10 76
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Vol. XXXV
No. 2
February 2017
Corporate Ofce: Plot 99, Sector 44 Institutional Area, Gurgaon -122 003 (HR), Tel : 0124-6601200 0124-6601200 e-mail :
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43
CONTENTS
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8
41
Maths Musing Problem Set - 170 23
10 Practice Paper - JEE Main e g d E n o i t i t e p m o C
Paper - JEE Advanced 19 Practice Paper Subscribe online at
23 You Ask We Answer 24 Practice Paper Paper - JEE Advanced
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2 yrs.
3 yrs.
Mathematics Today Today
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32 Mock Test Paper - WB JEE
Chemistry Today Today
330
600
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Physics For You You
330
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775
41 Math Archives
Biology Today Today
330
600
775
43 Brain@Work 63 Key Concepts on Conic Sections 73 MPP-8 I I X s s a l C
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MATHEMATICS TODAY
|
FEBRUARY‘17
7
10 76
19
Vol. XXXV
No. 2
February 2017
Corporate Ofce: Plot 99, Sector 44 Institutional Area, Gurgaon -122 003 (HR), Tel : 0124-6601200 0124-6601200 e-mail :
[email protected] website : www.mtg.in www.mtg.in Regd. Ofce: 406, Taj Apartment, Near Safdarjung Hospital, Ring Road, New Delhi - 110029. Managing Editor : Mahabir Singh Editor : Anil Ahlawat
32 87
73
43
CONTENTS
8
8
41
Maths Musing Problem Set - 170 23
10 Practice Paper - JEE Main e g d E n o i t i t e p m o C
Paper - JEE Advanced 19 Practice Paper Subscribe online at
23 You Ask We Answer 24 Practice Paper Paper - JEE Advanced
1 yr.
2 yrs.
3 yrs.
Mathematics Today Today
330
600
775
32 Mock Test Paper - WB JEE
Chemistry Today Today
330
600
775
Physics For You You
330
600
775
41 Math Archives
Biology Today Today
330
600
775
43 Brain@Work 63 Key Concepts on Conic Sections 73 MPP-8 I I X s s a l C
www.mtg.in
Individual Subscription Rates
Combined Subscription Rates
89 Maths Musing Solutions I X s s a l C
63
76 Ace Your Your Way Practice Practi ce Paper 87 MPP-8
1 yr.
2 yrs.
3 yrs.
PCM
900
1500
1900
PCB
900
1500
1900
1000
1800
2300
PCMB
Send D.D/M.O in favour of MTG Learning Media (P) Ltd. Payments should should be made directly to : MTG Learning Media (P) Ltd, Plot 99, Sector 44 Institutional Area, Gurgaon 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 (P) (P) Ltd., Okhla Industrial 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 incentives 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
|
FEBRUARY‘17
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.
JEE MAIN 1.
2.
3.
If the tangent at the point (a (a, b) on the curve 3 3 3 3 x + y + y = a + b meets the curve again at the point ( p, p, q), then (a) ap + bq + ab = 0 (b) bp + aq + ab = 0 ap + bq – pq (c) = 0 (d) bp + aq – pq = 0
5.
∆
u2 2 −u2 2 sin a (d) cos a, 2 g 2 g
=
R2 9 6 18 6 36 2 18 3 (a) (b) (c) (d) 35 35 35 35 3 −i If z = , then (i (i101 + z 101)103 = 2 (a) z (b) z (c) iz (d) iz Let m be the number of 5-element subsets that can be chosen from the set of the first 15 natural numbers numbers so that at least two of the five numbers are consecutive. Te sum of the digits of m is (a) 11 (b) 12 (c) 13 (d) 14
4 6 6 Let A = 1 3 2 . A value of l, such that −1 −5 −2 AX = = l X , where X where X is is a non-zero column vector, is (a) 0
(b) 1
(c) 2
Te parametric equations of a parabola are x = = u cos a⋅t , 1 2 y = = u sin a⋅t − gt , t is is a parameter. 2 7. Its latus rectum is 2u2 2 2u2 sin a cos2 a (a) (b) g g MATHEMATICS TODAY
|
FEBRUARY‘17
INTEGER MATCH 9.
I is the incentre of the triangle ABC . AI , BI , CI when produced meet the opposite sides at D, E, F respectively. Ten the value of AI BI CI AI BI CI is ⋅ ⋅ − + + ID IE IF ID IE IF MATRIX MATCH
10.
List-I contains S and List-II gives last digit of S. L ist-I
P. S = Q. S = R. S =
(d) 3
COMPREHENSION
8
u2 2 −u2 2 cos a (c) sin a, 2 g 2 g
If a1 = b1 = 1, then a22 = (a) 231 (b) 232 (c) 233 (d) 333 In a triangle ABC triangle ABC , if tan A tan A : : tan B : tan C = = 2 : 3 : 4,
JEE ADVANCED
6.
u2 −u2 cos 2a (b) sin 2a, 2 g 2 g
an+1 3 + 1 1 − 3 an , n ∈ N . Let = b 3 1 1 3 − + bn n+1
then
4.
8.
u2 u2 2 cos2 a (c) (d) sin a 2 g 2 g Its focus is u2 u2 s i n 2 , c o s 2 a a (a) 2 g 2 g
S. S = (a) (b) (c) (d)
P 4 4 2 1
L ist-II
11
∑ (2n − 1)2
1.
0
∑ (2n − 1)3
2.
1
∑ (2n − 1)2(−1)n
3.
5
∑ (2n − 1)3(−1)n−1
4.
8
n =1 10 n =1 18 n =1 15 n =1
Q 2 2 1 2
R 1 3 4 3
S 3 1 3 4
See Solution Set of Maths Musing 169 on page no. 89
Exam Dates OFFLINE : 2nd April ONLINE : 8th & 9th April
1.
2.
3.
4.
5.
6.
10
ree numbers are chosen at random from 1, 2, ..., 30. What is the probability that they will form a G.P.? 10 12 19 20 (a) (b) (c) (d) 4060 4060 4060 4060 If n balls are distributed into m boxes, so that each ball is equally likely to fall in any box, the probability that a specified box will contain r balls is n n Pr (m 1)n r Cr (m 1)nr (a) (b) n n C m C m n n r n Cr (m 1) Cr (m 1)nr (c) (d) mn nm e probability that at least one of the events A and B occur is 0.4. If A and B be occur simultaneously with probability 0.1, then probability of AC or BC is equal to (a) 1.2 (b) 1.4 (c) 1.5 (d) 1.8 In an attempt to land an unmanned rocket on the moon, the probability of a successful landing is 0.4 and the probability of monitoring system giving the correct information of landing is 0.9 in either case. e probability of a successful landing, it being known that the monitoring system indicated it correctly 6 4 9 2 (a) (b) (c) (d) 7 25 10 5 Each of N + 1 identical urns marked 0, 1, 2, ..., N contains N balls so that the ith urn contains i black and N – i white balls (0 i N ). An urn is chosen at random and n balls are drawn from it one aer another with replacement. If all the n balls turn out to be black, the probability that the next ball will also black [assume that N is large] is n n n 1 n 1 (a) (b) (c) (d) n2 n 1 n2 n 1 e probability that a teacher will give a surprise test during any class meeting is 3/5. If a student is absent on two days, the probability that he will miss at least one test is MATHEMATICS TODAY | FEBRUARY ‘17
3 20 21 2 (b) (c) (d) 5 25 25 5 If two independent events A and B are such 3 8 that P ( A BC ) and P ( AC B) and 25 25 1 P ( A) , the value of P ( A) + P (B) = 2 8 4 (a) (b) 5 5 7 (c) (d) None of these 5 If two events A and B are such that P ( AC ) = 0.3, P (B) = 0.4, P ( ABC ) = 0.5, then P (B| A BC ) is 4 1 1 1 (a) (b) (c) (d) 5 2 4 5 (a)
7.
8.
9.
10.
e independent probabilities that A, B, C can 1 1 1 solve the problem are , , respectively. e 2 3 4 probability that just two of them only solve the problem is 2 3 (a) (b) 3 4 1 (c) (d) None of these 4 A bag A contains 2 white and 3 red balls and bag B contains 4 white and 5 red balls. One ball is drawn at random from one of the bags and is found to be red. e probability that it was drawn from bagB is (a)
11.
0 (a) (b) (c) (d)
25 8 13 25 (b) (c) (d) 51 17 27 52 e set of equations x − y + (cos )z = 0 3x + y + 2 z = 0 (cos)x + y + 2z = 0 < 2, has non−trivial solutions for no value of and for all values of and for all values of and only two values of for only one value of and all values of .
12.
Orthogonal trajectories of family of parabolas y 2 = 4a(x + a) where ‘a’ is an arbitrary constant is (a) ax 2 = 4cy
20.
2
{ f (1)}
(b) x 2 + y 2 = a2
n
(c) y = (d) axy = c2, where c is a constant. If
n
21.
5z 1 2z1 + 3z 2 is purely imaginary, then 7 z 2 2z1 − 3z 2
(a) 5/7
(b) 7/5
=
(c) 25/29 (d) 1
14.
Te mean daily profit made by a shopkeeper in a month of 30 days was ` 350. If the mean profit for the first twenty days was ` 400, then the mean profit for the last 10 days would be (a) ` 200 (b) ` 250 (c) ` 800 (d) ` 300
15.
Consider points A(3, 4) and B(7, 13). If P be a point on the line y = x such that PA + PB is minimum, then coordinates of P are 13 13 12 12 (a) , (b) , 7 7 7 7 31 31 (c) , (d) (0, 0) 7 7
16.
lim x →0
log(2 + x 2 ) − log(2 − x 2 ) 2
x (b) 2
(a) –1 17.
Te sum of the series is equal to (a) 1
18.
(b) 0
= k, the value of k is
(c) 1 3 2
2
1 ⋅2
+
22.
23.
(d) 0 5 2
2
2 ⋅3
(c) –1
+
7 2
3 ⋅ 42
(d) 2
24.
∑= f (k) = k 1
(b) n(n + 1)(2n + 1)/6 (d) n(n + 1)/2
sin(a + 1)x + sin x , x c , 19. If f ( x ) = + 2 1/2 − 1/2 x (x bx ) , bx 3/2
for
x < 0
for
x = 0
forr
x > 0
is continuous at x = 0, then a = (a) 3/2 (b) –3/2 (c) 1/4 (d) –1/4
12
MATHEMATICS TODAY
|
FEBRUARY ‘17
2nC
n
2nC 1
–1
For the three events A, B and C, P (exactly one of the events A or B occurs) = P (exactly one of the events B or C occurs) = P (exactly one of the events C or A occurs) = p and P (all the three events occurs simultaneously) = p2, where 0 < p < 1/2. Ten the probability of at least one of the three events A, B and C occurring is 1 1 2 2 (a) (3 p + 2 p ) (b) ( p + 3 p ) 2 4 1 1 2 (c) ( p + 3 p2 ) (d) (3 p + 2 p ) 4 2 Te area of the smaller region bounded by the curves x 2 + y 2 = 5 and y 2 = 4x is π 1 π 1 + (a) (b) − 4 6 4 6 1 + 5 sin−1 2 1 5 −1 2 2 (c) 2 − sin (d) 3 2 3 2 5 5 x + a2
ab
ac
If a, b, c are real, then f (x ) =
ab
x + b2
bc
is decreasing in
ac
bc
x + c2
− 2 (a2 + b2 + c2 ), 0 3 2 (b) 0, (a2 + b2 + c2 ) 3 a2 + b2 + c2 (c) , 0 (d) None of these 3
+ ...∞
n
(b) (d)
(a)
If f : R → R, f (x + y ) = f (x ) · f ( y ). If f (1) = 1, then
(a) n (c) 1
2 2 f n (1) f ′(1) f ′′(1) + + + .... + n 1! 2 ! !
is given by (a) 2nC (c) 2nC + 1
x − ce 2a
13.
If f (x ) = x n, then the value of
25.
If P (n) : n2 + n is an odd integer. It is seen that truth of P (n) ⇒ the truth of P (n + 1). Terefore, P (n) is true for all (a) n > 1 (b) n (c) n > 2 (d) none of these 2 2 2 If y = cos x + cos x + cos x + ...... to ∞ ,
then (a)
dy is dx
− sin x 2 x(2 y − 1)
(b)
−2x sin x 2 2 y − 1
(c)
− sin x 2 y − 1
(d) none of these
26. Area lying between the curves y 2 = 4x and y = 2x is
(a) 2/3 (c) 1/4
(b) 1/3 (d) none of these
27. If f (x ) is a continuous function satisfying f (x ) f (1/x ) = f (x ) + f (1/x ) and f (1) > 0, then lim f ( x ) x →1
is equal to (a) 2 (c) 3
(b) 1 (d) none of these
28. For any integer n, the integral π
∫ e cos
2
x
cos 3 x( 2n + 1)x dx has
the value
0
(a) π (c) 0
(b) 1 (d) none of these
29. Let E and F be two independent events. Te probability that both E and F happen is 1/12 and the probability that neither E nor F happens is 1/2.
Ten (a) P (E) = 1/3, P (F ) = 1/4 (b) P (E) = 1/2, P (F ) = 1/6 (c) P (E) = 1/6, P (F ) = 1/2 (d) P (E) = 1/8, P (F ) = 1/3 30. If nC r denotes the number of combinations of n things taken r at a time, then the expression n−1 n n+ k C0 + C k +1 equals k =0
∑
(a) (c)
2nC n – 1 2nC n
(b) 2nC n + 1 (d) nC n – 1 SOLUTIONS
1. (c): Let A be the event that the numbers of the
triplets form a G.P. Since 3 numbers can be selected from 30 numbers in 30C ways, therefore total number of triplets = 30C 3 3 Now we count the triplets (arranged in increasing order) whose terms form a G.P. by listing them as follows. Common ratio
2 3 4 5 3/2
Triplet
{(i, 2i, 4i), 1 ≤ i ≤ 7} {(i, 3i, 9i), 1 ≤ i ≤ 3} (1, 4, 16) (1, 5, 25) (4, 6, 9), (8, 12, 18), (12, 18, 27)
5/2 4/3 5/3 5/4
(4, 10, 25) (9, 12, 16) (9, 15, 25) (16, 20, 25)
Tus among 30C 3 triplets, there are 19 triplets, whose terms form a G.P. 19 19 \ Required probability P(A) = 30 = C 3 4060 2. (c): Te total number of ways of distributing n balls into m boxes is mn which are assumed to be equally
likely. Since r balls which should go to the specified box can be chosen in nC r ways and for any such way the remaining (n – r ) balls can be distributed to the (m – 1) boxes in ( m – 1)n – r ways. n \ Te number of favourable cases = (m − 1)n−r r n n− r r (m − 1) \ P( A) = mn [where A is the event that a specified box will contain r balls] 3. (c): Given that P ( A ∪ B) = 0.4 and P ( A ∩ B) = 0.1 \ P ( A) + P (B) – P ( A ∩ B) = 0.4 or P ( A) + P (B) = 0.4 + 0.1 = 0.5 or 1 – P ( AC ) + 1 – P (BC ) = 0.5 or P ( AC ) + P (BC ) = 2 – 0.5 = 1.5 4. (d) : Let A denote the event of a successful landing of the rocket, B1 denote the event of the monitoring system indicating it correctly and B2 denote the event of its indicating unsuccessful landing. \ P ( A) = 0.4, P (B1| A) = 0.9, P (B2| AC ) = 0.9 P ( B1 | A) P( A) Te required probability = P ( A| B1) = P ( B1 ) C Now, P (B1) = P ( A ∩ B1) + P ( A ∩ B1) = P (B1| A)P ( A) + P (B1| AC ) P ( AC ) = P (B1| A)P ( A) + [1 – P (B1C | AC )] P ( AC ) = P (B1| A) P ( A) + [1 – P (B2| AC )] P ( AC ) = 0.9 × 0.4 + (1 – 0.9) × 0.6 = 0.42 0.9 × 0.4 6 Hence the required probability = = 0.42 7 th 5. (a) : Let Ai be the event of choosing i urn and B the event of choosing n black balls in succession and C the event of drawing (n + 1)th ball as black. Te required probability is MATHEMATICS TODAY
|
FEBRUARY‘17
13
P( B ∩ C ) P ( B)
P (C | B) =
Now, P (B) =
N
∑= P( A ) ⋅ P( B |A ) i
i
i 0 n
N
1
1 i ∞ = = x ndx when N → + N 1 n i =0 0 Similarly,
∑
∫
n+
1
1 i 1 P (B ∩ C ) = ∞ = x n+1dx when N → + N 1 N i =0 0 Hence the required probability N
∑
N
=
∑= ni i 0 N
∑= ni i 0
=
n+1
n
∫
=
1
∫ x + dx n 1
=
0 1
if n is large
∫ x dx n
P ( A ∪ BC )
=
P ( A) − P( A ∩ BC ) P ( A ∪ BC )
0.7 − 0.5 0.2 1 = = 0.8 0.8 4
1 1 1 , P( B) = , P(C) = 2 3 4 \ Probability that only two of them can solve the problem = P ( A ∩ B ∩ C C ) + P ( A ∩ BC ∩ C ) + P ( AC ∩ B ∩ C )
0
n +1 n+2
= 1 – P (no test is missed) = 1 – P (no test on his first day of absence and no test of his second day of absence) = 1 – P (no test of his first day of absence) × P (no test of his second day of absence) 21 = 1 − 2 2 = 5 5 25 3 7. (c): Given that P ( A ∩ BC ) = 25 8 1 and P ( AC ∩ B) = , P ( A) > , 25 2 Let P ( A) = x , P (B) = y \ We are given that P ( A ∩ BC ) = P ( A) – P ( A ∩ B ) 3 = x − P( A) P( B) 25 3 = x − xy [ Q A and B are independent] 25 Also, P ( AC ∩ B) = P (B) – P ( A ∩ B) 8 1 ⇒ = y − xy. Hence y = x + 25 5 1 2 \ Solving x and y we get x = , y = 5 5 3 4 or x = and y = 5 5 MATHEMATICS TODAY
=
P ( A ∩ B)
9. (c): Given that P ( A) =
6. (d) : Required probability
14
1 3 4 As P ( A) > , we must have P ( A) = and P( B) = 2 5 5 7 \ P( A) + P( B) = 5 8. (a) : P ( AC ) = 0.3, P (B) = 0.4, P ( ABC) = 0.5 \ P ( A ∪ BC ) = P ( A) + P (BC ) – P ( A ∩ BC ) = 0.7 + 0.6 – 0.5 = 0.8 Now, P (B | A ∪ BC ) P[ B ∩ ( A ∪ BC )] P[( B ∩ A) ∪ ( B ∩ BC )] = = P ( A ∪ BC ) P ( A ∪ BC )
|
FEBRUARY ‘17
1 1 3 2 3 4
1 2 1 2 3 4
1 1 1 6 1 = 2 3 4 24 4 1 10. (a) : P (E1 ) = P( E2 ) = 2 3 5 and P(R |E1) = , P( R| E2 ) = 5 9 By Bayes' theorem P( E2 ) ⋅ P( R | E2 ) P ( E2 | R) = P (E1 ) ⋅ P( R | E1) + P( E2 ) ⋅ P( R | E2 )
= ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ =
1 5 5 ⋅ 25 = 2 9 = 9 = 1 3 1 5 3 5 27 + 25 ⋅ + ⋅ + 2 5 2 9 5 9 25 P ( E2 | R) = 52 11. (a) : Determinant of coefficients λ −1 cos θ 1 2 = cos θ − cos2 θ + 6 = 3 cos θ 1 2 and this is positive for all θ since |cosθ| ≤ 1. Te only solution is therefore the trivial solution. dy dy 2a = 4a ⇒ = dx dx y For orthogonal trajectory, 12. (c): 2 y ⋅
MATHEMATICS TODAY
|
FEBRUARY‘17
15
dx 2a − = dy y
⇒
19. (b) : f (0 − 0) = lim f (0 − h )
dy = y
⇒ ∫
x ln y = − + k 2a
⇒
1 − dx 2a − x y = ce 2a
∫
h→0
=
z = Ki (K ∈ R), then 1 = 7K i z 2 5 z 2 1 + 3 14 K i + 3 2z1 + 3z 2 z 2 = 5 =1 = Consider 14 K 2z1 − 3z 2 z 1 i−3 2 −3 5 z 2 14. (b) : 30 × 350 = 20 × 400 + 10 × x 15.
⇒ 10x = 2500 ⇒ x = 250 (c): Consider a point A′, the image of A through
10x = 10500 – 8000
y = x
\
Coordinates of A′ = (4, 3) [Notice that A and B lie to the same side with respect to y = x ]. Ten PA = PA′ Tus, PA + PB is minimum, if PA′ + PB is minimum, if P , A′, B are collinear. 13 − 3 ( x − 4) ⇒ 3 y − 10 x + 31 = 0 Now, AB is y − 3 = 7−4
31 31 It intersects y = x at , , which is the required 7 7 point P . 1 16. (c): lim [log(2 + x 2 ) − log(2 − x 2 )] 2 x →0 x 2 + x 2 2 − x 2 − log log x →0 x 2 2 2 1/ x 2 1/ x 2 2 2 = lim log 1 + x − log 1 − x x →0 2 2 = lim
1
1 1 2 2 17. (a) : Te given series is, ∞ ∞ (2r + 1) 1 ∞
= log e1/2 − log e −1/2 = + = 1
∑= r (r + 1) = ∑= r − ∑= (r +11) r 1
2
2
18. (a) : f (1) = 1
r 1
2
2
r 1
⇒
f (2) = 1 f (r ) = 1 ∀ r = 1, 2, ...., n
⇒
∑= f (k) = n k 1
MATHEMATICS TODAY
|
=1
f (3) = 1
n
16
sin(a + 1)(−h) + sin(−h)
FEBRUARY ‘17
=
0−h
h →0
5z 1 13. (d) : Let 7 z 2
⇒
lim
h→0
f (0 + 0) = lim f (0 + h) = lim h→0 h→0
= lim
bh − 1 2 1/2 bh ⋅ h
b1/2 1 +
h→0
lim
=
(a + 1 + 1)(−h) (−h)
=
a +2
(h + bh2 )1/2 − h1/2 bh3/2
1 2
\
Te given function is continuous f (0 – 0) = f (0 + 0). 1 3 \ a+2= , a= − 2 2 f r (1) n n f x x ( ) = ⇒ = C r 20. (a) : r ! n
∑= ( C ) n
r
2
=
2n
C n
r 0
21. (a) : P( A ∪ B ∪ C ) = ∑ P( A) − P( AB) − P (BC ) − P( AC) + P( A ∩ B ∩ C )
P( A) − P( AB) + P( B) − P( BC) + P( B) 1 = − P( AB) + P(C) − P( BC) + P( A) − P( AC) + P( ABC) 2 + P P (C) − P( AC) 1 {P( AB ) + P( AB ) + P( AC) + P( BC) + P( BC) + P( AC)} 2 + P ( ABC ) 1 1 = (3 p) + p2 = (3 p + 2 p2 ) 2 2
WEST BENGAL at
MATHEMATICS TODAY
|
FEBRUARY‘17
17
⇒
(2x )2 = 4x ⇒ 4x 2 = 4x 4x 2 – 4x = 0 ⇒ 4x (x – 1) = 0
⇒ x = 0, 1
1
Required area =
22. (d) :
∫ (2
x − 2 x )dx
0
1
2 = 2 ⋅ x 3/2 − x 2 3 0 2
Area =
4 3
= −1 =
1 3
27. (a) : Since f is continuous function so,
2 2 y dy 5 y − − ∫ −2 4
lim f ( x) = f (1).
x →1
2 1 5 2 = 2 ∫ 5 − y 2 − y dy = 2 + sin−1 5 3 2 4 0
Put x = 1 in the given equation we have ( f (1))2 = 2 f (1), so f (1) = 0 or 2. Since f (1) > 0, so f (1) = 2.
23. (a) :
28. (c)
2
1
0
x + a2 ab
0
2 f ′( x ) = ab x + b
bc
ac
x + c2
bc
+
ac
0
1
ac
bc x + c2 x + a2
+ ab
0
ab
ac
x + b2
bc
0
1
0
= (x + b2)(x + c2) – b2c2 + (x + a2)(x + c2) – a2c2 + (x + a2)(x + b 2) – a2b2 = 3x 2 + 2x (a2 + b2 + c2) f (x ) will be decreasing when f ′(x ) < 0 ⇒ 3x 2 + 2x (a2 + b2 + c2) < 0 ⇒ x ∈ − 2 (a2 + b2 + c2 ), 0 3 24. (d) : Te sufficient condition for the statement :
P (n) to be true for n ≥ a is (i) truth of P (n) ⇒ truth of P (n + 1) (ii) P (a) is true. Hence answer is none of these. 25. (b) : y = cos x
\ ⇒
2
+ y ⇒ y2 − y = cos x 2
dy dy 2 y − = − sin x 2 ⋅ 2x dx dx dy 2 x sin x 2 =− dx 2 y − 1
1 12 1 1 (1 − x )(1 − y) = , or 1 − ( x + y ) + xy = 2 2 1 1 7 7 1 x + y = + = x + y = , xy = ; 2 12 12 12 12 1 1 P (E ) = and P( F ) = 3 4 1 1 or, P ( E) = and P(F ) = 4 3 29. (a) : Let P (E) = x , P (F ) = y , xy =
n−1
n
30. (c): C0
18
|
FEBRUARY ‘17
Ck +1 = (n C0 + nC1 ) + n+1C2
= ( n + 2C 2 + n + 2C 3) + .... + Similarly,
2n – 1C
n – 1 +
2n – 1 C
n
2n – 1 C = 2nC
n
n
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k =0
n+k
+ n + 2C 3 + .... + n + n – 1C n = ( n + 1C 1 + n + 1C 2) + n + 2C 3 + .... + 2n – 1C n
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SECTION-1
6.
SINGLE OPTION CORRECT TYPE
Let A, B, C , D be distinct points on a circle with centre O. If there exists non-zero real numbers x and y such that | x OA + y OB | = | x OB + y OC |
1.
Consider the two statements. S 1 : he number of solutions to the equation
2.
3
5
(c) 11 40 3.
4.
5.
(a) a square (c) a rectangle
21 (b) 40
4 persons are playing a game of toy-gun shooting. At exactly midnight, each person randomly chooses one of the other three and shoots him. Te probability that exactly 2 persons are shot is 1 (a) 8 (b) 4 (c) 2 (d) 4 27 9 3 In a certain lottery, 7 balls are drawn at random from n balls numbered 1 through n. If the probability that no pair of consecutive numbers is drawn equals the probability of drawing exactly one pair of consecutive numbers, then n = (a) 50 (b) 60 (c) 54 (d) 64 If B is an idempotent matrix satisfying the condition (I – aB)–1 = I – 3 B, where I is unit matrix of same order as B and a ∈ R then 2a = (a) 1 (b) 2 (c) –1 (d) 3
(b) a trapezium (d) none of these
An ellipse has foci at (9, 20) and (49, 55) in the xy plane and is tangent to x -axis. Te length of its major axis is (a) 70 (b) 75 (c) 80 (d) 85
8.
A rectangular box has faces parallel to co-ordinate planes. If two of its vertices are (1, –1, 0) and (2, 3, 6) then volume of the box is (a) 20 (b) 22 (c) 24 (d) 26
9.
Te number of distinct real number pairs ( a, b) such that a + b ∈ integers and a2 + b2 = 2 is/are (a) 10 (b) 6 (c) 2 (d) 4
10.
Let n ≥ 3 be an integer and z = cis
3
5 21 (d) 40
7.
Te probability that a random chosen divisor of 3039 is a multiple of 30 29 is 11 (a) 40
= | x OC + y OD | = | x OD + y OA |, then ABCD is
ex = x 2 is 1. S 2 : he number of solutions to the equation ex = x 3 is 1. (a) S1 is true, S2 is false. (b) S1 is true, S2 is true. (c) S1 is false, S2 is true. (d) S1 is false, S2 is also false.
2p . Consider n
the sets A = {1, z , z 2, ....., z n –1} and B = {1, 1 + z , 1 + z + z 2, ..., 1 + z + z 2 + ... + z n – 1 }. Te number of elements in the set ( A ∩ B) when n is even is (a) 0 (b) 1 (c) 2 (d) 3 11.
Let a and b be positive real numbers such that a[a] = 17 and b[b] = 11 then ( a – b) = (a) 3/7 (b) 4/7 (c) 7/12 (d) 5/12 ([·] denotes greatest integer function]
12.
Let a function f (x ), x ≠ 0 be such that 1 1 f (x) + f = f (x) ⋅ f then f (x ) can be x x (a) 1 – x 4 + x 3 (b) | x | −x + 1 p 2x (c) (d) − 1 1 + λ log | x | 2 tan | x |
By : Tapas Kr. Yogi, Mob : 9533632105. MATHEMATICS TODAY
|
FEBRUARY ‘17
19
13. Assume that a (> 1) is a root of the equation
x 3 – x – 1 = 0 then (a) 1 (b) 2
3
3a2 − 4a + 3 3a2 + 4 a + 2 = (c) 2a (d) a1/3
14. Te maximum and minimum of the function 6sinx cos y + 2sinx sin y + 3cosx is (a) 7 and –7 (b) 5 and –5
(c) 6 + 13 and 6 − 13 (d) 5 + 2 and 5 − 2 15. For a real number a, let p I(a) = ∫ log(1 − 2a cos x + a2 ) dx then I (a ) = 0
(a) I (a
1 2 (b) I (a ) 2 (d) I (–a2)
2)
(c) 2I (a2)
21. 4r 22 – 4r 24 = (a) r 12 (b) r 22
(c) r 32
(d) r 32 – r 12
(c) 4
(d) 8
22. r 12 + s22 =
(a) 1
(b) 2
SOLUTIONS
1. (a) : For x < 0, e x = x 2 has one solution and ex = x 3 has no solution. x For x > 0, consider the function f (x ) = log x log x − 1 \ f ′(x ) = (log x )2
So, for x > e, f (x ) increases and for x < e, f (x ) decreases and moreover x = 1 is an asymptote. So, graph of f (x ) is
16. Te minimum of the expression x 2 + 16 − 2 (1 + cos q)x + 4 (1 + sin q) x x 2
+ (3 + 2cos q + 2sinq) for x > 0 and (a) 2 2 − 1 (c) 2 2 + 1
q ∈ [0, 2p] is (b) 2 2 + 3 (d) none of these
17. Te area of the region contained by all the points (x , y ) such that x 2 + y 2 ≤ 100 and sin(x + y ) ≥ 0 is (a) 10p (b) 25p (c) 50p (d) 100p SECTION-2 COMPREHENSION TYPE
Passage-1 Let a be a fixed real number, a a
∈(0, p) and
1 − r cos u
I r = ∫ du 2 − + 1 2 r cos u r −a 18. I 1 = (a) 0 (b) a (c) 2a
(d) –a
19. lim I r = r →1+
(a) a – p (b) a + p (c) a
Hence, probability that exactly 2 persons were shot 4 8 = 4C 2 × = 81 27
(d) –a
4. (c) : Tere are n – 6C 7 ways to draw 7 balls so that no two balls are consecutive and (n – 6) × n – 7 C 5 ways to draw 7 balls so that there is exactly one pair of consecutive balls. Hence, n – 6C 7 = (n – 6) × n – 7C 5 gives n = 54
Let r 1 < 0 < r 2 < r 3 be the real roots of 8x 3 − 6x + 3 = 0 a n d l e t s 1 < 0 < s 2 < s 3 be the real roots of 8x 3 – 6x + 1 = 0, then MATHEMATICS TODAY
|
3. (a) : Let the 4 persons were A, B, C , D. First, we choose which two were shot = 4C 2. Let C , D were shot. Ten C D and A → C or D and B → C or D. Any person shooting any other person has probability = 1/3. So, probability that C , D were shot 1 1 2 2 4 = × × × = 3 3 3 3 81 [for C, D ] [for A, B]
Passage-2
20
2. (a) : 3039 = 239 × 339 × 539. So, 40 × 40 × 40 divisors in total. 2 a 3 b 5 c to be a multiple of 30 29 , we must have a ∈ [29, 39], b ∈ [29, 39], c ∈ [29, 39] i.e. 11 choices for each a, b, c. 3 11 × 11 × 11 11 = Hence, required probability = 40 × 40 × 40 40
(d) –a
20. lim I r = r →1−
(a) a – p (b) a + p (c) a
Now, y = 2 does not intersect the graph but y = 3 meets the graph at 2 different points. Hence, in total ex = x 2 intersects once and ex = x 3 intersects twice.
