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Vol. XXXIV
No. 3
March 2016
Corporate Ofce: Plot 99, Sector 44 Institutional Area, Gurgaon -122 003 (HR), Tel Tel : 0124-660 1200 e-mail :
[email protected] n website : www.mtg.in
41 8 29
Regd. Ofce: 406, Taj Apartment, Near Safdarjung Hospital, Ring Road, New Delhi - 110029. Managing Editor : Mahabir Singh Editor : Anil Ahlawat
CONTENTS
8
54
Maths Musing Problem Set - 159
10 Practice Paper - JEE Main
84
89
17 JEE Work Outs 21 Practice Paper - JEE Main
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MATHEMATICS TODAY | MARCH ‘16
7
17
70 32
Vol. XXXIV
No. 3
March 2016
Corporate Ofce: Plot 99, Sector 44 Institutional Area, Gurgaon -122 003 (HR), Tel Tel : 0124-660 1200 e-mail :
[email protected] n website : www.mtg.in
41 8 29
Regd. Ofce: 406, Taj Apartment, Near Safdarjung Hospital, Ring Road, New Delhi - 110029. Managing Editor : Mahabir Singh Editor : Anil Ahlawat
CONTENTS
8
54
Maths Musing Problem Set - 159
10 Practice Paper - JEE Main
84
89
17 JEE Work Outs 21 Practice Paper - JEE Main
Subscribe online at
www.mtg.in
Individual Subscription Rates
27 Math Archives 29 Olympiad Corner 32 Practice Paper - JEE Main
1 yr.
2 yrs.
3 yrs.
Mathematics Mathematic s Today
330
600
775
Chemistry Today Today
330
600
775
Physics For You
330
600
775
Biology Today
330
600 60 0
775
41 Ace Your Way CBSE XII
Combined Subscription Rates
Practice Paper
54 Mock Test Paper - WB JEE 2016
1 yr.
2 yrs.
3 yrs.
PCM
900
1500
1900
PCB
900
1500
1900
1000
1800
2300
PCMB
Paper - JEE Advanced 70 Practice Paper
82 Math Musing Solutions 84 Practice Paper - BITSAT 89 You Ask We Answer
Send D.D/M.O in favour of MTG Learning Media (P) Ltd. Payments should be made directly to : MTG Learning Media (P) Ltd, Plot 99, Sector 44 I nstitutional Area, Gurgaon - 122 003, Haryana. We have not appointed any subscription agent. Owned, Printed and Published by Mahabir Singh from 406, Taj Apartment, New Delhi - 29 and printed by Personal Graphics and Advertisers Advertisers (P) Ltd., Okhla Industrial Area, Area, Phase-II, New Delhi. Readers are advised to make appropriate thorough enquiries before acting upon any advertisements published in this magazine. Focus/Infocus features are marketing incenti ves MTG does not vouch or subscribe to the claims and representations made by advertisers. All disputes are subject to Delhi jurisdiction only. Editor : Anil Ahlawat Copyright© MTG Learning Media (P) Ltd. All rights reserved. Reproduction in any form is prohibited.
MATHEMATICS TODAY | MARCH ‘16
7
MMusing is to augment the chances of bright students seeking admission into IITs with additional study material.
aths Musing was started in January 2003 issue of Mathematics Today with the suggestion of Shri Mahabir Singh. The aim of Maths
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 benetting 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.
COMPREHENSION
JEE MAIN 1.
2.
3.
4.
5.
Let S = {2 , 2 , 2 , . . ., 2 } . Consider all possible positive differences of elements of S. If M If M is is the sum of all these differences, then the sum of the digits of M is is (a) 24 (b) 27 (c) 30 (d) 31 0
1
2
10
A fair coin is tossed 12 times. Te probability of getting atleast 8 consecutive heads is 3 3 5 5 (a) 7 (b) (c) (d) 2 28 28 27 If
7.
8.
1 1 1 1 m + + + ... + = , reduced 1⋅ 2 ⋅ 3 2 ⋅ 3 ⋅ 4 3 ⋅ 4 ⋅ 5 10 ⋅ 1 1 ⋅ 12 n
fraction, then the sum of the digits of (m ( m + n) is (a) 13 (b) 14 (c) 15 (d) 16 1 If a, b are the roots of x 2 + ax – – = 0, a ≠ 0, then 2 2a the minimum value of a4 + b4 is (a) 2 (b) 3 + 2 (c) 2 + 2 (d) 4 − 2 c c3 c 5 sin −1(c cos x )dx = If 0 < c < 1 and + + + ..., 0 a1 a2 a3 then (a (a1 + a2 + a3) is divisible by (a) 3 (b) 5 (c) 7 (d) 11 π2
Te distance of point D from the plane through the line and the point C is is (a)
3
(b)
(c)
7
(d)
Let A Let A ( (x x 1, y 1), B (x (x 2, y 2) and C (x (x 3, y 3) be the vertices of a ABC . A parallelogram AFDE AFDE is drawn with D, E and F on on the line segment BC , CA CA and AB AB respectively. Ten, maximum area of such a parallelogram is (a)
(b)
(c)
(d)
5 2 10
Te distance of the point D from the image of the point C in in the line is (a) (a)
15
(b) (b)
20
(c) (c)
29
(d) (d)
34
INTEGER MATCH 9.
Te values 15x 15x = = 8 y and and 3x 3x = = 10 y contain contain points P PQ is (8, 6), and Q respectively. If the midpoint of PQ is m then the length of PQ = reduced fraction, where n (m – 8n 8n) is MATRIX MATCH
10.
Let S = {1, 2, 3, ..., 10}
∫
JEE ADVANCED 6.
Consider the line r = − ˆ j + kˆ + λ( −2iˆ + 2 ˆ j + kˆ ), and the points C (1, (1, 2, 3) and D(2, 0, 0)
C olumn I
C olumn II
(a)
Te number of subsets {x , y , z } of S so that x , y , z are are in A.P.
(p)
15
(b)
Te number of subsets {x , y , z } of S so that no two of them are consecutive.
(q)
20
(c)
Te number of subsets {x , y } of S so that x 3 + y 3 is divisible by 3.
(r)
24
(d)
Te number of subsets {x , y } of S so that x 2 – y 2 is divisible by 3.
(s)
30
(t)
56
See Solution set of Maths Musing 158 on page no. 82 8
MATHEMATICS TODAY | MARCH ‘16
1.
2.
3.
If a, b are the roots of the equation x 2 – 2 px + 2 p4 = 0 ( p ∈ R), then maximum value of a2 + b2 = 1 1 (a) –1 (b) 1 (c) (d) − 2 2 Te plane a y +bz = 0 is rotated through an angle q about its line of intersection with the plane x = 0, then equation of the plane in new position is (a)
a y + bz ± a2 + b2 sin qx = 0
(b)
a y + bz ± a2 + b2 cos qx = 0
(c)
a y + bz ± a2 + b2 tan qx = 0
(d)
a y + bz ± a2 + b2 sec qx = 0
Te range of x so that in the expansion of (1 + x )12, the numerically greatest term has the greatest coefficient is 6 7 −7 −6 (a) , ∪ , (b) −8 , −7 ∪ 7 , 8 6 7 7 6 7 8 8 7 (c)
4.
−6 , −5 ∪ 5 , 6 (d) −9 −8 8 9 5 6 6 5 8 , 9 ∪ 9 , 8
cos(tan x ) − cos x Te value of lim = x →0 x 4 1 1 1 (a) (b) − (c) (d) 3 2 2 ln 2
5.
Te value of
∫ 1 ln
(a) ln 8 6.
−
(a) (c) 7.
(c) ln 2
(d) – ln 2
If x is such that [2 sin x ] + [cos x ] = – 3, for some x ∈ [0, 25] then the range of the function g (x ) = sin x + cos x , where [⋅] represent greatest integer function, is
10
3
− 2 , −1) −1 − 3 , −1 2
(d) 2 If f (x ) = {x } + {–x } & g (x ) = sin–1 x , then number of elements in the range of gof (x ) is (where {⋅} represents fractional part function) (a) 0 (b) 1 (c) 2 (d) 3 2,
9.
Let f (x ) = [x + 2] and g (x ) = cos x , where [⋅] denotes the greatest integer function. Ten the value of ( fog )′(p/2), is (a) 0 (b) 1 (c) –1 (d) Does not exist
10.
Number of points of non-differentiability of f (x ) = min {0, cos x , sin x } in (0, 2np) is (a) 3n – 1 (b) 3n (c) 3n + 1 (d) 3n + 2
11.
If the principle argument of a complex number np (–1 + i)50 is and the equation z 2 – az + b + 2i = 0 2 has one real root. If a = (1 + i)–n and b ∈ R, then other root of given equation is (a) 1
(b) – 1
(c) 1 + i
(d) – 1 + i
12.
If the equation f (x ) = x 3 + 3x 2 – 9x + c = 0, c ∈ R, has two equal roots and one distinct root, then the total number of values of c is (a) 0 (b) 1 (c) 2 (d) 3
13.
Let a, b, c be in G.P., a, b, c, ∈ C – {0}, i.e., set of complex numbers except zero and a ≠ b ≠ c. If a + b, b + c and c + a are in H.P., then the value of 2a + c is (a) 0 (b) 1 (c) 2 (d) 3
Alok Sir Classes, Bhilai, 9993541840 MATHEMATICS TODAY | MARCH ‘16
−1 −
(b)
Te number of real solution of (x, y ) where |y | = cos x & y = cos–1 (cos x ), –2p ≤ x ≤ 2p is/are (a) 1 (b) 2 (c) 3 (d) 4
2
(b) ln 4
2 , −1
8.
1 3
e sin x dx sin x sin x − e +e
− −
14.
15.
16.
17.
18.
19.
Let the roots of the equation f (x ) = x 5 – 10x 4 + ax 3 + bx 2 + cx – 32 = 0 are positive. If f (x ) is divided by x – 1, then the remainder is (a) –1 (b) 1 (c) 2 (d) –2 Te total number of words formed by the letters of the word 'PARABOLA', if only two A's should come together, is (a) 5040 (b) 3600 (c) 4320 (d) 2400 For all complex numbers z 1, z 2 satisfying |z 1| = 15 and |z 2 – 4 – 3i| = 5 then the minimum value of |z 1 – z 2| is (a) 0 (b) 5 (c) 10 (d) 15 n arithmetic means are inserted between two sets of numbers a, a 2 & b, b2 where a, b ∈ R. Suppose mth mean between these two sets of numbers is same, then a + b equals n +1− m n (a) (b) n n +1− m m + n −1 m − n −1 (c) (d) m m If a, b, c, are real and x 3 – 3 b2x + 2 c3 is divisible by (x – a) and ( x – b), then a + b + c is (a ≠ b) (a) –2 (b) – 1 (c) 0 (d) 1
22 2
10
C1 +
23 3
10
C2 +
211 + .... + 11 then bCa is, (a) 55 (b) 165
21.
12
10
24 4
10
C3
(d) 11
MATHEMATICS TODAY | MARCH ‘16
23.
If pair of tangents drawn from any point A(z 1) on the curve |z – (3 + 4i)| = 2 to the curve |z – (3 + 4i)| = 1,
meeting at points B(z 2) and C (z 3), then distance between circumcentre and orthocentre of D ABC is 1 1 3 (a) 0 (b) (c) (d) 2 3 4 24.
If (1 +
x + 2x 2)n =
2n
a0 + a2 + a4 ∑ ar xr , n ∈ N then
r =0
+ .... + a2n is equal to
(b) 2n (2n – 1) (d) 2n–1 (2n–1 –1)
(a) 0 (c) 2n–1 (2n – 1) 25.
sin2 x is equal to (where [⋅] x →0 x sin(sin x )
Te value of lim
represents greatest integer function) (a) 0 (b) 1 (c) 2
(d) D.N.E
26.
Te minimum value of 4x 2 – 4x |sin q| – cos2 q is equal to (x ∈ R) (a) – 2 (b) – 1 (c) –1/2 (d) 0
27.
If P n = cosn q + sinn q, then P 6 – P 4 = KP 2, where (a) K = 1 (b) K = – sin2 q ⋅ cos2 q (c) K =sin2 q (d) K = cos2 q
28.
29.
otal number of ways of selecting four letters from the word 'ELLIPSE' is (a) 18 (b) 21 (c) 42 (d) 44 Te equation of the plane containing the lines x − 4 y + 1 z − 0 and 4x – y +5z – 7 = 0 = = 1 1 1 − = 2x – 5 y – z – 3 is ax + by + cz – 2 = 0, then a + b – c is equal to (a) 0 (b) 1 (c) 2 (d) 4
a = li + j + 3k, b = 2i + j − l k and c = −2i + l j + 3k (a) 0 (b) 1 (c) 2 (d) 3
ab − 1 , C 10 = 11
(c) 330
Te number of values of l (l ∈ R) for which r = xi + y j + zk is at right angle to each of the vectors
Te value of 2 10C0 +
20.
22.
If ' a' is a root of the equation 4 x 2 +2 x – 1 = 0 and f (x ) = 4x 3 – 3x + 1, then 2[ f (a) + a] equal to (a) 0 (b) –1 (c) 1 (d) 2 If x 1 + y + y 1 + x = 0 then 1
(a)
30.
(b)
dy is dx −1
(1 + x )2 (1 + x ) 1 −1 (c) (d) 1 + x 1 + x n If Cr denotes the number of combination of n things taken r at a time, then the expression n
2
C0 +
(a)
n −1
∑ n+k C k +1 equal
k =0 2nC n–1
(b)
2nC n+1
(c)
2nC
n
(d) nCn–1
SOLUTIONS
1.
a + b = 2 p, ab (a + b)2 – 2ab
(b) : We have
a
2 +
b
2 =
= 2 p4
tan x + x tan x − x −2 = lim 4 x →0 x x 3
= 4 p2 – 4 p4 = – 4( p4 – p2)
x 3 x + − x 1 tan x 3 = − 1 = − lim + 1 2 x →0 x 3 x 3
2
= 2.
1 1 1 −4 p 4 − p2 + − = 1 − 4 p2 − ≤ 1 4 4 2
(c) : Equation of a plane in new position,
a y + bz + kx = 0, (k is any variable) DR's of normal to a y + bz + kx = 0 is k, a , b and DR's of normal to a y + bz = 0 is 0, a, b a2 + b2 ⇒ cos q = k 2 + a2 + b 2 a 2 + b2
ln 2
5.
⇒
3.
2
tan
2
q=
a2 + b2
T 7
I =
T 6
⇒
(d) :
T 8 T 7
b b ∫ f (x )dx = ∫ f (a + b − x )dx a a
>
+ e − sin x dx = ln 2 − (− ln 2) 2I = ∫ sin x − sin x e e + − ln 2 ⇒ 2I = 2 ln 2 ⇒ I = ln 2
=1+
k2
a 2 + b2
k2
⇒ k = ± a2 + b2 tan q a2 + b2
6.
e sin x
(b) : We have [2 sin x ] + [cos x ] = – 3
⇒ [2 sin x ] = – 2 and [cos x ] = – 1 ⇒ x ∈ 7p , 11p and x ∈ p , 3p 2 2 6 6 ⇒ x ∈ 7p , 3p 6 2 p Now g (x ) = sin x + cos x = 2 sin x + 4 7p 3 p Now when < x < 6
<1 ⇒
7 x 6
> 1,
6 x < 1 7
6 7 , x < 7 6
⇒ x ∈ − 7 , − 6 ∪ 6 , 7 6 7 7 6 tan x + x tan x − x −2 sin sin 2 2 lim
x
dx ∫ − sin x + e sin x e − ln 2
ln 2
= T 7
> 1,
e − sin x
ln 2
On adding,
(a) : Term with greatest coefficient is 12C x 6 6
4.
q=
sec
k 2 + a2 + b2
∫
− ln 2
a2 + b2 = k 2 + a2 + b2 ⇒
(c) : I =
e sin x dx e sin x + e − sin x
x 4
x →0
2
17 p p 7p < x + < 12 4 4
⇒ −1 ≤ sin x + p < − 1 4 2 ⇒ − 2 ≤ 2 sin x + p < −1 ⇒ g (x ) ∈[− 4
2 , −1)
tan x + x tan x − x −2 sin sin 2 2 = lim tan x + x tan x − x x →0 2
⋅
2 tan2 x − x 2 4 x 4 MATHEMATICS TODAY | MARCH ‘16
13
7.
(c) : gof (x ) = g [ f (x )] = g ({x } + {–x })
Now product of roots = b + 2i
= sin–1 ({x }+{–x }) = sin–1 (0) or sin –1 (1) = 0 or
p
So the other root is (–1 + i)
2 So, number of elements in range of gof (x ) = 2 8.
= – 2 + 2i = 2(–1 + i)
(d) :
12. (c) : f (x ) = x 3 + 3x 2 – 9x + c = 0
...(i)
⇒ f ′(x ) = 3x 2 + 6x – 9 = 0
...(ii)
D = 36 – 4 × 3 × (–9) > 0
⇒ f ′(x ) = 0 has two roots, but (i) has two equal roots, so the graph of (i) will be as shown. If k1 and k2 are roots of f ′(x ) = 0, then f (k1) ⋅ f (k2) = 0
9.
(d) : fog (x ) = f ( g (x )) = [ g (x ) + 2] = [cos x + 2]
= [cos x ] + 2 Function is not continuous at x = 10. (a)
p 2
⇒ f (–3) ⋅ f (1) = 0 ⇒ (– 27 + 27 + 27 + c) (1 + 3 – 9 + c) = 0 ⇒ (c + 27) (c – 5) = 0 ⇒ c = 5 or – 27 13. (a) : Let a, b, c are of the type a, ar, ar 2
No. of points of non-differentiability in the interval (0, 2 p) = 2
⇒
in the interval (0, 4 p) = 5 in the interval (0, 6 p) = 8 in the interval (0, 2np) = (3n – 1) 11. (d) : arg (–1 + i)50 = 50 arg (–1 + i)
p = 50(p – tan–1 (1))= 50 p − 4 p 3p 75p 76p − p = 50 × = = = 38p − 4
2
2
So the principle argument =
⇒ n =
−
–1
p
2
2
Now, a = (1 + i)–n = (1 + i) So equation becomes: z 2 – (1 + i) z + b + 2i = 0
14
Now a + b, b + c, c + a are in H.P. 2 1 1 = + ⇒ b+c a +b c + a 2 1 1 ⇒ = + ar + ar 2 a + ar ar 2 + a
⇒ ⇒ ⇒ ⇒
2 + 2r 2 = r 3 + r 2 + 2r r 3 – r 2 + 2r – 2 = 0 (r – 1) (r 2 + 2) = 0 r ≠ 1 so, r 2 + 2 = 0
Now, 2a + c = 2a + ar 2 = a (2 + r 2) = 0 14. (a) : Let the roots of equation are
x 1, x 2, x 3, x 4, x 5 Now, x 1 + x 2 + x 3 + x 4 + x 5 = 10 and x 1 x 2 x 3 x 4 x 5 = 32
Let suppose a is the real root of equation, then a2 – (1 + i) a + b + 2i = 0
So,
⇒ (a2 – a + b) + i(2 – a) = 0 ⇒ 2 – a = 0 and a2 – a + b = 0 ⇒ a = 2 and 4 – 2 + b = 0 ⇒ b = – 2 So the real root of the equation is a = 2.
⇒ ⇒ ⇒ \
MATHEMATICS TODAY | MARCH ‘16
2 1 + r 2 + 1 + r = r (1 + r ) (1 + r )(1 + r 2 )
x1 + x2 + x3 + x4 + x5
= A.M. = 2
5 = (x 1 ⋅ x 2 ⋅ x 3 ⋅ x 4 ⋅ x 5)1/5 = G.M. A.M. = G.M. x 1 = x 2 = x 3 = x 4 = x 5 = 2 Given equation is (x – 2)5 =0 Remainder = f (1) = – 1
15. (b) : Required no. of words = (total no. of words)
– (No. of words when three A's come together) – (No. of words when no two A's come together) = A – B – C 8 = 6720 Now A = 3 B = 6 = 720
AAA 6
Now, C =6 C 3 × 5
P R B O L can be arranged in 5 ways
6 ×5 33 = 20 × 120 = 2400
We have 3 A's those can be placed at 6 places i.e., 6C ways. Remaining can 3 be arranged in 5 ways
=
n +1
(n − m + 1)b + mb n +1
∑
1 10 11 1 10 11 r +1 ⋅2 = = C r +1 ⋅ 2r +1 11 r =0 r + 111 − (r + 1) 11 r =0 1 11 1 11 = ( C1 2 + C2 22 +11 C3 23 + ... +11 C11 211 ) 11 1 311 − 1 ab − 1 11 + − = = ( 1 2 ) 1 = 11 11 11 ⇒ a = 3, b = 11 11 9 ⋅10 ⋅11 = = 165 Now, bCa = 11C 3 = 38 2⋅3
∑
∑
Case - III : 2 alike and 2 alike = 1
So total numbers of ways = 5 + 12 + 1 = 18 21. (a) : Equation of plane passing through the intersection of planes 4x – y + 5z – 7 = 0 2x – 5 y – z – 3 = 0 is (4x – y + 5z – 7) + l(2x – 5 y – z – 3) = 0 Plane passes through (4, – 1, 0), therefore l = – 1 Terefore, equation of plane is x + 2 y + 3z – 2 = 0
22. (b) : a, b and c must be coplanar vectors
2
a b l
⇒ (n – m + 1) (a – b) = m(b2 – a2) n − m +1 m − n −1 = ⇒ a+b = m −m
c = 0
1 3 2 1 −l = 0 −2 l 3
18. (c) : Let f (x ) = x 3 – 3b2 x + 2c3
Now, f (a) = a3 – 3b2a + 2c3 = 0
r =0
2r +1 10 10 2r +1 = C r ⋅ r + 1 r =0 r 10 − r r + 1
= 2C 1 × 4C 2 = 2 × 6 = 12
Min. value of |z 1 – z 2| = AB = 15 – 2(5) = 5
=
∑
10
Case - II : 2 distinct and 2 alike
16. (b) :
17. (d) : (n − m + 1)a + ma
10
19. (b) :
20. (a) : Case - I :All the letters are distinct = 5C 4 ways = 5
Now, A – B – C = 6720 – 720 – 2400 = 3600
2
2a + 2b + 2c = 0 ⇒ a + b + c = 0
...(i)
⇒ l3 + 11l = 0 ⇒ l = 0
and f (b) = b3 – 3b3 + 2c3 = 0
⇒ – 2b3 + 2c3 = 0 ⇒ b = c
...(ii)
Putting in (i), we get a3 – 3b2 a + 2b3 = 0
⇒ ⇒ ⇒ ⇒ ⇒
(a – b) (a2 + ab – 2b2) = 0 a2 + ab – 2b2 = 0, a ≠ b
(a +2b) (a – b) = 0 a + 2b = 0, a ≠ b
a + 2c = 0 Adding (iii) and (iv), we get
... (iii) ...(iv)
MATHEMATICS TODAY | MARCH ‘16
15
23. (a) :
Now, 2[ f (a) + a] = 2 −a + 29. (b) : Here, x 1 + y
1 + a =1 2
= −y 1+ x
⇒ x 2 (1 + y ) = y 2 (1 + x ) ⇒ x 2 – y 2 = y 2 x – x 2 y ⇒ x + y = – xy ⇒ OA = 2, OB = 1, AB =
3
1 −x = −1 1 + x 1 + x
dy −1 = dx (1 + x )2 n −1 n + k 30. (c) : n C0 + C k +1 k = 0 n n = C + C + n+1C + n+2C + ... + 2n−1C 0 1 2 3 n So,
D ABC is equilateral triangle. Therefore, distance between orthocentre and circumcentre is zero. 24. (c) : Put x = 1, a0 + a1 + a2 +...+ a2n =
y =
4n
put x = – 1, a0 – a1 + a2 – ... a2n = 2n
\ 2(a0 + a2 +...+ a2n) = 4n – 2n ⇒ a0 + a2 +...+ a2n = 2n–1 (2n – 1)
x 25. (a) : Let f (x ) = sin x sin x − x cos x cos x(tan x − x ) f ′(x ) = = (sin x )2 (sin x )2 When x > 0, then tan x > x and cos x > 0,
∑
= n+1 C
n +1 C2 1
+
+ n+2C3 + ... + 2n−1Cn
.... =
2nC
n nn
Since f (x ) > f (sin x ) for x > sin x x sin x > \ ⇒ 1 > (sin x ) sin x sin(sin x ) x sin(sin x ) (sin x )2 <1 x sin(sin x )
26. (b) : Minimum value =
−16 cos2 q − 16 sin2 q 4×4
= – (cos2 q + sin2 q) = – 1 2 b 2 Using ax + bx + c = a x + − D ≥ −D for a > 0 2a 4a 4a
27. (b) : P 6 – P 4 = KP 2
28.
16
⇒ (sin6 q + cos6 q) – (sin4 q + cos4 q) = K (sin2 q + cos2 q) ⇒ K = (1 – 3 sin2 q ⋅ cos2 q ) – (1 – 2 sin2 q ⋅ cos2 q) ⇒ K = – sin2 q ⋅ cos2 q (c) : Given that, 4 a2 = 1 – 2a So, f (a) = 4a3 – 3a+1 = a(1 – 2a) – 3a + 1 = – 2a2 – 2a + 1 1 − 2a − 2a + 1 = −a + 1 = − 2 2 MATHEMATICS TODAY | MARCH ‘16
= n+2 C + n+2C + ... + 2n−1C 2 3 n
therefore, f ′(x ) > 0, f (x ) is increasing.
Similarly, x < 0,
0124-6601200 for further assistance.
PAPER-I 1.
Let f : R → R be a differentiable function satisfying 4
f (x ) = f ( y ) f (x – y )
∀
∫
x , y ∈ R and f ′(0) = {2 x}dx , 0
where {⋅} denotes the fractional part function and f ′(–3) = aeb. Ten, |a + b| is equal to _________
2.
If a , b and c are unit vectors, then find the
maximum value of a − b
2
ln x ln x 1 dx = − − x 3 ax 2 bx 2 a + b must be
∫
2
+ b − c + c − a 2.
+ c, then
3.
If
4.
TP and TQ are any two tangents to a parabola and the tangent at a third point R cuts them in P ′ and Q′, TP ′ TQ ′ + then the value of must be TP TQ If a, b, g , d are the solution of the equation
5.
(a)
the value of
If the quadratic polynomial P (x ) = ( p – 3)x 2 – 2 px + 3 p – 6 ranges from [0, ∞) for every x ∈ R, then the value of p can be 3 (a) (b) 4 (c) 6 (d) 7 2 11. All non-zero complex numbers on the complex 1 plane satisfying z + ∈R can lie on z (a) y = 0 (b) x = 0 (c) the line y = x (d) a unit circle with centre at the origin 10.
12.
Domain of f (x ) = sin–1[2 – 4x 2] is ([⋅] denotes the greatest integer function) (a)
π = 3 tan 3θ, no two of which have equal 4 tangents, then the value of tan a + tan b + tan g + tan d
tan θ +
(c)
is equal to 6.
Te number of solutions for the equation log4(2x 2 + x + 1) – log2 (2x – 1) = 1, is
7.
Find the value of
8.
2π + cos 4 π − cos 7 π − cos π . 4 cos 15 15 15 15 If ( y 2 – 5 y + 3) (x 2 + x + 1) < 2x for all x ∈ R, then number of integral values of y will be
9.
If a, b, c are real numbers and z is a complex number such that a2 + b2 + c2 = 1 and b + ic = (1 + a)z , then 1 + iz equals 1 − iz
b − ic 1+ a a + ib 1− c (b) (c) (d) 1 − ia b + ic 1+ c a − ib
n
13.
− − i
3 3 , − {0} (b) 2 2 3 ,0 2
(d)
− −
3 3 , 2 2 3 , 8 2
j
∑∑ ∑
i =1 j =1 k =1
1 is equal to n
n(n + 1)(n + 2) n2 (a) (b) 6 n=1 n(n + 1)(2n + 1) (c) (d) n+2C 3 6 2 14. Te line 3x + 6 y = k intersect the curve 2x + 2xy + 3 y 2 = 1 at points A and B. Te circle on AB as diameter passes through the origin. Te possible value(s) of k is (are) (a) 3 (b) 4 (c) –4 (d) –3
∑
By : Vidyalankar Institut e, Pearl Centre, Senapati Bapat Marg, Dadar (W), Mumbai - 28. Tel.: (022) 24306367
MATHEMATICS TODAY
|MARCH ‘16
17
15.
