Mathematics Today

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MatheMatics tODaY | MARCH ’15 7

Mathematics – is it an art, science

or Metaphysics?

W

hile calculation remained the core of Mathematics in the early days, as one started learning higher mathematics, mathematics had developed into many branches. Symbolic representation developed into algebra. Study of Euclid’s geometry in the westernised studies in the early days developed to co-ordinate geometry and analytical geometry. The study of symmetry in crystals and molecules developed to crystallography and group theory.

In modern studies of diffraction of crystals, group theory is very important. Now, quantum mechanics embraces everything from the determination of energy levels, intensities and widths of spectral lines. If atomic and molecular physicists study X - ray diffraction, nuclear physicists study spectroscopy and diffraction of g - rays. When mathematics is so interesting and exciting, why do many students and even professors try to avoid mathematics?

It is a purely psychological problem and some fear of punishment. The solution is novelty in packaging which should be so attractive that persons rush to study maths in schools. This is being done in many innovative modern schools. However, to make it accessible to everybody, the solution is simple – to publish popular books in science, mathematics, languages and so on. The key for making education a success is to publish popular books in a simple language so that all can understand these books on self studies on every topic. It is a success for European languages. Why not science and mathematics for every topic written in a popular style?

Anil Ahlawat

Editor

Vol. XXXIII No. 3 March 2015 Corporate Office

Plot 99, Sector 44 Institutional Area, Gurgaon, (HR). Tel : 0124-4951200

e-mail : [email protected] website : www.mtg.in Regd. Office

406, Taj Apartment, Near Safdarjung Hospital, Ring Road, New Delhi - 110 029.

Managing Editor : Mahabir Singh Editor : Anil Ahlawat (BE, MBA)

cONteNts

Maths Musing Problem Set - 147 8

Mock Test Paper 10

JEE Main - 2015

BITSAT - 2015

10 Challenging Problems 28

JEE Advanced - 2015

Math Archives 48

JEE Main - 2015

You Asked, We Answered 58

ISI - 2015

CBSE Board 2015 71

Sample Paper

Maths Musing - Solutions 82

Practice Paper 83

JEE (Main & Advanced) & Other PETs

Olympiad Corner 88

rial

edit

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8 MatheMatics tODaY | March ’15 jee main

1. In triangle ABC, if D = a2 – (b – c)2, then sin A = (a) 15₂₂ (b) ₁₁4 (c) 15 17 (d) 8 17 2. If y x x x ( ) = −

1 4 , then the sum of the digits of

d y dx 5 5 0 ( ) is (a) 3 (b) 4 (c) 5 (d) 0

3. The planes x + y = 1, y –z = 2, z + x = 3 form a

triangular prism with cross sectional area (a) 4₃ (b) 8₃ (c) 4 3 (d) 2 3

4. The area of the triangle, whose vertices are the roots of the equation x3 + ix2 + 2i = 0, is

(a) 2 (b) 3 (c) 5 (d) 7 5. x2 x dx 0 6 sin =

∫

π (a) π2 3 (b) − π2 3 (c) π2 6 (d) −π 2 6 jee advanced 6. If C_r =_r    10 and C C C C₀ ₁ ₂ ₃ C₁₀ 10 11 12 13− + − + +... 20 = 1 n , then n is divisible by (a) 11 (b) 13 (c) 17 (d) 19 comprehension Let S = {1, 2, 3, 4, ..., 25} and T = {x, y} ⊂ S.

Prof. Ramanaiah is the author of MTG JEE(Main & Advanced) Mathematics series

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.

Prof. Dr. Ramanaiah Gundala, Former Dean of Science and Humanities, Anna University, Chennai

7. The probability that x2 – y2 is divisible by 5 is (a) 1₅ (b) 1₄ (c) 1

3 (d)

2 5

8. The probability that x2 –y2 is divisible by 7 is (a) 1 3 (b) 730 (c) 71300 (d) 73 300 integer match 9. Let 88−_{1 7}8 8 8₂ 68 8₃ 58 ₄8 48    +     −     +     −    +     −     = 8 5 3 86 2 8 7 1

8 8 8 _{n.The sum of the}

digits of n is

matching list

10. The sequence of positive integers a₁, a2, a3, ... is

such that a1, a2, a3 are in G.P., a2, a3, a4 are in A.P., a3, a4, a5 are in G.P., a4, a5, a6 are in A.P. etc. Let a1 = 1 and a5 + a6 = 198. Then match the following:

Column-I Column-II

P. Sum of the digits of a8 is 1. 5

Q. Sum of the digits of a9 is 2. 9

R. Sum of the digits of a10 is 3. 15

S. Sum of the digits of a11 is 4. 19

P Q R S

(a) 3 2 1 4

(b) 2 1 4 3

(c) 1 4 3 2

(d) 4 3 2 1

See Solution set of Maths Musing 146 on page no. 82

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10 MatheMatics tODaY | March ’15 1. _r (r2 r )xr 1 1 3 − + − = = ∞

∑

(a) 3 + 2x_{(1 –}x₎–2 _{(b) 3} 2 1 3 x x + − ( ) (c) 3 1 4 1 2 3 ( ) ( ) x x x + − − (d) none of these

2. The tangent at (1, 7) to the curve x2₌_{y – 6}

touches the circle x2_{+ y}2_{+ 16x + 12y + c = 0 at}

(a) (6, 7) (b) (–6, 7) (c) (6, –7) (d) (–6, –7) 3. If log | | | | | | 3 2 ₁ 2 2 z z z − + +     

 < , then the locus of z is

(a) |z| = 5 (b) |z| < 5

(c) |z| > 5 (d) None of these

4. Let f (x) be differentiable on the interval (0, ∞)

such that f (1) = 1, and lim ( ) ( )

t x t f x x f t t x → − − = 2 2 1 for each x > 0. Then f(x) is (a) ₃1_x+2₃x2 (b) − 1 + 3 4 3 2 x x (c) − +1_{x x}2₂ (d) 1_x 5. lim tan x a x a x a →    −  2  2 = p (a) 2/p (b) e2/p (c) e–2/p _{(d) none of these}

6. A bag contains 8 red and 5 white balls. Three balls are drawn at random. Find the probability that I. All the three balls are white

II. All the three balls are red

III. One ball is red and two balls are white

I II III (a) 5 143 29 143 40 143 (b) ₁₄₃5 ₁₄₃28 ₁₄₃40 (c) ₁₄₃7 ₁₄₃28 ₁₄₃40 (d) None of these

7. Let f : (–1, 1) → B, be a function defined by

f x x x ( ) tan= −    −1 2 2

1 , then f is both one-one and onto when B is in the interval

(a) −_ p p_{2 2}, _ (b) −    p p 2 2, (c) 0 2 , p   (d) 0 2 , p

8. The locus of the centre of circle which touches (y – 1)2_{+ x}2_{= 1 externally and also touches x-axis, is}

(a) {x2_{= 4y, y ≥ 0} ∪ {(0, y), y < 0}}

(b) x2_{= y} (c) y = 4x2 (d) {y2_{= 4x} ∪ (0, y), y ∈ R} 9. If y z z x x y a b c      _ _ = 1 and A y z b c =   − 1 , B z x c a =   − 1 , C= _x_y_a b− 1 , then (a) A = B = C (b) ABC = 1 (c) A + B + C = 0 (d) none of these

Exam on

4th April

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12 MatheMatics tODaY | March ’15 10. If 1 4 4 1 4 4 1 4 4 0 2 2 2 2 2 2 + + + =

sin cos sin

q q q q q q q q q , then q is equal to (a) 7₂₄p,11₂₄p (b) 5₂₄p p,7₂₄ (c) 11_{24 24}p p, (d) ₂₄p ,7₂₄p