FEBRUARY ‘17
5. (d) : From given relation, I = ( I – 3B)(I – aB)
12. (2) : Rearranging the given fractional equation, we
i.e., I = I – aB – 3 B + 3aB = I + 2 aB – 3B Hence, 2a = 3
have
1 −1 = 1 , x f (x ) − 1
f
6. (a) : Squaring, | x OA + y OB |2 = | x OB + y OC |2 2
2
⇒ x OA ⋅ OA + y OB ⋅ OB + 2xy OA ⋅ OB
= x 2OB ⋅ OB + y 2OC ⋅ OC + 2xy OB ⋅ OC
i.e., x 2r 2 + y 2r 2 + 2xy OA ⋅ OB
13. (b) : Since, a3 – a – 1 = 0, we have 3
p
which is satisfied by only f (x ) = out of the −1 x , 2 tan four options. 3a2 − 4a = 3 3a2 − 4a − (a3 − a − 1) = 3 (1 − a)3 = 1 − a
= x 2r 2 + y 2r 2 + 2xy OB ⋅ OC and 3 2 3a + 4a + 2 + a3 − a − 1 = 3 (1 + a)3 = 1 + a Hence, ABCD is a square. 7. (d) : Let F 1 and F 2 be the foci and P be the point
of tangency. Let F 1′ be the image of F 1 in x -axis then F 1P + F 2P = F 1′P + F 2P = F 1′F 2 = 2a
= (40)2 + (75)2 = 85 8. (c) : Vertices are (1, –1, 0) and (2, 3, 6). Notice that
these must be diagonal vertices. Making (1, –1, 0) as origin, we have (2, 3, 6) as (1, 4, 6). Hence, dimensions of the box are 6 × 4 × 1 i.e. Volume = 24 cubic units 9. (b) : Using, R.M.S ≥ A.M., we have | a + b| ≤ 2 So, a + b ∈ {–2, –1, 0, 1, 2} and a2 + b2 = 2 gives 1 ± 3 1 3 , (a, b) = (1, –1), (–1, 1), , 2 2 −1 ± 3 −1 3 , 2 2 10. (c) : Clearly, 1
Hence, given expression = 2 14. (a) : Using Cauchy-Schwarz inequality \ (x 1 y 1 + x 2 y 2 + x 3 y 3)2 ≤ ( x 12 + x 22 + x 32) ( y 12 + y 22 + y 32) We have, (6sinx cos y + 2sin x sin y + 3cosx )2 ≤ (62 + 22 + 32) ((sinx cos y )2 + (sinx sin y )2 + cos2x ) i.e. (6sinx cos y + 2sin x sin y + 3cosx )2 ≤ 49 Hence, maximum = 7 and minimum = –7
∈ A ∩ B. Let w ∈ A ∩ B, w ≠ 1
As an element of B, for some k = 1, 2, 3, ... (n – 1) 1 − z k +1 k 2 w = 1 + z + z + .... + z = 1 − z For w ∈ A, we have |w| = 1 1 − z k +1 (k + 1)p p ⇒ = 1 ⇒ sin = sin n n 1 − z i.e. k = n – 2 1 − 1 / z −1 = So, w = z 1− z −1 So, A ∩ B = 1, z
{ }
11. (c) : a[a] = 17
⇒ a ∈ (4, 5). So, [a] = 4
11 17 . Similarly, b[b] = 11 ⇒ b = 3 4 7 Hence, a − b = 12 and a =
MATHEMATICS TODAY
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FEBRUARY ‘17
21
15. (b) : We have, I (a) = I (–a) by putting x = p – y.
Putting tan( u/2) = t and simplifying
Now, we have (1 – 2acosx + a2)(1 + 2acosx + a2) = 1 – 2a2 cos2x + a4 p So, I(a) + I(−a) = ∫ log(1 − 2a2 cos 2x + a4) dx
J r =
(1 − r 2 ) 4 1 + r a tan−1 ⋅ So, Ir = a + 2 1 − r tan 2 2 | 1 − r |
0
1 2p log(1 − 2a2 cos y + a 4 ) dy, ( y = 2 x ) ∫ 20 1 2 1 2p = I(a ) + ∫ log(1 − 2a2 cos y + a4) dy 2 2 p Putting y = 2 p – t we have 2p ∫ log(1 − 2a2 cos y + a4) dy = I (a2) p 1 1 Hence, I(a) + I (−a) = I( a2) + I(a2) 2 2
1 + r a So, lim Ir = a − 2 tan −1 r − 1 tan 2 r →1 r →1+
=
1 So, I(a) = I(−a) = I (a2) 2 16. (d) : Te given expression can be rearranged as
4 [x − (1 + cos q)]2 + − (1 + sin q) x
2
1 2 So, minimum distance = 2 2 − 1 + = 2 −1 2 17. (c) : Using the symmetry, required area of the region
1 × 100p = 50p sq. units 2 a a 1 1 − cos u 18. (b) : I 1 = ∫ du = ∫ du = a −a 2 −a 2 − 2 cos u is
1 2
2
19. (a) : For r > 0, Ir = a + (1 − r ) ⋅
(1 − r 2) Ir = a + ⋅ J r (let) 2 a
du J r = ∫ 2 −a 1 − 2r cos u + r 22
MATHEMATICS TODAY
2
20. (b) : lim Ir = a + 2 tan r →1−
2
21. (c) : Putting x = sinq in 8x 3 − 6x + 3 = 0 , we have
sin 3q =
FEBRUARY ‘17
3 2
p 2 p 7p 8p 13p 14p ⇒ q= , , , , , 9
Hence, r1 = sin
9
9
9
in [0, 2p]
9
p 13p 4p 2p = − sin , r2 = sin , r 3 = sin 9 9 9 9
Similarly, putting x = cosq in 8x 3 – 6 x + 1 = 0, 1 we have, cos3q = − 2 p 8p So, s1 = cos = − cos , and 9 9 4p 2p s2 = cos , s3 = cos 9 9
p
p
p
Hence, 4r22 − 4r 24 = 4 sin2 − 4 sin 4 = sin2 2 = r 32 9 9 9
= sin2
4p 4p + cos2 = 1 9 9
du
MPP-8 CLASS XII
3.
ANSWER
KEY
4.
(a)
5.
(c,d)
10. (a,b)
1.
(c)
2. (d)
6.
(c)
7. (a,b,c) 8. (a,b,c) 9.
(d)
(b)
11. (a,b,c,d)
12. (a,b)
13. (b,c)
14. (a)
15. (d)
17. (2)
18. (1)
19. (1)
20. (3) |
−1 1 + r tan a 1 − r 2 r →1
p = a + 2⋅ = a + p
22. (a) : r12 + s22
∫ 1 − 2r cos u + r 2 −a
p = a −2⋅ = a − p
9
4 which is square of distance between point x , x on hyperbola xy = 4 and point (1 + cos q, 1 + sinq) on a circle centre (1, 1) and radius 1. Clearly the minimum distance occurs at (2, 2) on the hyperbola 1 , 1 + 1 on the circle. and 1 + 2 2
a
−1 1 + r tan a tan 1 − r 2 |1 − r 2 |
4
16. (d)
Y
U ASK
WE ANSWER Do you have a question that you just can’t get answered? Use the vast expertise of our MTG team to get to the bottom of the question. From the serious to the silly, the controversial to the trivial, the team will tackle the questions, easy and tough. The best questions and their solutions will be printed in this column each month. 1. Te base of a D is divided into three equal parts. If t 1, t 2, t 3 be the tangents of the angles subtended by these
parts at the opposite vertex, prove that: 1 + 1 1 + 1 1 + 1 (Yogesh, Delhi) = 4
t1
t 2 t2
t 3
t 2 2
Ans. Let the points P and Q divide the side BC in three equal parts such that BP = PQ = QC = x Also let, ∠BAP = a, ∠PAQ = b, ∠QAC = g and ∠ AQC = q B
A
In D A1OB, and
\
A1B OB
r R
= cos ∠ A1OB = cos
= tan
π n
, i.e., A1B = r tan
A1 A2 = 2 A1B = 2r tan
n
...(1)
π n
...(2)
n r
π + r tan , (from (1)) π 2n 2n cos n 1 + cos π 1 − cos π q 1 − cos q n ⋅ n tan = = 2r π π 2 sin q cos sin n n π π 1 − cos2 sin2 n = 2r ⋅ n = 2r ⋅ π π π π cos ⋅ sin cos ⋅ sin Now, 2(R + r ) tan
π
π
π
=2
n
= 2r tan
π n
n
n
n
= A1 A2, (from (2)).
3.
In how many ways can two distinct subsets of the set A of k(k ≥ 3) elements be selected so that they have ( Akhil, Assam) exactly two common elements? θ Ans. Let the two subsets be called P and Q. Te elements P Q C for the two sets will be selected as follows: From question, (i) 2 elements out of k elements for both the sets. tan a = t 1, tan b = t 2, tan g = t 3. Tis can be done in kC 2 ways. Applying, m : n rule in triangle ABC we get, (2x + x ) cot q = 2x cot (a + b) – x cot g ...(1) (ii) r elements for the subset P from k – 2 elements and any number of elements for Q from the remaining From D APC , we get (x + x ) cot q = x cot b – x cot g ...(2) k – 2 – r elements. Here r can vary from 0 to k – 2. For a fixed r , the number of selections Dividing (1) by (2), we get = k – 2C r ⋅ 2k – 2 – r , (because the number of selections 2 cot (a + b) − cot g 3 of any number of things from n = 2 cot b − cot g things is 2n.) 4 (cot a ⋅ cot b − 1) If r varies from 0 to k – 2, the total number of selections or 3 cot b – cot g = k−2 cot b + cot a k−2 k − 2 − r C r ⋅ 2 − 1, or 4 + 4 cot2 b = cot2 b + cot a ⋅ cot b + cot b ⋅ cot g = ∑ r = 0 + cot g ⋅ cot a excluding the case when both the subsets are equal or 4(1 + cot2 b) = (cot b + cot a)(cot b + cot g ) having only the two common elements. 1 + 1 1 1 1 t But every pair of P, Q is appearing twice like {a1, a2, a3}, + + or 4 = t 22 t1 t 2 t 2 t 3 {a1, a2, a4, a5} and {a1, a2, a4, a5}, {a1, a2, a3}. Hence, the required number of ways 2. If a regular polygon of n sides has the k−2 circumradius R and inradius r then prove 1 k−2 k − 2 − r k C r ⋅ 2 − 1 = C 2 × ∑ that each side of the polygon is equal to 2 r = 0 π 2(R + r )tan . (Raman, Gujarat ) 1 k(k − 1) k – 2 2n ⋅ [( C 0 ⋅ 2k – 2 + k – 2C 1 ⋅ 2k – 3 = ⋅ Ans. Let A1 A2 be a side and O be the centre. Let 2 2 + k – 2C 2 ⋅ 2k – 4 + ... + k – 2C k – 2) – 1] OB ⊥ A1 A2. k(k − 1) k(k − 1) k − 2 Clearly, OA1 = R and OB = r . k−2 = {(2 + 1) 1} = ⋅ − ⋅ (3 − 1). 2π π 4 4 Also, ∠ A OA = and ∠ A OB = . 1
2
n
1
α γ β
n
MATHEMATICS TODAY
|
FEBRUARY‘17
23
*ALOK KUMAR, B.Tech, IIT Kanpur
3p/4
SINGLE OPTION CORRECT TYPE 1.
If f (x ) = 0 is a cubic equation with positive and distinct roots a, b, g such that b is the H.M. of the roots of f ′(x ) = 0. Ten a, b, g are in (a) A.P. (b) G.P. (c) H. P. (d) none of these
2.
All the roots of the equation 11z 10 + 10iz 9 + 10iz – 11 = 0 lie (a) inside |z | = 1 (b) on |z | = 1 (c) outside |z | = 1 (d) none of these
3.
Let y = f (x ) be a continuous and differentiable curve. Te normals at (1, f (1)), (2, f (2)) and (3, f (3)) make angles p/3, p/4 and p/6 with positive x -axis respectively. Ten value of 2
7.
8.
9.
(b) 2 f (2) – f (1) + 1 (d) none of these
sin(p cos2(tan(sin x )))
lim x →0 (a) p (c) p/2
2
If I 1 =
x
10.
(b) p/4 (d) none of these
If a = x + y + z + w and b = (xy + yz + z w + wx + w y + xz ), then which of the following statement(s), is/are true? (a) 8a2 ≥ 3b (b) 3a2 ≥ 8b (c) a2b ≥ 27 (d) none of these ^
^
^
^
^
If a = i + j + k , a ⋅ b = 1 and a × b = j − k , then b is equal to ^ ^ ^ ^ (a) 2 i (b) i − j + k
(c) ^i
^
(d) 2 j
− k^
11.
101
MATHEMATICS TODAY
|
FEBRUARY ‘17
−
1 2
dx
101
dx
I
∫ 5 + 2x − 2x 2 , then I 1 is 2 −100 (b) 2 (d) none of these
| x |, for 0 < | x | ≤ 3 Let f (x ) = , then at x = 0 f (x ) x = 0 1, for has (a) a local maximum (b) no local maximum (c) a local minimum (d) no extremum 15 5 3 If x 2 + 9 y 2 + 25z 2 = xyz + + , then x , y , z x y z are in (a) A.P. (b) G.P. (c) A.G.P. (d) H.P.
12.
Te set of values of ‘ r ’ for which 23 Cr + 2 ⋅ 23Cr +1+ 23 Cr +2 ≥ 25C15 contains (a) 3 elements (b) 4 elements (c) 5 elements (d) 6 elements
* Alok Kumar is a winner of INDIAN NATIONAL MATHEMATICS OLYMPIAD (INMO-91). He trains IIT and Olympiad aspirants. 24
− p /2
∫ (5 + 2x − 2x 2)(1 + e2− 4x ) and −100
(a) 1/2 (c) 0
is equal to
= k ∫ sec xdx ,
A line meets the co-ordinate axes at A and B. A circle is circumscribed about DOAB. If the distances from A and B of the tangent to the circle at the origin be m and n respectively, then diameter of the circle is (a) m(m + n) (b) m + n (c) n(m + n) (d) m2 + n2
I 2 =
2
(a) f ′(2) + f (1) + 1 (c) f (3) – f (2) + 1
6.
p/4 x − p/ 4 (e + e )(sin x + cos x)
3
1
5.
∫
p /2
then the value of k is 1 1 1 (a) (b) (c) (d) 2 2 2 2
∫ ( f (x) + xf ′(x)) dx + ∫ f ′(x) f ′′(x) dx is equal to
4.
If
e p/4dx
13.
14.
15.
In an equilateral triangle inradius( r ), circumradius (R) and exradius (r 1) are in (a) A.P. (b) G.P. (c) H.P. (d) none of these If tan A, tanB are the roots of the quadratic equation abx 2 – c2x + ab = 0 where a, b, c are sides of the D ABC then sin2 A + sin2B + sin2C is (a) 1 (b) 2 (c) 3 (d) 3/2
(a) parallel to plane containing a, b and c (b) perpendicular to the plane formed with the vectors AB and AC (c) perpendicular to the plane containing a, b and c
17.
18.
19.
If (1 + x )n = nC 0 + nC 1x + nC 2x 2 + .... + nC n x n where nC , nC , nC , ... are binomial coefficients. Ten the 0 1 2 value of 2(C 0 + C 3 + C 6 + ...) + (C 1 + C 4 + C 7 + ...)(1 + w) + (C 2 + C 5 + C 8 + ...)(1 + w2), where w is the cube root of unity and n is a multiple of 3, is equal to (a) 2n + 1 (b) 2n – 1 + 1 (c) 2n + 1 – 1 (d) 2n – 1 Te chords of contact of the pair of tangents drawn from each point on the line 2 x + y = 4 to x 2 + y 2 = 1 pass through the point (a) (1, 2) (b) (1/2, 1/4) (c) (2, 4) (d) none of these If
a
−2 =
bc
22.
20.
If f (x ) is continuous such that | f (x )| ≤ 1 " x ∈ R and g (x ) =
e f (x ) − e f (x )
e
+e
f (x ) f (x )
(a) [0, 1]
e2 − 1
e2 + 1
(c) 0, 21.
Let three non-collinear points A, B and C have position vector a, b and c , then the vector a × b + b × c + c × a is
23.
If 0 ≤ a, b ≤ (a) a + b ≥ (c) a + b ≤
24.
If lim
p 2
and cos a + cos b = 1 , then
p
(b) cos(a + b) ≤ 0
2
p
(d) cos(a + b) ≥ 0
2
ae2x − b cos 2x + ce −2x − x sin x
25.
= 1 and
x 2 f (t ) = (a + b)t 2 + (a – b)t + c, then (a) a + b + c = 1 (b) a + b + c = 2 (c) f (1) = 3/4 (d) f (1) = 1 x →0
If the conics whose equations are S1 : (sin2q)x 2 + (2htanq)xy + (cos2q) y 2 + 32x + 16 y + 19 = 0 2 2 S2 : (cos q)x + (2h′cotq)xy + (sin 2q) y 2 + 16x + 32 y + 19 = 0
p
intersect in four concyclic points, where q ∈ 0, , 2 then the correct statement(s) can be (a) l = 0 (b) l = 0 (c) q = p/4 (d) none of these 26. ABCD is
a square of side 1 unit. P and Q lie on the side AB and R lies on the side CD. Te possible values for the circumradius of triangle PQR is ? (a) 0.5 (b) 0.6 (c) 0.7 (d) none of these
, then range of g (x ) is
e2 + 1 (b) 0, 2 e − 1 1 − e2 (d) , 0 1 + e2
271 should be split into how many parts so as to maximize their product? (a) 99 (b) 10 (c) 105 (d) none of these MORE THAN ONE OPTION CORRECT TYPE
b c , where a, b, c > 0, then family + c b
of lines ax + b y + c = 0 passes through the fixed point given by (a) (1, 1) (b) (1, –2) (c) (–1, 2) (d) (–1, 1)
(d) none of these
Te real function
If a function satisfies f (x + 1) + f (x − 1) = 2 f (x ) , then the period of f (x ) can be (a) 2 (b) 4 (c) 6 (d) 8
f (x) = cos −1 x 2 + 3x + 1 + cos −1 x 2 + 3x is defined on the set (a) {0, 3} (b) (0, 3) (c) {0, –3} (d) (–3, 0) 16.
27.
^
^
^
^
If a = ^i + ^j + k, a ⋅b = 2 and a × b = 2 i + j − 3 k, then ^ ^ ^ (a) a + b = 5 i − 4 j + 2 k
^
^ (b) a + b = 3 i + 2 k
^
(c) b = 2 ^i − ^j + k
^
(d) b = ^i − 2 ^j − 3 k
28.
Te solution of x 1/3 + (2x – 3)1/3 = [3(x – 1)]1/3 is (a) 0 (b) 3/2 (c) 1 (d) none of these MATHEMATICS TODAY
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FEBRUARY ‘17
25
29. Let a1, a2, a3, ...,
an are in G.P. such that 3a1 + 7a2 + 3a3 – 4a5 = 0, then common ratio of G.P. can be (a) 2 (b) 3/2 (c) 5/2 (d) –1/2
30. Te diagonals of a square are along the pair
represented by 2x 2 – 3xy – 2 y 2 = 0. If (2, 1) is the vertex of the square, then the other vertices are (a) (–1, 2) (b) (1, –2) (c) (–2, –1)(d) (1, 2)
x ∫ (5 + |1 − t |) dt , if 31. Let f (x ) = 0 5x + 1, if
x > 2
then the
x ≤ 2
COMPREHENSION TYPE
Let ABCD be a parallelogram whose diagonals equations are AC : x + 2 y = 3; BD : 2x + y = 3. If length of diagonal AC = 4 units and area of ABCD = 8 sq. units. 32. Te length of other diagonal BD is 10 3
(b) 2
(c)
20 3
(d) none of these
4 58 3 58 2 58 5 58 (b) (c) (d) 9 9 3 9
34. Te length of BC is equal to
4 10 2 10 8 10 (a) (b) (c) (d) none of these 3 3 3 MATRIX MATCH TYPE
35. Match the following. Column I
(A) (B)
sin1 sin 2 tan
−
Column II
P.
positive
Q. negative
e − 1 (where [⋅] denotes R. 1 x S.
MATHEMATICS TODAY
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FEBRUARY ‘17
24 5 7 3
16 3 8 3
INTEGER TYPE
37. If the normal to the curve x = t – 1, y = 3t 2 – 6 at the
point (1, 6) make intercepts a and b on x and y -axes a + 12b respectively, then the value of is______ 146
do es not exist
n f (r ) , where S = lim ∑ n→∞ r =− n n
then the value of f (n) =
S(3n) , is ______ S(2n) − S(n)
39. If f (u, v , w, x ) = u2 + v 2 + w2 + x 2 – 2(u + v + w + x ) + 10
with u, w ∈ [–3, 3] and v , x ∈ [1, 3] then 1 max ⋅ f (u, v, w, x ) is 23
40. If a
= 2,
a −b
2
b
= 3, and 2
c
+ b −c + c −a 2
87
the greatest integer function)
26
(D) Te length of chord intercepted by S. the parabola y 2 = 4(x + 1) passing through its focus and inclined at 60° with positive x-axis is
3 9 − 2 4
lim x →0
(B) Te maximum value of the function Q. 7 f (x) = 3 sin x − 4 cos x − 3 will be given by
x
(C)
(A) he area enclosed between the P. curves |x | + | y | = 2 and x 2 = y (in sq. units) is
sin 5 sin 6
Column II
38. Let S(n) denotes sum of first n terms of an A.P.,
33. Te length of side AB is equal to
(a)
Column I
(C) Te length of common chord of R. two circles of radii 3 and 4 units which intersect orthogonally is
function is (a) continuous at x = 2 (b) differentiable at x = 2 (c) discontinuous at x = 2 (d) not differentiable at x = 2
(a)
36. Match the following.
= 4 then
cannot exceed _____
SOLUTIONS
1. (b) : f (x ) = (x – a)(x – b)(x – g )
⇒ f ′(x ) = 3x 2 – 2x (a + b + g ) + ab + bg + ga 2a b ⇒ b = 1 1 (where a1, b1 are the roots of f ′(x ) = 0) a1 + b1 2(ab + bg + ga) ⇒ b= ⇒ b2 = ga 2(a + b + g )
11 − 10iz 2. (b) : z (11z + 10i) = 11 − 10iz ⇒ z = 11z + 10i 11i + 10z ⇒ | z 9 | = 11z + 10i 9
Also a ⋅ b = 1 ⇒ a + b + g = 1 ⇒ b + 1 + b + b = 1 ⇒ b = 0 \ a = 1, g = 0 ^ \ b=i
9
⇒ |11i + 10z |2 − |11z + 10i |2 = 211 ( − | z |2 ) ⇒ if | z | < 1, then | z 9 | > 1 (not possible)
7. (c) : I =
and if | z | > 1 ⇒ | z 9 | < 1 (not possible) 2
∫
xf ′(x) dx = xf (x)|12 −
1
So,
2
∫ f (x) dx
I =
1
2
3
∫ ( f (x) + xf ′(x))dx + ∫ f ′(x) f ′′(x) dx =xf 1
2 (x ) 1
2
+
2
( f ′(x )) 2
3
2
1 2 1 = 2 f (2) − f (1) + [3 − 1] = 2 f (2) − f (1) + 1 2
= 2 f (2) − 1 f (1) + [( f ′(3))2 − ( f ′(2))2]
4. (a) : lim
x 2
3
2 2 2 Now, (x + y + z + w) = Sx + 2Sxy ≥ Sxy + 2Sxy 3 8 8 ⇒ (x + y + z + w)2 ≥ Sxy ⇒ a2 ≥ b 3 3 6. (c) : Let b = a i
^
i
^
^
+ b j + g k, ^
j
^
^
a ×b = j − k
a2 4
+
b2 4
(0, b)
AL =
a
2
a2 + b2
, BM =
B
m
Equation of circle x 2 + y 2 – ax – by = 0 \ Equation of tangent at origin ax + by = 0
M B(a, 0) n L
b
2
a2 + b2
,
AL + BM = m + n = a2 + b2 = diameter of circle. 9. (a) : I 1 =
⇒ I 1 =
101
dx
∫ 2 2 − 4 x ) −100 (5 + 2x − 2x )(1 + e
101
dx
∫ 2 2 − 4(1− x ) ) −100 (5 + 2(1 − x) − 2(1 − x) )(1 + e
⇒ 2I 1 =
101
dx ∫ 5 + 2x − 2x 2 = I 2 ⇒ −100
^
k
⇒ 1 1 1 = j^ − k^ a b g ⇒ b – g = 0, a – g = 1, a – b = 1 ⇒ b = g , a = 1 + g , a = 1 + b,
b 2
1 p /2 Adding, 2I = ∫ sec tdt 2 − p/ 2 1 p /2 \ I= ∫ sec xdx 2 2 − p /2
Circumradius =
tan2 (sin x ) sin2 x = p sin2 x x 2 5. (b) : Since, 3(x 2 + y 2 + z 2 + w2) – 2(xy + yz + zx + z w + wx + w y ) 2 2 2 = (x – y ) + (x – z ) + (x – w) + ( y – z )2 + ( y – w)2 + (z – w)2 ≥ 0 2 ⇒ 3Sx 2 – 2Sxy ≥ 0 ⇒ Sx 2 ≥ Sxy
^
a , 2
et dt 1 p /2 ∫ 2 −p/2 (et + 1)cos t
8. (b) : Circumcentre of DOAB ≡
sin ( p sin2 (tan(sin x ) )) p sin2 (tan(sin x )) = lim x →0 p sin2 (tan(sin x )) tan2(sin x )
p 4
−p/ 4 2(e x − p/4 + 1)cos x −
sin ( p (1 − sin2 (tan(sin x ) )))
x →0
dx
∫
dt 1 p /2 = ∫ t 2 −p/2 (e + 1)cos t
⇒ |z | = 1 3. (b) :
3p/4
I 1 I 2
=
1 2
(0, 1)
10. (a) : 3
3
11. (d) : We have, x 2 + 9 y 2 + 25z 2 = 15 yz + 5zx + 3xy
⇒ (x )2 + (3 y )2 + (5z )2 – (x )(3 y ) – (3 y )(5z ) – (x )(5z ) = 0 MATHEMATICS TODAY
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FEBRUARY ‘17
27
⇒
1 2(x)2 + 2(3 y) 2 + 2(5z)2 − 2(x)(3 y) − 2(3 y)(5 z) 2 − 2(x)(5z ) = 0
⇒ 1 (x − 3 y)2 + (3 y − 5z)2 + (x − 5z)2 = 0 2 ⇒ x – 3 y = 0, 3 y – 5z = 0, x – 5z = 0 1 1 5 1 1 1 = , = and = 5z x x 3y 3y z 1 1 1 5 2 ⇒ + = + = x z 3y 3y y 1 1 1 ⇒\ , , are in A.P. ⇒ x , y , z are in H.P. x y z 12. (d) : 23C r + 2 · 23C r + 1 + 23C r + 2 = 24C r + 1 + 24C r + 2
⇒
= 25C r + 2 ≥ 25C 15
\ (r + 2) can be 10, 11, 12, 13, 14 and 15. So, 6 elements. 13. (a) : In an equilateral triangle r = R/2 A B C Also ex-radius r1 = 4R sin cos cos 2 2 2 1 3 3 3 = 4R . . = R 2 2 2 2 ⇒ r , R, r 1, are in A.P. 2
c and tan A·tan B = 1 ab tan A ⋅ tan B = 1 ⇒ tan A = cot B ⇒ A = 90° – B ⇒ A + B = 90° ⇒ C = 90° a b ⇒ sin A = , sin B = c c 14. (b) : Here, tan A + tan B =
\ sin2 A + sin 2 B + sin 2 C =
a c
2
+ 2
b
2
c2
+1= 2
15. (c) : Obviously, 0 ≤ x 2 + 3x + 1 ≤ 1
and 0 ≤ x 2 + 3x ≤ 1 ⇒ x 2 + 3x = 0 ⇒ x = 0, –3 16. (d) : By replacing x = x + 1 and x = x – 1, we get ...(1) f (x + 2) + f (x) = 2 f (x + 1) ...(2) f (x) + f (x − 2) = 2 f (x − 1) Adding (1) & (2) gives, f (x + 2) + f (x − 2) + 2 f (x) = 2 [ f (x + 1) + f (x − 1)]
= 2 2 f (x )
f (x + 8) = – f (x + 4) = f (x ) " x \ f (x ) is periodic with period 8. 17. (d) : (1 + w)n = C 0 + C 1w + C 2w2 + ... + C nwn
(1 + 1)n = C 0 + C 1 + C 2 + ... + C n (1 + w)n + (1 + 1) n = 2C 0 + C 1(1 + w) + C 2(1 + w2) + C 3(1 + w3) + C 4(1 + w) + C 5(1 + w2) + C 6(1 + w3) + ... + C n(1 + wn) = 2(C 0 + C 3 + C 6 + ....) + (C 1 + C 4 + C 7 + ...)(1 + w) + (C 2 + C 5 + C 8 + ...)(1 + w2) = –w2n + 2n ⇒ 2n – 1 (Q n is a multiple of 3, wn = 1) 18. (b) : Let (h, k) be any point on the given line \ 2h + k = 4 and the chord of contact hx + ky = 1 hx + (4 – 2h) y = 1 ⇒ (4 y – 1) + h(x – 2 y ) = 0, p + lq = 0, it passes through the intersection of p = 0, q = 0 or (1/2, 1/4). a
19. (d) :
⇒ ⇒
−2 =
bc
a =( b
2 c)
+
b c ⇒
c b
+
⇒ a = b + c + 2 bc
( a − b − c )( a + b + c ) = 0
a − b − c = 0, a + b + c ≠ 0 (Q a, b, c > 0) Comparing with ax + b y + c = 0 Hence x = –1, y = 1. e f (x) − e| f (x )|
20. (d) : We have, g (x ) = f (x) | f (x )| , −1 ≤ f (x ) ≤ 1 +
e
e
For 0 ≤ f (x ) ≤ 1, g (x ) = 0 For –1 ≤ f (x ) < 0, e f (x) − e − f (x ) e 2 f (x ) − 1 2 g (x ) = f (x) − f (x ) = 2 f (x) = 1 − 2 f (x ) e e e +e +1 +1
1 − e2 , For –1 ≤ f (x ) < 0, g (x ) ∈ 1 + e2 1 − e2 For –1 ≤ f (x ) < 1, g (x ) ∈ , 2 + 1 e
thus is perpendicular to a, b and c . 22. (d)
p
a +b a −b 1 ⇒ cos = cos 2 2 2
FEBRUARY ‘17
⇒ (r − a) ⋅ (a × b + b × c + c × a) = 0 Tus a × b + b × c + c × a is perpendicular to r − a and
On replacing x by x + 2 we get f (x + 4) + f (x ) = 0 Finally replacing x by x + 4, we get f (x + 8) + f (x + 4) = 0, |
the plane of a, b and c ⇒ (r − a) ⋅ (b − a) × (c − a) = 0
23. (a, b) : 0 ≤ a, b ≤
MATHEMATICS TODAY
0 .