16.
17.
Let P (x ) = x 2 + bx + c, where b and c are integer. If P (x ) is a factor of both x 4 + 6x 2 + 25 and 3x 4 + 4x 2 + 28x + 5, then (a) P (x ) = 0 has imaginary roots (b) P (x ) = 0 has roots of opposite sign (c) P (1) = 4 (d) P (1) = 6 1 [( nC0 + nC3 + ...) − ( nC1 + nC2 + nC4 + nC5 + ...)]2 2 3 n + ( C1 − nC2 + nC 4 − nC5 + ...)2 = 4 (a) 3 (b) 4 (c) 2 (d) 1 For every pair of continuous functions f , g : [0, 1] → R, such that max{ f (x ) : x ∈[0, 1]} = max{ g (x ) : x ∈[0, 1]}, the correct statement(s) is (are): (a) ( f (c))2 + 3 f (c) = ( g (c))2 + 3 g (c) for some c ∈[0, 1] (b) ( f (c))2 + f (c) = ( g (c))2 + 3 g (c) for some c ∈[0, 1] (c) ( f (c))2 + 3 f (c) = ( g (c))2 + g (c) for some c ∈[0, 1] (d) ( f (c))2 = ( g (c))2 for some c ∈[0, 1]
18.
Let PR = 3i + j − 2k and SQ = i − 3j − 4k determine diagonals of a parallelogram PQRS and PT = i + 2 j + 3k be another vector. Ten the volume of the parallelopiped determined by the vectors PT , PQ and PS is (a) 5 (b) 20 (c) 10 (d) 30
19.
20.
Consider the system of equations a1x + b1 y + c1z = d 1 a2x + b2 y + c2z = d 2 a3x + b3 y + c3z = d 3 a1 b1 c1 d1 Now, let
D = a2
b2
c2 ,
a3
b3
c3
a1
d1
c1
D2 = a2
d2
c2 and
b1
c1
D1 = d2
b2
c2
d3
b3
c3
a1
b1
d 1
D3 = a2
b2
d 2
b3
d 3
a3 d3 c3 According to Crammer’s rule,
a3
D D D x = 1 , y = 2 , z = 3 D
D
D
Match the Following : Column I
(a) If D ≠ 0 and D1 = D2 = D3 = 0, then the solutions are (b) If D1 = D2 = D3 = D = 0, then the solutions are (c) If D ≠ 0 and atleast one of D1, D2, D3 is non-zero, then the solutions are (d) If atleast one of D1, D2, D3 is non-zero, and D = 0, then the solutions are
Column II
(p) Unique (q) Inconsistent (r) Non-rivial
(s) Infinitely many
Match the following : Column I
Column II
(a) Consider the differential equation, (p) 1 2
dy dy y = x + 1 + dx dx Its degreee is (b) Te differential equation represen- (q) 4 ting the family of curve y 2 = 2c(x + c ) , where c is a parameter is of degree (c) Consider differential equation,
(r) 2
2 3 dy 1 dy 1 dy 1+ + + dx 2 ! dx 3! dx + ...∞ = y
then its degree is (d) Consider the differential equation, (s) 3
2 2 2/3 dy d y = 1 + 2 . dx dx It’s degree is 18
MATHEMATICS TODAY
|MARCH ‘16
0124-6601200 for further assistance.
PAPER - II 1.
2.
3.
Let f (x , y ) be a periodic function satisfying the condition f (x , y ) = f ((2x + 2 y ), (2 y – 2x )) ∀ x , y ∈R. Now, define a function g by g (x ) = f (2x , 0). If period of g (x ) function is a, then value of |a – 7| is For a point P in the plane, let d 1(P ) and d 2(P ) be the distances of the point P from the lines x – y = 0 and x + y = 0 respectively. Te area of the region R consisting of all points P lying in the first quadrant of the plane and satisfying 2 ≤ d 1(P ) + d 2(P ) ≤ 4, is Te circle x 2 + y 2 – 4x – 4 y + 4 = 0 is inscribed in a triangle which has two of its sides along the coordinate axes. Te locus of the circumcentre of 2
4.
5.
6.
7.
8.
1 1 1 − tan x + tan −1 tan x + c 2 2 2 1 1 1 tan x + c tan −1 (b) − cot x + 2 2 2 1 1 tan x + c tan −1 (c) − cot x − 2 2 1 1 1 cot −1 tan x + c (d) − cot x − 2 2 2 (a)
12.
dy If y (t ) is a solution of (1 + t ) − ty = 1 and dt 3 y (0) = –1, then y (1) + is equal to 2 Suppose that f is differentiable for all x and that f ′(x ) ≤ 2 for all x . If f (1) = 2 and f (4) = 8, then f (2) has the value equal to ______.
wo whole numbers are randomly selected and multiplied. If the probability that the units place in their product is “Even” is p and the probability that the units place in their product is “odd” is q, then p q +1 is equal to
Te point of intersection of the plane r ⋅ (3i − 5j + 2k ) = 6 with the straight line passing through the origin and perpendicular to the plane 2x – y – z = 4 is (a) (1, –1, –1) (b) (–1, –1, 2) 4 −2 , −2 (c) (4, 2, 2) (d) , 3 3 3 10. If the median through A of a D ABC having vertices A ≡ (2, 3, 5), B ≡ (–1, 3, 2) and C ≡ (l, 5, m) is equally inclined to the axes, then (a) l = 7 (b) m = 10 (c) l = 10 (d) m = 7
2 2 5 9 2 5/2 2 9/2 (b) − cot x − cot x + c 5 9 2 2 (c) − cot 5/2 x − cot 3/2 x + c 5 9 (d) none of these (a)
13.
Te integral of
sin2 x + tan2 x
must be
Te line y = x + 5 touches (a) the parabola y 2 = 20x (b) the ellipse 9x 2 + 16 y 2 = 144
=1
14.
How many numbers between 5,000 and 10,000 can be formed using the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 each digit appearing not more than once in each number? (a) 1680 (b) 5 × 8P 8 (c) 30 × 8C 3 (d) 5! × 8C 3
15.
riplets ( p, q, r ) is chosen from a set {1, 2, 3, 4, 5, 6, 7, 8, 9} such that x < y ≤ z . Tere could be ______ such triplets.
11.
− cot3/2 x − cot9/2 x + c
x 2 y 2 − (c) the hyperbola 29 4 (d) the circle x 2 + y 2 = 25
9.
1
∫ sin11 x dx is equal to
I =
2
the triangle is x + y − xy + k x + y = 0 , find k. If f (x ) = x 3 + x 2 f ′(1) + x f ″(2) + f ′″(3) for all x ∈ R, then find the value of f (1). 1 3 (3x 4 + 2 x 2 )sin + x + 5 x Find lim 3 2 x →−∞ x + x + x + 1
cos3 x
(a) 143 (b) 200 16.
(c) 180
(d) 120
If in triangle ABC , CD is the angle bisector of the angle ACB, then CD is equal to a +b C cos 2ab 2 C 2ab cos (c) a +b 2
(a)
(b) (d)
a +b C cos ab 2 b sin A C sin B + 2
MATHEMATICS TODAY
|MARCH ‘16
19
Paragraph for Questions Nos. 17 and 18 Consider a hyperbola whose centre is at origin. A line x + y = 2 touches this hyperbola at P (1, 1) and intersect the asymptotes at A and B such that AB = 6 2 units. 17. Equation of asymptotes are (a) 2x 2 + 5xy + 2 y 2 = 0 (b) 2x 2 – 5xy – 2 y 2 = 0 (c) 2x 2 – 5xy + 2 y 2 = 0 (d) 2x 2 + 5xy – 2 y 2 = 0 7 18. Equation of tangent to the hyperbola at −1, 2 is
(a) 3x + 2 y = 2 (c) 3x + 2 y = 4
(b) 3x + 4 y = 11 (d) 3x + 2 y = 7
Paragraph for Q. No. 19 to 20 Let C 1 and C 2 be two curves passing through the origin as shown in the figure. A curve C is said to “bisect the C 2 area” between C 1 and C 2, if for Y C each point P of C , the two shaded y = f(x ) n regions A and B shown in the P y = x C 1 figure have equal areas. Let A n the equation of the curve C 1 be B y = x 2 x n n y = and that of C be y = x . X O 2 19. Area of the total shaded region ( A + B), if x -coordinate of P is x and n = 2 is
(a)
20
x 2
2
(b)
MATHEMATICS TODAY
x 3
2
|MARCH ‘16
(c)
x 3
x 3
(d) 6 3 20. Equation of the curve C 2 for n = 2 is y = 3x 2 (a) 4 9x 2 (c) 16
4 x 2 (b) 3 16x 2 (d) 9 ANSWER KEYS PAPER-I
1. 6. 10. 13. 15. 19.
4. (1) 5. (0) (4) 2. (9) 3. (6) 9. (b), (c) (1) 7. (2) 8. (2) 12. (a) (c) 11. (a), (d) 14. (a), (d) (a), (d) 16. (d) 17. (a), (d) 18. (c) (a), (c) (a) → (r), (b) → (s), (c) → (p), (d) → (q) PAPER-II
1. 6. 10. 13. 16. 20.
4. (4) 5. (5) 2. (6) 3. (1) 9. (d) (1) 7. (4) 8. (4) 11. (b), (d) 12. (a), (b) 14. (a), (c) 15. (a), (b), (c) 17. (a) 18. (c) 19. (c), (d) (d) For detailed solution to the Sample Paper, visit our website : www. vidyalankar.org
(2) (b) (d) (c)
nn
1.
If
1 − x 3 ( f (x ))2 f = x , x ≠ – 1, 1 and f (x ) ≠ 0, then 1 + x
7.
P (x ) =
f (x ) is
2.
3.
(a)
1 − x x 2 1 + x
(c)
1 − 2x x 2 1 + x
(b)
4.
1 + x x 2 1 − x
(d) None of these
Suppose a, b, c ∈ R and b ≠ c. If α, β are roots of x 2 + ax + b = 0 and γ , δ are roots of x 2 + ax + c = 0, (α − γ )(α − δ ) then the equation whose roots are (β − γ )(β − δ ) and 2 is
Let
∆(x ) =
9.
0
1
1
1
0
−1
6.
Let A be a 3 × 3 matrix such that det( A) = –2. Ten det (–2 A–1) + 2det( A) is equal to (a) –2 (b) – 4 (c) 7 (d) None of these Te number of ways in which we can arrange the digits 1, 2, 3, ……., 9 such that the product of five digits at any of the five consecutive positions is divisible by 7 is (a) 7! (b) 9P 7 (c) 8! (d) 5(7!)
(b) 36 C17 (d) 0
1 1 1 1 1 + − + − + ....... = a , then value of 3 5 7 9 11
10.
100
∑ a2r = α
If lim x →0
100
and
∑ a2r −1 =
r =1
β,
(b) 50 (d) None of these
(cos x )1/2 − (cos x )1/3 sin2 x
12a + 2 is (a) 1 (c) 2 11.
(d) a
Let an be the nth term of an A.P. with common
then (α - β) – 100d = (a) 0 (c) 100
∆′(x ) vanishes at least once in (a) (0, π/2) (b) (π/2, π) (c) (0, π/4) (d) (– π/2, 0) 5.
is
r =1
then
If 1 –
difference d . If
(b) x 2 – x + 1 = 0 (d) x 2 – 3x – 2 = 0 cos x sin 2 x + cos 2 x
∏ (x + 35C r )
1 1 1 + + + ........ is 1 ⋅ 3 5 ⋅ 7 9 ⋅11 a a a (a) (b) (c) 2 4 3
If (α + βi)11 = x + iy , where α, β, x , y ∈ R, then Re((β + αi)11) equals (a) y (b) – y (c) x (d) – x
sin x
17
r =0 (a) 234 (c) 235 – 36 C17 8.
(a) x 2 + x + 1 = 0 (c) x 2 – 3x + 2 = 0
Coefficient of x 17 in the polynomial
= a, then the value of
(b) –1 (d) None of these
Let f be a differentiable function such that
1 –x = 5(x ≠ 0) and y = x 2 f (x ), then dy at x dx
8 f (x ) + 6 f x = 1 is (a)
15 14
(c)
19 14
17 14 17 (d) − 14 (b)
Contributed by : Sanjay Singh, Mathematics Classes, Chandigarh, 9888228231 MATHEMATICS TODAY
|MARCH ‘16
21
12.
If f (x ) = x 2/3, then (a) (0, 0) is a point of maximum (b) (0, 0) is not a point of minimum (c) (0, 0) is a critical point (d) Tere is no critical point
13.
Te difference between the greatest and least values of the function
14.
3 8
(b)
2 3
(c)
8 7
π/ 4 ∫ 0 ln(1 + tan2 θ + 2 tan θ) d θ =
value of 2a + 3b is (a) 15 (c) 17 15.
19.
1 cos 2x – 1 cos 3x is 2 3
f (x ) = cos x +
(a)
eccentricity ‘ e’ of the hyperbola satisfies 2 3 (a) e = (b) e = 2 3 2 2 (c) e > (d) 1 < e < 3 3
(d)
π ln a b
(a) 9 4
then
20.
the
(b) 16 (d) None of these
dy 1 = Te solution of is given by dx 2x − y 2
(a) y = C e–2x + (b) x = C e– y +
1 2 1 1 y + y + 2 2 4 Te incentre of triangle with vertices (– 3 , –1) (0, 0) and (0, –2) is 1 , −1 (a) − 3 , − 1 (b) − 3 (c)
18.
If PQ is a double ordinate of the hyperbola x 2 y 2 − = 1 such that OPQ is an equilateral a 2 b2 triangle, O being the centre of the hyperbola, then
22
23.
24.
wo tangents are drawn from the origin to a circle with centre at (2, – 1). If the equation of one of the tangents is 3x + y = 0, the equation of the other tangent is (a) 3x – y = 0 (b) x + 3 y = 0 (c) x – 3 y = 0 (d) x + 2 y = 0
MATHEMATICS TODAY
|MARCH ‘16
1 8
(d)
2 3
53 units
(d)
66 units
x − 4 y − 2 z − k = = 1 1 2 lies in the plane 2x – 4 y + z = 7 is (a) 7 (b) 6 (c) no real value (d) – 7
If a variable takes the values 0, 1, 2, …….., n with frequencies proportional to the binomial coefficients nC 0, nC 1, ………., nC n then the mean of the distribution is n n(n + 1) (a) (b) 2 4 n(n + 1) n(n − 1) (c) (d) 2 2 Four natural numbers are selected at random and are multiplied. Te probability that the product is divisible by 5 or 10 is 369 6 49 256 (a) (b) (c) (d) 625 625 625 625
(d) None of these
17.
(c)
22.
)
2 − 3 , − 1
3 2
Te value of k for which the line
(d) x = C e2 y +
(
(b)
21.
1 2 1 1 x + x + 4 2 4 1 2 1 1 y + y + 4 4 2
1 2
Te length of the perpendicular drawn from the x y − 2 z − 3 point (3, – 1, 11) to the line = is = 2 3 4 (a) 29 units (b) 33 units (c)
1 1 (c) x = C e y + y 2 + y + 4 2
16.
Te slope of the line touching both the parabolas y 2 = 4x and x 2 = – 32 y is
25.
Each of two persons A and B toss three fair coins. Te probability that both get the same number of heads is 3 1 (a) (b) 8 9 5 (c) (d) None of these 16 3 If cos A = , then the value of 4 A A 5 A 16 cos2 –32 sin sin is 2 2 2 (a) – 4 (b) – 3 (c) 3 (d) 4
0 ≤ a, b ≤ 3 and the equation x 2 + 4 + 3 cos (ax + b) = 2x has at least one solution, then the locus of the point (a, b) is (a) x + y = 2π (b) x + y = π 2 2 2 (c) x + y = π (d) y 2 = 2x
SOLUTIONS
26. If
1.
1 − x = x 3 1 + x 2 − 1 x We have f 1 + x
If Aditya observes that the angle of elevation of the top of the tower at an extremity of the major axis of the field is α, at its focus is β and an extremity of the minor axis is γ , then (a) cot2 α = cot2 β – cot2 γ (b) cot2 β = cot2 γ – cot2 α (c) cot2 γ = cot2 α – cot2 β (d) None of these 2.
3.
(b) : (α + βi)11 = x + i y
(c) : x 2 + ax + c = (x – γ )(x – δ)
(α − γ )(α − δ) α2 + aα + c = (β − γ )(β − δ ) β2 + aβ + c −b + c = 1 [α, β are roots of x 2 + ax + b = 0] = −b + c Hence for the required equation sum of root = 1 + 2 = 3 and product of root is 1 × 2 = 2. So the equation is x 2 – 3x + 2 = 0 Tus,
Statement II : Te line
∆(x ) is continuous on [0, π/2] and differentiable on (0, π/2). Also ∆(0) = 0 and ∆ π = 0. Tus, by the Rolle’s theorem there exists 2 at least one c ∈ (0, π/2) such that ∆′(c) = 0.
4.
(a) : Te function
5.
(d) : As A–1 is a 3 × 3 matrix,
det (–2 A–1) = (–2)3 det ( A–1)= (–2)3 (det( A))–1
30. Suppose a, b, c ∈R Statement I : If z = a + (b + ic)2017 + (b – ic)2017 then
1 =4 2
= (–8) −
z is real.
Hence det(–2 A–1) + 2det A = 4 + 2 × (–2) = 0
Statement II : If z = z , then z is real.
(a) Statement I is true, State ment II is true; Statement II is a correct explanation for Statement I. (b) Statement I is tr ue; Stateme nt 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.
...(ii)
⇒ (α – βi)11 = x – iy ⇒ [(–i) (β + iα)]11 = – i( y + ix ) ⇒ (–i)11 (β + iα)11 = – i( y + ix ) We get (β + iα)11 = – ( y + ix ) = – y – ix Hence Re((β + iα)11) = – y
x y − 1 z − 2 = = . 1 2 3
x y − 1 z − 2 bisects the = = 1 2 3 line segment joining A(1, 0, 7) and B (1, 6, 3). (a) Statement I is true, State ment II is true; Statement II is a correct explanation for Statement I. (b) Statement I is tr ue; Stateme nt II is true; Statement II is not a correc t ex planation for Statement I. (c) Statement I is true, Statement II is false. (d) Statement I is false, Statement II is true.
3 1 − x f (x ) = 1 + x
3 x 3 2 1 − x f (x ) = 2 1 + x ( f (x )) 3 ( f (x ))3 1 + x 1 + x 2 ⇒ = f ( x ) = x ⇒ 1 − x 1 − x x 6
29. Statement I : Te point A(1, 0, 7) is the mirror image
of the point B (1, 6, 3) in the line
...(i)
Solving (i) and (ii), we have
Statement II : ~ ( p ↔ ~ q) is a tautology.
(a) Statement I is true, Stateme nt II is true ; Statement II is a correc t explanation for Statement I. (b) Statement I is tr ue, Statement II is true ; Statement II is not a c orrect explanation for Statement I. (c) Statement I is true, Statement II is false. (d) Statement I is false, Statement II is true.
1 − x in the equation 1 + x
( f (x ))2 f
27. A tower is standing at the centre of an elliptic field.
28. Statement I : ~ ( p ↔ ~ q) is equivalent to p ↔ q
(b) : Replacing x by
6.
(c) : Let an arrangement of 9 digit number be
x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9. Note that we require product of 5 consecutive digits. Tis is possible in the following ways : (x 1 , x 2 , x 3 , x 4 , x 5 ); (x 2 , x 3 , x 4 , x 5 , x 6 ); …. ..; (x 5, x 6, x 7, x 8, x 9) MATHEMATICS TODAY
|MARCH ‘16
23
if the product of 5 consecutive digits is divisible by 7. Tis is possible if x 5 is 7. Terefore, we can arrange the 9 digits in desired number of ways in 8! ways. 7.
(a) : P ( x ) is a polynomial of degree 18, and
coefficient of x 17 is 35C + 35C + 35C + ……… + 35C 0 1 2 17 n n Using C r = C n – r , we can write N as N = 35C 35 + 35C 34 + …….. + 35C 18
N =
...(1) ...(2)
Adding (1) and (2), we get 35C + 35C + 0 1 34 N = 2
2N =
⇒ 8.
…….. +
35C = 35
235
(a) : We have
2 2 2 + + + ...... 1 ⋅ 3 5 ⋅ 7 9 ⋅11 1 1 1 a + + + ........ = 1 ⋅ 3 5 ⋅ 7 9 ⋅11 2
100
100
r =1
r =1
⇒ (α – β) – 100d =
0
(cos x )1/2 − (cos x )1/3 0 form 10. (a) : a = lim 0 x →0 sin2 x
\
dy 3 1 19 =2× + = dx x =1 7 2 14
dy 2 −1/3 = x . Tis derivative is never zero, but dx 3 there is no derivative for x = 0. So (0, 0) is a critical dy dy point. If x < 0 then < 0 and if x > 0 then > dx dx 0. Tus (0, 0) is a point of minimum.
12. (c) :
3x sin x 2 2
2π , π and 2 π are the critical points. 3 2π 1 1 7 13 A l s o , f ( 0 ) = 1 + − = , f = – , 3 2 3 6 12 7 1 f (π) = – and f (2π) = . 6 6 7 Hence the greatest value is and the least value 6 13 is − . 12 7 −13 27 9 Hence the difference is − = = 6 12 12 4 π/4 14. (b) : Let I = ∫ 0 ln(1 + tan2 θ + 2 tan θ) d θ π/4 π/ 4 ∫ 0 ln(1 + tan θ)2 d θ = ∫ 0 2ln (1 + tan θ)d θ π/ 4 Let I1 = ∫ 0 ln(1 + tan θ)d θ
1 1 (cos x )−1/2 sin x + (cos x)−2/3 sin x 3 a = lim 2 2 sin x cos x x →0
−
1 1 (cos x )−1/2 + (cos x )−2 /3 3 a = lim 2 2 cos x x →0 1 1 1 1 a = − − = − 2 2 3 12
−
π/4 π I1 = ∫ 0 ln 1 + tan − θ d θ 4 π/4 1 − tan θ d θ = ∫ 0 ln 1 + 1 + tan θ π /4 = ∫ 0 [ln 2 − ln (1 + tan θ)]d θ
12a + 2 = – 1 + 2 = 1
11. (c) : Differentiating the given expression, we get
1 − 1 – 1 = 0 x x 2 1 ⇒ 8 f ′(1) + 6 f ′(1) (–1) = 1 ⇒ f ′(1) = 8 f ′(x ) + 6 f ′
dy Also = 2xf (x ) + x 2 f ′(x ) dx 24
3 7
Hence x = 0,
α – β = ∑ (a2r − a2r −1 ) = ∑ d = 100d
⇒
Hence
= – 4 sinx sin
then
⇒
14 f (1) = 6 ⇒ f (1) =
f ′(x ) = – (sin x + sin 2x – sin 3x )
(a) : Let d be the common difference of the A.P.,
⇒
⇒
2π. So the difference between the greatest and least valu es of the functi on is the difference between these values on the interval [0, 2 π]. We have
=
9.
8 f (1) + 6 f (1) – 1 = 5
13. (d) : Te given function is periodic, with period
1 1 1 1 1 a = 1 − + − + − + ...... 3 5 7 9 11
⇒
dy = 2 f (1) + f ′(1) dx x =1 Putting x = 1 in the given equation, we obtain So,
MATHEMATICS TODAY
|MARCH ‘16
2
⇒
2I 1 =
π 4
ln 2 = I
Hence
π ln a
=
π
ln 2. Hence a = 2 and b = 4.
4 b So 2a + 3b = 16
dx = 2x – y 2 (Tis is a linear equation in x ). dy − 2dy Te integrating factor is e ∫ = e −2 y (xe −2 y ) = − y 2 e −2 y So dy Integrating, we have − y 2e −2 y –2 y xe = − ∫ ye −2 y dy + constant −2 y 2 −2 y ye −2 y 1 −2 y e + − ∫ e dy + constant = 2 2 2 2 y −2 y y −2 y e −2 y = e + e + + C 2 2 4 2 y 1 2 y y \ x = Ce + + + 2 2 4
Equation of OP is y =
is an equilateral triangle and hence its incentre is the centroid
−
1 3
, − 1 of the triangle.
17. (c) : Let the equation of the other tangent from the
origin be y = mx , then length of the perpendiculars from the centre (2, – 1) on the two tangents is same.
⇒
2m + 1 1 + m2
=
6 −1 9 +1
=
3 y 2 a
2
−
y 2 b
2
=1 ⇒
(3b2 – a2) y 2 = a2b2
For real values of y , 3b2 > a2
⇒
3a2(e2 – 1) > a2
⇒
3e2 > 4
⇒
e>
2
3 19. (a) : Equation of the tangent at (t 2 , 2t ) to the parabola y 2 = 4x is ty = x + t 2 ……(1) And the tangent at ( – 16t ′, – 8t ′ 2) to the parabola x 2 = –32 y is t ′x = y – 8t ′ 2 …….(2) Since (1) and (2) represent the same line, comparing we get
−1 t t 2 ⇒ = = −t ′ 1 8t ′2
t = 2
Hence the required slope of the tangent is 1 1 = t 2 20. (c) : Any point on the given line is
P (2r , 3r + 2, 4r + 3) and the given point be A Direction ratios of AP are 2r – 3, 3r + 2 + 1, 4r + 3 – 11 or 2r – 3, 3r + 3, 4r – 8
5
AP is perpendicular to the given line
10
if
2(2r – 3) + 3(3r + 3) + 4(4r – 8) = 0
⇒
r = 1, so coordinates of point are (2, 5, 7)
\
AP =
⇒ ⇒ ⇒
10(2m + 1)2 = 25(1 + m2) 3m2 + 8m – 3 = 0 ⇒ (3m – 1) (m + 3) = 0 m = – 3 or 1/3 m = – 3 represents the given tangent hence the slope of the required tangent is 1/3 and its equation is y = (1/3) x ⇒ x – 3 y = 0 18. (c) : As
x , which meets the
3 hyperbola at points for which
15. (d) :
16. (b) : Length of each side of the triangle is 2 so it
1
∠POQ = 60º. OP makes an angle of 30º
with the positive side of x -axis.
(2 − 3)2 + (5 + 1)2 + (7 − 11)2
=
53 units
21. (a) : Since (2) (1) + (– 4) (1) + (1) (2) = 0 the line
is parallel to the plane. It will lie in the plane if the point (4, 2, k) lies on the plane.
⇒
2(4) – 4(2) + k = 7
22. (b) : N =
∑ f i
⇒
k = 7
= k[nC 0 + nC 1 + …… + nC n]
= k(1 + 1)n = k2n
∑ f i x i =
k[1.nC 1 + 2 . nC 2 + ……. + n nC n]
=k Tus x
=
1 2n
n
∑ r . nCr = k ⋅ n ⋅ 2n−1
r = 0
(n2n−1 ) =
n 2
MATHEMATICS TODAY
|MARCH ‘16
25
23. (b) : Te product will be divisible by 5 or 10 if at
minor axis respectively and S be a focus, then
least one of the number has last digit as 0 or 5. Tus required probability = 1 – P (none of the number has last digit 0 or 5) 4 4 8 4 369 =1− =1− = 10 5 625
24. (c) : Let X be the number of heads obtained by
A and Y be the number of heads obtained by B. Note that both X and Y are binomial variate with parameters n = 3 and p = 1/2. Probability that both A and B obtain the same number of heads is \ P ( X = 0) P ( Y = 0) + P ( X = 1) P ( Y = 1) + P ( X = 2) P (Y = 2) + P ( X = 3) P (Y = 3)) 2
2
2
3 1 3 3 1 3 3 1 3 = C0 + C1 + C 2 2 2 2 3 1 3 2 + C 3 2 20 5 1 6 = = [1 + 9 + 9 + 1] = 2 64 16 25. (c) : Te given expression is equal to
⇒
cot2 γ = cot2 α – cot2
β
28. (c) : able for basic logical connectives :
↔ ~q
↔ ~q) ( p ↔ q)
p
q ~q
F
F
F
F
F
F
F
F
F
F
F
F
p
Note that ~ ( p
\
~( p
↔ ~ q) is not a tautology.
