | G1)P(G3 |(G1 ∩ G2 )) + P(G1)P(G2 | G1)P(G3 |(G1 ∩ G2 ))

3 2 5 5 3 2 3 5 2 15 ⇒ P(X = 2) = × × + × × + × × = 8 7 6 8 7 6 8 7 6 56 and P(X = 3) = Probability of getting three green balls in the three draws. P(X = 3) = P (G1 ∩ G2 ∩ G3) = P(G1)P(G2 |G1)P(G3|G1 ∩ G2) 3 2 1 1 ⇒ P(X = 3) = × × = 8 7 6 56 Thus, the probability distribution of the number of green balls is given by X : 0 1 2 3

Now, we evaluate Z at the corner points A(2000, 10000), B(8000, 4000) and C(2000, 4000). 2x y Corner Point Z= + 25 10

5 15 15 1 28 28 56 56 12. Let Anil invest ` x in bond A and ` y in bond B, then 8 2x =` Interest on bond A = ` x × 100 25

13. Let E1, E2, E3 and E4 be the events of losing a card of spades, clubs, hearts and diamonds respectively. Let E be the event of drawing 2 spade cards from the remaining 51 cards. 13 1 Now P(E1) = P(E2) = P(E3) = P(E4) = = 52 4 P(E|E1) = Probability of drawing 2 spade cards given that a card of spade is missing.

P( X ) :

Y 12000

A(2000, 10000) B(8000, 4000) C(2000, 4000)

A(2000, 10000)

=

9000 8000

x = 2000

7000

x+

6000

y=

=

12

0

C

(2000, 4000)

y = 4000

B (8000, 4000)

x = 2000

1000

O

1000 2000 3000 4000 5000 6000 7000 8000 –1000

X

–2000 Y

10 y =` 100 10 2 x y   His total annual interest = `  25 + 10  Thus, our L.P.P. is to 2x y ... (i) + Maximise Z = 25 10 Subject to constraints x ≥ 2000 ... (ii) y ≥ 4000 ... (iii) and x + y ≤ 12000 ... (iv) On plotting inequalities (ii) to (iv), we have the required region shown shaded in the figure. and interest on bond B =` y ×

51

C2

C2

=

12 × 11 22 = 51 × 50 425

13

51

C2

=

13 × 12 26 = 51 × 50 425

\

2000 X –2000 –1000

12

C2 By Bayes' theorem, P(E ) ⋅ P(E | E1) P(E1 | E) = 4 1

5000

3000

Maximum

P(E|E2)= P(E|E3) = P(E|E4) 00

4000

←

\ Z is a maximum i.e., `1160 when x = 2000 and y = 10000. So, Anil should invest `2000 in Bond A and `10000 in Bond B to earn maximum profit of ` 1160.

11000 10000

1160 1040 560

∑ P(Ei ) ⋅ P(E | Ei ) i =1

1 22 × 4 425 = 1 22 1 26 1 26 1 26 × + × + × + × 4 425 4 425 4 425 4 425 22 22 = = = 0.22 22 + (3 × 26) 100 14. In this case the sample space contains 5 × 4 = 20 sample points of the type (x, y), x ≠ y, x, y ∈ {1, 2, 3, 4, 5} \ Sum of the numbers on the two cards can take values from 3 to 9. 2 1 P(X = 3) = P({(1, 2),(2, 1)}) = = , 20 10 2 1 P(X = 4) = P({(1, 3),(3, 1)}) = = , 20 10 4 1 P(X = 5) = P({(1, 4),(4, 1),(2, 3),(3, 2)}) = = , 20 5 mathematics today | JANUARY ‘16

69

P(X = 6) = P({(1, 5),(5, 1),(2, 4),(4, 2)}) =

4 1 = , 20 5

P(X = 7) = P({(2, 5),(5, 2),(3, 4),(4, 3)}) =

4 1 = , 20 5

2 1 = and 20 10 2 1 P(X = 9) = P({(4, 5),(5, 4)}) = = . 20 10 P(X = 8) = P({(3, 5),(5, 3)}) =

Thus, the probability distribution of X is as follows : X

3 1 P( X ) 10

4 1 10

5 1 5

6 1 5

7 8 1 1 5 10

9 1 10

Hence mean = SX P(X) 1 1 1 1 1 = 3× + 4× +5× +6× +7× 10 10 5 5 5 +8×

`40x `20

y 00 – `30 (x + y)

300

70

Q 20000

1 1 +9× 10 10

R 15000

`20 00 – x) `60 20000 – y 0 `40 – 1500 y) + (x

(150


+ 40[(x + y) – 15000]) 1 (1800000 + 30 x − 30 y ) = 1000 Now, we have to minimise the objective function 1 (1800000 + 30 x − 30 y ) (the cost of transZ= 1000 portation) which satisfies all the constraints (1) to (3). The graph of inequations of the constraints given in (1) –(3) gives the feasible region ABCDE, where the vertices are A(15000, 0) B(15000, 15000), C(10000, 20000), D(0, 20000) and E(0, 15000). x = 15000

15. Let x and y units be shipped to buildings P and Q respectively from depot A. Then, the units shipped to other buildings from two depots A and B are shown in the figure. Therefore, there is the following constraints to the problem : x ≥ 0, 15000 – x ≥ 0, i.e., 0 ≤ x ≤ 15000 ...(1) y ≥ 0, 20000 – y ≥ 0, i.e., 0 ≤ y ≤ 20000 ...(2) 30000 – (x + y) ≥ 0, (x + y) – 15000 ≥ 0 i.e. 15000 ≤ x + y ≤ 30000 ...(3) The total cost of transportation Z, for the distribution given in figure is

A 30000

1 (40 x + 20 y + 30[30000 − (x + y )] 1000 + 20(15000 – x) + 60(20000 – y)

Y

3 + 4 + 10 + 12 + 14 + 8 + 9 60 = = =6 10 10 2 and variance = SX P(X) – (mean)2 1 1 1 1 1 = 32 × + 4 2 × + 52 × + 6 2 × + 7 2 × 10 10 5 5 5 1 1 + 82 × + 92 × − (6)2 10 10 9 + 16 + 50 + 72 + 98 + 64 + 81 = − 36 10 390 = − 36 = 39 − 36 = 3 10

P 15000

Z=

B 20000

(0, 30000) D(0, 20000) E(0, 15000)

X

y = 20000

C(10000, 20000)

x+

0

B(15000, 15000)

y=

x + y = 30000

15

00

0

A(15000, 0)

(30000, 0)

