34 10 48
Vol. XXXIII
No. 11
Novembe r 2015
Corporate Ofce: Plot 99, Sector 44 Institutional Area, Gurgaon -122 003 (HR), Tel : 0124-4951200 0124-4951200 e-mail :
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Regd. Ofce: 406, Taj Apartment, Near Safdarjung Hospital, Ring Road, New Delhi - 110029. Managing Editor : Mahabir Singh Editor : Anil Ahlawat
63
CONTENTS 82 19
8 Maths Musing Problem Set - 155 23
10 Jee Work Outs
16 72
16 You Ask, We Answer 18 Math Musing Solutions
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Your Way CBSE XI 23 Ace Your
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Physics For You You
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MATHEMATICS TODAY | NOVEMBER ‘15
7
34 10 48
Vol. XXXIII
No. 11
Novembe r 2015
Corporate Ofce: Plot 99, Sector 44 Institutional Area, Gurgaon -122 003 (HR), Tel : 0124-4951200 0124-4951200 e-mail :
[email protected] website : www.mtg.in www.mtg.in
8
Regd. Ofce: 406, Taj Apartment, Near Safdarjung Hospital, Ring Road, New Delhi - 110029. Managing Editor : Mahabir Singh Editor : Anil Ahlawat
63
CONTENTS 82 19
8 Maths Musing Problem Set - 155 23
10 Jee Work Outs
16 72
16 You Ask, We Answer 18 Math Musing Solutions
Subscribe online at
www.mtg.in
Individual Subscription Rates
19 Olympiad Corner
1 yr.
2 yrs.
3 yrs.
Your Way CBSE XI 23 Ace Your
Mathematics Today Today
330
600
775
Chemistry Today Today
330
600
775
(Series 6)
Physics For You You
330
600
775
Biology Today Today
330
600
775
34 Concept Boosters (XI) 48 Concept Boosters (XII) 63 Mock Test Paper WB JEE 2016
Your Way CBSE XII 72 Ace Your (Series 6)
82 Math Archives
Combined Subscription Rates 1 yr.
2 yrs.
3 yrs.
PCM
900
1500
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PCB
900
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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 | NOVEMBER ‘15
7
M
aths Musing was started in January 2003 issue of Mathematics Today with the suggestion of Shri Mahabir Singh. The aim of Maths Musing is to augment the chances of bright students seeking admission into IITs with additional study material. During the last 10 years there have been several changes in JEE pattern. To suit these changes Maths Musing also adopted the new pattern by changing the style of problems. Some of the Maths Musing problems have been adapted in JEE 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. Prof. Dr. Ramanaiah Gundala, Former Dean of Science and Humanities, Anna University, Chennai
Set 155 JEE MAIN 1.
2.
3.
4.
5.
8.
Let 0 < a < b < 125. e number of pairs of integers (a, b), such that the A.M. of a and b exceeds their G.M. by 2, is (a) 8 (b) 9 (c) 10 (d) 7 e area bounded by the curves y = = | 1 – x 2| and y = 5 – |x | is (a) 18 (b) 19 (c) 20 (d) 21 Let {a, b} be a subset of {1,2,3,…,100}. e probability of a2– b2 is divisible by 5 is 31 32 34 35 (a) (b) (c) (d) 99 99 99 99 In a triangle ABC , if A : B : C = = 1 : 2 : 4, then tan B tan C + + tan C tan tan A + tan A tan B = (a) –4 (b) –5 (c) –6 (d) –7 P is is a point on the line r = 5i + 7 j − 2k + s(3i − j + k) and Q is a point on the line r = −3i + 3j + 6k + t (−3i + 2 j + 4 k).If PQ is parallel to the vector 2i + 7 j − 5k, then | PQ | =
(a) (c)
9.
78 (b) 2 78 (c) 4 78 (d) 3 78 CED
e number of non-decreasing onto functions for {1, 2, 3, 4, 5, 6, 7, 8} to {1, 3, 5, 7, 9} is divisible by (a) 2 (b) 3 (c) 5 (d) 7
Let Z be a complex number such that Z = reiq and 1 Z + = 2 Z
(a) (c)
(d)
3 +1 2 5 −1 2
7− 3 2
In a quadrilateral quadrilateral ABCD, BC = 8, CD = 12, AD = 10,
p . If AB = m + n , then the sum of the
10. C olumn-I
P.
1/ x
f (x ) If lim 1 + x + x →0 x f (x ) then lim 2 =
MATHEMATICS TODAY | NOVEMBER ‘15
1. 0 3
=e ,
Q. R.
S.
(a) (b) (c) (d)
x e number number of roots roots of the equation equation 2. tan x + + sec x = = 2 cos x in [0, 2p] is 2 2 Let x + y – 4x – 2 y – 11 = 0 be a 3. circle. e area of the quadrilateral formed by the tangents from the point (4, 5) with a pair of radii is Let S and S1 be the foci of the ellipse 4. 9x 2+ 5 y 2 = 30 y and a point P be (3, 3). e area of triangle PSS1 is 5. P Q R S 2 2 5 4 1 1 3 4 5 4 3 2 4 2 3 5
2 4
6
8
See Solution set of Maths Musing 154 on page no. 18
Prof.. Ramanaiah is the a uthor of MTG JEE(Main & Advanced) Mathematics series Prof 8
Column-II
x →0
p , then r = =
(b)
(d)
MATCHING LIST
COMPREHENSION
6 7+ 3 2 5 +1 2
5 −1 2
JEE ADVANCED
If q =
(b)
3 digits of (m + n) is
7.
3 7+ 3 2 3 −1 2
∠ A= ∠B =
6.
p , then r = =
INTEGER MATCH
(a)
If q =
PAPER -1 SECTION-1
SECTION-2
This section contains 8 questions. The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive. For each question, darken the bubble corresponding to the correct integer in the ORS. Marking scheme : +4 If the bubble corresponding to the answer is darkened, 0 In all other cases.
This section contains TEN questions. Each question has FOUR options (a), (b), (c) and (d). ONE OR MORE THAN ONE of these four option (s) is (are) correct. For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS. Marking scheme : +4 If only the bubble(s) corresponding to all the correct option(s) is (are) darkened, 0 If none of the bubbles is darkened, 2 In all other cases.
1.
2.
3.
4.
3 e area bounded by y = 2 – |2 – x | and y = is x | | k − 3 log e (3) , where k is 2 2
Let f (x ) = [x ] + { x } , f : R → R, then area of figure bounded by y = f –1(x ), y = 0 between the ordinates x = 1/2 and x = 5 is ([·] = G.I.F) and {} = fractional 40 2 − 1 function) , where k is k 2 Area bounded by the curve y = g ( f (x )), x -axis, x = –3 and x = 4, if g (x ) = ex , f (x ) = { x }, is k(e – 1), where k is
p sin( p ln x ) dx is e value of x 1 3 p/ 2
6.
If A =
0
cos x cos x − sin x
dx and B =
0
sin x cos x − sin x
10.
8.
(1 − x 2 )dx
1 + e x + 1 x
1+ e −1
MATHEMATICS TODAY
|
NOVEMBER ‘15
1 + e x + 1
(d) f (x ) = 2(x – 2)
If l = sec2 x cosec4 xdx
12.
27 2 6 2 A curve g (x ) = x (1 + x + x ) (6 x + 5x + 4)dx is passing through origin, then
27 37 g (a) g (1) = (b) (1) = 7 7 1 37 g g − = − = ( 1 ) ( 1 ) (c) (d) 7 14 8 x + 4 dx and f (0) = 0, then If f (x ) = 4 2 x − 2x + 2 (a) f (x ) is an odd function
By : Vidy re l Cen tre ,S n e ap ta i Ba p ta Ma r g,D ad ar (W ) ,M umbai - 8.2T l anka a r I nstitute ,P a 10
1 + e x − 1
= A cot x + B tan x + C cot x + D, then (a) A = –1/3 (b) B = 2 (c) C = –2 (d) none of these
dx
1 If ln(b), then find the value of b. = 2 4 2 x x 1 + + 0
(b) g (x ) =
3
e area bounded by y = xe|x | and lines |x | = 1, y = 0 is 1
(1 + e )
(c) g (x ) =
then the value of A + B is 7.
x
dx = f (x ) (1 + e x ) − 2 ln g (x ) + C ,
(a) f (x ) = x – 1
11.
3 p/2
If then
If y = 2sin(x ) + sin(2x ), 0 x p, then the area enclosed by the curve and the x -axis is e 37
5.
9.
xe x
le .:( 20)
24 7630
13.
SECTION-3
(b) f (x ) has range R (c) f (x ) has atleast one real root (d) f (x ) is a monotonic function esin x Let f (x ) = , x 0. x
This section contains TWO (A) P Q S R T questions. Each question Q S R T contains two columns, (B) P Column I and Column II. Q S R T Column I has four entries (C) P (A), (B), (C) and (D), Column (D) P Q S R T II has five entries (P), (Q), (R), (S) and (T). Match the entries in Column I with the entries in Column II. One or more entries in Column I may match with one or more entries in Column II. The ORS contains a 4 × 5 matrix whose layout will be similar to the one shown below. For each entry in Column I, darken the bubbles of all the matching entries. For example, if entry (A) in Column I matches with entries (Q), (R) and (T), then darken these three bubbles in the ORS. Similarly, for entries (B), (C) and (D). Marking scheme : For each entry in Column I, +2 If only the bubble(s) corresponding to all the correct match(es) is (are) darkened, 0 If none of the bubbles is darkened, –1 In all other cases.
3
4
3esin( x ) dx = f (k ) − f (l ) , then If x
1
(a) k = 64 (c) l = 2 14.
(b) k = 32 (d) l = 1
If f (x ) = ae2x + bex + cx satisfies the conditions f (0) = –1, f (loge2) = 31,
then (a) a = 5 (c) c = 2 15.
loge (4 )
( f (x) − cx)dx = 0
39 , 2
(b) b = –6 (d) a = 3
A function y = f (x ) satisfying the differential
19.
dy sin2 ( x ) equation sin x − y cos x + 2 = 0 is such dx x that y → 0 as x → then p /2 p (a) lim f (x ) = 1 (b) f (x )dx 2 x →0 0 p/2 (c) f (x )dx 1 (d) f (x ) is even 0
16.
Column I
(A)
x
(B)
It is defined on the interval [–1, 1] It is an increasing function It is an odd function e point (0, 0) is the point of inflection 1
dt
1/ x
(c) I 2 > I 1
(d) I2 = cot −1 x − 4
(C)
2p 4 x 5 x 4 + 3 sin x dx (b) sin cos dx ln 2 2 0 4 + 3 cos x p/5
12
− x (d) e (cos x − sin x)dx 0
MATHEMATICS TODAY | NOVEMBER ‘15
q.
1
If
x 4 + 1
dx
x(x 2 + 1)2
r.
3
s.
4
m
+ n, 1 + x 2 where n is the constant of integration, then mk is greater than
p
3 p/10
p sin 4 x dx (c) 0 sin x
0
x k
= k ln | x | +
Which of the following definite integrals when simplified reduces to zero? (a)
dx If ( x )7 + x 6
1
(b) I 1 > I 2
p.
+ c, x k + 1 then ak is less than
dt
(a) I 1 = I 2
dx
( x )5
= a ln
1 + t 2 and I 2 = 1 + t 2 for x > 0, then
If I 1 =
x
18.
1− 4
x
then k is greater than
0
17.
If
2x
Column II
= ksin–1( f (x )) + C ,
e function f (x ) = 1 − t 4 dt is such that (a) (b) (c) (d)
Match the following:
(D)
dx If 5 + 4 cos x x = k tan −1 m tan + C then 2
k/m is greater than
20.
Match the following: Column I
p
(A)
x ln(sin x )dx =
p.
p − ln(2)
p
q.
p 1 + ln 2
4
ln(1 + tan q)d q =
cos x dx
(C)
2
0 1 + cos x + sin x
= r.
p 8
ln 2
2
0
(B)
p /2
Column II
4
sin−1 x dx
0
(1 − x 2 )3/2
(D)
2
1/ 2
s.
=
p 1 − ln 2 4
2
0
PAPER -2 SECTION-1
9.
This section contains EIGHT questions. The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive. For each question, darken the bubble corresponding to the correct integer in the ORS. Marking scheme : +4 If the bubble corresponding to the answer is darkened, 0 In all other cases. 1.
2.
3.
10.
e order of the differential equation of all circles, having centre on y -axis -axis and passing through the origin is Area bounded by y by y = x sin x and x -axis -axis between x = 0 and x = 2p is kp, where k is e area enclosed between the curve y curve y = loge(x + x + e) and the coordinate axes is 11.
1 1 Evaluate cosec 101 x − dx x 1/2 x
p
5.
If
2 x − p
0
k
dx =
p
−5
dx dx + 3
x
1
0
x
2/ 3
e
9( x −2/ 3)
2
dx
12.
x
8.