11. If a, b, c are the sides of a DABC such that x2_{– 2(a + b + c)x + 3l(ab + bc +}_{ca) = 0 has real}

roots, then (a) l < 4 3 (b) l > 5 3 (c) l ∈_4₃,5₃_ _{(d) l ∈}   1 3 5 3 , 12. The expression 3 3 2 3 4 4 sin _ p a− _+sin ( p a+ )    _ −  _ + _+ −   _ 2 2 5 6 6

sin p a sin ( p a is equal to)

(a) 0 (b) 1

(c) 3 (d) sin 4a + cos 6a

13. 5-digit numbers are to be formed using 2, 3, 5, 7, 9 without repeating the digits. If p be the number of such numbers that exceed 20000 and q be the number of those that lie between 30000 and 90000, then p : q is

(a) 6 : 5 (b) 3 : 2

(c) 4 : 3 (d) 5 : 3

14. The area bounded by the curves y = (x –1)2_, y = (x + 1)2_{and y =} 1

4 is

(a) 1₃sq. unit (b) 2₃sq. unit (c) 1₄sq. unit (d) 1₅sq. unit

15. Given a= + −2i j^ ^ 2k b^,= +2i j c^ ^, | | = ⋅a c , |c a − |= 4cos₄p and angle between   c & ( × ) a b

is p/6, then | (a b c × ×) | ?= is equal to (a) 5 (b) 1 (c) 3/2 (d) none of these 16. If tan−1 +sin−1 = , 2 x x p then x = (a) ± 5 1− 2 (b) ± ± 5 1 2 (c) ± 5 1+ 2 (d) none of these

17. For 0≤ ≤x p , the value of₂

sin ( ) cos cos ( )

sin ₋ ₋ +

_∫

∫

1 1 0 0 2 2 t dt x t dt x is equal to (a) − p 4 (b) 0 (c) 1 (d) p 4

18. For each real x : –1 < x < 1. Let A(x) be the

matrix (1 ) 1 and 1 1 1 − − −         = ++ − x x x z x y xy , then

(a) A(z) = A(x) – A(y)

(b) A(z) = A(x) A(y) (c) A(z) = A(x) [A(y)]–1

(d) A(z) = A(x) + A(y)

19. A variable plane x_a+ + = 1 at a unit _by z_c distance from origin cuts the coordinate axes at

A, B and C. Centroid (x, y, z) satisfies the equation

1 1 1

2 2 2

x + y +z = . The value of K isK

(a) 9 (b) 3 (c) 1₉ (d) 1₃

20. If a_r be the coefficient of x_r in the expansion of (1 – x)2011_{, then sum of which of the following pair}

vanishes ?

(a) a₇₇₇, a₁₂₃₄ (b) a₁₁₁₁, a₉₀₀ (c) a₆₅₄, a₁₃₅₇ (d) All of these

21. The value of /2_/₂ sin2 1 2+ x dxx −

∫

p p is (a) p (b) p₂ (c) 4p (d) p₄

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MatheMatics tODaY | March ’15 13 22. The locus of the orthocentre of the triangle

formed by the lines (1 + p)x –py + p(1 + p) = 0, (1 + q)x –qy + q(1 + q) = 0 and y = 0, where p ≠ q, is

(a) a hyperbola (b) a parabola

(c) an ellipse (d) a straight line

23. If a variable x takes values x_i such that a ≤ x_i ≤ b, for i = 1, 2, ..., n, then

(a) a2_{≤ var(x) ≤ b}2 _(b) _{a ≤ var(x) ≤ b}

(c) a2 x

4 ≤ var( ) (d) (b – a)2 ≥ var(x)

24. If _{2 +}3_e_iq =lx i y+ m , then locus of P(x, y) will represent a/an

(a) Ellipse if l = 1, m = 2

(b) Pair of straight lines if m = 1, l = 0 (c) Circle if l ≠ m ≠ 1

(d) None of these

25. Domain of definition of the function

f x x x x ( )= log ( ), is − + − 3 4 2 10 3 (a) (1, 2) (b) (–1, 0) ∪ (1, 2) (c) (1, 2) ∪ (2, ∞) (d) (–1, 0) ∪ (1, 2) ∪ (2, ∞)

26. The set of all points, where the function

f x x x ( ) | | = + 1 is differentiable, is (a) (– ∞, ∞) (b) [0, ∞) (c) (–∞, 0) ∪ (0, ∞) (d) (0, ∞)

27. A particle is projected vertically upward and reaches at a height of h after time t seconds. It further takes t′ seconds to reach the ground. Let the greatest height attained be H, then

(a) A.M. of t and t′ = 1 g

(b) G.M. of t and t′ = 2h_g (c) A.M. of t and t′= 2h

g

(d) None of these

28. Tangent is drawn to ellipse x₂₇2 + y2=1 at (3 3cos , sin )q q (where q ∈ (0, p/2)).

Then, the value of q such that the sum of intercepts on axes made by this tangent is minimum, is (a) p₃ (b) p₆ (c) p₈ (d) p₄

29. The complex numbers z = x + iy which satisfy

the equation z_z−₊5i_i =

5 1, lie on (a) x-axis

(b) straight line y = 5

(c) a circle passing through the origin (d) None of these

30. If x₁, x₂, x₃ and x₄ are the roots of the equation

x4_–x3_{sin 2b + x}2_{cos 2b –x cos b –sin b = 0, then}

tan–1_x

1 + tan–1 x2 + tan–1 x3 + tan–1 x4 is equal to

(a) b (b) p b₂ − (c) p – b (d) – b SolutionS 1. (c) : Let S r r xr r = − + − = ∞

∑

( 2 ) 1 1 3 \ = + + + + ∞ − ⋅ = + + + ∞ − = + S x x x x S x x x x S x 3 5 9 15 3 5 9 1 3 2 2 3 2 3 ... ( ) ... ( ) to to ++ + + ∞ = + + + + ∞ 4 6 3 2 1 2 3 2 3 2 x x x x x ... ( ... ) to to = + − = − + − = + − − − 3 2 1 3 1 2 1 3 1 4 1 2 2 2 2 2 x x x x x x x x ( ) ( ) ( ) ( ) ( ) \ = + − − S x x x 3 1 4 1 2 3 ( ) ( )

2. (d) : The tangent to the parabola x2_{= y – 6 at}

(1, 7) is y = 2x + 5

which is also a tangent to the given circle.

i.e., x2_{+(2x + 5)}2_{+ 16x + 12 (2x + 5) + c = 0}

⇒ 5x2_{+ 60x + 85 + c = 0 must have equal roots.}

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14 MatheMatics tODaY | March ’15 \ a a+ = −60 ⇒ a= − 5 6 \ x = –6 and y = 2x + 5 = –7 3. (b) : log | | | | | | 3 2 ₁ 2 2 z z z − + +    _ < ⇒ − + + < | | | | | | ( ) z z z 2 2 1 2 3 ⇒ |z|2_{– |z| + 1 < 3(2 + |z|)} ⇒ |z|2_{–4|z| – 5 < 0} ⇒ (|z| + 1) (|z| –5) < 0 ⇒ –1 < |z| < 5 ⇒ |z| < 5 as |z| > 0 \ Locus of z is |z| < 5.

4. (a) : Given, lim ( ) ( )

t x t f x x f t t x → − − = 2 2 1 ⇒ x2_{f ′(x) –2xf (x) + 1 = 0} ⇒ x f x′ − xf x + 1 = x x 2 2 22 4 0 ( ) ( ) ( ) ⇒    = − d dx f x x x ( ) 2 14

On integrating both sides, we get

f x cx x ( ) = 2+ 1 3 Also ( )f 1 1 c 2 3 = ⇒ = Hence ( )f x x x =2 + 3 1 3 2

5. (b) : Q Given limit is in 1∞_form

\  −    →     lim tan x a x a x a 2 2 p = → = → − −      −    ex a e x a x a ax x a x a

lim 2 1 tan ₂ limcot 1 2 p p   = → −    − _ _ e x a a x a a lim 1 2 2 2 cosec p p [L-Hospital’s Rule] = → =    ex a e x a lim sin2 _/ 2 2 2 p p p

6. (b) : I. P(All three balls are white) = ₁₃5 3 3 C C = × × = × × × × = 5 10 2 13 5 4 3 13 12 11 5 143 ! ! ! !