X

Y

The values of the objective function at the corner points are given in the following table : Corner points Value of the objective (x, y) 1 function Z = × 1000 (1800000 + 30x – 30y) A(15000, 0) Z = 2250 B(15000, 15000) Z = 1800 C(10000, 20000) Z = 1500 D(0, 20000) Z = 1200 E(0, 15000) Z = 1350 We find that minimum cost of transportation is ` 1350, when x = 0 units and y = 15000 units. Thus, the units transported from depot A and B to three buildings P, Q and R will be as follows so as to minimize the cost. Depot A B

Bricks shipped to buildings Total bricks P Q R 0 20000 10000 30000 15000 0 5000 20000 nn

The entire syllabus of Mathematics of WB-JEE is being divided in to six units, on each unit there will be a Mock Test Paper (MTP) which will be published in the subsequent issues. The syllabus for module break-up is given bellow: Unit- I: Algebra, Unit-II: Trigonometry, Unit-III: Co-ordinate geometry of two dimension & three dimensions, Unit-IV: Calculus, Unit-V: Vector, Unit-VI: Statistics & probability.

cateGoRy 1 Category 1 shall have questions of 1 mark each. For incorrect response, 25% of full mark (1/4) would be deducted.

1. Given 5 line segments of the length 2, 3, 4, 5, 6 units. Then, the no. of triangles that can be formed by joining these segments is (a) 5C3 – 3 (b) 5C3 5 (c) C3 – 1 (d) 5C3 – 2 2. How many numbers greater than 10,00,000 be formed from 2, 3, 0, 3, 4, 2, 3? (a) 420 (b) 360 (c) 400 (d) 300   4   1  tan  cos −1  − sin −1  is  5 2   17    3 29 3 29 (a) (b) (c) (d) 29 3 29 3 n 4. If in the expansion of (1 + px) , n ∈ N, then the coefficient of x and x2 are 8 and 24 respectively, then (a) n = 3, p = 2 (b) n = 5, p = 3 (c) n = 4, p = 3 (d) n = 4, p = 2 3.

5. If z =

( 3 + i)3 (3i + 4)2 (8 + 6i)2 (b) 3

(a) 1 2nd

5th

then | z | is equal to (c) 0

(d) 2

6. If the and terms of a G.P. are 24 and 3 respectively, then the sum of the first six terms is 2 189 179 (a) (b) (c) 189 (d) 189 5 2 2 7. If a and b are the roots of x2 – ax + b2 = 0, then a2 + b2 is equal to

(a) 2a2 – b2 (c) a2 – 2b2

(b) a2 + b2 (d) a2 – b2

8. If sin(q + f) = nsin(q – f), n ≠ 1, then the value of tan q = tan f n n +1 n n −1 (b) (c) (d) n −1 n −1 1− n n +1

(a)

9. If tanq = 5tanf then the maximum value of tan2(q – f) is 4 4 16 (a) (b) (c) (d) 1 9 5 25 2 − sin x − cos x = sin x − cos x

10.

x p (a) tan  −  2 8

x p (b) tan  +  2 8

x p (c) tan  −  2 4

x p (d) tan  +  2 4

(n !)1/n = n→∞ n (a) e (c) 1

(b) e–1 (d) none of these

11. lim

12. Given that, x 2dx

∞

∫

0

2

2

2

2

2

2

(x + a )(x + b )(x + c ) ∞

then ∫

0

dx 2

2

(x + 4 )(x 2 + 92 )

=

p , 2(a + b)(b + c)(c + a)

is

By : Sankar Ghosh, HOD(Math), Takshyashila. Mob : 09831244397 mathematics today | JANUARY ‘16

71

p 20 p (c) 80 (a)

(b)

p 40

(a)

(d) none of these

13. The length of the perpendicular drawn from x −6 y −7 z −7 (1, 2, 3) to the line is = = 3 2 −2 (a) 4 (b) 5 (c) 6 (d) 7 14. For any integer n, the integral p

∫e

cos2 x

0

cos3 (2n + 1)x dx = (b) p (d) none of these

(a) 1 (c) 2p

15. If [x] denotes the greatest integer ≤ x, then the value of lim | x |[cos x ] is x →0

(a) 0 (c) –1

16. If the angles of a triangle are in the ratio 4 : 1 : 1, then the ratio of the longest side and perimeter is (a) 3 : (2 + 3 ) (b) 1 : 6 (c) 1 : (2 + 3 ) (d) 2 : 6  x 3 + x 2 − 16 x + 20 , if x ≠ 2  17. If f (x ) =  is continuous (x − 2)2  b, if x = 2  for all x, then b is equal to (a) 7 (b) 3 (c) 2

19. The area bounded by the curve y = f (x), x-axis and the ordinates x = l and x = b is (b – 1)sin(3b + 4). Then f (x) is (a) (x – 1)cos(3x + 4) (b) sin(3x + 4) (c) sin(3x + 4) + 3(x – 1)cos(3x + 4) (d) none of these  x2 − y2  dy = log a, then = 20. If cos −1  2 2 dx x + y  72

(a) y


(c)

(b) xy

22. If I = ∫

x2

(d)

y2

y2 x2

dx (1 − x )(x − 2)

(c)

1 y 4

(c)

xy

, then I is equal to

(a) sin–1(2x – 3) + c (b) sin–1(2x + 5) + c (c) sin–1(3 – 2x) + c (d) sin–1(5 – 2x) + c 23. ∫

sin 2 x

(3 + 4 cos x )3

(c)

dx =

3 cos x + 8 2

(3 + 4 cos x ) 3 + cos x

2

(3 + 4 cos x )

+ c (b) + c (d)

3 + 8 cos x 16(3 + 4 cos x )2 3 − 8 cos x 16(3 + 4 cos x )2

+c +c

24. If the algebraic sum of the deviations of 20 observations from 30 is 20, then the mean of observations is (a) 30 (b) 30.1 (c) 29 (d) 31 25. The variance of first n natural numbers is (a)

(n2 − 1) 12

(b)

(n2 − 1) 6

(c)

(n2 + 1) 6

(d)

(n2 + 1) 12

(d) 5

18. An integrating factor of the differential equation dx y log y + x − log y = 0 is dy (a) log(logy) (b) logy 1 1 (c) (d) log y log(log y )

x y

(b)

 d 2 y 1 dy  x − x , then  x + 21. If y = e + e =  dx 2 2 dx 

(a) (b) 1 (d) does not exists

y x

8  26. If 1, log 81 (3x + 48), log 9  3x −  are in A.P., then  3 the value of x equals (a) 9 (b) 6 (c) 2 (d) 4 27. If the arithmetic mean of two positive numbers a and b (a > b) is twice their G.M., then a : b is (a) 6 + 7 : 6 − 7