If lim 0 x →
1 + x 2
=
p2 k
MATHEMATICS TODAY | NOVEMBER ‘15
= e2 tan
2 tan −1 y (b) xe
= e2 tan
−1 y
−1 y
+k +k
−1 (d) (x − 2) = ke tan y
SECTION-2
14
−1 (a) 2 xe tan y
tan −1 y = tan−1 y + k (c) xe
, then k then k is
This section contains EIGHT questions. Each question has FOUR options (a), (b), (c) and (d). ONE OR MORE THAN ONE of these four option (s) is (are) correct. For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS. Marking scheme : +4 If only the bubble(s) corresponding to all the correct option(s) is (are) darkened, 0 If none of the bubbles is darkened, –2 In all other cases.
e solution of differential di fferential equation −1 dy (1 + y 2 ) + ( x − e tan y ) = 0 is dx
1/3
If f (t )dt = x + tf (t )dt , then find the value of 2 f (1). (1).
(tan−1 t )2 dt
3 2 + 1)
(a) f (x ) is an even function (b) f (x ) is a bounded function (c) e range of f of f (x ) is (0, 1] (d) f (x ) has two points of extrema
,
−4
is 7.
2
Sum of the two integral e( x +5)
x 2 − x + 1 x e dx = e x f (x ) + c, then If (x
then k is 6.
2 dx = tan−1 ( f (x)) )) − tan −1 ( g( x )) + C , then 3 x + 1 6
2
2 p x sin 2 x sin 2 cos x
If
x 4 + 1
(a) both f both f (x ) and g and g (x ) are odd functions (b) f (x ) is monotonic function (c) f (x ) = g = g (x ) has no real roots f (x ) 1 3 dx = − + +c (d) g (x ) x x 3
2
4.
If sin x d(sec x ) = f (x ) − g (x ) + c , then (a) f (x ) = secx secx (b) f (x (x ) = tanx tanx (c) g (x ) = 2x 2x (d) g (x ) = x
13.
Which of the following is the integrating factor of dy x log x + y = 2 log x ? dx (a) x (b) ex (c) loge(x )
(d) loge(logex )
2
−n dy
14. e value of min(x − [x], − x − [− x ])dx is ([·] −2 denotes the greatest integer function) (a) 0 (b) 1 (c) 2 (d) none of these
Put the value of y
15. Let f (a) > 0, and let f (x ) be a non decreasing 1 b continuous function in [a [a, b]. en is f (x )dx is b a − a (a) less than equal to f to f (b) (b) greater than equal to f to f (a) (c) maximum value bf (b) f (a) (d) minimum value b−a = sin 2 sin xdx = 16. e values of which satisfy p/2 ( [0, 2p]) are equal to (a) p/2 (b) 3p/2 (c) 7p/6 (d) 11p/6
= C + (1 − n)Q e (1−n)Pdx dx (1−n)Pdx i.e. i.e. y1−n e = C + (1 − n)Q e (1−n )Pdx dx
SECTION-3
This section contains TWO paragraphs. Based on each paragraph, there will be TWO questions. Each question has FOUR options (a), (b), (c) and ( d). ONE OR MORE THAN ONE of these four option(s) is(are) correct. For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS. Marking scheme : +4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened, 0 If none of the bubbles is darkened, -2 In all other cases
(1−n)Pdx
z e
is form is called Bernoulli’ B ernoulli’s equation. dy y ( x) − y 2 19. Solution of = , where (x ) is a (x ) dx function of x is (a) (x ) = x + C (arbitrary constant) (x ) (b) = x + C (arbitrary constant) y (c) y (x ) = –x + C (arbitrary constant) (d) y (x ) = x + C (arbitrary (arbitrary constant) 20. Solution of the differential equation
dy 1 = is dx xy[x 2 sin y 2 + 1] 1 y 2 1 2 2 (a) e (cos y − sin y ) − 2 = C x 2 2 1 1 (b) e y (cos y 2 + sin y 2 ) − = C 2 x 2 2 1 1 (c) e y (cos y 2 − sin y 2 ) + = C x 2 2 1 y 2 1 2 2 (d) e (cos y + sin y ) + 2 = C x 2
x
Also function g (x ) = f (t )dt and g (1) = A. 0
Function g (x ) is 17. Function g (a) odd (b) even (c) neither even nor odd (d) can’t can’t be determined determi ned
−n – n Let y Let y 1 – n = z (1 − n) y
dy dz = dx dx
in (i), we get
dz + (1 − n)Pz = (1 − n)Q which is linear in z . dx So, solution is
Paragraph for Q. No. 17 & 18 An even function f function f is defined and integrab i ntegrable le everywhere and is periodic with period 2.
of g (2) (2) in terms of A of A is 18. Value of g (a) 2 A A (b) A/2 A/2 (c) 4 A (d) A/4 Paragraph for Q. No. 19 & 20 Sometimes equations which are not linear can be reduced to the linear form by suitable transformations. transformations. dy + Py = Qyn , For example an equation of the form dx where P and Q are functions of x only or constants and n ( 0 and 1) is a constant. dy dy + Py = Qyn y −n + Py1−n = Q ... (i) dx dx [Dividing by y by y n on both sides]
dx
ANSWERS KEY
PAPER-1
1. 6. 11. 15. 18. 19. 20.
(4) 2. (3) 3. (7) 4. (4) 5. (2) (0) 7. (2) 8. (3) 9. (b, d) 10. (a, c) (a, c) 12. (a, b, c, d) 13. (a, d) 14. (a, b) (a, b, c, d) 16. (a, b, c, d) 17. (a, d) (a, b, c, d) (A) → (p, q); (B) → (r, s); (C) → (p); (D) → (p, q) (A) → (p), (B) → (r), (C) → (s), (D) → (s) PAPER-2
1. 6. 11. 15. 19.
(1) 2. (4) 3. (1) 4. (0) 5. (8) (0) 7. (1) 8. (4) 9. (b, d) 10. (a, c, d) (a, b, c) 12. (a) 13. (c) 14. (b) (a, b) 16. (a, b , c, d) 17. (a) 18. (a) (b) 20. (a) For detailed solution to the Sample Paper, Paper, visit our our website website : www www.. vidyalankar.org. vidyalankar.org.
nn
MATHEMATICS TODAY | NOVEMBER ‘15
15
Y
U ASK
which gives two values of x and hence the coordinates of two points Q and R (say), so that the chords PQ and PQ and PR are PR are bisected by x -axis. -axis. If the
WE ANSWER
chords PQ and PQ and PR are PR are distinct, the roots of (i) are real and distinct.
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.
y P(p, q)
R
Integrate tan( x − ) ta tan( x + ) ta tan 2 xdx
the discriminant discriminant p p2 – 8q 8q2 > 0
p2 > 8q 8q2. n
tan 2x = tan( x + + x − )
= or
3.
tan(x + ) + tan( x − ) 1 − tan(x + ) tan(x − )
If Sn = t n Sn
= tan (x + + ) + tan (x (x – – )
n
C r
Given, Sn =
r =0
n
I = tan 2 x − tan(x + ) − tan( x − ) dx
= log 2.
sec 2x cos(x + ) cos(x − ) + c .
If two distinct chords, drawn from the point are bisected by the x -axis, -axis, then prove that p that p2 > 8q 8q2 – Nalini Sharma, Bangalore
PQ be a chord of the given circle passing Let PQ through P ( ( p, p, q) and the coordinates of Q be (x (x , y ). ). Since PQ is PQ is bisected by the x -axis, -axis, the mid-point mid-point of PQ lies PQ lies on the x -axis -axis which gives y gives y = = –q –q. Now Q lies on the
16
circle x 2 + y + y 2 – px – px – – qy = =
so
x 2 + q2 – px – px + + q2 = 0
x 2 – px – px + + 2q 2q2 = 0 MATHEMATICS TODAY | NOVEMBER ‘15
r =0
1 n
C r
r n
C r
, then prove that
– Ramesh Ahuja, Ludhiana n
and t n =
r =0
r
n
C r
1 n
C n−r
n n nS = n n r =0 C n−r n
( p, p, q) on the circle x 2 + y + y 2 = px = px + + qy (where pq (where pq 0)
Ans.
Sn =
1 = log sec 2x + log cos(x + ) + log cos(x − ) + c 2
r =0
n 2
= tan 2x – – tan(x tan(x + + ) – tan (x (x – – ) Let I = tan( x − ) tan( x + ) tan 2 x dx
and t n =
n
Ans.
tan(x – – )tan(x )tan(x + + )tan2x )tan2x tan(x
=
n
1
r =0
tan 2x – – tan 2x 2x tan(x tan(x + + )tan(x )tan(x – – )
Q
– Rakesh Avasthi, Patna Ans.
x
O
r n − r +n nSn = n C n−r r =0 C n−r n
n n − r r + nSn = n n r =0 C n−r r =0 C r
n −1 1 n r n nSn = n + n + ... + n + n Cn−1 C 1 r =0 C r C n
nSn = t n + t n
0
…(i)
t n Sn
=
n 2 nn
MATHEMATICS TODAY | NOVEMBER ‘15
17
0
8. (a) : Area =
−xe x dx = 1 −
y
SOLUTION SET 154
1. (c) : Let . stand for H or T . e 4 consecutive heads st
nd
rd
start with 1 , 2 , 3 , ... 7 toss HHHH.........., THHHH .......... .THHHH .........., ..THHHH.......... .......... ..........THHHH 1 6 1 Prob. = + = 24 25 4 2. (d) : f (–q) = f (q), f (q + p) = f (q)
f (q) is an even function periodic with period p. Maximum value = f (0) = 1 + sin 21
p
Minimum value = f = cos21 2 e difference = 1 + sin 21 – cos21 = 1 – cos2 3. (b) : Additional digits 0, 4 or 1, 3 or 2, 2 4! 00124 → 3 = 36 numbers 2! 4! 001123 → 4 ! + 2 = 48 numbers 2! 4! 4! 01222 → + = 16 numbers 3! 2! Total numbers = 36 + 48 + 16 = 100 4. (c) : Let c = kb, 20, b, kl are in H.P. 20 b = 40 − , k = 2, 4, 5, 10, 20 → k b = 30, 35, 36, 38, 39 Number of value of b = 5
5. (b) : Let x + y = z → dx = 1 −
2 dz z + 2
x + c1 = z – 2ln(z + 2) x + y + z = ce y /2 6. (a,b) : p = (x 1, y 1) → t 12 = x 12 + y 21 – a2, t 22 = x 12 + y 12– 2ax 1, t 32 = x 12 + y 12 – 2ay 1, t 14 = t 22t 32 + a4 → locus of p is (x + y ) (x 2 + y 2 – ax – ay ) = 0, straight line and circle. x y 7. (b) : f (x + y ) = e f ( y ) + e f (x ) x = y = 0 → f (0) = 0
f ( x + y ) = f (x + y ) x y f ( x ) − f ( x ) f ( y) − f ( y ) = = k, constant x y
18
1
th
9. (9) : 162 = (a + b + c)(b + c – a)(c + a – b)(a + b – c)
= [(b + c)2 – a2] [a2 – (b – c)2] = (9b2 – 27) (27 – b2) = 9(b2 – 3) (27 – b2)
162 = – 81 + 30 b2 – b4 = 144 – ( b2 – 15)2 144 9 9, maximum area = 9 10. (b) : P → 4, Q → 5, R → 3, S → 2 z 0
z n
z 2
z 1
z n z 1 z 2 = q
z − z q z0 − z1 = (z2 − z1 ) 0 1 cis z2 − z 1 2
q 1 = 1 + i tan (z2 − z 1 ) 2 2 1 q z0 = Az1 + Az2 , A = 1 − i tan 2 2 1 2 A = 2(1 + cos q) n = 6, | A|2 = 1, n = 8, | A|2 = 1 + n = 10, | A|2 =
1 2
3+ 5 , n = 12, | A|2 = 2 + 3 2
Solution Sender of Maths Musing SET-153
1. Harpal Singh (Punjab) 2. Gouri Sankar Adhikari (W.B.) 3. Khokon Kumar Nandi (W.B.) 4. Gajula Ravinder (Karimnagar) 5. N. Jayanthi (Hyderabad) SET-154
e e x f (x ) – f (x ) = ke → f (x ) = (kx + c)ex f (0) = 0 → c = 0, f (0) = 1 → k = 1, f (x ) = xex
1. Jayanthi (Hyderabad)
It has mininum at x = –1.
3. Gouri Sankar Adhikari (W.B.)
MATHEMATICS TODAY | NOVEMBER ‘15
x
O
2. Khokon Kumar Nandi (W.B.)
nn
1.
Prove that a+b+c 3
2.
3.
4.
5.
1
3
Using the arithmetic-geometric mean inequality again, we then have
(b + c)2 (c + a)2 (a + b)2
,
abc
4
3 4
A hexagon is inscribed in a circle with radius r . Two of its sides have length 1, two have length 2 and the last two have length 3. Prove that r is a root of the equation 2r 3 – 7r – 3 = 0. Prove that sin2(q + ) + sin2(q + ) – 2cos( – ) sin(q + ) sin(q + ) = sin2( – ). …(1)
(a + b + c).