II. P(All the three balls are red) = ₁₃8 3 3 C C = × × = × × × × = 8 10 5 13 8 7 6 13 12 11 28 143 ! ! ! !

III. P(One ball is red and two balls are white)

= × = _×× _× × = 8 1 5 2 13 3 8 10 13 12 11 3 2 40 143 C C C

7. (a) : Since, x ∈ (–1, 1) ⇒ tan−1 ∈ −_ , _ 4 4 x p p ⇒ 2 − ∈ −_ _ 2 2 1 tan x p p, Given that, f x x x x x ( ) tan= tan ( ) −    = < −1 − 2 1 2 2 1 2  1 So, f x( )∈ −_ p p, _ 2 2

Hence, function is one-one onto.

8. (a) : Let the locus of centre of circle be (h, k)

touching (y –1)2_{+ x}2_{= 1 and x-axis shown as}

C (0, 1) O B x′ y′ x y | |k A ( , )h k 1

distance between O and A is always 1 + |k|,

i e. ., (h−0)2+ −(k 1)2 = +1 | |,k ⇒ h2_{+ k}2_{– 2k + 1 = 1 + k}2_{+ 2|k|} ⇒ h2_{= 2|k| + 2k} ⇒ x2_{= 2|y| + 2y} where, | | , , y =₋y y_{y y}≥_<  0 0

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16 MatheMatics tODaY | March ’15

\ x2_{= 2}_{y + 2y, y ≥ 0 and x}2_{= –2y + 2y, y < 0}

⇒ x2_{= 4y when y ≥ 0 and x}2_{= 0 when y < 0}

\ {(x, y) : x2_{= 4}_{y, y ≥ 0} ∪ {(0, y), y < 0}} 9. (a) : We have, __xz_{ ⋅}b _x_y_c = z_y a _xz x_y a _xz a x_y    = ⋅     =    ⋅_ _ a ⇒     =    − − x y z x c a a b ⇒    =   ⇒ = − − x y z x C B a b c a 1 1

Similarly, it can be proved that A = B \ A = B = C.

10. (a) : Applying R₁ → R₁ – R₃ and R₂ → R₂ – R₃ we get, 1 0 1 0 1 1 1 4 4 0 2 2 − − + =

sin q cos q sin q

⇒ sin2_{q + cos}2_{q + 1 + 4 sin 4q = 0}

⇒ sin4 = −1 2 q ⇒ q=7p p 24 11 24 ,

11. (a) : Since, roots are real, therefore D ≥ 0.

⇒ 4(a + b + c)2_{–12l (ab + bc + ca) ≥ 0}

⇒ (a + b + c)2_{≥ 3l (ab + bc +}_ca) ⇒ a2_{+ b}2_{+ c}2_{≥ (ab + bc + ca) (3l – 2)} ⇒ − ≤ + + + + 3l 2 a2 b2 c2 ab bc ca ...(i) Also, cos A b c a bc = 2+ −2 2 < 2 1 ⇒ b2_{+ c}2_–a2_{< 2bc} Similarly, c2_{+ a}2_{– b}2_{< 2ca} and a2_{+ b}2_{– c}2_{< 2ab} ⇒ a2_{+ b}2_{+ c}2_{< 2(ab + bc + ca)} ⇒ + + + + < a b c ab bc ca 2 2 2 2 ...(ii)

\ From (i) and (ii), we get 3l – 2 < 2 ⇒ l 4< 3 12. (b) : 3 3 2 3 4 4 sin _ p a− _+sin ( p a+ )    _ −  _ + _+ −   _ 2 2 5 6 6 sin p a sin ( p a)

= 3(cos4_{a + sin}4_{a) –2(cos}6_{a + sin}6_a)

= 3(1 – 2sin2_{a cos}2_{a) –2(1 – 3 sin}2_{a cos}2_a)

= 3 – 6 sin2_{a cos}2_{a –2 + 6 sin}2_{a cos}2_{a = 1} 13. (d) : p = The number of such numbers that

exceeds 20000 =5! = 120

q = The number of those that lie between 30000

and 90000 = 5! – 4! – 4! = 120 – 24 –24 = 72 \ p= = q 120 72 5 3

14. (a) : The curves y = (x – 1)2_,_{y = (x + 1)}2_and y = 1/4 are shown as 1 –1 –1/2 O y = 1/4 x 1/4 y y x= ( + 1)2 y x= ( – 1)2 P Q R

where point of intersection are (x −1) =1 4 2 _⇒ _{x 1}₌ 2 and (x+1) =1 ⇒ x = − 4 1 2 2 \ Q_1 _ R_− _ 2 1 4 1 2 1 4 , and , \ Required area =  − −   

∫

2 1 1 4 2 0 1 2 ( ) / x dx =  − −        2 1 3 1 4 3 0 1 2 (x ) _x / = − − − − −_ _      = 2 1 8 3 1 8 1 3 0 1 3 . sq . unit

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MatheMatics tODaY | March ’15 17 15. (a) : | c a− =| 4cos = ⇒ | c a− | = 4 2 2 2 8 p ⇒ (c a c a   − ⋅ −) ( )=8 ⇒ | |c 2+| |a 2 −2(a c ⋅ =) 8 ⇒ | |c 2−2| |c = −8 9 [ a = 3| | ] ⇒ | |c 1= Now, a b i j k i j × = − = − ^ ^ ^ ^ ^ 2 1 2 2 1 0 2 4 ⇒ |a b × |= 20 \ | (a b c × ×) | |= ×a b c  || | sinp 6 =    = 20 1 1 2 5 ( )

16. (d) : We have, tan−1 = −sin−1 =cos−1 2 x p x x ⇒ sec−1 1+x2 =sec−1_1_ x ⇒ 1+x2 = 1₂ ⇒ 4+ 2− =1 0 x x x \ x2= − ±1 1 4 1− ⋅ ⋅ − = − ±1 2 1 5 2 ( )  x2 0 x2 5 1 2 </ \ = − \ x = ± 5 1− 2

But, if x < 0 then L.H.S. of given equation becomes

–ve and given equation is not satisfied.

\ x = 5 1−

2

17. (d) : Put t = sin2_{z in 1st integral and t = cos}2_u

in 2nd_integral.

dt = 2 sin z cos zdz and dt = – 2 cos u sin udu

\ I =

∫

₀x2zsin cosz zdz+

∫

_px_/₂−2ucos sinu udu

=

_∫

xz z dz−

_∫

x u udu I IIsin2 /2 sin2 0 p = − +      − − +      z z z u u u x x

cos sin cos sin

/ 2 2 2 4 2 2 2 4 0 p2 = − + − +      xcos2x sin x { } 2 2 4 0 0 − − + −_ + _   xcos₂2x sin₄2x p₄ 0  =_ p₄ 18. (b) : We have, A(x) A(y) = − − −         − − −         − − (1 ) 1 ( ) 1 1 1 1 1 1 x x x y y y = + − + + − + − + +         − [( ) ( )] ( ) ( ) 1 1 1 1 xy x y xy x y x y xy = − + +     − + +     − + +               − 1 1 1 1 1 1 1 x y xy x y xy x y xy     = A(z) 19. (a) : Since, x a y b z c

+ + = 1 cuts the coordinate axis at A(a, 0, 0), B(0, b, 0), C(0, 0, c) and its distance from origin = 1 \ + + = 1 1 1 1 1 2 2 2 a b c or 1₂ 1₂ 1 1₂ a +b +c = ...(i)