(b) 2 + 3 : 2 − 3

(c) 5 + 6 : 5 − 6

(d) none of these

28. If S, P and R are the sum, product and reciprocal of n n k n terms of an increasing G.P. and S = R · P , then k is equal to (a) 1 (b) 2 (c) 3 (d) none of these

29. If the difference between the roots of x2 + 2p x + q = 0 is two times the difference between the roots of p x 2 + qx + = 0, where p ≠ q, then 4 (a) p – q + 1 = 0 (b) p – q – 1 = 0 (c) p + q – 1 = 0 (d) p + q + 1 = 0 30. The roots of the equation are (a) 2, 1, –9 (c) –1, 1, –9

1+ x 3 2 2+x 2 3

5 5 =0 x+4

(b) 1, 1, –9 (d) –2, 1, –8

 0 3 2b  31. If  2 0 1  is singular, then the value of b is    4 −1 6  equal to (a) –3 (b) 3 (c) –6 (d) 6 32. If A and B are square matrices of the same order and if A = AT and B = BT, then (ABA)T = (a) BAB (b) ABA (c) ABAB (d) ABT 33. The distance between the point (1, 2) and the point of intersection of the lines 2x + y = 2 and x + 2y = 2 is 16 (c) 17 (d) 19 3 5 3 34. ABCD is a square with side a. If AB and AD are along the coordinate axes, then the equation of the circle passing through the vertices A, B and D is (a)

17 3

(b)

(a) x 2 + y 2 = 2a(x + y ) 2 2 (b) x + y =

a

(x + y ) 2 (c) x2 + y2 = a(x + y) (d) x2 + y2 = a2(x + y) 35. If the semi-major axis of an ellipse is 3 and the latus 16 rectum is , then the standard equation of the 9 ellipse is (a)

x2 y2 + =1 9 8

(b)

x2 y2 + =1 8 9

(c)

x2 3 y2 + =1 9 8

(d)

3x 2 y 2 + =1 8 9

36. The one end of the latus rectum of the parabola y2 – 4x – 2y – 3 = 0 is at (a) (0, –1) (b) (0, 1) (c) (0, –3) (d) (3, 0) x2

y2

5 4 a 2 b2 and 2x + 3y – 6 = 0 is a focal chord of the hyperbola, then the length of transverse axis is equal to

37. The eccentricity of the hyperbola

(a)

24 5

(b)

5 24

(c)

12 5

(d)

−

= 1 is

6 5

38. The solution of the differential equation y′(y2 – x) = y is (a) y3 – 3xy = C (b) y3 + 3xy = C (c) x3 – 3xy = C (d) x3 – xy = C  ^  ^ ^ ^ ^ ^ 39. Let OB = i + 2 j + 2 k and OA = 4 i + 2 j + 2 k . The distance of the point B from straight line passing through A and parallel to the vector 2 i^ + 3 ^j + 6 k^ is (a)

7 5 9

(b)

5 7 9

(c)

3 7 7

(d)

9 5 7

40. The angle between the straight line  ^ ^ ^ ^ ^ ^ r = ( i + 2 j + k ) + s( i − j + k ) and the plane  ^ ^ ^ r × (2 i − j + k ) = 4 is −1  2 2  (a) sin    3 

−1  2  (b) sin    6 

 2 (c) sin −1    3 

−1  2  (d) sin    3

41. Let A and B be two events. Then 1 + P(A ∩ B) – P(B) – P(A) is equal to (a) P(Ac ∪ Bc) (b) P(A ∩ Bc) c (c) P(A ∩ B) (d) P(Ac ∩ Bc)   42. If a and b are two unit vectors inclined at an angle   p/3 then the value of | a + b | is (a) equal to 1 (b) greater than 1 (c) equal to 0 (d) less than 1       43. If | a × b |2 + | a ⋅ b |2 = 144 and | a | = 4, then | b | is equal to (a) 12 (b) 3 (c) 8 (d) 4 mathematics today | JANUARY ‘16

73

44. Let S be the set of real numbers. A relation R has been defined on S by a R b ∀ |a – b| ≤ 1, then R is (a) symmetric and transitive but not reflexive (b) reflexive and transitive but not symmetric (c) reflexive and symmetric but not transitive (d) an equivalence relation n + 1 , if n is odd  45. Let f : N → N defined by f (n) =  2  n , if n is even  2 then f is (a) (b) (c) (d)

onto but not one-one one-one and onto neither one-one nor onto one-one but not onto

48. The set of values of x for which tan 3x − tan 2 x = 1 is 1 + tan 3x ⋅ tan 2 x p (a) f (b)   4 p   (c) np + , n = 1, 2, 3 4   p   (d) 2np + , n = 1, 2, 3 4   3 x −1 dx = 49. ∫ 3 x +1 (a) x – log x + log(x2 + 1) – tan–1x + c (b) x + logx + log(x2 + 1) – tan–1x + c 1 (c) x + log x + log(x 2 + 1) − tan −1 x + c 2 (d) none of these

(c) 0

(d) 2

(b)

ln x (ln x )2 + 1

+c

(d) none of these

54. The value of

 3   1    2  log 3     6  (81) log5 9 + 3  log 25 7 log 25 6  − (125)  is   ( 7 ) 409        

(b) 1

(c) 2

(d) 3

55. ABCD is a trapezium such that AB, DC are parallel p and BC is perpendicular to them. If ∠ADB = , 4 BC = 3, CD = 4 then AB = (a)

14 3

(b)

7

3

(c)

13 4

(d)

25 7

  3 1   56. The value of tan −1  cos  2 tan −1  + sin  2 cot −1      4 2   is p p p p (a) (b) (c) > (d) < 4 4 4 3 57. Let z = cosq + isinq, then the value of 15

2m−1 ) at q = 2° is ∑ Im(z

2


(b) –1

2x 52. The function f (x ) = log(1 + x ) − is increasing 2+x in (a) (–1, ∞) (b) (–∞, 0) (c) (–∞, ∞) (d) none of these

(a) 0

 dy  3 3 50. If x = a cos q, y = a sin q, then 1 +   is  dx 

74

(a) 1

2

47. A fair die is rolled. Consider the events A = {1, 3, 5}, B = {2, 3} and C = {2, 3, 4, 5}. Then the conditional probability P((A ∪ B) | C) is (a) 1/4 (b) 5/4 (c) 1/2 (d) 3/4

(b) 1 (d) sec2q

51. If a, b, c, d > 0, x ∈ R and (a2 + b2 + c2)x2 – 2(ab + bc + cd)x + b2 + c2 + d2 ≤ 0 then 33 14 log a 65 27 log b is equal to 97 40 log c

 ln x − 1  53. ∫   dx =  (ln x )2 + 1  x (a) +c 2 x +1 x (c) +c (ln x )2 + 1

x −2 y −3 z −4 46. The lines and = = 1 1 −k x −1 y − 4 z − 5 are coplanar if = = k 2 1 (a) k = 2 (b) k = 0 (c) k = 3 (d) k = –1