21. (c) : Let r be the position vector of any point on
\ f (x + 2) + f (x – 2) = 0
28
0
2
and cos a + cos b = 1
MATHEMATICS TODAY
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FEBRUARY ‘17
29
a+b a − b ≥ 1 ⇒ atleast one of cos and cos 2 2 2 p ⇒ atleast one of a + b and a − b ≥ 2
p
a−b ≥
but
2
24. (a, c) : lim
and
p
a+b ≥
2
x 2
2 x
→0
⇒ a – b + c = 0 ⇒
2a – 2c = 0 2a – 2b + 2c = 2 a = c and b = 2a
⇒
a + 2a + a = 1
a=
1 4
=c
\
b=
=2
1 2
x 2(sin2 q + l cos2 q) + 2(h tan q – lh′ cot q)xy
+ (cos2 q + l sin2 q) y 2 + (32 + 16 l)x + (16 + 32 l) y + 19(1 + l) = 0 Te above equation will represent a circle if sin2 q + l cos2 q = cos2 q + l sin2 q ⇒ sin2 q – l sin2 q = cos2 q – l cos2 q ⇒ (1 – l)sin2 q = (1 – l)cos2 q p ⇒ (1 – l)(sin2 q – cos2q) = 0 ⇒ l = 1 or q = 4 h tan q – lh′ cot q = 0 ⇒ h tan q = lh′ cot q which is satisfied if l = 1 and q = p/4 ⇒ h = h′ 26. (a, b, c) : Let O be the circumcentre. hen
OP + OR ≥ PR ≥ AD = 1, so the radius is at least 1/2. P , Q, R always lie inside or on the circle through A, B, C , D 1 1 which has radius , so the radius is at most . 2 2 27. (b, c) : We have 2
(a × b) × a = (a ⋅ a) b − (b ⋅ a) a = | a | b − 2a
30
MATHEMATICS TODAY
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FEBRUARY ‘17
| a |2
^
^
^
^
^
Now, (a × b) × a = 4 i − 5 j + k and | a | = 3
^
^
^
(4 i − 5 j + k) + 2(i + j + k) ^ ^ ^ = 2 i − j + k. b= 3
29. (b, d) : Given 3a1 + 7a2 + 3a3 – 4a5 = 0
3 2 1 1 3t 2 − t + 1 = ( ) f t t − t + = Now, 4 4 4 4 25. (b, c) : Curve through the intersection of S1 and S2 is given by S1 + lS2 = 0
⇒
when either a = 0 or b = 0 or a + b = 0 3 ⇒ x = 0, , 1 2
...(1) ...(2) ...(3)
⇒
b=
28. (a, b, c) : a1/3 + b1/3 = (a + b)1/3 is true,
=1
2 (2x)2 (2x )4 (2x ) + + + x 2 ... − b 1 + 2 ! + 4 ! − ... 2! 2 (2x ) +c 1 − 2x + − ... 2!
x
\
\
and cos(a + b) ≤ 0
a 1
⇒ lim
^
ae 2x − b cos 2x + ce −2x − x sin x
x →0
(a × b) × a + 2a
⇒ ⇒ ⇒ ⇒
7(a1 + a2 + a3) = 4(a1 + a3 + a5) 7(1 + r + r 2) = 4(1 + r 2 + r 4) 7 = 4(r 2 – r + 1)
⇒ 4r 2 – 4r + 1 = 4
(2r – 1)2 = 4
3 1 r = , − 2 2 30. (a, b, c) : Since the diagonals intersect at origin and are at right angles. Let B and D be the points adjacent to A.
⇒ 2r – 1 = ±2 ⇒
2
Also OA = 2
D O
A
2
+1 = 5
(0, 0) B
(2, 1)
Let affix of B and D are z 2 and z 1 respectively. z p \ arg 2 = 2 + i 2 ⇒ z 2 = (2 + i)i = –1 + 2i ⇒ B(–1, 2) z p Also, arg 4 = − 2 + i 2 ⇒ z 4 = (2 + i)(–i) = 1 – 2i ⇒ D(1, –2) x
31. (c, d) : x > 2 (5 + |1 − t |) dt
∫ 0
1
x
= ∫ (6 − t )dt + ∫ (4 + t ) dt 0
=1+ 4 +
1
2
x 2
x 2 + + ⇒ f (x ) = 1 4x 2 , x > 2 5x + 1, x ≤ 2 4 + x, x > 2 f ′(x ) = 5, x ≤ 2 f (2+) = f (2–) = 11 continuous at x = 2 f ′(2+) ≠ f ′(2–) ⇒ not differentiable at x = 2
1 3 3 32. (c) : tan θ = 2 = , sin θ = , 1+1 4 5 1 20 area o ∆CPB = × PC × PB sin θ = 2 ⇒ BD = 2 3
A
− +2
33. (a) : cos(π − θ) =
AP
2
2
+ PB − AB 2 AP ⋅ PB
100 − AB2 4 2 58 9 ⇒ AB = ⇒ − = 10 5 3 2×2× 3 PC 2 + PB2 − BC2 34. (a) : ln ∆CPB, cos θ = 2PC ⋅ PB 2 10 3
sin1 sin 5 sin x , Take f (x ) = − sin 2 sin 6 sin(x + 1) 1 sin(x + 1)cos x − cos( x + 1)sin x sin f ′(x ) = = 2 >0 2 sin (x + 1) sin (x + 1) ⇒ f (x ) is increasing ⇒ f (1) < f (5). (B) Take f (x ) = tanx – x 2 f ′(x ) = sec2x – 2x , f ′′(x ) = 2sec2x tanx – 2 f ′(1) > 0 and f ′′(x ) > 0 or x > 1 π ⇒ f (x ) is increasing in 1, ⇒ tan 3 > 9 2 2 4 (A)
e x − 1 lim does not exist as lef hand and right x →0 x
hand limits are not equal. 36. (A) - (Q), (B)-(S), (C)-(P), (D)-(R) (A) Required area
(B) 5 −
B
Substituting parametric coordinates in (1) 2
3 r + 1 , 3r 2 − 2r − 4 = 0 = r 4 2 4 2 2
Length o AB = PA − PB = ( PA + PB) − 4 PAPB 2
8 16 16 = − 4 × = 3 3 3 37. (1) : Given point is corresponding to t = 2 and
35. (A)-(Q), (B)-(P), (C)-(S)
(C)
(1, 0)
2
4+
⇒ BC =
60°
1 1 = 2 (2 + 1) × 1 − ∫ x 2dx 2 0 3 1 7 = 2 − = sq. units 2 3 3
7 8 = 3 3
dy = 6t dx
⇒ slope o normal at t = 2 is −
\ Equation o normal is y − 6 = − ⇒ a = 73, b =
1 12
1 (x − 1) 12
73 ⇒ a + 12b = 146 12
38. (6) : Let a be the first term o A.P. and d be the
common difference. S(3n) 3n S(3n) = {2a + (3n − 1) d }, = 3 = f (n) S(2n) − S(n) 2 . 1 n 2n + 1 \ S = lim ∑ f (r ) = lim 3 = 6 n→∞ n r =−n n→∞ n 39. (2) : f (u, v , w, x ) = (u – 1)2 + (v – 1)2 + (w – 1)2 + (x – 1)2 + 6 Clearly f is max when u = w = –3 and v = x = 3 ⇒ f max =(–3 – 1)2 + (–3 – 1) 2 + (3 – 1)2 + (3 – 1)2 + 6 = 46 2
40. (1) : | a − b |2 + | b − c |2 + c − a
= 2(a2 + b2 + c2) −2(a ⋅ b + b ⋅ c + c ⋅ a) = 2(4 + 9 + 16) − 2(a ⋅ b + b ⋅ c + c ⋅ a) = 58 − 2(a ⋅ b + b ⋅ c + c ⋅ a)
Now (a + b + c )2 ≥ 0
2 × 3 × 4 24 = = (C) Length o chord = 2 2 5 5 r1 + r 2 (D) Parabola is y 2 = 4(x + 1) ... (i), ocus is (0, 0) x − 0 y − 0 = = r Equation o AB is 1/ 2 3 /2 2r1r 2
⇒ a2 + b2 + c2 ≥ −2(a ⋅ b + b ⋅ c + c ⋅ a) ⇒ − 2(a ⋅ b + b ⋅ c + c ⋅ a) ≤ 29
⇒ | a − b |2 + | b − c |2 + | c − a |2 ≤ 87
MATHEMATICS TODAY
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FEBRUARY ‘17
31
Series-7 The entire syllabus of Mathematics of WB-JEE is being divided in to eight units, on each unit there will be a Mock Test Paper (MTP) which will be published in the subsequent issues. The syllabus for module break-up is given bellow: Unit No.
Topic
Syllabus In Details
Co-ordinate Geometry-3D
7 . O N T I N U
Direction ratios and direction cosines. Angle between two intersecting lines. Skew lines, the shortest distance between them and its equation. Equation of a line and a plane in different forms, intersection of a line and a plane, coplanar lines. Differential Rolle’s and Lagrange’s Mean value theorem theorems, Applications of derivatives: Rate of calculus change of quantities, monotonic-increasing and decreasing functions, Maxima and minima of functions of one variable, tangents and normals. Integral calculus Integral as an anti-derivative. Fundamental integrals involving algebraic, trigonometric, exponential and logarithmic functions. Integration by substitution, by parts and by partial fractions. Integration using trigonometric identities. Evaluation of simple integrals of type: dx dx dx dx dx dx ( px + q)dx , ∫ , ∫ ∫ 2 2 , ∫ 2 2 , ∫ 2 2 , ∫ 2 2 , ∫ 2 2 x ± a a − x ax + bx + c x ± a a − x ax 2 + bx + c ax + bx + c
Time : 1 hr 15 min.
Full marks : 50 CATEGORY-I
4.
Te equation of the line joining the points (2, –1, 4) and (1, 1, –2) is x − 2 y + 1 z − 4 = = (a) 1 2 6 x − 2 y + 1 z − 4 (b) = = 2 −1 −6 x − 1 y − 1 z + 2 (c) = = 1 2 6 x − 1 y + 1 z + 2 (d) = = 2 −1 −6
5.
A plane meets the coordinate axes at P , Q, R such that the centroid of the triangle PQR is (a, b, c). If x y z the equation of the plane is + + = m , then the a b c value of m is (a) 2 (b) 3 (c) 1 (d) 6
6.
Te equation of the plane passing through the points (0, 1, 0) and (3, 4, 1) and parallel to the line x + 3 y − 3 z − 2 is = = 2 7 5
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.
p p and with positive 4 3 direction of x -axis and z -axis respectively. Ten the acute angle made by the line with y -axis is (a) p/6 (b) p/4 (c) p/3 (d) cos–1(1/3)
1.
A straight line makes angles
2.
Te angle between the lines x +1 y − 2 z + 4 x − 3 2 y + 3 z − 2 and = = = = is 3 1 1 5 2 −2 (a) p/3 (b) p/2 –1 (c) cos (3/5) (d) cos–1(4/5)
3.
Te coordinates of the foot of perpendicular drawn from the point A(2, 4, –1) on the line x + 5 y + 3 z − 6 = = are 1 4 −9 (a) (1, –3, 4) (b) (–4, 1, –3) (c) (4, 1, 3) (d) none of these
By : Sankar Ghosh, S.G.M.C, Kolkata, Ph: 09831244397.
32
MATHEMATICS TODAY |
FEBRUARY ‘17
7.
(a) 4x – 13 y + 15z = 13 (b) 8x – 13 y + 15z = 15 (c) 8x – 13 y + 15z + 13 = 0 (d) none of these Te vector equation of the line passing through the points (2, –3, 1) and (–4, 3, 6) is ^ ^ ^ ^ ^ ^ (a) r = −4 i + 3 j + 6 k + l(2 i − 3 j + k)
(a) 7 cm2/min (c) 17.5 cm2/min 15.
^
^
^
^
^
^
(b) r = 2 i − 3 j + k + l(−4 i + 3 j + 6 k)
^
^
^
^
^
^
(c) r = −6 i + 6 j + 5 k + l(2 i − 6 j − 5 k)
^
^
^
^
^
^
(d) r = 2 i − 3 j + k + l(−6 i + 6 j + 5 k)
8.
^
^
^
^
^
16.
^
If the line r = 2 i − j + 3 k + l(2 i + j + 2 k) is parallel ^ ^ ^ to the plane r ⋅ (3 i − 2 j + p k ) = 4 , then the value of p is (a) 2 (b) –2 (c) 3 (d) –3 Te vector equation of a plane passing through the point (1, –1, 2) and having 2, 3, 2 as direction number of normal to the plane is ^ ^ ^ (a) r ⋅ (2 i + 3 j + 2 k ) + 4 = 0
9.
17.
^
^
^
^
^
^
^
^
^
(b) r ⋅ (i − j + 2 k ) + (i − j + 2 k ) ⋅ (2 i + 3 j + 2 k ) = 0
^
^
^
^
^
^
(c) r ⋅ (2 i + 3 j + 2 k ) + 5 = 0
(d) r ⋅ (2 i + 3 j + 2 k ) = 3 Te distance of the point (–3, 2, 4) from the plane ^ ^ ^ r ⋅ (2 i − 3 j + 6 k ) + 7 = 0 is 29 (a) 5 units (b) units 7 18 (c) (d) 19 units units 7 7
10.
18.
In the mean value theorem f (b) – f (a) = (b – a) f ′(c) (a < c < b), if a = 4, b = 9 and f (x) = x , then the value of c is (a) 8 (b) 5.25 (c) 4 (d) 6.25 12. If the function f (x ) = 4x 3 + ax 2 + bx – 1 satisfies 1 all the conditions of Rolle’s theorem in − ≤ x ≤ 1 4 1 and if f ′ = 0 , then the values of a and b are 2 (a) a = 2, b = –3 (b) a = 1, b = –4 (c) a = –1, b = 4 (d) a = –4, b = –1
19.
11.
13.
14.
Let f (x ) be a continuous in (– ∞, ∞) and f ′(x ) exists in (–∞, ∞). If f (3) = –6 and f ′(x ) ≥ 6 for all x ∈ [3, 6] then (a) f (6) ≥ 24 (b) f (6) ≥ 12 (c) f (6) ≤ 24 (d) f (6) ≤ 12 A spherical balloon is being inflated at the rate of 35 cm3/min. Ten the rate of increase of surface area of the balloon when its diameter is 14 cm, is
(b) 10 cm2/min (d) 28 cm2/min
If h(x ) = f (x ) – ( f (x ))2 + ( f (x ))3 and f ′(x ) > 0 " x then (a) h(x ) is an increasing function of x in some specified interval (b) h(x ) is an increasing function " x ∈ R (c) h(x ) is a decreasing function " x ∈ R (d) h(x ) is a decreasing function of x in some specified interval . Electric current C , measured by a galvanometer, is given by the equation C = k tanq, where k is a constant. Ten the percentage error in the current corresponding to an error 0.7 percent in the measurement of q when q = 45° is (a) 1.4 (b) 2.8 (c) 1.1 (d) 2.2 Te point on the parabola 2 y = x 2, which is nearest to the point (0, 3) is 1 (a) (± 4, 8) (b) ± 1, 2 9 (c) (±2, 2) (d) ± 3, 2 If m be the slope of the tangent to the curve e y = 1 + x 2, then (a) |m| > 1 (b) |m| < 1 (c) m < 1 (d) |m| ≤ 1 Te normal to the curve x = a(cosq + qsinq), y = a(sinq – q cosq) at any point q is such that (a) it passes through the origin (b) it passes through (a, –a) (c) it is at a constant distance from origin
p
(d) it makes angle + q with the x -axis. 4 20. values of a for which the function ( a + 2)x 3 – 3ax 2 + 9 ax – 1 decreases monotonically throughout for all real values of x , are (a) a < –2 (b) a > –2 (c) –3 < a < 0 (d) a ≤ –3 21.
22.
sin x dx = Ax + B log |sin(x − a)| + c, then the If ∫ sin(x − a) value of ( A, B) is (a) (cosa, sina) (b) (–sina, cosa) (c) (sina, cosa) (d) (–cosa, sina)
∫ e
x
dx is
(a) e x + c 1 (c) e x + c 2
(b) 2( x − 1)e
x
+c
(d) 2( x + 1)e
x
+c
MATHEMATICS TODAY
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FEBRUARY ‘17
33
23.
−1 e tan x
∫ 1 + x 2
dx is
(a) tan–1x + C
(c) e 24.
tan −1 x
(b)
+ C
1 1 + x 2
(d)
+ C
−1 2xe tan x
(1 + x 2 )2
+ C
CATEGORY-II
∫ f (x)dx = f (x), then ∫ { f (x)}2dx is equal to (a)
1 { f (x)}2 + C 2
{ f (x )}3 (c) + C 3 25.
x − a 2 dx 30. If ∫ = f (a)log + C , then x + a cos x + cos a cos 2 the value of f (a) is (a) sina (b) cosa (c) coseca (d) seca cos
(b) { f (x )}3 + C
27.
r
1 x + 2 + C (b) log(x 2 + 4x + 13) + C tan −1 3 3
(c)
1 x + 5 log 6 x − 1
+ C (d)
x + 2 (x 2 + 4x + 13)2
33.
+ C
1 + sin x dx = 1 + cos x x + C 2
34.
n
(b) n(logx )n + C (d) x (logx )n + C
e x dx
(a) log x e +2 (c)
(b) log x e +1
x
+1 + C x e +2 e
+ C
35.
e x + 2
(d)
MATHEMATICS TODAY
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x
+2 + C e x + 1
e
FEBRUARY ‘17
5
units 14 (d) none of these
(a + b)2 2
(d)
a2 + ab + b2 3
Te four common tangents to the ellipses
∫
x
+ C
(b)
x 2 y 2 x 2 y 2 and + = 1 form + =1 4 9 9 4 (a) a rectangle of area 13 unit2 (b) a parallelogram which is neither a square nor a rectangle (c) a rhombus (d) none of these
In = ∫ ( log x ) dx, then the value of (In + nI n−1) is
e x + 1
unit
Te minimum area of the triangle formed by a x 2 y 2 tangent to the ellipse + 2 = 1 and the coordinate 2 a b axes, is a 2 + b2 (a) ab (b) 2 (c)
(d) ex tanx + C
∫ (e x + 2)(e x + 1) is equal to
3
14 (c) 5 units
Te value of ∫ e x
(b) e x sec
= ^i + l (2 ^i + 5 ^j + 3 k^) , is
(a)
(a)
(a) (logx )n–1 + C (c) (x logx )n + C
34
Te distance between the plane whose equation is ^ ^ ^ r ⋅ (2 i + j − 3 k) = 5 and the line whose equation is
Te value of the integral ∫ 2 is x + 4x + 13
x + C 2 x x (c) e tan + C 2
29.
32.
dx
(a) e x sec2
28.
If the lines
dx Te value of ∫ is 2 x (x + 1) 1 −1 tan−1( x ) + C (b) 2 tan ( x ) + C 2 −1 (c) tan−1( x ) + C (d) tan (2 x ) + C
x y z x −1 y − 2 z − 3 and = = , = = 1 2 3 3 4 −1 x + k y −1 z − 2 are concurrent. Ten, = = h 3 2 (a) h = –2, k = –6 (b) h = 1/2, k = 2 (c) h = 6, k = 2 (d) h = 2, k = 1/2
31.
(d) { f (x )}2 + C
(a)
26.
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.
(a) (b)
x 2 − 1 3
4
2
⋅ 2x − 2x + 1 1 2
x 1
dx =
⋅ 2x 4 − 2x 2 + 1 + C
4 2 x x ⋅ − + 1 + C 2 2 3
x
p
1 . 2x 4 − 2x 2 + 1 + C x 1 . 2x 4 − 2x 2 + 1 + C (d) 2 2x (c)
(c)
1.
CATEGORY-III
of the planes P 1 and P 2 through the origin. Te plane ^
(c): Let the line makes an acute angle q with the y -axis.
Terefore, the direction cosines are
36. Let A be a vector parallel to the line of intersection ^
^
^
P 1 is parallel to the vectors 2 j + 3 k and 4 j − 3 k ^
^
while the plane P 2 is parallel to the vectors j − k and
^
f′(1) = me m
SOLUTIONS
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)
^
m 3 3p +1 e (d) f′( 3) = m
^
^
^
3 i + 3 j . Te angle between A and 2 i + j − 2 k is (a) p/2 (b) p/4 (c) p/6 (d) 3p/4 37. f (x ) is a polynomial of third degree which has a
local maxima at x = –1. If f (1) = –1, f (2) = 18 and f ′(x ) has a local minimum at x = 0 then (a) f (0) = 5 (b) f (x ) has local minimum at x = 1 (c) f (x ) is increasing in [1, 2 5] (d) Te distance between (–1, 2) and (a, f (a)) is 2 5, where a is a point of local minimum.
p
p
l = cos , m = cos and n = cos q 4 3 2 2 2 But l + m + n = 1
p p ⇒ cos2 + cos2 + cos2 q = 1 4
3
1 1 3 1 ⇒ cos2 q = 1 − + = 1 − = 2 4 4 4
⇒ cos q =
1 [ 2
q is acute]
So, q = p/3 2. (b) : Te given equations of the lines are x +1 y − 2 z + 4 x − 3 2y + 3 z − 2 and = = = = −2 3 1 1 5 2 Let the angle between them be q. Ten 5 3 ⋅1 + (−2) ⋅ + 1 ⋅ 2 2 cos q = =0
2
5 + 22 2
32 + (−2)2 + 12 12 +
Hence, q = p/2 3. (b) : Te general point on the straight line 38. A tangent to the curve y = f (x ) at P (x , y ) cuts the x + 5 y + 3 z − 6 ... (1) = = = l (say) x -axis and y -axis at A and B, respectively, such that −9 1 4 BP : AP = 3 : 1. If f (1) = 1 then is (l – 5, 4l – 3, –9l + 6) dy Let the foot of perpendicular L ≡ (l – 5, 4l – 3, –9l + 6) (a) the equation of the curve is x + 3 y = 0 dx \ Te direction ratios of AL are l – 7, 4l – 7, –9l + 7 Since AL is perpendicular to the line (1), therefore 1 (b) the curve passes through 2, 1(l – 7) + 4(4 l – 7) – 9(–9 l + 7) = 0 ⇒ l = 1 8 Tus coordinates of foot of perpendicular are (–4, 1, –3) dy x 3 y 0 − = (c) the equation of the curve is 4. (b) : Te equation of the line joining the points dx (2, –1, 4) and (1, 1, –2) is (d) the normal at (1, 1) is x + 3 y = 4 x − 2 y + 1 z − 4 x − 2 y + 1 z − 4 i.e., = = = = 11 cos x − 16 sin x − − − + − 2 1 1 1 4 2 1 2 6 dx 39. If ∫ 2 cos x + 5 sin x x − 2 y + 1 z − 4 i.e., = = = −lx + m log l cos x + d sin x + C, then −1 −6 2 (a) l = 2 (c) d = l + m 40. If f(x ) =
−1 e 2 tan x (1 + x )2
∫
(a) m = 2
(b) m = 3 (d) d = m – l 1 + x 2 (b)
−1 dx =xem tan x + C then
f′ (0) = 1
5. (b) : Let the vertices of the triangle PQR be
P (x , 0, 0), Q(0, y , 0) and R(0, 0, z ) x y \ a = ⇒ x = 3a, b = ⇒ y = 3b 3 3 z ⇒ z = 3c. and c = 3 MATHEMATICS TODAY
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FEBRUARY ‘17
35
Tus, the equation of the plane is x y z x y z + + =1⇒ + + = 3 a b c 3a 3b 3c 6. (c) : Te equation of the plane passing through the
point (0, 1, 0) is ax + b( y – 1) + cz = 0 ... (1) Since (1) is also passing through the point (3, 4, 1) \ 3a + 3b + c = 0 ... (2) Again the plane (1) is parallel to the line x + 3 y − 3 z − 2 = = 2 7 5 ... (3) \ 2a + 7b + 5c = 0 Now solving (2) and (3), we get a b c a b c = = ⇒ = = = l (say) 15 − 7 2 − 15 21 − 6 8 −13 15 \ a = 8l, b = –13l, c = 15l Now (1) becomes 8lx –13l y + 15lz = –13l ⇒ 8x – 13 y + 15z + 13 = 0 is the required equation of plane 7. (d) : Te vector equation of the line passing through the points (2, –3, 1) and (–4, 3, 6) ^
^
i.e. a = 2 i − 3 j
^
^
^
^
^
^
^
^
^
^
⇒ r = 2 i − 3 j + k + l( −6 i + 6 j + 5 k)
8. (b) : Here the given equation of the line is
r
^
^
^
^
^
^
... (1)
^
^
^
And equation of the plane is r ⋅ (3 i − 2 j + p k) = 4 ... (2) Since the line (1) is parallel to the plane (2) \ 2(3) + 1(–2) + 2 × p = 0 ⇒ p = –2 9. (d) : Te position vector of the point having ^
^
^
coordinates (1, –1, 2) is i − j + 2 k and normal to the ^
^
^
plane is 2 i + 3 j + 2 k Tus, the equation of the plane is ^
^
^
^
^
^
^
i.e. r ⋅ (2 i
^
^
+ 3 j + 2k) = 3
10. (d) : Te position vector of the point having ^
^
coordinates (–3, 2, 4) is −3 i + 2 j + 4 k ^
^
^
^
^
^
Let a = −3 i + 2 j + 4 k and n = 2 i − 3 j + 6 k Ten the distance of the point (–3, 2, 4) from the plane
^
^
^
r ⋅ (2 i − 3 j + 6 k) + 7 = 0 is
a ⋅n + 7 n
36
^
=
⇒
1 2 c
=
^
^
^
^
^
|(−3 i + 2 j + 4 k) ⋅ (2 i − 3 j + 6 k) + 7 | 22 + (−3)2 + 62
MATHEMATICS TODAY
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FEBRUARY ‘17
1
f (9) − f (4) 3 − 2 1 = = 9−4 5 5
⇒ 2 c = 5 ⇒ 4c = 25 ⇒ c =
25 = 6.25 (4 < 6.25 < 9) 4
12. (b) : Here, the given function is
f (x ) = 4x 3 + ax 2 + bx – 1 ⇒ f ′(x ) = 12x 2 + 2ax + b 1 Given that f ′ = 0 ⇒ 3 + a + b = 0 2
⇒ a + b = –3
a b 65 15 5 65 +b+ =− ⇒ a+ b=− 16 4 4 16 4 16 ⇒ 3a + 4b = –13 Solving (1) and (2) we get a = 1 and b = –4 13. (b) : Given that f (x ) is continuous in (– ∞,
... (2)
∞) and
f ′(x ) exists in (– ∞, ∞) f (6) − f (3) where c ∈(3, 6) \ f ′(c) = 6−3 f (6) − f (3) ≥6 3
⇒ f (6) − (−6) ≥ 18 ⇒ f (6) ≥ 12
14. (b) : Let the volume of the spherical balloon be V
4 3
dV dr = 4pr 2 dt dt dS dr Again S = 4pr 2 ⇒ = 8pr dt dt dS dS 2 dV 2 2 Now dt = ⇒ = ⋅ = × 35 = 10 dV r dt r dt 7 dt Tus, the rate of increase of the surface area is 10 cm2/min
\ V = pr 3 ⇒
^
⇒ f ′(c) =
2 x 2 c Now by mean value theorem, we have f (b) − f (a) f ′(c) = b−a
⇒
r ⋅ (2 i + 3 j + 2 k ) = (i − j + 2 k) ⋅ (2 i + 3 j + 2 k)
1
⇒ a−
= 2 i − j + 3 k + l(2 i + j + 2 k)
⇒ f ′(x ) =
... (1) Since f (x ) satisfies all the conditions of Rolle’s Teorem 1 a b 1 \ f (1) = f − 4 ⇒ 4 + a + b − 1 = − 16 + 16 − 4 − 1
+ k and b = −4 i + 3 j + 6 k
is r = a + l(b − a)
−6 − 6 + 24 + 7 19 units = 7 4 + 9 + 36 11. (d) : Here f (x) = x =
15. (b) : h(x) = f (x) − ( f (x))2 + ( f (x))3
\ h′(x) = f ′(x) − 2 f (x) f ′(x) + 3( f (x))2 f ′(x) 2 1 1 1 1 2 = 3 f ′(x) ( f (x)) − 2 f (x ) ⋅ + + − 3 3 3 9 2 1 2 = 3 f ′(x) f (x ) − + 3 3 It is given that f ′(x ) > 0 " x Tus h′(x ) > 0 for all real values of x .