Let P be the centroid of triangle

\ P x y z( , , )=_a+ +0 0, + +b , + +c_ 3 0 0 3 0 0 3 ⇒ x a y b z c= = = 3, 3, 3 ...(ii)

\ From (i) and (ii), we get 1 9 1 9 1 9 1 2 2 2 x + y + z = ⇒ + + = = 1 1 1 ₉ 2 2 2 x y z K \ K = 9 20. (d) :  t_{r + 1} = 2011_C r(–x)r = (–1)r · 2011Cr · xr \ According to problem, a_r = (–1)r_·2011_C r Also, a₇₇₇ + a₁₂₃₄ = (–1)7772011_C 777 + (–1)1234_·2011_C 1234 = – 2011_C 777 + 2011C7 77 = 0

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18 MatheMatics tODaY | March ’15 Also, a₁₁₁₁ + a₉₀₀ = (–1)1111_·2011_C 1111 + (–1)900_·2011_C 900 = – 2011_C 1111 + 2011C1111 = 0 Similarly, a₆₅₄ + a₁₃₅₇ = (–1)6542011_C 654 + (–1)13572011C1357 = 2011_C 654 – 2011C654 = 0 21. (d) : Let I = x dx_x + −

∫

/ sin / 2 2 2 1 2 p p ⇒ = + −

∫

I 2x _xxdx 1 2 2 2 2 sin / / p p ⇒ 2I =

∫

₋p_p/2_/₂sin2xdx=

∫

₀p/22sin2xdx =

∫

( cos )− / 1 2 0 2 x dx p ⇒ = −    = 2 2 2 ₀ 2 2 I x sin x p/ p _⇒ _I₌ p 4 22. (d) : (1 + p)x – py + p(1 + p) = 0 ...(i) (1 + q)x – qy + q(1 + q) = 0 ...(ii)

On solving (i) and (ii), we get

x = pq, y = (1 + p) (1 + q)

\ Coordinates of C are {pq, (1 + p)(1 + q)}. \ Equation of altitude CM passing through

C and perpendicular to AB is

x = pq ...(iii)

Q Slope of line (ii) is 1 + 

 

q q

\ Slope of altitude BN (as shown in figure) is − + q q 1 . C M B O A N x p,( 0) Line (ii) Line (i) _{H h k}_{( , )} y Equation of BN is y− = −q_q x p + + 0 1 ( ) ⇒ = − + + y q q x p (1 )( ) ...(iv)

Let orthocentre of triangle be H(h, k) which is the

point of intersection of (iii) and (iv). On solving (iii) and (iv), we get

x = pq and y= –pq ⇒ h = pq and k = –pq \ h + k = 0 \ Locus of H(h, k) is x + y = 0. 23. (d) : Since, S.D. < Range ⇒ s ≤ (b – a) ⇒ s2_{≤ (b –}_a)2 24. (a) : 2+ = 3 + e x i y iq l m ⇒ + + = + 2 cosq sinq 3 l m i x i y ⇒ + = − − + cosq sinq ( l ) m l m i x i y x i y 3 2 2

On taking modulus to both sides,

⇒ = − + − + 1 3 2 2 2 2 2 2 ( ) ( ) ( ) ( ) l m l m x y x y ⇒ l2_x2_{+ m}2_y2_{= 9 – 12lx + 4l}2_x2_{+ 4m}2_y2 ⇒ l2_x2_{+ m}2_y2_{– 4l}_{x + 3 = 0} _...(1)

This is the locus of P(x, y)

If l = 1, m = 2, then (1) becomes x2_{+ 4y}2_{– 4x + 3 = 0} ⇒ ( − ) + = ⇒( − ) + = / , x 2 4y 1 x 2 y 1 1 4 1 2 2 2 2 2 which is an ellipse. 25. (d) : Since, f x x x x ( )= log ( ) − + − 3 4 2 10 3 For domain of f(x), x3_{–x > 0 ⇒ x(x – 1)(x + 1) > 0} 0 1 + + Region is (–1, 0) ∪ (1, ∞) And 4 –x2_{≠ 0} ⇒ x ≠ ± 2 Region is (–∞, –2) ∪ (–2, 2) ∪ (2, ∞). \ Common region is (–1, 0) ∪ (1, 2) ∪ (2, ∞)

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MatheMatics tODaY | March ’15 19 26. (b) : Given, f x x_x x x x x x x ( ) | | , , = + = + ≥ − <       1 1 0 1 0 ⇒ ′ = + ≥ − <       f x x x x x ( ) ( ) , ( ) , 1 1 0 1 1 0 2 2 \ RHD at x x x = ⇒ + = → 0 1 1 1 0 2 lim ( ) and LHD at x = 0 ⇒ − = → lim ( ) x 0 _x 2 1 1 1

Hence, f(x) is differentiable for all x.

27. (b) :

Since total time = t + t′= 2u g H h t t \ t t+ ′ = u g 2 ⇒ u g t t= ( + ′) 2 ...(i)  H u= _g2 = _g

{ }

g t t+ ′ 2 = g t t+ ′2 2 1 2 2 8 ( ) ₍ ₎ \ t t+ ′ = H g 2 2 Also, h ut= −1gt = g t t t gt gtt+ ′ − = ′ 2 2 1 2 2 2 ( ) 2 _[_From_(i)] \ tt′ = h ⇒ ′ = g tt h g 2 2

28. (b) : Given, tangent is drawn

at (3 3cos , sin ) to .

27 1 1

2 2

q q x + y =

\ Equation of tangent is xcosq ysinq .

3 3 + 1 =1

Thus, sum of intercepts

= +



 =

3 3 1

cosq sinq f( ) (say)q

⇒ f ( )′ = sin −cos sin cos q q q q q 3 3 3 3 2 2 For maxima/minima f ′(q) = 0 ⇒ sin3 = _{3 2}1_/ cos3 3 q q ⇒ tanq= 1 , . ., q p= 3 i e 6 and atq p= , ′′( )> , 6 f 0 0

\ Hence, tangent is minimum at q p= 6.

29. (a) : Given, z_z−₊5i_i =

5 1

⇒ |z – 5i| = |z + 5i|

(if |z –z₁| = |z –z₂|, then it is a perpendicular bisector of z₁ and z₂) y x x y (0, 5) (0, –5) O

\ Perpendicular bisector of (0, 5) and (0, –5) is

x-axis.

30. (b) : We have, Sx₁ = sin 2b, Sx₁x₂ = cos 2b, Sx₁x₂x₃ = cos b and x₁x₂x₃x₄ = –sin b \ tan–1_x

1 + tan–1 x2 + tan–1 x3 + tan–1 x4

= − − +     − tan 1 1 1 2 3 1 2 1 2 3 4 1 S S S x x x x x x x x x x = − − −     −

tan sin cos

cos sin 1 2 1 2 b b b b = − −       = − −

tan ( sin ) cos

sin ( sin ) tan (cot )

1 2 1 1 2 1 b b b b b =  _ − _    = − − tan 1 tan 2 2 p b p b nn

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20 MatheMatics tODaY | MARCH ’15 1. If f x( )=_xx ,x ,

−1 ≠1 then ( fofo of x_{19 times}... )( ) is  equal to (a) _{x −1}x _(b) __{x −}x₁_ 19 (c) 19 1 x x − (d) x

2. If a, b, c are in H.P., b, c, d are in G.P. and c, d, e are in A.P., then ab

a b 2 2 2 ( − ) is equal to (a) b (b) a (c) e (d) d

3. The imaginary part of

(z – 1)(cosa – isina) + (z – 1)–1_{(cosa + isina)}

is zero, if

(a) |z – 1| = 2 (b) arg (z – 1) = 2a (c) arg (z – 1) = a (d) |z| = 1

4. The values of 'a' for which

(a2_{– 1)x}2_{+ 2(a – 1)x + 2 is positive for any x, are}

(a) a > 1 (b) a ≤ 1

(c) a > –3 (d) a < 1

5. _{If cosq =}−3₅ _{and p < q <}3

2π, then the value of

cosecθ θ θ θ + − cot sec _tanis (a) 1/6 (b) 1/7 (c) 1/5 (d) 1/2 6. Solution of (2x + 1)(x – 3) (x + 7) < 0 is (a) (– ∞, –7) ∪ _−1 _ 2,3 (b) (– ∞, –7) ∪ _1₂,3_ (c) (– ∞, 7) ∪ −_ 1₂, 3_ (d) (– ∞, – 7) ∪ (3, ∞)

7. How many numbers lying between 100 and 1000 can be formed with the digits 0, 1, 2, 3, 4, 5, if the repetition of the digits is not allowed?