(a) tan2q (c) tanq

cateGoRy 2 Category 2 shall have questions of 2 marks each. For incorrect response, 25% of full mark (1/2) would be deducted.

m=1

1 sin 2° 1 (c) 2 sin 2° (a)

1 3 sin 2° 1 (d) 4 sin 2°

(b)

58. The equation of the plane containing the lines 2x – 5y + z = 3, x + y + 4z = 5, and parallel to the plane, x + 3y + 6z = 1 is (a) x + 3y + 6z = 1 (b) 2x + 6y + 12z = –13 (c) 2x + 6y + 12z = 13 (d) x + 3y + 6z = –7 59. Locus of the image of the point (2, 3) in the line (2x – 3y + 4) + k(x – 2y + 3) = 0, k ∈ R is a (a) circle of radius 2 (b) circle of radius 3 (c) straight line parallel to x-axis (d) straight line parallel to y-axis. (1 − cos 2 x )(3 + cos x ) = x tan 4 x x →0 (a) 2 (b) 1/2 (c) 4

60. lim

(d) 3

61. The area (in square units) of the region described by {(x, y) : y2 ≤ 2x and y ≥ 4x – 1} is 7 9 5 15 (a) (b) (c) (d) 32 32 64 64 62. If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of the boxes contains exactly 3 balls is 11

12

1 (b) 22   3 10 55  2  (d)   3 3

1 (a) 220   3

11

55  2    3 3

(c)

k x + 1, 0 ≤ x ≤ 3 63. If the function g (x ) =  is  mx + 2, 3 ≤ x ≤ 5 differentiable, then the value of k + m is (a) 10/3 (b) 4 (c) 2 (d) 16/5 64. Let f ′(x ) = 1

192 x 3

1 for all x ∈ R with f   = 0. 2 2 + sin px 4

If m ≤ ∫ f (x )dx ≤ M then the possible values of 1/2

m and M are (a) m = 13, M = 24 (c) m = –11, M = 0 x

du

ln 2

u

65. If ∫

(a) 4 (c) ln4

e −1

=

1 1 (b) m = , M = 4 2 (d) m = 1, M = 12

p , then the value of x is 6 (b) ln8 (d) none of these

cateGoRy 3 In category 3, questions may have more than one correct option. All correct answers only will yield two marks. Candidates must mark all the correct options in order to be awarded full marks

66. A line L is a tangent to the parabola y2 = 4x and a normal to the parabola x 2 = 2 y . The distance of the origin from L is 2 3 (a) 0 (b) (c) (d) 2 3 5 67. Let f (x) = ln|x| and g(x) = sinx. If A is the range of f (g(x)) and B is the range of g(f (x)), then (a) A ∪ B = (– ∞, 1] (b) A ∩ B = (– ∞, ∞) (c) A ∩ B = [–1, 0] (d) A ∩ B = [0, 1] 68. The general solution of the equation 3 – 2cosq – 4sinq – cos2q + sin2q = 0 is (a) np (b) 2np p p (c) 2np + (d) 2np + 2 4 69. A ship is fitted with three engines E1, E2 and E3. The engines function independently of each other 1 1 1 with respective probabilities , and . For the 2 4 4 ship to be operational at least two of its engines must function. Let X denote the event that the ship is operational and let X1, X2 and X3 denote respectively the events that the engines E1, E2 and E3 are functioning. Which of the following is(are) true? 3 c (a) P ( X1 | X ) = 16 (b) P(exactly two engines of the ship are 7 functioning |X) = 8 7 5 (c) P ( X | X2 ) = (d) P ( X | X1 ) = 16 16       70. Let DPQR be a triangle. Let a = QR, b = RP and       c = PQ. If | a | = 12, | b | = 4 3 and b ⋅ c = 24 then which of the following is true?   | c |2  | c |2  + | a | = 30 (a) (b) − | a | = 12 2 2     (c) | a × b + c × a | = 48 3   (d) a ⋅ b = −72 mathematics today | JANUARY ‘16

75

71. Let E1 and E2 be two ellipses whose centres are at the origin. The major axes of E1 and E2 lie along the x-axis and the y-axis respectively. Let S be the circle x2 + (y – 1)2 = 2. The straight line x + y = 3 touches the curves S, E1 and E2 at P, Q and R respectively. 2 2 Suppose that PQ = PR = . If e1 and e2 are the 3 eccentricities of E1 and E2 respectively, then the correct expression(s) is (or) 2 2 (a) e1 + e2 =

(c)

| e12

− e22 |

7

43 40

(b) e1e2 =

5 = 8

3 (d) e1e2 = 4

2 10

 e|x|+[ x ] − 2  72. If f (x ) = x   , then  | x | +[x]  (a) lim f (x ) = −1 (b) lim f (x ) = 0 x →0+

x →0−

(c) lim f (x ) = −1

(d) lim f (x ) = 0

x →0

x →0

73. The value of x satisfying the equation 2

  p  x − 2  x sin  x   + 1 = 0 is  2   (a) 1 (b) –1 (c) 0 (d) no value of x 4

74. Let f (x) be real valued function such that f (x + y) = f (x) f (a – x) + f (y) + f (a – x) ∀ x , y ∈ R then for some real a

76


(a) f (x) is a periodic function (b) f (x) is a constant function 1 1 (c) f (x ) = (d) f (x ) = cos x 2 2 75. System of equations x + 3y + 2z = 6, x + ly + 2z = 7 and x + 3y + 2z = m has (a) unique solution if l = 2, m = 6 (b) infinitely many solution, if l = 4, m = 6 (c) no solution if l = 5, m = 7 (d) no solution if l = 3, m = 5 ANSWER KEYS

1. 6. 11. 16. 21. 26. 31. 36. 41. 46. 51. 56. 61. 66. 69. 72. 75.

(a) 2. (c) 7. (b) 12. (a) 17. (c) 22. (c) 27. (c) 32. (a) 37. (d) 42. (b) 47. (c) 52. (c) 57. (b) 62. (a, b) (b, d) (a, b) (b, c, d)

(b) (c) (d) (a) (a) (b) (b) (a) (b) (d) (a) (d) (c)

3. 8. 13. 18. 23. 28. 33. 38. 43. 48. 53. 58. 63. 67. 70. 73.