ABCD is a rhombus with ∠ A = 60°. Suppose that E, F are points on the sides AB, AD, respectively and that CE, CF , meet BD at P , Q respectively. Suppose that BE2 + DF 2 = EF 2. Prove that BP 2 + DQ2 = PQ2
e numbers 2, 4, 8, 16, ..., 2 n are written on a chalkboard. A student selects any two numbers a and b, erases them and replaces them by their average, namely (a + b)/2. She performs this operation (n – 1) times until only one number is le. Let Sn and T n denote the maximum and minimum possible value of this final number, respectively. Determine a formula for Sn and T n in terms of n.
By the arithmetic-geometric mean inequality, we have 2
2
2
2
2
a(b + c) + b(c + a) + c(a + b)
2
a b + ab + b c + bc + c a + ca
6 6 a6b6c 6
= 6abc,
2
2
2
2
2
2
8(a b + ab + b c + bc + c a + ca + 3abc),
or 9(a + b)(b + c)(c + a) 8(a + b + c)(ab + bc + ca) = 4(a + b + c)(a(b + c) + b(c + a) + c(a + b)).
abc(a + b)(b + c)( c + a)
…(i)
From (i), it follows that 1
(a + b)2 (b + c)2 (c + a)2
3
2.
abc
4
a+b+c 3
.
Equal chords subtend equal angles at the centre of a circle; if each of the sides of length i subtends an angle i (i = 1, 2, 3) at the centre of the given circle, then 21 + 22 + 23 = 360°,
whence and
1
2
+
= 90 −
2
2
3
2
,
+ = cos 90 − = sin . 2 2 2 2 1
cos
3
2
3
Next we apply the addition formula for the cosine:
1
2
cos
2
2
− sin
1
2
sin
2
2
= sin
, 3
2
…(i)
where (see figures) sin
which implies 9(a2b + ab2 + b2c + bc2 + c2a + ca2 + 2abc)
3
(a + b + c)
cos
SOLUTIONS 1.
(a + b)(b + c)(c + a)
where a, b, c > 0. Equality holds if a = b = c.
sin sin
1
2
2
2
3
2
=
1/ 2
r
, cos
1
r
2
= , cos =
1
2
2
4r
=
2r 2
2
=
−1 ,
r
−1
r
;
3/ 2
r MATHEMATICS TODAY | NOVEMBER ‘15
19
Finally,
α α1
α
r
r
1
1
2
x 1 − x BP + DQ = x + 1 + 2 2
r
=
We substitute these expressions into (i) and obtain, aer multiplying both sides by 2r 2 , 4r
2
2
Now write it in the form (4r 2 −1)(r 2 − 1) = 3r + 1, and square, obtaining (4r 2 – 1)(r 2 – 1) = 9r 2 + 6r + 1, which is equivalent to r (2r 3 – 7r – 3) = 0. Since r 0, we have 2r 3 – 7r – 3 = 0, which was to be shown. 3.
4.
Let AB = 1 and put EB = x and FD = y en DB = 1, AE = 1 – x and AF = 1 – y By applying the Cosine Law to AEF (∠EAF = 60°) we transform the condition x 2 + y 2 = EF 2 to x 2 + y 2 = (1 – x )2 + (1 – y )2 – 2.
1 2
.(1 – x ) (1 – y ),
which simplifies to 1 – x – y – xy = 0, 1 − x and so, y = …(i) 1 + x Furthermore, BP is the angle-bisector of ∠EBC , in EBC so, by a well-known formula, 2 EB.BC . cos 60 x BP = . = EB + BC x + 1 Similarly, DQ =
y .Using (i), DQ = y + 1
Now , PQ = DB – BP – DQ =1– 20
x 1 − x − x + 1 2
=
+ x . 2(1 + x ) 1
2
MATHEMATICS TODAY | NOVEMBER ‘15
1
− x 2
.
+ 2 x + x (1 + x ) = = PQ , 4(1 + x ) 4(1 + x ) 2
4
2
2
2
2
2
We shall prove by induction that if we perform the operation on the numbers a1, a2, ..., an–1 , an, where a1 < a2 < ...< an, then a a a a a a T n = 1 + 2 + 3 + ... + nn−−22 + nn−−11 + nn−1 . 2
2(sin2(q + ) + sin2 (q + )) = 2 – (cos (2q + 2) + cos (2q + 2)) = 2 – 2cos(q + + ) cos ( – ), – 4 cos ( – ) sin (q + ) sin (q + ) = –2 cos ( – ) (cos( – )–cos(2q + + )), so that, 2 × L.H.S. of (1) = 2 (1 – cos (2 q + + ) cos ( – ) – cos2( – ) + cos ( – ) cos(2q + + )) = 2 (1 – cos 2( – )) = 2 sin2( – ).
1
which completes the proof. 5.
− 1 r − 1 − 1 = 3r
2
2
4
8
2
2
2
For n = 2, we have two numbers, a1 and a2 and thus T 2 must be (a1 + a2)/2. us the claim holds for n = 2. Assume that for n = 2, 3, . . . , k – 2, k – 1, the identity holds. en for the case n = k, that is, when we start off with the numbers a1, a2, a3, ..., ak, we will perform the operation k – 2 times and be le with two numbers, x and y . Say x comes from performing the operation on a set of p numbers from the set. ey comes from performing the operation on the remaining set of k – p numbers. To illustrate this, say we start off with 2, 4, 8, 16, and 32. en we can replace 2 and 8 by 5, 16 and 32 by 24, then 4 and 24 by 14. en we are le with two numbers, 5 and 14. We got 5 from performing the operation on the numbers 2 and 8, and we got 14 from performing the operation on the other three numbers, namely 4, 16, and 32. Let b1, b2, b3, ..., b p be the set of numbers that we used to get the number x , where the bi’s are in increasing order and let c1, c2, c3, . . . , ck – p be the other numbers from the set that we used to get y , where the ci’s are in increasing order. en, by our induction hypothesis, we have b p− 2 b p −1 b p b b x 1 + 2 + .... + p − 2 + p −1 + p −1 , 2
4
2
2
2
MATHEMATICS TODAY | NOVEMBER ‘15
21
and similarly, y
c1 2
+
c2 4
ck
+ .... +
2
− p− 2
k − p−2
ck
+
2
− p −1
k − p−1
+
ck 2
−p
Adding a1 a2 a3
. 1
k − p−
4
Note that the bi’s and the ci’s are just some permutation of the set a1, a2, a3, ..., ak. Hence, to minimize the value of (x + y )/2, the average of the two numbers will be minimized when the denominators are as large as possible. Hence, we want either p or k – p to be 1, since we will then have a denominator of 2k–1 in one of the terms. Without loss of generality, assume that k – p = 1. us, we want to now show that bk 3 bk 2 bk 1 x + y c1 b1 b2 + + + ... + k 2 + k 1 + k 1 2
2
4
a1 2
+
8
a2 4
+
a3 8
2
+ ... +
−
−
−
−
ak 2
2
−2
k −2
+
ak 2
−
2
−1
k −1
+
−
ak 2
k −1
,
because this will prove the case for n = k. Note that the bi’s are in increasing order, so if c1 = ar for some r , then b1 = a1, b2 = a2, ...., br–1 = ar–1, br = ar + 1, br + 1 = ar + 2, . . ., bk–1 = ak. Noting that the ai’s are in increasing order, we have ar ar ar ar = + + + ... 2
4
ar 4
+
8
ar 8
+
16
ar 16
+ ... +
ar 2
r
a1 a2 4
+
8
+
a3 16
+ .... +
ar 1 −
2
r
+
+
8
ar 1 −
+ ... +
16
2
+
r
ar 1 +
2
r +1
+
ar 2 +
2
r + 2
+ ... +
ak 2
−1
k −1
+
ak 2
k −1
to both sides, we get the desired inequality. us, the induction is proved and hence, for our question, we take ai = 2i for i = 1, 2, ..., k and we have T n =
2
+
2
4 4
+
8 8
n− 2
+ ... +
n −1
2
+
n− 2
2
2
n −1
2
+
2 2
n
n−1
= (n – 1). 1 + 2 = n + 1 us, our formula for T n is obtained by taking the ai’s in increasing order. In contrast, to get Sn, we want to take the ai’s in decreasing order, namely a1 = 2n, a2 = 2n–1, . . . , an–1 = 4, an = 2. is is so that we can get the maximum possible value at the end. us, we have Sn =
2
n
2
n −1
+
2
4
n− 2
+
2
+ ... +
8
2 2
3
n− 2
+
2
2
n −1
2
1
+
2
n −1
2
= 2n–1 + 2n–3 + 2n–5 + ... + 25–n + 23 –n + 22–n = 23–n . (1 + 22 + 24 + ... + 22n–4) + 22 – n = 23–n.
4
n −1
−1
3
n 1 = 8(4 − 1) + −
3 2
n
+2
4
2 −n
= n
2
2
2n + 1
+4 n
3 2
nn
22
MATHEMATICS TODAY | NOVEMBER ‘15
STRAIGHT LINES
DEFINITION
SLOPE OR GRADIENT OF A LINE
A path traced by a point in a constant direction and endlessly in its opposite direction is known as a straight line.
If q is the inclination of a non-vertic al line, the n m = tan q is called the slope or gradient of the line. q is measured from x -axis and it is taken as positive or negative according as it is measured in anticlockwise or clockwise direction.
Distance Formula
Distance between the points A(x 1, y 1) and B(x 2, y 2) is given by 2
Note :
2
AB = ( x2 − x1 ) + ( y2 − y 1 ) Section Formula Let a point M divides a line segment joining the points (x 1, y 1) and (x 2, y 2) in the ratio m : n, then
e inclination of a line parallel to the x -axis or of a horizontal line is 0°, i.e. q = 0°. Hence, slope of a line parallel to x -axis is 0. e inclination of a line parallel to the y -axis or of a vertical line is 90°. i.e., q = 90°. Hence, slope of a line perpendicular to y -axis is not defined.
Types of division
Coordinates of M
Internal
mx2 + nx1 my2 + ny1 m + n , m + n
Slope m of a non-vertical line passing through the points ( y − y ) (x 1, y 1) and(x 2, y 2) is given by m = 2 1 . (x2 − x 1 )
mx2 − nx1 my2 − ny1 m − n , m − n
Condition for Parallelism and Perpendicularity Let L1 and L2 be two lines whose slopes are m1 and m2
External
Mid-point
x1 + x2 , 2
y1 + y2
2
Slope of a Line Passing through Two Given Points
respectively. en, (i) L1|| L2 m1 = m2 (ii) L1 L2 m1m2 = –1 ANGLE BETWEEN TWO LINES
Area of Triangle Area of ABC whose vertices are A(x 1, y 1), B(x 2, y 2) and C (x 3, y 3) is given by = 1 . | x1( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 ) | sq. unit. 2 Not e : The points A, B , C are collinear area of ABC = 0.
If is the acute angle between two non-vertical lines L1 and L2 with slopes m1 and m2, then tan =
m2 − m1
1 + m1m2
,
where 1 + m1m2 MATHEMATICS TODAY | NOVEMBER ‘15
23
Case I : If tan > 0, then is the acute angle between two lines and is the obtuse angle. Case II : If tan < 0, then is the obtuse angle between two lines and is the acute angle. Collinearity of three points : Three points A, B, C in a plane are said to be collinear, if and only if slope of AB = slope of BC .
respectively is
VARIOUS FORMS OF THE EQUATION OF A LINE
Equation of the x -axis : Since ordinate of any point on the x -axis is zero. Hence, the equation of the x -axis is y = 0 Equation of the y -axis : Since abscissa of any point on the y -axis is zero. Hence, the equation of the y -axis is x = 0. Equation of a line parallel to the x - axis : Let AB be any line parallel to the x -axis and at a distance c from it. The ordinate of any point P on the line AB is c. Hence, the equation of a line parallel to the x -axis is of the form y = c. Equation of a line parallel to the y - axis : Let CD be any line parallel to the y - axis and at a distance d from it. The abscissa of any point P on the line CD is d . Hence, the equation of a line parallel to the y -axis is of the form x = d .