Statement – II is false.
8(1 + cos A) – 16(cos 2 A – cos 3 A)
From table ~ ( p ↔ q) is equivalent to ( p
= 8(1 + cos A) – 16[2cos2 A – 1 – cos A
Tus, statement-I is true.
(4cos2 A – 3)] 9 3 9 3 = 8 1 + − 16 2 × −1 − 4 × − 3 4 4 16 16 = 14 – (18 – 16 – 27 + 36) = 3 26. (b) : We have
⇒ (x –
x 2 –
2x + 4 = – 3 cos (ax + b)
1)2 + 3 = – 3 cos (ax + b)
...(1)
As (x – 1)2 ≥ 0 and – 1 ≤ cos ( ax + b) ≤ 1 Equation (1) is possible if cos ( ax + b) = –1 and x – 1 = 0
⇒ a + b = π, 3π, 5π, ………… Since a + b ≤ 6 and 3π > 6, a + b = π Hence the locus of ( a, b) is x + y = π 27. (c) : Let the equation of the elliptic field be
x 2 a
2
+
y 2 b
2
= 1.
OP be the tower of height h at O, the centre of the field, Let A and B be extremities of major and 26
OA = h cot α = a OS = h cot β = ae OB = h cot γ = b Since b2 = a2(1 – e2)
MATHEMATICS TODAY
|MARCH ‘16
↔ q)
29. (b) : Mid point of the segment AB is
1+1 , 0 + 6 + 7 + 3 2 2 = (1, 3, 5) which lies on the 2 x y − 1 z − 2 line = So, Statement II is true. = 1 2 3 Direction ratios of AB are 0, 6, – 4 and the given line are 1, 2, 3.
Since 0 × 1 + 6 × 2 + ( –4) × 3 = 0, the line AB is perpendicular to the given line so, statement I is also true, but statement II is not a correct explanation for Statement I. 30. (a) : Statement II is true.
We have,
z
= a + (b + ic)
2017
+ (b − ic)
2017
= a + (b – ic)2017 + (b + ic)2017 = z ⇒ z is real \ Statement I is true and Statement II is correct explanation for it. nn
M
th rchives 10 Best Problems
Math Archives, as the title itself suggests, is a collection of various challenging problems related to the topi cs 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. Te equation of the tangent line of slope 1 to the x
6. Te value of
curve f (x) = ∫ | t |dt is (a) y = x + 2 −1 (b) y = x – 2 (c) y = x – 1 (d) none of these 1
(a) less than e – 1 (c) less than e – e2
1 + x 8
(a) I 1 > 1, I 2 < 1 (c) 1 < I 1 < I 2
(b) I 1 < 1, I 2 > 1 (d) I 2 < I 1 < 1
∫ x ⋅( f (cos x))2 ⋅ dx = 0
(b)
∫ f (cos 0 b−c
(c)
∫
0
∫
π
x) dx = n ∫ f (cos x ) dx , n ∈N 0
c
b
x. f (sin x )dx =
a
1
4.
d
1
π π−a 2
∫
f (sin x )dx
a
−1
e
e +1 2 (c) 1+ e
(b)
1 e +1
(d) none of these x
dt satisfies the equation t
∫
5. Te function L(x ) =
1
(a) L(x + y ) = L(x ) + L( y ) (b) L(x / y ) = L(x ) + L( y ) (c) L(xy ) = L(x ) + L( y ) (d) None of these
f ′′(x ) = – f (x ) and f ′(x ) = g (x ) if h(x ) = ( f (x ))2 + ( g (x ))2 and h(5) = 2, then h(10) = (a) 0 (b) 1 (c) 2 (d) none of these
10. If the tangent to the curve xy + ax + by = 0 at (1, 1)
∫ dx 1 + e1/x dx is equal to (a)
(a) degree is 1 (b) order is 1 (c) degree and order both are 1 (d) none of these
9. Let f be a twice differentiable function such that
2
f (x + c )dx = ∫ f (x) dx
π−a
(d)
2
7. A differential equation is called linear, if its
lines is a family of (a) parallel lines (b) concurrent lines (c) concentric circles (d) concentric ellipses
a
−a nπ
(b) greater than e (d) 1 – e
8. Te orthogonal trajectories of a family of parallel
3. Which of the following is not correct ?
(a)
2
∫ e x dx is 0
1 1 + x 9 dx and dx , then I 2 = ∫ ∫ 1 + x 4 3 0 0 1 + x
2. If I 1 =
1
makes an angle tan –12 with x -axis, then a + b = ab (a) 0 (b) 1/2 (c) –1/2 (d) None of these SOLUTIONS
1. (d) : f ′(x ) = |x | = 1
⇒ x = –1, 1
1
f (–1) = 0, f (1) = ∫ | t |dt = 1. −1 Tus required tangents have points of contact as (–1, 0), (1, 1). angents are y = x + 1, y – 1 = x – 1 ⇒ y = x + 1 and y = x . 2. (d) : Obviously I 1, I 2 < 1, as
1 + x 8 1 + x 4
< 1, " x ∈(0, 1)
By : Prof. Shyam Bhu shan, Director, Narayana IIT Academy, Jamshedpur. Mob. : 09334870021
MATHEMATICS TODAY
|MARCH ‘16
27
9 and 1 + x < 1, "x ∈(0, 1) . 1 + x 3 Now 1 + x 8 > 1 + x 9 and 1 + x 4 < 1 + x 3, " x ∈ (0, 1) 1 + x 8 1 + x 9 , " x ∈(0, 1) ⇒ I 1 > I 2. > Tus 4 3 1 + x 1 + x b−c
3. (c) : Note that
∫
c
f (x + c) dx = − ∫ f (x) dx
0
b
1
1 4. (d) : Te given integral = 1 + e1/ x −1 1 1− e e = − = 1+ e 1+ e 1+ e 5. (c) : L(x ) = ln|x |
⇒ L(xy ) = ln|xy | = ln |x | | y | = ln|x | + ln| y | = L(x ) + L( y )
6. (a) : e
x2
x
⇒
1
∫ e
x2
1
dx < ∫ e xdx = e − 1
0
0
7. (d) : A differential equation is called linear if its
dy d 2 y , ,..... are not multiplied degree is 1 and y , dx dx 2 among themselves.
8. (a) : If y = mx + c is a family of parallel lines ( m being
fixed) then orthogonal trajectory is y = −
which is again a family of parallel lines. 9. (c) : g (x ) = f ′(x ) \ g ′(x ) = f ′′(x ) = – f (x ) h′(x ) = 2 f (x ) f ′(x ) + 2 g (x ) g ′(x ) = 2 f (x ) f ′(x ) + 2 f ′(x ) · {– f (x )} = 0 \ h(x ) = constant = 2. …(1) 10. (b) : Given curve is xy + ax + by = 0 dy dy \ y + x + a + b = 0 dx dx dy −a + y = dx b + x dy a +1 =− = 2 at (1, 1) \ a + 2b = –3 dx b +1 Also (1, 1) lies on (1) \ a + b = –1 b = –2, a = 1 a + b − 1 1 ⇒ = = −2 2 ab
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[email protected] | www.mtg.in |MARCH ‘16
1 x + c1, m
Your logo here
1.
m +n
2
m 2.
3.
4.
5.
2
If m and n are positive integers such that
+ mn + n
=
2
49
then m + n must be
An equilateral triangle of side length 2 units is inscribed in a circle. Te length of a chord of this circle which passes through the midpoints of two sides of this triangle is How many ways are there of walking up a flight of 10 stairs if you take either one or three stairs with each step? A straight line joins two opposite vertices P and Q of a cube of side length one metre and M is any other vertex. What is the distance, in metres, from M to the closest point on the line PQ? In a soccer tournament eight teams play each other once, with two points awarded for a win, one point for a draw and zero for a loss. How many points must a team score to ensure that it is in the top four (i.e. has more points than at least four other teams)?
2
m + n − mn = 4k 0≤ − (16k2 − 49k ) 2 2 = 49k − 12k , from which k < 4 follows. Hence k = 4 (since it must be an integer), and from 2
2
+ n = 16 and mn = 6, n = 10, and m + mn + n = 196
m
follow. Indeed, m +n 2
m 2.
+ mn + n2
=
16 4 = . 196 49
Let the circumcircle of the equilateral Δ ABC have centre O, (0, 0) and radius r , Join A to D, the midpoint of BC , then AD passes through O and is perpendicular to BC . Draw OB. Let the chord LM cut the sides AB and AC of Δ ABC at X and Y . Ten LXYM is parallel to BC .
SOLUTIONS 1.
Assume that m + n = 4k, m2 + mn + n2 = 49k. Ten 2
2
2
2
2
m + 2mn + n = (m + n) = (4k) = 16k ,
hence mn = 16k2 – 49k. Since mn > 0, from 16k2 – 49k > 0 we find that k > 3. Since we also have the identity 2
m + n
mn =
2
m −n − 2
2 2 m−n ≥ 0, 2
we also find that MATHEMATICS TODAY
|MARCH ‘16
29
1
OD = tan30° =
Ten D = 0,
−
4.
3 1
, 3
are 1, 2 and 3 , and is right angled at M . We need to know the length o the altitude MR. Since triangles MPQ and RPM are similar, it can be seen
2 and 3 −1, − 1 B= and the equation o the circle is 3
A = 0,
4 3 Now X is the mid-point o AB so 1 1 2 1 1 1 − = − , − X = − , 2 2 3 3 2 2 3 o find the x – coordinates o L and M , substitute 4 the y - coordinate of X in the equation x 2 + y 2 = , i.e. 3 2 1 4 x 2 + =3 2 3 1 5 2 4 − = ⇒ x = 3 12 4 ⇒
x = ±
5 2
5 and the 2 5 x-coordinate o M is , so the length o LM is 2 5 5 + = 5. 2 2 Tus the x -coordinate o L is
3.
Te problem is equivalent to that of writing 10 as a sum of a 1s and 3s, for example, 10 = 1 + 3 + 1 + 3 + 1 + 1. No. No. of 3s of 1s
30
−
0
10
1
7
2
4
3
1
Pattern
1111111111; 31111111, 13111111, ..., 11111113; 331111, 133111, ..., 111133; 313111, 131311, 113131, 111313; 311311, 131131, 113113; 311131, 131113; 311113; 1333, 3133, 3313, 3331; otal
MATHEMATICS TODAY
|MARCH ‘16
Number of Expressions
1 8 5 4 3 2 1 4 28
6 . 3 2 3 3 Since there are 8 teams, there are 7 rounds o our matches and thus a total o 7 × 8 = 56 points available. that
5.
x 2 + y 2 = .
If we draw the triangle MPQ we see that its sides
MR
=
1
, so MR =
2
=
Consider a team with 10 points. It is possible to have 5 teams on 10 points and 3 teams on 2 points when each o the top 5 draws with each other, each o the bottom 3 draws with each other and each o the top 5 wins against each o the bottom 3. So 10 points does not guarantee a place in the top 4. Consider a team with 11 points. I this team was fifh then the number o points gained by the top 5 teams is > 55. Tis is impossible as the number o points shared by the bottom 3 teams is then 1, as these 3 teams must have at least 3 × 2 = 6 points between them or the games played between themselves. Hence 11 points is sufficient to ensure a place in the top 4. Tus 11 points are required. nn
Form IV
1. Place of Publication 2. Periodicity of its publication 3. Printer’s and Publisher’s Name Nationality Address
: : : : :
4. Editor’s Name Nationality Address
: : :
New Delhi Monthly Mahabir Singh Indian Mathematics Today, 406, Taj Apartment, New Delhi - 110029. Anil Ahlawat Indian Mathematics Today, 19, National Media Centre, Gurgaon Haryana - 122002 Mahabir Singh 406, Taj Apartment New Delhi
5. Name and address of : individuals who own the newspapers and partners or shareholders holding more than one percent of the total capital I, Mahabir Singh, hereby declare that particulars given above are true to the best of my knowledge and belief. Mahabir Singh Publisher
MATHEMATICS TODAY | MARCH ‘16
31
*ALOK KUMAR, B.Tech, IIT Kanpur
1.
2.
−4 3 Let A = then A482, A700, A345 are −7 5 respectively (a) A – I , A, – A + I (b) A, – A, I (c) A – I , – A, –I (d) A – I , – A + I , –I If Sn =
n
∑ r =0
(a) n/2 3.
1 n n (c) n +1
5.
6.
n
, t n =
C r (b) n
n
7.
r
n
(b)
(−1)K −1 10 ∑ K .( C K ) = K =1 1 1 1 1 (a) 1 + + + + .... + 2 3 4 11 1 1 1 1 (b) 1 + + + + .... + 2 3 4 10 1 1 1 1 (c) 1 + + + + ..... + 2 3 4 9 1 1 1 1 (d) 1 + + + + ..... + 2 3 4 12 Te greatest and the least value of | z 1 + z 2| if z 1 = 24 + 7i and |z 2| = 6 are respectively (a) 31, 19 (b) 25, 19 10
, then t n = Sn r = 0 C r (c) 2n (d) n/3
∑
If C 0, C 1, ..... are the binomial coefficients in the n C r expansion of (1 + x )n, and ∑ (−1)r = 2 ( 1 ) r + r =0 n 1 ∑ r + 1 , then k is equal to r =0 (a)
4.
1
f (x ) = ax 3 + bx 2 + cx + d then the value of 3(a + b + c + d ) equals (a) 45 (b) –15 (c) –45 (d) none of these
8.
1 n +1
(c) 31, 25
(d) none of these
(d) none of these 9.
Assume that f is continuous on [a, b], a > 0 and differentiable on an open interval ( a, b). If f (a) f (b) , then there exist x 0 ∈ (a, b) such that = a b (a) x 0 f ′(x 0) = f (x 0) (b) f ′(x 0) + x 0 f (x 0) = 0 (c) x 0 f ′(x 0) + f (x 0) = 0 (d) f ′(x 0) = x 02 f (x 0)
10.
Let f (x ) = min{1, cosx , 1 – sin x } –p ≤ x ≤ p then (a) f (x ) is derivable at x = 0 (b) f (x ) has local maximum at x = 0 (c) f (x ) is derivable at x = p/2 (d) none of these
11.
Te remainder when (2222)5555 is divided by 7 is (a) 2 (b) 3 (c) 4 (d) 5
* Alok Kumar is a winner of INDIAN NATIONAL MATHEMATICS OLYMPIAD (INMO-91).
He trains IIT and Olympiad aspirants. 32
MATHEMATICS TODAY | MARCH ‘16
x − 1 y − 2 z − 3 and = = 2 3 K
x − 2 y − 3 z − 1 intersect at a point then the = = 3 2 K integer K = (a) –2 (b) –5 (c) 5 (d) 2
Number of ordered pairs of real numbers such that (a + ib)2008 = (a – ib) holds good, is (a) 2007 (b) 2008 (c) 2009 (d) 2010 For a curve y = f (x ), f ″(x ) = 4x at each point (x , y ) on it and it crosses the x -axis at (–2, 0) an angle of 45° with positive direction of x -axis. If
If the straight lines
It is given that complex numbers z 1 and z 2 satisy |z 1| = 2 and |z 2| = 3. I the included angle o their
12.
corresponding vectors is 60° then
18.
I the lengths o medians o a triangle are 2 units, 3 units and 4 units, then the area o the triangle is (a) greater than 4 3 (b) less than 4 3 (c) less than or equal to 4 3 (d) greater than or equal to 4 3
19.
Consider a curve ax 2 + 2 hxy + by 2 = 1 and a point P not on the curve. A line drawn rom P intersects the curve at points Q and R. I the product PQ·PR is independent o the slope o the line, then the curve is (a) parabola (b) circle (c) ellipse (d) hyperbola
z1 + z 2 can be z1 − z 2
x where ‘ ’ is a natural number x 7 then ‘x ’ equals to (a) 19 (b) 119 (c) 126 (d) 133
expressed as
Let p and q are positive integers. f is a unction defined or positive numbers and attains only positive values such that f (x f ( y )) = x p y q then (a) q = p2 (b) p = q2
13.
(c) p = q 14.
(d) p = 2q
In a D ABC , AB = AC , P and Q are points on AC and AB respectively such that CB = BP = PQ = QA. I ∠ AQP = q, then tan2q is a root o the equation (a) y 3 + 21 y 2 – 35 y – 12 = 0
1 −2 0 1 1 by observing orthogonally 20. Let A = 1 1 1 −1 among the column vectors o A one may obtain the inverse o A as 1 1 1 3 3 3 1 1 1 2 1 1 (a) A = −2 1 1 (b) − 6 6 6 0 1 −1 1 1 0 − 2 2
(b) y 3 – 21 y 2 + 35 y – 12 = 0 (c) y 3 – 21 y 2 + 35 y – 7 = 0 (d) 12 y 3 – 35 y 2 + 35 y – 12 = 0 15.
16.
Consider a amily o circles which are passing through the point (–1, 1) and are tangent to x -axis. I (h, k) is the centre o the circles, then 1 1 1 (a) − ≤ k ≤ (b) k ≤ 2 2 2 1 1 (c) 0 ≤ k ≤ (d) k ≥ 2 2 Direction cosine o normal to the plane containing z − 1 lines x = y = z and x − 1 = y − 1 = (where d d ∈ R – {1}), are 1 1 1 , 0, 1 (a) ,− , 0 (b) 2 2 2 2 (c)
17.
x
0, − 1 , 1 2 2
21.
1
−2 0 1 3 1 −3
10
)
2 + 3 3 + 6 5 is (a) 2 (b) 3
(c) 4
(d) 0
22.
Number o ways o giving away 10 different gifs to 5 students so that each get at least one and a particular student gets at least 5 gifs is (a) 5040 (b) 60480 (c) 65520 (d) 10080
23.
Te shortest distance between the lines
x
= 3i − 15j + 9k + l(2i − 7j + 5k) and r = (−i + j + 9k) + µ(2i + j − 3k) is r
x
∫ f (t)dt + x f ′′′(3) = ∫ x 3dx + f ′(1)∫ x 2dx + f ′′(2)∫ x dx −2
2 1 (d) 2 6 2
Te number o rational terms in the expansion o
(
(d) none o these
I f (x ) satisfies the relation x
2 2 2 (c) −2 1 1 0 −3 3
0
(c) x 3 – 5x 2 + 2x – 6 (d) 2 x 3 – 5x 2 – 2x + 6
3
then f (x ) = (a) x 3 + 5x 2 + 3x + 6 (b) x 3 + 5x 2 – 2x – 6
(a) 24.
34 (b)
3
(c) 3 3 (d) 4 3
How many real solutions does the equation x 7 + 14x 5 + 16x 3 + 30x – 560 = 0 have ? (a) 5 (b) 7 (c) 1 (d) 3 MATHEMATICS TODAY | MARCH ‘16
33
25.
(a)
2
1
0
0 2
(c) 2 ∫ loge (1 + x 2 )dx (d) ∫ loge (x
27.
2ab c4
(c)
− p , 0 2 3p (c) p, 2
30.
c
3
2ab 35.
2 5 1 r 1 If z + = 3 , then ∑ z + r = z z r =1 (a) 8 (b) 10 (c) 12 (d) 15
36.
37.
For a twice differentiable function f (x ), g (x ) is defined as g (x ) = ( f ′(x ))2 + f (x ) f ′′(x ) on (a, e). If a < b < c < d < e, f (a) = 0, f (b) = 2, f (c) = –1, f (d ) = 2, f (e) = 0, then the minimum number of roots of the equation g (x ) = 0 is (a) 4 (b) 5 (c) 6 (d) 7 32. If A, B, C are angles in a D ABC and
1 log 2 2
(b)
(c)
p 1 − log 2
(d)
4
2
p p p 1 , sin B − 4 sin C − 4 = 4 2 2
S tan A tanB = 34
MATHEMATICS TODAY | MARCH ‘16
then
2
p 2
− log 2 + log 2
4x 2 + 3x + 5 If lim exists then x →∞ x + 1 + x k −1 (a) k = 2 (b) k < 2 (c) k ≥ 3 (d) k < –3 OA is the perpendicular drawn from the centre x 2 y 2
(a) a2
(b) a a2 + b2
(c) b2
(d) b a2 + b2
If three distinct normals can be drawn to the parabola y 2 – 2 y = 4x – 9 from the point (2a, 0) then range of values of a is (a) no real values possible (b) (2, ∞) (c) (–∞, 2) (d) none of these xp + y
x
y
Te determinant yp + z
y
z
= 0 , if
0
31.
sin A −
p
+ = 1 , to the tangent at any a 2 b2 point P on the ellipse. If the normal to the ellipse at the point P meets the X -axis at B , then (OA) · (PB) is
p 2
From a point (h, k) three normals are drawn to the parabola y 2 = 4ax . angents are drawn to the parabola at the feet of the normals to form a triang le. Ten the centroid of the triangle is (a) 2a − h , − k (b) (a, –k) 3 2 2a − h , 0 3a − h (c) (d) ,0 3 2
(a)
‘O’ of the ellipse
(b) (0, p)
0,
then the value of
sin x
0
34.
c5
(d)
dx
∫ 1 + sin x + cos x = log 2 ,
∫ 1 + sin x + cos x dx is equal to
− 4x + 5)dx
(d) 2ab 2ab Te equation sinx + x cosx = 0 has at least one root in (a)
29.
(b)
S cot A (d) (S cot A)–1 (b)
0
If a2x 4 + b2 y 4 = c6, then the greatest value of xy is c
If p/2
0
2
S tan A (S tan A)–1 p /2
0 Te number of ordered pairs of positive integers (a, b) such that LCM of a and b is 23571113 is (a) 2385 (b) 2835 (c) 3825 (d) 8325
(a)
28.
2
(c) 33.
2 ∫ loge (2 + x 2 )dx (b) ∫ loge (1 + x )dx 1
26.
(a)
1 2n 2 2 1/n Suppose S = log e S is ∏ (n + r ) , then nlim →∞ n4 r =1 equal to
xp + y
yp + z
(a) x , y , z are in A.P. (b) x , y , z are in G.P (c) x , y , z are in H.P. (d) xy , yz, zx are in A.P. 38.
∞) satisfying x x 2 2 g (1) = 1 and if ∫ 2xg (t )dt = ∫ 2 g (x − t )dt , then 0 0 g (x ) is A function g (x ) is continuous in [0,
(a) (c) x
x
1/ 2 (b) x 1+ 2 (d) x
39. Let f (x ) = cos2x + cos22x + cos23x . Number of values
of x ∈ [0, 2p] for which f (x ) equals the smallest positive integer is (a) 3 (b) 4 (c) 5 (d) none of these
40. Te radius of the largest circle with centre at
47. Ifn ∈ N , then the remainder when 37n + 2 + 16n + 1 + 30n
is divided by 7 is (a) 0 (b) 1
(c)
11a 3
(b)
11 a 3
(d)
internal and external angle bisectors of the triangle ABC at C and the side AB produced. If CP = CQ, then the value of (a2 + b2) is (where a and b and R have their usual meanings for D ABC )
11 a 3 11 a 3
2 41. If (2 − 3 sin x) sec x dx = 2 f (x) g( x) + c and f (x )
∫
is non constant function, then (a) f 2(x ) + g 2(x ) = 1 (b) f 2(x ) – g 2(x ) = 1 (c) f (x ) g (x ) = 1 (d) f (x ) = g (x ) 42. Tree distinct vertices are randomly chosen among
(c)
8 2 9
3 5
(d)
8 4 9
44. A license plate consists of 8 digits out of 10 digits
0, 1, 2, 3, ...., 9. It is called even if it contain an even number of 0’s.Te number of even license plates is 108 − 88 (a) 2
(b) 108 – 810
108 + 88 (c) 2
108 + 810 (d) 2
45. Te shape of surface of a curved mirror such that
light from a source at origin will be reflected in a beam of rays parallel to x -axis is (a) circle (b) parabola (c) ellipse (d) hyperbola 46.