(a) 90 (b) 50 (c) 40 (d) 100

8. Assuming that straight line work as the plane mirror for a point, find the image of the point (1, 2) in the line x – 3y + 4 = 0. (a) (–2, –1) (b) (–1, –2) (c) 6 5 7 5 ,   _ (d) −₅6, −₅7 9. The equation x a y a 2 2 14− +9− =1 represents a/an (a) ellipse if a > 9 (b) hyperbola if 9 < a < 14 (c) hyperbola if a > 14 (d) ellipse if 9 < a < 14

10. lim sin sin

x x x x → − 0 3 2 _{2 is equal to} (a) –1 (b) 1 (c) 2 (d) –2

11. For any vector a, prove that |a i × | |2+ ×a j | |2+ ×a k |2 is equal to

(a) 2| |a (b) | |a2 2 (c) 3| |a (d) 42 | |a 2

12. If the scalar product of the vector i j k  + + with the unit vector in the direction of the resultant of the vectors 2 4 5i+ −j kand λi+ +2 3j k is unity, then l equals

(a) 5 (b) 2 (c) 1 (d) –1

Exam on 14th_{to 29}th_May

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22 MatheMatics tODaY | MARCH ’15

13. The first 12 letters of English alphabet are written in a row at random. The probability that there are exactly four letters in between A and B is (a) ₆₆5 _(b) 1 22 (c) 7 66 (d) 1 11

14. The probability of the birth dates of all 6 persons to fall in only two different months is (a) 341 126 (b) 341 125 (c) 341 124 (d) 541 126

15. If x, y, z are positive real numbers and x + y + z = 1, then prove that the minimum value of 4 9 16

x y z+ + is

(a) 80 (b) 81 (c) 85 (d) 82

16. In DABC, if 8R2_{= a}2_{+ b}2_{+ c}2_{, then the triangle}

is

(a) isosceles (b) right angled

(c) equilateral (d) scalene

17. The number of solutions of the equation 1

2 0

2

+sin sinx x= in [–p, p] is

(a) zero (b) one (c) two (d) three

18. [ ]x dx2 0 2

∫

is equal to (a) 2− 2 (b) 2+ 2 (c) 2 1− (d) − −2 3 5+ 19. Let f : (2, 3) → (0, 1) be defined by f(x) = x – [x]. Then, f–1_{(x) equals to} (a) x – 2 (b) x + 1 (c) x – 1 (d) x + 2

20. Which of the following is/are true? (i) The principal value of cos–1₃

2 is π 6. (ii) The principal value of cosec–1_{(2) is} π

4. (iii) The principal value of tan–1₍_{− 3 is}₎ −π

3 . (a) (i), (ii) (b) (ii), (iii)

(c) (i), (iii) (d) (i), (ii), (iii)

21. If A is a square matrix such that A2_{= A, then}

(I + A)3_{– 7A is equal to} (a) A (b) I – A (c) I (d) 3A 22. The determinant x x x sin cos sin cos θ θ θ θ − − 1 1 is (a) independent of q only

(b) independent of x only (c) independent of both q and x (d) none of the above

23. If f x x x x x x ( ) ( ), , , = − ≤ 0 + >     λ 2 2 4 1 0 if

if then which one

of the following is correct ?

(a) f(x) is continuous at x = 0 for any value of l (b) f(x) is discontinuous at x = 0 for any value of l (c) f(x) is discontinuous at x = 1 for any value of l (d) None of the above

24. The point on the curve y = (x – 2)2_{at which the}

tangent is parallel to the chord joining the points (2, 0) and (4, 4), is

(a) (1, 3) (b) (3, 1) (c) (1, 0) (d) (0, 1)

25. Solution of the differential equation

ye dx xe y dy y x y ₌₍ xy ₊ 2_{) (} _≠_{0 is}₎ (a) e y C x y _{= +} _{(b) e} _{x C} x y _{= +} (c) ex_{+ y + C} _{(d) e}y_{= x + C}

26. Find the equation of the plane, which is perpendicular to the plane 5x + 3y + 6z + 8 = 0 and which contains the line of intersection of the planes

x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0.

(a) 51x – 15y + 50z + 173 = 0 (b) 51x + 15y – 50z + 173 = 0 (c) 15x + 15y + 50z + 173 = 0 (d) none of these

27. How many times must a man toss a fair coin, so that the probability of having atleast one head is more than 80% ?

(a) Atleast 3 (b) Atleast 5

(c) Atmost 3 (d) Atmost 5

28. Maximize Z = 10x₁ + 25x₂, subject to 0 ≤ x₁ ≤ 3, 0 ≤ x₂ ≤3, x₁ + x₂ ≤ 5.

(a) 80 at (3, 2) (b) 75 at (0, 3) (c) 30 at (3, 0) (d) 95 at (2, 3)

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29. The value of lim cos( )

( ) x ax bx c x → − + + − α _α 1 2 2 , where

a and b are the roots of the equation ax2_{+ bx + c = 0,}

is

(a) (α β−₂ )2 (b) (a – b)2

(c) α2(a b₂− )2 (d) a2(α β₂− )2

30. The mean of five numbers is 0 and their variance is 2. If three of these numbers are –1, 1 and 2, then the other two numbers are

(a) – 5 and 3 (b) – 4 and 2

(c) – 3 and 1 (d) – 2 and 0

31. If the third term in the expansion of

1 ₁₀ 5 x+x x    log is 1000, then x = (a) 102 _{(b) 10}3 _{(c) 10} _{(d) 10}4

32. If the latus rectum of an ellipse with axis along

x-axis and centre at origin is 10, distance between

foci = length of minor axis, then the equation of the ellipse is

(a) x_{25 10}2+y2 = 1 (b) x_{50 100}2 + y2 =1 (c) x₅2+y₅2 = 1 (d) _{100 50}x2 +y2 =1

33. For all n ∈ N, 41n_{– 14}n_{is a multiple of}

(a) 26 (b) 27

(c) 25 (d) none of these

34. Number of ways in which 3 boys and 3 girls (all are of different heights) can be arranged in a line so that boys as well as girls among themselves are in decreasing order of their heights (from left to right), is

(a) 6! (b) 3! × 3! × 2!

(c) 10 (d) 20

35. _{If cos}−1 +cos−1 = 2

x y π_{and tan}–1_{x – tan}–1_{y = 0,}

then x2_{+ xy + y}2_{is equal to} (a) 0 _(b) 1 2 (c) 3 2 (d) 1 8

36. Using cofactors of elements of third row, evaluate ∆ = + + + 1 1 1 x y z y z x z x y. (a) 1 (b) –1 (c) 0 (d) 2

37. Let f(x) = sinx, g(x) = x2_{and h(x) = log}

ex.