Slope-intercept form : e equation of the straight line whose slope is m and which cuts an intercept c on the y -axis i.e. which passes through the point (0, c) is y = mx + c. Note : If the line passes through the origin, then c = 0 and hence equation of the line will become y = mx . Point slope form : e equation of the straight line having slope m and passing through the point (x 1, y 1) is y – y 1 = m(x – x 1). Two-point form : e equation of the straight line which passes through the point (x 1, y 1) and (x 2, y 2) y − y 1 (x − x 1 ). is y − y 1 = 2 x2 − x 1
24
MATHEMATICS TODAY | NOVEMBER ‘15
Intercept form : e equation of the straight line making intercepts a and b on x - axis and y -axis
x y + =1 a b
Normal form : e equation of the straight line upon which the length of the perpendicular from the origin is p and this normal makes an angle with the positive direction of x -axis is
x cos + y sin = p
Distance form or parametric form or symmetric form : e equation of the straight line passing through the point (x 1, y 1) and making an angle q with the positive direction of x -axis is x − x1 y − y1 = = r cos q sin q
GENERAL EQUATION OF A LINE
Any equation of the form Ax + By + C = 0, where A, B and C are real constants such that A and B are not simultaneously zero is called general equation of the line. Reduction of the general equation to the standard form: (i) Reduction to slope-intercept form : If B 0, then the equation Ax + By + C = 0 can be written as
A C x − , which is of the form y = mx + c B B C whose slope, m = − A and y - intercept, c = − B B (ii) Reduction to intercept form : If C 0, then y = −
the equation Ax + By + C = 0 can be written as y x + =1 −C / A −C / B which is of the form
(provided A 0, B 0) x y + =1 a b
where x -intercept, a = −
C C and y -intercept, b = − A B
(iii)Reduction to the normal form : If the length of the
perpendicular from the origin on Ax + By + C = 0 be p and if be the angle which the perpendicular makes with the axis, then Ax + By + C = 0 must be same as
x cos + y sin – p = 0
...(i)
Length of perpendicular from a given point P (x 1, y 1) on a line Ax + By + C = 0, is given by | Ax1 + By1 + C | . d = 2 2 A + B Note : Length of the perpendicular from the origin on | C | . the straight line Ax + By + C = 0 is 2 2 ( A + B )
Comparing both equations, we get cos =
and p =
A
( A2 + B2 )
, sin =
B
( A2 + B2 )
C
( A2 + B2 )
Hence, equation (i) can be written as
A
( A2 + B2 )
x
B
( A2 + B2 )
y =
DISTANCE OF A POINT FROM A LINE
DISTANCE BETWEEN TWO PARALLEL LINES
C
( A2 + B2 )
which is the required form. Sign of the equation is taken as so that p should be
Distance between the parallel lines Ax + By + C 1 = 0 and Ax + By + C 2 = 0 is given by | C − C 1 | . d = 2 2 2 A + B
positive. CONIC SECTIONS
From figure, we have PS = constant = e PK
DEFINITION
A conic section or conic is the locus of a point which moves in a plane so that its distance from a fixed point is in a constant ratio to its distance from a fixed straight line. e fixed point is called focus, the fixed straight line is called directrix, and the constant ratio is called the eccentricity, which is denoted by e.
If e = 0, the curve is a circle. If e = 1, the curve is a parabola. If e < 1, the curve is an ellipse. If e >1, the curve is a hyperbola. The straight line passing through the focus and perpendicular to the directrix is cal led the axis of the conic and the point of intersection of conic with its axis is called the vertex of the conic.
CIRCLES
DEFINITION
A circle is the locus of a point which moves in a plane such that its distance from a fixed point remains constant. The fixed point is called the centre of the circle and the constant distance is called the radius of the circle.
x 2 + y 2 = a2
General Form : Equation x 2 + y 2 + 2 gx + 2 fy + c =
0 represents general form of circle with centre at (– g , – f ) and radius
=
g 2 + f 2 − c ( g 2 + f 2
c).
Equation of Circle
Standard Form :
Equation of circle with centre ( h, k) and radius a is given by (x – h)2 + ( y – k)2 = a2 If the centre of the circle is at origin, then h = 0 and k = 0. So, equation of circle becomes MATHEMATICS TODAY | NOVEMBER ‘15
25
PARABOLA
DEFINITION
A Parabola is the set of all points in a plane that are equidistant from a fixed line and a fixed point.
Standard equation of parabola
y 2 = 4ax
y 2 = – 4 ax
x 2 = 4ay
x 2 = – 4 ay
e=1
e=1
e=1
e=1
S(a, 0)
S(–a, 0)
S(0, a)
S(0, –a)
x + a = 0
x – a = 0
y + a = 0
y – a = 0
y = 0
y = 0
x = 0
x = 0
A(0, 0) (a, ±2 a)
A(0, 0) (–a, ±2 a)
A(0, 0) (±2a, a)
A(0, 0) (±2a, –a)
4a
4a
4a
4a
x – a = 0
x+a=0
y – a = 0
y + a = 0
Graph
Eccentricity Focus Equation of directrix Equation of axis Vertex Extremities of latus rectum Length of latus rectum Equation of latus rectum
is symmetric with respect to the axis of the parabola. If the equation has y 2 term, then the axis of symmetry is along the x -axis and if the equation has x 2 term, then the axis of symmetry is along the y -axis.
Note : Parabola
ELLIPSE
DEFINITION
An ellipse is the set of all points in a plane, the sum of whose distances from two fixed points in the plane is constant.
26
MATHEMATICS TODAY | NOVEMBER ‘15
Fundamental terms Standard equation
Ellipse (Horizontal ellipse) x 2 y 2 a
+ 2
b2
= 1, a b
Conjugate Ellipse (Vertical ellipse) x 2 a
+ 2
y 2 b
2
= 1, a b
Graph
Centre
(0, 0)
(0, 0)
Vertices
(±a, 0)
(0, ±b)
Length of major axis
2a
2b
Length of minor axis
2b
2a
(±ae, 0)
(0, ±be)
Foci Equation of directrices
Eccentricity Length of latus rectum Ends of latus rectum Focal distance or radii
x =
a e
b e
y =
b2 e = 1− 2 a
a2 e = 1− 2 b
2b2 a
2a b
b2 ae , a
a2 , be b
|SP | = (a – ex 1) and |SP | = (a + ex 1) |SP | = (b – ey 1) and |SP | = (b + ey 1)
Sum of focal radii = |SP| + |SP |
2a
2b
Distance between foci
2ae
2be
Note : If length of major axis is 2 a, length of minor axis is 2b and distance between the foci is 2 c, then a2 = b2 + c2 Ellipse is symmetric w.r.t. both the coordinate axes.
larger denominator and it is along the y -axis if the coefficient of y 2 has the larger denominator.
e foci always lie on the major axis. i.e, major axes is along the x -axis if the coefficient of x 2 has the
If both the foci merge together with ellipse and a = b, the ellipse become circle. When distance between foci ( SS) = a then b = 0. e ellipse is reduced to the line segment SS. MATHEMATICS TODAY | NOVEMBER ‘15
27
HYPERBOLA
DEFINITION
A hyperbola is the set of all points in a plane, the difference of whose distance from two fixed points is constant.
Fundamental terms Standard equation
Hyperbola x 2 y 2 a2
−
b2
=1
Conjugate Hyperbola y 2 b2
−
x 2 a2
=1
Graph
Centre Vertices Length of transverse axis Length of conjugate axis Foci
(0, 0) (±a, 0) 2a 2b (±ae, 0)
Equation of directrices
x =
Eccentricity Length of latus rectum Ends of latus rectum Difference of focal radii = |SP| – |SP | Distance between foci
a e
b2 e = 1+ 2 a
28
MATHEMATICS TODAY | NOVEMBER ‘15
b e
y =
e = 1+
a2 b2
2b2 a
2a 2 b
b2 ae, a
a2 , be b
2a 2ae
2b 2be
Note : A hyperbola in which a = b is called an equilateral hyperbola. Hyperbola is symmetric w.r.t. both the axes. e foci are always on the transverse axis.GTRY
(0, 0) (0, ±b) 2b 2a (0, ±be)
INTRODUCTION TO THREE DIMENSIONAL GEOMETRY
COORDINATE AXES
L e t XO X , YOY , ZOZ be the three mutually perpendiculars lines, intersecting at O. e position of a point is located with reference to these lines. The lines XOX , YOY , ZOZ are called the rectangular coordinate axes. The point of intersection O of these lines is known as origin. Distance along OX , OY , OZ are considered as positive, while distance in the direction OX , OY , OZ are negative. COORDINATE PLANES
(i) XOY is called the xy -plane. (ii) YOZ is called the yz -plane. (iii) ZOX is called the zx -plane.
ese three, taken together, are known as coordinates of P . us the coordinates of any point in space are the perpendicular distances of the point from the yz , zx and xy planes respectively. Signs of Coordinates in Eight Octants :
The octants XOYZ , X OYZ , X OY Z , XOY Z , XOYZ , X OYZ , X OY Z and XOY Z are denoted by I, II, III, ..., VIII respectively. I
II
III
IV
V
VI
VII VIII
x
+
–
–
+
+
–
–
+
y
+
+
–
–
+
+
–
–
z
+
+
+
+
–
–
–
–
DISTANCE BETWEEN TWO POINTS
The distance between the points P ( x 1 , y 1 , z 1 ) and Q(x 2, y 2, z 2) is given by PQ = ( x2 − x1 )2 + ( y2 − y1 )2 + ( z2 − z1 ) 2
ese planes, called the coordinate planes, divides the space into eight parts known as octants. ese octants could be named as OXYZ , X OYZ , XOY Z , X OY Z , XOYZ X OYZ , XOY Z and X OY Z .
SECTION FORMULA
Let M be a point which divides the line joining the points A ( x 1 , y 1 , z 1 ) and B ( x 2 , y 2 , z 2 ) in the ratio m : n, then
COORDINATES OF A POINT IN SPACE
Let P be any point in space. Draw PL , PM , PN perpendicular to the yz , zx and xy planes respectively then
Types of division
Internal
External
(i) LP is called the x -coordinate of P . (ii) MP is called the y -coordinate of P . (iii) NP is called the z -coordinate of P .
Mid-point
Coordinates of M
mx2 + nx1 , my2 + ny1 , mz2 + nz1 m+n m+n m + n mx2 − nx1 , my2 − ny1 , mz2 − nz1 m−n m−n m − n x1 + x2 , y1 + y2 , z1 + z 2 . 2 2 2
MATHEMATICS TODAY | NOVEMBER ‘15
29
COORDINATES OF CENTROID OF A TRIANGLE
e coordinates of the centroid of a triangle whose vertices are (x 1, y 1, z 1), (x 2, y 2, z 2) and (x 3, y 3, z 3) are x1 + x2 + x 3 y1 + y2 + y3 z1 + z2 + z 3 , , 3 3 3
12.
Find the length of axes of the ellipse whose eccentricity is 4/5 and whose foci coincide with those of the hyperbola 9x 2 – 16 y 2 + 144 = 0.
13.
Find the equation of the circle passing through the points (3, 7), (5, 5) and having its centre on the line x – 4 y = 1.
14.
(i) If the point (h, 0), (a, b) and (0, k) lie on a line, show that a + b = 1. h k (ii) Using slopes, show that the points (5, 1), (1, –1) and (11, 4) are collinear.
15.
Find the perpendicular distance of the line joining the points A(cos q, sin q) and B(cos, sin) from the origin.
Very Short Answer Type 1.
What is the inclination of a line whose slope is
− 3. 2.
Find the co-ordinates of the focus and equation of directrix of the parabola x 2 = – 16 y
3.
Find the value(s) of k for which the line (k – 3)x + (k2 – 4) y + (k – 1)(k – 6) = 0 is parallel to the x -axis,
4.
5.
Find the equation of a circle with centre (3, –2) and radius 5. Find the coordinates of the point which divides the join of P (2, –1, 4) and Q(4, 3, 2) in the ratio 2 : 3 externally.
SOLUTIONS
We have, m = − 3
2.
We have, equation of parabola, x 2 = –16 y
Short Answer Type 6.
7.
8.
Reduce the lines 3x – 4 y + 4 = 0 and 4x – 3 y + 12 = 0 to the normal form and hence determine which line is nearer to the origin.
9.
Find the equation of the hyperbola whose foci are (0, 10 ) and which passes through the point (2, 3).
10.
e centroid of a triangle ABC is at the point (1, 1, 1). If the coordinates of A and B are (3, –5, 7) and (–1, 7, –6), respectively, find the coordinates of the point C . Long Answer Type
11.
30
Comparing with x 2 = –4ay , we get a = 4
If A(–2, 1), B(2, 3) and C (–2, –4) are three points, find the angle between BA and BC . (i) Find the equation of a line that has y -intercept 4 and is perpendicular to the line joining (2, –3) and (4, 2). (ii) Find the equation of the line which cuts off equal and positive intercepts from the axes and passes through the point ( , ).
e equation of a straight line which passes t hrough the point (acos3q, asin3q) and perpendicular to x sec q + y cosec q = a is MATHEMATICS TODAY |
NOVEMBER ‘15
tan q = − 3 tanq = tan (180 – 60°) q = 120°
1.
Coordinates of focus are (0, –4). Equation of directrix is y – 4 = 0 3.
Given equation of the line is (k – 3)x + (k2 – 4) y + (k – 1)(k – 6) = 0 e given line will be parallel to thex -axis, when the coefficient of y is zero, i.e., when k2 – 4 = 0 k = 2.
4.
We know that the equation of a circle with centre C (h, k) and radius r is given by (r – h)2 + ( y – k)2 = r 2 . Here, h = 3, k = –2 and r = 5. e required equation of the circle is (x – 3) 2 + ( y + 2)2 = 52 x 2 + y 2 – 6x + 4 y – 12 = 0.
5.
We have, P (2, –1, 4), Q(4, 3, 2) and m1 : m2 = 2 : 3 Let R(x , y , z ) be the required point. en, x=
(2 4) − (3 2) ( 2 3) − (3 −1) (2 2) − (3 4) , y= , z = 2−3 2−3 2−3
x = –2, y = –9, z = 8. So, required point is R (–2, –9, 8).
6.
7.
8.