(d) 4. (b) 9. (d) 14. (b) 19. (b) 24. (b) 29. (a) 34. (a) 39. (b) 44. (a) 49. (c) 54. (a) 59. (c) 64. (a, c) (a, c, d) (a, b)

(d) (a) (d) (c) (d) (d) (c) (d) (c) (d) (b) (a) (d)

5. (d) 10. (a) 15. (b) 20. (a) 25. (a) 30. (b) 35. (c) 40. (a) 45. (a) 50. (d) 55. (d) 60. (a) 65. (c) 68. (b, c) 71. (a, b) 74. (a, b, c) nn

and the other from row 3. 1 6 11 2 7 12 3 8 13 4 9 14 5 10 15

soLUtioN set-156

2ab = 2016 a +b \ (a – 1008)(b – 1008) = (1008)2 = 28⋅34⋅72. The 1 number of pairs = (The number of divisors – 1) 2 9⋅5⋅3 −1 = = 67 2 1 1 1 2. (c) : + = a b 2016 (a – 2016)(b – 2016) = (2016)2 = 210⋅34⋅72 1 Number of pairs = (11 ⋅ 5 ⋅ 3 − 1) = 82 2 2016 3. (b) : (5x – 2) = (5x)2016 – 2016(5x)2015 × 2 2016 ⋅ 2015 + (5x )2014 × 22 + ... = 0 2 2016 ⋅ 2015 (5)2014 × 22 The sum of the roots = 2 ⋅ 2016 (5)2015 × 2 = 403 1. (d) :

5 5 \ The number of pairs (a, b) = 5   + 2   2   1  = 100

\

5. (b) :

a 4

0

x 5 (2016 − x )6 dx

Put x = 2016t

1

I = 201612 ∫ t 5 (1 − t )6 dt = 201611 ⋅ 2016 ⋅ 0

5!6 ! 12 !

4 ⋅ 201611 11 π 6. (a, b, c) : In triangle ABC if C = 2 2 2 2r + c = a + b, (a + b – 2r) = a + b2 (a – 2r)(b – 2r) = 2r2 = 2(2016)2 = 211⋅34⋅72 1 Number of triangles is (number of divisors of 2r2) 2 = 12⋅5⋅3 = 22⋅32⋅5 =

7. (b) : a2 – b2 = (a – b)(a + b) is divisible by 5 if both a and b are from any of the five rows, or one from row 1 and the other from row 4, or one from row 2 78


2

b c 5 16 6 15  10 11

a b c a b c a b c 5 6 14 6 7 12 7 8 10 7 13 8 11 8 12 9 10 9 11 Total (a, b, c) are 6 + 4 + 3 + 1 = 14 and taking the permutations 14 × 3! = 84. a + b + c = 25, where a, b, c ≥ 4 ⇒ a + b + c = 16, where a, b, c ≥ 1. The number of 15  positive integer solutions =   = 105 2 84 4 m Prob. = = = , n−m =1 105 5 n

(104030 − 1) (102015 − 1) +4 +1 (10 − 1) (10 − 1) = 444...4 + 444...4 + 1 S = (4 × 4030) + (4 × 2015) + 1 = 24181, with digit sum 16.

∫

2

5   4  5 \ The number of pairs (a, b) = 5   +      2   2  1  = 200 9. (1) : a < b < c, a + b + c = 25

N =4

2016

21 22 23 24 25

8. (b) : a4 – b4 = (a2 – b2)(a2 + b2) is divisible by 5 if both a and b over from way of the five rows or one each from any of the first 4 rows.

4. (a) : (102016 + 5)2 = 225 N 9N = (2⋅102015 + 1)2 = 4⋅104030 + 4⋅102015 + 1 \

16 17 18 19 20

10. (a) : (P) → 3, (Q) → 5, (R) → 4, (S) → 1   2 2 (P) 3(4 + 5 ) − 40∑ a ⋅ b ≤ 123 + 60 = 183   since −2∑ a ⋅ b ≤ 3 (Q) t1 + t2 + t3 = – 1 – 1 + 8 = 6. Sum = 33 × 6 – 1 = 197 (R) Using Leibnitz rule : y4(1)= 1 × 24 × (–1)(–8) = 192 (S) (a + b)(r + s) = – 15 ⇒ p = 15 Product of roots = 158 = q, \ p + q = 173 nn Solution Sender of Maths Musing set-155 1. 2.

Khokon Kumar Nandi (W.B.) Gajula Raninder (Karim Nagar (TS))

1. We are considering triangles ABC in space. (a) What conditions must be fulfilled by the angles a, b, g of triangle ABC in order that there exists a point P in space such that ∠APB, ∠BPC, ∠CPA are right angles? (b) Let d be the maximum distance among PA, PB, PC and let h be the longest altitude of triangle ABC. Show that

(

6 / 3) h ≤ d ≤ h.

2. Two circles intersect at A and B. P is any point on an arc AB of one circle. The lines PA, PB intersect the other circle at R and S, as shown below. If P′ is any other point on the same arc of the first circle and if R′, S′ are the points in which the lines P′A, P′B intersect the other circle, prove that the arcs RS and R′S′ are equal. P

S

S

3. For any positive integer n, evaluate an/bn, where k =1

n kp kp , bn = ∏ tan2 . 2n + 1 2n + 1 k =1

4. There are 9 regions inside the 5 rings of the Olympics. Put a different whole number from 1 to 9 in each so that the sum of the numbers in each ring is the same. What are the largest and the smallest values of this common sum?

∑ ai = 0. i =1

1 + a a + a + a i i +1 ) ( i i +1 i +2 ) ... (ai + ai +1 + ⋅⋅⋅ + ai +n−2 )

i =1 i

where an+1 = a1, an+2 = a2, etc., assuming that the denominators are non zero. solutions

1. (a) By Pythagoras theorem, we have AB2 = PA2 + PB2, AC2 = PA2 + PC2 and BC2 = PB2 + PC2. P

d A

R B

n

n

n

∑ a (a

R

A

P

an = ∑ tan2

5. Given are real numbers a1, a2, ..., an with Determine

2

h 2

2

B

MH

C

Hence AB + AC – BC = 2PA2 > 0, thus ∠BAC < 90° , i.e., a < 90°. Similarly, we get b < 90° and g < 90°. Conversely, if a, b and g are all acute angles, we may prove that there exists a point P such that ∠APB, ∠ABC, ∠CPA are right angles. (b) We may without loss of generality assume that PA ≥ PB ≥ PC, so PA = d, Because AB2 = AP2 + BP2 ≥ AP2 + CP2 = AC2 and AC2 = AP2 + CP2 ≥ BP2 + CP2 = BC2, we have AB ≥ AC ≥ BC. Let H be the foot of the perpendicular from A to BC, then AH is the longest altitude of DABC, so AH = h. As AP ^ BP and AP ^ CP, AP is perpendicular to the plane of BPC. Thus AP ^ BC and AP ^ PH so that AP < AH, i.e., d < h. ...(1) Because AP ^ BP and AH ^ BC, we get BC is perpendicular to the plane of APH. Thus, we have mathematics today | JANUARY ‘16