Terefore h(x ) is an increasing function for all real values of x . 16. (c) : We have, C = k tanq dC dC ⇒ = k sec2 q ⇒ dq = k sec2 qdq d q d q dC ⇒ dC = k sec2 qdq dC = dq d q
dC
Now,
C
= 2×
× 100 =
k sec2 qdq k tan q
× 100 =
sec2 q dq × 100 tan q
p dq p × 100 = × 0.7 = 1.57 × 0.7 = 1 .099 4 q 2
\ Percentage error in C is 1.1
17. (c) : Let the point on the parabola 2 y = x 2 be
x 2 x , 2 x 2 Now the distance between the point x , and 2 (0, 3) be L (say)
x 2 2 So, L = x + − 3 2 x 2 1 dL2 = 2x + 2 − 3 × × 2x dx 2 2 x 2 x 2 = 2x 1 + − 3 = 2x − 2 2 2 2
d 2(L2 ) 2
dx
2
= 3x 2 − 4
dL2 = 0 gives x 3 – 4x = 0 dx x (x 2 – 4) = 0 ⇒ x = 0 or x = ±2
Now
⇒ \
d 2(L2 ) dx 2 x =± 2
=8>0
So L2 is minimum when x = ±2 Terefore the required point is (±2, 2). 18. (d) : Te given equation of the curve is ... (1) e y = 1 + x 2 Differentiating both sides w.r.t. x , we get 2x dy dy 2x = y = = m (using (1)) e y = 2x ⇒ 2 dx dx e 1 + x
⇒ |m| =
2 | x |
... (2)
1+ | x |2
But (1 – |x |)2 ≥ 0 ⇒ 1 + |x |2 – 2|x | ≥ 0 ⇒ 1 + |x |2 ≥ 2|x | Tus, from (2) we get | m| ≤ 1 19. (c) : Given that x = a(cosq + qsinq) dx ⇒ = a(− sin q + sin q + q cos q) = aq cos q d q and y = a(sinq – qcosq) dy ⇒ = a(cos q − cos q + q sin q) = aq sin q d q dx cos q \ = dy sin q Terefore, the equation of the normal to the given curve at q is cos q y − a(sin q − q cos q) = − {x − a(cos q + q sin q)} sin q ⇒ x cosq + y sinq = a(cos2q + qsinqcosq + sin2q – qsinq cosq) ⇒ x cosq + y sinq = a ... (1) ⇒ x cosq + y sinq – a = 0 Now the distance of the normal from the origin is |0 − a | = a(constant) cos2 q + sin2 q Terefore the normal (1) is at a constant distance from the origin. 20. (d) : Given that f (x ) = (a + 2)x 3 – 3ax 2 + 9ax – 1 \ f ′(x ) = 3(a + 2)x 2 – 6ax + 9a 2 2 a a a2 = 3(a + 2) x − 2 ⋅ x ⋅ + − + 9a 2 2 2 a a + + ( 2 ) a + 2 a 3a2 = 3(a + 2) x − + 9a − a + 2 (a + 2) 2
a 6a(a + 3) = 3(a + 2) x − + a + 2 a+2
Given that the function decreases monotonically throughout for all real values of x . MATHEMATICS TODAY
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FEBRUARY ‘17
37
2 a 6a(a + 3) \ f ′(x ) < 0 ⇒ 3(a + 2) x − + a+2 <0 a + 2
Clearly the above condition is satisfied for all real values of x when a ≤ –3 sin x dx 21. (a) : Let I = ∫ sin(x − a) Put x – a = z ⇒ dx = dz sin(z + a) sin z cos a + cos z sin a dz = ∫ dz \ I = ∫ sin z sin z = cos a ∫ dz + sin a∫ cot zdz = z cosa + sina⋅log(sinz ) + c = (x – a)cosa + sina⋅log|sin(x – a)| + c = x cosa + sina⋅log|sin(x – a)| + k (where k = c – acosa) \ ( A, B) = (cosa, sina) 22. (b) : Let I =
Put x
\
= z ⇒
∫ e 1
2 x
x x 1 = ∫ e x sec2 + tan dx 2 2 2 x = e x tan + C ∫ e x { f (x) + f ′(x)} dx = ex f (x) + C 2
∫
1 = x(log x)n − ∫ n(log x)n− 1 ⋅ x dx x = x(log x)n − n ∫ (log x)n−1dx + C In = x(log x)n − nIn−1 + C ⇒ In + nIn −1 = x(log x)n + C 29. (a) : Let I =
Put e x
x
dx
\
dx = dz ⇒ dx = 2zdz
z + 1 = log z + 2
Put x
\
I = ∫
∫ 2
dx
∫ cos x + cos a
1 − z 2 (z + 1) cos a + 1 + z 2 2
dz
= 2
(1 + cos a) − (1 − cos a)z
dx
∫ x 2 + 4x + 13
= ∫
x 2 + 4x + 4 + 9
dx
MATHEMATICS TODAY
|
dx
FEBRUARY ‘17
2dz
= ∫
(1 + z 2 )cos a + 1 − z 2
2 ∫ (1 − cos a)
cot
dz dz = = tan−1 z + C = tan−1 x + C ∫ 2 2 2z(z +1) 1 + z
1 −1 x + 2 = tan 3 + C (x + 2)2 + 32 3 x 1 + sin x dx 27. (c) : We have ∫ e 1 + cos x x x 2 sin cos 1 2 2 dx = ∫ e x + 2 x 2 x 2 cos 2 cos 2 2 38
I = ∫
+ C
2z
26. (a) : We have,
= ∫
\
= 2 ∫
= z ⇒ dx = 2zdz
e x + 2
x 2dz Put tan = z ⇒ dx = 2 1 + z 2 2dz
dx x (x + 1)
e x + 1
+ C = log
30. (c) : Let I =
∫ f (x) dx = f (x) \ f (x) = f ′(x) 2 Now ∫ { f (x)} dx =∫ f (x) f (x)dx =∫ f (x) f ′( x) dx 1 = ∫ f (x) d { f (x)} = { f (x)}2 + C 2 25. (c) : Let I =
dz (z + 2) − (z + 1) = ∫ dz (z + 2)(z + 1) (z + 2)(z + 1)
= ∫
−1x = z ⇒ 1 dx = dz dx put tan 23. (c) : ∫ 1 + x 2 1 + x 2
24. (a) :
∫ (e x + 2)(e x + 1)
dz dz − ∫ = log | z + 1 | − log | z + 2 | + C z + 1 z + 2
I = ∫ 2ze zdz = 2(ze z − e z ) + c = 2e x ( x − 1) + c
−1 I = ∫ e z dz = e z + C = e tan x + C
e x dx
= z ⇒ e x dx = dz
I = ∫
−1 e tan x
\
n
28. (d) : We have, In = (log x) dx
=
1 sin
2
a 2
⋅
1 + z tan
1 2 cot
a 2
log 1 − z tan
a 2
a
+ C
2
x a 1 + tan tan 2 2 + C = coseca ⋅ log x a 1 − tan tan 2 2 x − a cos 2 + C = coseca log x + a cos 2 \ f (a) = cosec a
dz 2
a − z 2 2
31. (d) : Te coordinates of any point on the line
34. (d)
x y z = = = l (say ) 1 2 3 are given by x = l, y = 2l and z = 3l Tus the coordinates of a general point on first line are (l, 2l, 3l) Te coordinates of any point on the line x − 1 y − 2 z − 3 = = = m (say ) 3 4 −1 are given by x = 3m + 1, y = –m + 2 and z = 4m + 3 Tus the coordinates of a general point on second line are (3m + 1, –m + 2, 4m + 3) If the lines intersect, then they have a common point. So, for some values of l and m, we must have l = 3m + 1, 2l = –m + 2 and 3l = 4m + 3 or l – 3m = 1, 2l + m = 2, 3l – 4m = 3 Solving two of these equations, we get l = 1 and m = 0. So, the point of intersection of the first two lines is (1, 2, 3). It is also lies on the third line. 1+ k 2 −1 3 − 2 So, = = h 3 2 1+ k 1 1 1 1 ⇒ = ⇒ k = and = ⇒ h = 2 3 2 2 2 h 32. (a) : Te vector equations of the plane and the
straight line are respectively ^ ^
^
... (1)
r ⋅ (2 i + j − 3 k) = 5
and r
^
^
^
^
= i + l(2 i + 5 j + 3 k)
... (2)
Here, b ⋅ n = (2 ^i + 5 ^j + 3 k^) ⋅ (2 ^i + ^j − 3 k^) = 0
So, b is perpendicular to n . Hence, the given line is parallel to the given plane. Let the required distance be d
^
( i
+ 0^j + 0 k^) ⋅ (2 ^i + ^j − 3 k^) − 5
3 unit 14
1
− 3
2
1
−1 x 5 dx dx = ∫ x 2 1 x 3 2x 4 − 2x 2 + 1 2− 2 + 4 x
∫
35. (d) : Let I =
x
2 Put 2 − 2 x
+
x
1 − 1 dx = 2zdz 2 z = ⇒ 4 3 5 x 4 x x 1
1 zdz 1 1 1 2 1 = ∫ dz = z + C = 2 − 2 + 4 ∫ 2 z 2 2 2 x x 1 = 2 2x 4 − 2x 2 + 1 + C 2x
\ I =
+ C
36. (a, d) : A vector along the line of intersection of two
planes one of them being parallel to the vectors a and b and the other plane being parallel to the vectors c and d , is (a × b) × (c × d ).
^
^
^
^
^
^
^
^
\ A = {(2 j + 3 k) × (4 j − 3 k)} × {(j − k) × (3 i + 3 j)}
^
^
^
^
^
^
= −18 i × (3 i − 3 j − 3 k) = 54(k − j) Let q be an angle between them, then ^
cos q =
^
54(k − j )
54 2 p 3p So, angle , 4 4
^
^
^
.(2 i + j − 2 k)
=−
1 2
=
1 2
37. (b, c) : Let f (x ) = ax 3 + bx 2 + cx + d
f (1) = –1 ⇒ a + b + c + d = –1 f (2) = 18 ⇒ 8a + 4b + 2c + d = 18 f ′′(0) = 0 ⇒ b = 0 and f ′(1) = 0 ⇒ 3a + 2b + c = 0 19 57 17 Solving, a = , b = 0, c = − , d = 4 4 2 1 \ f (x) = (19x 3 − 57x + 34) 4
⇒ 57x 2 − 57 = 0 ⇒ x = 1, −1. 22 + 12 + 32 f ′′(x) = 57 × 2x, f ′′(−1) < 0 and f ′′(1) > 0. 33. (a) : Te equation of tangent to the ellipse \ f (x ) has a local minimum at x = 1. f (x ) is increasing ⇒ f ′(x ) ≥ 0 x 2 y 2 x cos q y sin q 1 + = at , ) is (acos bsin q q + = 1 . ⇒ x 2 – 1 ≥ 0 ⇒ x ≤ –1 or x ≥ 1 a b a 2 b2 \ In [1, 2 5], f (x ) is increasing. Te intercepts of the tangent on the axes are asecq and bcosecq. Te distance of (–1, 2) from (1, –1) is not 2 5. 1 38. (a, b) : Te equation of tangent to the curve at \ Te area of the triangle = × a sec q × bcosecq d =
=
2
ab = ≥ ab sin2q
f ′(x) = 0
P (x , y ) is
Y − y =
dy ( X − x) dx
dy dy − Y = x − y dx dx
⇒ X
MATHEMATICS TODAY
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39
X Y ⇒ + =1 dy dy x − y y − x dx dx dy dx dx dy \ A x − y , 0 and B 0, y − x dy dx BP 3 Now, = . AP 1 3 dx 1 dy So, P = x − y , y − x = (x, y ) dy 4 dx 4 dy dy 1 ⇒ y − x = y ⇒ 3 y + x = 0 dx dx 4 dx dy dx dy ⇒ 3 + = 0 ⇒ 3 ∫ + ∫ = 0 x y x y 3 ⇒ 3logx + log y = logc ⇒ x y = c Now, f (1) = 1 ⇒ (1, 1) is on it. \ c = 1. Hence, x 3 y = 1 1 Also 2, lies on it. 8 Te normal to x 3 y = 1 at (1, 1) has the equation −1 y − 1 = (x − 1) dy dx (11,)
−1 (x − 1) = x − 3 y + 2 = 0 −3 11 cos x − 16 sin x dx (a, b, c) : Let I = ∫ 2 cos x + 5 sin x
⇒ y − 1 = 39.
40
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Let, 11cosx – 16sinx = l (5cosx – 2sinx ) + m(2cosx + 5sin x ) Now equating the coefficient of cos x and sinx , we get 5l + 2m = 11 .... (1) and –2l + 5m = –16 ⇒ 2l – 5m = 16 .... (2) Solving (1) and (2), we get l = 3 and m = –2 3(5 cos x − 2 sin x) − 2(2 cos x + 5 sin x) dx \ I = ∫ 2 cos x + 5 sin x (5cos x − 2 sin x ) dx − 2∫ dx = 3 ∫ 2 cos x + 5 sin x = 3log|2cosx + 5sin x | – 2x + C = –2x + 3log|2cosx + 5sinx | + C \ l = 2, m = 3 and d = 5 40. (a, b, c, d) : We have, f(x ) =
−1 e2 tan x (1 + x )2
∫
= ∫
e
= xe
2 tan −1 x
2
(1 + x + 2x )
dx
2
1 + x 2 tan
−1 x
− ∫ 2e
tan
−1 x
×
−1
= xe2 tan x + C \ m=2 Now f′(x) = xe2 tan
−1 x
×
= ∫e
2 tan
1 2
1 + x
2
−1
1 + x 2 x
dx +
xdx +
∫
2xe
∫
+ e2 tan 2
2xe
dx
2 tan−1 x 2
1 + x
2 tan
−1 x
2
1 + x
dx
dx + C
−1 x
1 + x
\ f′(0) = 1, f′ (1) = 2e p/2 = me p/m 2p
3 3 3 and f′( 3) = + 1 e = + 1 e 2 m
mp 3
10 B E S T
P R OB L EM S
Math Archives, as the title itself suggests, is a collection of various challenging problems related to the topics of IIT-JEE Syllabus. This section is basically aimed at providing an extra insight and knowledge to the candidates preparing for IIT-JEE. 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.
− x 3
+ x 2 sin6 3 2 (a – 8a + 17) then f ′(sin8) is 1.
If f ( x ) =
(a) < 0
(b) > 0
– x sin4· sin8 – 5sin–1
(c) 12
(d) –12
7.
Let A and B be two sets containing 3 and 4 elements respectively. Te number of subsets of A × B having 2 or more elements is (a) 4083 (b) 4096 (c) 4046 (d) 4076
1 2 2. If 2 < x < 3, and if {x } = , where {⋅} represents x 1 fractional part function, then the value of x − is (a) –1
(a)
(b) 0
(c) 1
(d) 2
x
If f (x ) = a + bcos–1x , (b > 0) has same domain and range then the value of a + b is 2 p p (d) 2 − 1 (a) 1 − (b) 1 − (c) −1 p p 2 2 3.
4.
If x 2 + pxy +
4 y 2 + x –
2 y + q = 0 represent the pair of parallel lines then p + q may be (a) –5 (b) 1 (c) 2 (d) 5 Consider the circle x 2 + y 2 = 8. If from point P (2, 2), two chords PA and PB drawn of length 1 unit then the equation of AB is (a) 4x + 4 y + 15 = 0 (b) 4x + 4 y – 15 =0 (c) x + y + 15 = 0 (d) x + y – 15 = 0 5.
6.
Range of the function f ( x ) =
(a)
0, ∞) 3 , ∞ − 3 4 { }
(c)
(b)
3
x − 1 is x − 1
3 , ∞ 4
(d) R
^
^
^ ^
^
If the vectors AB = 2 i − 2 k and AC = 4 i + j+ 2 k are sides of a D ABC , then the length of the median through A is 8.
9.
33 (b) 4
33 (c) 2
37 (d) 4
37 2
5
∫ (x − 1) (x − 2) (x − 3) ( x − 4) (x − 5) dx = 1
(a) 3
(b) –3
(c) 0
(d) 1
10. Te number of all possible positive triplets(x , y , z )
such that (2 y – x )sinq + (x – z )cos2q – 2x cos2q = 0 for all q is (a) infinite (b) one (c) zero (d) greater than 10 but less than 100 SOLUTIONS
1. (b) : f ( x ) =
− x 3 3
+ x 2 sin6 – x sin4 · sin8 – 5sin–1 (a2 – 8a + 17)
f ′(x ) = – x 2 + 2x · sin6 – sin4 · sin8 f ′(sin8) = –(sin8)2 + 2sin8 · sin6 – sin4 · sin8 = –sin28 + 2sin8 · sin6 – sin4 · sin8 = –sin8[sin8 + sin4 – 2sin6] = –sin8[2sin6 · cos2 – 2sin6] = –2sin6 · sin8[cos2 – 1] = 2sin8 · sin6(1 – cos2) > 0
By : Prof. Shyam Bhushan, Director, Narayana IIT Academy, Jamshedpur. Mob. : 09334870021 MATHEMATICS TODAY
| FEBRUARY ‘17
41
⇒ ⇒ ⇒ ⇒
1 – 1 = 0 ⇒ x 3 – 2x – x – 1 = 0 x 3 + x 2 – x 2 – x – x – x – 2 x (x + + 1) – x (x + + 1) – 1(x 1( x + + 1) = 0 2 (x + + 1)(x 1)(x – x – – 1) = 0 – 1 = 0 x ≠ –1, x 2 – x – x 2 − 2 =
⇒
1± 5 1− 5 = x = But x ≠ 2 2
⇒
1 2 5 −1 5 −1 = × = x 2 5 +1 5 −1
\
x −
8. (c) :
B
⇒
0≤
...(i)
≤ p ⇒ 0 ≤ ⇒ a ≤ a + bcos–1x ≤ a + bp
bcos–1x
≤ bp
⇒ ...(ii)
From (i) and (ii), a = –1
⇒
a + bp = 1 ⇒ bp = 2
⇒ b =
2
((x ((x – – 2)2 + ( y – – 2)2 – 1) – (x ( x 2 + y 2 – 8) = 0 + 4 y – – 15 = 0 ⇒ 4x +
)
2 x 3 − 1 ( x − 1) x + x + 1 6. (b) : f ( x ) = = x − 1 ( x − 1) 2 + 1, x ≠ 1 ⇒ f (x ) = x + x +
f(x) = x2 + x + 1
AM = 9 +
1 4
=
37 4 = dt ⇒ dx =
5
∫ (x − 1)(x − 2)(x − 3)(x − 4)(x − 5)dx
2
⇒
5. (b) : Required equation of AB of AB is S – S′ = 0
1
4. (a) : For pair of lines to be parallel, h2 – ab = ab = 0
p = p = –4 x – 2 y ) + q = 0, lines to be parallel Now (x (x – – 2 y )2 + (x – root is real and distinct 1 ⇒ 1 − 4q > 0 ⇒ q < 4 ⇒ p + p + q = – 5
9. (c) : Put x – – 3 = t
\
p
(
A C − AB ⇒ BC = AC − AB ⇒ BM = AC 2 Again, AB + BM + MA = 0 AC AC − AB \ AM = AB + BM = AB + 2 ^ ^ ^ AB + AC 6 i + j j = = = 3 ^i + 2 2 2
1 5 +1 5 −1 5 +1− 5 +1 = − = =1 2 2 2 x cos–1x
C M
Since, AB + BC + CA = 0
1+ 5 x = 2
3. (d) : Domain : x ∈ [–1, 1] Q
A
1 x
2 2 < x < 3 and {x } =
2. (c) : Since,
(t + 2)(t + 1)t ⋅ (t − 1) ⋅ (t − 2) dt ∫ −
=
2
= 0 (since function is odd). 10. (c) : 2 y sin sinq – x sin sinq + (x – – z )(1 )(1 – sin2q) – 2x 2 x (1 – 2sin2q) = 0 ⇒ 2 y sin x – x sin sinq – x sin sinq + x – sin2q – z + + z sin sin2q – 2x 2 x + + 4x 4 x sin sin2q = 0 ⇒ (3x (3x + + z )sin )sin2q + (2 y –x – x )sin )sinq – x – – z = = 0 2 (3x + + z )sin )sin q + (2 y – – x )sin )sinq – (x ( x + + z ) = 0 ⇒ (3x Tis equation is zero for all values of q. Tus, 3x 3x + + z = = 0, 2 y 2 y – – x = = 0, x + = 0 x + z = − z x So, x = , y = , x = −z 3 2 −z , z and −z , −z , z So the triplet formed is −z , 3 6 2 Tus no positive triplet is formed.
3
3
3/4
MPP-8 CLASS XI
ANSWER
KEY
5.
x = 1
7. (a) : n ( A ( A × B) = 12
1. (a)
2. (a)
3.
(a)
4.
(a)
No. of subset each contain 2 or more elements is = ( 12C 2 + 12C 3 + ... + 12C 12)
6.
7.
8.
(a, b)
9.
(a, b, c, d)
= (12C 0 + 12C 1 + 12C 2 + 12C 3 + ... + 12C 12) – (12C 0 + 12C 1) = 2 12 – (1 + 12) = 4096 – 13 = 4083 42
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FEBRUARY ‘17
(d)
10. (b) 15. (d) 20. (4)
(b, c)
11. (a, b, c) 12. (a, c) 16. (c) 17.. (5) 17
(d)
13. (a, b, c)14. (b) 18.. (4) 18 19. (0)
ELLIPSE This article is a collection of shortcut methods, important formulas and MCQs along with their detailed solutions which provides an extra edge to the readers who are preparing for various competitive exams like JEE(Main & Advanced) and other PETs. DEFINITION AND STANDARD EQUATION OF ELLIPSE
An ellipse is the locus of a point in a plane which moves in such a way that the ratio of its distance from a fixed point (called focus) in the same plane to its distance from a fixed straight line (called directrix) is always constant which is less than unity. Te constant ratio is generally denoted by e (0 < e < 1) and is known as the eccentricity of the ellipse. Te general equation ax 2 + 2hx 2hxy + by 2 + 2 gx + + 2 fy + c = 0 y + bc + 2 fgh fgh – af 2 – bg 2 – ch2 ≠ 0 represents an ellipse, if D = abc + ab. and h2 < ab. EQUATION OF THE ELLIPSE z
z
Te equation of the ellipse, whose focus is the point (h, k) k ) and directrix is lx + my + n = n = 0 and having eccentricity e, is (lh + mk + n) (x − h)2 + ( y − k)2 = e 2 ⋅ (l 2 + m2 ) ae, 0) and Equation of the ellipse whose focus is (ae, directrix is is x = a/e a/e and having eccentricity e is given by 2
x a
2
2
+
2
y 2
2
a (1 − e )
=1
x
or
a
2
y b
2
(0, b)B
X
x i r t c e r i P(x1 , y 1) D
x = –a/e –a/e
z
S
S
(–ae, 0) (ae, 0)
b2
= 1−
(PN )2 b2
=
x 2 a2
y 2 b2 =
= 1 can also be defined as
a2 − x 2 a2
A ′N ⋅ AN a2
y 2
⇒
b2
=
(a + x )( a − x) a2
b2 (PN )2 = AN ⋅ A′N a2
or
p2
M
A (a, 0)
y 2
a2
+
B
B (0, –b) Y
x 2
Ellipse
=
( BC )2 ( AC )2
Y
C A (–a, 0)
(a) wo ellipses are said to be similar, if they have the same value of eccentricity. (b) Distance of every focus from the extremity of minor axis is equal to a, as b2 + a2 e2 = a2 .
or Y
M
Note :
=1
where b2 = a2 · (1 – e2) x i r t c e r i D
...(1) + =1 a 2 b2 Te foci S and S′ are (ae (ae,, 0) and (–ae (–ae,, 0). he equation of its directrices are x = a/e and = –a/e. x = a/e. Let P (x 1, y 1) be any point on (1), Now, SP = = ePM = e (a/e ( a/e – – x 1) = a – ex 1 a/e + x 1) = a + ex 1 and S′P = = ePM ′ = e(a/e + = (a – ex 1) + (a + ex 1) = 2a = AA = AA′ \ SP + S′P = = Length of major axis Te sum of focal distances of any point on the ellipse is equal to the length of major axis.
⇒
2
+
Te ellipse is
y 2
x 2
x = a/e
X
X
A (–a, 0)
P p1 N
C
A (a, 0)
X
B
Y
Sanjay Singh Mathematics Classes, Chandigarh, Ph : 9888228231, 9216338231 MATHEMATICS TODAY
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Equation of Ellipse Referred to Two Perpendicular Lines
z
(a) Ellipse
x 2 a2
+
y 2 b2
= 1 can also be defined as
Let Q be a point on auxiliary circle x 2 + y + y 2 = a2 such that QP produced produced is perpendicular to the x -axis. -axis. Ten P and and Q are the corresponding points on the ellipse and the auxiliary circle respectively. Y
2
(L.P. (L.P. from from any any poin pointt on elli ellips pseto eto the the majo majorr axis axis))
Q
(Leng (Length th of semi semi min minor o r axis) axis)2
P
+
(L.P. (L.P. from from any any poin pointt on elli ellips pseto eto the the min minor axis axis))2 2
(Length of semi emi major ajor axis) axis)
X
=1
⇒
b2
+
p22 a2
x 2 + y 2 = 1 a2 b2
x2 + y2 = a2 Y
=1
(b) If L1 :a : a1x +b + b1 y +c + c1 = 0 and L2 :b : b1x –a – a1 y +c + c2 = 0, then 2
2
a x + b y + c b x − a y + c 1 1 1 1 1 2 a12 + b12 a12 + b12 the equation + =1 2 2
b a represents an ellipse in the plane such that (i) he centre of the ellipse is the poi nt of intersection of the lines L1 = 0 and L2 = 0. (ii) Te major major axis lies along along L2 = 0 and the minor axis along L1 = 0, if a > b. If a < b, then the major axis is along L1 = 0 and minor axis is along L2 = 0. P(x, y) p1 p2
L1
a
b
L2
Let ∠QCA = QCA = q (0 ≤ q < 2p) i.e., i.e., the eccentric angle of P on on an ellipse is the angle which the radius (or radius vector) through the corresponding corresponding point on the auxiliary circle makes with the major axis. ( a cos q, a sin q) and P ≡ (a ( a cos q , b sin q) \ Q ≡ (a Te equation x = a cos q and y and y = b sin q taken together 2 2 are called parametric equation of ellipse x + y = 1 a 2 b2 where q is real parameter and P ≡ (a cosq , b sin s inq ) is any point on the ellipse also known as P (q) or ‘q’ point on the ellipse. Note: (i) (i) Eccentri c angles of the extremities of latus
x 2
Parametric Equation of the Ellipse Te circle described on the major axis of an ellipse as diameter is called the auxiliary circle of the ellipse.
a2
2
x
x 2
+
y 2 b2
y 2
= 1 according as
h2 a2 Y
(outside) P (outside)
2
( AA′ is diameter of the circle)
MATHEMATICS TODAY
|
b2
>, = , < 1
B
X
P (inside) (inside)
A
C
A B P (on) (on)
Y
44
+
k2
y
+ = 1 (a > b). a 2 b2 Equation of its auxiliary circle is x 2 + y 2 = a 2
Let the equation of ellipse be
x 2
+ = 1 and P ≡ (h, k) be any a 2 b2 point then P will will lie outside, on or inside the ellipse
Let the ellipse be
z
y 2
rectum of the ellipse + 2 = 1 are given by 2 a b b tan−1 ± . ae x 2 y 2 (ii) Area of of the ellipse ab sq. units. + 2 = 1 is pab sq. 2 a b Position of Points w.r.t. an Ellipse
(iii) If a > b, then the lengths of the major and minor b , then axes are 2a 2a and 2b 2b respectively and if a < b, the lengths of major and minor axes are 2b 2 b and 2a respectively.
\
X A (a, 0)
N S
(–a, 0)
(where L.P. means length of perpendicular) p12
C
A
FEBRUARY‘17
X
Important Points Te equation of the ellipse whose axes are parallel to the coordinate axes and whose centre is at the origin, is
x 2 a2
+
y 2
= 1 with the following properties :
b2
S. No.
x 2
Ellipse
(i)
Basic Fundamentals Coordinates of the centre
(ii) (iii iii) (iv) (v) (vi) (vii) vii) (vii (viii) i) (ix)
Coordinates of the vertices Coo Coordinates of the the foci oci Length of the major axis Length of the minor axis Equation ion of the the major axis Equ Equatio tion of the the mino inor axis axis Equa Equati tio ons of the the dir direc ectr tric ices es Eccentricity
(x) (xi) (xii) (xiii)
a2
Equation of Chord z Equation of the chord joining the points whose
is given by
2
x
a2
+
2
y
b2
=1
x a + b + y sin a + b = cos a − b . cos 2 2 b 2 a z
x a
+
2
tan z
y
a z
2
2
b
= 1 ( a > b ) at the point (ae ( ae,, 0), then b e −1 . = z e + 1
If a and b are the eccentric angles of extremities of a focal chord of the ellipse
a −b 2 e = ⇒ a +b cos 2
x 2 a
a
2
+
y 2 2
b
b
a2
z
= 1 (a > b) , then
e −1 tan ⋅ tan = or 2 2 e +1
e +1 . e −1
+
b2
= 1, a < b
(0, 0)
(±a, 0) (±ae, ae, 0) 2a 2b y = = 0 x = = 0 x = = ±a/e
(0, ±b) (0, ±be) be) 2b 2a x = = 0 y = = 0 y = = ± b/e
e = 1−
a2 e = 1− a2
(± ae, ae, ± b2/a) 2b2/a (a cos q, b sin q) x 2 + y 2 = a2
(± a2/b, ± be) be) 2 2a /b (a cos q, b sin q) x 2 + y 2 = b2
Locus of mid-points of focal chords of an ellipse x 2 a2
+
y 2 b2
= 1 (a > b) is
x 2
y 2
a
b2
+ 2
=±
ex . a
Equation of Tangents in Different Forms Point Form : Equation of tangent to the ellipse
y 2
xx 1 yy 1
+ 2 = 1. a2 b a 2 b2 Parametric Form : Equation of tangent to the +
ellipse
tan
cos
= 1, a > b
y 2
(0, 0)
x 2
If the chord joining two points whose eccentric angles are a and b, cut the major axis of the ellipse 2
b2
x 2
b2 a2
Ends of the latus rectum Length of the latus rectum Parametri tric coordina inates Auxill illiary iary circle
eccentric angles are a and b on the ellipse
+
y 2
(x 1, y 1) is = 1 at the point (x
x 2 2
+
y 2 2
= 1 at the point (a (a cos q, b sinq)
a b y x is cos q + sin q = 1. a b Slope Form : Equation of tangent of slope m to the ellipse
x 2 a2
+
y 2 b2
= 1 are given by
y = mx ± (a2m2 + b2 ) and the coordinates of the points of contact are
a2m b2 ,± . 2 2 2 (a2m2 + b2 ) (a m + b ) MATHEMATICS TODAY
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Slope Form : Equation of normal of slope m to the
Results Related to Tangents
(i) he equations of the tangents to the ellipse at points P (a cos q1, b sin q1) and Q(a cos q2, b sin q2) y y x x are cos q1 + sin q1 = 1 and cos q2 + sin q2 = 1 a b a b and these two intersect at the point
q1 + q2 q1 + q2 a cos 2 b sin 2 . , cos q1 − q2 cos q1 − q2 2 2 (ii) he line joining two points on an ellipse, the difference of whose eccentric angle is constant, touches an other ellipse of same eccentricity. (iii) Te locus of the point of intersection of tangents to an ellipse at the points whose eccentric angles differs by constant is an ellipse of the same eccentricity. If the eccentric angles differs by a 2 2 right angle then the locus is x + y = 2. a 2 b2 (iv) Te locus of the point of intersection of tangents x 2
to the ellipse
2
+
y 2 2
= 1 at the points the sum
a b of whose eccentric angles is constant is a straight line passing through the centre of the ellipse. (v) If the tangent at P on an ellipse meets the directrix in F , then PF will subtend a right angle at the corresponding focus i.e., ∠PSF = p/2. (vi) If SM and S ′M ′ are perpendiculars from the foci upon the tangent at any point of the ellipse, then SM ·S ′M ′ = b2 and the point M & M ′ lie on the auxiliary circle (where a > b). (vii) Te tangents drawn from any point on the director circle of a given ellipse to the ellipse are always at right angle. Equations of Normals in Different Forms Point Form : Equation of normal to the 2
x
ellipse
2
+
2
y
2
= 1 at the point ( x 1 , y 1 ) is
x 2
ellipse
a2
b2
= 1 are given y = mx
m(a2 − b2 )
(a2 + b2m2 )
a2 b2 m ,± at the points ± 2 2 2 a +b m a2 + b2 m2 Results Related to Normals (i) If the line lx + my + n = 0 be a normal to the
ellipse x 2
y 2
a2
b2
(a2 − b2 )2
+ = 1 then 2 + 2 = a2 b2 m n2 l (ii) Locus of mid-points of normal chords of an ellipse x 2
2
y 2
x 2 y 2 a6 b6 2 2 2 + = 1 is + + = (a − b ) a 2 b2 a2 b2 x 2 y 2 (iii) Four normals can be drawn from a point to an ellipse. Intersection of a Circle and an Ellipse (i) Te circle x 2 + y 2 + 2 gx + 2 fy + c = 0 and ellipse x 2 y
+ = 1 can intersect each other at maximum a2 b2 of four points. (ii) Te maximum number of common chord of a circle and an ellipse is six. (iii) Te maximum number of common tangent of a circle and an ellipse is four. Director Circle Te locus of the point of intersection of the tangents to x 2
y 2
+ = 1 which are perpendicular to each a2 b2 other is called director circle and its equation is given by x 2 + y 2 = a2 + b2. Sub-tangent and Sub-normal an ellipse
a2 Length of sub-tangent NT = x 1
Length of sub-normal GN =
b2 a
a b a x b y 2 − = a − b2 . x 1 y 1
2
− x 1 | x 1| = (1 – e2) | x 1|
Y
2
2
T
P(x1, y 1)
Parametric Form : Equation of normal to 2
ax sec
q
x
2
y
+ = 1 at ( a co s q , b si n q ) is a2 b2 – by cosecq = a2 – b2.
the ellipse
46
+
y 2
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X
C G
Y
N
T
X
Reflection Property of an Ellipse
If an incoming light ray passes through the focus ( S) strike the concave side of the ellipse then it will get reflected towards other focus ( S′) and ∠SPS′ = ∠SQS′ Y B
Q Y X
S
A
T an g en t P
C
N S B
Reected ray Y
Te tangent and normal at any point of an ellipse bisect the external and internal angles between the focal radii to the point. PROBLEMS
Te line y = mx + c, will be a tangent to the ellipse
x 2 a
2
+
y 2 b
2
= 1 (|b| > |a|) , provided
c2 =
a2 + b2m2 (a) (c) c2 = (a2 + b2)m2
(a)
(b) c2 = b2 + a2m2 (d) c2 = (b2 – a2)m2
angents are drawn to the ellipse x 2 + 2 y 2 = 4 from any arbitrary point on the line x + y = 4, the corresponding chord of contact will always pass through a fixed point, whose coordinates are 1 1 (a) 1, (b) , 1 2 2 1 1 (c) 1, − (d) , − 1 2 2 3. Equation of the ellipse whose axes are along the coordinate axes and whose length of latus rectum and eccentricity are equal and equal to 1/2 each , is (a) 6x 2 + 12 y 2 = 1 (b) 12x 2 + 6 y 2 = 1 (c) 3x 2 + 12 y 2 = 1 (d) 9x 2 + 12 y 2 = 1 2.