We have, A(–2, 1), B(2, 3) and C (–2, –4) Let m1 and m2 be the slopes of BA and BC respectively. en, 3 −1 = 2 = 1 , and m2 = −4 − 3 = 7 m1 = −2 − 2 4 2 − (−2) 4 2 Let q be the angle between BA and BC . en, 7 1 5 − m2 − m1 tan q = = 4 2 = 4 =2 15 3 1 + m1m2 1 + 7 1 4 2 8 q = tan−1 2 3 (i) Let m be the slope of the required line. Since, the required line is perpendicular to the line joining A(2, –3) and B(4, 2). m × slope of AB = – 1 m 2 + 3 = −1 m = − 2 4−2 5 Also, the required line has y -intercept 4. So, c = 4 Substituting the value of m and c in y = mx + c, we get 2 y = − x + 4 x + 5 y − 20 = 0 w h i c h i s t h e 5 required equation of line x y (ii) Let the equation of the line be + = 1 which a b cuts off intercepts a and b with the coordinate axes. It is given that a = b. erefore the equation of the line is x y ... (i) + =1 x + y = a a a But, it passes through (, ). So, + = a. Putting the value of a in (i), we get x + y = + which the required equation of line We have, 3x – 4 y + 4 = 0 –3x + 4 y = 4 3x
−
(−3)2 + 42
+
4 x
−
+
3 y
9.
e foci of hyperbola are at (0, 10 ). ese are on the y -axis. Clearly, centre of the hyperbola will be (0, 0). Let equation of the hyperbola be y 2
2
− x 2 = 1,
...(i) a2 b where a, b > 0 and b2 = a2(e2 – 1) The foci of this hyperbola are at (0, ± ae )
ae = 10. Now, b2 = a2 (e2–1) = a2e2 – a2 = 10 – a2. From (i), the equation of hyperbola is y 2
x 2 − =1 a2 10 − a2
(−3)2 + 42
2
2
5
5
From(i), the equation of hyperbola is y − x = 1
4
Let the coordinates of C be (x , y , z ). Coordinates of the centroid G = (1, 1, 1). en y − 5 + 7 z + 7 − 6 x + 3 − 1 = 1, = 1, = 1 3 3 3 x = 1, y = 1, z = 2. Hence, coordinates of C are (1, 1, 2).
(−3)2 + 42
− 3 x + 4 y = 4
5 5 5 is is the normal form of 3x – 4 y + 4 = 0 and the length of the perpendicular from the origin to it is given by p1 = 4/5 Now, 4x –3 y + 12 = 0 –4x + 3 y = 12
…(ii)
Since (2, 3) lies on hyperbola (ii), 92 − 4 2 − = 1 a 10 − a 90 – 9a2 – 4a2 = 10a2 – a4 a4 – 23a2 + 90 = 0 a2 = 18, 5 a2 = 18 b2 = 10 – 18 = – 8 not possible. a2 = 5 b2 = 10 – 5 = 5
10.
=
12
(−4)2 + 32 (−4)2 + 32 (−4)2 + 32 − 4 x + 3 y = 12 5 5 5 is is the normal form of 4x – 3 y + 12 = 0 and the length of the perpendicular from the origin to it is 12 given by p2 = . 5 Clearly, p2 > p1. erefore, the line 3x – 4 y + 4 = 0 is nearer to the origin.
or y 2 – x 2 = 5. 4 y
=
11.
We have x secq + y cosecq = a y cosecq = a – x secq a − sec q y = x + cosec q cosec q y = (– tanq)x + a sinq MATHEMATICS TODAY | NOVEMBER ‘15
31
slope = –tanq Slope of line 1 = cot q tan q
13.
Let the equation of the required circle be ... (i) x 2 + y 2 + 2 gx + 2 fy + c = 0 It passes through the point (3, 7) 9 + 49 + 6 g + 14 f + c = 0 or 6 g + 14 f + c + 58 = 0 ... (ii) Also point (5, 5) lies on (i) 25 + 25 + 10 g + 10 f + c = 0 or 10 g + 10 f + c + 50 = 0 ... (iii) Centre of (i), i.e., (– g , – f ) lies on line x – 4 y = 1 – g + 4 f = 1 or g – 4 f + 1 = 0 ... (iv) Subtracting (iii) from(ii), we get –4 g + 4 f + 8 = 0 or – g + f + 2 = 0 ... (v) Adding (iv) and (v), we get –3 f + 3 = 0 f = 1 Putting f = 1 in (iv), we have g – 4 + 1 = 0 g = 3 Putting g = 3, f = 1 in (ii), we get 6(3) + 14(1) + c + 58 = 0 or c + 18 +14 + 58 = 0 or c + 90 = 0 c = – 90 Substituting the values of g = 3, f = 1 and c = – 90 in (i), we get x 2 + y 2 + 6x + 2 y – 90 = 0 which is the required equation of the circle.
14.
(i) Let A (h, 0), B(a, b) and C (0, k) be the given points. Since the given point A, B, C are collinear, we have slope of AB = slope of BC b − 0 = k − b b = (b − k) a−h 0−a a (a − h)
to x secq + y cosecq = a is
Equation of line through (a cos3q, a sin3q) and slope cot q is y – a sin3q = cotq (x – a cos3q) cos q = a sin3 q − a cos3 q cot q y − x sin q cos 4 q 3 q − q = q − q + a x cos y sin sin a sin sin q x cosq – y sinq = a[– sin4q + cos4q] = a(sin2q + cos2q)(–sin2q + cos2q) = a(cos2q – sin2q) = a cos2q x cosq – y sinq = a cos2q 12.
e equation of the hyperbola can be written as 2 y 2 x 2 x 2 y − = 1 or − + = 1 Here a = 3, b = 4 9 16 16 9 e eccentricity of the hyperbola is given by
e = 1+
b2 a
2
= 1 + 16 = 5 9
3
Since transverse axis of the hyperbola is along y -axis, its foci are (0, ± ae), i.e., (0, ± (5/3) × 3) i.e., (0, ± 5). e ellipse whose foci are S(0, 5) and S(0, –5) has its major axis along y -axis. Its centre is (0, 0) [mid-point of SS] Let 2a and 2b be the lengths of major and minor axes of the ellipse; then its equation will be x 2 b
2
+
y 2 a
2
=1
... (1)
Since 2ae = SS = 10 2
2
a . 4 = 5 a = 25 5
4
2
Now, b = a (1 – e ) 225 b2 = 625 1 − 16 = 16
25
16
Hence, equation of the required ellipse is 16 2 16 2 x + y = 1 225 625 25 Length of major axis of the ellipse = 2a = and 2 15 . Length of minor axis = 2b = 2 32
MATHEMATICS TODAY | NOVEMBER ‘15
ab = (a – h)(b – k) ak + hb = hk a + b =1 h
k
[on dividing both sides by hk]
(ii) Let A(5, 1), B(1, –1) and C (11, 4) be the given points. en, (−1 − 1) −2 1 = = Slope of AB = (1 − 5) −4 2 4 − (−1) 5 1 = = 11 − 1 10 2 Slope of AB= Slope of BC AB || BC and have a point B in common. A, B, C are collinear. Hence, the given points are collinear. and slope of BC =
15.
We know that the equation of the line AB is given by y − sin q sin − sin q = x − cos q cos − cos q
+ q sin − q 2 2 y − sin q = x − cos q −2 sin + q sin − q 2 2 +q cos 2 y − sin q = x − cos q − sin + q 2 2 cos
+ q + y sin + q 2 2
x cos
− cos q.cos + q + sin q sin + q = 0 2 2 x cos + q + y sin + q − cos + q − q = 0 2
2
2
+ q + y sin + q − cos − q = 0 2 2 2
x cos
... (i) Let d be the perpendicular distance from the origin to line (i). en, d =
+ q + 0 sin + q − cos − q 2 2 2 + q + sin2 + q cos2 2 2
0 cos
= cos − q 2
−q . 2
Hence, the required distance is cos
nn
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33
*
ALOK KUMAR, B.Tech, IIT Kanpur
This column is aimed at Class XI students so that they can prepare for competitive exams such as JEE Main/Advanced, etc. and be also in command of what is being covered in their school as part of NCERT syllabus. The problems here are a happy blend of the straight and the twisted, the simple and the difficult and the easy and the challenging.
CARTESIAN & POLAR COORDINATES Y
II quadrant (–, +)
P(x, y )
I quadrant (+, +)
X
r X
O
III quadrant (–, –)
Y
X
O
x
y X
IV quadrant (+, –) Y
P (x , y ) - Cartesian coordinates P (r , q) - Polar coordinates y r = x 2 + y 2 , tan q = x DISTANCE FORMULA, SECTION FORMULA & AREA Distance formula : Distance between the points P (x 1, y 1) and Q(x 2, y 2) is PQ = (x2 − x1 )2 + ( y2 − y 1 )2 Section formula : The coordinates of a point dividing the line segment joining the points ( x 1, y 1) and (x 2, y 2) internally, in the ratio m : n are mx2 + nx1 my2 + ny1
m+n
,
m+n
and externally in the ratio m : n are
mx2 − nx1 my2 − ny1 m − n , m − n
1 [x ( y − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 )] 2 1 2 x1
Y
Area of a Triangle : Let ( x 1, y 1), (x 2, y 2) and ( x 3, y 3) respectively be the coordinates of the vertices A, B, C of a ABC . en the area of triangle ABC , is
Mid-Point formula : In particular, if m = n ,
the coordinates of the mid-point of the line segment joining the points ( x 1, y 1) and (x 2, y 2) is
x1 + x 2 y1 + y2 2 , 2 . *
1 = x2 2 x3
1 y 2 1 y 3 1 y 1
While using formulas, order of the points (x 1, y 1), (x 2, y 2) and (x 3, y 3) has not been taken into account. If we plot the points A( x 1, y 1), B(x 2, y 2) and C (x 3, y 3), then the area of the triangle as obtained by using formulas will be positive or negative as the points A, B, C are in anticlockwise or clockwise directions.
So, while finding the area of triangle ABC , we take modulus. Remark :
In case of polygon with vertices ( x 1 , y 1 ), (x 2, y 2), ...., (x n, y n) in order, then area of polygon is given by 1 .... |(x y − y x ) + (x2 y3 − y2 x3 ) + 2 1 2 1 2 + ( x n – 1 y n – y n – 1 x n) + (x n y 1 – y nx 1)|
Alok Kumar is a winner of INDIAN NATIONAL MATHEMATICS OLYMPIAD (INMO-91). He trains IIT and Olympiad aspirants.
34
MATHEMATICS TODAY | NOVEMBER ‘15
SLOPE & EQUATION OF LINE If q is the inclination of a line L, (0 q < p) then tanq is called the slope or gradient of the line L.
e slope of a line y whose inclination is 90° is not defined. e slope of a line is denoted by m. x us, m = tanq, O q 90° It may be observed that the slope of x -axis is zero and slope of y -axis is not defined.