If F(x) = (hogof)(x), then F″(x) is equal to

(a) a cosec3_x _{(b) 2cotx}2_{– 4x}2_cosec2_x2

(c) 2x cot x2 _{(d) –2 cosec}2_x 38. Evaluate tan sin cos x x⋅ xdx

∫

(a) 3 tan x c+ (b) tan x c+

(c) 2 tan x c₊ (d) tan x c

2 +

39. Find the order and degree of the differential equation, if defined y x dy dx dy dx = + +     1 2. (a) 1, 2 (b) 2, 2 (c) 1, 3 (d) 2, 3

40. Two integers are selected at random from integers 1 to 11. If the sum is even, then find the probability that both the numbers selected are odd. (a) 1₅ (b) 4₅ (c) 2₅ (d) 3₅

41. The relation S defined on the set N × N by (a, b)S(c, d) ⇒ a + d = b + c is an

(a) equivalence relation (b) reflexive but not symmetric

(c) only transitive (d) only symmetric

42. If a i j k b= + +^ ^ ^,=4^i−2^j+3 andk^

c i j k= −^ 2^ ^+ ,then find a vector of magnitude 6 units which is parallel to the vector 2a b − + .3c

(a) ± − +(2 4 4i j k ) (b) ± − +(i 2 2j k) (c) ± − +(i j k   ) (d) ± + +(i j k  )

43. If the trace of the matrix

A x x x x = − − − − − −             1 0 2 5 3 2 4 1 1 2 3 1 2 0 4 6 2 2 is 0, then x is equal to (a) {2, 3} (b) {–2, –3} (c) {–3, 2} (d) {1, 2}

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44. If f(x) defined by the following is continuous at

x = 0, then the values of and c are f x a x x x x c x x bx x b x x ( ) sin( ) _{sin ,} , , = + + _< = + − _>         1 ₀ 0 0 2 3 2 if if if  (a) − 3 2 1 2 , (b) 1₂, (c) 1₂ 3₂_{, (d) −}1 1₂,₂1

45. An open box with a square base is to be made out of a given quantity of metal sheet of area c2_{, then}

the maximum volume of the box is

(a) _{6 3}c3 (b) c2₃ (c) c₆3 (d) 3 3c3 sOlutiOns 1. (a) : (fof x f) x x x x x x x = −   = − −   − = 1 1 1 1 ⇒ (fofof)x = f(fof)x = f(x) =_{x 1}x₋ \ (fofof...19 times)(x) =_{x 1}x₋ 2. (c) : Q a, b, c are in H.P. \ b ac a c = + 2 ...(i) b, c, d are in G.P. \ c2_{= bd} _...(ii)

and c, d, e are in A.P. \

d c e= +2 ...(iii) From (i), ab + bc = 2ac

⇒ c ab a b = − 2 ...(iv) Also, d ab a b e = − +    1

2 2 (Using (iii) and (iv)) ...(v) On putting the values of c and d from (iv) and (v) in (ii), we get a b a b b ab a b e 2 2 2 2 2 2 ( − ) = − +     ⇒ = − e ab a b 2 2 2 ( )

3. (c) : Let z – 1 = r(cosq + isinq) = reiq

\ Given expression = re e re e i i i i θ α θ α ⋅ − + 1 ⋅ =rei(θ α− )+re− −i(θ α) 1

Since, imaginary part of given expression is zero, we have r r sin(θ α− −) 1sin(θ α− )=0 ⇒ r2_{– 1 = 0 ⇒ r = 1 ⇒ |z – 1| = 1} or, sin(q – a) = 0 ⇒ q – a = 0 ⇒ q = a ⇒ arg(z – 1) = a

4. (a) : (a2_{– 1)x}2_{+ 2(a – 1)x + 2 is positive for}

all x, if

a2_{– 1 > 0 and 4(a – 1)}2_{– 8(a}2_{– 1) < 0}

⇒ a2_{– 1 > 0 and (a – 1)(a + 1) > 0} ⇒ a < –1 or a >1 5. (a) 6. (a) : (2x + 1)(x – 3)(x + 7) = 0 ⇒ x = – 1 2, 3, –7 Clearly, (2x + 1) (x – 3) (x + 7) < 0 when x < – 7 or − 1 2< x < 3 \ x ∈ (– ∞, – 7) ∪ −_ 1 _ 2, .3

7. (d) : Every number between 100 and 1000 is a 3-digit number. We first have to count the permutations of 6 digits taken 3 at a time. This number would be 6_P

3. But, these permutations will

include those also where 0 is at the 100th_place.

To get the number of such numbers, we fix 0 at the 100th_{place and rearrange the remaining 5 digits}

taking 2 at a time. This number is 5_P 2.

So, the required number = 6P₃− 5P₂=₃6!_!−5₃!_! = (4 × 5 × 6) – (4 × 5) = 100

8. (c) : Let Q(h, k) be the image of the point

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MatheMatics tODaY | MARCH ’15 25 Y Q h k( , ) X Y X P(1, 2) x– 3y+ 4 = 0 O Hence, slope of line PQ

= − − + = 1 3 4 0 Slope of line x y so that k h − − = − 2 1 1 1 3 or 3h + k = 5 ...(ii) Also, h₂+ −1 3 _k+2_{ + =} 2 4 0 or h – 3k = –3 ...(iii) Solving (ii) and (iii), we get h = 6

5 and k = 75. 9. (b) : Given, x a y a 2 2 14− +9− =1

The equation will represent an ellipse if 14 – a > 0 and 9 – a > 0

⇒ a < 14 and a < 9 ⇒ a < 9

a hyperbola if 14 – a > 0 and 9 – a < 0 ⇒ – a > –14 and a > 9 ⇒ 9 < a < 14

10. (b) : lim sin sin

x x x x → − 0 3 2 2 = − + −       − − + −    → lim ! ! .... ( ) ! ( ) ! .... x x x x x x x 0 3 5 3 5 2 3 5 2 2 3 2 5  x3 = − ₊ ₋  ₋ ₊      → lim ! ! .... ! ! .... x x x x x x 0 3 5 3 5 3 2 3 2 5 8 3 32 5 = − + −2 _ − + _= − + = = 3 0 8 3 0 2 6 8 6 6 6 1 ! .... ! ....

11. (a) : _Leta a i a j a k= ₁+ ₂+ ₃. Then, we have

       a i i j k a a a a j a k × = ₁ ₂ ₃ = ₃ − ₂ 1 0 0 Therefore, |a i × =|2 a₂2+a₃2 Similarly, |a j × |2=a₁2+a₃2 and |a k × |2=a₁2+a₂2 Hence, | | | | | | ( ) ( ) (       a i a j a k a a a a a × + × + × = + + + + + 2 2 2 22 32 12 32 12 aa22) =2(a12+a22+a32)=2| |a2 12. (c)

13. (c) : A and B can be arranged in 12_P 2

= 11 × 12 ways. Since we want 4 letters in between

A and B.

\ A and B can take the following places.

Place for A Place for B

1 2 3 4 5 6 7 6 7 8 9 10 11 12

A and B can be interchanged. Therefore required

probability is 14_{11 12}_× =₆₆7

14. (b) : Since the birth date of any person can fall in anyone of the 12 months, the number of total outcomes is 126_.

Let E : Event that the birth dates of all 6 fall in two different months. P E( )=12C2(2₆6−2)= ×₆ = ×₅ = ₅ 12 66 62 12 11 31 12 341 12 15. (b) : Since x + y + z = 1, we have 4 9 16 4 9 16 x y z+ + = + +x y z x y z+ +     ( ) = + + + +  + +    + +     (4 9 16) 4z 16 9 16 4 9 x x z z y y z y x x y ≥29 2 64 2 144 2 36 81+ + + =

Equality holds if and only if x=2 y= z= 9 3 9 4 9 , ,

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16. (b) : 8R2_{= 4R}2_(sin2_{A + sin}2_{B + sin}2_C)

⇒ 2 1= − 2 + − + − 2 1 2 2 1 2 2

cos A cos B cos C

⇒ –1 = cos2A + cos2B + cos2C

= –1 – 4cosAcosBcosC

Hence, cosA cosB cosC = 0

So one of the angles of the triangle is a right angle.