Te area of the loop of the curve y 2 = x 4(x + 2) is [in square units ] 32 2 105 128 2 (c) 105 (a)
64 2 105 256 2 (d) 105
(d) 4 2R2
" x ≥ 0 be a non-negative continuous function. If f ′(x )cosx ≤ f (x )sinx " x ≥ 0, then the 5p value of f is 3 −1 3 −1 3 (a) e
(b)
3 +1 2
(c)
2
(b)
(c) 4R2
50. Let f (x ) ≥ 0
divisor of 1099 is an integer multiple of 10 88 is
3 25
(b) 2 2R2
y 2 = 4ax (a > 0) and x 2 + y 2 – 6ax + a2 = 0 is (a) (a, 0) (b) (–a, 0) (c) (–a, –a) (d) (–2a, 0)
43. Te probability that a randomly chosen positive
(a)
(a) 2R2
49. Te point of intersection of common tangents of
the vertices of a cube. Te probability that they are vertices of an equilateral triangle is (a) 3/7 (b) 4/7 (c) 1/7 (d) 6/7
2
(d) 5
48. Let P and Q be the respective intersections of the
(a, 0), (a > 0) that can be inscribed in the ellipse x 2 + 4 y 2 = 16a2 is (a)
(c) 2
2
(d) 0 SOLUTIONS
1. (c) : Using characteristic equation
−4 − l 3 =0 ⇒ −7 5 − l ⇒ l2 – l + I = 0 ⇒ A2 – A + I = 0 ⇒ A3 = –I
–20 – l +
l2 + 21 = 0
So, A482 = ( A3)160 A2 = (–I )160 ( A – I ) = A – I A700 = ( A3)233 · A = – A A345 = ( A3)115 = –I
2. (a) : Sn =
⇒
nSn =
⇒
t n Sn
=
n
n
∑n
r = 0
∑n
r = 0
n 2
n
1 C n−r
=
n
C n−r r =0 C n−r
n 3. (b) : We have (1 + x) =
⇒
(b)
⇒
n
r +1
x
∑ C r r + 1 =
r =0 n
n − r
∑ n
n +1
+n
C n−r r
n
∑ Cr x r
r = 0
(1 + x ) − 1 (n + 1)
x r 1 (1 + x )n+1 − 1 = C ∑ r r + 1 n + 1 (1 + x ) − 1 r = 0 MATHEMATICS TODAY | MARCH ‘16
35
⇒
2 26 \ 3(a + b + c + d) = 3 + 0 − 7 − = – 45 3 3
C r 0 r ∑ (r + 1) ∫ x dx r = 0 −1 n
0 2 n−1 ) ( ( 1 ) ( 1 ) ... ( 1 ) x x x dx = + + + + + + ∫ n + 1 −1 1
= ⇒
1
n
0
r = 0
−1
(1 + x)r dx ∑ ∫ n +1 n
(−1)r C r
1
n
1 = ∑ ∑ 2 n + 1 r =0 r + 1 r = 0 (r + 1)
f (x ) x as f ( x ) and x are differentiable hence g ( x ) is also differentiable. f (b) f (a) Now, g (a) = and g (b) = b a f (a) f (b) Since, = a b \ g (a) = g (b) Hence Rolle’s theorem is applicable for g (x ) \ ∃ some x 0 ∈ ( a, b) Where g ′(x ) = 0 x f ′(x0 ) − f (x 0) xf ′(x) − f (x) g ′(x 0 ) = 0 =0 But g ′(x ) = ; x 02 x 2 x 0 f ′(x 0) = f (x 0) 4. (a) : Consider a function g (x ) =
= a − ib , hence z 2008 = z | z |2008 = | z | = | z | ⇒ | z |(| z |2007 − 1) = 0 ⇒ |z | = 0 or |z | = 1 If | z | = 0 ⇒ z = 0 ⇒ (0, 0) If | z | = 1, z 2009 = zz = | z |2 = 1 ⇒ 2009 values of ‘z ’ \ otal values of ‘z ’ are 2010 6. (c): f ′′ (x ) = +4x f ′(x ) = +2x 2 + c 5. (d) : Let z = a + ib
⇒ \
⇒
z
1 = +8 + c1 (slope is 45°) c1 = –7 f ′(x ) = +2x 2 – 7
2x 3 f (x ) = + − 7x + c2 (passes through (–2, 0)) 3 26 ⇒ c2 = − 3 3 2x 26 − 7x − f (x ) = + 3 3 and given f ( x ) = ax 3 + bx 2 + cx + d 2 26 ⇒ a = + , b = 0, c = −7, d = − 3 3 36
MATHEMATICS TODAY | MARCH ‘16
7. (b) : Required value is
10 C 10 C1 10C 2 10C3 ...... − + 1 2 3 10 to find which, consider (1 – x )10 = 10C 0 – 10C 1x + 10C 2x 2 .... + 10
⇒ ⇒
10C x 10 10
(1 − x )10 − 1 = −[10C1 − 10C2x + ....... − 10C10 x 9] x
1
1 1 − (1 − x )10 ∫ x dx = ∫ [10C1 − 10C2x + .... − 10C10 x9] dx 0 0
=
10
C 1 −
10
C 2 2
+
10
C3 3
− .... −
10
C 10 10
o find L.H.S. consider 1
1 − (1 − x )n I n = ∫ dx x 0
\ \
I n+1 =
1 n +1
⇒
1
1 In+1 − In = ∫ (1 − x)n dx = n +1 0
+ I n
1
1 − (1 − x )10 1 1 1 I 10 = ∫ dx = + I9 = + + I8 ≈ ..... 10 10 9 x 0
=
1 1 1 + + + ...... + 1 10 9 8
8. (a) : |z 1 + z 2| = |z 2 – (–24 – 7i)|
|z | = 6, C (–24 – 7i) OC = 25, CA = OC – OA = 25 – 6 = 19 and CB = OC + OB = 25 + 6 = 31 y B O C(–24
–7i )
x
A
9. (b) : Te lines intersect
⇒ they are coplanar
1 1 −2 ⇒ k 2 3 =0 3 k 2 ⇒ 1(4 – 3k) – 1(2k – 9) – 2(k2 – 6) = 0 ⇒ 2k2 + 5k – 25 = 0 ⇒ 2k2 + 10k – 5k – 25 = 0 ⇒ 2k(k + 5) – 5(k + 5) = 0 5 k = −5, 2
cos x −p ≤ x ≤ 0 10. (b) : f (x ) = 1 − sin x 0 < x < p / 2 cos x p / 2 ≤ x ≤ p − sin x −p ≤ x ≤ 0 f ′(x ) = − cos x 0 < x < p / 2 − sin x p / 2 ≤ x ≤ p Lf ′(0) = 0, Rf ′(0) = –1 ⇒ f ′(0) does not exist Lf ′(p/2) = 0, Rf ′(p/2) = –1 ⇒ f ′(p/2) does not exist f ′(x ) changes sign from positive to negative at x = 0. 11. (d) : 2222 ≡ 3 mod 7 (2222)3 ≡ 27 mod 7 ≡ –1 mod 7 ( 27 ≡ –1 mod 7) 5553 1851 ⇒ (2222) ≡ (–1) mod 7 ≡ –1 mod 7 (2222)2 ≡ 9 mod 7 (2222)5555 ≡ –9 mod 7 ≡ 5 mod 7 ( –9 ≡ 5 mod 7)
12. (d) : By Cosine rule
= | z1 |2 + | z2 |2 +2 | z1 || z2 | cos 60°
| z1 + z2 |
= 19 = | z1 |2 + | z2 |2 −2 | z1 || z2 | cos 60°
| z1 − z2 |
= 7 z1 + z 2 19 133 = = 7 7 z1 − z 2 13. (a) : For x =
⇒
f ( y ) =
1 f ( y )
y q / p 1/ p
⇒
f (1) =
y q
( f ( y )) p
, y = 1 ⇒ f (1) = 1
⇒ f ( y ) = y q/ p
( f (1)) f (xy q/ p) = x p y q For y = z p/q ⇒ f (xz ) = x pz p ⇒ f (x ) = x p q = p ⇒ q = p2 p q 14. (c): ∠QAP = ∠QPA = 90° − 2 B
Q
A
P C
∠PQB = ∠PBQ = 180° – q q ∠BCA = ∠ABC = ∠BPC = 45 °+ 4 q q Now 90° − + (2q − 180°) + 45° + = 180° 2 4
5p ⇒ 7q = 5p ⇒ 4q = 5p – 3q ⇒ q= 7 ⇒ tan4q = –tan 3q 15. (d) : Let the circle will be (x – h)2 + ( y – k)2 = k2
⇒ (–1 – h)2 + (1 – k)2 = k2 ⇒ h2 + 2h + 2 – 2k = 0
1 k ≥ . 2 16. (a) : Let l , m, n be direction ratios of the normal \ l+m+n=0 and l + m + nd = 0 ⇒ n(1 – d ) = 0 ⇒ n = 0 \ Direction cosines are 1 , − 1 , 0 or − 1 , 1 , 0 2 2 2 2 ... (1) 17. (c) : f (x ) + f ′′′(3) = x 3 + f ′(1)x 2 + f ′′(2)x Diff. successively w.r.t ‘x ’, we have ... (2) f ′(x ) = 3x 2 + 2xf ′(1) + f ′′(2) ... (4) f ′′(x ) = 6x + 2 f ′(1) ... (3) f ′′′(x ) = 6 Put x = 1 in (2) x = 2 in (3), x = 3 in (4) f ′′′ (3) = 6, f ′(1) = 3 + 2 f ′(1) + f ′′(2) f ′′(2) = 12 + 2 f ′(1) Solving we get f ′′′(3) = 6, f ′(1) = –5, f ′′(2) = 2 18. (b) : If length of medians be ma, mb, mc then 2(ma + mb + mc) < 3(a + b + c) < 4(ma + mb + mc) ⇒ 6 < a + b + c < 12 ⇒ s < 6 s − a + s − b + s − c Also, [(s − a )(s − b)( s − c)]1/3 < <2 3 \ (s – a)(s – b)(s – c) < 8 \ s(s – a)(s – b)(s – c) < 48 ⇒ D < 4 3 19. (b) : Coordinate of any point on line may be taken as (a + r cosq, b + r sinq). If point lies on the curve then a ( a + r cosq ) 2 + 2h (a + r cosq ) ( b + r sin q) + b(b + r sinq)2 = 1 Q
h is real D ≥ 0
⇒
PQ ⋅ PR = r1 ⋅ r2 =
aa2 + 2hab + bb2 − 1 a cos2 q + b sin2 q + h sin2 q
It will be independent of q if a = b and h = 0. 20. (b) : If it is column wise orthogonal then take 1 1 1 1 −2 0 3 0 0 AT A = −2 1 1 1 1 1 = 0 6 0 0 1 −1 1 1 −1 0 0 2 Prefactor matrix will become inverse of A if 1 st row is divided by 3, 2nd by 6, 3 rd by 2. 21. (b) : Using multinomial theorem, x
x
x
2 3 1 10 ! 3 6 2 ∑ x ! x ! x ! 2 ⋅ 3 ⋅ 6 1 2 3 where x 1 + x 2 + x 3 = 10 MATHEMATICS TODAY | MARCH ‘16
37
For rational terms possibilities are x 1 = 4, x 2 = 6, x 3 = 0 x 1 = 10, x 2 = 0, x 3 = 0 x 1 = 4, x 2 = 0, x 3 = 6 So, three rational terms. 22. (c): otal number o ways o giving 10 gifs to 5 students so that each get atleast one and a particular student gets atleast 5 gifs. (2 + 1 + 1 + 1)! (1 + 1 + 1 +1)! 4 ! = 10C5 × 4 ! + 10C 6 = 65520 2 !1!1!1!× 3 ! 1!1!1 !1 !4 !
4i
16j
(a2 − a1) ⋅ (b1 × b2 ) b1 × b2
23. (d) : S.D =
(a2 − a1) = −
max{a1, a2} = 3, max{b1, b2} = 7 and max( c1, c2) = 13. Hence {a1, a2} can be (0, 3), (1, 3), (2, 3), (3, 3), (3, 2), (3, 1), (3, 0) (one o the number is 3 and other number can be any where rom 0 to 3 ) giving us 7 choices. Similarly (b1, b2) has 15 choices and ( c1, c2) has 27 choices. Hence total number o choices = 7 × 15 × 27 = 2835. 27. (b) : Let z = xy
z 4 = (xy )4 =
1 2
(a2 x 4 )(b2 y 4 )
(ab) z is maximum when z 4 is maximum. i.e. when a2x 4 and b2 y 4 are equal. a 2 x 4 b2 y 4 c 6 = = 1 1 2 4 Maximum value o z =
+ (b1 × b2 ) = 16i +16j + 16k (b1 × b2 ) 1 = (i + j + k) | b1 × b2 | 3
1 12 S.D. = (−4 + 16) = =4 3 3 3 24. (c): f (x ) = x 7 + 14x 5 + 16x 3 + 30x – 560 \ f' (x ) = 7x 6 + 70x 4 + 48x 2 + 30 > 0 " x \ f (x ) is increasing " x \ It is odd degree polynomial. It has only one real root. 1 2n 2 2 1/n 25. (d) : S = ∏ (n + r ) n4 r =1 2n 1 log S = ∑ log(n2 + r 2 ) − 4 log n r =1 n
= ∑ log =1 1 2n
n
r
2
n
2 1 + − 4 log r
n
n
2n 1 2n r = ∑ 2 log n + ∑ log 1 + n r =1 n r =1
1 2n r = 4 log(n) + ∑ log 1 + n r =1 n
2
2
− 4 log(n)
− 4 log(n) r 2 1 2n \ lim log S = lim ∑ log 1 + n→∞ n→∞ n r =1 n 2
2
0
0
= ∫ log(1 + x 2) dx = ∫ log(x 2 − 4x + 5)dx a a ( ) ( ) ∫ f x dx = ∫ f a − x dx 0 0
a b c a b c 26. (b) : a, b are actors o the orm 2 15 111 1 , 2 2 5 211 2
where a1, b1, c1, a2, b2, c2 are non negative integers. Since LCM o a, b is 23 57 1113, 38
MATHEMATICS TODAY | MARCH ‘16
Maximum value o z = 28. (b) : Let g (x) =
c6 c6 (ab)2 2 2 1
c3
2ab
x (sin x + x cos x) dx 0
∫
g (0) = 0 = g (p) By Rolle’s Teorem at least one root ∈ (0, p) For g ′(x ) = 0 i.e. sin x + x cos x = 0. 29. (d) : Equation o normal rom ( h, k) is am3 + (2a – h) m – k = 0 2a − h k , Πt 1 = St1 = 0 , St1t 2 = a a Te vertices o the triangle ormed by the tangents are A(at 1t 2, a(t 1 + t 2)), B(at 2t 3, a(t 2 + t 3)), C (at 3t 1, a(t 3 + t 1)) 2a a Centroid G = St1t 2, St 1 3 3 2a − h ,0 = 3 30. (a) : Let z = pcosq 1 1 z + = 3 ⇒ p + cos q = 3 p z p − 1 sin q = 0 ⇒ q ≠ 0 and p = 1
p p
2 1 pr n ⇒ q = ⇒ z + n = 4 cos2 6 6 z z n 1 n 4 cos2 p cos2 p cos2 p ∑ + n = 6 + 3 + 2 z 2p 5p + cos2 + cos2
3
=8
6
31. (c): f (b) f (c) < 0, f (c) f (d ) < 0
f (x ) = 0 has atleast 4 roots a, c1, c2, e [by intermediate value property, f (x ) = 0 has at least one root c1 in (b, c)] And at least one root c2 in (c, d ) By Rolle’s theorem f ′(x ) has 3 roots in (a, c1), (c1, c2), (c2, e) \ f (x ) f ′(x ) = 0 has at least 4 + 3 = 7 roots d ⇒ g (x ) = ( f (x) f ′(x )) = 0 has at least 6 roots dx 32. (a) : Given : (sin A – cos A)(sinB – cosB)(sinC – cosC ) = 1 ⇒ (tan A – 1)(tanB – 1)(tanC – 1) = sec A secB secC ⇒ (∑ tan A − ∑ tan A tan B + ∑ tan A − 1) = p sec A ⇒ 2 ∑ tan A − ∑ tan A tan B = 1 + p sec A ... (i) But in a D ABC , S tan A tanB = 1 + psec A ... (ii) ⇒ 2 ∑ tan A = 2 ∑ tan A tan B ⇒ ∑ tan A tan B = ∑ tan A p /2 sin x 33. (c): If a = ∫ dx then x x + + 1 sin cos 0 p /2 cos x a = ∫ dx 1 sin x cos x + + 0 p /2 p \ log 2 + 2a = ∫ dx = 2 0 4x 2 + 3x + 5 34. (c): lim exist when in denominator x →∞ x + 1 + x k −1 power of x is more. i.e., k – 1 ≥ 2 ⇒ k ≥ 3. 35. (c): P = (acosq, bsinq)
x y Te tangent at P is cos q + sin q = 1 a b ab OA = a2 sin2 q + b2 cos2 q ax by Normal at P is − = (a2 − b2 ) cos q sin q
(a2 − b2 ) B= cos q, 0 a 2
(PB)
2
=b
2
sin
2 (a2 − b2 )cos q q + a cos q − a
b a2 sin2 q + b2 cos2 q PB = a (OA)·(PB) = b2 36. (b) : Given parabola can be rewritten as
( y – 1)2 = 4(x – 2)
Equation of any normal to the given parabola is y – 1 = m(x – 2) – 2m – m3 ⇒ y = mx – 4m – m3 + 1 Since, this passes through (2a, 0) \ m3 + 2m(2 – a) – 1 = 0 will have three distinct and real values of m. Also, 3m2 + 2(2 – a) = 0 will have all real and distinct root ⇒ a > 2 37. (b) : Using C 1 → C 1 – pC 2 – C 3 0
x
y
0
y
z
−( p2x + py) − ( py + z)
px + y
py + z
⇒
( y 2 – xz )( p2x + py + py + z ) = 0 x , y , z are in G.P. 2 x ... (1) x ∫ g 2(t )dt = 2 ∫ g (t )dt 0 0 x x g ( t ) dt g ( x t ) dt = − ∫ ∫ 0 0 x
38. (d) :
x
x
Differentiating, ∫ g 2(t )dt + xg 2(x) = 4 g (x)∫ g (t)dt
⇒
x
0
x
0
x ∫ g 2(t)dt + x 2 g 2 (x) = 4 xg(x)∫ g(t)dt 0
... (2)
0
From (1) & (2), x x 2 2 ∫ g (t )dt − 4xg (x)∫ g (t)dt + x 2 g 2(x ) = 0 0 0 x 2± 2 xg (x ) ⇒ ∫ g (t )dt = 0
Differentiating,
2
g ′(x ) 1 ± 2 = g (x) x
g (x) = c ⋅ x 1± 2 g (1) = 1 ⇒ c = 1 39. (d) : f (x ) = cos2x + cos22x + cos23x = 1 + cos2x + cos22x – sin23x = 1 + cos2x + cos5x ·cosx = 1 + cosx (cosx + cos5x ) = cosx cos2x cos3x = 0. 40. (b) : Any point on given ellipse is (4acosq, 2asinq),
1 cos q . 2 sin q Tis is a tangent to circle with centre at (1, 0) if 1 cos q 2 sin q 1 − × = −1 ⇒ cos q = 2 sin q 4 cos q − 1 3 \ Radius of the required circle is slope of tangent at which is
−
MATHEMATICS TODAY | MARCH ‘16
39
dy 2 tan f y dx = tan q = tan 2f = = 2 2 x 1 − tan f dy 1− dx Solving gives
a (4 cos q − 1)2 + 4 sin2 q
=a
2
4 − 12 + 4 1 − 1 = a 3 9
11 3
41. (a)
dy − x ± x 2 + y 2 or = dx y
42. (c): Te 3 vertices can be chosen among 8 is 8C 3 =
56 ways. For an equilateral triangle to be formed, sides must be chosen along face diagonals. Since through each 8×3 vertex, 3 face diagonals can be drawn, totally =8 3 (each equilateral triangle is counted 3 times here ) such triangles . 8 1 \ Required Probability = = 56 7 43. (a) : 1099 = 299 · 599 has totally (100) 2 divisors of the form 2a · 5b (0 ≤ a, b ≤ 99). Of these multiples of 1088 are of the form 2 a · 5b (88 ≤ a, b ≤ 99). So there are 12 · 12 = 12 2 choices for both a and b. 2
Hence, the probability required is
3 = 1002 25 12
⇒ ±
∑(
8−2K
C2K ) 9
=
8
8
C0 9
+
8
6
C2 9
+
8
C4 9
4
+
8
C6 9
+
C8
0
−2
−2
= 4 ∫ (z 2 − 2)2 z 2dz
x + 2 = z)
(where
0
z 7 4z 5 4z 3 2 256 2 = 4 − + = 7 5 3 0 105 47. (a) : 37n + 2 = (5 × 7 + 2)n + 2
= a multiple of 7 + 2n + 2 16n + 1 = (2 × 7 + 2)n + 1 = a multiple of 7 + 2n + 1 and 30n = (4 × 7 + 2)n = a multiple of 7 + 2 n 48. (c): a2 + b2 = 4R2(sin2(45° – q) + sin2(135° – q)
= 4R2(sin2(45° – q) + cos 2(45° – q)) = 4R2 C
K =0
(9 + 1)8 + (9 − 1)8 = 2
+ y
= ±dx
∫ x 2 x + 2dx
2
.
8
0
∫
2
2
2
= x +c
46. (d) : Area = 2 ydx = 2
(0 ≤ K ≤ 4) and 8-2K non zero digits, each of which has 9 choices. Tere are 8C 2K ways to choose 2K places for 0’s and ( 8C 2K )98 – 2K plates have exactly 2K zeroes. Terefore the required answer is 8
x
2
squaring gives y 2 = 2cx + c2 which is a parabola.
44. (c): For an even license plate, there must be 2K zeros
4
x 2 + y2
xdx + ydy
b
108 + 88 = 2
9 0° –
a 45 ° –
45. (b) : Let the shape of surface of a curved mirror be
A
as shown.
45° +
45°
P
B
45°
Q
49. (b) : Solving y 2 = 4ax , x 2 + y 2 – 6ax + a2 = 0, we B
P(x, y )
\
Points of contact are (a, 2a), (a, –2a) Hence tangents intersect at (–a, 0)
A
50. (d) : f (x ) ≥ 0 " x ≥ 0
y
(0, 0) x
By law of reflection, a = b, but f = Also q = a + f = 2b = 2f y Since tan q = , we get x 40
get
MATHEMATICS TODAY | MARCH ‘16
b
.... (1) 0
f ′(x )cosx – f (x ) sinx ≤ 0 ⇒ ( f (x )cosx )′ ≤ Let G(x ) = f (x ) cosx is a decreasing function. p 5p 5p ⇒ G ≥ G ⇒ G ≤ 0 2 3 3 5p ⇒ f ≤ 0 .... (2) 3 5p From (1) and (2), f = 0 3 nn
PRACTICE PAPER 2016 Time Allowed : 3 hours
Maximum Marks : 100 GENERAL INSTRUCTIONS
(i) (ii) (iii) (iv) (v) (vi) (vii)
All questions are compulsory. Please check that this Question Paper contains 26 Questions. Marks for each question are indicated against it. Questions 1 to 6 in Section-A are Very Short Answer Type Questions carrying one mark each. Questions 7 to 19 in Section-B are Long Answer I Type Questions carrying 4 marks each. Questions 20 to 26 in Section-C are Long Answer II Type Questions carrying 6 marks each. Please write down the serial number of the Question before aempting it. 9.
SECTION-A 1.
2. 3.
Evaluate : ∫ 0 [x]dx (where [x ] is greatest integer function).
dy + y sec2 x = tan x sec2 x ; y (0) = 1 dx
x + 3 y y 4 −1 If 7 − x 4 = 0 4 , find the values of x and y . Find the value of p, for which a = 3i + 2 j + 9k and
Solve the following differential equation : dy x + y = x log x ; x ≠ 0 dx Find the equation of the line passing through the point P (4, 6, 2) and the point of intersection of the line x − 1 = y = z + 1 and the plane x + y – z = 8. 3 2 7 By using properties of determinants, prove the following : x + 4 2x 2 x
1.5
OR
b = i + p j + 3k are parallel vectors. Find the integrating factor of the linear differential dx + (tan y )x = sec2 y equation dy cos x dx . Evaluate : ∫ x z − 5 x − 3 y + 2 If the equation of a line AB is = = , 4 1 −2 find the direction ratios of a line parallel to AB.
10.
4.
5.
6.
SECTION-B 7. 8.
Solve the differential equation :
Evaluate : ∫ x sin–1 xdx If y = 3 cos(log x ) + 4 sin(log x ), then show that 2
d y dy x 2 2 + x + y = 0. dx dx
11.
x + 4 3 dx 2 2 Prove that ∫ 2 = + log . 1 x ( x + 1) 3 3 Find a vector of magnitude 5 units perpendicular 2x
12. 13.
2 x = ( 5x + 4 ) ( 4 − x )2
x+4
2x
2x
to each of the vectors (a + b ) and (a − b ) where
i
j
a = + + k and b = i + 2 j + 3k .
MATHEMATICS TODAY
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41
OR
SECTION-C
Te scalar product of the vector i + j + k with a unit vector along the sum of vectors 2i + 4 j − 5k and li + 2 j + 3k is equal to one. Find the value of l. Prove the following : 1 2 1 3 tan−1 + tan −1 = cos −1 . 4 9 2 5 Let X be a non-empty set and * be a binary operation on P ( X ) (the power set of set X ) defined by A * B = A ∪ B for all A, B ∈ P ( X ) Prove that '*' is both commutative and associative on P ( X ). Find the identity element with respect to '*' on P ( X ). Also, show that f ∈ P ( X ) is the only invertible element of P ( X ).
14.
15.
OR
Show that the relation R in the set of real numbers, defined as R = {(a, b) : a ≤ b2} is neither reflexive, nor symmetric, nor transitive.
1 3 −2 16. Find the inverse of matrix −3 0 −5 , if exists, 2 5 0 17. 18.
using elementary row transformation. dy If (cos x ) y = (sin y )x , find . dx Te probability that student entering a university will graduate is 0.4. Find the probability that out of 3 students of the university: (i) none will graduate, (ii) only one will graduate, (iii) all will graduate.
Find the probability distribution of the number of white balls drawn in a random draw of 3 balls without replacement from a bag containing 4 white and 6 red balls. Also find the mean and variance of the distribution. 21. Every gram of wheat provides 0.1 gm of proteins and 0.25 gm of carbohydrates. Te corresponding values for rice are 0.05 gm and 0.5 gm respectively. Wheat costs ` 4 per kg and rice ` 6 per kg. Te minimum daily requirements of proteins and carbohydrates for an average child are 50 gms and 200 gms respectively. In what quantities should wheat and rice be mixed in the daily diet to provide minimum daily requirement of proteins and carbohydrates at minimum cost. Frame a L.P.P. and solve it graphically. What are the two uses of proteins? 22. Find the equation of the plane passing through the points (3, 4, 1) and (0, 1, 0) and parallel to the line x + 3 y − 3 z − 2 = = 2 7 5 20.
OR
Find the distance of the point (–2, 3, –4) from the x + 2 2 y + 3 3z + 4 line measured parallel to = = 3 4 5 the plane 4x + 12 y – 3z + 1 = 0 23.
OR
A football match may be either won, drawn or lost by the host country’s team. So there are three ways of forecasting the result of any one match, one correct and two incorrect. Find the probability of forecasting at least three correct results for four matches.
x − 5 | x − 5 | + a, if x < 5 if x = 5 is a continuous a + b, 19. If f (x ) = x − 5 + b, if x > 5 | x − 5 | function. Find the value of a and b. 42
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MARCH ‘16
A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m 3. If building of tank costs ` 70 per sq. metre for the base and ` 45 per sq. metre for sides, what is the cost of least expensive tank? OR
Show that the height of the cone of maximum volume that can be inscribed in a sphere of radius 12 cm is 16 cm. 24. 25.
26.
dy Solve : x + y – x + xy cot x = 0 (x ≠ 0) dx Find the ratio of shaded to unshaded region into which curve y 2 = 6x divides the region bounded by x 2 + y 2 = 16.
Consider f : R + → [–5, ∞) given by f (x ) = 9x 2 + 6x – 5. Show that f is invertible. Find the inverse of f .
SOLUTIONS 1.
1.5
1
1.5
∫ 0 [x]dx = ∫0 0dx + ∫ 1 1dx = 0 +
[ x]11. 5
= 1.5 – 1 = 0.5
2.
x + 3 y y 4 −1 Given = 0 4 x 7 − 4 ⇒
4.
5.
Since a || b , therefore a = lb ⇒ 3i + 2 j + 9k = l(i + p j + 3k ) ⇒ l = 3, 2 = l p, 9 = 3l 2 or l = 3, p = 3 I.F. = e∫ tan y dy = elog |sec y | = sec y
cos x
I = 2 sin
∫ ye = ∫ tan x sec2 x e tan dx tan x
7.
−1 Let I = x sinI x dx
∫ II
− ∫
2
dx
2 1 − x x 1 1 − x 2 − 1 –1 dx ⇒ I = sin x + 2 2 2 1 − x 2 x 1 1 = sin–1 x + 1 − x2 dx − sin−1 x 22 2 2 x 1 –1 –1 = sin x – sin x + 2 2 1 −1 1 x 2 1 − + sin x + C x 2 2 2 2 x 1 1 = sin−1 x − sin−1 x + x 1 − x 2 + C 2 4 4 1 2 2 −1 = (2 x − 1)sin x + x 1 − x + C 4 We have y = 3cos(log x ) + 4 sin(log x ) ...(i) Differentiating (i) w.r.t.x, we get dy −3 sin(log x ) 4 cos(log x ) 2
∫
∫
8.
⇒
=
+
dx x x dy ⇒ x = –3 sin(log x ) + 4 cos(log x ) dx Differentiating again (ii) w.r.t.x, we get
x
= t ⇒ tan x = log t ⇒ e sec2 x dx = dt \ y etan x = log t dt = t log t − t (Integrating by parts) x x x ⇒ y etan = etan . tan x – etan + C …(i) Now, from given value x = 0, y = 1, we get 1 . e0 = e0. 0 – e0 + C ⇒ C = 2 Putting the value of C in (i), we get y etan x = etanx . tan x – etan x + 2. ⇒ y = tan x – 1 + 2e – tan x Let e
Te direction ratios of line parallel to AB is 1, –2 and 4.
2
Given differential equation, is
tan x
6.
x 2
dx
y × I.F. = Q × I.F. dx
x + C
x 2 I = sin–1x .
dy d 2 y dy = − y ⇒ x 2 2 + x + y = 0 dx dx dx
+ x 2
Solution is given by
1
∫
d 2 y
2 \ I.F. = e ∫ Pdx = e ∫ sec xdx = e tan x
dx Let x = t ⇒ dx = dt x 2 x I = ∫ cos t . 2dt ⇒I = 2 sin t + C
Let I =
⇒ ⇒
⇒
x 2
dy + y sec2x = tan x sec2 x ; y (0) = 1 dx dy It is linear form of the type + Py = Q dx Here, P = sec2 x and Q = tan x sec2 x
y = –1 and 7 – x = 0 x = 7, y = –1
3.
9.
⇒
d 2 y dy −3 cos(log x ) 4 sin(log x ) x 2 + = − x x dx dx
tan x
∫
OR
dy y dy + y = x log x ⇒ + = log x dx x dx
Given, x
Tis is linear differential equation of the type dy + Py = Q dx
1 Here P = , Q = log x x
Soution is given by
∫ y . x = ∫ x ⋅ log x dx x2 x 2 1 xy = log x . − . dx 2 ∫ 2 x
y × I.F. = Q × I.F. dx
⇒ ⇒
...(ii)
1
∫ dx Integrating factor I.F. = e x = e log x = x
⇒
xy =
x 2 log x 1 x 2
2
−
2 2
+ C
2 y = x log x − 1 + C
2
2
MATHEMATICS TODAY
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MARCH ‘16
43
10.