79

BC ^ PH. Let M be the midpoint of BC, then PH ≤ PM. As 1 ∠BPC = 90°, we have PM = BM = MC = BC. 2 Hence, 2PH ≤ 2PM = BC, so that 4PH2 ≤ BC2 = PB2 + PC2 ≤ 2PA2 As ∠APH = 90°, we get PH2 = AH2 – AP2 = h2 – d2 From (2) and (3), we have 4(h2 – d2) ≤ 2d2, i.e., 2h2 ≤ 3d2, from which, we have 6 h ≤ d. 3 From (1) and (4) we obtain

k =0

...(3)

So tan2

jp ,1 ≤ j ≤ n, are the roots of 2n + 1

n

...(4)

 n  n−2k −1 cos q sin2k +1 q  2 k 1 +  

n k  2n + 1 2n−2k sin (2n + 1) q = ∑ ( −1)  q sin2k +1 q  cos 2 k 1 +  k =0 n k  2n + 1 2k = tan q cos2n+1 q ∑ ( −1)   tan q. 2 1 k +  k =0

80


k

∑ (−1)

(−1)k 

So

k =0

 2n + 1 2k  2k + 1 tan q = 0

jp ,1 ≤ j ≤ n 2n + 1

3. Using De Moivre’s theorem cosnq + i sinnq = (cosq + i sinq)n, one finds easily that

∑

k

∑ (−1)

for q =

(Notice that this also gives the ‘power of the point’ result for P, PA ⋅ PR = PB ⋅ PS) Similarly P ′A = P ′B = AB P ′S P ′R ′ R ′S ′ Consider now traingles APS and AP′S′. We have ∠APS = ∠APB = ∠AP′B = ∠AP′S′, because P, P′ lie on the same arc of chord AB of the one circle. From the fact that S and S′ lie on the same arc of chord AB of the second circle ∠AS′P′ = ∠ASP. But then DAPS and DAP′S′ are similar. Thus PA P ′A AB AB = . So = PS P ′S ′ RS R ′S ′ Thus RS = R′S′ and the arcs are equal.

sin nq =

n

...(2)

6 h ≤ d < h, as required. 3 2. Because opposite angles of a cyclic quadrilateral are supplementary, we have that ∠PBA = p – ∠ABS = ∠ARS. Similarly ∠PAB = ∠BSR. Thus DPAB and DPSR are similar, from which PA PB AB = = PS PR RS

 n−1   2 

Thus

k =0 n

k

∑ (−1)

k =0

 2n + 1 k  2k + 1 x = 0 and thus also of  2n + 1 n−k = 0. Since an and bn are the  2k  x

sum and product of the roots, respectively, we have  2n + 1  2n + 1 ( ) an =  = n n + b = 2 1 and n  2n  = 2n + 1,  2  an =n and so bn

4. For the five rings, we have a+b=b+c+d=d+e+f=f+g+h = h + i = N. ...(1) Since we are dealing with the nine non-zero decimal digits, we have

9

∑ j = 9 (10) / 2 = 45. 1

The

five regions sum to a common N for 45/5 = 9 but then one pair must be 9 + 0 or one triplet 9 + 0 + 0, which isn’t allowed. So N > 9. Since a + b = h + i, there must be at least two pairs of decimal digits that sum to N. For 10 ≤ N ≤ 15, we have N = 9 + a = 8 + (1 + a) = ..., for 1 ≤ a ≤ 6 while N = 16 = 9 + 17 and N = 17 = 9 + 8 only. So N ≤ 15. From (1) a + b = b + c + d or a = c + d ...(2) and h + i = f + g + h or i = f + g ...(3) The five central digits must equal 45 – 2N (c + d) + e + (f + g) = a + e + i = 45 – 2N So, we have N

2N

45 – 2N

a, e, i

10

20

25

9, 8 – no digit available

11

22

23

9, 8, 6;...

12

24

21

9, 8, 4;...

13

26

19

9, 8, 2;...

14

28

17

9, 7, 1;...

15

30

15

9, 5, 1;...

So 11 ≤ N ≤ 15.

1

i =1 ai +1

(ai+1 + ai+2 )... (ai+1 + ai+2 + ⋅⋅⋅ + ai+n−1 ) i

be the given sum. Let bi = ∑ ar so bn = 0. Also, let n−1

p = ∏ br . Since for all j ≠ i

r =1

r =1

bj – bi = bn – bi + bj = ai+1 + ai+2 + ⋅⋅⋅ + an + a1 + a2 + ⋅⋅⋅ + aj = ai+1 + ai+2 + ⋅⋅⋅ + an + an+1 + ⋅⋅⋅ + an+j, We have 1 1 n−1 1 = +∑ ∏ p i =1 1≤ j≤n b j − bi i =1 1≤ j ≤n b j − bi n

S=∑

∏ j ≠i

=

n

1 S=∑ i =1 ai (ai + ai +1 ) ... (ai + ai +1 + ⋅⋅⋅ + ai +n−2 )

• • • • •

j ≠i

1 n−1  1 1  +∑ − . ∏  p i =1 bi 1≤ j≤n−1 b j − bi    j ≠i

Using Lagrange’s interpolation, we let n−1  bj − x  1 F (x ) = ∑  − ∏  i =1  p 1≤ j ≤n−1 b j − bi  j ≠i   So that F(bk) = –1/p for all 1 ≤ k ≤ n – 1. Note that F(x) is a polynomial in x of degree at most n – 2, yet it is a constant at n – 1 distinct values. Hence F(x) is a constant, F(x) = –1/p. Since bj 1 1 1 − =− , ∏ ∏ p 1≤ j≤n−1 b j − bi bi 1≤ j≤n−1 b j − bi j ≠i

5. Let

Do you want yourself to be updated about

n

=∑

j ≠i

we get F(0) = S – 1/p and thus 1 S = F (0) + = 0. p

nn

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Like us on Facebook https://www.facebook.com/pcmbtoday mathematics today | JANUARY ‘16

81

10 Best Proble

M

th rchives 10 Best Problems

Prof. Shyam Bhu

Math Archives, as the title itself suggests, is a collection of various challenging problems related to the topics of JEE (Main & Advanced) Syllabus. This section is basically aimed at providing an extra insight and knowledge to the candidates preparing for JEE (Main & Advanced). In every issue of MT, challenging problems are offered with detailed solution. The readers' comments and suggestions regarding the problems and solutions offered are always welcome.