17. (a) : Since, 1 2 0 2 +sin sinx x = ∴ 1+ _1− _{ =} 2 0 sinx cosx

⇒ 2 + sinx – sinx cosx = 0 ⇒ sin 2x – 2 sinx = 4

which is not possible for any x in [–p, p]

18. (d) : [ ]x dx2 [ ]x dx [ ]x dx 0 2 2 0 1 2 0 2

∫

=

∫

+

∫

+

∫

[ ]x dx2 +

∫

[ ]x dx 2 3 2 3 2 =

_∫

0 +

_∫

1 +

_∫

+

_∫

3 0 1 1 2 2 3 3 2 dx dx dx dx = +0 1 2 1 2 3( − +) ( − 2 3 2) (+ − 3) = −5 3− 2

19. (d) : The given function is f : (2, 3) → (0, 1) defined by f(x) = x –[x]

Let y ∈ (0, 1) be such that y = x – 2

{Q 2 < x < 3 ⇒ [x] = 2} ⇒ x = y + 2 ⇒ f–1_{(x) = x + 2} 20. (c) 21. (c) : Here, A2_{= A} \ (I + A)3_{– 7A = I}3_{+ A}3_{+ 3I}2_{A + 3A}2_{I – 7A} = I + A3_{+ 3A + 3A}2_{– 7A} = I + A·A – A = I + A – A = I (Q A2_{= A)} 22. (a) : Let ∆ = − − x x x sin cos sin cos θ θ θ θ 1 1

⇒ ∆ x= −₁x _x1−sinθ−_cossin_θθ _x1+cosθ−_cossin_θθ −₁x

= x(–x2_{– 1) – sinq (–x sinq – cosq) + cosq}

(–sinq + x cosq) = –x3_{– x + x sin}2_{q + sinq cosq – sinq cosq}

+ x cos2_q

= –x3

23. (b)

24. (b) : Slope of chord joining (2, 0) and (4, 4) is

y y x2₂ x1₁ 4 0 4 2 4 2 2 − − = −− = = ...(i)

Equation of given curve is y = (x – 2)2

⇒ dy= −

dx 2(x 2)

Now, from (i), we get

2(x – 2) = 2 ⇒ x = 3 ⇒ y = (3 – 2)2_{= 1}

Hence, the required point is (3, 1).

25. (a)

26. (b) : Since (x + 2y + 3z – 4) + l(2x + y – z + 5) = 0 ⇒ x(1 + 2l) + y(2 + l) + z(3 – l) – 4 + 5l = 0 ...(i) This is perpendicular to the plane

5x + 3y + 6z + 8 = 0

\ 5(1 + 2l) + 3(2 + l) + 6(3 – l) = 0 ⇒ 7l + 29 = 0 ⇒ λ = − 29

7

On putting l = –29/7 in (i), we have the equation of the required plane as

x 1 58 y z 7 2 297 3 297 4 1457 0 −   +  − +  + − − = ⇒ 51x + 15y – 50z + 173 = 0

27. (a) : Let man tosses the coin n times.

P(X ≥ 1) > 80₁₀₀, where, X is the number of heads.

\ 1 0 80 100 −P X( = >) ⇒ 1 1 2 1 2 8 10 0 0 −nC _ _ _ _n> ⇒ 1−₂1_n >4₅_⇒ 1 2 1 5 n < ⇒ 2n > 5 ... (i)

Inequality (i) is satisfied for n ≥ 3.

Hence, coin must be tossed 3 or more times.

28. (d) 29. (d)

30. (d) : Let the other two numbers be x and y. According to question,

− + + + + = ⇒ + = −1 1 2

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MatheMatics tODaY | MARCH ’15 27 Also, s2_{= 2} ⇒ (− −1 0) (2+ −1 0) (2+ −2 0₅ ) (2+ −x 0) (2+ −y 0)2=2 ⇒ 1 + 1 + 4 + x2_{+ y}2_{= 10 ⇒ x}2_{+ y}2_{= 4 ...(ii)} ⇒ (x + y)2_{– 2xy = 4 ⇒ xy = 0} _...(iii) Now, (x – y)2_{= x}2_{+ y}2_{– 2xy = 4 – 0 = 4} ⇒ x – y = ±2 ...(iv) ⇒ x = 0, y = –2 or x = –2, y = 0

(from (i) and (iv)

31. (a) : Given, T₃ = 1000 \ 5 2 3 ₂ 1 ₁₀ C x x x  

(

log

)

= 1000 ⇒ 10 x– 3_{× x}2log10x_{= 1000} ⇒ 2_log₁₀_x 3 _log 102 x −

(

)

= ⇒ (2log₁₀_{x – 3) =} 2 10 log x ⇒ 2t – 3 =2_t , where t = log10x ⇒ (2t + 1) (t – 2) = 0 ⇒ t = 2 (t ≠ – 1/2, neglected) \ log₁₀x = 2 ⇒ x = 102_{= 100} 32. (d)

33. (b) : Let P(n) be the statement given by

P(n) : 41n_{– 14}n_{is a multiple of 27.}

For n = 1, i.e., P(1) = 411_{– 14}1_{= 27 = 1 × 27}

which is a multiple of 27 \ P(1) is true.

Let P(k) be true i.e., 41k_{– 14}k_{= 27l ... (i)}

For n = k + 1, 41k+1_{– 14}k+1

= 41k_{41 – 14}k_{14 = (27l + 14}k_{) 41 – 14}k₁₄

[using (i) ] = (27l × 41) + (14k_{× 27) = 27(41l + 14}k₎

which is a multiple of 27.

Therefore, P(k+1) is true when P(k) is true. Hence, from the principle of mathematical induction, the statement is true for all natural numbers n.

34. (d) : Since order of boys and girls are to be maintained in any of the different arrangements. \ Required number = 6_{3 3}_{! !}! = 20. 35. (c) 36. (c) 37. (d) 38. (c) 39. (a) : y x dy dx dy dx = + +     1 2 ⇒ y x− dy_dx= + dy_dx    1 2 ⇒ (x ) dy dx xy dydx y 2_{− }₁ 2 ₂ 2 _{1 0}   − + − =

(squaring on both sides) which represents a quadratic polynomial in dy

dx.

\ Order of the differential equation = 1 and degree of the differential equation = 2

40 (d) : Let E = Event of selecting both odd numbers.

F = Event that the sum of chosen number is even in

integers from 1 to 11, here 5 even and 6 odd integers.

∴ = = × ×    × ×    = × × = P E C C ( ) ₁₁6 2 2 6 5 2 1 11 10 2 1 6 5 11 10 3 11 and P(F) = P(both numbers are even)

+ P(both numbers are odd)

= + = × ×    × ×    + × ×    6 2 11 2 5 2 11 2 6 5 2 1 11 10 2 1 5 4 2 1 11 C C C C ×× ×    = 10 2 1 5 11 and P E F C C ( ∩ =) = × ×    × ×    = 6 2 11 2 6 5 2 1 11 10 2 1 3 11 \ Required probability =P E_ _= ∩ = F P E F P F ( ) ( ) 3 5 41. (a)

42. (a) : Let the vector be r=λ(2a b − +3c) ⇒ r=λ(2^i+2j^+2k^−4i^+2^j−3k^+3^i−6^j+3k^) ⇒ r=λ(i^−2j^+2k^)

∴ | |r = ±λ|i^−2j^+2k^|

⇒ ±λ 1 4 4 6+ + = ⇒ ± ⋅ = ⇒ = ±2λ 3 6 λ Therefore, r= ±2(^i−2^j +2k^)= ±(2 4i^− ^j+4k^)

43. (c) : Trace of matrix is defined as

a_ii x x x i n =

∑

= + − = ⇒ = − 1 2 2 2 12 0 3 2, 44. (a) 45. (a) _nn

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28 MatheMatics tODaY | MARCH ’15 1. Given that iz z z z z 2 2 3 4 1 2 3 4 5 = + + + + + ...

and z n= ± − . Find the value of n.i

2. Find the value of x by evaluating the given series 1 1 5 1 3 5 10 1 3 5 5 10 15 + + × × + × ×× × +...∞ = x x Q, ∈

3. If r₁, r₂, r₃, r₄, r₅ are the complex roots of the equation x5_{– 3x}4_{– 1 = 0. Find the value of}

1 1 1 1 1

19 29 39 49 59

r +r +r +r +r .