\
x − 4 = 4+8 x − 4 = 12
y − 6 z − 2 = 6 + 6 2 + 22 z − 2 y − 6 z − 2 . or x – 4 = y – 6 = = 2 12 24
x+4
11.
2x 2 x L.H.S. = 2x x + 4 2 x 2x 2 x x + 4 Applying C 1 → C 1 + C 2 + C 3, we get 5x + 4 2 x 2 x = 5x + 4 x + 4 2 x 5x + 4 2 x x + 4 aking 5x + 4 common from C 1, we get 1 2x 2 x = (5x + 4) 1 x + 4 2 x 1 2 x x + 4 Applying R2 → R2 – R1; R3 → R3 – R1, we get
12.
1 2x 2 x = (5x + 4) 0 4 − x 0 0 0 4 − x Expanding along C 1, we get = (5x + 4) (4 – x )2 = R.H.S. 3 dx Let L.H.S. = I = 1 2 x ( x + 1) Using partial fraction, we have 1 A B C = + + x 2 ( x + 1) x x 2 x + 1
∫
⇒
1
=
Ax( x + 1) + B (x + 1) + Cx 2
x 2 ( x + 1) x 2 ( x + 1) ⇒ 1 = ( A + C )x 2 + ( A + B)x + B Equating coefficient of similar terms, we have A + C = 0, A + B = 0, B = 1 44
On solving these three equations, we have A = –1, B = 1 and C = 1 3 3 1 1 1 dx = − + 2 + dx \ I = 1 2 1 x x x + 1 x ( x + 1)
x − 1 y z + 1 …(i) = = =l 3 2 7 Coordinates of any general point on line (i) is of the form ≡ (1 + 3l, 2l, –1 + 7l) For point of intersection (1 + 3l) + 2l – (7l – 1) = 8 ⇒ 1 + 3l + 2l – 7l + 1 = 8 ⇒ –2l = 6 ⇒ l = –3 Point of intersection ≡ (–8, –6, –22) \ Required equation of line passing through P (4, 6, 2) and Q (–8, –6, –22) is
Let
MATHEMATICS TODAY
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MARCH ‘16
∫
∫
3
1 = − log | x | − + log | x + 1 | x 1
1 + log 4 + 0 + 1 – log 2 3
= – log 3 –
2 + (2 log 2 – log 2 – log 3) ( log 4 = log 22) 3 2 2 2 = + (log 2 – log 3) = + log = R.H.S. 3 3 3 We have, a = i + j + k and b = i + 2 j + 3k =
13.
a + b = 2i + 3j + 4k and a − b = − j − 2k
\
i
Tus, (a + b) × (a − b) = 2 0
j
k
3 −1
4 −2
= i (–6 + 4) – j (–4 – 0) + k (–2 – 0)
= −2i + 4 j − 2k
\ |(a + b) × (a − b)| = (−2)2 + (4)2 + (−2)2 =
4 + 16 + 4 =
24 = 2 6
(a + b) × (a − b) \ Required vector = 5 |(a + b) × (a − b)| 5(−2i + 4 j − 2k) −5 10 5 = = i+ j− k
2 6
6
6
6
OR
Let a = i + j + k, b = 2i + 4 j − 5k and c = li + 2 j + 3k
Tus, b + c
\
= (2 + l ) i + 6j − 2k
b +c
= (2 + l )2 + ( 6 )2 + ( −2 )2
= 4 + l2 + 4l + 36 + 4 = l 2 + 4l + 44 b +c Now, According to question, a ⋅ = 1 b +c ( 2 + l ) i + 6 j − 2k ⇒ (i + j + k ) ⋅ =1 2 l + 4l + 44 ( )+6−2 ⇒ 2+l = 1 ⇒ l + 6 = l2 + 4l + 44 l2 + 4l + 44
Squaring both sides, we have
l2 + 36 + 12l = l2 + 4l + 44 ⇒ 8l = 8 ⇒ l = 1
14.
15.
2 1 L.H.S. = tan + tan–1 9 4 1 + 2 17 = tan–1 4 9 = tan–1 36 1 2 34 1 − . 4 9 36 1 1 1 = tan–1 = 2 tan −1 2 2 2 1 − 1 1 1 − A2 −1 4 1 1 − − = cos 2 tan A = cos 1 2 1 + A2 1 + 4 1 −1 3 = cos = R.H.S. 5 2 For any A, B, C ∈ P ( X ), we have A ∪ B = B ∪ A and ( A ∪ B) ∪ C = A ∪ (B ∪ C ) ⇒ A * B = B * A and ( A * B) *C = A* (B * C ) Tus, '*' is both commutative and associative on P ( X ). Now, A ∪ f = A = f ∪ A for all A ∈ P ( X ) ⇒ A * f = A = f * A for all A ∈ P ( X ). So, f is the identity element for '*' on P ( X ). Let A ∈ P ( X ) be an invertible element. Ten, there exists S ∈ P ( X ) such that A * S = f = S * A ⇒ A ∪ S = f = S ∪ A ⇒ S = f = A. Hence, f is the only invertible element of P ( X ).
Transitivity: Let a, b, c real numbers aRb ⇒ a < b2 and bRc ⇒ b ≤ c2 Considering aRb and bRc ⇒ a < c4 ⇒ / aRc For example, if a = 2, b = –3, c = 1 aRb ⇒ 2 ≤ 9 bRc ⇒ –3 ≤ 1 aRc ⇒ 2 ≤ 1 is not true. Hence R is not transitive
–1
OR
Given relation is R = {(a, b): a ≤ b2} Reflexivity: Let a ∈ real numbers. aRa ⇒ a ≤ a2 but if a < 1. 1 1 For example a = a ⇒ a2 = 4 2 Hence R is not reflexive. Symmetry: Let a, b ∈ real numbers aRb ⇒ a ≤ b2 But then b < a2 is not true \ aRb ≠ bRa For example, a = 2, b = 5 then 2 < 52 but 5 < 22 is not true. Hence R is not symmetric.
16.
Consider A =
1 −3 2
3 0 5
We write A = IA
1 3 −2 1 ⇒ −3 0 −5 = 0 2 5 0 0 1 3 −2 1 0 9 −11 = 3 2 5 0 0
−2 −5 0 0 0
1 0 A 0 1 0 0
1 0 0 1
[By performing R2 → R2 + 3R1]
1 3 −2 1 ⇒ 0 9 −11 = 3 0 −1 4 −2
0 0 1 0 A
1 0 10 −5 ⇒ 0 9 −11 = 3 0 −1 4 −2
0 3
1 0 10 −5 ⇒ 0 1 − 11 = 1 9 3 0 −1 4 −2
0
0 1 [By performing R3 → R3 – 2R1]
1 0 A
0 1
[By performing R1 → R1 + 3R3]
a
2
1 9 0
3
0 A 1
[By performing R2 →
1 ⇒ 0 0
0 1 0
10 −5 11 1 − = 9 3 25 5 − 9 3
0
3 0 A 1
1 R] 9 2
1 9 1 9 [By performing R3 → R3 + R2]
MATHEMATICS TODAY
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45
1 0 10 −5 0 3 11 1 1 ⇒ 0 1 − = 0 A 9 3 9 0 0 1 3 1 9 − 5 25 25
[By performing R3 →
5 0 3 − 1 0 10 2 4 11 A ⇒ 0 1 0 = − 5 25 25 0 0 1 3 1 9 − 5 25 25
[By performing R2 → R2 +
1 − 2 − 3 5 5 1 0 0 2 4 11 A ⇒ 0 1 0 = − 5 25 25 0 0 1 3 1 9 − 5 25 25
⇒ \ 18.
25
R3]
11 R] 9 3
[By performing R1 → R1 – 10R3] Hence, we obtain I = BA ⇒ B is inverse of A. 1 − 2 − 3 5 5 25 −10 −15 2 4 11 1 = −10 4 11 Hence A–1= − 5 25 25 25 15 1 9 − 3 1 9 − 5 25 25 y x 17. Given, (cos x ) = (sin y ) aking log on both sides, we get \ log (cos x ) y = log(sin y )x ⇒ y log(cos x ) = x log(sin y ) Differentiating both sides w.r.t.x , we get 1 d dy 1 d . = x . y . (cos x ) + log(cos x ). sin y dx dx cos x dx (sin y ) + log sin y .1 cos y dy sin x dy + log(cos x ). = x + log sin y ⇒ – y sin y dx cos x dx dy dy ⇒ – y tan x + log (cos x ) = x cot y + log sin y dx dx dy dy ⇒ log (cos x ). – x cot y = log (sin y )+ y tan x dx dx 46
MATHEMATICS TODAY
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MARCH ‘16
y [log (cos x ) – x cot y ] = log (sin y )+ y tan x dx dy log (sin y ) + y tan x
=
dx log (cos x ) − x cot y Let X denote the number of students who graduated.
Now, the probability that a student graduates, p = 0.4 q = 1 – p = 1 – 0.4 = 0.6 (i) P (none will graduate) = P ( X = 0) = nC 0 p0qn–0 = 3C 0 (0.4)0 . (0.6)3 = (0.6)3 = 0.216 (ii) P (only one will graduate) = P ( X = 1) = nC 1 p1qn–1 = 3C 1(0.4)1 . (0.6)2 = 3 × (0.4) × (0.6)2 = 0.432 (iii) P (all will graduate) = P ( X = 3) = nC 3. p3.qn–3 = 3C 3 (0.4)3 . (0.6)0 = (0.4)3 = 0.064 OR
We have, Te probability of correct result =
1 3
2 3 \ Probability of forecasting at least three correct results for four matches = Probability of 3 correct and one incorrect or all correct 4 1 2 1 = 4× × + 3 3 3 4 4 1 1 1 1 = (8 + 1) = × 32 = = 3 3 32 9 and probability of incorrect result =
x − 5 | x − 5 | + a , if , if a +b 19. We have f (x ) = x − 5 + b , if | x − 5 |
x < 5 x = 5 x > 5
Since f is continuous, it must be continuous at x = 5 x − 5 + a = –1 + a L.H.L. = lim f (x ) = lim x →5− x →5− −( x − 5) x − 5 R.H.L. = lim+ f (x ) = lim+ + b = 1 + b x →5 x → 5 ( x − 5) and f (5) = a + b Since, f (x ) is continuous at x = 5
\
L.H.L. = R.H.L. = f (5) –1 + a = 1 + b = a + b ⇒ –1 + a = a + b ⇒ b = –1 and 1 + b = a + b ⇒ a = 1 Hence, a = 1, b = –1 20.
Let X be the number of white balls. Terefore, we have 6
When X = 0, p1 =
C 3
10 4
When X = 1, p2 =
C 3
When X = 2, p3 =
1 6
C1 ×6 C 2 10
4
=
C 3
C2 ×6 C 1 10
4
C 3
=
1 2
=
3 10
Here, the shaded region is the required region on the basis of constraints and which is unbounded (shown in figure). Now, we find the value of Z at each corner point.
C 3
1 When X = 3, p4 = 10 = C 3 30 X
0
1
2
3
pi
1 6
1 2
3 10
1 30
Now, Mean(µ) = = 0×
= 0 × =
i
i
1 6 1 + 2 10 10
= +
=
6 5
2 i i
2
i i
1 1 3 1 36 + 1 × + 4 × + 9 × − 6 2 10 30 25
22.
36 14 = 25 25
Let x grams of wheat and y grams of rice should be mixed. Terefore, we have 0.1 x + 0.05 y ≥ 50
…(i)
0.25 x + 0.5 y ≥ 200
…(ii)
x , y ≥ 0
…(iii)
and Z min =
4 x 6 y + 1000 1000
On plotting equations (i), (ii) and (iii), we have the following graph.
A (800, 0)
3.2
B (400, 200)
2.8
C (0, 1000)
6
Minimum
Hence, cost will be minimum when 400 grams of wheat and 200 grams of rice are mixed in the daily diet, and minimum cost is ` 2.8 per gram. Proteins are used as a building material. It stores and transport molecules.
∑ X p − (∑ X p )
1 + 12 + 9 − 36 2 10 30 25
= 2− 21.
∑ p X
1 1 3 1 +1× + 2 × + 3 × 6 2 10 30
Variance =
Corner Point Value of Z
Equation of the plane passing through (3, 4, 1) is a(x – 3) + b( y – 4) + c(z – 1) = 0 …(i) Since this plane passes through (0, 1, 0) also \ a(0 – 3) + b(1 – 4) + c(0 – 1) = 0 or –3a – 3b – c = 0 or 3a + 3b + c = 0 …(ii) Since (i) is parallel to the line x + 3 y − 3 z − 2 = = 2 7 5 \ 2a + 7b + 5c = 0 From (ii) and (iii), we get
…(iii)
a b c = = =k 15 − 7 2 − 15 21 − 6
⇒
a = 8k, b = –13k, c = 15k Putting in (i), we have 8k(x – 3) – 13k( y – 4) + 15k(z – 1) = 0 MATHEMATICS TODAY
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51
⇒ ⇒
Differentiating (iii) again w.r.t. L, we get d 2C 1440 = 3 > 0 ∀ L > 0 dL2 L \ Cost is minimum when L = 2 From (i), B = 2 4 Minimum cost = 280 + 180 + 2 (from (ii)) 2 = 280 + 720 = ` 1000
8(x – 3) – 13( y – 4) + 15(z – 1) = 0 8x – 13 y + 15z + 13 = 0 which is the required equation of the plane. OR
x + 2 2 y + 3 3z + 4 = = =l 3 4 5 Any general point on the line is given by Let
4 l − 3 5l − 4 , 2 3 Now, direction ratios if a point on the line is joined to (–2, 3, –4) are 3l – 2,
OR
Let O be the centre of sphere of radius 12 cm and a cone ABC of radius R cm and h cm inscribed in the sphere. AP = AO + OP ⇒ h = 12 + OP ⇒ OP = h – 12 Now in right angled Δ OBP , we have BO2 = BP 2 + OP 2 ⇒ (12)2 = R2 + (h – 12)2 ⇒ 144 = R2 + h2 + 144 – 24h ⇒ R2 = 24h – h2
4 l − 9 5l + 8 , 2 3 Now the distance is measured parallel to the plane 4x + 12 y – 3z + 1 = 0
⇒
3l,
4 l − 9 – 3 × 5l + 8 = 0 2 3 ⇒ 12l + 24l – 54 – 5l – 8 = 0 ⇒ 31l – 62 = 0 ⇒ l = 2 \ Te point required is 4, 5 , 2 . 2 \
4 × 3l + 12 ×
\
Distance =
2
(4 + 2)
5 + − 3 + (2 + 4)2 2
1 289 17 units = = 4 4 2 Let the length and breadth of the tank be L and B. 4 …(i) \ Volume = 8 = 2 LB ⇒ B = L Te surface area of the tank, S = Area of Base + Area of 4 Walls = LB + 2(B + L)2 = LB + 4B + 4L Te cost of constructing the tank is C = 70(LB) + 45(4B + 4L) =
23.
2
1 3 1 dV p ⇒ V = p ( 24h2 − h3 ) ⇒ = ( 48h − 3h2 ) 3 dh 3 For maximum/minimum value of V , we have
36 + 36 +
4 + 180 4 + L L L 4 ⇒ C = 280 + 180 + L L
Volume of cone, V
dV = 0 ⇒ 48h − 3h2 = 0 dh ⇒ h(48 – 3h) = 0 ⇒ Either h =0 or h = 16 But height of cone cannot be zero. Terefore h = 16 cm.
= 70 L.
Now,
Differentiating (ii) both sides w.r.t. L, we get
52
MATHEMATICS TODAY
Hence, volume of cone is maximum when h = 16 cm …(iii)
For minimisation, dC = 0 dL 720 = 180 ⇒ L2 = 720 = 4 ⇒ 2 180 L |
MARCH ‘16
24.
⇒ L = 2
d dV p = ( 48 − 6h ) dh dh 3
2 p ⇒ d V = ( 48 − 6 × 16 ) = −16p < 0 dh2 h=16 3
…(ii)
720 dC = − 2 + 180 dL L
1 3
= pR2h = p ( 24h − h2 ) h
dy We have x + y – x + xy cot x = 0, x ≠ 0 dx dy 1 we have + cot x + y = 1 dx x
…(i)
2 × 4 3 − 2 3 + 8p 3 3 2 3 8p + = 2 3 3 4 3 16 p + =
dy Tis is a linear equation of the form + Py = Q dx 1 Here, P = cot x + , Q = 1 x
= 2
1
∫ cot x + x dx
So, I.F. = e
= elog | sin x | + log | x |
= elog | x sin x | = | x sin x | = x sin x (x sin x is always +ve for any x ) Solution is given by
= ∫ Q × I.F. dx ⇒ y × x sin x = ∫ 1× x sin x dx ⇒ xy sin x = ∫ x sin x dx ⇒ xy sin x = x ( − cos x ) − ∫ − cos x × 1 ⇒ xy sin x = – x cos x + sin x + c
3
\ Area of unshaded region = 16p – 4 3 + 16p 3 3 \ Required ratio 4 3 16p 4 3 16p 16 p − + + = 3 3 3 3 1 4p 4 3 16 p + + 3 3 3 = = 3 32 p 4 3 8 p − 1 −
y × I.F.
∴
dx
xy sin x = sin x – x cos x + c
3
1 C Hence, y = – cot x + is the required solution. x x sin x 25.
2
We have, y = 6x
Now, Area of the shaded region = 2(Area of OAM + Area of AMB) = 2
∫ 2 16 − x2 dx ∫0 4 x 3/2 2 x 16 x 2 1 − = 2 6 16 − x + sin + 2 4 2 3 / 2 0 2 p p 2 3/2 = 2 6 (2 ) + 8. − 12 − 8 × 2 6 3 4p 2 = 2 48 − 12 + 4 p − 3 3 2
6 xdx +
=
…(i)
and x 2 + y 2 = 16 …(ii) Area of the circle = p × 16 = 16 p sq. units On plotting these curves, we get the following figure.
4
3
26.
3
3
3
3 + 4 3p
8 3p − 3 Hence required ratio is (3 + 4 3 p) : (8 3 p – 3) Given f (x ) = 9x 2 + 6x – 5 Let x 1, x 2 ∈ R+ , such that f (x 1) = f (x 2) ⇒ 9x 21 + 6x 1 – 5 = 9x 22 + 6x 2 – 5 ⇒ 9(x 21 – x 22) + 6(x 1 – x 2) = 0 ⇒ (x 1 – x 2) {9(x 1 + x 2) + 6} = 0 ⇒ x 1 – x 2 = 0 as 9( x 1 + x 2) + 6 > 0 \ x 1 = x 2 So, f (x ) is one-one and we see that for every x ∈ R+, there is f (x ) ∈ [–5, ∞) So, f (x ) is onto. i.e., f (x ) is one-one and onto. Hence f (x ) is invertible. Now, let y = 9x 2 + 6x – 5 ⇒ 9x 2 + 6x – 5 – y = 0
⇒
x =
−6 ± 36 + 4 × 9(5 + y ) 2×9
As x ∈ R+
\
x =
=
\
−6 + 6 1+ 5 + y −6 + 6 6 + y = 18
18
−1 + 6 + y
f –1( y ) =
3
−1 + 6 + y 3 nn
MATHEMATICS TODAY
|
MARCH ‘16
53
CATEGORY-I
For each correct answer one mark will be awarded, whereas, for each wrong answer, 25% of total marks (1/4) will be deducted. If candidates mark more than one answer, negative marking will be done. 1.
en different letters of an alphabet are given. Words with five letters are formed from these given letters. Ten the number of words which have at least one letter repeated is (a) 69,760 (b) 24,320 (c) 99,777 (d) none of these
2.
Te interior angles of a regular polygon measure 120° each. Te number of diagonals of the polygon is (a) 9 (b) 15 (c) 44 (d) 33
3.
If (cot–1x )2 – 7(cot–1x ) + 10 > 0, then x lies in the interval (a) (cot5, cot2) (b) (–∞, cot5) ∪ (cot2, ∞) (c) (–∞, cot5) (d) (cot2, ∞)
4.
7.
5.
6.
(c) 6
8.
If 0 < a < p then the quadratic equation cos(a – 1)x 2 + x cosa + sina = 0, has (a) both roots imaginary (b) only one root imaginary (c) only one root irrational (d) none of these
If
p p q ∈ , and f (q) = sec 2q − tan 2q, then 4 2 p f − q = 4
(a) tanq (c) sec2q 9.
(d) 5
If z 1, z 2, z 3 represent the vertices of a triangle, then the centroid of the triangle is given by az1 + bz 2 + cz 3 z + z + z (a) (b) 1 2 3 3 a +b+c z z z (c) 1 2 3 (d) none of these 3
(sin x − cos x + 1) 1 (d) 2 2(sin x − cos x + 1)
(c)
(x + (x 3 − 1)1/2)5 + (x − ( x 3 − 1)1/ 2)5 is (b) 7
(b) cotq (d) tan2q
a cos A + b cos B + c cos C , a +b+c where a, b, c and A, B, C have their usual meaning, is
Te maximum value of
(a) sin A (c) sin 10.
11.
(b) 4 sin
C 2
MATHEMATICS TODAY | MARCH ‘16
A B C sin sin 2 2 2
(d) none of these
f (x ) lim = 2 where f (x) = min {sin [m]x,| x |} x → 0 x and [·] greatest integer, then (a) m ∈ {4} (b) m ∈ [4, 5] (c) m ∈ [4, 5) (d) m ∈ {5} x
1
2
0
x
1
If ∫ f (z)dz = x + ∫ zf (z) dz then ∫ f (x)dx equals
By : Sankar Ghosh, HOD(Math), Takshyashila. Mob : 09831244397 54
equivalent to 1 (sin x + cos x − 1) (a) (b) 2(sin x + cos x − 1) 2
Te degree of the expansion (a) 15
1 Te expression tan x + cot x + sec x + cosec x
2 3 3 (c) log3 (d) log 2 12. A line makes the same angle q with each of the x and z axis. If the angle b, which it makes with y -axis is such that sin 2b = 3 sin2q then cos2q equals (b) log
(a) 1 + x
(a) 3/5 p /3 13.
14.
(b) 1/5
(c) 2/3
sin x k dx = , then the value of k equals 4 p/6 sin x + cos x (a) p/12 (b) p/3 (c) p/2 (d) none of these log(1 + x 3) equals lim x → 0 sin3 x (a) 1 (b) 0 (c) –1 (d) none of these
In a G.P. of positive terms , any term is equal to the sum of next two terms, then common ratio of this G.P. is (a) cos18° (b) sin18° (c) 2cos18° (d) 2sin18°
24.
Solution set of inequality log10(x 2 – 2x – 2) ≤ 0 is (a) [−1, 1 − 3] (b) [1 + 3,
3]
(c) [−1, 1 − 3) ∪ (1 + 3, 3] (d) none of these 25.
In a survey it is to be found that 70% of employees like bananas and 64% like apples. If x % like both bananas and apples, then (a) x ≥ 34 (b) x ≤ 64 (c) 34 ≤ x ≤ 64 (d) all of these
26.
If a relation R is defined from a set A = {2, 3, 4, 5} to a set B = {3, 6, 7, 10} as follows (x , y ) ∈ R ⇔ x divides y . Expression of R–1 is represented by (a) {(6, 2), (10, 2), (3, 3)} (b) {(6, 2), (10, 5), (3, 3)} (c) {(6, 2), (10, 2), (3, 3), (6, 3), (10, 5)} (d) none of these
27.
If two arithmetic means A1, A2, two geometric means G1, G2 and two harmonic means H 1, H 2 are A + A2 inserted between any two numbers then 1 H1 + H 2 is GG (a) 1 2 (b) G1G2 H1H 2 H H (c) 1 2 (d) none of these G1G2
28.
(c) 2( x sin x + cos x ) + c
A test consist of 6 multiple choice questions each having four alternative answers of which only one is correct. Also only one of the alternatives must be marked by each candidate. Te number of ways of getting exactly four correct answers by a candidate answering all the questions is
(d) 2( x sin x − cos x ) + c
(a) 46 – 32 (b) 135
16.
Te domain of the function f (x ) = log4(log5(log3(18x – x 2 – 77))) is (a) x ∈ (4, 5) (b) x ∈ (0, 10) (c) x ∈ (8, 10) (d) none of these
17.
Te order and degree of the differential equation of all tangent lines to the parabola x 2 = 4 y is (a) 2, 2 (b) 3, 1 (c) 1, 2 (d) 4, 1
18.
Te area enclosed between the curves y 2 = x and y = |x| is (a) 1/6 (b) 1/3 (c) 2/3 (d) 1 Let f be the differentiable for " x . If f (1) = –2 and f ′(x ) ≥ 2 for [1, 6], then (a) f (6) < 8 (b) f (6) ≥ 8 (c) f (6) = 5 (d) f (6) < 5 y
If xy = e − e then
d 2 y dx 2
(a) 1/e (c) 1/e2 21.
23.
∫
cos2 A + cos2(B – A) – 2cos A cosB cos( A – B) = (a) cos2 A (b) sin2 A (c) sin2B (d) cos2B
20.
If f (x + y ) = f (x ) + f ( y ) – xy – 1 " x , y ∈ R and f (1) = 1, then the number of solutions of f (n) = n, n ∈ N is (a) 0 (b) 1 (c) 2 (d) none of these
(d) 2/5
15.
19.
22.
∫ cos (a) (b)
−
equals when x = 0
(b) 1/e3 (d) none of these
xdx is equal to sin x +c 2 x x sin x + cos x + c
(c) 55
(d) 120
MATHEMATICS TODAY | MARCH ‘16
55
29.
30.
31.
32.
2 2 An ellipse of eccentricity is inscribed in 3 a circle and a point within the circle is chosen at random. Te probability that this point lies outside the ellipse is (a) 1/9 (b) 4/9 (c) 1/3 (d) 2/3 Consider points A(3, 4) and B(7, 13). If P be a point on line y = x such that PA + PB is minimum, then co-ordinate of P is 13 13 2 12 (a) , (b) , 7 7 7 7 23 23 (c) , (d) none of these 7 7
If the difference between the roots of the equation ax 2 + ax + b = 0 is equal to the difference between the roots of the equation x 2 + bx + a = 0 (a ≠ b), then (a) a + b = 4 (b) a + b = –4 (c) a – b = 4 (d) a – b = –4
If
D=
q 2
1
− cos
1 cos
q
2
− cos
q 2
lies in the interval
35.
Te values of a for which the system of equations ax – 3 y + z = 0, x + a y + 3z = 1, 3x + y + 5z = 2, does not have unique solution are 11 −11 (a) −1, (b) −1, 5 5 −11 11 (c) 1, (c) 1, 5 5 Te equation of the plane perpendicular to the line x − 1 y − 2 z + 1 and passing through the point = = 1 2 −1 (2, 3, 1) is ^
^
+ 2 k^ ) = 1 (b) r ⋅ (^i − ^j + 2 k^ ) = 1 ^ ^ ^ (c) r ⋅ (i − j + 2 k ) = 7 (d) none of these (a) r ⋅ (i + j
56
MATHEMATICS TODAY | MARCH ‘16
(a) 2
(b) –1
(c) 1
(d)
1 6 3
37.
Te coordinates of a point on the parabola y 2 = 8x whose focal distance is 4 are 1 (a) , ± 2 (b) (1, ± 2 2 ) 2 (c) (2, ± 4) (d) (± 2, 4)
38.
Te derivative of the function 1 (2 cos x − 3 sin x ) f ( x ) = cos −1 13 1 + sin−1 (2 sin x + 3 cos x ) 13 with respect to 1 + x 2 is (a) 2x 2 (c) 1 + x 2 x
(b) 2 1 + x 2 2x (d) 1 + x 2
39.
Let f ′′(x ) be continuous at x = 0 and f ′′(0) = 4. Te 2 f (x) − 3 f (2x) + f (4x ) value of lim is equal to x →0 x 2 (a) 11 (b) 2 (c) 12 (d) none of these
40.
If b1, b2, b3, ... belongs to A.P. such that b1 + b4 + b7 + ... + b28 = 220 then the value of b1 + b2 + b3 + ... + b28 = (a) 616 (b) 308 (c) 2,200 (d) 1,232
41.