1. If An is the area bounded by y = x and y = xn, n ∈ N, then A2 ⋅ A3 …… An = (a) (c)

1 n (n + 1) n−1

2

(b)

1 n (n + 1)

(d)

n

1

2 n (n + 1) n−2

2

1 n (n + 1)

cos–1(cos

2. The area bounded by y = x), x-axis and the lines x = 0, x = 2p is (a) p2 (b) 2p2 2 (c) 4p (d) Area is not bounded 3. The area bounded by y = x2, the lines y = 1 and y = 2, on the left of y–axis is 4 4 1− 2 2 2 2 −1 (a) (b) 3 3 2 2 1− 2 2 (c) (d) 2 2 −1 3 3

( (

( (

) )

) )

cos 4 x + 1 ∫ cot x − tan x dx = acos22x + bcos2x + c ln|cos2x| + l, l being a constant of integration, then 1 1 (a) a = − (b) b = − 8 8 5 (c) c = (d) a + b + c = 0 8 4.

5.

If

dx

∫ x (1 + x 4 ) =

(a) 3x2 + c

(b) ln

1 (c) ln | x | − ln(1 + x 4 ) + c 4 (d) None of these

x 1 + x2

+c

6. Let f(x) be a polynomial of degree three such that f(0) = 1, f(1) = 2 and 0 is a critical point of f having no f (x ) dx . local extreme. Find ∫ x2 + 7 7.

 1  Evaluate ∫ sin −1  2 x 1 − x 2  dx , x ∈  , 1 .    2 

8. Using derivatives up to order of f(x) = x1/3 + x2/3 and giving main points, draw the graph of y = f(x). Also find the area bounded by the curve y = f(x) and the x – axis. 9.

Solve

d3 y

d2 y

= 0 , given that dx 3 dx 2 1 y(0) = , y ′(0) = 0 and y ′′(0) = 1. 8 10. Draw the graph of the function f defined by f(x) = cot–1x, giving main points. Hence or otherwise evaluate

1

∫

−8

[cot −1 x]dx ([.] denotes the greatest inte-

− 3

ger function). soLUtioNs 1

1  x 2 x n+1  1. (d) : An = ∫ (x − x n ) dx =  −   2 n + 1  0 0 1 1 n −1 y = − = 2 n + 1 2(n + 1) Thus A2 ⋅ A3 ⋅ A4 ⋅ ..... An

=

1 1 2 3 n −1  . . ....... +  3 4 5 n 1 2 1 n−1

O

= n−2 2 n(n + 1)

By : Prof. Shyam Bhushan, Director, Narayana IIT Academy, Jamshedpur. Mob. : 09334870021 82


1

x

2. (a) : Obviously area = p2

=∫

y



O

2

3. (c) : The required area 2

y

2

2  = − ∫ (− y ) dy =  y 3/2  3 1 1

2

y = x2

2 = (2 2 − 1) sq. units 3

1 O

x

4 4. (b) : 1 (cos x + 1)sin 2 x dx 2∫ cos 2 x

(

)

Hence ∫ sin −1 (2 x 1 − x 2 ) dx = ∫ (p − 2 sin −1 x ) dx = px − 2 ∫ sin −1 x dx

2

1 1 (1 + t ) + 4 =− . ∫ dt t 8 2 1  5 = − ∫  t + 2 +  dt 16  t

  x = px − 2  x sin −1 x − ∫ dx   1 − x 2 

1 1  = − .  cos2 2 x + 2 cos 2 x + 5 ln| cos 2 x | + l 16  2  5. (c) :

x

1 1

1 

Let I = ∫

∫ x 4 (1 + x 4 ) dx = 4 ∫  t − t + 1  dt

1 = ln | x | − ln(1 + x 4 ) + c 4 6. Let f(x) = ax3 + bx2 + cx + d. Since f(0)=1, so d=1. Moreover f ′ (0) = 0 i.e., c = 0. Also f ′′(x) = 6ax + 2b. Since f(x) doesn’t have extremum at 0, so f ′′(0) = 2b = 0 Hence f(x) = ax3 + 1 f(1) = 2 ⇒ a = 1 ⇒ f(x) = x3 + 1

∫

=∫

f (x ) x2 + 7 x3 2

x +7

dx = ∫ dx + ∫

x3 + 1 x2 + 7 dx 2

x +7

dx

2x

= px − 2 x sin −1 x + ∫

= −∫

1 t 1  x4  = ln + c = ln  +c 4 t +1 4  x4 +1

Now,

(Put y2 = x2 + 7)

3

p = –sin–1(sin(2q – p)), − ≤ 2q − p ≤ 0 = p– 2q = p– 2 sin–1x 2

2 1 (1 + cos 2 x ) + 4 sin 2 x = ∫ dx 8 cos 2 x

3

y

+ log | x + x 2 + 7 | +C

y − 7 y + log| x + x 2 + 7 | + C 3 1 = (x 2 + 7)3/2 − 7(x 2 + 7)1/2 + log | x + x 2 + 7 | +C 3 p p 7. Let sin–1 x = q, then x = sinq and ≤ q ≤ 4 2 sin −1  2 x 1 − x 2  = sin–1(2sinq| cosq|)   p p = sin–1(2 sinq cosq), as ≤q≤ 4 2 = sin–1(sin 2q) =

x

( y 2 − 7) ydy

2x

1− x

2

1 − x2

dx

dx

2t dt , Put 1 − x 2 = t 2 t

= −2 1 − x 2 + c Hence

∫ sin

−1

(2 x 1 − x 2 ) dx = px − 2 x sin −1 x − 2 1 − x 2 + c

8.

Let f (x ) = x1/3 + x 2/3

−2 x1/3 + 1 . 2 5 9 3 3⋅ x x3 Obviously f(x) = 0 at x = 0, f(x)>0 for x ∈(−∞, − 1) ∪ (0, ∞) and f(x)<0 for x ∈ (–1, 0) 1 Further f ′(x ) > 0, ∀ x ∈  − , ∞  , where f increases and  8  ⇒ f ′(x ) =

1 + 2 ⋅ x1/3

⇒ f ′′(x ) =

1  f ′(x ) < 0, ∀ x ∈  −∞, −  , where f decreases.  8 mathematics today | JANUARY ‘16

83

y

f ′(x ) < 0, ∀ x ⇒ f is decreasing f ′′(x ) < 0, ∀ x < 0 and f ′′(x ) > 0, ∀ x > 0 .