4. Find the sum of all (distinct) complex values of

c for which the polynomial

f_c(x) = x4_{– (c}2_{– 7c + 11)x}2_{+ (18 – 21c + 8c}2_{– c}3₎

has strictly less than four distinct complex zeroes.

5. The minimum possible perimeter of a triangle with one vertex at (3, 9), one anywhere on the y-axis and one anywhere on the line y = x is

y x= x y O 6. lim n r n _r r →∞ = − + ∑ 3₃ = 3 8 8

7. Inside a square ABCD points P and Q are positioned so that DP || QB and DP = PQ = QB. Of all configurations that satisfy these requirements, what is the minimum possible value of ∠ADP, (in degrees)?

8. Find the value of

max{sin ,sin (sin )} .x x dx

n −

∫₀2 π 1 Draw the graph of

the same.

9. Let P be a point (other than the origin) lying on the parabola y = x2_{. The normal line to the parabola}

at P will intersect the parabola at another point Q. The minimum possible value for the area bounded by the line PQ and the parabola is

10. Find the value of lim ( ( ))

/ x x x by a y dy → ∫ + −     0 0 1 1 1 where b > a. sOlutiOns

1. Multiplying both sides of the equation by z, we get iz z z z 3 2 2 3 4 = + + + + ...

and subtracting the original equation from this one, we get iz z z z z z 2 2 3 1 1 1 1 1 ( − = + + +) + +...

Using the formula for an infinite geometric series, we find iz z z z z z 2 ₁ 2 1 1 1 ( − =) − = − Rearranging, we get iz z z z i z i 2₍ ₋₁₎2₌ 2_{⇒ −}₍ ₁₎2_{= ⇒ = ± −}1 ₁ _. Thus n = 1

2. Notice that each term is of the following form:

( ) ( ) 2 1 5 1 1 k k k n k n − ∏ ∏ = = = − ∏ ⋅ ∏ ∏ = = = = = ( ) ( ) ( !) ( ) ( )! ( !) ( !) 2 1 2 5 2 2 5 2 1 1 1 1 1 k k n k n n n k n k n n k n n n ₀₀ 2 n n n     Heren_r nC_r   =

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Hence we need to find s _n _nn

n =     ∑ = ∞ ₁ 10 2 0

Now since the result is the square root of a rational number, let’s find s2_{. Using the Cauchy Product}

(with 1/10 as the independent variable), we get the following formula. s _kk _{n k}n k n k n n 2 0 0 1 10 2 2 =     − −     ∑ ∑ = = ∞ ₍ ₎

Now it can be shown that for all whole numbers

n we have 2 2 4 0 k k n k n k n k n _    − −    = ∑ = ( ) Hence, we have s _n n n n n 2 0 0 1 10 4 2 5 1 1 2 5 5 3 = ∑ = = _ _ = − = ∑ = ∞ = ∞ Thus, x = 5/3

3. From the given polynomial, we have

r_i =

∑ _{3 and r}∑ _i−1=0

The sum of the reciprocals of the roots come from the fact that the polynomial with reciprocal roots has its coefficients reversed.

x5_{– 3x}4_{– 1 = 0 ⇒ x}4_{(x – 3) = 1} x4_{= (x – 3)}–1_{⇒ x} x x −9_{= −}( 3) 2 r r r r r r i i i i i i − ∑ =∑ − =  − +      ∑ 9 ( 3)2 2 6 9 =∑(ri − +6 9ri−1)=∑ri− ∑ + ∑6 9 ri−1 = 3 –5(6) + 0 = –27.

4. The polynomial f_c(x) will fail to have four distinct complex zeroes when the quadratic polynomial

g_c(x) = x2_{– (c}2_{– 7c + 11)x + (18 – 21c + 8c}2_{– c}3₎

either has repeated roots or has 0 as a root.

Case1: One of the roots of g_c(x) is zero precisely when

c3_{– 8c}2_{+ 21c – 18 = (c – 2)(c – 3)}2_{= 0}

and so precisely when c = 2, 3

Case 2 : The roots of g_c(x) are repeated when its discriminant is zero, so that

(c2_{– 7c + 11)}2_{– 4(18 – 21c + 8c}2_{– c}3_{) = 0} c4_{– 10c}3_{+ 39c}2_{– 70c + 49 = 0}

(c2_{– 5c + 7)}2_{= 0}

and this happens for two distinct values of c which sum to 5.

Thus the sum of the possible values of c is 2 + 3 + 5 = 10.

5. Let A be the vertex (3, 9), B be the vertex on the y-axis and C be the vertex on the line y = x. Also let D (–3, 9) be the reflection of A in the y-axis and

E(9, 3) be the reflection of A in the line y = x. Then AB = BD and AC = CE, and thus the perimeter of

DABC is equal to DB + BC + CE.

But the shortest distance between two points is a straight line, so

DB + BC + CE ≥ DE = ( ( )) (9− −3 2+ −3 9)2

= 180 6 5=

This minimum can be obtained by then choosing

B and C as the points of intersection of the line DE

with the y-axis and the line y = x, respectively. This gives us the points B 0 15

2 , 

  and C(5, 5). This will yield a perimeter for DABC of 6 5.

6. lim ... n n n →∞ − +       − +       − +       3 8 3 8 4 8 4 8 8 8 3 3 3 3 3 3 = − + + + + −       −₊ + + + →∞ lim . ( ) ( ) . ( ) n 3 2 3 2 3 4 2 3 3 4 2 3 4 2 4 2 4 4 2 4 4 4 2 2 2 2 ₋₋        −₊   2 4 2 2 ( ) ... . n n n n n n = − + + + + −       −₊ + + + →∞ lim . ( ) ( ) . ( ) n 3 2 3 2 3 4 2 3 3 4 2 3 4 2 4 2 4 4 2 4 4 4 2 2 2 2 ₋₋       −₊ + + + −       2 4 2 2 4 2 4 2 2 2 ( ) ... . ( ) ( ) n n n n n n = − + − + − + − +       →∞ lim . . ... n n n 3 2 3 2 4 2 4 2 5 2 5 2 2 2 3 4 2 3 3 4 2 3 4 4 2 4 4 4 2 4 4 2 2 2 2 2 2 2 + + + − + + + − + + + ( ) ( ). ( ) ( )... ( ) n n n 44 2−         ( )n = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅       ⋅ ⋅ ⋅ ⋅ →∞ lim .... .... .... n 1 2 3 4 5 6 5 6 7 8 19 28 39 52 63 7⋅⋅ ⋅ ⋅ ⋅ ⋅ ⋅       12 19 28 39 42 63.... = + + + ⋅ ⋅ ⋅ − + + ⋅    →∞ lim ( )( )( ) ( ) ( )( ) n n n n n n n n 2 ₅ 2 ₂ _{4 1 2 3 4} 1 1 2 1 7 12 == 27

7. Without loss of generality let the corners of the square be

A(0, 2), B(2, 2), C(2, 0) and D(0, 0). Now let point P have coordinates (a, b); then by symmetry the

coordinates of point Q must be (2 – a, 2 – b). Then since DP = PQ, we have that