Te middle term in the expansion (1 – 3x + 3x 2 – x 3)2n is 6n ! n 6n ! 3n x (a) (b) x 3n ! 3n ! 3n ! 6n ! (−x )3n (c) (d) none of these 3n ! 3n !
42.
Te solution of the (x 2 + y 2)dx – 2xydy = 0 is
1 1 2 (a) [2, 4] (b) [0, 4] (c) [1, 3] (d) [–2, 2] 34.
1
q
| a + b + c | = 6 then | a | is equal to
Te equation (x + y – 6)(xy – 3x – y + 3) = 0 represents the sides of a triangle then the equation of the circumcircle of the triangle is (a) x 2 + y 2 – 5x – 9 y + 20 = 0 (b) x 2 + y 2 – 4x – 8 y + 18 = 0 (c) x 2 + y 2 – 3x – 5 y + 8 = 0 (d) x 2 + y 2 + 2x – 3 y – 1 = 0
cos 33.
36.
a, b , c are three vectors of equal magnitude. Te angle between each pair of vectors is p/3 such that
(a)
x x 2 + y 2
y 2 − x 2 (c) x
differential
=c
x 2 + y 2 (b) x
=c
=c
x 2 − y 2 (d) x
=c
of
equation
43.
Te number of vectors of unit length perpendicular to vector a ≡ (5, 6, 0) and b ≡ (6, 5, 0) is (a) 1 (b) 4 (c) 3 (d) 2
51.
44.
If the eccentricity of the hyperbola x 2 – y 2 sec2 q = 4 is 3 times the eccentricity of the ellipse x 2sec2q + y 2 = 16 the value of q equals (a) p/6 (b) 3p/4 (c) p/3 (d) p/2
45.
46.
47.
Angle subtended by common tangents of two ellipses 4(x – 4)2 + 25 y 2 = 100 and 4( x + 1)2 + y 2 = 4 at origin is (a) p/3 (b) p/4 (c) p/2 (d) none of these 5z 2 2z + 3z 2 is purely imaginary, then 1 is 11z 1 2z1 − 3z 2 11 5 37 (a) (b) (c) 1 (d) 5 11 33
52.
53.
If
54.
0 2
x
2
4
1 − 10 6 3 x − 7 1 2 0 −2
Let f(x ) = f (x ) + f (1 – x ) and f ′′(x ) < 0 in 0 ≤ x < 1, then (a) f(x ) decreases in (0, 1) (b) f(x ) increases in (0, 1) (c) f(x ) decreases in (0, 1/2) (d) none of these
If z ≠ 0, then
50.
Te number of tangents to the curve x 2/3 + y 2/3 = a2/3 which are equally inclined to the axes is (a) 4 (b) 3 (c) 2 (d) 1 Te odds against A solving a certain problem are 3 to 2 and the odds in favour of B solving the same are 2 to 1. Te probability that the problem will be solved if they both try, is 2 11 2 4 (a) (b) (c) (d) 3 15 5 5
∫ arg(− | z |) dx = (b) not defined (d) 50p
55.
Te number of values of q in [0, 2 p] that satisfy the equation 3cos2q + 13sinq – 8 = 0 is (a) 1 (b) 2 (c) 3 (d) 4
56.
If A = { a, b, c, d } and B = {x , y , z }, then which one of the following relations from A to B is not a mapping? (a) {(a, x ), (b, y ), (c, z ), (d , x )} (b) {(a, y ), (b, y ), (c, x ), (d , z )} (c) {(b, x ), (c, x ), (d , z ), (a, y )} (d) {(b, x ), (a, y ), (b, z ), (c, z )}
57.
In a moderately asymmetrical distribution, the mean is 18 and median 22, the value of mode is (a) 30 (b) 10 (c) 4 (d) none of these ^
If P = 2 ^i + ^j − k and Q = ^i + 3 k^ . If R is a unit vector then the maximum value of the scalar triple product [R P Q] is (a) –1 (b) 10 + 6
49.
50
(a) 50 (c) 0
assumes the value zero, then the value of x equals to, (a) –6, –4 (b) –6, 4 (c) 6, 4 (d) 6, –4 48.
Let P(n) : 2n > n " n ∈ N and 2k > k, " n = k , then which of the following is true? ( k ≥ 2) (a) 2k > 5k > 1 (b) 2k + 1 > 2k > k + 1 (c) 2k > 2(k + 1) > k (d) none of these
0
If the trace of the matrix
x − 5 3 A = −2 1
Six coins are tossed simultaneously. Te probability atleast one tail turns up is 1 63 (a) (b) 64 64 3 (c) (d) none of these 32 It is given that event A and B are such that 1 1 2 P(A) = , P(A B) = , P(B A) = then P (B) is 4 2 3 equal to (a) 1/3 (b) 2/3 (c) 1/2 (d) 1/6
58.
(c) 59.
59
(d)
60
If the coefficient of variation of some observations is 60 and their standard deviation is 20, then their mean is (a) 35 (b) 34 (c) 38.3 (d) 33.33 MATHEMATICS TODAY | MARCH ‘16
57
60.
If w is a cube root of unity then
66.
1 p tan w200 + 200 p + = w 4 1 (a) 1 (b) 2 (c) 0 (d) none of these CATEGORY-II
Every correct answer will yield 2 marks. For incorrect response, 25% of full mark (1/2) would be deducted. If candidates mark more than one answer, negative marking will be done. 61.
62.
63.
64.
65.
If m is A.M. of two distinct real numbers l and n (l , n > 1) and G1, G2, and G3 are three geometric means between l and n, then G14 + 2G24 + G34 equals (a) 4lmn2 (b) 4l 2m2n2 (c) 4l 2mn (d) 4lm2n Let a and b be the roots of the equation x 2 – 6x – 2 = 0. If an = an – bn, for n ≥ 1, then the a − 2a8 value of 10 = 2a9 (a) 3 (b) –3 (c) 6 (d) –6
67.
68.
(c)
58
1 − 3x 2
(d)
1 − 3x 2
MATHEMATICS TODAY | MARCH ‘16
∫ x22(x 7 − 6) = A{ln( p)6 + 9 p2 − 2 p3 − 18 p} + c then
x 7 − 6 1 , p= 7 (d) A = 9072 x 69.
Solution of
−1
x + y − 1 dy x + y + 1 given x + y − 2 dx = x + y + 2 ,
that y = 1 when x = 1, is
| x | <
3x + x 3
dx
x 7 1 , p= 7 (c) A = 54432 x − 6
2x , where 1 − x 2
3x − x 3
, ([·] denotes the greatest 5 + [x ]2 integer function), Ten f (x ) is (a) discontinuous at some x (b) continuous at all x , but the derivative f ′(x ) doesn’t exist for some x (c) f ′′(x ) does not exist for all x (d) none of these
x 7 − 6 1 , p= 7 (b) A = 54432 x
Let tan−1 y = tan−1 x + tan−1
1 . Te value of y is 3 3x − x 3 3x + x 3 (a) (b) 1 + 3x 2 1 + 3x 2
If f (x ) =
sin(2p[p2 x ])
x 7 − 6 1 (a) A = , p= 7 9072 x
Te relation R in N × N such that (a, b ) R (c, d ) iff a + d = b + c, is (a) reflexive but not symmetric (b) reflexive and transitive but not symmetric (c) an equivalence relation (d) none of these In a test, an examinee either guesses or copies or knows the answer to a multiple choice question with four choices and only one correct option. Te probability that he makes a guess is 1/3. Te probability that he copies the answer is 1/6. Te probability that the answer is correct, given that he copied it, is 1/8. Te probability that he knows the answer to the question , given that he correctly answered it is 23 25 27 24 (a) (b) (c) (d) 29 29 29 29
Let a and b be two non-zero reals such that a ≠ b. Ten the equation of the line passing through origin x y x y and point of intersection of + = 1 and + = 1 a b b a is (a) ax + by = 0 (b) bx + ay = 0 (c) y – x = 0 (d) x + y = 0
(a) log
(x − y )2 − 2 2
= 2(x + y )
(b) log
(x − y )2 + 2 2
= 2(x − y )
(x + y )2 + 2 (c) log 2
= 2(x − y )
(d) none of these 70.
Te minimum value of px + qy when xy = r 2 is equal to (a) 2r pq (b) 2 pr r (c)
−2r
pq
(d) none of these
CATEGORY-III
In this section more than 1 answer can be correct. Candidates will have to mark all the correct answers, for which 2 marks will be awarded. If, candidates marks one correct and one incorrect answer then no marks will be awarded. But if, candidate makes only correct, without making any incorrect, formula below will be used to allot marks. 2 × (no. of correct response/total no. of correct options)
71. If the vectors a, b, c, d are any four vectors then
(a × b) ⋅ (c × d ) is equal to (a) a ⋅ {b × (c × d )} (b) (a ⋅ c )(b ⋅ d ) − (a ⋅ d)(b ⋅ c ) (c) {(a × b ) × c } ⋅ d (d) (d × c) ⋅ (b × a)
72. Let z 1, z 2 be two complex numbers represented by
points on the circle |z 1| = 1 and | z 2| = 2 respectively then (a) max |2z 1 + z 2| = 4 (b) min |z 1 – z 2| = 1 (c) z 2 +
1 ≤3 z 1
(d) none of these 73. In D ABC , if cos
A b+c , then = 2 2c
1 ab 2 1 (b) circumradius is equal to c 2 1 (c) area of the triangle is bc 2 1 (d) circumradius is equal to a 2 (a) area of the triangle is
74. In the expansion of ( x + y + z )25
(a) every term is of the form 25C · r C · x 25 – r · y r – k z k r k (b) the coefficient of x 8 y 9 z 9 is zero (c) the number of terms is 325 (d) none of these x2
( y + z)2
yz
2 75. y
(z + x)2
zx is divisible by
z2
(x + y)2
xy
(a) x 2 + y 2 + z 2 (c) x – y – z
loge[x], 1 ≤ x < 3 f (x ) = | log e x |, 3 ≤ x < 4 Te graph of the function f (x ) (a) is broken at two points (b) is broken at exactly at one point (c) does not have a definite tangent at two points (d) does not have a definite tangent at more than two points.
follows:
76. A function f (x ) is defined in the interval [1, 4] as
(b) x – y (d) x + y + z
77. A coin is tossed repeatedly. A and B call alternately
for winning a prize of ` 30. One who calls correctly first wins the prize. A starts the call. Ten the expectation of (a) A is ` 10 (b) B is ` 10 (c) A is ` 20 (d) B is ` 20 78. Te function f (x) = x 2 +
l
has a x (a) minimum at x = 2 if l = 16 (b) maximum at x = 2 if l = 16 (c) maximum for no real value of l (d) point of inflection at x = 1 if l = –1
79. Let In =
p/ 4
n
∫ tan
x dx, n ∈ N , Ten
0
(b) In + I n−2 =
(a) I 1 = I 3 + 2I 5 (c) In + I n−2 =
1 n −1
1 n
(d) none of these
80. Let f (x ) = x 2 + xg ′(1) + g ′′(2) and g (x ) = f (1)·x 2
+ xf ′(x ) + f ′′(x ) then (a) f ′(1) + f ′(2) = 0 (b) g ′(2) = g ′(1) (c) g ′′(2) + f ′′(3) = 6 (d) none of these SOLUTIONS
1. (a) : otal number of 5 letter word when any letter
can be chosen any number of times is (10) 5. Now these numbers include those which never comes in the picture and such numbers are 10P 5. \ Required number of words which have at least one letter repeated are 105 – 10P 5 = 1,00,000 – 30,240 = 69,760 2. (a) : Let the number of sides of the polygon be n. \ (2n – 4)90° = 120°n ⇒ 180°n – 120°n = 360° ⇒ 60°n = 360° \ n = 6 MATHEMATICS TODAY | MARCH ‘16
59
\ number of diagonals = nC 2 – n
3.
8. (a) : f (q) = sec2q – tan2q
= 6C 2 – 6 = 15 – 6 = 9 (b) : (cot–1x )2 – 7(cot–1x ) + 10 > 0 (cot–1x – 2)(cot–1x – 5) > 0
1 − sin 2q (cos q − sin q)2 = = cos 2q (cos q + sin q)(cos q − sin q)
⇒ ⇒ cot–1x < 2 and cot–1x > 5 But cot–1x is decreasing " x ∈ R ⇒ x > cot 2 and x < cot 5 ⇒ x ∈ (–∞, cot 5) ∪ (cot 2, ∞)
p − q = tan p − p + q = tan q 4 4 4
\
4. (b) : [x + (x 3 – 1)1/2]5 + [x – (x 3 – 1)1/2]5
= 2[x 5 + 10x 3((x 3 – 1)1/2)2) + 5x ((x 3 – 1)1/2)4 [Note that : (x + a)n + (x – a)n = 2[nC 0 x n a0 + nC 2x n – 2 a2 + nC 4 x n – 4 a4 + ...] = 2[x 5 + 10x 3(x 3 – 1) + 5x (x 3 – 1)2] = 2[x 5 + 10x 6 – 10x 3 + 5x (x 6 – 2x 3 + 1)] \ degree is 7. 5. (b) : D is the mid-point of BC .
\
cos q − sin q 1 − tan q p = = tan − q cos q + sin q 1 + tan q 4
=
BD = DC z + z D= 2 3 2
f
a cos A + b cos B + c cos C a +b+c a b c = = = 2R sin A sin B sin C
9. (b) :
2R sin A cos A + 2R sin B cos B + 2R sin C cos C 2R sin A + 2R sin B + 2R sin C R(sin 2A + sin 2B + sin 2C) = 2R(sin A + sin B + sin C)
=
=
4 sin A sin B sin C A B C = 4 sin sin sin A B C 2 2 2 2 4 cos cos cos 2 2 2
A(z 1)
f (x ) = 2, x →0 x
10. (c) : We have, lim
2 G
1 D
B(z 2)
C(z 3)
We know that centroid divides median internally in 2 : 1 ratio. z + z 2 ⋅ 2 3 + z 1 z + z + z 2 \ G= = 1 2 3 2 +1 3
Given that f (x) = min(sin [m]x, | x |) min(sin [m]x, | x |) = lim =2 x x →0 sin [m]x = 2 [Q sinkx ≤ |x | " k, x ∈ R] x x →0
= lim
sin [m]x ⋅ [m] = 2 x →0 [m]x
= lim
6. (d) : Here the given equation is
\
(cosa – 1)x 2 + x cosa + sina = 0 D = B2 – 4 AC = cos2a + 4sina(1 – cosa)
2
− sec x cos x(1 + tan x − sec x) sin x + cos x − 1 = = 2
60
MATHEMATICS TODAY | MARCH ‘16
1
∫ f (z)dz = x + ∫ zf (z) dz 0
sin x(1 + tan x − sec x ) (1 + tan x)
⇒ [m] = 4 m ∈ [4, 5) i.e. 4 ≤ m < 5.
11. (d) :
1 7. (b) : f (x ) = tan x + cot x + sec x + cosecx sin x cos x sin x = = 1 + sin x + cos x (1 + tan x) + sec x 2
[m] = 2
x
> 0 " 0 < a < p \ roots are real and distinct.
=
⇒ ⇒
2
x
Differentiating (i) both sides, we get d 1 d x ( ) ( ) f z dz x z f z dz = + ∫ ∫ dx 0 dx x
⇒ f (x ) = 1 + (0 – x f (x )) ⇒ f (x ) = 1 – x f (x ) ⇒
f (x ) = 2
1 1 + x
2 1 3 ( ) f x dx dx = log \ ∫ = ∫ 2 1 x 1 1 +
...(i)
12. (a) : We know that l 2 + m2 + n2 = 1
... (i) Since the line makes the same angle q with each of x and z-axis and b with y -axis. \ l 2 = cos2q, n2 = cos2q and m2 = cos2b (i) becomes 2l 2 + m2 = 1 or 2n2 + m2 = 1 ⇒ 2cos2q = 1 – cos2b ⇒ 2cos2q = sin2b ⇒ 2cos2q = 3sin2q ⇒ 5cos2q = 3 p/3 sin x k ... (i) dx = 13. (b) : ∫ 4 p/6 cos x + sin x
⇒
p + p − x 3 6 k = p p p p 4 cos + − x + sin + − x 3 6 3 6
∫
p /3
Now putting the value of m in my +
14. (a) : lim
x →0
sin3 x
log(1 + x 3) x 3
1 = x , we get m
2
dy dy ⇒ − x + y = 0 dx dx
a
a
∫ f (x)dx = ∫ f (a + b − x) dx =
k 4
... (ii)
Adding (i) and (ii), we get p/3 k p k ∫ 1dx = 2 ⇒ 6 = 2 p /6
x →0
\
⇒ order 1 and degree 2.
b
p/6 sin x + cos x
log(1 + x 3 )
1 = x, m is arbitrary constant m 1 dy m =1 ⇒ m = dx dy / dx my +
y dy + = x dy / dx dx
b
cos x
∫
= lim 3
17. (c) : Equation of tangent to parabola x 2 = 4 y is
sin
p/3 p /6
⇒
⇒ log3(18x – x 2 – 77) > 50 ⇒ 18x – x 2 – 77 > 31 ⇒ x 2 – 18x + 80 < 0 ⇒ (x – 8)(x – 10) < 0 ⇒ 8 < x < 10 \ x ∈ (8, 10)
= lim
⇒
log(1 + x 3 ) x 3
x →0
x ⋅ lim x →0 sin x
log(1 + u) x = lim ⋅ lim u u→ 0 x → 0 sin x
p
k=
Te equation of the given curves are y 2 = x and y = |x |
3
⋅
18. (a) :
x 3
sin3 x
1
Required area = ∫ ( x − x)dx 0
2 3/2 x 2 1 2 1 1 = x − = 3 − 2 = 6 3 2 0
3
3
(Q x → 0 \ x 3 → 0. Put x 3 = u) lim sin x = 1 =1 x →0 x 15. (c) : cos2 A + cos2(B – A) – 2cos A cosB cos( A – B) = cos2 A + cos2(B – A) – cos( A – B)[cos( A + B) + cos( A – B)] 2 2 2 2 = cos A + cos ( A – B) – (cos A – sin B) – cos2( A – B) = sin2B 16. (c) : Here f (x ) = log4(log5(log3(18x – x 2 – 77))) Te given function is defined when log5(log3(18x – x 2 – 77)) > 0 [Q logx is defined " x > 0]
19. (b) : By Lagrange's mean value theorem, we have
f (b) − f (a) = f ′(c) b−a
⇒
f (6) − f (1) ≥2 6 −1
⇒ f (6) – f (1) ≥ 10 ⇒ f (6) + 2 ≥ 10 [Q f (1) = –2] ⇒ f (6) ≥ 8 20. (c) : Given that, xy = e – e y
⇒
dy dy y + x = −e y dx dx
⇒
dy dy x + e y + y = 0 dx dx
... (i)
1 dy =− dx e Now, differentiating (i) with respect to x , we get When x = 0 then y = 1.
\
MATHEMATICS TODAY | MARCH ‘16
61
y
e
d 2 y dx 2
y dy
+ e dx
Using x = 0, y = 1,
2
⇒
x ≥ 34 Again A ∩ B ⊆ B. \ n( A ∩ B) ≤ n(B) ⇒ x ≤ 64 From (i) and (ii), we get 34 ≤ x ≤ 64
d 2 y dy + x 2 + 2 = 0 dx dx
d 2 y 1 dy 1 we get = =− , dx e dx 2 e2
∫ cos xdx. Let x = t ⇒ dx = 2tdt = 2 ∫ t cos tdt = 2 [t sin t − ∫ sin t dt]
21. (c) :
Given that (x , y ) ∈ R ⇔ x divides y \ (2, 6) ∈ R, (2, 10) ∈ R, (3, 3) ∈ R, (3, 6) (5, 10) ∈ R ⇒ R = {(2, 6), (2, 10), (3, 3), (3, 6), (5, 10)} \ R–1 = {(6, 2), (10, 2), (3, 3), (6, 3), (10, 5)}
22. (b) : Given that
f (x + y ) = f (x ) + f ( y ) – xy – 1 " x , y ∈ R and f (1) = 1 f(2) = f(1 + 1) = f(1) + f(1) – 1 – 1 = 0 f(3) = f(2 + 1) = f(2) + f(1) – 2·1 – 1 = –2 f(n + 1) = f(n) + f(1) – n – 1 = f (n) – n f(n + 1) < f(n) Tus, f(1) > f(2) > f(3) > .... and f(1) = 1 \ f(1) = 1 and f(n) < 1, for n > 1 Hence f (n) = n, n ∈ n has only one solution n = 1.
23. (d) : Let the G.P. be a, ar , ar 2, ....., ar n – 1
Given that, t n = t n + 1 + t n + 2 \ a = ar + ar 2 ⇒ r 2 + r = 1 ⇒ r 2 + r – 1 = 0
⇒
r =
−1 ±
12 − 4 ⋅ 1 ⋅ (−1) 2 ⋅1
=
−1 ± 2
5 −1 2 −1 − 5 r = 2 is rejected as terms are positive r =
\ ⇒
a, A1, A2, b are in A.P. A1 + A2 = a + b a, G1, G2, b are in G.P. ⇒ G1G2 = ab Again, a, H 1, H 2, b are in G.P. 1 1 1 1 \ + = + H1 H 2 a b
⇒
H1 + H 2 H1H 2
=
a + b A1 + A2 = ab G1G2
⇒
A1 + A2 H1 + H 2
=
G1G2 H1H 2
chosen in 6C 4 ways. Since only one of the four alternatives is correct, the wrong answers can be given in 3 different ways for each of the two remaining questions attempted unsuccessfully by the candidate. Te desired number of ways = 6C 4·32 = 135.
... (i)
... (ii) From (i) and (ii), common values of x are given by −1 ≤ x < 1 − 3 or 1 + 3 < x ≤ 3 25. (d) : Let A and B denote the set of employees who
like bananas and apples respectively. Further, let the total number of employees be 100. \ n( A) = 70, n(B) = 64, n(U ) = 100 and n( A ∩ B) = x Now n( A ∪ B) ≤ 100 \ n( A) + n(B) – n( A ∩ B) ≤ 100 70 + 64 – x ≤ 100 MATHEMATICS TODAY | MARCH ‘16
27. (a) : Let a and b are two numbers.
29. (d) : Let a is the radius of the circle.
⇒ r = 2
62
∈ R,
28. (b) : Four questions with correct answers can be
5
5 − 1 = 2sin18° 4 24. (c) : Given that log10(x 2 – 2x – 2) ≤ 0 ⇒ x 2 – 2x – 2 ≤ 100 ⇒ x 2 – 2x – 3 ≤ 0 ⇒ –1 ≤ x ≤ 3 For logarithm to be defined, x 2 – 2x – 2 > 0 ⇒ x > 1 + 3 and x < 1 − 3
... (ii)
26. (c) : Here A = {2, 3, 4, 5} and B = {3, 6, 7, 10}
= 2(t sint + cost ) = 2( x sin x + cos x ) + c
⇒
... (i)
\
Te length of the major axis of the ellipse is 2a. Area of the region between circle and ellipse is
p
a2 –
b pab = pa − pa a
= pa2 − pa2
2
b2 1 − 1 − 2 a
2
2
= pa − pa
2
b2 a2
= pa2 − pa2
1 − e2
e = 1−
b2 a
2
\
Probability that the randomnly chosen point in the circle will fall outside of the ellipse is Probability =
pa2 − pa2 1 − e2 pa2
=1−
1 − e2
=1−
1−
8 9
=1−
2 2 3
1 3
2 3
30. (d) : Let A′ be the image of A in y = x
\
32. (b) : Let
2
1−
=1− =
Co-ordinate of A′ ≡ (4, 3)
Te shaded triangle is right angled at (1, 3). \ Te circumcircle is the circle on (3, 3) and (1, 5) as ends of a diameter. \ Equation of the circle whose extremities of the diameter (3, 3) and (1, 5) is (x – 3)(x – 1) + ( y – 3)( y – 5) = 0 i.e. x 2 + y 2 – 4x – 8 y + 18 = 0
... (i)
a, b
are the roots of the equation x 2 + ax + b = 0 and that of x 2 + bx + a = 0 be g and d. \ a + b = –a and ab = b g + d = –b and gd = a We also have a – b = g – d ⇒ (a – b)2 = ( g – d)2 ⇒ (a + b)2 – 4ab = ( g + d)2 – 4 gd ⇒ a2 – 4b = b2 – 4a ⇒ a2 – b2 + 4(a – b) = 0 ⇒ (a + b)(a – b) + 4(a – b) = 0 ⇒ (a – b)(a + b + 4) = 0 a ≠ b \ a + b + 4 = 0 i.e. a + b = – 4. Q cos
Now PA′ + PB is minimum if PA′B are collinear \ equation of PA′B is 10 13 − 3 y − 3 = (x − 4) ⇒ y − 3 = (x − 4) ... (ii) 3 7−4 Solving (i) and (ii), we get the point P whose co 31 31 ordinate is , 7 7 31. (b) : Te given equation is
(x + y – 6)(xy – 3x – y + 3) = 0 \ x+y–6=0 or xy – 3x – y + 3 = 0 or (x – 1)( y – 3) = 0 So, the equations of the sides of the triangle are x + y = 6, y = 3, x = 1
33. (a) :
D=
q
1
− cos = cos
q 2
1
2
cos
q
1
q 2
− cos
1
2
(0) − 1 −1 − cos2
q 2
−1 q q + 1 1 + cos 2 2 2
q D = 2 cos2 + 2
2 Now, Dmax = 2 + 2 = 4 and Dmin = 2 × 0 + 2 = 2
0 ≤ cos2
q
≤ 1 2
\ D ∈ [2, 4] 34. (a) : Te given system of equations is
ax – 3 y + z = 0, x + a y + 3z = 1, 3x + y + 5z = 2 a −3 1 Now, D = 1 a 3 3
1
5
= a(5a – 3) + 3(5 – 9) + 1(1 – 3a) = 5a2 – 6a – 11 = (5a – 11)(a + 1) Te system does not have a unique solution iff D = 0 11 or − 1 ⇒ a= 5 MATHEMATICS TODAY | MARCH ‘16
63
35. (b) : Te given line is parallel to the vector ^
n=i
^
^
− j + 2k
Given that, the required plane passes through the ^ ^ ^ point (2, 3, 1) i.e. 2 i + 3 j + k Also given that the plane is perpendicular to
^
^
n= i −j
^
+ 2k
^
^
^
^
^
So, its equation is [(r − (2 i + 3 j + k)] ⋅ ( i − j
⇒
^
^
r ⋅ (i − j
+ 2 k^) = 0
2
2
1 2
= | a |2 = c ⋅ a
2 2 3 3 | a |2 + 2 × | a |2 = 6 2 '
2
⇒ 6 | a |2 = 6. \ | a |2 = 1 ⇒ | a | = 1
37. (c) : Te equation of the parabola is
y 2 = 8x = 4 · 2 x \ Focus (2, 0) Let the point on the parabola be (2 t 2, 4t )
(2t 2 − 2)2 + (4t )2
=4
4(t 2 – 1)2 + 16t 2 = 16 (t 2 – 1)2 + 4t 2 = 4 (t 2 + 1)2 = 4 ⇒ t 2 + 1 = ±2 [Q t 2 ≠ –3] t 2 = 2 – 1 = 1 t=±1 Te required points are (2, 4) and (2, –4). 1
1 (2 cos x − 3 sin x ) 13 1 + sin −1 (2 sin x + 3 cos x ) 13
− 38. (c) : Here f (x) = cos
Let
\
64
39. (c) : lim
2 f (x) − 3 f (2x) + f (4x )
form 0 0
x 2 2 f ′(x) − 6 f ′(2x) + 4 f ′(4x ) = lim 2x x →0 [By L-Hospital's rule] x →0
f ′(x) − 3 f ′(2x) + 2 f ′(4x ) x x →0
= lim
0 0
f ′′(x) − 6 f ′′(2x) + 8 f ′′(4x ) 1 x →0 = f ′′(0) – 6 f ′′(0) + 8 f ′′(0) = 3 f ′′(0) = 3 × 4 = 12
= lim
1 1 1 ⇒ 3 | a |2 +2 | a |2 + | a |2 + | a |2 = 6
\ ⇒ ⇒ ⇒ \ \ \
2 1 + x 2 = x
2 x 1 + x 2
3
⇒
\
^
=6 i.e. (a + b + c )⋅ (a + b + c ) = 6 ⇒ | a |2 + | b |2 + | c |2 +2(a ⋅ b + b ⋅ c + c ⋅ a) = 6 Given that | a | = | b | = | c | and angle between each pair is p/3. p 1 \ a ⋅ b = | a || b | cos = | a |2 and b ⋅c
df (x ) = dg (x )
+ 2 k) = 1
36. (c) : | a + b + c |
= cos–1cos(x + a) + sin–1sin(x + a) d { f (x )} = 2 f (x ) = 2x + 2a dx 1 2x d { g (x )} = Let g (x) = 1 + x 2 \ 2 1 + x 2 dx
2 3 = cos a and = sina 13 13 f (x ) = cos–1(cosx cosa – sinx sina) + sin–1(sinx cosa + cosx sina) MATHEMATICS TODAY | MARCH ‘16
[Q f ″(0) = 4]
40. (a) : Given that b1 + b4 + b7 + ... + b25 + b28 = 220
1, 4, 7, 10, ..., 25, 28 are in A.P. \ 1 + (n – 1)3 = 28 ⇒ 3n = 30 i.e. n = 10 and b1 + b28 = b4 + b25 = b7 + b22 = .... = b13 + b16 \ 5(b1 + b28) = 220 ⇒ b1 + b28 = 44 Now, b1 + b2 + b3 + .... + b28 = 14 (b1 + b28) = 14 × 44 = 616 41. (c) : (1 – 3x + 3x 2 – x 3)2n = (1 – x )6n Since the power of the expansion is even, so there will be a unique middle term, which is t (3n + 1) = 6nC 3n (–x )3n 6n ! (−x )3n = 3n ! 3n ! Q
42. (d) : (x 2 + y 2)dx – 2xy dy = 0
⇒ 2xydy = (x 2 + y 2)dx dy x 2 + y 2 ⇒ = dx
[Putting y = vx ]
2xy
⇒
dv 1 + v 2 v + x = 2v dx
⇒
dv 1 − v 2 x = 2v dx
⇒
⇒
dv 1 + v 2 x = 2v dx
2v
− v
dx dv = ∫ 1 − v 2 ∫ x
⇒ log(1 – v 2) = –logx + logc
⇒ 1 − v 2 = ⇒
x 2 − y 2 x
c x
⇒
x 2 − y 2 x 2
=
c x
5z 2 is purely imaginary. 11z 1
46. (c) : Given that
\
=c
5z 2 = il 11z 1
43. (d) : Vector of unit length perpendicular to given
a × b vectors = ± | a || b |
Now,
2z1 + 3z 2 2z1 − 3z 2
Hence 2 such vectors.