O

1

Thus x = 0 is a point of inflection and f is concave down for x < 0, f is concave up for x > 0.

x

f ′′(x ) < 0, ∀ (−∞, − 1) ∪ (0, ∞) , where f is concave down.

⇒

Lastly f (x ) → ∞ as | x | → ∞ .

⇒

The graph of y = f(x) is shown in adjacent figure. 2  1 3 3  dx = 3 sq.units  = − x + x Required area ∫ 20  −1  0

⇒

9. Let yn = dny / dxn, then the given differential equation is y3 – 8y2 = 0 or y3/y2 = 8 which implies log|y2| = 8x + C1 (integrating both sides) Putting x = 0, we get C1 = log y2(0) = log 1 = 0. \ log|y2| = 8x or y2 = ± e8x, i.e., y1 = ± e8x/8 + C2. Again putting x = 0, we have C2 = ± 1/8.  1 8x 1  e8 x − x  + C3 So y1 = ± (e − 1) ⇒ y = ±  8 8 8  Putting x = 0, we have

Thus

n

1 e 9 or y = −  −x−  8 8 8

lim f (x ) = 0 and

n

84

n

n

lim f (x ) = p . Thus y = 0 and n

y = p are asymptotes. f(0) = p/2 ⇒ (0, p/2) is a point on the graph.

cot 3

2x

(1 + x 2 )2

x →−∞

y  3 2 /2 1 cot 2

O cot 1

n

n

y=

1


x

cot −1 x  dx =  

cot 2

∫

− 3

2dx +

cot 1

∫

1dx +

cot 2

1

∫

0 dx

cot 1

Your favourite MTG Books/Magazines available

8x

x →∞

∫

2, x ∈[− 3 , cot 2]  x] = 1, x ∈(cot 2,cot 1] 0, x ∈(cot 1,1] 

nn

n

1+ x

1

−1

= 2 (cot 2 + 3 ) + (cot 1 − cot 2) = cot 2 + cot 1 + 2 3

n

⇒ f ′′(x ) = 2

[cot

− 3

n

1  e8 x 7 Thus y =  −x+  8 8 8

−1

cot 3 < − 3 < cot 2 < 0 < cot 1 < 1.

in ASSAM

C3 = (1 / 8) + (1 / 64) = 7 / 64, 9 / 64.

10. f (x ) = cot −1 x ⇒ f ′(x ) =

p p 5p <1< < 2 < <3 4 2 6 5p p p cot 3 < cot < cot 2 < cot < cot 1 < cot 6 2 4

We have

f ′′(x ) > 0, ∀ x ∈(−1, 0), where f concave up and

& NORTH EAST

at

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Y U ASK

and L.H.S. of (5) = 2n + i(1 + i)n + i2(1 + i2)n + i3(1 + i3)n

WE ANSWER Do you have a question that you just can’t get answered? Use the vast expertise of our mtg team to get to the bottom of the question. From the serious to the silly, the controversial to the trivial, the team will tackle the questions, easy and tough. the best questions and their solutions will be printed in this column each month.

1.

Prove that

{

}

1 n−1 n/2 np 2 − 2 sin 2 4 – Chinmay, Assam Ans. Q In given series difference in suffixes is 4. n

C3 + nC7 + nC11 + ... =

Now (1)1/4 = (cos 0 + i sin 0)1/4 = (cos2rp + i sin 2rp)1/4 rp rp = cos + i sin , where r = 0, 1, 2, 3 2 2 Four roots of unity are 1, i, –1 and –i = 1, a, a2and a3 (say) n

n Also (1 + x ) = ∑ nCr x r r =0

Putting x = 1, a, a2 and a3, we get n

n

r =0

r =0

n n r 2n = ∑ nCr ... (1) , (1 + a ) = ∑ Cr a n

(1 + a2 )n = ∑ nCr a2r and (1 + a

3

r =0 n

)

... (2)

... (3)

n

= ∑ nCr a3r ... (4 ) r =0

Multiplying (1) by 1, (2) by a, (3) by a2 and (4) by a3 and adding, we get 2n + a(1 + a)n + a2(1 + a2)n + a3(1 + a3)n n

= ∑ nCr (1 + ar +1 + a2r +2 + a3r +3 ) r =0

... (5)

For r = 3, 7, 11, ... R.H.S. of (5) is nC (1 + a4 + a8 + a12) + nC (1 + a8 + a16 + a24) 3 7 + nC11(1+ a12 + a24 + a36) + ... (... a4 = 1) = 4(nC3 + nC7 + nC11 + ...)

np 4 np \ 4 nC3 + nC7 + nC11 + ... = 2  2n−1 − 2n/2 sin   4  np 1 ⇒ nC3 + nC7 + nC11 + ... =  2n−1 − 2n/2 sin  2 4  3 Prove that ≤ sin2 q + cos 4 q ≤ 1 for all real q. 4 – Nandini, Gujarat = 2n + i{(1 + i)n – (1 – i)n} = 2n − 2n/2 ⋅ 2 sin

(

2.

)

Ans. sin2q + cos4q = cos4q – cos2q + 1 2 1 1 3 = (cos2 q) − 2. cos2 q + + 2 4 4 2 1 3 3  2 =  cos q −  + ≥  2 4 4 2 4 Also, sin q +cos q = cos4q – cos2q + 1 = cos2q (cos2q – 1) + 1 = 1 – sin2q cos2q ≤ 1

{

}

3 ≤ sin2 q + cos 4 q ≤ 1. 4 sin −1 x − cos −1 x Evaluate ∫ dx sin −1 x + cos −1 x – Raman, A.P.

\ 3.

Ans. Let

p − 2 cos −1 x 2 I = ∫ −1 dx = ∫ dx p x + cos −1 x sin 2 4  4  −1 −1 x  dx = x − ∫ 1 ⋅ cos = ∫ 1 − cos x dx  p  p 4 −1 1  . = x − cos −1 x ⋅ x − ∫ x ⋅ dx  2 2 x p   1− ( x ) 4x 2 x \ I = x − cos −1 x − ∫ dx p p 1− x Put x = cos2q; then dx = –2cosq ⋅ sinq dq \

sin −1 x − cos −1 x

∫

x 1− x

dx = ∫

cos q 1 − cos2 q

(−2 cos q ⋅ sin q) dq

= − ∫ 2 cos2 qdq = − ∫ (1 + cos 2q)dq

{

= − q+

}

sin2q + c = − cos −1 x − x . 1 − x + c 2

\ I = x − 4 x cos −1 x − 2 {− cos −1 x − x 1 − x + c} p p 2 2 = x + (1 − 2 x ) cos −1 x + x (1 − x ) + c. p p nn mathematics today | JANUARY ‘16

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Mathematics Today - January 2016

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