Te line y = x - 1 touches the ellipse 3 x 2 + 4 y 2 = 12, at 1 1 (a) , − (b) (3, 2) 2 2 (c) (–1, –2) (d) None of these 4.
a
One foot of normal of the ellipse 4x 2 + 9 y 2 = 36, that is parallel to the line 2 x + y = 3, is
9 8 (b) − , 5 5 (d) None of these
+
+ 2
x
y 2 b2 2 b
2
y
=2
(b)
=4
(d)
a2 x 2 x 2 a
2
+ +
b2 y 2 y 2 2
b
Normals drawn to the ellipse
7.
y 2 2
= 1 at
=2 =4 x 2 2
+
y 2 2
= 1 at
a b point ‘P ’ meets the coordinate axes at points A and B respectively. Locus of mid point of segment AB is (a) 4x 2a2 + 4 y 2b2 = ( a2 – b2)2 (b) 4x 2b2 + 4 y 2a2 = ( a2 – b2) (c) 16x 2a2 + 16 y 2b2 = (a2 – b2)2 (d) 16x 2b2 + 16 y 2a2 = (a2 – b2) 8. If the line y = mx + c is a tangent to the ellipse x 2 y 2
a2
+
b2
= 1 then corresponding point of contact is
2 (a) a m , c a2 m (c) , c
b2 c b2 − c
a2 m (b) − , c a2 m (d) − , c
b2 − c b2 c
If the line y = mx + c is a normal to the ellipse
9. 2
x
a2
(a) (b) (c) (d)
5.
9 8 (a) , 5 5 9 8 (c) − , − 5 5
a2 2
X
Normal
1.
x 2
(c) A
2
+
a b point ‘P ’ meets the coordinate axes at points A and B respectively. Locus of mid-point of segment AB is
Light ray
x 2
angent drawn to the ellipse
6.
+
y 2 b2
= 1, then corresponding foot of normal is
a2 c b2c 2 2 , 2 2 m(b − a ) (b − a ) a2 c b2c 2 2 , 2 2 m(a − b ) (a − b ) a2 m b2c 2 2 , 2 2 c(b − a ) (b − a ) a2 m b2c 2 2 , 2 2 c(a − b ) (a − b )
10. Locus of the mid-point of chords of the ellipse
x 2
y 2
+ = 1 that are parallel to the line y = 2x + c, is a2 b2 (a) 2b2 y – a2x = 0 (b) 2a2 y – b2x = 0 (c) 2b2 y + a2x = 0 (d) 2a2 y + b2x = 0 MATHEMATICS TODAY
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11.
Te locus of the moving point P (x, y ) satisfying
(x − 1)2 + y 2
+ ( x + 1)2 + ( y − 12)2 = a
will be an ellipse if (a) a < 4 (b) a > 2
(c) a > 4
(d) a < 2
Foci of the ellipse 16x 2 + 25 y 2 = 400 are (a) (± 2, 0) (b) (±3, 0) 13.
Equation of an ellipse whose focus is S ≡ (1, 0), corresponding directrix being the line x + y = 2 and 1 eccentricity being is 2 (a) 3x 2 + 3 y 2 – 2xy + 4x – 4 y = 0 (b) 3x 2 + 3 y 2 + 2xy – 4x + 4 y = 0 (c) 3x 2 + 3 y 2 + 2xy + 4x – 4 y = 0 (d) 3x 2 + 3 y 2 – 2xy – 4x + 4 y = 0 Te equation 3(x + y – 5)2 + 2(x – y + 7)2 = 6 represents an ellipse, whose centre is (a) (–1, 6) (b) (6, –1) (c) (1, –6) (d) (–6, 1) 15.
16.
2
+
y 2 2
= 1, centered at
a b point ‘O’ and having AB and CD as it’s major and minor axes respectively. If S1 be one of the focus of the ellipse, radius of incircle of triangle OCS1 be 1 unit and OS1 = 6 units, then area of the ellipse is equal to 65 (a) 16 p sq. units (b) p sq. units 4 (c) 65 p sq. units (d) 65 p sq. units 2 x 2 y 2 17. Normal drawn to the ellipse + = 1 at the 42 32 point P (q) meets the ellipse again at point Q(2q), then value of cos q can be 1 16 (a) − (b) − (c) − 19 (d) − 21 2 23 23 23 2
18.
P is any variable point on the ellipse
x
2
+
2
y
2
=1
a b having the points S1 and S2 as its foci. Ten maximum 52
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P 1 and P 2 are the foot of altitudes drawn from the
foci S1 and S2 respectively of the ellipse
14.
x 2
incentre of triangle PS1S2 will be (a) a straight line (b) a circle (c) a parabola (d) an ellipse 20.
48 (d) ± , 0 5
Consider an ellipse
2 y 2 P is any variable point on the ellipse x + =1 2 2 a b having the points S1 and S2 as its foci. Ten Locus of 19.
y 2 x 2 12. Te equation + = 1 will represent an 6 2 − − a a ellipse if (a) a ∈ (1, 3) (b) a ∈ (1, 6) (c) a ∈ (– ∞, 2) ∪ (6, ∞) (d) a ∈ (2, 6) ~ {4}
24 (c) ± , 0 5
area of the triangle PS1S2 is equal to (a) b2e sq. units (b) a2e sq. units (c) ab sq. units (d) abe sq. units
x 2 2
+
y 2 2
=1
a b on one of its variable tangent. Maximum value of (S1P 1) (S2P 2) is equal to (a) b2
(c) a2
(b) b
(d) a
Te equation of the ellipse centered at (1, 2) having the point (6, 2) as one of its focus and passing through the point (4, 6) is 21.
(a)
(x − 1)2 36
3( y − 2)2 + 64
(b)
(x − 1)2 18
+
(x − 1)2 (c) 72
( y − 2)2 32
=1
=1
7( y − 2)2 + 128
=1
2 2 (d) (x − 1) + ( y − 2) = 1 45 20 22. Te angle between the tangents drawn to the ellipse 3x 2 + 2 y 2 = 5, from the point P (1, 2) is equal to (a) tan−1 (24 5 ) (b) tan−1 (12 5 )
24 5
(c) tan −1 23.
12 5
(d) tan −1
Te line 5x – 3 y = 8 2 is a normal to the ellipse
x 2 y 2 + = 1. If ‘ q’ be the eccentric angle of the foot of 25 9 this normal then ‘q’ is equal to
p
(a)
p
(b)
(c)
p
(d) None of these
6
4
3
Te distance of a point P (lying in the first quadrant) on the ellipse x 2 + 3 y 2 = 6 from the center of the ellipse is 2 units. Eccentric angle of the point ‘P ’ is 24.
p
(a)
(b)
p
3 (d) None of these
6
(c) p 4
25.
2
x Te tangent drawn to the ellipse 16
2
11 y + 256
p
(a)
(b)
6
p (c) 26.
x 2 2
31.
3
x
2
= 1 whose distance from the center of the b a2 + 2b2
a ellipse are equal and equal to . Eccentricity 2 of this ellipse is equal to 1 1 3 2 (a) (b) (c) (d) 2 3 3 2 27.
For all admissible values of the parameter ‘a’ the 2
straight line 2ax + y 1 − a = 1 will touch an ellipse whose eccentricity is equal to 3 (b) 1 (c) 1 (d) 2 (a) 3 2 2 3 ellipse 4x 2 + 5 y 2 = 20 at the point
Te normal to the ‘P ’ touches the parabola y 2 = 4x , the eccentric angle of the point ‘P ’ is 1 p 1 p (a) + cos−1 (b) + tan−1 5 2 2 5 28.
(c)
1 p − tan−1 5
(d)
1 p − cos −1 5 x 2
Te tangent drawn to the ellipse
+
y 2
= 1 , at a 2 b2 the point ‘P ’ meets the circle x 2 + y 2 = a2, (a > b) at the points A and B. Te line segment AB subtends an angle p/2 at the center of the ellipse. If ‘e’ be the eccentricity of the ellipse and ‘q’ be the eccentric angle of point ‘P ’ then e2(1 + cos2q) = 1 (a) e2(1 + sin2q) = 1 (b) e2(2 + cos2q) = 1 (c) e2(2 + sin2q) = 1 (d) 29.
a2 (a)
y 2 2
2a2b a2 sin2 q + b2 cos2 q
a2 cos2 q + b2 sin2 q 2a2b
(d)
a2 cos2 q + b2 sin2 q
Te line y = mx + c, will be a normal to the ellipse ,
+
y 2 b2
= 1, if
2 (b2 − a2 )2 (b) c +b = m2 m2 c2
a2
2
2 (c) a + 32.
2b2 a
(b)
a2 sin2 q + b2 cos2 q
(c)
p
Tere are exactly two points on the ellipse
+
2b2 a
= 1 at
(d) None of these
4
Length of the focal chord of the ellipse
(a)
the point P (q), touches the circle (x – 1)2 + y 2 = 16. ‘q’ is equal to
y 2
x 2
+ = 1, a 2 b2 that is inclined at an angle ‘q’ with the x -axis, is equal to 30.
b2 m2
=
(b2 − a2 )2 c2
+ b2 =
b2 (d) c 2 + m2 y 2
+
and Q(q2), such that q1 – q2 =
2p , is 3
a2 2
x (c)
c2 (b2 − a2 )2 a2
Locus of the point of intersection of tangents
2 x drawn to the given ellipse a2
(a)
=
(b2 − a2 )2
x 2 a
2
+ +
b2 2
y
y 2 2
b
=4
(b)
=4
(d)
b2
= 1, at points P (q1)
a2
b2
x
y 2
x 2
y 2
+ 2
a2
+
b2
=2 =2
y 2
x 2
+ = 1, having it’s a 2 b2 eccentricity equal to e. P is any variable point on it and P 1, P 2 are the foot of perpendiculars drawn from P to the x and y -axis respectively. Te line P 1P 2 will always be a normal to an ellipse whose eccentricity is equal to 33.
Consider the ellipse
(a) e2
(b)
e
(c)
2e (d) e 1+ e x 2
y 2
+ =1 a 2 b2 (a > b) at the extremity of the latus rectum passes 34.
Te normal drawn to the ellipse
through the extremity of the minor axis. Eccentricity of this ellipse is equal to MATHEMATICS TODAY
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(a)
(c)
5 −1 2 3 −1 2
(b)
5 −1 2
(a)
(d)
3 −1 2
(c)
(a) (b) (c) (d)
Te equation of common tangents of the curves 4 y 2 = 8 and y 2 = 4x are y – 2x – 4 = 0, y + 2x + 4 = 0 y – 2x – 2 = 0, y + 2x + 2 = 0 2 y – x – 4 = 0, 2 y + x + 4 = 0 None of these
36.
If L is the length of perpendicular drawn from the
35.
x 2 +
2 y 2 x origin to any normal of the ellipse + = 1, then 25 16 maximum value of L is (a) 5 (b) 4 (c) 1 (d) None of these 37.
Te maximum distance of the centre of the ellipse
x 2 y 2 + = 1 from the chord of contact of mutually 9 4 perpendicular tangents of the ellipse is 3 6 36 9 (a) (b) (c) (d) 13 13 13 13 38.
angents PA and PB are drawn to the ellipse
x 2 y 2 + = 1 from the point P (0, 5). Area of triangle 16 9 PAB is equal to 256 16 sq. units (a) (b) sq. units 25 5 (c)
32 sq. units 5
(d)
1024 sq. units 25
Te straight line x – 2 y + 4 = 0 is one of the common x 2 y 2 2 tangents of the parabola y = 4x and + = 1. Te 4 b2 equation of another common tangent of these curves is (a) x + 2 y + 4 = 0 (b) x + 2 y – 4 = 0 (c) x + 2 y + 2 = 0 (d) x + 2 y – 2 = 0 39.
x 2 y 2 + = 1 at the 36 9 point ‘P ’ meets the y -axis at A and normal drawn to the ellipse at point ‘P ’ meets the x -axis at B. If area of 27 triangle OAB is sq. units, then eccentric angle of 4 point ‘P ’ is 40.
54
angent drawn to the ellipse
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p
p
(b)
3
6 p
(d) None of these 4 41. Te tangent and normal drawn to the ellipse x 2 + 4 y 2 = 4 at the point P , meet the x -axis at A and B respectively. If AB = 2 then cosq is equal to, (‘q’ being the eccentric angle of point P ) 8 2 5 (a) 1 (b) (c) (d) 3 3 3 3 42. Te chord joining the points P (a cosq , b sinq ) 1 1 1 and P 2(a cosq2, b sinq2) meets the x -axis at A. If OA = c(‘O’ being the origin), then the value of q q tan 1 ⋅ tan 2 is equal to 2 2 c −a c −a (a) (b) (c) c +b c +a
c − b (d) c − b c+a c +b 2 2 x y 43. A tangent to the ellipse + = 1, having 18 32 4 slope − cuts the x and y -axis at the points A and B 3 respectively. If O is the origin then area of triangle OAB is equal to (a) 12 sq. units (b) 24 sq. units (c) 36 sq. units (d) None of these 44.
Which of the following is a common tangent to the
ellipses
x 2 a2 + b2
+
y 2 b2
= 1 and
x 2 a2
+
y 2 a 2 + b2
= 1?
(a) by = ax − a4 + a2 b2 + b4 (b) ay = bx − a4 + a2 b2 + b4 (c) by = ax − a4 − a2 b2 + b4 (d) ay = bx − a4 − a2 b2 + b4 45.
x 2 a2
Maximum length of the chord of the ellipse +
y 2 b2
differ by
= 1, such that eccentric angles of its p
2
extremities
is (a > b)
(a) a 2 (b) b 2 (c) ab 2 (d) None of these 46. Consider an ellipse having its axes along the coordinate axes and passing through the point (4, –1).
If the line x + 4 y – 10 = 0 is one of its tangent, then area of the ellipse is equal to (a) 20 p sq. units (b) 30 p sq. units (c) 15 p sq. units (d) 10 p sq. units 47. S1 and S2 are the foci of an ellipse. ‘B’ be one of
the extremity of its minor axis. If triangle S1S2B is right angled then eccentricity of the ellipse is equal to 1 3 (a) (b) 2 2 3 (c) (d) None of these 2 48. If the chord of contact of tangents drawn from a x 2 y 2 point P on the ellipse + = 1 touches the circles a2 b2 x 2 + y 2 = c2, then locus of P is (a)
(c)
x 2 a2 x 2
+
+ 4
y 2 b2 y 2 b4
= =
a2
(b)
c2
(d)
c2 1
x 2 a2 x 2
+ +
y 2 b2 y 2
= =
b4 c4 a4
a2 b2 c 4 49. Te point on the ellipse 4x 2 + 9 y 2 = 1 such that tangent drawn to the ellipse at this point is perp endicular to the line 8x – 9 y = 0, is 27 16 27 16 ,− , − (a) (b) 2 985 2 985 2 985 3 985 (c)
a
27 16 , 2 985 3 985
(d) None of these
2 y 2 x 50. angents drawn to the ellipse + = 1, from 16 9 the point ‘P ’, meets the coordinate axes at concyclic points, then locus of point ‘P ’ is (a) x 2 + y 2 = 7 (b) x 2 – y 2 = 7 (c) x 2 + y 2 = 25 (d) x 2 – y 2 = 25 SOLUTIONS
1. (b) : If the point of contact is P (a cosq, b sinq),
x y cos q + sin q = 1 a b and given line y = mx + c will be identical. Comparing the coefficients, we get then equation of tangent
⇒
cos q sin q 1 = =− −b am c am b cos q = − , sin q = c c
⇒
a2m2 + b2 = c2
2. (a) : Let the aribitrary point be (a, 4 – a), then
equation of corresponding chord of contact is, x · a + 2 y (4 – a) = 4 ⇒ a(x – 2 y ) + 8 y – 4 = 0 It is indeed a family of concurrent lines. Te point of concurrency being the intersection point of the lines x – 2 y = 0 and 8 y – 4 = 0 1 \ Required point is 1, 2 2b2 3. (d) : a e = 1−
b2 a2
=
b2 1 = ⇒ a 2
=
1 4
...(i)
(Given)
1 2
...(ii)
1 2 1 From (i) and (ii), we get, a = , b = 3 12 2 2 x y \ Equation of ellipse is + = 1 1 1 9 12 i.e., 9x 2 + 12 y 2 = 1 4. (d) : Given ellipse is,
x 2 4
+
y 2 3
=1
and line is y = x – 1 Let the point of contact be (2 cosq,
\ Equation of tangent becomes cos q sin q ⇒ = =1 2 − 3 ⇒ cosq = 2, sinq = – 3
3 sinq). x y cos q + sin q = 1 2 3
which is not possible. Tus, the given line can’t be a tangent to the given ellipse. 2
y 2 + =1 9 4 Let the foot of normal be (3 cos q, 2 sinq) Equation of normal at this point is, 3x 2 y − =5 cos q sin q 3 Slope = tanq = – 2 2 ⇒ tanq = – 4 3 4 3 or sin q = , cos q = − 5 5 9 8 Tus foot of normal be − , . 5 5 5. (b) : Given ellipse is, x
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6. (c) : Let P ≡ (a cosq, b sinq)
\ Equation of tangent is x cos q + y sin q = 1
a b ⇒ A ≡ (a secq, 0), B ≡ (0, b cosecq) If the mid-point of segment AB be Q(h, k), then 2h = a secq, 2k = b cosecq
⇒
a2 2
b2
+
2
=1
4h 4k a 2 b2 \ Locus of Q is 2 + 2 = 4 x y 7. (a) : Let P ≡ (a cosq, b sinq) \ Equation of normal is ax − by = a2 cos q sin q
i.e.,
⇒
xh yk
+
a2 h 2a
2
b2
=−
=
h2 a2
k b
2
+
=−
k2 b2
h2 k2 2+ 2 a b c
Tus the required locus is y = −
− b2
a 2 − b2 a 2 − b2 ⇒ A ≡ cos q, 0 and B ≡ 0, sin q b a
⇒ k =− b
2 2
Tus, locus of the point is 4a2 x 2 + 4b2 y 2 = (a2 – b2)2 8. (d) : Let the point of contact be P (a cosq, b sinq).
y x cos q + sin q = 1 a b cos q sin q 1 = =− hus on comparing, am c −b
Ten equation of tangent is
⇒
am b cos q = − , sin q = c c
a2m b2 Tus the point of contact is − , c c 9. (a) : Let the foot of normal be P (a cosq, b sinq)
by ax Equation of normal at P is − = a 2 − b2 m cos q sin q c cos q sin q \ = = 2 2 a b (b − a )
⇒ cos q =
ac m(b2 − a2 )
, sin q =
bc (b2 − a2 )
a2 c b2c Tus the foot of normal is , m(b2 − a2 ) (b2 − a2 ) 10. (d) : Let the mid-point of chord be P (h, k). Equation
of this chord will be T = S1 56
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2a
2
⋅h
x or 2a2 y + b2x = 0
2a 11. (c) : Let A ≡ (1, 0), B ≡ (–1, 12 ) \ Given expression is, PA + PB = a Here, AB = 4 + 12 = 4 hus P ( x, y ) will lie on an ellipse having foci as A and B, provided a > 4 12. (d) : Clearly, 6 – a > 0, a – 2 > 0 and 6 – a ≠ a – 2 ⇒ a ∈ (2, 6) ~ {4}
If (h, k) be the mid-point of segment AB, then a 2 − b2 a 2 − b2 2h = cos q, 2k = sin q a b ⇒ 4h2a2 + 4k2b2 = (a2 – b2)2
b2
13. (b) : Equation of the ellipse is
x 2 y 2 + 25 16
=1
⇒ e2 = 1 − 16 = 9 ⇒ e = 3 25
25 5 3 3 Tus, foci are 5 ⋅ , 0 and −5 ⋅ , 0 5 5 i.e. (± 3, 0) 14. (d) : If P (x, y ) be any point on the ellipse, then
PS = e · PM, where PM is perpendicular distance of P from directrix.
⇒
1 x + y − 2 (x − 1) + y = 2 2 2
2
2
Tus required equation is 3x 2 + 3 y 2 – 2xy – 4x + 4 y = 0 15. (a) : Given equation can be rewritten as
x + y − 5 2
2
2
x − y + 7 2
+ =1 1 3/2 It is clearly an ellipse whose major and minor axes are along the lines x – y + 7 = 0 and x + y – 5 = 0 Tus, centre of this ellipse is the intersection point of these lines i.e., the point (–1, 6) 16. (b) : OS1 = ae = 6, OC = b (say), CS1 = a 1 DOCS1 = ⋅(OS1 )(OC) = 3b 2 Semi perimeter of triangle OCS 1 1 1 = (OS1 + OC + CS1) = (6 + a + b) 2 2
Now, inradius of triangle OCS 1 3b = = 1 (given) 1 (6 + a + b) 2 ⇒ 5b = 6 + a Also b2 = a2 (1 – e2) = a2 – 36 Tus, 25b2 = 36 + a2 + 12a ⇒ 25(a2 – 36) = 36 + a2 + 12a 13 5 ⇒ 2a2 – a – 78 = 0 ⇒ a = 2 , b = 2 65p Hence, area of ellipse = pab = sq. units. 4 17. (a) : Equation of normal at P (q) is, 3 y 4x − =7 cos q sin q Since it passes through the point Q(2q), thus 16 ⋅ cos 2q 9 ⋅ sin 2q − =7 cos q sin q 16(2 cos2 q − 1) ⇒ − 18 cos q = 7 cos q ⇒ 23 cos2q – 7 cosq – 16 = 0 [ as cosq ≠ 1] ⇒ cosq = – 16 23 18. (d) : In this case, base S1S2 is fixed and PS1 + PS2 is fixed. Hence area will be maximum when PS1 = PS2 1 \ Maximum area = (2ae)b = abe sq. units. 2 19. (d) : Here, S1S2 = 2ae, PS1 + PS2 = 2a If the normal to ellipse at P , meets the x -axis at Q, then incentre of DPS1S2 will lie on segment PQ and PS1 + PS2 1 will divide it in the ratio i.e., S1S2 e Let P ≡ ( a cosq, b sin q) Equation of normal at ‘ P ’ is ax by − = a2 − b2 = a2 e2 cos q sin q ⇒ Q ≡ (ae2 cosq , 0) Let R(h, k) be the incenter, then a cos q ⋅ e + ae2 cos q b sin q ⋅ e + 0 ⋅1 h= , k= 1 1 1+ 1+ e e
1 1 ⇒ h 1 + = (ae + ae2 ) cos q and be ⋅ sin q = k 1 + e e Tus locus of R is
2 2 1 1 2 x 1 + y 1 + e e + =1 2 2 2 2 2 a e (1 + e) b e 2
which is clearly an ellipse. 20. (a) : S1 = (ae, 0), S2 = (–ae, 0) Let the tangent be xb cosq + ay sin q – ab = 0 S1P 1 =
⇒ S1P 1 =
| abe cos q − ab |
b2 cos2 q + a2 sin2 q ab(1 − e cos q) b2 cos2 q + a2 sin 2 q ab(1 + e cos q)
Similarly, S2P 2 =
⇒ S1P 1·S2P 2 = =
b2 cos2 q + a2 sin2 q a2b2 (1 − e 2 cos2 q) b2 cos2 q + a2 sin 2 q
a2b2 (1 − e2 cos2 q) a
2
2
2
2
+ (b − a )cos q
=
a2b2 (1 − e2 cos2 q) a
2
2 2
2
− a e cos q
= b2
which is clearly a constant Tus, (S1P 1) (S2P 2) = b2 21. (d) : Let the equation of ellipse having centre (1, 2) be
(x − 1)2
a2 We have, ae = 5
Also,
9
16
a
b2
+ 2
+
( y − 2)2
⇒
=1 b2 a2e2 = 25
=1
Now, b2 = a2(1 – e2) = a2 – 25 9 16 ⇒ 2+ 2 = 1 ⇒ a4 – 50a2 + 225 = 0 a a − 25 2 ⇒ a = 5, 45 But a2 = 5 is clearly rejected \ a2 = 45 ⇒ b2 = 20 (x − 1)2 ( y − 2)2 Tus, required equation is + =1 45 20 22. (d) : angent to the ellipse is
5m2 5 y = mx ± + 3 2 5m2 5 If it passes through P (1, 2), then (2 – m)2 = + 3 2 2 ⇒ 4m + 24m – 9 = 0 If it’s roots are m1 and m2 then MATHEMATICS TODAY
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57
m1 + m2 = – 6, m1m2 = –
9 4
⇒ | m1 − m2 | = 36 + 9 = 3 5 \ tan q =
m1 − m2
1 + m1m2
=
3 5 12 = 9 5 1− 4
⇒ A2 ⋅
23. (c) : Equation of normal at point ‘ q’ is
3 y 5x − = 16 cos q sin q On comparing with 5x – 3 y = 8 2, we get p 1 ⇒ q= ⇒ cos q = sin q = 4 2 24. (c) : Let the point ‘P ’ be ( 6 cos q,
⇒
2 sin q)
1
⇒ 6 – 4 sin2q = 4 ⇒ sin2q = 2 p ⇒ q = ( q lies in first quadrant)
1 − a2
+ B2 =
1 1 − a2
a2 (4 A2 − B2 ) + B2
=
1
⇒ A2 =
1 4
5 tanq – sin q 2 2 Since (i) is tangent to y 2 = 4x 1 \ y = mx + m
25. (b) : Equation of tangent at point ‘ q’ is,
x 11 y sin q = 1 cos q + 4 16
It will touch the given circle, if cos q −1 4 =4 2 2 cos q 11sin q + 16 256 ⇒ 4 cos2q + 8 cosq – 5 = 0 1 ⇒ cos q = ⇒ q = p , 5p 2 3 3 26. (c) : Te given distance is clearly the length of semi major axis a2 + 2b2 =a hus, 2 ⇒ 2b2 = a2 ⇒ 2 a2 (1 – e2) = a2 1 1 ⇒ e2 = ⇒ e = 2 2
⇒ m= ⇒
1 5 sin2 q ⋅ = −1 ⇒ cos q = − , and cos q = 5 5 4 cos q
1 ⇒ q = p − cos−1 or p + cos−1 1 5 5 29. (a) : Let line AB be bx cosq + ay sin q – ab = 0 Now, combined equation of lines OA and OB is x 2 + y 2
=
y x a2 cos q + sin q
a
b
27. (a) : Let the equation of ellipse be 2 +
A
y 2 B
2
...(i)
5 1 sin q tan q, = − m 2 2
i.e., x 2 · sin2q + y 2 1 −
x 2
FEBRUARY‘17
4a 2
i.e., y = x .
4
|
1 − a2
1 − a2 1 − a2 If ‘e’ be the eccentricity of the ellipse, then A2 = B2(1 – e2) 1 3 ⇒ 1 − e2 = ⇒ e= 4 2 28. (d) : Any normal to the ellipse is 5x 2 y − =1 cos q sin q
A..Q. 6 cos2q + 2 sin2q = 4
MATHEMATICS TODAY
1 − a2
1
x +
2 2 2 Comparing it with y = mx ± A m + B , we get 2a m= − 1 − a2 1 A2 m2 + B2 = 1 − a2
12 ⇒ q = tan−1 5
58
2a
We have, y = −
2
a sin2 q − sin 2q · xy = 0 b2 b a2
Since the lines OA and OB are mutually perpendicular, hence
=1
sin2 q + 1 −
a2 2
b
sin2 q = 0
a2 1 ⇒ sin q 2 − 1 = 1 ⇒ sin2 q 2 − 1 = 1 1 − e b ⇒ e2 · sin2q = 1 – e2 ⇒ e2 (1 + sin2q ) = 1 2
30. (a) : Let the extremities of the focal chord be
P 1(q1) and P 2(q2). Equation of chord P 1P 2 is x q1 + q2 y q1 + q2 q − q cos + sin = cos 1 2 2 b 2 2 a It should pass through (a e, 0) and its slope should be tan q
q + q q − q ⇒ e cos 1 2 = cos 1 2 2 2 b q + q and − cot 1 2 = tan q 2 a q + q a tan q ⇒ cot 1 2 = − 2 b Now, P 1P 2 = P 1S1 + P 2S1 = a – ea cosq1 + a– ea cosq2 = 2a – ea(cosq1 + cosq2)
q − q q + q = 2a – 2ae cos 1 2 cos 1 2 2 2
q + q = 2a – 2ae2 cos2 1 2 2 2 2 = 2a – 2ae2 · a tan q a2 tan2 q + b2 2a3e 2 sin2 q = 2a − 2 2 a sin q + b2 cos2 q 2a3 sin2 q(1 − e 2 ) + 2ab2 cos2 q = a2 sin2 q + b2 cos2 q 2ab2 2ab2 sin2 q + 2ab2 cos2 q = = 2 2 a2 sin2 q + b2 cos2 q a sin q + b2 cos2 q 31. (a) : Let the foot of normal is (a cosq, b sinq), then
by ax − = a2 − b2 cos q sin q
and the given lines will be identical, m cos q sin q c = = 2 2 hus, a b (b − a ) ac bc ⇒ cos q = 2 2 , sin q = 2 2 (b − a ) m(b − a )
⇒
a2 2
m
2
+b =
(b2 − a2 )2
32. (c) : If the point of intersection be R(h, k), then
q + q a cos 1 2 2
q + q b sin 1 2 2
1 2
2
, k= q1 − q2 q − q cos cos 1 2 2 2 q + q q + q b sin 1 2 a cos 1 2 2 2 k = , ⇒ h= 1 h=
hus,
h2 a2
+
k2 b2
=4
Hence, locus is,
x 2 a
2
+
y 2 b
2
=4
33. (d) : Let P ≡ (a cosq, b sinq)
⇒ P 1 ≡ (a cosq, 0), P 2 ≡ (0, b sinq)
y x + =1 Tus, equation of line P 1P 2 is a cos q b sin q x /a y /b ⇒ − =1 cos(−q) sin(−q) which is clearly a normal to the ellipse of the form x 2
+ 2
y 2
λ = and
A
= 1 where,
B
=
λ
A B2 A2 − B2 a A2 − B2 b If a > b, then B > A Let the eccentricity of the second ellipse be e1
⇒
1 − e12
=
A2 B
2
=
b2 a
2
= 1 − e2 ⇒ e1 = e
b2 34. (a) : Equation of normal at ae, is a a( x − ae) 2 = (ay – b ) e It should pass through (0, – b)
⇒ ⇒ ⇒
a(0 − ae) = – a2 – b2 e b2 b 2 2 a = ab + b ⇒ 2 + − 1 = 0 a a b −1 + 5 = 2 a
⇒ e2 =
5 −1 2
⇒
b2 a
⇒ e=
2
=
3− 5 2
= 1 − e2
5 −1 2
35. (c) : angent for y 2 = 4x is y
= mx +
c2 MATHEMATICS TODAY
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1 m
FEBRUARY‘17
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x 2 y 2 + = 1 is y = mx ± 8m2 + 2 8 2 For common tangents we must have 1 = ± 8m2 + 2 ⇒ m = ± 1 m 2 Tus, common tangents are x x y = + 2 and y = − − 2 2 2 i.e. 2 y – x – 4 = 0 and 2 y + x + 4 = 0 36. (c) : Any normal to the ellipse is 5x 4 y − = 9 i.e., 5 secq x – 4 cosecq y – 9 = 0 cos q sin q 9 \ L= 25 sec2 q + 16 cosec2q 9
and for
= =
25(1 + tan2 q) + 16(1 + cot 2 q) 9
25 tan2 q + 16 cot2 q + 41 25 tan2 q + 16 cot2 q Now, ≥ 20 2 ⇒ 25 tan2q + 16 cot2q + 41 ≥ 81 9 ⇒ ≤ 1 ⇒ L ≤ 1 2 2 25 tan q + 16 cot q + 41 \ Maximum value of L is 1. 37. (a) : Clearly, the tangents have been drawn by taking any point on the director circle of the ellipse. Any point on the director circle can be taken as ( 13 cos q, 13 sin q). Equation of corresponding chord of contact is 13 13 .cos q. x + sin q y = 1 9 4 1 \ Distance from the origin = 13 2 13 cos q + sin2 q 81 16 36 36 9 = = = 208 13 1053 − 845 cos2 q 38. (b) : Equation of AB is T = 0,
5 y 9 = 1 ⇒ y = 9 5 9 Putting y = in the equation of ellipse, we get 5 2 x 81 16 32 + = 1 ⇒ x = ± . hus, AB = 16 25 ⋅ 9 5 5 i.e.