Then its equation is x y + =1 a b
(x2, y 2)
P
Q
Conditions for parallelism and perpendicularity of lines in terms of their slopes : If lines L1 || L2, then m 1 = m2
x – b y x = b =
O
x
y – a =
Point slope form Equation of line L having slope m and passing through (x 1, y 1) is y – y 1 = m(x – x 1) Two point form : Equation of line L passing through P (x 1, y 1) and Q(x 2, y 2) is y − y 1 y − y y − y 1 = 2 (x − x 1 ) or y − y 2 = 2 1 (x − x 2 ) x2 − x 1 x2 − x 1
Slope-intercept form If line L having slope m and
intercept on y - a x i s i s c , then equation o f l i n e L is y = mx + c
O
L
O
x
p P
General Equation of a Line Ax + By + C = 0
Intercept form If C 0, then Ax + By + C = 0
x y + = 1 , we get a b C a = Intercept on x -axis = − A C b = Intercept on y -axis = − B Normal form Let Ax + By + C = 0 can be written as Ax + By = –C Compare (ii) with x cos + y sin = p
Compare (ii) with
L
x
y
can be written as y x + =1 −C − C A B
y
(0, c)
Normal form
....(i) ....(ii)
Compare (ii) with y = mx + c, we get A Slope of line = m = − B −C Intercept on y -axis = c = B
•
x
x
a
can be written as A C y = − x − B B
VARIOUS FORMS OF THE EQUATION OF A LINE: Line Parallel to x -axis Line Parallel to y -axis
O
O
DIFFERENT FORMS OF Ax + By + C = 0 Slope Intercept Form If B 0, then Ax + By + C = 0
If L1 L2, then m1·m2 = –1
y = a
y
Let length of perpendicular from origin to line L is OP = p and is the angle which is made by OP with +ve x -axis. e equation of line L is x cos + y sin = p
Slope of a line when coordinates of any two points on the line are given : y − y 1 dy mPQ = 2 = tan q = x2 − x 1 dx (x1, y 1)
Intercept form Suppose a line L makes intercept ‘ a’ on x -axis a n d ‘ b ’ o n y -axis. b
MATHEMATICS TODAY | NOVEMBER ‘15
....(i) ...(ii)
....(iii)
....(i) ....(ii)
35
so
• m AB = mBC = m AC • | AB| + | BC | = | AC | or |BC | + | CA| = |BA| or |CA| + | AB| = | CB|
A B −C = = cos sin p
− pA − pB cos = and sin = C
LENGTH OF PERPENDICULAR AND REFLECTION Length of perpendicular from ( x 1, y 1) to a straight line ax + by + c = 0 is given by
C
Using sin 2 + cos2 = 1
− pA 2 + − pB 2 = 1 C C C 2
2
p =
2
A
+B
2
p =
ax1 + by1 + c a2 + b2 C 2
If two lines ax + by + c1 = 0 and ax + by + c2 = 0
2
A + B Proper choice of signs is made so that p should be +ve. A B and sin = So, cos = A2 + B2 A2 + B2
are parallel, then d =
PARAMETRIC EQUATIONS OF A STRAIGHT LINE
In figure given below, y let BAP be a straight P (x, y ) + line through a given point A ( x 1 , y 1 ), the (x1, y 1) A angle of slope being q. – e positive direction Q ( X, Y ) of the line is in the sense BAP . (Direction x O B of increasing ordinate is called the positive direction of the line). For the points P (x , y ) and Q ( X , Y ) shown in the figure, AP is regarded as a positive vector and AQ as a negative vector, as indicated by the arrows. From the general definitions of cos q and sinq, we have x − x 1 y − y 1 cos q = , sin q = AP AP or x – x 1 = AP cosq, y – y 1 = AP sinq x − x1 y − y1 or = = r cos q sin q ANGLE BETWEEN TWO STRAIGHT LINES AND COLLINEAR POINTS Angle between two lines e acute angle between lines L1 and L2 with slopes m1 and m2 respectively, is given by m − m2 , as m1m2 –1 tan q = 1 1 + m1m2
36
us obtuse angle = 180° – q Collinearity of three points If A, B, C are collinear, then
MATHEMATICS TODAY | NOVEMBER ‘15
c1 − c2
a2 + b2 Reflection of a point about a line Let the image of a point ( x 1, y 1) about the line ax + by + c = 0 is ( x 2, y 2), then x2 − x 1 y2 − y 1 ax + by + c = = −2 1 2 12 a b a + b and let the foot of perpendicular from a point (x 1, y 1) on the line ax + by + c = 0 is ( x 2, y 2) then x2 − x 1 y2 − y 1 ax + by + c = = − 1 2 12 a b a + b
CENTRES CONNECTED WITH A TRIANGLE (w.r.t. ABC , where A ( x 1 , y 1), B ( x 2, y 2), C ( x 3, y 3), BC = a, CA = b & AB = c ). Centroid : e point A
of concurrency of the medians of a triangle is called the centroid E 2 F : of the triangle. The G 1 centroid of a triangle divides each median B C D 1:1 in the ratio 2 : 1. e coordinatesofcentroid x + x + x y + y + y3 are given by G 1 2 3 , 1 2 3 3 Orthocentre : The point of concurrency of the altitudes of a triangle is called the F E orthocentre of the triangle. H The coordinates of the orthocentre are given by D x tan A + x2 tan B + x3 tan C B C H 1 , tan A + tan B + tan C y1 tan A + y2 tan B + y3 tan C tan A + tan B + tan C
The triangle formed by joining the feet of altitudes in a triangle is called the orthic triangle. Here DEF is the orthic triangle of ABC . Incentre : The point of concurrency of the internal bisectors A of the angles A/2 A/2 of a triangle N M is called the + I incentre of the : triangle. The C / 2 B / 2 coordinates of B C/2 B/2 C L the incentre are c:b given by ax + bx2 + cx3 ay1 + by2 + cy3 I 1 a+b+c , a + b + c Excentre : Coordinate of excentre opposite to A A is given by BL = c , c LC b b L AI 1 b+c B Also =− C I1L a
Remarks : • Circumcentre O, centroid G and orthocentre H of a ABC are collinear. G divides OH in the ratio 1 : 2, i.e. OG : GH = 1 : 2 • In an isosceles triangle centroid, orthocentre, incentre and circumcentre lie on the same line and in an equilateral triangle all these four points coincide.
c
b a
−ax1 + bx2 + cx3 , −a + b + c −ay1 + by2 + cy3 −a + b + c
I 1
POSITION OF A GIVEN POINT RELATIVE TO A GIVEN LINE & CONCURRENCY e figure shows a point P (x 1, y 1) lying above a given line. If an ordinate is dropped from P to meet the line L at N , then the x-coordinate of N will be x 1. Putting x = x 1 in the equation ax + by + c = 0 (ax + c) gives y-coordinate of N = − 1 b If P (x 1, y 1) lies above the line, then we have
y 1 − i.e.
I 1
and similarly for excentres ( I 2 & I 3 ) o p p o s i t e t o ΑB and ΑC are given by ax − bx2 + cx3 ay1 − by2 + cy3 I 2 1 , a−b+c a − b + c
ax1 + bx2 − cx3 ay1 + by2 − cy3 , a +b−c a + b − c
I 3
Circumcentre : The point of concurrency of C 2 n i s the perpendicular + C 2 B 2 bisectors of n i O : s the sides of a A 2 n i 2 B s triangle is called B L circumcentre of sin2C : sin2B the triangle. The coordinates of the circumcentre are given by x sin 2 A + x2 sin 2 B + x3 sin 2C O 1 sin 2 A + sin 2 B + sin 2C ,
A
C
y1 sin 2 A + y2 sin 2 B + y3 sin 2C sin 2 A + sin 2 B + sin 2C
(ax1 + c) (ax + c) 0 i.e. y 1 + 1 b b
(ax1 + by1 + c) b
0 i.e.
L(x1 , y 1 ) b
0 .... (i)
Hence, if P (x 1, y 1) satisfies equation (i), it would mean that P lies above the line ax + by + c = 0, L(x1 , y 1 ) and if 0 , it would mean that P lies b below the line ax + by + c = 0. Remark : If (ax 1 + by 1 + c) and (ax 2 + by 2 + c) have same signs, it implies that (x 1, y 1) and (x 2, y 2) both lie on the same side of the line ax + by + c = 0. If the quantities ax 1 + by 1 + c and ax 2 + by 2 + c have opposite signs, then they lie on the opposite sides of the line. Concurrency of straight lines : e condition for three lines a1x + b1 y + c1 = 0, a2x + b2 y + c2 = 0, a3x + b3 y + c3 = 0 to be concurrent is a1 b1 c1 (i)
a2
b2
c2
a3
b3
c3
=0
MATHEMATICS TODAY | NOVEMBER ‘15
37
If tanq > 1 then the bisector taken is the bisector of the obtuse angle and the other one will be the bisector of the acute angle. If 0 < tanq < 1 then the bisector taken is the bisector of the acute angle and the other one will be the bisector of the obtuse angle. If two lines are a1x + b 1 y + c1 = 0 and a2x + b 2 y + c2 = 0, then
(ii) ere exist three constants l , m, n (not all zero at the same time) such that lL1 + mL2 + nL3 = 0, where L1 = 0, L2 = 0 and L3 = 0 are the three given straight lines. (iii) The three lines are concurrent if any one of the lines passes through the point of intersection of the other two lines. ANGLE BISECTORS OF TWO STRAIGHT LINES 0 Angle bisector = c + is the locus of y b a point which + x moves in such a Q M a way so that P D its distance O fro m t wo a 2 x + b intersecting N y + 2 c 2 = lines remains 0 same. The equations of the two bisectors of the angles between the lines a1x + b1 y + c1 = 0 and a2x + b2 y + c2 = 0 are a1x + b1 y + c1 a2 x + b2 y + c2
a1x + b1 y + c1
1
1
=
If the two given lines are not perpendicular i.e., a 1 a 2 + b 1 b 2 0 and not parallel i.e., a 1 b 2 a 2 b 1 , then one of these equations is the equation of the bisector of the acute angle between two given lines and the other is the obtuse angle between two given lines. Remark : Whether both given lines are perpendicular or not but the angular bisectors of these lines will always be mutually perpendicular. The bisectors of the acute and the obtuse angles : Take one of the lines and let its slope be m 1 and take one of the bisectors and let its slope be m 2 . If q be the acute angle between them, then find tan q =
38
a22 + b22
m1 − m2 1 + m1m2
MATHEMATICS TODAY | NOVEMBER ‘15
a2 x + b2 y + c2
a12 + b12
1
a12 + b12
=
a22 + b22
will represent the equation of the bisector of the acute or obtuse angle between the lines according as a1a2 + b1b2 is negative or positive. e equation of the bisector of the angle which contains a given point : e equation of the bisector of the angle between the two lines containing the point (a, b) is a1x + b1 y + c1 a2 x + b2 y + c2 =
a12 + b12 or
a1x + b1 y + c1
a22 + b22 =−
a2 x + b2 y + c2
a12 + b12
a22 + b22
according as a1a + b1b + c1 and a2a + b2b + c2 are of the same signs or of opposite signs. For example, the equation of the bisector of the angle containing the origin is given by a1x + b1 y + c1 a2 x + b2 y + c2 =+
a12 + b12
a22 + b22
for same sign of c1 and c2 (for opposite sign take –ve sign in place of +ve sign) Remark : (i) If c1c2(a1a2 + b1b2) < 0, then the origin will lie in the acute angle and if c1c2(a1a2 + b1b2) > 0, then the origin will lie in the obtuse angle. (ii) Equation of straight lines passing through P (x 1, y 1) and equally inclined with the lines a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 are those which are parallel to the bisectors between these two lines and passing through the point P . LOCUS When a point moves in a plane under certain geometrical conditions, the point traces out a path. is path of the moving point is called its locus. Equation of locus : e equation to a locus is the relation which exists between the coordinates of
any point on the path, and which holds for no other point except those lying on the path. In other words equation to a curve (or locus) is merely the equation connecting the x and the y coordinates of every point on the curve. Procedure for finding the equation of the locus of a point : (i) If we are finding the equation of the locus of a point P , assign coordinates (h, k) or (x 1, y 1) to P . (ii) Express the given conditions in terms of the known quantities to facilitate calculations. We sometimes include some unknown quantities known as parameters. (iii) Eliminate the parameters. So that the eliminant contains only h, k and known quantities. If h and k coordinates of the moving point are obtained in terms of a third variable ‘t ’ called the parameter, eliminate ‘t ’ to obtain the relation in h and k and simplify this relation. (iv) Replace h by x , and k by y , in the eliminant. e resulting equation would be the equation of the locus of P .
TRANSFORMATION OF AXES Changes of Axes (Shifting of origin without rotation of axes) : Let P (x , y ) with respect to axes OX and OY . L e t O (a , b ) with respect to axes OX and OY and let P (x , y ) with respect to axes O X and O Y where OX a n d O X a r e parallel and OY and OY are parallel. then x = x + a, y = y + b or x = x – a, y = y – b Thus if origin is shifted to point (a, b) without rotation of axes, then new equation of curve can be obtained by putting x + a in place of x and y + b in place of y . Rotation of the axes (To change the direction of the axes of coordinates, without changing the origin, both systems of co ordinates being rectangular) : Let OX , OY be given rectangular axes with respect to which the coordinates of a point P are
(x , y ) . Suppose t h a t O U , O V are the two perpendicular lines obtained by rotating OX , OY respectively through an angle q in the counter-clockwise sense. We take OU , OV as a new pair of coordinate axes, with respect to which the coordinates of P are (x , y ), then x = x cosq – y sin q y = x sin q + y cosq PAIR OF STRAIGHT LINES The general equation of second degree ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0 represents a a h g
pair of straight lines if h g
b
f
f
c
=0
abc + 2 fgh – af 2 – bg 2 – ch2 = 0 The homogeneous second degree equation ax 2 + 2hxy + by 2 = 0 represents a pair of straight lines through the origin. If lines through the origin whose joint equation i s a x 2 + 2h xy + by 2 = 0 , a r e y = m 1 x and y = m2x , then y 2 – (m1 + m2)xy + m1m2x 2 = 0 and a 2h y 2 + xy + x 2 = 0 are identical. If θ is the angle b b between the two lines, then
tan q =
(m1 + m2 )2 − 4m1m2 1 + m1m2
2 h2 − ab = a +b
Note : (i) (a) a + b = 0 lines are perpendicular. (b) h2 > ab lines are real & distinct. (c) h2 = ab lines are coincident. (d) h2 < ab lines are imaginary with real point of intersection i.e. (0, 0). (ii) If y = m1x & y = m2x be the two equations represented by ax 2 + 2hxy + by 2 = 0, then 2h a m1 + m2 = − & m1m2 = b b . (iii) e equation to the straight lines bisecting the angle between the straight lines
ax 2 + 2hxy + by 2 = 0 is
x 2 − y 2 a −b
=
xy h
MATHEMATICS TODAY | NOVEMBER ‘15
39
(iv) A homogeneous equation of degree n represents n straight lines (in general) passing through origin.
Joint equation of pair of lines joi ning the ori gi n and the points of intersection of a line and a curve If the line lx + my + n = 0, ((n 0) i.e .