=
=
= 1 [Q |a + ib| = |a – ib|]
44. (b) : Te equation of the hyperbola is
x 2 – y 2sec2q = 4
11 il 5 3 z 33 1+ ⋅ 2 1 il + 2 z 1 10 = 3 z 2 33 1 − il 1− ⋅ 10 2 z 1 z 2 z 1
⇒
⇒
x2 4
−
m
y 2
4 cos
2
47. (b) : Given that
=1
q
⇒ ⇒ ⇒
Te equation of the ellipse is x 2sec2q + y 2 = 16
⇒
x2
y 2 + 16 cos2 q 16
=1
45. (c) : Te given equations of the ellipse are
(x − 4)2 100 or 25 (x + 1) 1
–7–2=0 (x – 4)(x + 6) = 0 ⇒
∑ aii
i =1
48. (d) : Given that f (x ) = f(x ) + f (1 – x ) and f ′′(x ) < 0
16 − 16 cos2 q 4 + 4 cos2 q ⇒ = 3 4 16 ⇒ 1 + cos2q = 3(1 – cos2q) 1 ⇒ 4cos2q = 2 ⇒ cos q = ± 2 p 3p \ q= , 4 4
and 4(x + 1)2 + y 2 = 4 or
x – 5 + x 2 + 2x – 24 = 0 x = –6, 4
Note that : race of a matrix A = (aij)m × m is
16 cos2 q 4 cos2 q 1+ = 3 1 − 4 16
2
i =1 2 x – 10 + x
m
Now, by the problem,
4 (x – 4)2 + 25 y 2 =
∑ aii = 0
+
2
y 4
+
y 2 4
=1
=1
in 0 ≤ x ≤ 1 \ f′(x ) = f ′(x ) – f ′(1 – x ) f ′′ (x ) < 0 in 0 ≤ x ≤ 1 Q ⇒ f ′(x ) decreases in 0 < x < 1. 1 Now, if x > 1 – x i.e. x > 2 f′(x ) < 0 in 1 < x < 1 ⇒ f(x ) decreases in 2 1 < x < 1 2 1 and if x < 1 – x i.e. x < . f′(x ) > 0 2 1 ⇒ f(x ) increases in 0 < x < 2
49. (a) : Since
the tangent is equally inclined to the axes, it must be either the line AB or B′ A as shown in the diagram. \ Slope = ±1 1/3
dy y i.e. = − dx x
2 2 × = −1 −1 4 OP and OQ are perpendicular to each other.
Slope of OP × slope of OQ
\
=
⇒
y = ± x = ±
= ±1 a
2 2 \ Tere are four points of contact, one each in the 4 quadrant MATHEMATICS TODAY | MARCH ‘16
65
50. (c) : Here, probability that A will solve the problem
is 2/5. Probability that B will solve the problem is 2/3. Problem will be solved if atleast any one can solve the problem. i.e. P ( A ∪ B) is the required probability. Now, P ( A ∪ B) = 1 – P ( A ∪ B)C = 1 – P ( AC ∩ BC ) = 1 – P ( AC ) · P (BC ) [Q Te two events are independent] 3 1 1 4 =1− × =1− = 5 3 5 5 51. (a) : Let X denotes the number of tails in tossing 6
coins.
1 2
Here, X ~ B 6,
1 \ P(X = x) = Cx 2 6
x
1 6− x = 6C 1 6 x 2 2
1 \ P(X = 0) = C 0 ⋅ 6 = 1 × 6 = 64 2 2 We want P ( X ≥ 1) = 1 – P ( X < 1) = 1 – P ( X = 0) 1 63 =1− = 64 64 6
1
1
52. (a) : Given that
1 1 2 P(A) = , P( A | B) = , P( B | A) = 4 2 3 P(A ∩ B) Now, P(A | B) = P(B)
⇒
P(B) =
P(A ∩ B) = 2P(A ∩ B) P(A | B)
P(A ∩ B) Again, P(B | A) = P(A)
3 \ P(A) = P(A ∩ B) 2 1 ⇒ P(A ∩ B) = 6 1 1 Now, P(B) = 2 × = 6 3 53. (b) : P (n) = 2n > n
⇒
⇒
P(A ∩ B) P(A) = P(B | A)
1 2 × = P( A ∩ B) 4 3
⇒ 2 · 2k > 2k ⇒ 2 · 2k > 2k > k + 1 as k ≥ 2 54. (d) : Q – |z | ∈ R, \ arg(–|z |) = p
66
3cos2q + 13sinq – 8 = 0 ⇒ 3(1 – 2sin2q) + 13sinq – 8 = 0 ⇒ 3 – 6sin2q + 13sinq – 8 = 0 ⇒ 6sin2q – 13sinq + 5 = 0 ⇒ 6sin2q – 10sinq – 3sinq + 5 = 0 ⇒ 2sinq(3sinq – 5) – 1(3sinq – 5) = 0 ⇒ (3sinq – 5)(2sinq – 1) = 0 5 1 p 5p . \ sin q = gives q = , sin q ≠ 3 2 6 6 56. (d) : Te relation {(b, x ), (a, y ), (b, z ), (c, z )} has two
ordered pairs (b, x ) and ( b, z ) whose first elements are same. Hence it is not a mapping. 57. (a) : We know, for moderately asymmetrical distribution, mode = 3 median – 2 mean = 3(22) – 2(18) = 66 – 36 = 30
50
50
0
0
50 ∫ pdx = p [x]0 = 50p
MATHEMATICS TODAY | MARCH ‘16
^
^ ^ ^ ^ j − k and Q = i + 3 k + Q ] = | R || P × Q | cos f where R is unit
58. (c) : We are given P = 2 i
Now, [R P vector. between R and P × Q f is angle ⇒ | P × Q | cos f ≤ | P × Q |
\ Minimum value of [R P Q ] = | P × Q | = 59 ^ ^ ^ i j k ^ ^ ^ P × Q = 2 1 −1 = i (3) − j(7) + k(−1) 1 0 3 \ | P × Q | = 32 + 72 + (−1)2 = 59 59. (d) : Given that C.V. = 60, S.D. = 20
S.D. × 100 Mean 20 60 = × 100 . \ Mean = 33.33 (nearly) Mean
But C.V. =
\
p + p , w200 4 w is a cube root of unity p 1 = tan (w3)66 ⋅ w2 + 3 66 2 p + 4 (w ) ⋅ w 1 p = tan w2 + 2 p + [ w3 = 1] w 4 p = tan (w2 + w) p + 4
60. (a) : tan w
⇒ P (k) = 2k > k
Now, ∫ arg(− | z |) dx =
55. (b) : Te given equation is
200
+
1
p = tan −p +
[Q 1 + w + w2 = 0]
p = − tan p −
[Q tan(–q) = –tanq]
4
= tan
p
4
[Q tan(p – q) = –tanq]
4
=1 Let the common ratio be r . \ G1 = lr , G2 = lr 2, G3 = lr 3, n = lr 4 4
4
Now, G1 + 2G2 + G3 = (lr )4 + 2(lr 2)4 + (lr 3)4 = l 4r 4 + 2l 4r 8 + l 4r 12 = l 3(lr 4) + 2l 2(lr 4)2 + l (lr 4)3 = l 3n + 2l 2n2 + ln3 = ln(l + 2nl + n2) = ln(l + n)2 l + n = m = 4m2nl 2 62. (a) : Given that
a and b are the two roots of the
equation x 2 – 6x – 2 = 0.
\ a2 – 6a – 2 = 0 and b2 – 6b – 2 = 0
Multiplying the above two equations by an and bn respectively, we get an + 2 – 6an + 1 – 2 = 0 and bn + 2 – 6bn + 1 – 2 = 0 Subtracting, we get (an + 2 – bn + 2) – 6(an + 1 – bn + 1) – 2(an – bn) = 0 [Q an = an – bn] i.e. an + 2 – 6an + 1 – 2an = 0 ⇒ an + 2 – 2an = 6an + 1 a − 2an ⇒ n +2 =3 2an+1 a − 2a8 Put n = 8 to get 10 2a9
a + d + c + f = b + c + d + e i.e. a + f = b + e. Hence (a, b) R (e, f ) Hence if (a, b) R (c, d ) and ( c, d ) R (e, f ) then (a, b) R (e, f ) " (a, b), (c, d ), (e, f ) ∈ N × N \ R is transitive. Since R is reflexive, symmetric and transitive, hence R is an equivalence relation on N × N . 64. (c) : Let us define the events first.
61. (d) : Given that l , G1, G2, G3, n are in G.P.
4
\
E1 = the examinee guesses the answer E2 = the examinee copies the answer E3 = the examinee knows the answer and A = the examinee has answered the question correctly. 1 1 1 1 1 \ P(E1) = , P(E2) = and P(E3) = 1 − − = 3 6 3 6 2
[Q E1, E2, E 3 are mutually exclusive and exhaustive events, therefore P (E1) + P (E2) + P (E3) = 1] 1 1 Now, P(A | E1) = , P(A | E2) = , P( A | E3) = 1 4 8 Now, P(E3) P(A | E3 ) P(E3 | A) = P(E1 )P(A | E1 ) + P(E2) P( A | E2 ) + P(E3) P( A | E3 )
=
− − − 65. (c) : tan y = tan x + tan 1
1
N × N ,
(a, b) R (a, b) since a + b = b + a. Hence (a, b) R (a, b) " (a, b) ∈ N × N \ R is reflexive. Symmetric property : If (a, b) R (c, d ) then a + d = b + c. Now, when a + d = b + c then c + b = d + a . Hence (c, d ) R (a, b). Hence, if (a, b) R (c, d ) then (c, d ) R (a, b) " all (a, b), (c, d ) ∈ N × N . \ R is symmetric. Transitive property: Let (a, b) R (c, d ) and (c, d ) R (e, f ). Ten a + d = b + c and c + f = d + e
2x 1 − x 2
3x − x 3 1 | | < x 1 − 3x 2 3
tan−1 y = tan −1
∈
1
= tan–1x + 2tan–1x = 3tan–1x
=3
63. (c) : Reflexive property : For all (a, b)
1 ⋅1 24 2 = 1 1 1 1 1 ⋅ + ⋅ + ⋅ 1 29 3 4 6 8 2
\
y =
3x − x 3
1 − 3x 2
66. (c) : Equation of the line passing through the point
x y of intersection of x + y = 1 and + = 1 is b a a b x + y − 1 + l x + y − 1 = 0 ... (i) a b b a Since (i) is passing through (0, 0) \ –1 + l(–1) = 0 ⇒ l = –1 \ (i) becomes x y x y + − 1 − 1 + − 1 = 0 a b b a MATHEMATICS TODAY | MARCH ‘16
67
1 1 1 1 ⇒ − x + − y = 0 a b b a ⇒ x – y = 0 ⇒ y – x = 0 67. (d) : Here [p2x ] is an integer " x ∈ R and so 2 p[p2x ] is an integral multiple of p. Tus, sin(2p[p2x ]) = 0 and 5 + [ x 2] ≠ 0 " x ∈ R. \ f (x ) = 0 " x ∈ R Tus, f (x ) is a constant function and so it is continuous and differentiable any number of times " x ∈ R. dx
1 [ln z 6 + 9z 2 − 2z 3 − 18z] + c 54432
r 2 f (x) = px + q [Q xy = r 2] x qr 2 f ′(x) = p − x 2
f (x)min = pr
= 2r
q p
+
q p
z 2
From the figure, |z 1 – z 2| is least when O, z 1, z 2 are collinear. Ten |z 1 – z 2| = 1 Again,
z 1 A O 1
1
B
73. (a, b) : cos
⇒ cos2
q r p
=
pq ⋅ r + pq ⋅ r
A b+c = 2 2c
A b + c = 2 2c
∆=
⇒ 2 cos2
b+c c
A b + c = 2 c
⇒ cos A =
1 1 a bc sin A = bc × 2 2 c a a = sin A a / c
b2 c2
=
b+c b −1 = c c
=
c 2 − b2 c
=
a c
1 ab 2
=c
26 × 27 = 13 × 27 = 351 2
1 1 =2+ =3 | z 1 | 1
1 R= c 2
=
≤ | z 2 | +
= 25C 0 x 25 + 25C 1 x 24( y + z ) + ....... + 25C r x 25 – r ( y + z )r + ... = .... + 25C r x 25 – r (.... + r C k y r – k z k + ...) + .... 8 9 9 Q 8 + 9 + 9 ≠ 25, so there is no term like x y z . Te number of terms = 1 + 2 + 3 + ... + 26
q p
qr 2
1 z 1
74. (a, b) : (x + y + z )25 = {x + ( y + z )}25
× d ) = scalar triple product of a, b , c × d = a ⋅ {b × (c × d)} (a × b) ⋅ (c × d ) = scalar triple product of a × b , c , d = {(a × b) × c)} ⋅ d
68
≤ 2 × 1 + 2 ≤ 4
Again, 2R =
pq
71. (a, b, c, d) : (a × b) ⋅ (c
72. (a, b, c) : |2z 1 + z 2| ≤ |2z 1| + |z 2| ≤ 2|z 1| + |z 2|
\
and f ′′(x) > 0 for x = r
Now,
x 3
= (b ⋅ d)(a ⋅ c ) − (b ⋅ c )(a ⋅ d)
2qr 2
Now, f ′(x ) = 0 gives x = ± r
\ sin A = 1 − cos2 A = 1 −
70. (a) : We have, f (x ) = px + qy
⇒ 1 + cos A =
69. (d)
\
z 2 +
1 (1 − z)3dz 1 (1 − 3z + 3z 2 − z 3) dz = ∫ = 42 (42) ⋅ (216) ∫ z (63)z
f ′′(x ) =
× d) = a ⋅ {b × (c × d)} = a ⋅ {(b ⋅ d ) c − (b ⋅ c ) d}
dx
Let 1 − 6 = z ⇒ 42 dx = dz and x 7 = 6 1 − z x 7 x 8
\
∫ x22(x 7 − 6) = ∫ 29 6 x 1 − 7 x
68. (b) :
=
(a × b) ⋅ (c
x2
y2 + z 2
yz
∆ = y 2
z2 + x2
zx
z2
x2 + y2
xy
75. (a, b, d) :
MATHEMATICS TODAY | MARCH ‘16
=
x 2 + y2 + z 2
y2 + z 2
yz
x 2 + y2 + z 2
z2 + x2
zx [C 1′ = C1 + C 2]
x 2 + y2 + z 2 x 2 + y 2
xy
1 y 2 + z 2
= (x 2 + y 2 + z 2) 1
yz
z2 + x2
zx
1 x2 + y2
xy
0 y 2 − x 2
z( y − x )
= (x 2 + y 2 + z 2) 0
z2 − y2 2
2
= (x 2 + y 2 + z 2)( y – x )(z – y )[1(xy + x 2 – z 2 – yz )] = (x 2 + y 2 + z 2)( y – x )(z – y )(x – z )(x + y + z )
log e[x], 1 ≤ x < 3 76. (a, c) : f (x ) = | loge x |, 3 ≤ x < 4 Te function can be re-written as when 1 ≤ x < 2 0, f (x ) = log e 2, when 2 ≤ x < 3 loge x , when 3 ≤ x < 4
\
f(x ) is continuous and differentiable everywhere except at 2, 3. f (2 – 0) = 0, f (2 + 0) = loge2 ⇒ f (x ) is not continuous at x = 2 f (3 – 0) = loge2, f (3 + 0) = loge3 ⇒ f (x ) is not continuous at x = 3 \ f (x ) is not continuous at x = 2, 3, i.e. the graph is broken at x = 2, 3 ⇒ f (x ) is not differentiable at x = 2, 3 , i.e. the graph does not have a definite tangent at x = 2, 3. 77. (b, c) : Te probability of A winning
= P ( A) + P ( AC ∩ BC ∩ A) + P ( AC ∩ BC ∩ AC ∩ BC ∩ A) + ... 1 1 1 1 = + 1 − 1 − + .... 2 2 2 2 1 1 = + 2 2
3
+ ... =
1/ 2 1 1− 4
2 78. (a, c, d) : Here f (x) = x +
\
f ′(x) = 2x −
x(z − y)
xy +y [R1′ = R1 – R2 and R2′ = R2 – R3] 0 y + x z 2 2 2 = (x + y + z )( y − x)(z − y) 0 z + y x 1 x 2 + y 2 xy
1 x
2 1 3 3 2 \ Te expectation of A = ` 30 × = ` 20 3 1 and the expectation of B = ` 30 × = ` 10 3
\ Probability of B winning = 1 − =
=
2 3
l x
l
2l and f ′′(x ) = 2 + x 2 x 3 1/3
l Now, f ′(x ) = 0 gives x = 2 If l = 16 then x = 2 and f ′′(x )|x = 2 = 2 + 4 = 6 > 0. Hence, f (x ) has a minimum value at x = 2. Tus, f (x ) has a maximum value for no real value of l. When l = –1, f ′′(x ) = 0 if x = 1, so f (x ) has a point of inflection at x = 1. p/ 4 n 79. (a, c) : In = ∫ tan x dx 0
p/4
= ∫ tann−2 x(sec2 x − 1)dx
0
p/4
n −2
= ∫ tan
2
x sec x dx −
0
p/4
p/ 4
∫ tann−2 xdx
0
n −2
= ∫ tan
xd(tan x) −
0
p/4
∫ tann−2 xdx
0
p/4 tann−1 x 1 = − I n−2 = − I n −1 0 n − 1 n− 2
\
I n =
1 n −1
− I n−2
1 1 − I3 and I3 = − I1 ⇒ I 1 = I 3 + 2I 5 4 2 80. (a, b) : Let g ′(1) = a, g ′′(2) = b. Ten f (x ) = x 2 + ax + b \ g (x ) = (1 + a + b)x 2 + (2x + a)x + 2 = (3 + a + b)x 2 + ax + 2 \ g ′(x ) = 2(3 + a + b)x + a Hence a = 2(3 + a + b)·1 + a i.e. 3 + a + b = 0 and b = 2(3 + a + b) i.e. b + 2a + 6 = 0 Hence, b = 0, a = –3. So f (x ) = x 2 – 3x and g (x ) = –3x + 2
\
I5 =
MATHEMATICS TODAY | MARCH ‘16
nn
69
*ALOK KUMAR, B.Tech, IIT Kanpur 7.
SINGLE ANSWER CORRECT TYPE 1.
2.
x 2 + 3 xy + y 2
In the triangle ABC the medians from B and C are perpendicular. Te value of cotB + cotC cannot be 7 5 (b) 3 3 4 (c) (d) none of these 3 Te position vectors of the vertices A, B, C of
(a) 18 8.
(a) 0 4.
(b) 2
(c) 8
Te area of the loop of the curve y 2 = x 4 (x + 2) is [in square units] (a) 32 2 105
5.
(d) 4
(b)
^
^
^
^
^
9.
6.
70
(a) 8
(b) 4
(c) 1
(d) none of these
(c) 30
10.
+k
Te solution of the differential equation 2x 3 ydy + (1 – y 2)(x 2 y 2 + y 2 – 1)dx = 0 [Where c is a constant]
(c) x 2 y 2 = (cx – 1)(1 – y 2) Area of the region (in sq. units) in which point
16 p 3
(b)
8p +8 3 3
(c) 4 3 − p
(d)
3 − p
(a)
A circle of radius 4 cm is inscribed in D ABC , which touches the side BC at D. If BD = 6 cm, DC = 8 cm then which is false ? (a) the triangle is necessarily obtuse angled triangle A (b) tan = 2 7 (c) perimeter of the triangle ABC is 42 cm (d) area of ABC is 84 cm2
11.
Area bounded by the curves y = e x , y = logex and the lines x = 0, y = 0, y = 1 is
Te reflection of the hyperbola xy = 1 in the line y = 2x is the curve 12x 2 + rxy + sy 2 + t = 0 then the value of 'r ' is (a) – 7
(b) 25
(c) –175
(a) e2 + 2
(b) e2 + 1
(c) e + 2
(d) e – 1 * Alok Kumar is a winner of INDIAN NATIONAL MATHEMATICS OLYMPIAD (INMO-91).
MATHEMATICS TODAY
|MARCH ‘16
(d) 36
2 P (x, y ), {x > 0} lies; such that y ≤ 16 − x and y p tan −1 ≤ is x 3
^ ^
i + p2 j + p2 k , p 2 ^i + j + p 4 k and p 4 ^i + p 4 j are coplanar is
(b) 24
(d) none of these
64 2 105
256 2 128 2 (c) (d) 105 105 Te number of integer values of p for which the vectors
= 9 and x 2 + xz + z 2 xy + 2 yz + 3 xz is
(b) x 2 y 2 = (cx + 1)(1 + y 2)
2 2 tetrahedron is , then the length of DE is 3 (a) 1 (b) 2 (c) 3 (d) 4 Number of solutions of the equation 4sin 2x + tan2x + cot2x + cosec2x = 6 in [0, p] is
y2 + z2
(a) x 2 y 2 = (cx + 1)(1 – y 2)
^
3.
= 25;
= 16. Ten the value of
(a)
a tetrahedron ABCD are ^i + ^j + k, ^i and 3 ^i respectively and the altitude from the vertex D to the opposite face ABC meets the face at E. If the length of the edge AD is 4 and the volume of the
Tree positive real numbersx , y , z satisfy the equations
12.
(d) 90
An ellipse with major and minor axes length 10 3 and 10 respectively slides along the co-ordinate axes
He trains IIT and Olympiad aspirants
MATHEMATICS TODAY
|MARCH ‘16
71
and always remains confined in the first quadrant. Te length of the arc of the locus of the centre of the ellipse is
13.
14.
15.
16.
17.
18.
19.
20.
72
5p (a) 10p (b) 5p (c) 5p (d) 3 4 Te equation of common tangent at the point of contact of two parabolas y 2 = x and 2 y = 2x 2 – 5x + 1 is (a) x + y + 1 = 0 (b) x + 2 y + 1 = 0 (c) x – 2 y – 1 = 0 (d) –x + 2 y – 1 = 0 Te range of ‘a’, for which a circle will pass through the points of intersection of the hyperbola x 2 – y 2 = a2 and the parabola y = x 2, is (a) a ∈ (–3, –2) (b) a ∈ (–1, 1) (c) a ∈ (2, 4) (d) a ∈ (4, 6) x 2 y 2 wo tangents are drawn to an ellipse + =1 a 2 b2 from a point P (h, k); if the points at which these tangents meet the axes of the ellipse be concyclic, then the locus of P is (a) an ellipse (b) rectangular hyperbola (c) parabola (d) circle
(a) A = B (c) B ⊂ A 21.
Eight straight lines are drawn in the plane such that no two lines are parallel and no three lines are concurrent. Te number of parts into which these lines divide the plane is (a) 29 (b) 32 (c) 36 (d) 37 How many combinations can be made up of 3 hens, 4 ducks and 2 geese so that each combination has hens, ducks and geese? (birds of same kind all different) (a) 305 (b) 315 (c) 320 (d) 325 Let ‘C ’ denote the set of complex numbers and define A & B by A = {(z, w); z, w, ∈ C and |z | = |w|} B = {(z, w); z, w, ∈ C and z 2 = w2} then MATHEMATICS TODAY
|MARCH ‘16
i =3
3
i =1
i =1
ai < ai+1, ∏ (x − ai ) + ∑ ai − 3 x, where
then f (x ) = 0 has (a) only one real root (b) three real roots of which two of them are equal (c) three distinct real roots (d) three equal roots MULTIPLE ANSWER CORRECT TYPE 22.
Te projection of line 3x – y + 2z – 1 = 0 = x + 2 y – z – 2 on the plane 3x + 2 y + z = 0 is
(a)
23.
Te inclination to the major axis of the diameter of an ellipse the square of whose length is the harmonic mean between the squares of the major and minor axes is p 2p p (a) p (b) (c) (d) 2 3 3 4 Te number of the functions f from the set X = {1, 2, 3} to the Y = {1, 2, 3, 4, 5, 6, 7} such that f (i) ≤ f ( j) for i < j and i, j ∈ X is (a) 6C 3 (b) 7C 3 (c) 8C 3 (d) 9C 3
If f (x ) =
(b) A ⊂ B (d) none of these
24.
x + 1 y − 1 z − 1 = = 11 −9 −15 (b) 3x – 8 y + 7z + 4 = 0 = 3x + 2 y + z
(c) x + 12 = y + 8 = z + 14 11 15 −9 8 x + 12 y + z + 14 (d) = = 11 −9 −15 Which of the following functions will not have absolute minimum value? (a) cot (sinx ) (b) tan (log x ) (c) x 2005 – x 1947 + 1 (d) x 2006 + x 1947 + 1 x y z + + = 1 intersects the co-ordinate axes at a b c points A, B and C respectively. If DPQR has midpoints A, B and C then (a) centroids of D ABC and DPQR coincide (b) foot of normal to D ABC from O is circumcentre of DPQR (c) ar (DPQR) = 2 a2b2 + b2c 2 + c 2a 2 (d) incentres of D ABC and DPQR coincide
25.
If in a triangle ABC , cos A cosB + sin A sinB = Sin3C = 1, then, with usual notation in D ABC , (a) the triangle is isosceles (b) the triangle is right angled (c) R : r = ( 2 + 1) : 1 (d) r1 : r2 : r 3 = 1 : 1 : ( 2 + 1)
26.