60
MATHEMATICS TODAY
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FEBRUARY‘17
9 16 Distance of P from AB = 5 – = 5 5 1 32 16 256 Tus, area of D = ⋅ ⋅ = sq. units PAB 2 5 5 25 39. (a) : Other common tangent will clearly be the reflection of x – 2 y + 4 = 0 in the x -axis. Tus other common tangent will be x + 2 y + 4 = 0 40. (c) : Let the eccentric angle of point ‘ P ’ be q , then equation of tangent and normal at point are respectively y 3 y x 6x cos q + sin q = 1 and − = 27 6 3 cos q sin q
3 and B ≡ 27 cos q, 0 Tus, A ≡ 0, 6 sin q 1 3 27 27 Area of DOAB = ⋅ ⋅ ⋅ cos q = (given) 2 sin q 6 4 p 5p ⇒ cotq = 1 ⇒ q = , 4 4 41. (c) : Let P ≡ (2 cosq, sinq) x Equation of tangent at ‘ P ’ is cosq + y sin q = 1 2 ⇒ A ≡ (2 secq, 0) y 2x − =3 Equation of normal at ‘ P ’ is q q cos sin 3 ⇒ B ≡ cos q, 0 2 3 Now, AB = | 2 sec q – cosq | = 2 (given) 2 2 ⇒ |4 – 3 cos q| = 4 |cosq| ⇒ 3 cos2q ± 4 cosq – 4 = 0 2 2 ⇒ cosq = , − 3 3 42. (a) : Equation of chord is x q1 + q2 y q1 + q2 q − q + sin = cos 1 2 cos 2 b 2 2 a
q1 − q2 a cos 2 ,0 ⇒ A ≡ cos q1 + q2 2 Since OA = c, therefore,
q − q q + q a cos 1 2 = c.cos 1 2 2 2 q q q .cos 2 (a − c) = sin 1 .sin 2 (−c − a) 2 2 2 2 q q c −a ⇒ tan 1 . tan 2 = 2 2 c +a ⇒ cos
q1
MATHEMATICS TODAY
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FEBRUARY‘17
61
43. (b) : Equation of tangent at point
x
( 18 cos q, 32 sin q) , is 32
Its slope = –
18
y
cos q +
18
32
sin q = 1
4 (given) 3
. cot q = –
⇒ cotq = 1 Also, A ≡ ( 18 sec q, 0), B ≡ (0, 32 cosec q ) 1 2
⇒ DOAB = 3 2 sec q ⋅ 4 2cosec q
a
x 2
2
2
+b
+
y 2 b
2
= 1 is,
y = mx ± (a2 + b2 )m2 + b2 It will also be a tangent to second ellipse if a2m2 + (a2 + b2) = (a2 + b2) m2 + b2 a ⇒ b2m2 = a2 ⇒ m = ± b Tus common tangents are 2 a 2 2 a y = ± x ± (a + b ) ⋅ b b2
+b
45. (a) : Let the extremities of the chord be
P 1 ≡ (a cos q, b sinq) and P 2 ≡ (–a sinq, b cos q) Now, P 1P 22 = a2 (cosq + sinq)2 + b2(sinq – cosq)2 = a2 + b2 + (a2 – b2)2sinqcosq ≤ a2 + b2 + a2 – b2 = 2a2 ⇒ P 1P 2 ≤ a 2 x 2 2
+
y 2 2
=1
xh yk a2
+
i.e., xb2h + ya2k – a2b2 = 0 It will touch the circle x 2 + y 2 = c2 , if 4 2
b2
=c
4 2
b4x 2 + a4 y 2 =
a 4b4
i.e.,
x 2
y 2
=
x 2
+
y 2 b2
= 1 is − b cot q a
9 8
2 3
− = − cot q ⇒ tan q =
⇒ cos q = −
27
or cos q = −
27
985
, sin q = , sin q = −
985 Tus, required point is 27 16 2 985 , 3 985 or
16 27
16 985 16 985
27 16 − 2 985 , − 3 985
50. (b) : Let P ≡ (h, k)
m c 1 1 5 25 a2 =− = ⇒ m = − , c = \ = + b2 1 4 −10 4 2 4 16 2 2 ⇒ a + 16b = 100 We also have P (4, –1) to be a point on the ellipse, \ 162 + 12 = 1 a b ⇒ a2 + 16b2 = a2b2 ⇒ a2b2 = 100 ⇒ ab = 10 \ Area of ellipse = pab = 10p sq. units 47. (b) : If ‘O’ be the center of ellipse then for triangle S1S2B to be right angled, we must have
If it passes through ‘P ’, then (k – mh)2 = 16m2 + 9 ⇒ m2(h2 – 16) – 2mhk + (k2 – 9) = 0 If its roots are m1 and m2, then
p ∠OBS1 = ∠OBS2 = .
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Any tangent to the ellipse is y = mx ± 16m2 + 9
m1m2
=
k2 − 9
h2 − 16
Since, the drawn tangents meet the coordinate axes at concyclic points, thus, m1m2 = 1 ⇒ k2 – 9 = h2 – 16 Hence required locus is, x 2 – y 2 = 7.
4
62
1
b4 c2 9 49. (c) : Slope of tangent must be – . 8 Slope of tangent at any point P ( q ) on the ellipse c2
a4
+
a b If the line y = mx + c be its tangent, then c2 = a2m2 + b2 angent is given to be x + 4 y – 10 = 0
⇒
=1
b h +a k Hence, locus of point ‘P ’ is
hus,
i.e., by = ± ax ± a4 + b4 + a2b2
46. (d) : Let the ellipse be
Equation of chord of contact is
a2
2
⇒ ae = b ⇒ a2e2
48. (c) : Let P ≡ (h, k)
a 2b 2
= 12 ⋅ 2 ⋅ 2 = 24 sq. units 44. (a) : Any tangent to
Tus, OS1 = OB = OS2 = a2 – a2e2 1 1 ⇒ e2 = ⇒ e = 2 2
|
FEBRUARY‘17
on
Conic Sections *
ALOK KUMAR, B.Tech, IIT Kanpur
IMPORTANT FACTS AND FORMULAE z
z
z
z
z
z
z
z
z
z
z
z
Te area of the triangle inscribed in the parabola 1 y 2 = 4ax is |( y − y )( y − y )( y − y )|, where 8a 1 2 2 3 3 1 y 1, y 2, y 3 are the ordinates of the vertices. Te length of the side of an equilateral triangle inscribed in the parabola y 2 = 4ax is 8a 3. (one angular point is at the vertex). y 2 = 4a(x + a) is the equation of the parabola whose focus is the origin and the axis is x -axis. y 2 = 4a(x – a) is the equation of parabola whose axis is x -axis and y -axis is directrix. x 2 = 4a( y + a) is the equation of parabola whose focus is the origin and the axis is y -axis. x 2 = 4a( y – a) is the equation of parabola whose axis is y -axis and the x -axis is directrix. Te equation of the parabola whose vertex and focus are on x -axis at a distance a and a′ respectively from the origin is y 2 = 4(a′ – a)(x – a). Te equation of the parabola whose axis is parallel to x -axis is x = Ay 2 + By + C and y = Ax 2 + Bx + C is a parabola with its axis parallel to y -axis. If the straight line lx + my + n = 0 touches the parabola y 2 = 4ax , then ln = am2. If the line x cosa + y sina = p touches the parabola y 2 = 4ax then pcosa + asin2a = 0 and point of contact is (atan2a , –2atana). x y If the line + = 1 touches the parabola l m y 2 = 4a(x + b), then m2(l + b) + al 2 = 0. If the two parabolas y 2 = 4x and x 2 = 4 y intersect at point P , whose absiccae is not zero, then the tangent to each curve at P , make complementary angle with the x -axis.
z
z
z
z
z
angents at the extremities of any focal chord of a parabola meet at right angle on the directrix. Area of the triangle formed by three points on a parabola is twice the area of the triangle formed by the tangents at these points. If the tangents at the points P and Q on a parabola meet in T , then ST is the geometric mean between SP and SQ, i.e., ST 2 = SP ·SQ angent at one extremity of the focal chord of a parabola is parallel to the normal at the other extremity. Te angle of intersection of two parabolas y 2 = 4ax and x 2 = 4by is given by tan−1
z
. 2(a2/3 + b2/3) Te equation of the common tangent to the parabolas y 2 = 4ax and x 2 = 4by is
z
z
z
z
z
3a1/ 3b1/3
1 1 2 2 a 3 x + b3 y + a 3b3
=0.
Te line lx + my + n = 0 is a normal to the parabola y 2 = 4ax , if al (l 2 + 2m2) + m2n = 0. If the normals at points ( at 12, 2 at 1) and (at 22, 2 at 2) on the parabola y 2 = 4ax meet on the parabola, then t 1t 2 = 2. If the normal at a point P (at 2, 2at ) to the parabola y 2 = 4ax subtends a right angle at the vertex of the parabola then t 2 = 2. If the normal to a parabola y 2 = 4ax , makes an angle f with the axis, then it will cut the curve again at an 1 angle tan−1 tan f . 2 If the normal at two points P and Q of a parabola y 2 = 4ax intersect at a third point R on the curve. Ten the product of the ordinate of P and Q is 8a2.
*
Alok Kumar is a winner of INDIAN NATIONAL MATHEMATICS OLYMPIAD (INMO-91). He trains IIT and Olympiad aspirants. MATHEMATICS TODAY
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FEBRUARY ‘17
63
z
z
z
z
Te chord of contact joining the point of contact of two perpendicular tangents always passes through focus. If tangents are drawn (at 12 , 2 at 1) Q from the point ( x 1, y 1) to the parabola y 2 = 4ax , (x1, y1)P then the length of their (at 22 , 2 at 2)R chord of contact is 1 ( y12 − 4ax1)( y12 + 4a2) |a| Te area of the triangle formed by the tangents drawn from (x 1, y 1) to y 2 = 4 ax and their chord of
( y12 − 4ax 1)3/2 contact is . 2a If one extremity of a focal chord is ( at 12, 2at 1), then
a −2a the other extremity ( at 22, 2at 2) becomes , t 12 t 1
z
x 2 y 2 a
z
z
z
z
z
z
z
z
a 64
2
+
b
2
MATHEMATICS TODAY
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FEBRUARY‘17
b2
he locus of the foot of the perpendicular drawn from centre upon any tangent to the ellipse 2
z
+
2
= 1 is (x 2 + y 2)2 = a2x 2 + b2 y 2 or r 2 =
a b 2 2 a cos q + b2sin2q. (In terms of polar coordinates) Te locus of the mid points of the portion of the x 2 y 2
+ = 1 intercepted a 2 b2 between the axes is a2 y 2 + b2x 2 = 4x 2 y 2. tangents to the ellipse
z
If y = mx + c is the normal to the ellipse 2
then condition of normality is c = z
z
z
x 2 y 2 a2
+
b2
= 1,
m2(a 2 − b2 )2
. (a2 + b2m2 ) Te straight line lx + my + n = 0 is a normal to the ellipse
x 2 y 2
+ 2
2 2 2 2 2 = 1, if a + b = (a − b ) . 2 2 2
a b2 l m n Four normals can be drawn from a point to an ellipse. Y Normal If S be the focus P(x1, y 1) and G be the point where the normal at X X S C G S T P meets the axis of an ellipse, then Y SG = e. SP , and the tangent and normal at P bisect the external and internal angles between the focal distances of P . Any point P of an ellipse is joined to the extremities of the major axis then the portion of a directrix intercepted by them subtends a right angle at the corresponding focus. Te equations to the normals at the end of the latus rectum and that each passes through an end of the minor axis, if e4 + e2 – 1 = 0.
z
= 1 , if a2l 2 + b2m2 = n2.
+
x 2 y 2
the harmonic mean between the segments of any focal chord of the parabola. Te straight line lx + my + n = 0 touches the ellipse x 2 y 2
= 1 , if a2cos2a + b2sin2a = p2 and that
= 1 , then the common tangent is inclined r 2 − b2 − 1 to the major axis at an angle tan 2 2. a − r
z
z
b
2
A circle of radius r is concentric with the ellipse a2
2
on the parabola y 2 = 4ax is a(t1 − t2 ) (t1 + t 2)2 + 4 Te length of intercept made by line y = mx + c between 4 the parabola y 2 = 4ax is a(1 + m2 )(a − mc) . 2 m Locus of mid-points of all chords which is inscribed a right angle on the vertices of parabola is parabola. Te focal chord of parabola y 2 = 4ax making an angle a with the x -axis is of length 4 acosec2a and perpendicular on it from the vertex is asina. Te length of a focal chord of a parabola varies inversely as the square of its distance from the vertex. If l 1 and l 2 are the length of segments of a focal chord 4l l of a parabola, then its latus-rectum is 1 2 . l1 + l 2 Te semi-latus rectum of the parabola y 2 = 4ax is
+
x 2 y 2
1 = a t + . t Te length of the chord joining two points 't 1' and 't 2'
2
2 a 2 a point of contact is a cos , b sin . p p
by virtue of relation t 1t 2 = –1 and length of chord
z
he line x cos a + y sin a = p touches the ellipse
z
Te area of the triangle formed by the three points, on 2
the ellipse
2
x
y
a
b2
+ 2
= 1 , whose eccentric angles are q,
f −y y −q q−f f and y is 2absin sin sin 2 2 2 z
z
z
If the point of intersection of the ellipses x 2 y 2 2
+
2
= 1 and
x2 2
+
y 2 2
z
z
z
z
z
b2
a
b
+ 2
a 2 b2 lie on a 2 y (x – h) + b2 x ( y – k) = 0.
Te length of chord cut off by hyperbola 2 a from the line y = mx + c is 2ab [c 2 − (a 2m2 − b2)](1 + m2)
hyperbola z
x 2 y 2
− 2
= 1 , then a2l 2 – b2m2 = n2.
touches the hyperbolas a2 cos2 a − b2 sin2 a = p2 .
2
x
y
a
2
− 2
b
z
normal to the hyperbola a2
−
b2
=
(a2 + b2 )2
x 2 y 2 a
2
−
b
2
z
z
hyperbola from any point. If a, b, g are the eccentric angles of three points on
x1x 2 y1 y 2
+
a4 b
4
z
=1
.
q1
q 1− e tan 2 = . 2 2 1+ e
x 2 y 2 a2
−
= 1 are at right angles,
b2
=0.
Te parallelogram formed by the tangents at the extremities of conjugate diameters of a hyperbola has its vertices lying on the asymptotes and is of constant area. Te product of length of perpendiculars drawn
the asymptotes is
a2b2 a 2 + b2
x 2 y 2 a
2
−
b
2
= 1 to
.
c on the curve xy = c2 meets t
If the normal at ct ,
1 the curve again in 't ', then t ′ = − 3 . t z
z
x 2 y 2
− = 1 , the normals at which a 2 b2 are concurrent, then sin(a + b) + sin(b + g ) + sin( g + a) = 0 the hyperbola
b2
= 1 , then tan
from any point on the hyperbola
= 1 , then
l 2 m2 n2 In general, four normals can be drawn to a
−a m )
b2
If the polars of (x 1, y 1) and (x 2, y 2) with respect
then
= 1 , then
If the line l x + my + n = 0 will be
− 2
to the hyperbola
z
2 2
x 2 y 2 a
z
2
−
If the chord joining two points (asecq1, btanq1) (asecq2, btanq2) passes through the focus of the hyperbola
a b2 I f t h e s t r a i g h t l i n e x c o s a + y s i n a = p 2
z
(b z
Te sum of the squares of the reciprocal of two perpendicular diameters of an ellipse is constant. In an ellipse, the major axis bisects all chords parallel to the minor axis and vice-versa, therefore major and minor axes of an ellipse are conjugate diameters of the ellipse but they do not satisfy the condition m1 · m2 = –b2/a2 and are the only perpendicular conjugate diameters. Te foci of a hyperbola and its conjugate are concyclic. wo tangents can be drawn from an outside point to a hyperbola. If the straight line lx + my + n = 0 touches the
= 1 from (h, k) x 2 y 2
= 1 be at the extremities
=2. 2
−
Te feet of the normals to
a b a b of the conjugate diameters of the former, then a2
x 2 y 2
z
A triangle has its vertices on a rectangular hyperbola; then the orthocentre of the triangle also lies on the same hyperbola. All conics passing through the intersection of two rectangular hyperbolas are themselves rectangular hyperbolas. An infinite number of triangles can be inscribed in the rectangular hyperbola xy = c2 whose all sides touch the parabola y 2 = 4ax .
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FEBRUARY ‘17
65
PROBLEMS
(a) x 2 + y 2 = a2 – b2 (c) x 2 + y 2 = a2 + b2
Single Correct Answer Type
Te equation of the common tangent of the parabolas x 2 = 108 y and y 2 = 32x , is (a) 2x + 3 y = 36 (b) 2x + 3 y + 36 = 0 (c) 3x + 2 y = 36 (d) 3x + 2 y + 36 = 0 1.
Te line x cosa + y sina = p will touch the parabola 4a(x + a), if (a) pcosa + a = 0 (b) pcosa – a = 0 (c) acosa + p = 0 (d) acosa – p = 0 2.
y 2 =
Te angle of intersection between the curves 4x and x 2 = 32 y at point (16, 8), is 4 3 (a) tan −1 (b) tan −1 5 5 p (c) p (d) 2 If y 1, y 2 are the ordinates of two points P and Q 4. on the parabola and y 3 is the ordinate of the point of intersection of tangents at P and Q, then (a) y 1, y 2, y 3 are in A.P. (b) y 1, y 3, y 2 are in A.P. (c) y 1, y 2, y 3 are in G.P. (d) y 1, y 3, y 2 are in G.P. 3.
y 2 =
Te equation of the common tangent touching the circle (x – 3)2 + y 2 = 9 and the parabola y 2 = 4x above the x -axis, is (a) (b) 3 y = 3x + 1 3 y = −(x + 3) 5.
3 y = x + 3
(c)
(d)
3 y = −(3x + 1)
Te angle between the tangents drawn from the points (1,4) to the parabola y 2 = 4x is 6.
(a)
p
p
p (c)
p
(b) (d) 2 3 6 4 Let a circle tangent to the directrix of a parabola 7. y 2 =2ax has its centre coinciding with the focus of the parabola. Ten the point of intersection of t he parabola and circle is (a) (a, –a) (b) (a/2, a/2) (c) (a/2, ± a) (d) (± a, a/2) 8.
An ellipse passes through the point (–3, 1) and its
2 Te equation of the ellipse is . 5 (a) 3x 2 + 5 y 2 = 32 (b) 3x 2 + 5 y 2 = 25 (c) 3x 2 + y 2 = 4 (d) 3x 2 + y 2 = 9 eccentricity is
9.
Te locus of the point of intersection of
2 2 perpendicular tangents to the ellipse x + y a2 b2 66
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FEBRUARY‘17
= 1 , is
(b) x 2 – y 2 = a2 – b2 (d) x 2 – y 2 = a2 + b2
2 2 x y If the eccentricity of the two ellipse 10. + 169 25 2 2 x y and + 2 = 1 are equal, then the value of a/b is 2 a b (a) 5/13 (b) 6/13 (c) 13/5 (d) 13/6
=1
Te eccentricity of the curve represented by the equation x 2 + 2 y 2 – 2x + 3 y + 2 = 0 is (a) 0 (b) 1/2 (c) 1 / 2 (d) 2 11.
Te equation of the hyperbola whose conjugate axis is 5 and the distance between the foci is 13, is (a) 25x 2 – 144 y 2 = 900 (b) 144x 2 – 25 y 2 = 900 (c) 144x 2 + 25 y 2 = 900 (d) 25x 2 + 144 y 2 = 900 12.
Te vertices of a hyperbola are at (0, 0) and (10, 0) and one of its foci is at (18, 0). Te equation of the hyperbola is 13.
(a)
x 2 y 2 − =1 25 144
2 (b) (x − 5) 25
y 2 − 144
=1
x 2 ( y − 5)2 (x − 5)2 ( y − 5)2 − = 1 (d) − =1 25 144 25 144 14. Te equation of the hyperbola whose foci are (6, 4) and (–4, 4) and eccentricity 2 is given by (a) 12x 2 − 4 y 2 − 24 x + 32 y − 127 =0 (c)
(b)
12x 2 + 4 y 2 + 24 x − 32 y − 127 =0
(c)
12x 2 − 4 y 2 − 24 x − 32 y + 127 =0
(d) 12x 2 − 4 y 2 + 24 x + 32 y + 127 =0 Te latus rectum of the hyperbola 9x 2 – 16 y 2 + 72x – 32 y – 16 = 0 is 32 9 9 32 (a) (b) − (c) (d) − 3 2 2 3 16. If m1 and m2 are the slopes of the tangents to the 15.
x 2 y 2 hyperbola − 25 16 (6, 2), then
= 1 which pass through the point
24 11 48 (c) m1 + m2 = 11
10 11 11 (d) m1m2 = 20 2 2 x y 17. Let E be the ellipse + = 1 and C be the circle 9 4 2 2 x + y = 9. Let P and Q be the points (1, 2) and (2, 1) (a) m1 + m2 =
(b)
m1m2 =
respectively. Ten (a) Q lies inside C but outside E (b) Q lies outside both C and E (c) P lies inside both C and E (d) P lies inside C but outside E
Comprehension Type
18. Te value of m, for which the line y = mx + 2 2 is a normal to the conic x − y = 1, is 16 9 2 3 (a) (c) − (d) 1 3 (b) − 3 2
25 3 , 3
19. If q is the acute angle of intersection at a real point of intersection of the circle x 2 + y 2 = 5 and the parabola y 2 = 4x then tanq is equal to 1 (a) 1 (b) (c) 3 (d) 3 3 20. Te equation of the hyperbola in the standard form (with transverse axis along the x -axis) having the length of the latus rectum = 9 units and eccentricity = 5/4 is
(a)
x 2 y 2 − =1 16 18 2
2
(c)
x y − =1 64 36
(e)
x 2 y 2 − =1 16 9
(b) (d)
x 2 y 2 − =1 36 27 2
2
x y − =1 36 64
Paragraph for Q. No. 24 to 26 Consider the circles x 2 + y 2 = a2 and x 2 + y 2 = b2 where b > a > 0. Let A(–a, 0), B(a, 0). A parabola passes through the points A, B and its directrix is a tangent to x 2 + y 2 = b2 . If the locus of focus of the parabola is a conic then 24. Te eccentricity of the conic is (a) 2a/b (b) b/a (c) a/b
(d) 1
25. Te foci of the conic are (a) (±2a, 0) (b) (0, ±a) (c) (±a, 2a) (d) (±a, 0) 26. Area of triangle formed by a latusrectum and the lines joining the end points of the latusrectum and the centre of the conic is a 2 2 (a) (b) 2ab (b − a ) b (c) ab/2 (d) 4ab/3 Paragraph for Q. No. 27 to 29
P is any point of an ellipse
x 2 2
+
y 2 2
= 1. S and S′ are
a b foci and e is the eccentricity of ellipse. ∠PSS′ = a and ∠PS′S = b 27. tan
a
b tan is equal to 2 2
21. Te equations of the common tangents of the curves x 2 + 4 y 2 = 8 and y 2 = 4x are (a) x + 2 y + 4 = 0 (b) x – 2 y + 4 = 0 (c) 2x + y = 4 (d) 2x – y + 4 = 0
(a)
2e 1− e
(c)
1− e 1+ e
22. A straight line touches the rectangular hyperbola 9x 2 – 9 y 2 = 8 and the parabola y 2 = 32x . An equation of the line is (a) 9x + 3 y – 8 = 0 (b) 9x – 3 y + 8 = 0 (c) 9x + 3 y + 8 = 0 (d) 9x – 3 y – 8 = 0
28. Locus of incentre of triangle PSS′ is (a) an ellipse (b) hyperbola (c) parabola (c) circle
23. A hyperbola having the transverse axis of length 1 unit is confocal with the ellipse 3x 2 + 4 y 2 = 12, then 2 x 2 y 2 1 (a) Equation of the hyperbola is − = 15 1 16 (b) Eccentricity of the hyperbola is 4 (c) Distance between the directrices of the hyperbola 1 is units 8 15 (d) Length of latus rectum of the hyperbola is 2 units
1+ e 1− e 2e (d) 1+ e
(b)
29. Eccentricity of conic, which is locus of incentre of triangle PSS′
(a)
e 1+ e
(b)
2e 1+ e
(c)
2e 1− e
(d)
e 1− e
Paragraph for Q. No. 30 to 32 A point P moves such that sum of the slopes of the normals drawn from it to the hyperbola xy = 16 is equal to the sum of ordinates of feet of normals. Te locus of P is a curve C . MATHEMATICS TODAY
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FEBRUARY ‘17
67
30. Te equation of the curve C is
x 2 =
36. Te equation of the curve on reflection of the
x 2 =
(a) 4 y 2 (c) x = 12 y
(b) 16 y 2 (d) y = 8x
31. If the tangent to the curve C cuts the coordinate
axis in A and B, then the locus of the middle point of AB is (a) x 2 = 4 y (b) x 2 = 2 y (c) x 2 + 2 y = 0 (d) x 2 + 4 y = 0 32. Area of the equilateral triangle inscribed in a curve
C having one vertex is the vertex of curve C . (a) 772 3 sq. units (b) 776 3 sq. units (c) 760 3 sq. units
(d) 768 3 sq. units
Column II P.