Y
(– , 4) (7, ) (– , 5) (7, ) (– , 4) (5, 7) (7, ) (– , 4) (4, 5) (7, )
A line has intercepts a, b on axes. When the axes are rotated through an angle , the line makes equal intercepts on axes then tan = a +b a −b b a (a) (b) (c) (d) a −b a +b a b 6. e values of m for which the system of equations 3x + my = m and 2x – 5 y = 20 has a solution satisfy the conditions x > 0, y > 0 are given by the set 5.
A
B
X
the line not passing O through origin) cuts the curve ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0 at two points A and B, then the joint equation of straight lines passing through A and B and the origin is given by homogenizing the equation of the curve by the equation of the line i.e. ax 2 + 2hxy + by 2 + (2 gx + 2 fy )
(a) (b) (c) (d)
lx + my + −n 2 lx + my c −n = 0
17 m : m −13 (b) m : m 2 2 −13 17 (c) m : m or m 2 2 −15 (d) m : m 30 or m 2 7. If f (x + y ) = f (x ) f ( y ) x , y R and f (1) = 2, then area enclosed by 3|x | + 2| y | 8 is (in sq.units) (a)
is the equation of the lines OA and OB. PROBLEMS SECTION-I Single Correct Answer Type
e line x + y = 1 meets x -axis at A and y -axis at B. P is the mid-point of AB. P 1 is the foot of the perpendicular from P to OA; M 1 is that from P 1 to OP ; P 2 is that from M 1 to OA; M 2 is that from P 2 to OP ; P 3 is that from M 2 to OA and so on. If P n denotes the nth foot of the perpendicular on OA from M n – 1, then OP n = 1 1 1 1 (a) (b) (c) (d) 2 2 2n 2n/2 1.
2. e orthocentre of the triangle formed by the lines x + y = 1, 2x + 3 y = 6 and 4x – y + 4 = 0 lies in
(a) I quadrant (c) III quadrant
(b) II quadrant (d) IV quadrant
A line passes through (2, 0). e slope of the line, for which its intercept between y = x – 1 and y = –x + 1 subtends a right angle at the origin, is/are 1 (a) (d) 1 3 (b) − 3 (c) 3 3.
A straight line passes through the point of intersection of x – 2 y – 2 = 0 and 2 x – by – 6 = 0 and the origin then the complete set of values of b for which the acute angle between this line and y = 0 is less than 45° are 4.
40
MATHEMATICS TODAY | NOVEMBER ‘15
(a) f (4)
(b)
1 1 1 f (6) (c) f (6) (d) f (5) 2 3 3
If 9x 2 + 2hxy + 4 y 2 + 6x + 2 fy – 3 = 0 represents two parallel lines, then (a) h = 6, f = 2 (b) h = –6, f = 2 (c) h = 6, f = –2 (d) none of these 8.
9. If the area of the rhombus enclosed by the lines lx ± my ± n = 0 be 2 square units, then (a) l , 2m, n are in G.P (b) l , n, m are in G.P (c) lm = n (d) ln = m 10. If P is a point which moves inside an equilateral triangle of side length ‘ a’ such that it is nearer to any
angular bisector of the triangle than to any of its sides, then the area of the region in which P lies is __ sq. units.
3 − 1 (a) a 3 + 1 3 − 1 (c) 3a2 3 + 1 2
(b)
3a2 2
3 − 1 3 + 1
(d) a2
11. e area of the triangle formed by the line x + y = 3 and the angular bisectors of pair of straight lines x 2 – y 2 + 2 y = 1 is
(a) 8 sq.units (c) 4 sq.units
(b) 6 sq.units (d) none of these
Triangle is formed by the lines x + y = 0, x – y = 0 and lx + my = 1. If l and m vary subject to the condition l 2 + m2 = 1, then the locus of its circumcentre is (a) (x 2 – y 2)2 = x 2 + y 2 (b) (x 2 + y 2)2 = x 2 – y 2 (c) (x 2 + y 2)2 = 4x 2 y 2 (d) (x 2 – y 2)2 = (x 2 + y 2)2 12.
A piece of cheese is located at (12, 10) in a coordinate plane. A mouse is at (4, –2) and is running up the line y = –5x + 18. At the point (a, b), the mouse starts getting farther from the cheese rather than closer to it. e value of (a + b) is (a) 6 (b) 10 (c) 18 (d) 14 13.
If h denote the A.M., k denote the G.M. of the intercepts made on axes by the lines passing through (1, 1) then (h, k) lies on (a) y 2 = 2x (b) y 2 = 4x (c) y = 2x (d) x + y = 2xy 14.
A straight rod of length 3l units slides with its ends A, B always on the x and y axes respectively then the locus of centroid of OAB is (a) x 2 + y 2 = 3l 2 (b) x 2 + y 2 = l 2 (c) x 2 + y 2 = 4l 2 (d) x 2 + y 2 = 2l 2 15.
ABC , the coordinates of B are (0, 0) p AB = 2, ABC = and the mid point of BC is (2,0). 16.
In a
3 e centroid of triangle is
1 3 (a) , 2 2 4 + 3 1 (c) , 3 3
5 1 (b) , 3 3 (d)
4 − 3 1 3 , 3
A ray travelling along the line 3x – 4 y = 5 aer being reflected from a line 'l ' travels along the line 5x + 12 y = 13. en the equation of the line 'l ' is (a) x + 8 y = 0 (b) x – 8 y = 0 (c) 32x + 4 y + 65 = 0 (d) 32x – 4 y + 65 = 0 17.
All points inside the triangle formed by A(1, 3), B(5, 6), C (–1, 2) will satisfy (a) 2x + 2 y 0 (b) 2x + y + 1 0 (c) 2x + 3 y – 12 0 (d) –2x + 11 0 18.
A system of line is given as y = mix + ci, where mi can take any value out of 0, 1, –1 and when mi is positive then ci can be 1 or –1, when mi equal 0, ci can be 0 or 1 and when mi equals – 1, ci can take 0 or 2. en the area enclosed by all these straight lines is 3 3 (a) ( 2 − 1) sq. units (b) sq. units 2 2 19.
(c)
3 sq. units 2
(d)
3 sq. units 4
SECTION-II Multiple Correct Answer Type
P is a point inside a ABC of area K (K > 0). e lengths of perpendiculars drawn to the sides BC , CA, AB a b c of lengths a, b, c are respectively P 1, P 2, P 3. + + P 1 P 2 P 3 is minimum when (a) P is incentre of ABC (b) P is equidistant to all the 3 sides 2K (c) P1 = P2 = P 3 = a +b+c 20.
(d) P is orthocentre of ABC Equations (b – c)x + (c – a) y + (a – b) = 0 and – c3)x + (c3 – a3) y + (a3 – b3) = 0 will represent the same line if (a) b = c (b) c = a (c) a = b (d) a + b + c = 0 21.
(b3
e equation of the line passing through (2, 3) and making an intercept of 2 units between the lines y + 2x = 5 and y + 2x = 3 is (a) 5x – 4 y + 2 = 0 (b) 3x + 4 y = 18 (c) x = 2 (d) y = 3 22.
e equation of the diagonal of the square formed by the pairs of lines xy + 4x – 3 y – 12 = 0 and xy – 3x + 4 y – 12 = 0 is (a) x – y = 0 (b) x + y + 1 = 0 (c) x + y = 0 (d) x – y + 1 = 0 23.
Under rotation of axes through q, x cos + y sin = P changes to x cos + y sin = P, then (a) cos = cos( – q) (b) cos = cos( – q) (c) sin = sin( – q) (d) sin = sin( – q) 24.
Sides of a rhombus are parallel to the lines x + y – 1 = 0 and 7x – y – 5 = 0. It is given that diagonals of the rhombus intersect at (1, 3) and one vertex ‘ A’ of the rhombus lies on the line y = 2x . en the coordinates of the vertex A are 7 14 8 16 (a) , (b) , 15 15 5 5 6 12 4 8 (c) , (d) , 5 5 15 15 25.
Equations of the diagonals of a rectangle are y + 8x – 17 = 0 and y – 8x + 7 = 0. If the area of the rectangle is 8 sq. units, then the equation of the sides of the rectangle is/are 26.
MATHEMATICS TODAY | NOVEMBER ‘15
41
(a) x = 1 (c) y = 9
SECTION-IV
(b) x + y = 1 (d) x – 2 y = 3
Matrix-Match Type
27. e lines (m – 2)x + (2m – 5) y = 0 ; (m – 1)x + (m2 – 7) y – 5 = 0 and x + y – 1 = 0 are (a) concurrent for three values of ‘m’ (b) concurrent for one value of ‘m’ (c) concurrent for no value of ‘m’ (d) are parallel for m = 3 28. If x 2 + 2hxy + y 2 = 0 (h 1) represents the equations
of the straight lines through the origin which make an angle with the straight line y + x = 0, then (b) cos =
1+ h 2h
(c) m1 + m2 = –2sec2 (d) cot =
h +1 h −1
(a) sec2 = h
Paragraph for Question No. 29 to 31 In a PQR, with PQ = r , QR = p, PR = q, the cosine
values of the angles are given by q 2 + r 2 − p2 p2 + r 2 − q2 cos P = ; cos Q = 2qr 2 pr p2 + q2 − r 2 cos R = and the area of PQR is 2 pq
1 1 1 pq sin R = qr sin P = pr sin Q 2 2 2 Let ABCD be a parallelogram whose diagonal equations are AC x + 2 y – 3 = 0 ; BD 2x + y – 3 = 0. If AC = 4 units, and area of ABCD = 8 sq units, and BPC is acute where P is point of intersection of diagonals AC and BD, then =
29. e length of other diagonal BD is
11 2 10 20 (b) 2 (c) (d) 3 3 3 30. e length of side AB is equal to (a)
(a)
58
4 10 2 10 (a) (b) (c) 8 10 (d) 3 3 3
10 3
MATHEMATICS TODAY | NOVEMBER ‘15
Column II
(A) The distance between the lines (p)
2
(x + 7 y )2 + 4 2 (x + 7 y ) − 42 = 0 is (B) If the sum of the distance of a (q) point from two perpendicular lines in a plane is 1, then its locus is |x | + | y | = k, where k is equal to (C) If 6x + 6 y + m = 0 is acute angle (r) bisector of lines x + 2 y + 4 = 0 and 4x + 2 y – 1 = 0, then m is equal to
7
3
1
33. Match the following:
Comprehension Type
42
Column I
(D) Area of the triangle formed by (s) the lines y 2 – 9xy + 18x 2 = 0 and y = 6 is
SECTION-III
4 58 2 58 58 (b) (c) (d) 3 3 3 31. e length of BC is equal to
32. Match the following Column I with Column II
Column I
Column II
(A) T h e v a lu e o f k for which 4x 2 + 8xy + ky 2 = 9 is the equation of a pair of straight lines, is (B) If the sum of the slopes of the lines given by x 2 – 2Cxy – 7 y 2 = 0 is four times their product, then the value of C is (C) If the gradient of one of the lines x 2 + hxy + 2 y 2 = 0 is twice that of the other, then h = (D) If the lines ax 2 + 2hxy + by 2 = 0 are equally inclined to the lines ax 2 + 2hxy + by 2 + (x 2 + y 2) = 0, then the value of can be
(p)
3
(q)
–3
(r)
2
(s)
4
1 (i = 1, 2, 3) represent three straight mi lines whose slopes are the roots of the equation 2m3 – 3m2 – 3m + 2 = 0, then Match the following: 34. If y = mi x +
Column I
(A) A l g e b r a i c s u m intercepts made by on x -axis is (B) A l g e b r a i c s u m intercepts made by on y -axis is
Column II
o f t h e (p) (4 2 + 9 5 ) the lines 4 o f t h e (q) the lines
3/2
(C) Sum of the distances of the (r) lines from the origin is
OP 1 = P 1P = 1/2
–21/4
(D) Sum of the lengths of the (s) (5 2 + 9 5 ) lines intercepted between the 10 coordinate axes is SECTION-V
Integer Answer Type
35. ABCD is a square of side length 1 unit. P and Q are
points on AB and BC such that PDQ = 45°. Find the perimeter of PBQ.