If the orthocenter of an isosceles triangle lies on the incircle of the triangle then 2 (a) the base angle of the triangle is cos −1 3 (b) the triangle is acute
5 (c) the base angle of the triangle is tan−1 2 (d) If S, I are the circumcentre and incentre and R SI 1 is circumradius then = R 3 1 2 27. A circle touches the parabola y = 2 x at P , 1 2 and cuts the parabola at its vertex V. If the centre of the circle is Q, then
(a) 3790
29. If a1, a2, a3, .... an is sequence of positive numbers which are in A.P. with common difference ‘ d ’ and a1 + a4 + a7 + ...... + a16 = 147 then
(a) a1 + a6 + a11 + a16 = 98 (b) a1 + a16 = 49 (c) a1 + a4 + a7 + ..... a16 = 6a1 + 45d (d) Maximum value of a1a2.....a16 is
49 1 2
30. Suppose three real numbers a, b, c are in G.P. Let a + ib z = then c − ib ib ia (a) z = (b) z = c b ia (c) z = (d) z = 0 c 31. Te number of ways in which we can choose 2 distinct integers from 1 to 200 so that the difference between them is atmost 20 is
200C – 180C 2 2
19 × 20 (c) 180 C 1 × 20 + (d) 180C 2 2 32. In a gambling between Mr. A and Mr. B a machine continues tossing a fair coin until the two consecutive throws either H or are obtained for the first time. If it is H, Mr. A wins and if it is , Mr. B wins. Which of the following is (are) true? (a) probability of winning Mr.A is
(a) Te radius of the circle is 5 / 2 (b) Te radius of the circle is the maximum value 7 1 of sin 3x + cos 3x 2 2 15 (c) Area of DPVQ is 16 (d) Slope of PQ is –2 28. Tirteen persons are sitting in a row. Number of ways in which four persons can be selected so that no two of them are consecutive is equal to (a) number of ways in which all the letters of the word “M A R R I A G E” are permutated if no two vowels are never together. (b) number of numbers lying between 100 and 1000 using only the digits 1, 2, 3, 4, 5, 6, 7 without repetition. (c) number of ways in which 4 alike chocolates can be distributed among 10 children so that each child getting at most one chocolate. (d) number of triangles can be formed by joining 12 points in a plane, of which 5 are collinear.
(b)
4 1 (b) Probability of Mr.B winning is 4 (c) Given first toss is head probability of Mr. A winning is 1 (d) Given first toss is tail, probability of Mr.A 1 winning is 2 COMPREHENSION TYPE
Paragraph for Question No. 33 to 35
Let the curves S1 : y = x 2, S2 : y = – x 2, S3 : y 2 = 4x – 3 33. Area bounded by the curves S1, S2, S3 is
(a)
(c)
sq. units
(b)
1 sq. units 6
(d)
3
3
sq. units
1 sq. units 3
34. Area bounded by the curves S1, S3 and the line x = 3 is
5 (a) 13 sq. units (b) sq. units 4 3 (c) 8 sq. units (d) 7 sq. units 4 3 35. Area bounded by the curve S 3, y ≤ –1 and the line x = 3 is 11 7 (a) sq. units (b) sq. units 3 3 9 13 (c) sq. units (d) sq. units 2 4 Paragraph for Question No. 36 to 38 Let PQRS be a rectangle of size 9 × 3, if it is folded along QS such that plane PQS is perpendicular to plane QRS and point R moves to point T . 36. Distance between the points P and T will be
(a) (c) 4 5
90
3 205 5 (d) none of these
(b)
MATHEMATICS TODAY
|MARCH ‘16
73
37. If q is angle between the line QP and QT then tanq
is equal to 10 3 (a) (b) 3 10 (c) 91 (d) none of these 3 38. Shortest distance between the edges PQ and TS is (a) 3
10 19
(b)
10 (c) 2 19
10 19
43. From the data of the above problem the radius of the circle passing through P, B, C is
9 (a) 5 units (b) units 4 3 16 (c) (d) none of these units 3 44. Te eccentricity of the hyperbola whose transverse axis lies along the line through B, C and the parabola passes through B, C and (0, 2) is
(d) none of these
Paragraph for Question No. 39 to 41
At times the methods of coordinates becomes effective in solving problems of properties of triangles. We may choose one vertex of the triangle as origin and one side passing through this vertex as x -axis. Tus without loss of generality, we can assume that every triangle ABC has a vertex situated at (0,0) another at ( x ,0) and third one at (h,k).
(a)
7
(b)
11
(c) 2 2
(d) 2 11
40. Suppose the bisector AD of the interior angle A of D ABC divides side BC into segments BD = 4; DC = 2
then (a) b > c and c < 4 (b) 2 < b < 6 and c < 1 (c) 2 < b < 6 and 4 < c < 12 (d) b < c and c > 4 41. If in the above question, altitude AE > 10 and suppose lengths of AB and AC are integers, then b
will be (a) 3 (c) 4 or 5
(b) 6 (d) 3 or 6
In a D ABC B(2, 4) , C (6, 4) and A lies on a curve S B C 1 such that tan tan = 2 2 2 42. Let a line passing through C and perpendicular to BC intersects the curve S at P and Q. If R is the mid point of BC then area of DPQR is 18 sq.units 3 32 (c) sq.units 3 74
MATHEMATICS TODAY
(b)
(c)
7 3
(d)
17 2 2 3
45. Match the following Column I
Column II
x Number of solutions of sin = x (A) 10 (p) is
1
Number of ordered pairs (x, y ) satisfying (B)
(q)
4
Number of solution of the equation (C) (r) px 2 sin 2 3 4 = − + x x 2 3
7
he number of ordered pairs ( x, y ) satisfying the equation (D) (s) sin x + sin y = sin(x + y ) and |x | + | y | = 1 is
6
px 2 x + y = 2,sin = 1 is 3
46.
Paragraph for Question No. 42 to 44
(a)
19 4
MATRIX MATCH TYPE
39. If in D ABC , AC = 3, BC = 4 medians AD and BE are perpendicular then area of D ABC (in sq. units) is
(a)
8 sq.units 3 26 (d) sq. units 3
(b)
|MARCH ‘16
Column I
Column II
If slope of tangents from (–1, 2) to parabola y 2 = 8x are (A) (p) 1 1 m1 and m2 then + − 1 is m1 m2 Let S1, S2 are foci of an ellipse whose eccentricity is e. If P is (B) (q) extremity of minor axis such that ∠S1PS2 = 90°, then 4e2 is If foot of perpendicular from focus upon any tangent of (C) (r) parabola y 2 – 4 y + 4x + 8 = 0 lies on the line x + k = 0 then k is
2
4
0
INTEGER TYPE
Let hyperbola has eccentricity (D) 2 and its conjugate hyperbola (s) having e then 3e2 is 47.
We are given M urns, numbered 1 to M and n balls (n < M ) and P ( A) denote the probability that each of the urns numbered 1 to n, will contain exactly one ball. Column I
Column II
If the balls are different and any number of balls (p) (A) can go to any urn then P ( A)=
1 ( M +n−1) C
a + b + c a b c value of + + is r1 + r2 + r 3 r 1 r 2 r 3 51.
54.
Te point P (1,2,3) is reflected in the xy -plane, then its image Q is rotated by 180° about the x -axis to produce R, and finally R is translated in the direction of the positive y -axis through a distance d to produce S (1, 3, 3). Te value of d is
55.
Shortest distance between the z -axis and the line x + y + 2z – 3 = 0 = 2x + 3 y + 4z – 4 is ___
56.
Te minimum distance of 4x 2 + y 2 + 4x – 4 y + 5 = 0 from the line –4x + 3 y = 3 is _____________
57.
Te tangents drawn from the origin to the circle x 2 + y 2 – 2rx – 2hy + h2 = 0 are perpendicular then sum h of all possible values of is _____________ r wo lines zi − zi + 2 = 0 and z(1 + i) + z (1 − i) + 2 = 0 intersect at a point P . Tere is a complex number a = x + iy at a distance of 2 units from the point P which lies on line z (1 + i) + z (1 − i) + 2 = 0. Find [|x |] (where [.] represents greatest integer function).
n! M
C n
n! M n
(A)
8n − 4n
(B)
the last digit is 2, 4, (q) 6, 8
5n − 4n 10n 59.
4
n
10 the last digit is zero (s)
58.
10n
n
10n − 8n − 5n + 4n 10n
1 + x p sin−1 = sec(x − 1) are 2 x 2 Let R = {x, y : x 2 + y 2 ≤ 144 and sin(x + y ) ≥ 0}. And S be the area of region given by R, then find S/9p. Number of real values of x , satisfying the equation [x ]2 – 5[x ] + 6 – sinx = 0 ([.] denoting the greatest integer function) is
Column II
(r)
Te number of solutions of the equation
53.
M −1
the last digit is 1, 3, (p) 7 or 9
(D)
With usual notation in triangle ABC , the numerical
52.
‘n’ whole numbers are randomly chosen and multiplied, then probability that
the last digit is 5
50.
2
C n
If the balls are different and at most one ball can (D) (s) be put in any box, then P ( A)=
(C)
If the area bounded by the curves y = –x 2 + 6x – 5, 73 y = –x 2 + 4x – 3 and the line y = 3x – 15 is , then l the value of l is
M
If the balls are identical but at most one ball can (C) (r) be put in any box, then P ( A)=
Column I
49.
1
If the balls are identical and any number of balls (B) (q) can go to any urn then P ( A)=
48.
1
Let x be in radians with 0 < x <
p . If sin(2sinx )
2 = cos(2 cosx ); then tanx + cotx can be written a as where a, b, c N . Ten the value of c p −b a + b + c is 25
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SOLUTIONS
y y + 2 x x = 3xy 1. (d) : tan B = y 2 2 x 2 − y 2 1− 2 x 2 2 x 2 − y 2 2 y 2 − x 2 cot B = , cot C = 3xy 3xy x 2 + y 2 cot B + cot C = 3xy
2 ≥ 3
e
∫
Area = area of rectangle OABC − log e x dx
∫ = e − 1
e − ( x ln x − x ) 1 7. (b) : Given x 2 + y 2 − 2 xy cos and y 2 + z 2 = 9.
1 × 2 ×2 = 2 2
1 2 2 Volume = × 2 × DE = 3 3 3. (b) : tan x = t ⇒
⇒ ⇒ ⇒
4t 2 2
1 + t
2
+ t +
⇒ DE = 2
2 2
t
=5
(t 2 – 1) (t 4 + t 2 – 2) = 0 t 2 = 1 x =
5p 2p = 25; x 2 + z2 − 2 xz cos = 16 6 3
^ BC
2. (b) : BC is the x -axis and AB
Area of D ABC =
1
e
Area =
\
p 3p 0
0
∫
4. (d) : Area = 2 ydx = 2
−2
∫ x2
x + 2dx
−2
=6
3 zx = 24 2 y dy y 2 1 1 8. (c) : ⋅ + − = (1 − y 2 )2 dx (1 − y 2 ) x x 3 Put
, 4 4
1 1 1 1 3 xy ⋅ + yz ⋅ + xz ⋅ 2 2 2 2 2
xy + 2 yz +
y 2 1 − y 2
= t ⇒
2 y
dy dt = 2 2 dx dx (1 − y )
⇒
dt t 1 + = dx x x 3
⇒
x 2 y 2 = (cx – 1)(1 – y 2)
⇒ t ⋅ x = ∫
1
x 2
dx + c
9. (b) :
2
= 4 ∫ (z 2 − 2)2 z 2dz (where
x + 2 = z)
0
z 7 4z 5 4z 3 2 256 2 = 4 − + = 105 7 5 3 0 1
p2
p2
5. (d) : p2
1
p4
p 4
p4
1
y y p ≤ is equivalent to − 3 ≤ ≤ 3 x x 3 Required area is the area of shaded region ( APOQ) = area of DOAQ + area of sector (OAP ) tan −1
=0
1 2
= ×4×4
6. (d) :
76
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3+
p(4 × 4) 6
8p + 8 3 3
B 2 C 1 A 4 = , tan = ⇒ tan = 2 3 2 2 2 7 B C s−a tan .tan = ⇒ 2s = 3a = 42 2 2 s Perimeter = 42 cm
10. (a) : tan
⇒
15. (b) : Te equation of pair of tangents from P (h, k )
\ \ D = r.s. = 84 cm2 \
to the ellipse
A B C tan , tan , tan all are less than 1. All angles 2 2 2 are acute.
4b − 3a , 4 a + 3b Also, a b = 1 1 1 5 5 ⇒ 12a2 – 7ab – 12b2 + 25 = 0
xh h2 k 2 − 1 + 2 2 − 2 + 2 = 0 a2 b2 a a b
\ OA1 . OA2 =
h2 k 2 2 2 + 2 a a b 1−
k
2
...(iii)
b2 If these tangents meet the y -axis at B1, B2 then putting x = 0 in (i), we get
5 p − 2 tan −1 = 2 5 3 6
p 5p \ arc length = 10. =
1)2
...(ii)
OA1 & OA2 are the roots of (ii)
p
OB1 . OB2 =
3
x 2 y = 2x 2 –5x + 1 Solving (i) & (ii)
= 1 is
x 2 k 2
circle i.e. x 2 + y 2 = 100 C 1OC 2 = q
13. (b) : y 2 =
b
2
Let these tangents meet the x -axis at the points A1, A2 putting y = 0 in (i), we have
(a1 , b1 ) =
12. (d) : Te locus of the centre of the ellipse is director
+
2
y 2
x 2 y 2 h2 k2 xh yk 2 2 + 2 − 1 2 + 2 − 1 = 2 + 2 − 1 ...(i) a b a b a b
in the line ax + by + c = x − a y − b −2 (aa + bb + c ) . 0 is given by = = 2 2 a b a +b Te reflection of ( , ) in the line y = 2x is
6
x 2 a
11. (b) : Te reflection of ,
⇒
Moreover this equation represents a real circle, if g 2 + f 2 – c > 0. 0 + 1 – a2 > 0 ⇒ a ∈ (–1, 1)
h2 k 2 2 2 + 2 b a b 1−
a2 OA1, OA2, OB1, OB2 are concyclic. OA1 . OA2 = OB1 . OB2
...(i) ..(ii)
h2 2 2 k2 1 − 2 a = b 1 − 2 a b
(2 y 2
( y + – 4 y + 1) = 0 \ y = –1 is the repeated root which is the y co-ordinate of point of contact. So point of contact = (1, –1). \ Common tangent at point of contact x + 1 or x + 2 y + 1 = 0. y (−1) = 2 14. (b) : Family of curves passing through the intersection of the parabola and hyperbola is
h2
or
x 2 – y 2 = a2 – b2. 2
16. (a) : 4(a cos
2
2
2
q + b sin q) =
2(4a2 )(4b2 )
4a2 + 4b2 (Standard formula) 17. (d) : 7C 3 + 2 × 7C 2 + 7C 1 = 9C 3. 18. (d) : Find t n in 2, 4, 7, 11……. n(n + 1) 2 3 19. (b) : (2 – 1) (24 – 1) (22 – 1) 20 (c) : z 2 = w2 taking modulus on both sides |z |=|w| Tus, B A 21. (c) : f (x ) = (x – a1) (x – a2) (x – a3) + (a1 – x ) + (a2 – x ) + (a3 – x ) Now f (x ) → – ∞ as x → – ∞ and f (x ) → ∞ as x → ∞. Again f (a1) = (a2 – a1) + (a3 – a1) > 0 a1 < a2 < a3 t n = 1 +
\
x 2 – y 2 – a2 + l ( x 2 – y ) = 0. i.e.(1 + l)x 2 – y 2 – l y – a2 = 0 For this equation to represent a circle 1 + l = –1 ⇒ l = –2 x 2 + y 2 – 2 y + a2 = 0
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77
⇒ One root belongs to (–∞, a1)
Also, f (a3) = (a1 – a3) + (a2 – a3) < 0 ⇒ One root belongs to (a1, a3) So f (x ) = 0 has three distinct real roots. 22. (a) : Equation of a plane passing through the line 3 x – y + 2 z – 1 = 0 = x + 2 y – z – 2 is 3x – y + 2z –1 + λ(x + 2 y – z – 2) = 0 Since it is perpendicular to the given plane 3 \ l=− 2 (ake the dot product of d.c. of both normal) Equation of the line of projection is 3x – 8 y + 7z + 4 = 0 = 3 x + 2 y + z Its direction ratios are <11, –9, –15 > and the point (–1, 1, 1) lies on the line x + 1 y − 1 z − 1 \ = = is also the equation of the 11 −9 −15 line of projection. 23. (d) : Even degree polynomial with leading coefficient +ve will have absolute minimum. 24. (a, b, c) :
27. (a, b, c, d) : Te circle is the tangent to the parabola
at
1 ,1 and the equation of tangent is 2
2x –2 y + 1 = 0 we can write the family of circle in the form x − 1 2 + ( y − 1)2 + l(2 x − 2 y + 1) = 0 2 28. (b, c, d) : x 1 + x 2 + x 3 + x 4 + x 5 = 9, x 1, x 5 ≥ 0 x 2, x 3, x 4 ≥ 1, number of solutions are 210. (a) 5 × 12 × 12 = 720
(b) 7P 3 = 210
= 210 (d) 12C 3 – 5C 3 = 210 29. (a, b, c, d) : a1 + a4 + a7 + .... a16 = 147 ⇒ 3(a1 + a16) = 147 ⇒ a1 + a16 = 49 Again a1 + a4 + a7 + a10 + ...... + a16 = a1 + a1 + 3d + a1 + 6d + ...... + a1 + 15d = 6a1 + 45d = 147 ⇒ 2a1 + 15d = 49 a1 + a6 + a11 + a16 = a1 + a1 + 5d + a1 + 10d + a1 + 15d
(c)
10C 4
= 4a1 + 30d = 2(2a1 + 15d ) = 2(49) = 98 Now using AM ≥ GM a1 + a 2 + ... + a16
16
1
≥ (a1a2a3...a16 )16 1
AC ||PR and 2 AC = PR So, ABPC is a parallelogram comparing the coordinates of mid-point of diagonals, we get P (–a, b, c ) and Q(a, –b, c) and R(a, b, –c ) Also, AD and AP are median of D ABC and DPQR respectively. So, centroids are coinciding. he perpendicular bisector of PR is also perpendicular to AC . Terefore circumcentre of DPQR is orthocenter of D ABC. ar DPQR = 4 ar D ABC
= 4 (OAB )2 + (OBC )2 + (OAC )2
where OAB is the area of the projection of D ABC on the plane XOZ etc. 25. (a, b, c, d) : Te given relation implies cos (A – B) =1 and so, A = B and C = 90° 26. (a) : Let ABC be the triangle in which AB = AC . Let I , P respectively be the incentre and the orthocenter of the triangle. A (b) AI = r cosec , AP = 2 R cos A 2 A (c) r cosec = 2R cos A + r 2 78
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|MARCH ‘16
8(a1 + a16 ) ≥ (a1a2a3...a16 )16 16 16
49 ≥ a a a ...a 2 1 2 3 16 30. (a, b) : Let r be common ratio of G.P., we have
a 1 +i +i i ib ia b r z = = = ⇒ z = or c c b − i r − i r b 31. (a, b, c) : For any no. choosen from [1,180] there are 20 ways to select the second no. and from [181,199] there are 19, 18, …..1, ways resp. to select the second no. hence required no.of ways = 20 × 180 + (19 + 18 + …….+1) = 3790 32. (a, b, d) : If comes in first toss then Mr. B can win in only one case that is . 1 Probability of Mr. B winning = 4 3 Probability of Mr.A winning = 4 Given first toss is head, Mr. A can win in successive tosses are ,H, HH, …..
38. (a) : S h o r t e s t d i s t a n c e b e t w e e n t h e l i n e s
1 2
Probability =
=1 1 1− 2 Given first toss is head, Mr.A can win in the following cases HT, HHT, HHHT,….. 1 It is a G.P. with first term = , 4 1 common ratio = 2 1 1 Probability = 4 = 1 2 1− 2
r = a + la and r = b + µb is
given by
|(a − b ).(a × b)| |a ×b|
=3
10 19
39. (b) : Take B as origin, BC as x -axis and take A as (h, k)
C (4,0).
1 ...(i) × 4 × k = 2k 2 h + 4 , k D = (2,0) and E 2 2 AD ^ BE slope of AD × slope of BE = –1 ...(ii) k2 + (h + 4)(h – 2) = 0 2 2 Also AC = 3 ⇒ (h – 4) + k = 9 ...(iii) (ii)-(iii) gives Area of D ABC =
⇒
3 11 and k 2 = 2 4 11 ⇒ k= 2 From (i), Area of D ABC = 11 AB BD 40. (c) : Now AD is the bisector = ⇒ c = 2b AC DC b+c>a⇒ b+c>6 b>2 b2 + 4b2 − 3b Again <1 ⇒ b < 6 2 4b 2 < b < 6 and consequently 4 < c < 12
⇒ h=
33. (d) : Area OAB 1 2
1 y + 3 = 2 ∫ − y dy = sq. units 4 3 0 3
2 34. (a) : Area APQ = (x
∫
− 4x − 3 dx = 13/3 sq.units
1
41. (c) : Now c2 = h2 + k2 and b2 = (6 – h)2 + k2
b2 + 12 c − b = 12h − 36 ⇒ h = 4 Given that k2 > 10 ⇒ c2 – h2 > 10 2
35. (a) : Area BTR = Area of rectangle LMTR
−
y 2 + 3 dy = 7/3 sq. units 4
∫
−Area LMBR = 6 −
−3 36. (b) : Equation of line QS in 2-D will be x + 3 y – 9 = 0, RE =
9 10
81 , 3 , 0 , 10 10
and E ≡
2
b2 + 12 ⇒ 4b − > 10 4 2
⇒ b2 ∈(20 − 96 , 20 + 96 ) B is either 4 or 5
so point T will be
81 3 9 10 , 10 , 10 , 3 205 5 37. (c) : Direction ratio of QP ≡ 9, 0, 0 9 −3 −9 Direction ratio of QT ≡ , , 10 10 10 3 So, cosq = ⇒ tanq = 91 10 3
2
(42-44)
s−a 1 = ⇒ b + c = 3a ⇒ CA + BA = 12 2 s A lies on ellipse whose foci are B and C Centre of ellipse = (4, 4) and major axis parallel to x -axis Length of major axis = 12 units 1 12e = 4 ⇒ e = 3 Length of minor axis = 8 2 units
Given
Hence PT =
⇒ \ ⇒
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|MARCH ‘16
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42. (c) : PQ is the latus rectum of the ellipse.
47. A
→ s;
1 k22 32 Area = 2 × k1e × = ek22 = sq. units k1 3 2 43. (d) : DPBC is right angled at C .
(A)
44. (d) : Equation of Hyperbola is
⇒
2
(x − 4) 4
−
2
( y − 4) 4(e
2
− 1)
=1
It passes through (0, 2) 45. A
→ r;
⇒ e=
2 3
→ q; C → p; D → s
B
(A)
(B)
px 2
= (4n + 1)
3 x2
=
(C) sin
2
, n ∈ z
3 (4n + 1), n ∈ z 2
x = ±
p
px
3 2
= (x −
→ q; C → p; D → p n(s) = M n n( A) = n! ⇒ P( A) = B
M +n−1)
( (B) n(s) = P ( A) =
C M −1 n(A)=1
1 M +n−1
C M −1
1 M (C) n(s) = Cn n( A) = 1 ⇒ P( A) = M C n
(D) n(s) = M Cn n! n( A) = n ! ⇒ P( A) =
1
M
→ r;
C n
B → p; C → q; D → s (A) Te required event will occur if last digit in all the chosen numbers is 1, 3, 7 or 9. n 4 \ Required probability = 10 (B) Required probability = P (that the last digit is 2, 4, 6, 8) = P (that the last digit is 1, 2, 3, 4, 6, 7, 8, 9) 8n − 4n – P (that the last digit is 1, 3, 7, 9) = . 10n (C) Rquired prob. =P (1, 3, 5, 7, 9) – P (1, 3, 7, 9) 5n − 4n
48. A
=
3 )2 + 1
=
(10n − 8n ) − (5n − 4n ) 10n − 8n − 5n + 4n 5
4
(C) ( y – 2) 2 = –4 (x + 1) foot of ^ will lie on tangent at vertex x + 1 = 0 ⇒ k = 1. 1 1 1 1 (D) + = = 1⇒ 1 − ⇒ 3e 2 = 4 4 e12 e22 e2 MATHEMATICS TODAY
|MARCH ‘16
2
10n 3
− 5) dx − ∫ ( 4x − x2 − 3) dx
2
1
5
+ ∫ (4 x − x − 3)dx + ∫ (3 x − 15)dx = 50. (4) :
B → p; C → s; D → q (A) Let equation of tangent 2 2 y = mx + ⇒ 2 = –m + ⇒ m2 + 2m – 2 = 0 m m \ m1 + m2 = –2 1 1 m1m2 = –2 ⇒ + −1 = 0 m1 m2 (B) m1m2 = –1 ⇒ b2 = a2e2 ⇒ 4e2 = 2
∫ (6x − x 1
3
→ r;
=
10n
(6) : Area = 49.
x + y = 2n , x = 2m , y = 2k
80
M n
10n (D) Required prob. = P (0, 5) – P (5)
2 3 x = 3 x + y cos x − y = 2 sin x + y .cos x + y (D) 2 sin 2 2 2 2
46. A
n!
4
73 6
A / 2 cos A / 2 ∑ r a1 = 2R ∑ 4R sin2 sin A / 2 cos B / 2 cos C / 2 B C = ∑ tan + tan 2 2
r r + r + r A = 2∑ tan = 2 ∑ 1 = 4 1 2 3 a + b + c 2 s 1 + x 2 51. (1) : ≤ 1 ⇒ x = 1 ⇒ x = ±1
2 x
(0, 0, a ) (say) from plane (ii) is 2 56. (9) : Te given curve represents the point \ Minimum distance = 1.
57. (0) : Combined equation of the tangents drawn
from (0, 0)to the circle is
But x = = –1 will not satisfy the equation.
(x 2 + y 2 – 2rx – – 2hy + + h2)h2 = (–rx – – hy + + h2)2, here coefficient of x 2 + coefficient of y 2 = 0
≤ 144 and sin(x + y ) ≥ 0 + y ≤ (2 n + 1) ; n ∈ I ⇒ 2np ≤ x +
52. (8) : x 2 + y 2
Hence, we get the area S p ⋅144 ⇒ =8 S= 2 9p 53. (1) : [ x ] = 5 ± 25 + 4 sin x − 24 = 5 ± 1 + 4 sin x 2 2 ⋅1 –1 ≤ sinx ≤ 1 ⇒ –4 ≤ 4sinx ≤ 4 ⇒ –3 ≤ 1 + 4sinx ≤ 5 ⇒ 0 ≤ 1 + 4 sin x ≤ 5 ⇒ [x ] is an integer ⇔ sinx = 0 = p ⇒ [x ] = 3 ⇒ x = 54. (5) : Reflecting the point (1, 2, 3) in the xy -plane -plane produces (1,2,–3) . A half turn about the x -axis -axis yields (1, –2, 3). Finally translation 5 units will produce (1, 3, 3) 55. (2) : Equation of any plane; continuing the general plane is + 2z – – 3 + l(2x + + 3 y + + 4z – 4) = 0 ...(i) x + y + 1 if plane (i) is parallel to z -axis -axis ⇒ l = − 2 Terefore plane, parallel to z -axis -axis is ...(ii) y + 2 = 0 Now, Now, shortest distance between any point on z -axis -axis
− 1 , 2 2 .
⇒ (h2 – r 2) + (h2 – h2) = 0 ⇒
h = ±1 r
58. (1) : Solving the equation of the lines, we get
z = −z
⇒z =i a − 1 = 2; a = 2eiq + i, put it in the equation of the second line, we get cosq – sinq = 0 ip a = i ± 2e 4
59. (2) :
\ x = ± 2 ⇒ [|x |]|] = 1 p sin (2 sin x ) = sin − 2 cos x 2 p sin x + cos x = , implies 4
1 + sin 2 x =
p2 16
tan x + cot x =
2 2 × 16 32 = 2 = 2 sin 2 x p − 16 p − 16
a = 32, b = 16, c = 2 So, a + b + c = 2 25
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|MARCH ‘16
81