(0, 0)
C. AB and CD are the chords of a parabola which intersect at a point E on the axis. he radical axis of the two circles described on AB and CD as diameter always passes through the point
(12, 0)
R.
the curve y = 2 x . If m is slope of the normal then m + 6 = 4x 4 y + 2 289 λ
to
an
ellipse
17 = 1 λ < passes through foci F 1 and F 2 2
cuts the ellipse at ‘P ’ such that area of triangle P F 1 F 2 is 30 sq.units. If F 1F 2 = 13K where K ∈ Z then K = 68
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FEBRUARY‘17
y 2 + = 1 is 3K . Ten K is equal to 5 38. Te equation of Asymptotes of xy + 2x + 4 y + 6 = 0 is xy + 2x + 4 y + C = 0, then C = ___
x 2 y 2
− 2
π = 1 is . Ten the eccentricity of conjugate
3 a b2 hyperbola is
≡ x 2 – 108 y = 0
x 12 T ≡ xx1 − 2a( y + y1) = 0 ⇒ xx1 − 54 y + = 0 108 S2 ≡ y 2 – 32x = 0 y 22 T ≡ yy2 − 2a(x + x2 ) = 0 ⇒ yy2 − 16 x + = 0 32 x 1 54 − x 12 54 \ = = 2 = r ⇒ x1 = 16r and y2 = 16 y 2
−(16r )2
= r ⇒ r = − 2
(54 / r )
r
y 2
9 4
(−36)2 x1 = −36, y2 = −24, y1 = 108
34. A line passing through (21, 30) and normal to
concentric
x2 9
\
Integer Answer Type
circle
tangents at the end point of latus rectum to the ellipse
1. (b) : S1
(3, 0)
2
37. Te area of the quadrilateral formed by the
SOLUTIONS
B. Variable chords of the parabola Q. passing through a fixed point K on the axis, such that sum of the reciprocals of two parts of the chord through K , is a constant. Coordinates of K are
2
is 16x 2 + 9 y 2 + ax – 36 y + b = 0 then the value of a + b – 125 =
25e is equal to 13 40. If the angle between the asymptotes of hyperbola
Consider the parabola y 2 = 12x
35. A
= 1 about the line x – y – 2 = 0
(5 y + 15)2 = (12x – 5 y + 1)2 then
33. Match the following
A. a ng e nt a nd n or m al a t th e extremities of the latus rectum intersect the x - axis at T & G respectively. Te coordinates of middle point of T & G are
( y − 3)2 + 9
39. If e is the eccentricity of the hyperbola (5x – 10)2 +
Matrix – Match Type
Column I
(x − 4)2 ellipse 16
(−24)2 = 12, x2 = 32
= 18
\ Equation of common tangent is −36 ( y − 12) = (x + 36) ⇒ 2x + 3 y + 36 = 0
54 2. (a) : x cosa + y sina – p = 0 2ax – yy 1 + 2a(x 1 + 2a) = 0 cos a sin a − p From (i) and (ii), = = 2a − y 1 2a(x1 + 2a)
...(i) ...(ii)
⇒ y 1 = –2a tana and x 1 = – pseca – 2a \ y 2 = 4a(x + a) ⇒ 4a2 tan2a = –4a( pseca + a) ⇒ pcosa + a = 0
3. (a) : Using formula
q = tan−1
3a1/ 3b1/3 2(a2/3 + b2/3)
\ q = tan−1
, where a = 1 and b = 8
6 3 = tan−1 5 2(1 + 4)
4. (b) : L e t t h e c o o r d i n a t e s o f P a n d Q b e
(at12 , 2at1) and (at 22 , 2at 2 ) respectively. Ten y 1 = 2at 1 and y 2 = 2at 2. Te coordinates of the point of intersection of the tangents at P and Q are {at 1t 2, a(t 1 + t 2)} \ y 3 = a(t 1 + t 2) y + y ⇒ y 3 = 1 2 ⇒ y1, y3 and y 2 are in A.P. 2 1 5. (c) : Any tangent to y 2 = 4x is y = mx + . It touches m 1 3m + m the circle, if 3 = 1 + m2 1 ⇒ 9(1 + m2) = 3m + m 1 ⇒ 2 =3
8. (a) : Let the equation of ellipse be Q
It passes through (– 3, 1)
a2
+
b2
=1
9 1 a2 So, 2 + 2 = 1 ⇒ 9 + 2 = a2 a b b Given eccentricity is 2 / 5 So,
2 b2 =1− 2 5 a
⇒
...(i)
b2 a
3 = 2
...(ii)
5
32 32 From equation (i) and (ii), a2 = , b2 = 3 5 2 Hence, required equation of ellipse is 3x + 5 y 2 = 32 be
x 2 y 2 h2 k 2 hx yk 2 2 + 2 − 1 2 + 2 − 1 = 2 + 2 − 1 a b a b a b Pair of tangents will be perpendicular, if coefficient of x 2 + coefficient of y 2 = 0
1 3
For the common tangent to be above the x - axis, 1 m= 3
\ Common tangent is, y =
1 3
x+ 3
⇒
3 y = x + 3
⇒
k2 2
+ 2
a b
h2 2
= 2
ab
1
1
a
2
+ 2
b
⇒ h2 + k2 = a2 + b2
Replace (h, k) by (x , y ) ⇒ x 2 + y 2 = a2 + b2 10. (c) : In the first case, e = 1 − (25 / 169)
6. (b)
In the second case, e′ = 1 − (b2 / a2)
7. (c) : Given parabola is y 2 = 2ax
According to the given condition,
\ Focus (a/2, 0) and directrix is given by x = –a/2, as circle touches the directrix. \ Radius of circle = distance from the point (a/2, 0) y to the line x = –a/2
1 − b2 / a 2
= 1 − (25 / 169) ⇒ b / a = 5 / 13 (Q a > 0, b > 0) ⇒ a / b = 13 / 5 11. (c) : Equation x 2 + 2 y 2 – 2x + 3 y + 2 = 0 can be
written as
2
a a = + = a 2 2
x 2 y 2
9. (c) : Let point be (h, k). Teir pair of tangent will
2
m
\ m= ±
Putting these values in y 2 = 2ax , we get, a y = ±a at x = 2 3a and y = a − 3 at x = − 2 [Which is imaginary value of y ] \ Required points are (a/2, ±a)
(–a/2, 0)
(a/2, 0)
(x − 1)2 2
2
\ Equation of circle be x − a + y 2 = a2 2 Also y 2 = 2ax
a 3a Solving (i) and (ii) we get, x = , − 2 2
y + 3 4
x
...(i) ...(ii)
2
2
3 1 (x − 1)2 + y + = ⇒ + = 1, 4 16 (1 / 8) (1 / 16)
1 1 which is an ellipse with a2 = and b2 = 8 16 1 1 1 1 1 \ = (1 − e 2) ⇒ = (1 − e2) ⇒ e = 16 8 16 8 2 MATHEMATICS TODAY
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69
12. (a) : Conjugate axis is 5 and distance between
foci = 13 ⇒ 2b = 5 and 2ae = 13 Now, also we know for hyperbola
⇒
\ Required equation is 25x 2 – 144 y 2 = 900 13. (b) : 2a = 10, ⇒ a = 5
\ b=5
132 2
8 13 = 5 5
−1 = 5 ×
12 = 12 5
5 and centre of hyperbola
≡ (5, 0)
\ Required equation of hyperbolais
2
(x − 5) 25
y 2 4
+
= 1. he value
x 2 y 2 + − 1 is positive for x = 1, 9 4 y = 2 and negative for x = 2, y = 1. Terefore P lies outside E and Q lies inside E. Te value of the expression x 2 + y 2 – 9 is negative for both the points P and Q. Terefore P and Q both lie inside C . Hence P lies inside C but outside E. 18. (b) : Any normal to the hyperbola is ax by ...(i) + = a2 + b2 sec q tan q But it is given by lx + m ′y – n = 0 ...(ii) Comparing (i) and (ii), we get a n b n sec q = and tan q = l a2 + b2 m′ a2 + b2
⇒
ae − a = 8 or e = 1 +
x2 9
of the expression
25 (13)2 2 = 2 (e − 1) 4 4e 25 169 169 169 13 = − 2 or e2 = ⇒ e= 4 4 144 12 4e 5 a = 6, b = 2
b2 = a2(e2 − 1)
⇒
17. (d) : he given ellipse is
−
2
( y − 0) 144
=1
Hence eliminating q, we get
a2 l 2
−
b2
(a2 + b2 )2
m′
n2
= 2
14. (a) : Foci are (6,4) and (–4,4), e = 2
Since a2 = 16, b2 = 9, l = m, m′ = –1 and n =
6 − 4 , 4 + 4 = (1, 4) \ Centre is 2 2
\ On substituting in (iii), we get m = ±
5 2
5 2
⇒ 6 = 1 + ae ⇒ ae = 5 ⇒ a = and b = ( 3) 2 − ( x 1 ) Hence, the required equation is (25 / 4) 2 2 or 12x – 4 y – 24x + 32 y – 127 = 0
( y − 4)2 − (75 / 4)
15. (a) : Given equation of hyperbola is
9x 2 – 16 y 2 + 72x – 32 y – 16 = 0 ⇒ 9(x 2 + 8x ) – 16( y 2 + 2 y ) – 16 = 0 ⇒ 9(x + 4)2 – 16( y + 1)2 = 144 (x + 4)2 ( y + 1)2 − =1 16 9 2b2 Terefore, latus rectum = a
⇒
9 4
= 2× =
16. (a) : Te line through (6,2) is
y – 2 = m(x – 6) ⇒ y = mx + 2 – 6m Now from condition of tangency, (2 – 6m)2 = 25m2 – 16 ⇒ 36m2 + 4 – 24m – 25m2 + 16 = 0 ⇒ 11m2 – 24m + 20 = 0 If m1, m 2 are its roots then 24 20 m1 + m2 = and m1m2 = 11 11 70
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FEBRUARY‘17
9 2
=1
...(iii)
25 3 3
2
. 3 19. (c) : Solving equations x 2 + y 2 = 5 and y 2 = 4x we get, x 2 + 4x – 5 = 0 i.e., x = 1, – 5 For x = 1; y 2 = 4 ⇒ y = ± 2 For x = –5 ; y 2 = –20 (imaginary values) \ Points are (1, 2)(1, –2) m1 for x 2 + y 2 = 5 at (1, 2) is dy x 1 =− =− dx y (1, 2) 2 Similarly, m2 for y 2 = 4x at (1, 2) is 1 1 − −1 m1 − m2 2 \ tan q = = =3 1 1 + m1m2 1− 2 20. (c) : Q 2b
a Now b
2
2
2
2 2
= 9 ⇒ 2b2 = 9a2
= a (e − 1) =
9 2 a 16
...(i) 4 3
⇒ a= b
...(ii)
Q e = 5 4 From (i) and (ii), b = 6, a = 8 x 2 y 2 − Hence, equation of hyperbola 64 36
=1
x2 21. (a, b) : 8
+
y 2 2
2
= 1, y = 4x
Any tangent to parabola is y = mx +
1 m
If this line is tangent to ellipse then 1 = 8m2 + 2 ⇒ 8m4 + 2m2 − 1 = 0 2 m \ m2 = −2 ± 4 + 32 = −2 ± 6 16 16 1 1 ⇒ m2 = ⇒ m = ± 4 2 x x y = + 2 or y = − − 2 2 2 x – 2 y + 4 = 0 or x + 2 y + 4 = 0
Let (h, k) be a point on the locus. Any tangent to circle x 2 + y 2 = b2 is x cosq + y sinq = b \ Equation of parabola is (x − h)2 + ( y − k)2 =| x cos q + y sin q − b | i.e., (x – h)2 + ( y – k)2 = (x cosq + y sinq – b)2 Te points (± a, 0) satisfy this equation \ (a – h)2 + k2 = (acosq – b)2 (a + h)2 + k2 = (acosq + b)2 Subtracting (1) from (2), we get h = bcosq 2 2 2 ax \ Required locus is (a + x) + y = + b b x 2 y 2 i.e., + = 1 which is an ellipse. b2 b 2 − a 2 27. (c) :
22. (b, c) : y 2 = 32x
8 Let equation of tangent be y = mx + m 64 8 2 8 = m − 9 m2 9 m = ± 3, y = ± 3 x ± 8/3 23. (b, c, d) : Ellipse is
x 2 4
+
y 2 3
=1
3 1 = 1 − e2 ⇒ e = 4 2 Foci are (± 1, 0) Now the hyperbola is having same focus i.e. (± 1, 0) Let e1 be the eccentricity of hyperbola 2ae1 = 2 1 But 2a = So, e1 = 4 2 1 15 b2 = a2 e 2 − 1 = (16 − 1) = 16 16 So, the equation of the hyperbola is Here,
(
2
x 1 16
−
2
y 15 16
)
1
=1⇒
2
x 1
−
2
y 1 = 15 16
Its distance between the directrices 2a 1 1 = = = units e1 2 × 4 8 2b2 2 × 15 × 4 15 = = units \ Length of latusrectum = 16 × 1 2 a (24 - 26) : 24. (c)
x 2
+ y 2
= B = (a, 0)
25.
a2
;
(d)
x 2
+ y 2
26. (a)
= b 2 ; b > a > 0, A = (– a, 0);
or
...(1) ...(2)
2ae PS PS ′ = = sin b sin a sin( π − (a + b) 2a
y
2ae sin a + sin b sin(a + b)
=
a+b
P β
a−b
α
S′ O
1 2 sin 2 .cos 2 ⇒ = a+b a+b e 2 sin cos 2 2 1− e a b \ = tan tan 1+ e 2 2
S
x
b
28 (a) : y − 0 = tan (x + ae)
y − 0 = − tan
... (i)
2
a
(x − ae) 2 1 − e [x 2 − a2e2] or, y 2 = − 1 + e
... (ii)
2
2
y 1− e 2 1 − e a2e2 or, x + =1 ⇒ x + y 2 = 2 2 1 − e a2e 2 1 + e 1 + e a e 1 + e
which is clearly an ellipse.
1− e 2e = 1+ e 1+ e 4 30. (b) : Any point on the hyperbola xy = 16 is 4t , of the normal passes through P (h, k), then t k – 4/t = t 2(h – 4t ) ⇒ 4t 4 – t 3h + tk – 4 = 0 h \ t 1 = and St 1t 2 = 0 4 k t1t 2t 3 = − and t1t2t 3t 4 = −1 4 1 1 1 1 k \ + + + = ⇒ y1 + y 2 + y 3 + y 4 = k t1 t2 t 3 t 4 4 29. (b) : e ′ = 1 −
∑ ∑
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71
2 2 2 2 2 h Now, t1 + t 2 + t 3 + t 4 = = k 16 ⇒ Locus of (h, k) is x 2 = 16 y . 31. (c) : x 2 = 16 y Equation of tangent of P is
(4t, t 2)
2
16( y + t ) 2 4tx = 8 y + 8t 2 tx = 2 y + 2t 2 A = (2t , 0), B = (0, –t 2) M (h, k) is the middle point of AB x ⋅ 4t =
meets the curve will be (am2 , –2am) where the curve does not exist. Terefore m = –5 \ m + 6 = 1 35. (1) : Since F 1 and F 2 are the ends of the diameter 1 1 Area of ∆PF1F2 = ( F1P )( F2 P ) = x (17 − x ) = 30 2 2
P A B M
⇒ x = 5 or 12 ⇒ F 1F 2 = 13 \ K = 1 36. (7) : Let P (4, 0) and Q(0, 3) are two points on given
ellipse E1 P 1 and Q1 are images of P ,Q w.r.t x – y – 2 = 0 \ P 1(2, 2) Q1(5, –2) lies on E2 \ a = –160, b = 292 ⇒ a + b – 125 = 7
t 2 h = t, k = − ⇒ 2k = −h2 2 Locus of M(h, k) is x 2 + 2 y = 0. 4t
1 32. (d) : tan30° = 2
⇒
1
4 = 3 t 1
t 1
=
4 t 1
37. (9) : Eccentricity of given ellipse e = 2/3
Equation of tangent at L is Y
A (–4t, t 21)
⇒ t 1 = 4 3
B
30° O (0,
(+4t, t 21)
(0, 3) B
0)
AB = 8t 1 = 32 3 Area of ∆OAB =
3 4
2x y + = 1 . It meets 9 3
(–2, 0)
L(2,
5 3)
(0, 0) (2, 0)
× 32 3 × 32 3 = 768 3 sq. units
9 ( 2 , 0)
X
33. A–Q, B–Q, C–P (A) Equation of tangent at (3, 6) : y = x + 3
\ T ( – 3, 0)
Equation of normal at (3, 6) : y = –x + 9 \ G( 9, 0) Hence middle point is (3, 0) (B) Point is obviously focus (3, 0) (C) Let A(t 1) and B(t 2), C (t 3) and D(t 4) If AB and CD intersect at a point E on the axis, then by solving the equations of AB and CD we get the relation t 1t 2 = t 3t 4 Now equations of the circles with AB and CD as diameters are (x − at12 )(x − at22) + ( y − 2at1)( y − 2at2 ) = 0
9 x -axis at A , 0 and y axis at B(0, 3) 2 1 9 \ Area = 4 ⋅ ⋅ 3 = 27 \ K = 9. 2 2 38. (8) : xy + 2x + 4 y + C = 0 represents pair of lines ⇒ C = 8
(x − at32 )(x − at 42) + ( y − 2at3)( y − 2at4 ) = 0 If we solve these two circles, then the equation of their radical axis is of the form y = mx . So it passes through the origin.
\
34. (1) : Equation of the normal is y = mx – 2m –
m3
If it pass through (21, 30) we have 30 = 21m – 2m – m3 ⇒ m3 – 19m + 30 = 0 Ten m = –5, 2, 3 But if m = 2 or 3 then the point where the normal 72
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FEBRUARY‘17
39. (5) : Equation can be rewritten as
(x − 2)2 + ( y + 3)2 25e 13
=
13 12x − 5 y + 1 13 So, e = . 5 13 5
=5 b = π a 3
40. (2) : 2 tan −1
1 b = , e2 = 1 + 1 = 4 a 3 3 3 Let eccentricity of conjugate hyperbola be e′. 1 1 1 3 1 1 1⇒ 1⇒ \ + = + = = ⇒ e′ = 2 e′2 e2 e ′2 4 e′2 4
8
Class XI
T
his specially designed column enables students to self analyse their extent of understanding of specified chapters. Give yourself four marks for correct answer and deduct one mark for wrong answer. Self check table given at the end will help you to check your readiness.
STATISTICS AND PROBABILITY Total Marks : 80
Time Taken : 60 Min.
Only One Option Correct Type 1.
A five digit number be selected at random. Te probability that the digits in the odd places are odd and the even places are even (repetition is not allowed) 5 5 P2 × 5 P 3 P2 × 5 P 3 (a) (b) 104 × 9 105 5
5
6.
C2 × 5C 3
× 2 (d) 9 × 104 104 × 9 Te mean of the data is 30 and data are given as
(c) 2.
C2 × 5C 3
7.
Te probabilities that a student pass in Mathematics, Physics and Chemistry are a, b and g respectively. Of these subjects, student has a 75% chance of passing in atleast one, a 50% chance of passing in atleast two and a 40% chance of passing in exactly two subjects. Which of the following relations are true? (a) a + b + g = 19/20 (b) a + b + g = 27/20 (c) abg = 1/10 (d) abg = 1/4
8.
Te variable x takes two values x 1 and x 2 with frequencies f 1 and f 2, respectively. If s denotes the standard deviation of x , then
C.I. 0-10 10-20 20-30 30-40 40-50 f 1 f 2 Frequency 8 10 15 If total of frequencies is 70, then the missing numbers are (a) 14, 23 (b) 25, 21 (c) 24, 20 (d) none of these 3.
4 , B speaks 5 truth is 3 . Te probability they contradict each Te probability that A speaks truth is 4
other is
4 3 (b) 1 (c) (d) 5 20 20 5 In a series there are 2 n observations half of them equal to ‘a’ and half equal to – a. If the standard deviation of the observations is 2, then | a| equals
(a) 4.
5.
7
2 1 (a) 2 (b) 2 (c) (d) n n If ten rupee coins which are 10 in number and five rupee coins which are 5 in number are to be placed in a line, then the probability that the extreme coins are five rupee coins is
1 1 (b) 10! 15! 5!10 ! (c) (d) none of these 15! In an experiment with 15 observations on x , the following results were available : Σx 2 = 2830 and Σx = 170. One observation i.e., + 20 was found to be wrong and was replaced by the correct value 30, then the corrected variance is (a) 188.66 (b) 177.33 (c) 8.33 (d) 78.00 One or More Than One Option(s) Correct Type (a)
2 (a) s
f1x12 + f 2x 22 = f1 + f 2
(b) s 2 =
f x + f x − 1 1 2 2 f1 + f 2
2
f1 f 2 (x1 − x 2 )2 2 ( f1 + f 2 )
(x1 − x 2 )2 (c) s = (d) None of these f1 + f 2 If A and B are two events such that P ( A) = 1/2 and P (B) = 2/3, then (a) P ( A∪B) ≥ 2/3 (b) P ( A∩B′)≤1/3 (c) 1/6 ≤P ( A∩B) ≤ 1/2 (d) 1/6 ≤P ( A′∩B)≤1/2 2
9.
MATHEMATICS TODAY
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73
For two data sets, each of size 5, the variance are given to be 4 and 5 and the corresponding means are given to be 2 and 4, respectively. Te variance of the combined data set is 13 11 5 (a) (b) (c) 6 (d) 2 2 2 11. Te letters of the word PROBABILIY are written down at random in a row. Let E1 denotes the event that two I’s are together and E2 denotes the event that two B’s are together, then 2 2 (a) P (E1) = P (E2) = (b) P (E1∩E2) = 11 55 18 (c) P (E1∪E2) = (d) none of these 55 12. Tere are 60 students in a class. Te following is the frequency distribution of the marks obtained by the students in a test Marks 0 1 2 3 4 5 2 2 Frequency x –2 x x (x + 1) 2x x + 1 where, x is a positive integer. Ten, (a) Mean = 2.8 (b) Variance = 1.12 (c) Standard Deviation = 1.12 (d) None of these 13. For two events A and B, P ( A∩B) is (a) not less than P ( A) + P (B) – 1 (b) not greater than P ( A) + P (B) (c) equal to P ( A) + P (B) – P ( A∪B) (d) equal to P ( A) + P (B) + P ( A∪B) 10.
16.
Matrix Match Type Match the following: Column I
Column II
P. wo balls are drawn from an urn containing
8 2 white, 3 red and 4 black balls one by one 1. = 15 without replacement. If the probability that both the balls are of same colour is l and at least one ball is red is m, then Q. A bag contains 2 white and 4 black balls while 11 another bag contains 6 white and 4 black balls. 2. = 12 A bag is selected at random and a ball is drawn. If l be the probability that the ball drawn is of white colour and m be the probability that the ball drawn is black colour, then R. Bag A contains 4 red and 5 black balls and 5 bag B contains 3 red and 7 black balls. One 3. l = 18 ball is drawn from bag A and two from bag B. If l be the probability that out of 3 balls drawn two are black and one is red and m be the probability that out of three balls drawn two are red and one is black, then 4. m =
(a) (b) (c) (d)
Comprehension Type
P 2 3 3 1
Q 1 4 1 2
11 45
R 3 2 4 3
Integer Answer Type
A JEE aspirant estimates that she will be successful with an 80% chance if she studies 10 hours per day, with a 60% chance if she studies 7 hours per day and with a 40% chance if she studies 4 hours per day. She further believes that she will study 10 hours, 7 hours and 4 hours per day with probabilities 0.1, 0.2 and 0.7, respectively: 14. Given that she is successful, the probability that she studied for 4 hours, is 6 7 8 9 (a) (b) (c) (d) 12 12 12 12 15. Given that she does not achieve success, the probability she studied for 4 hours, is 18 21 19 20 (a) (b) (c) (d) 26 26 26 26
17.
18.
If four squares are chosen at random on a chess board. If the probability that they lie on a diagonal xyz line is 64 , then the value of x + y – z must be C 4 If the mean deviation about the median of the numbers a, 2a, ......., 50a is 50, then |a| equals
19.
wo integers x and y are chosen (without random) at random from the set {x : 0 ≤ x ≤ 10,} x is an integer}. If the probability for |x – y | ≤ 5 is p, then the value of 121 p – 91 must be
20.
If the variance of first 50 even natural numbers is abc. Ten find the value of a + b – c. Keys are published in this issue. Search now !
J
Check your score! If your score is No. of questions attempted
……
No. of questions correct …… Marks scored in percentage …… 74
MATHEMATICS TODAY
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> 90%
EXCELLENT WORK !
You are well prepared to take the challenge of final exam.
90-75%
GOOD WORK !
You can score good in the final exam.
74-60%
SATISFACTORY !
You need to score more next time.
NOT SATISFACTORY!
Revise thoroughly and strengthen your concepts.
< 60%
FEBRUARY ‘17
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75
Exam on th
20 March 2017
PRACTICE PAPER 2017 ime Allowed : 3 hours
Maximum Marks : 100 GENERAL INSTRUCTIONS
(i) (ii) (iii) (iv) (v) (vi)
All questions are compulsory. Tis question paper contains 29 questions. Questions 1-4 in Section-A are very short-answer type questions carrying 1 mark each. Questions 5-12 in Section-B are short-answer type questions carrying 2 marks each. Questions 13-23 in Section-C are long-answer-I type questions carrying 4 marks each. Questions 24-29 in Section-D are long-answer-II type questions carrying 6 marks each. SECTION-A
1.
=
x 3 − x 2 + x − 1 dx x − 1
∫
Evaluate :
3.
If the binary operation * on the set of integers Z is defined by a * b =a + 3b2, then find the value of 2 * 4. x
If − sin q
− x
1
1
x
= 8 , then find the value of x .
SECTION-B
f (x ) =
1 + −1 1 + −1 1 + −1 1 = p tan tan tan . 3 5 7 8 4
2 0 If A = 2 1 1 −1
1 3 , find A2 – 5 A + 4I. 0
MATHEMATICS TODAY
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FEBRUARY ‘17
k(x 2 + 2), 3x + 1,
if x ≤ 0
if x > 0 is continuous at x = 0 ? 9.
Form the differential equation of the family of curves y = a cos (x + b), where a and b are arbitrary constants.
10.
If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, then find the approximate error in calculating its surface area.
11.
Let * be a binary operation on the set of rational numbers given as a * b = (2a – b)2, a, b ∈ Q. Find 3 * 5 and 5 * 3. Is 3 * 5 = 5 * 3 ?
Prove that : tan −1
76
For what value of k, the function defined by
sin q cos q
cos q
6.
8.
y − 1 z + 1 . = 2 −2
2.
5.
Show that f : N → N , defined by n + 1 , if n is odd f (n) = 2 n , if n is even 2 is a many-one onto function.
Write the vector equation of a line passing through the point (1, –1, 2) and parallel to the line x −3 1
4.
7.
12.
If xy + y 2 = tan x + y , find
dy . dx
SECTION-C 13.
Using integration, find the area o the region bounded by the curves y = x 2 and y = x .
14.
Solve the differential equation
15.
dy cosec x log y + x 2 y 2 = 0. dx In a parliament election, a political party hired a public relations firm to promote its candidates in three ways-telephone, house calls and letters. Te cost per contact (in paise) is given in matrix A as
140 elephone A = 200 House call 150 Letters
Find the equation o the plane passing through the intersection o the planes x + 3 y + 6 = 0 and 3x – y – 4z = 0 and whose perpendicular distance rom the origin is unity.
20.
Evaluate :
OR
Using properties o definite integrals prove that p 2 x tan x
∫ sec x cosec x dx = 21.
p
4
500
5000 City X
1000
10000 City Y
22.
23.
Find the intervals in which the ollowing unction is (a) increasing (b) decreasing : f (x ) = x 4 – 8x 3 + 22x 2 – 24x + 21 Using properties o determinants, show that x+y
x
5x + 4 y
4x
2 x = x 3 .
10 x + 8 y
8x
3x
x
SECTION-D 24.
OR
Prove that : 9 p 9 −1 1 9 −1 2 2 − sin = sin . 3 4 3 8 4
Find the total amount spent by the party in the two cities. What should one consider beore casting his/ her vote-party’s promotional activity or their social activities?
1 I A = 2 2
cos x
∫ (1 − sin x)(2 − sin x ) dx
0
Te number o contacts o each type made in two cities X and Y is given in matrix B as elephone House call Letters
1000 B = 3000
19.
2 2
1 2 , then veriy that A2 – 4 A – 5I = O.
Find the equation o the plane through the line o intersection o the planes 2x + y – z = 3 and 5x – 3 y + 4z + 9 = 0 x − 1 y − 3 z − 5 . = = 2 4 5 Also, find the distance o the plane rom origin.
2 1
and parallel to the line
Hence find A–1.
16.
I a ,b and c are three vectors such that each one is perpendicular to the vector obtained by sum o
OR
Find the equation o the plane passing through the point A(1, 2, 1) and perpendicular to the line joining the points P (1, 4, 2) and Q(2, 3, 5). Also, find the distance o this plane rom the line
the other two and a = 3 b = 4 and c = 5 then prove that a + b + c = 5 2
,
,
.
17.
18.
An experiment succeeds twice as ofen as it ails. Find the probability that in the next six trials, there will be at least 4 successes. I y = 3cos(logx ) + 4sin(logx ), then show that x 2
d 2 y
dy + y = 0 dx
+ x 2
dx
i x ≤ 0 i x > 0
25.
=
y−5 −1
=
z − 7 . −1
Find the particular solution o the differential dy equation x + y – x + xy cot x = 0; x ≠ 0, dx given that when x =
OR
For what value o l, the unction defined by
l(x 2 + 2), f ( x ) = 4 x + 6,
x +3 2
is
continuous
at
x = 0? Hence check the differentiability o f (x ) at x = 0.
1 −1 3 26. I A = 2 0 1
p 2
, y = 0.
2 2 −4 4 and B = −4 2 −4 are two 2 −1 5 2 0
square matrices, find AB and hence solve the system o equations x – y = 3 , 2x + 3 y + 4z = 17 and y + 2z = 7. MATHEMATICS TODAY
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FEBRUARY ‘17
77
OR
27.
wo schools P and Q want to award their selected students on the values of discipline, politeness and punctuality. Te school P wants to award ` x each, ` y each and ` z each for the three respective values to its 3, 2 and 1 students with a total award money of ` 1,000. School Q wants to spend ` 1,500 to award its 4, 1 and 3 students on the respective values (by giving the same award money for the three values as before). If the total amount of awards for one prize on each value is ` 600, using matrices, find the award money for each value. A dealer in rural area wishes to purchase a number of sewing machines. He has only ` 5760 to invest and has space for at most 20 items. An electronic sewing machine costs him ` 360 and a manually operated sewing machine costs ` 240. He can sell an electronic sewing machine at a profit of ` 22 and a manually operated sewing machine at a profit of ` 18. Assuming that, he can sell all the items that he can buy, how should he invest his money in order to maximize his profit. Make it as a linear programming problem and solve it graphically. Keeping the rural background in mind justify the ‘values’ to be promoted for the selection of manually operated machine.
28.
If the length of three sides of a trapezium other than base is 10 cm each, then find the area of the trapezium when it is maximum.
29.
From a well shuffled pack of 52 cards, 3 cards are drawn one-by-one with replacement. Find the probability distribution of number of queens.
2.
x 3 − x 2 + x − 1 dx x − 1
∫
Let I =
= ∫
x 2 (x − 1) + 1(x − 1) (x 2 + 1)(x − 1) dx = x − 1 x − 1 1 (x 2 + 1)dx = x 3 + x + C 3
∫
= ∫ 3.
Here, a * b = a + 3b2 \ 2 * 4 = 2 + 3(4)2 = 2 + 3 × 16 = 50 x
4.
Given, − sin q cos q
5.
x − 3 y − 1 z + 1 is = = 1 2 −2 r = (i − j + 2k ) + l (i + 2 j − 2k )
78
MATHEMATICS TODAY
|
FEBRUARY ‘17
1
x
=8
1 1 + tan −1 + tan −1 7 8 1 + 1 1 + 1 = tan−1 3 5 + tan −1 7 8 1 1 1 1 1 − × 1 − × 3 5 7 8 8 15 = tan−1 15 + tan−1 56 14 55
1
1
−1 + tan −1 L.H.S. = tan 3 5
4+ 3 65 = tan−1 7 11 = tan−1 77 65 1 − 4 × 3 77 7 11 p = tan −1 1 = = R.H.S. 4
SOLUTIONS
and parallel to the line
1
15 56 = tan −1 4 + tan−1 3 11 7
Suppose the probability for A to win a game against B is 0.4. If A has an option of playing a ‘best of 3 games’ or a ‘best of 5 games’ match againstB, which option should A choose so that the probability of his winning the match a higher? (No game ends in draw).
Vector equation of the line passing through (1, –1, 2)
− x
⇒ x (–x 2 – 1) –sin q(–x sin q – cos q) + cos q (–sin q + x cos q) = 8 3 2 ⇒ –x – x + x sin q + sin q cos q – sin q cos q + x cos2q = 8 ⇒ –x 3 – x + x (sin2q + cos2q) = 8 ⇒ –x 3 – x + x = 8 ⇒ x 3 + 8 = 0 ⇒ (x + 2) (x 2 – 2x + 4) = 0 ⇒ x + 2 = 0 [Q x 2 – 2x + 4 > 0 "x ] ⇒ x = –2
OR
1.
sin q cos q
6.
2
We have, A – 5 A + 4I
2 0 1 2 0 1 2 0 1 1 0 0 = 2 1 3 2 1 3 −5 2 1 3 + 4 0 1 0 1 −1 0 1 −1 0 1 −1 0 0 0 1
5 = 9 0 −1 = −1 −5 7.
−1 2 −2 5 − −1 −2 −1 −3 −3 −10 4 2
10 0 5 4 0 0 10 5 15 + 0 4 0 5 −5 0 0 0 4
dS = 8 pr dr dS \ DS = Dr = 8pr Dr dr
⇒
= 8p × 9 × 0.03 = 2.16 p cm2 Tis is the approximate error in calculating surface area. 11.
(1 + 1) 2 2 = = 1 and f (2) = = 1 2 2 2 Tus, f (1) = f (2) while 1 ≠ 2 We have, f (1) =
\ f is many-one. In order to show that f is onto, consider an arbitrary element n ∈N . If n is odd, then (2n – 1) is odd (2n − 1 + 1) 2n \ f (2n − 1) = = =n 2 2 If n is even, then 2n is even 2n \ f (2n) = = n 2 Tus, for each n ∈ N (whether even or odd) there exists its pre-image in N . \ f is onto. Hence, f is many-one onto function. 8.
12.
dy dy dy x + y + 2 y = sec2 x + dx dx dx 2 dy dy sec x − y ⇒ (x + 2 y − 1) = sec2 x − y ⇒ = dx dx x + 2 y − 1 13.
x →0−
x →0
h →0
+
Te point of intersection of (i) and (ii) are A (1, 1) and O(0, 0). \ Required area = Area of shaded region
h→0
As f (x ) is continuous at x = 0 \
x → 0−
3 2 1 1 1 = ∫ (x − x )dx = x − x = − = sq.unit. 3 0 2 3 6 2 0
x → 0+
1 2
14.
Here, y = a cos (x + b) Differentiating (i) w.r.t. x , we get
...(i)
dy = − a sin ( x + b ) dx
Again differentiating w.r.t. x , we get d 2 y 2
dx
⇒
d 2 y dx 2
= − a cos ( x + b ) = − y ⇒
d 2 y dx 2
1
1
lim f (x ) = lim f (x ) = f (0 )
⇒ 2k = 1 ⇒ k =
10.
Te given curves are y = x 2 ... (i) and y = x ...(ii)
We have, lim f (x ) = lim k(h2 + 2) = 2k lim f (x ) = lim (3h + 1) = 1 and f (0) = 2k
9.
We have, a * b = (2a – b)2 \ 3 * 5 = (2 × 3 – 5) 2 = (6 – 5) 2 = 1 5 * 3 = (2 × 5 – 3) 2 = (10 – 3) 2 = 49 Tus, 3 * 5 ≠ 5 * 3 Given, xy + y 2 = tan x + y Differentiating w.r.t. x , we get
+ y = 0.
Let r be the radius of sphere and Dr be the error in measuring the radius. Ten, r = 9 cm, Dr = 0.03 cm Now, surface area (S) of the sphere is given by 4pr 2
2
dy 2 2 + x y = 0 dx
We have, cosec x log y
x 2 dy + dx = 0 cosec x y 2
log y
Integrating both sides, we get
∫
log y dy + ∫ x 2 sin x dx = 0 2 y
Put log y
1 = t ⇒ dy = dt and y = et y
Solution Sender of Maths Musing SET-169
1. N. Jayanthi
Hyderabad
2. Ravinder Gajula
Karimnagar
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FEBRUARY ‘17
79
⇒
∫
⇒ t .
Now, A2 – 4 A – 5 I = O Pre-multiplying by A–1 both sides, we get ( A–1 A) A – 4 ( A–1 A) – 5( A–1 I ) = O ⇒ IA – 4I = 5 A–1
t . e −t dt + x2 sin x dx = 0
e −t
−1
∫
− ∫ 1.
e −t dt + x 2 (− cos x ) − 2x ( − cos x ) dx = C −1
∫
⇒ –t e–t –e–t –x 2 cos x + 2x sin x −2 ∫ 1 ⋅ sin x dx = C ⇒ −
1 + log y 2 − x cos x + 2 x sin x + 2 cos x = C y
−3 1 1 A−1 = (A − 4 I ) = 2 5 2 2
⇒
is the required solution. 15.
Te total amount spent by the party in two cities X and Y is represented in the matrix equation by matrix C as, C = BA
⇒ ⇒
140 200 1000 10000 150 X 1000 ×140 + 500 × 200 + 5000 ×150 Y = 3000 ×140 + 1000 × 200 +10000 ×1150 990000 = 2120000 X 1000 Y = 3000
500
\ ⇒
80
9 8 0 0 0
+ c ) =
0 b ⋅ (c
,
+ a ) =
a ⋅b
0 c ⋅ ( a + b ) = 0
,
=
a
2
+
b
2
+
c
2
+ 2 ( a ⋅ b ) + 2 ( b ⋅ c ) + 2 ( c ⋅ a )
= 32 + 42+ 52 + 0 = 50
\ 17.
a +b
+c =5
[Using (i) and (ii)]
2.
2 2 = Here, p = P (success of the experiment) = 2 +1 3 2 1 \ q = 1− p =1− = 3 3 Also n = 6 Let X denote the number of successes. \ Required probability P ( X ≥ 4) = P ( X = 4) + P ( X = 5) + P ( X = 6) 2
4
1
5
0
1 2 1 2 1 2 = C4 + 6C5 + 6C 6 3 3 3 3 3 3 6
1 2 2 1
= 15 ×
2 2 1 2 2
1 2 2 1 2 2 1 2 2 1
MATHEMATICS TODAY
,
+ a ⋅c + b⋅ c + b⋅ a + c ⋅ a+ c ⋅ b =0 ...(ii) 2(a ⋅ b + b ⋅ c + c ⋅ a ) = 0 2 Now, a + b + c
2 2
4 8 − 8 9 8 0 0 = O 0
...(i)
,
OR
8 8
and a ⋅ (b
Tus, amount spent by the party in city X and Y is ` 990000 and ` 2120000 respectively. One should consider about the social activity before casting his/her vote.
9 = 8 8 0 = 0 0
Given, a = 3 b = 4 c = 5
5000
\ A2 – 4 A – 5I 9 8 8 1 = 8 9 8 − 4 2 2 8 8 9
2
16.
\ X = ` 990000 and Y = ` 2120000
1 Given, A = 2 2 1 Now, A2 = 2 2
−3 2 2 −3 2
9 = 8 8
2 2
1 1 2 − 5 0 0 2 1 8 8 5 0 4 8 − 0 5 8 4 0 0
|
FEBRUARY ‘17
8 8
18.
9 8 8 9 0 0
1 0 0 1
+6× 6
3
32
64
3
6
+ 6
3
=
240 + 192 + 64 6
3
=
496 729
We have, y = 3 cos (log x ) + 4 sin (log x ) Differentiating w.r.t. x , we get
⇒
dy 1 1 = −3 sin(log x ) × + 4 cos(log x ) × x x dx dy x = −3 sin(log x ) + 4 cos(log x ) dx
Again differentiating w.r.t. x , we get
0
0 5
16
6
d 2 y dy 1 1 x 3 cos(log x ) × − 4 sin(log x ) × + = − x x dx 2 dx 2
d 2 y
dy = −[3 cos(log x ) + 4 sin(log x )] dx
⇒
x
⇒
dy x x =− y + dx dx 2 2
2
dx
d 2 y
+ x
2
⇒ x
d 2 y 2
dx
dy dx
+ x + y = 0
OR
l(x 2 + 2), if x ≤ 0 Here, f (x ) = if x > 0 4 x + 6, At x = 0, f (0) = l(02 + 2) = 2l L.H. L. = lim f (x ) = lim l[(−h)2 + 2] = 2l − R.H.L. = lim f (x ) x →0 +
=
I =
= log p
∫ sec x cosec x dx
lim[4(h) + 6] = 6 h→0
Using
Hence, the function becomes if x ≤ 0
p 0
19.
3(h
Q
\
2
21.
=1
22.
cos x dx (1 − sin x )(2 − sin x )
∫
Let I =
Put sinx = t ⇒ cosxdx = dt
\
⇒
Distance of plane from origin is unity. 6
dt
∫ (1 − t )(2 − t )
I =
We write,
1 A B = + (1 − t )(2 − t ) 1 − t 2 − t
⇒ 1 = A (2 – t ) + B(1 – t ) Putting t = 1 in (1), we get 1 = A(2 – 1) ⇒ A = 1 Putting t = 2 in (1), we get 1 = B(1 – 2) ⇒ B = –1
...(1)
0
I =
...(2)
p tan x dx = p sin2 x dx sec x cos ec x
∫
0
p sin 2 x p 1 − cos 2 x dx = x − 2 2 2 0
= p ∫ 0 2
=
p2 2
p
4 9 p 9 −1 1 − sin 3 8 4
L.H.S. =
=
+ (3 − l) + 16 l ⇒ 26 = 26l2 ⇒ l2 = 1 ⇒ l = ±1 \ Equation of the required plane is r ⋅ (4i + 2 j − 4 k ) + 6 = 0 or r ⋅ (−2i + 4 j + 4 k ) + 6 = 0 20.
∫
p
(1 + 3l)
p (p − x )tan(p − x ) (p − x )tan x dx = dx sec(p − x )cosec (p − x ) sec x cos ec x
0
2
0
2 I = p
r ⋅[(1 + 3 l) i + (3 − l )j − 4 l k ] + 6 = 0
2
0
∫ f (x)dx = ∫ f (a − x)dx , we get
∫
2
Equation of plane passing through intersection of given planes is x + 3 y + 6 + l(3x – y – 4z ) = 0
⇒
a
Adding (1) and (2), we get p
+ 2) − 6 = lim (−3h) = 0 −h h→0 f (0 + h) − f (0) 4h + 6 − 6 and f ′(0+ ) = lim = lim =4 0+h h h →0 h→0 ⇒ f ′(0–) ≠ f ′(0+) \ f is not differentiable at x = 0. lim h→0
0 a
...(1)
∫
I =
if x > 0
f (0 − h) − f (0) f ′ (0 − ) = lim 0−h h→0
=
2 − t 2 − sin x + C = log + C 1 − t 1 − sin x
x tan x
Let I =
\ For f to be continuous at x = 0, 2l = 6 ⇒ l = 3. 3(x 2 + 2), f ( x ) = 4 x + 6,
−dt
OR
h→0
x →0
dt
∫ 1 − t + ∫ 2 − t = − log 1 − t + log 2 − t + C
\
9 1 cos −1 3 4
9 p = − sin−1 1 3 4 2 Q sin−1 x + cos −1 x = p 2
Q cos −1 x = sin −1 1 − x 2 for 0 ≤ x ≤ 1 9 = sin −1 2 2 = R.H.S. 3 4
=
9 −1 1 sin 1− 4 9
=
9 −1 8 sin 4 9
Te given function is f (x ) = x 4 – 8x 3 + 22x 2 – 24x + 21 ⇒ f ′(x ) = 4x 3 – 24x 2 + 44x – 24 = 4(x 3 – 6x 2 + 11x – 6) = 4(x – 1)(x 2 – 5x + 6) = 4(x – 1)(x – 2)(x – 3) Tus f ′(x ) = 0 ⇒ x = 1, 2, 3. Hence, possible disjoint intervals are (–∞, 1), (1, 2), (2, 3) and (3, ∞). In the interval (–∞, 1), f ′ (x ) < 0 In the interval (1, 2), f ′(x ) > 0 In the interval (2, 3), f ′(x ) < 0 In the interval (3, ∞), f ′(x ) > 0 \ f is increasing in (1, 2) ∪ (3, ∞) and f is decreasing in (–∞, 1) ∪ (2, 3). MATHEMATICS TODAY
|
FEBRUARY ‘17
81
x+y 23.
x
⇒ x – y + 3z – 2 = 0
x
x +3
L.H.S. = 5x + 4 y 4 x 2 x . 10 x + 8 y
8x
3x
Since each element in the first column of determinant is the sum of two elements, therefore, determinant can be expressed as the sum of two determinants given by x
x
x
5x
4 x 2 x
10 x
8 x 3 x
y
+
x
x
d =
4 y 4 x 2 x 8 y
8 x 3 x
aking xcommon from R1,R2,R3 in first determinant and x common from C 2, C 3, y common from C 1 in second determinant, we get 1 1 1 x
3
5 4 2
1 1 1 2
+ yx
10 8 3
4 4 2
1 1 1
=x
8 8 3
3
5 4 2
+ yx 2 ⋅ 0
10 8 3
(Q C 1 and C 2 are identical in the second determinant) Applying C 1 → C 1 – C 3 and C 2 → C 2 – C 3, we get 0 0 1 3 3 x 3 3 2 2 = x ⋅1⋅(15 – 14) = x = R.H.S.
24.
Te equation of the plane passing through the line of intersection of the planes 2x + y – z = 3 and 5x – 3 y + 4z + 9 = 0 is (2x + y – z – 3) + l(5x – 3 y + 4z + 9) = 0 ⇒ x (2 + 5l) + y (1 – 3l) + z (4l – 1) + 9l – 3 = 0 ...(i) Since, plane (i) is parallel to the line x −1 y − 3 = 2 4
=
z − 5 5
OR
Te line joining the given points P (1, 4, 2) and Q(2, 3, 5) has direction ratios <1 –2, 4 –3, 2 – 5> i.e., <– 1, 1, –3> Te plane through (1, 2, 1) and perpendicular to the line PQ is –1(x – 1) + 1( y – 2) – 3(z – 1) = 0 |
−3 − 5 + 3(7 ) − 2 ⇒ 1+1+ 9
d =
11 11
=
11 units.
dy + y − x + xy cot x = 0,(x ≠ 0) dx dy ⇒ x + (1 + x cot x) ⋅ y = x dx dy 1 + x cot x . y = 1 ⇒ + dx x dy Tis is linear D.E. of the form + Py = Q dx 1 + x cot x 1 where P = = + cot x , Q = 1 x x
We have, x
Now, I.F. = e ∫
P dx
...(i)
= elog (x sin x ) = x sin x
Te solution of (i) is
= ∫ 1⋅ x sin x dx + C = x(− cos x ) + ∫ 1 ⋅ cos x dx + C ⇒ x y sin x = − x cos x + sin x + C p Putting x = , y = 0 in (ii), we get 2 p p p 0 = − cos + sin + C ⇒ C = −1 y ⋅ x sin x
2
7x + 9 y – 10z – 27 = 0 Tis is the equation of the required plane. Now, distance of the plane 7x + 9 y – 10z – 27 = 0 from the origin is, 27 −27 unit = 72 + 92 + 102 230
MATHEMATICS TODAY
z − 7
2
...(ii)
2
\ xy sinx = sinx – x cos x – 1 is the required
\ 2(2 + 5l) + 4(1 – 3 l) + 5(4l – 1) = 0 1 ⇒ 18l + 3 = 0 ⇒ l = − 6 Putting the value of l in (i), we obtain
82
25.
\
7 5 3
y−5
Now, direction ratios of line = = 2 1 − −1 are 2, –1, –1. Now, 2(1) + (–1) (–1) + (3) (–1) = 2 + 1 – 3 = 0 \ Line is parallel to the plane. Since, (–3, 5, 7) lies on the given line. \ Distance of the point (–3, 5, 7) from plane is
FEBRUARY ‘17
particular solution.
1 −1 0 2 2 −4 26. Here, A = 2 3 4 ; B = −4 2 −4 0 1 2 2 −1 5 1 −1 0 2 2 −4 6 0 0 \ AB = 2 3 4 ⋅ −4 2 −4 = 0 6 0 = 6I 0 1 2 2 −1 5 0 0 6 1 1 ⇒ A ⋅ B = I ⇒ A is invertible and A−1 = B 6 6 Now, the given system of equations is x – y = 3 2x + 3 y + 4z = 17 y + 2z = 7 Te system of equations can be written as AX = P
MATHEMATICS TODAY
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FEBRUARY ‘17
83
1 −1 0 x 3 where, A = 2 3 4 ; X = y ; P = 17 0 1 2 z 7 Since A–1 exists, so system of equations has a unique solution given by X = A–1 P x 2 2 −4 3 1 1 ⇒ y = BP = −4 2 −4 17 6 6 z 2 −1 5 7
1 3 = −5 ≠ 0
1 1 1
OR
According to question, we have, 3x + 2 y + z = 1000
MATHEMATICS TODAY
3 2 1 A = 4
12 2 1 = −6 = −1 6 24 4 ⇒ x = 2, y = – 1, z = 4.
84
4x + y + 3z = 1500 ...(ii) x + y + z = 600 ...(iii) Te given system of equations can be written as AX = B 3 2 1 x 1000 where, A = 4 1 3 , X = y and B = 1500 1 1 1 z 6000
|
FEBRUARY ‘17
...(i)
\ A is invertible and system of equations has a unique solution given by X = A–1 B Now, A11 = –2, A12 = –1, A13 = 3, A21 = –1, A22 = 2, A23 = –1, A31 = 5, A32 = –5, A33 = –5
\
Te corner points of feasible region are : O(0, 0), C (16, 0), P (8, 12) and B(0, 20). Now, we calculate Z at each corner point.
−2 −1 5 adj A = −1 2 −5 3 −1 −5
Corner points Value of Z = 22x + 18 y 22(0) + 18(0) = 0 O(0, 0) C (16, 0) 22(16) + 18(0) = 352 22(8) + 18(12) = 392 (Maximum) P (8, 12) B(0, 20) 22(0) + 18(20) = 360 \ Maximum value of Z is 392 which occurs at the point P (8, 12). Hence, the maximum profit is ` 392 when 8 electronic machines and 12 manually operated sewing machines are purchased. Manually operated machine should be promoted, to save energy and increase employment for rural people.
− 2 −1 5 A adj( ) − 1 \ A−1 = = −1 2 −5 | A | 5 3 −1 −5 Now, X = A–1B x −2 −1 5 1000 −500 − 1 − 1 ⇒ y = −1 2 −5 1500 = −1000 5 5 z 3 −1 −5 600 −1500 x 100 ⇒ y = 200 z 300 ⇒ x = 100, y = 200, z = 300
Hence, the money awarded for discipline, politeness and punctuality are ` 100, ` 200 and ` 300 respectively. 27.
Suppose, number of electronic sewing machines purchased = x and number of manually operated sewing machines purchased = y Mathematical formulation of given problem is : Maximize Z = 22x + 18 y Subject to constraints : x + y ≤ 20 ...(i) 360x + 240 y ≤ 5760 or 3x + 2 y ≤ 48 ... (ii) x ≥ 0, y ≥ 0 Solving equations x + y = 20 and 3x + 2 y = 48, we get x = 8, y = 12 Let P ≡ (8, 12) Now, we plot the graph of system of inequalities.
28.
Let ABCD be the given trapezium. Ten AD = DC = CB = 10 cm In D APD and DBQC, DP = CQ = h AD = BC = 10 cm ∠DPA = ∠CQB = 90° \ D APD @ DBQC (by R.H.S. congruency) ⇒ AP = QB = x cm (say) \ AB = AP + PQ + QB = x + 10 + x = (2x + 10) cm Also from D APD, AP 2 + PD2 = AD2 ⇒ x 2 + h2 = 102 ⇒ h = 100 − x 2 Now, area S of this trapezium is given by 1 1 S = ( AB + DC ) ⋅ h = (2 x + 10 + 10)h 2 2 = (x + 10) ⋅ 100 − x 2
...(ii)
...(i)
[Using (i)]
Differentiating (i) w.r.t. x, we get dS 1 = 1 ⋅ 100 − x 2 + (x + 10)⋅ ⋅ (−2x ) 2 dx 2 100 − x
Te shaded portion OCPB represents the feasible region which is bounded.
100 − x 2 − x 2 − 10 x −2( x 2 + 5x − 50) = = 100 − x 2 100 − x 2 −2(x + 10)(x − 5) = 100 − x 2 dS For max. or min. value of S, =0 dx ⇒ (x + 10)(x – 5) = 0 ⇒ x = 5 (Reject x = – 10 as x
|
FEBRUARY ‘17
85
For this value of x ,
A will win the best of 3 games match if he wins in 2 or 3 games E2 = the event that A wins ‘a best of 5 games’ match. A will win a best of 5 games match if he wins in 3 or 4 or 5 games. Now P (E1) = P ( X = 2 or X = 3) = P ( X = 2) + P ( X = 3) = 3C 2 p2q + 3C 3 p3= 3(0.4)2(0.6) + (0.4)3 = (0.4)2[1.8 + 0.4] = (0.4)2(2.2) = 0.352 P (E2) = P ( X = 3 or X = 4 or X = 5) = P ( X = 3) + P ( X = 4) + P ( X = 5) =5C 3 p3q2 + 5C 4 p4q + 5C 5 p5 = (10 p3q2 + 5 p4q + p5) = p3(10q2 + 5 pq + p2) = (0.4)3 × [10 × (0.6)2 + 5 × (0.4) × (0.6) + (0.4)2] = (0.064) × (3.6 + 1.2 + 0.16) = 0.064 × 4.96 = 0.317 Since P (E1) > P (E2), hence A should choose the first option.
dS <0 dx
\ S is maximum at x = 5. From (ii), the max. value of S = (5 + 10) ⋅ 100 − 52
= 15 29.
75 = 75 3 sq.cm.
Let Ei(i = 1, 2, 3) be the event of drawing a queen in the ith draw. Let X denote the discrete random variable “Number of Queens” in 3 draws one by one with replacement. Here X = 0, 1, 2, 3 4 Now P(Ei ) = ; P(Ei ) = 48 (i = 1, 2, 3) 52 52 \ P ( X = 0) = P(E1E2 E3) = P(E1)P(E2)P( E3) 3
3
48 12 1728 = = = 52 13 2197 P ( X = 1) = P(E1E2 E3 or E1E2 E3 or E1E2 E3)
= P(E1E2 E3 ) + P(E1E2 E3) + P( E1E2 E3) = P(E1)P(E2)P(E3 ) + P(E1)P( E2)P( E3) + P(E1)P(E2 )P(E3) 2 2 4 48 1 12 432 = 3 × × = 3 × × = 52 52 13 13 2197 P ( X = 2) = P(E1E2 E3 or E1E2 E3 or E1E2 E3) = P(E1E2 E3) + P (E1E2 E3 ) + P ( E1E2 E3) = P(E1)P(E2 )P(E3) + P (E1)P( E2)P( E3 ) + P(E1)P(E2 )P(E3) 2
2
4 48 1 12 36 = 3 × × = 3 × × = 52 52 13 13 2197 P ( X = 3) =P (E1E2E3) = P (E1)P (E2)P (E3)
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3
4 1 = = 52 2197
•
\ Te required probability distribution of X is X P ( X )
0
1
2
3
1728 2197
432 2197
36 2197
1 2197
OR
Let E = the event that A wins a game against B. Let occurrence of the event E be called a success and X denote the number of successes. Let E1 = the event that A wins ‘a best of 3 games’ match. 86
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FEBRUARY ‘17
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8
Class XII
T
his specially designed column enables students to self analyse their extent of understanding of specified chapters. Give yourself four marks for correct answer and deduct one mark for wrong answer. Self check table given at the end will help you to check your readiness.
VECTORS & THREE DIMENSIONAL GEOMETRY Total Marks : 80
Time Taken : 60 Min.
Only One Option Correct Type 1.
2.
3.
4.
5.
If the volume of parallelopiped with a × b , b × c , c × a as co-terminous edges is 9 cu. units, t hen the volume of the parallelopiped with (a × b ) × (b × c ), (b × c ) × (c × a), (c × a ) × (a × b) as co-terminous edges is (a) 9 (b) 729 (c) 81 (d) 27 If the plane 2ax – 3ay + 4az + 6 = 0 passes through the midpoint of the line joining the centres of the spheres, x 2 + y 2 + z 2 + 6x – 8 y – 2z =13 and x 2 + y 2 + z 2 – 10x + 4 y – 2z = 8, then a equals (a) 1 (b) – 1 (c) 2 (d) – 2
6.
5 , the plane x + 2 y + 3z = 4 is cos −1 14 then equals (a) 2/5 (b) 5/3 (c) 2/3 (d) 3/2 One or More Than One Option(s) Correct Type 7.
If p × q = r and q × r = p, then (a) r = 1, p = q (b) p = 1, q = 1 (c) r = 2 p, q = 2 (d) q = 1, p = r
(a) i + 3j + k (b) 3i + 7 j + 3k (c) j + t(i + 2j + k), where t is any scalar
x − 2 y − 1 z + 2 Let the line lie in the plane = = 3 2 −5 x + 3 y – az + b = 0. Ten ( a, b) equals (a) (–6, 7) (b) (5, –15) (c) (–5, 5) (d) (6, –17)
(d) i + (t + 3)j + k, where t is any scalar
i
j
i
j
a = a1
a
a
a
2
1 2 3 p b is , then b1 b2 b3 is equal to
6
8.
If OABC is a tetrahedron such that OA2 + BC 2 = OB2 + CA2 = OC 2 + AB2 , then (a) OA ⊥ BC (b) OB ⊥ CA (c) OC ⊥ AB (d) AB ⊥ BC
If r is equally inclined to the co-ordinate axes and magnitude of r = 6 then r equals 2(i + j + k) (a) 3(k + i + j) (b) 3 6(i + j + k) (c) (d) − 2 3(i + j + k) 3 10. If the unit vectors a and b are inclined at an angle 2q and | a − b|< 1, then if 0 ≤ q ≤ p, q lies in the interval (a) [0, p/6) (b) (5p/6, p] (c) [p/6, p/2] (d) [p/2, 5p/6] 9.
that c is a unit vector perpendicular to both the vectors a and b and if the angles between a and
+ a2 + a3 k , b = b1 + b2 + b3 k and c = c1 i + c2 j + c3 k be three non-zero vectors such
Let
l
If a vector r satisfies the equation r × (i + 2j + k) = i − k, then r may be equal to
3 2 (a1 + a22 + a32 ) (b12 + b22 + b32 ) ( c12 + c22 + c32 ) 4 (d) none of these y − 1 z − 3 If the angle between the line x = and = 2 l (c)
c1 c2 c3 (a) 1 1 (b) (a12 + a22 + a32 ) (b12 + b22 + b32 ) 4
MATHEMATICS TODAY
| FEBRUARY‘17
87
A and B be perpendicular and Let the unit vectors the unitvector at an angle q to both C be inclined A and B . If C = a A + b B + g ( A × B ) then (a) a = b (b) g 2 = 1 – 2a2 1 + cos 2q (c) g 2 = – cos 2q (d) b2 = 2 12. Te equation of a sphere which passes through (1, 0, 0), (0, 1, 0) and (0, 0, 1) and whose centre lies on the curve 4xy = 1 is (a) x 2 + y 2 + z 2 – x – y – z = 0 (b) x 2 + y 2 + z 2 + x + y + z – 2 = 0 (c) x 2 + y 2 + z 2 + x + y + z = 0 (d) x 2 + y 2 + z 2 – x – y – z – 2 = 0 x − 1 y − 1 z + 3 = = 13. Te points in which the line 1 1 1 − cuts the surface x 2 + y 2 + z 2 – 20 = 0 are (a) (0, 2, 4) (b) (0, 2, – 4) (c) (4, 2, 0) (d) (0, – 2, – 4) Comprehension Type
r ⋅ a b⋅ a c⋅ a
∆1 = r ⋅ b r⋅c
b⋅ b
b⋅ c
∆2 =
c⋅ b ,
a⋅c
r⋅c
c⋅c
14.
r ⋅ b and ∆ = a ⋅ b
b⋅ c
r⋅c
b⋅ c
c⋅c
a⋅ c
(a) a =
a⋅ a
[a b c]
(b
× c) +
a⋅b
[a b c]
+
[ a b c ]
(a
15.
b
c
2 2
3.
2
× ( k × a ) = 0 , then g =
P 2 3 3 1
Q R 1 3 4 2 2 4 2 3 Integer Answer Type
Te line passes through the points (5, 1, a) and
If [a × b b × c c × a] = l[abc ]2 , then l is equal to
19.
Te plane x + 2 y – z = 4 cuts the sphere x 2 + y 2 + z 2 – x + z – 2 = 0 in a circle of radius
20.
b⋅ q
2.
18.
Te value for a ⋅ p b ⋅ p c ⋅ p , is a⋅ q
5 3
17 13 (3,b, 1) crosses the yz -plane at the point 0, , − . 2 2 then a – b =
× b)
(b) a = a ⋅ a(b ×c) + a ⋅ b(c × a) + c ⋅ a( a × b) (c) a = [ a b c ](b × c) + [ a b c ](c × a ) + [ a b c ](a × b) (d) None of these. a
17.
(c × a)
1.
(a) (b) (c) (d)
c⋅a
Column II
Point (a , b , g ) lies on the plane x + y + z = 2. Let a = ai + bj + g k,
R.
If vector r is expressible as, r = xa + yb + zc, then
and is perpendicular to two planes 2x – 2 y + z = 0 and x – y + 2z = 4. Te distance of the plane from the point (1, 2, 2) is
c ⋅ b ,then
Q. A plane passes through (1, –2, 1)
b⋅b
2z = 0 and passes through (0, 1, 0). Te perpendicular distance of this line from the origin is
a⋅ a b⋅ a c ⋅ a
b⋅ b
P. A line is perpendicular to x + 2 y +
a⋅ a b⋅ a r ⋅ a
∆3 = a ⋅ b a⋅ c
Column I
r⋅b
Match the following:
k
a⋅b
c⋅c
16.
a⋅ a r ⋅ a c ⋅ a
c⋅ b ,
(d) ( p × q) [ a × b b × c c × a]. Matrix Match Type
If a, b, c are three given non-coplanar vectors and r be any arbitrary vector in space, where
(a) ( p × q)[ a × b b × c c × a] (b) 2( p × q)[ a × b b× c c × a] (c) 4( p × q)[ a× b b × c c × a]
11.
c⋅ q
Let u = i + j, v = i − j , w = i + 2 j + 3k . If n is a unit vector such that u . n = 0 and v . n = 0 then w . n equals
Keys are published in this issue. Search now!
J
Check your score! If your score is > 90%
EXCELLENT WORK !
You are well prepared to take the challenge of final exam.
No. of questions attempted
……
90-75%
GOOD WORK !
You can score good in the final exam.
No. of questions correct
……
74-60%
SATISFACTORY !
You need to score more next time.
Marks scored in percentage ……
< 60%
NOT SATISFACTORY!
Revise thoroughly and strengthen your concepts.
88
MATHEMATICS TODAY
|
FEBRUARY‘17
E 4
60° F 60° SOLUTION SET-169
1. (c) : z 1, z 2, z 3 are roots
⇒ Sz 1 = – p, Sz 1z 2 = q 60° 60° A 1 B
2
\ p = q + 2q = 3q 1 1 1 + = a b 2013 (a – 2013)(b – 2013) = 2013 2
⇒
= 32 ⋅ 112 ⋅ 612 Te number of all pairs is the number of divisors of 32 ⋅ 112 ⋅ 612, which is 33 = 27. 27 − 1 = 13 Number of pairs (a, b), a < b is 2 3. (a) : x 2 – ix – 1 = 0. Solving, x = cis
π and x = cis 5π
6 6 π π x = cis (2013) and x 2013 = cis (10065) 6 6 2013 2013 –2013 ⇒ x = –i, x + x = –i + i = 0 2 100 100 9 (r +1) { x }dx = xdx − rdx 4. (c) : r = 0 r 2 0 0 2013
∫
=
∫
∑ ∫
2000 9 2000 2 − (2r 2 + r ) = − 615 = 51 3 3 3 r =1
∑
5. (b) : Te parabola is open upwards with vertex (2, – 1). Te focus is (2, –1 + 1) = (2, 0). Te reflected ray passes through the focus (2, 0). 6. (a, d) : Te equation represents a pair of planes. 3 1 a 2 2 3 ⇒ 1 1 = 0 ⇒ 4ab – 4a – 9b + 5 = 0 …(i) 2 1 1 b 2 For perpendicular planes a + 1 + b = 0 …(ii) Eliminating b from (i) and (ii), we get 7 4a2 – a – 14 = 0 ⇒ a = 2, − 4 7. (c) : ABEF and BCDE are trapeziums. BE = 2 + 2 · 4 cos60° = 6 AF = BE – 2cos60° = 6 – 1 = 5 Perimeter = 1 + 4 + 2 + 4 + 1 + 5 = 17
2
120° C
p2 = Sz 21 + 2q For equilateral triangle, Sz 21 = Sz 1z 2
2. (b) :
D
120°
4
8. (d) : Area = Area of BCDE + Area of ABEF 1 1 = (6 + 2)4 sin 60° + (6 + 5)sin 60° 2 2 11 4 3 =8 3+ 3= 3 4 4 9. (4) : N is the coefficient of x 30 in
(1 + x + x 2 + … + x 10)3 (1 + x + x 2 + … + x 20) (1 – x 11)3 (1 – x 21)(1 – x )–4 4 5 = (1 – 3x 11 + 3x 22 – x 21 …) 1 + x + x 2 1 2
+ ...
33 22 11 12 30 − 3 19 + 3 8 − 9 33 22 11 12 = − 3 + 3 − = 1111, 3 3 3 3
which is
with digit sum 4. 10. (a)-(s, t); (b)-(p), (c)-(r), (d)-(r)
(a) f ′(0) = lim x →0 if a > 1 x
f ( x ) − f (0) x
1 = lim x a −1 sin = 0 x
x → 0
1 1 ≠ 0 ⇒ f ′(x) = ax a − 1 sin − x a −2 cos
lim f ′(x ) = 0 if a
x →0
>2
x \ a = 3, 4.
x
(b) y (1) = 2 ⇒ 2 = a + b + c, y (0) = 0, y ′(0) = 1 ⇒ c = 0, b = 1, a = 1 \ y = x 2 + x , y (–1) = 0 π /2 cos tdt = 1 (c) f (x ) = sinx (1 + a), where a = 0 π /2 \ f (x ) = 2 sin x , f ( x )dx = 2.
∫
∫
2
0
x y 2 2 (d) Te hyperbolas − y = 1 and − x 2 = 1 have 2 2 common tangents with slopes ±1. Te four tangents are x + y ± 1 = 0, x – y ± 1 = 0. Tey form a square of area 2 × 2 = 2 MATHEMATICS TODAY
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FEBRUARY‘17
89