a OAB formed by the points O(0, 0), A(2, 0), B(1, 3 ) , P (x , y ) be any arbitrary interior point of OAB moving in such a way that d (P , OA) + d (P , AB) + d (P , OB) = 3 , where d (P , OA), d (P , AB), d (P , OB) represents perpendicular distances of P from the sides OA, AB and OB respectively. If area of the region representing all possible positions of P is 'k' then k 3 = 36. Consider
37. In a ABC , AB is parallel to y-axis, BC is parallel to
x-axis, centroid is at (2, 1). If median through C is x – y = 1, then the slope of median through A is 38. If the orthocentre of the triangle formed by
2x + 3 y – 1 = 0, x + 2 y – 1 = 0, ax + by – 1 = 0 is at the b−a = origin then 4 39. e area of the rhombus ABCD is 24. e equation of the diagonal BD is 4x + 3 y + 2 = 0 and A = (3, 2). e length of the side of the rhombus is 40. In ABC , the equation of altitudes AM and BN are
x + 5 y – 3 = 0, x + y – k = 0.If the altitude CL is given by 3x – y – 1 = 0, then k = 41. e coordinate axes are rotated through an angle
q
about the origin in anticlockwise sense. If the equation 2x 2 + 3xy – 6x + 2 y – 4 = 0 changes to aX 2 + 2hXY + bY 2 + 2gX + 2 fY + c = 0, then (a + b) is equal to 42. Origin is shied to (1, 2) then the equation
y 2 – 8x – 4 y + 12 = 0 changes as y 2 = 4ax . en a = 43. e centroid of the triangle formed by (a, b), (b, c),
(c, a) is the origin and a3 + b3 + c3 = kabc, then k is SOLUTIONS
y B
Equation of line OP is y = x We have, (OM n – 1)2 = (OP n)2 + (P n M n – 1)2
P
M 1 M
O P P P = 2(OP n)2 = 2 pn2 (say) Also, (OP n – 1)2 = (OM n – 1 )2 + (P n – 1 M n – 1)2
1
A
x
(OP n – 1)2 = (OM n – 1)2 + (P n – 1 M n – 1 )2 1 1 1 = 2 pn2 + pn2−1 pn2 = pn2−1 pn = pn−1 4 2 2 1 1 1 1 Opn = pn = pn−1 = 2 pn−2 = ..... = n−1 p1 = n 2 2 2 2 2. (a) : Coordinates of A and
B are (–3, 4) and
y
− 3 , 8 . 5 5
Let orthocentre be P (h, k)
0 = 4 +
y
–
A
C x 4 B
2 x +
3
y = en, (slope of PA) × (slope 6 x x + O of BC ) = –1 y = 1 k−4 4 = −1 h+3 4k – 16 = –h –3 h + 4k = 13 ... (i) and slope of PB × slope of AC = – 1 8 k− 5 − 2 = −1 5k − 8 2 = 1 3 3 5h + 3 3 h+ 5 10k – 16 = 15h + 9 15h – 10k + 25 = 0 ... (ii) 3h – 2k + 5 = 0
3 22 Solving eqs. (i) and (ii), we get h = , k = 7 7 Hence, orthocentre lies in I quadrant. 3. (c) : e joint equation of straight lines
y = x – 1 and y = –x + 1 is (x – y – 1)(x + y – 1) = 0 ... (i) x 2 – y 2 – 2x + 1 = 0 Let equation of line passes through (2, 0) be ... (ii) y = m(x – 2) By homogenizing equation (i) with the help of line (ii), 2
mx − y mx − y 0 we get x − y − 2 x 2m + 2m = 2
Q
2
coefficient of x 2 + coefficient of y 2 = 0 1 m= 3
1. (b) : x + y = 1 meets x -axis at A(1, 0) and y -axis at
B(0, 1). The coordinates of P are (1/2, 1/2) and PP 1 is perpendicular to OA.
4. (d) : As line passes through the point of intersection
of x – 2 y – 2 = 0 and 2x – by – 6 = 0 MATHEMATICS TODAY | NOVEMBER ‘15
43
It can be represented as (x – 2 y – 2) + (2x – by – 6) = 0 As it passes through the origin –2 – 6 = 0 = –3 Equation of the line is –x + (6 – b) y = 0 1 Its slope is 6−b As its angle with y = 0 is less than p/4 1 −1 1 6−b
0, −n , −n , 0 , 0, n , n , 0 m l m l Area = 1 2n 2n = 2 n2 = lm 2 m l 10. (b) : Shaded area is the region traced by P
= ABC – 3 ABD
6 – b > 1 or b – b < –1 b < 5 or b > 7 But b 4 (as the lines intersect) b (– , 4) (4, 5) (7, ) x y 5. (b) : Equation of the line is + = 1 a
3 2 3 − 1 a 3 + 1 2
25m 2m − 60 , y = 2m + 15 2m + 15 But x > 0, y > 0 25m > 0, 2m + 15 > 0, 2m – 60 > 0 or 25 m < 0, 2m + 15 < 0, 2m – 60 < 0
Let G(x , y ) =
−15 y
a , 3
(–8/3, 0)
O
x 2 + y 2 = l 2
16. (b) : Let A(h, k) then cos60 = (8/3, 0)
x
(0, –4) f (x ) = 2x 8. (a) : Since the given equation represents a pair of parallel lines, we have h2 = ab h = ±6 9 h 3
Condition for pair of lines h 4 3 f
f
=0
−3
108 ± 36 f – 9 f 2 – 144 = 0 f = 2 and h = 6 f = –2, h = –6 9. (b) : By solving the sides of the rhombus, the vertices
are MATHEMATICS TODAY | NOVEMBER ‘15
mouse
b , a = 3x , b = 3 y 3
(0, 4)
7. (c) : Area = 4
C
12. (a)
a2 + b2 = 9l 2
2
I
B
b=8 a + b = 10 14. (a) : a = x -intercept, b = y -intercept 2h = a + b, k2 = ab x y + = 1, substitute (1, 1) a b 1 1 + = 1 a + b = ab a b 2h = k2 y 2 = 2x 15. (b) : Let OA = a, OB = b, AB = 3l A = (a, 0), B = (0, b)
x =
44
=
D
13. (b) : a = 2,
b
6. (d) : Solving the equations, we get
1 8 4 2 3 64 26 = = 3 3
=
11. (d)
Transformed equation is 1 1 (x cos − y sin ) + (x sin + y cos ) = 1 a b Since, intercepts are equal x -coefficient y -coefficient a −b tan = a +b
m 30 or m
A
3 2 3 a a − a tan 15 4 2 2
h 2
h =1
k k= 3 2 5 3 A(1, 3 ) and Centroid = , 3 3 17. (b) : e line ‘l ’ can be any one of the bisectors of the angles between the lines 3x – 4 y = 5 and 5x + 12 y = 13 sin60 =
3x − 4 y − 5 5x + 12 y − 13 = 5 13 x – 8 y = 0, 32x + 4 y – 65 = 0 18. (b) : L1 2x + 2 y = 0 L1(1, 3) > 0 (a) is wrong. L2 2x + y + 1 = 0 L2(1, 3) > 0 L2(5, 6) > 0 (b) is true. L2(–1, 2) > 0
Angular bisectors,
y
19. (c) : Lines are y = 1,
=
y = 0, y = –x , y = –x + 2, y = x +1, y = x – 1 Area of OABCDEO = area of OBGF 3 3 = 1 = sq. units 2 2 20. (a, b, c, d) : Given,
y =
a b + P 1 P 2
When y = minimum.
+
23. (a, b) : (x – 3)( y + 4) = 0, (x + 4)( y – 3) = 0
1 +
x
y
C G
B A
y = 1
D (3/2, 1/2) x F
E
x
y = x – 1
O
y = –x + 2 y = –x
1 (aP1 + bP2 + cP3 ) = K 2
c is minimum. P 3
1 a b (aP1 + bP2 + cP3 ) + 2K P 1 P 2
c + is P 3
2 2 2 P 2 P 3 P 1 P 2 + + + + + a b c ab bc but y = P P P + P 2K 2 1 3 2 P P ca 1 + 3 P 3 P 1 1 2 2 2 (a + b + c + 2ab + 2bc + 2ac) 2K
(a + b + c)2 y 2K
y is minimum when
P 1 P 2
=
P 2 P 1
=
P 2 P 3
=
P 3 P 2
=
P 1 P 3
=
P 3 P 1
=1
i.e., when P 1 = P 2 = P 3 P is incentre of ABC .
21. (a, b, c, d) : e two lines will be identical if t here
exists some real number k, such that b3 – c3 = k(b – c), c3 – a3 = k(c – a) and a3 – b3 = k(a – b) b – c = 0 or b2 + c2 + bc = k, c – a = 0 or c2 + a2 + ca = k and a – b = 0 or a2 + b2 + ab = k i.e., b = c or c = a or a = b Next b2 + c2 + bc = c2 + a2 + ca b2 – a2 = c(a – b) Hence, a = b or a + b + c = 0 1 2 Let slope of required line be m Slope of given lines = –2
24. (a, c) : x cos + y sin = P
Axis rotated through an angle ‘q’ Transformed equation is cos(x cosq – y sin q) + sin(x sin q + y cosq) = P or x cos( – q) + y sin( – q) = P x cos + y sin = P cos = cos( – q), sin = sin( – q) 25. (a, c) : It is clear that diagonals of the rhombus will be parallel to the bisectors of the given lines and will pass through (1, 3). Equations of bisectors of the given lines are 7 x − y − 5 x + y −1
= 5 2
2
2x – 6 y = 0 and 6x + 2 y = 5 erefore, the equations of diagonals are x – 3 y + 8 = 0 and 3x + y – 6 = 0. us the required vertex will be the point where these lines meet the line y = 2x . Solving 8 16 these lines, we get possible coordinates as , 5 5 6 12 and , . 5 5 26. (a, c) : e intersection point of the given diagonals
3 is P , 5 2 Equation of angular bisectors of the diagonals are y + 8 x − 17 65
=
y − 8 x + 7
y – x + 7 = C
D
P (3/2, 5)
B
A
y + 8x – 1 7 = 0
65
3 2 Let length of BC be a and that of CD be b
x = and y = 5
a/2 a = =8 b/2 b Also ab = 8 a = 8, b = 1 So, equations of sides are y = 1, y = 9, x = 1 and x = 2. en tan q =
22. (b, c) : tan q =
−3 m+2 1 or = m= 1 − 2m 2 4 e lines are 3x + 4 y – 18 = 0, x – 2 = 0
e vertices are A (–4, –4), B (–4, 3), C (3, 3), D (3, –4) Diagonal AC is x = y ,Diagonal BD is x + y + 1 = 0
2
m − 2 2m − 5
2 5
27. (c, d) :
0
= m − 1 m2 − 7 −5 = 0 −1 1 1
(m – 3)(m2 – m + 2) = 0 For m = 3, the lines become parallel. MATHEMATICS TODAY | NOVEMBER ‘15
45
28. (a, b, c, d) : Let lines of x 2 + 2hxy + y 2 = 0 be given by y = m1x and y = m2x m1 + m2 = –2h Slope of y + x is –1
m1 + 1 m +1 , tan = 2 1 − m1 1 − m2
tan =
m + 1 m +1 tan = 1 and tan = − 2 1 − m1 1 − m2 (for +ve signs, both gives the same value but m1 m2) tan − 1 tan + 1 m1 = , m2 = tan + 1 tan − 1 m1 + m2 = –2sec2 h = sec2 1 1 cos2 = 2 cos2 − 1 = h h 1/2 h +1 1+ h cos = cot = 2h h −1
29. (c) 30. (a) 31. (b) Given, P be the point of intersection
−1
+2 3 3 2 tan q = = sin q = 1+1 4 5 Area of CPB 1 = PC PB sin q = 2 2 P 10 20 PB = BD = 3 3 A 100 4+ − AB2 2 58 −4 9 AB = cos( p − q) = = 10 5 3 22 3 2 10 Again from CPB, BC = 3 32. A - p; B - s; C - q; D - r 2
(A) (x + 7 y )
+ 4 2 (x + 7 y) − 42 = 0 (x + 7 y )[x + 7 y + 7 2 ] − 3 2 (x + 7 y ) − 42 = 0 (x + 7 y)[x + 7 y + 7 2 ] − 3 2 ( x + 7 y + 7 2 ) = 0 (x + 7 y + 7 2 )(x + 7 y − 3 2 ) = 0 x + 7 y + 7 2 = 0 and x + 7 y − 3 2 = 0
d = 46
7 2 +3 2 1 + 49
=
10 2 50
=2
MATHEMATICS TODAY | NOVEMBER ‘15
C
B
(B) Let two perpendicular lines
are coordinate axes. en, PM + PN = 1 h +k=1 Hence, the locus is x + y =1 But if the point lies in other quadrants also, then |x | + | y | = 1 Hence, value of k is 1. (C) Angle bisector between the lines x + 2 y +4 = 0 and 4x + 2 y – 1 = 0 is (−4 x − 2 y + 1) x + 2y + 4 = 1+ 4 16 + 4 (−4 x − 2 y + 1) x + 2 y + 4 = 2 2(x + 2 y + 4) = (−4 x − 2 y + 1) Since aa 1 + bb1 < 0, so +ve sign gives acute angle bisector. Hence, 2x + 4 y + 8 = –4x – 2 y + 1 6x + 6 y + 7 = 0 m = 7 (D) We have, y 2 – 9 xy + 18 x 2 = 0 or y 2 – 6xy – 3xy + 18x 2 = 0 y ( y – 6x ) – 3 x ( y – 6 x ) = 0 y – 3x = 0 and y – 6 x = 0 e third line is y = 6. erefore, area of the triangle formed by these lines, 0 0 1 1 1 = 1 6 1 = | 6 − 12 | = 3 sq. units 2 2 2 6 1 33. A – s; B – r; C – p, q; D – p, q, r, s (A) e equation represents pair of lines if (4)(k)(–9) – (–9)(4) 2 = 0 k = 4 (B) m1 + m2 = 4 m1m2 −2h 4a −2(−C ) 4 1 = = C = 2 −7 −7 b b (C) 2m2 + m2
=
− h
2m22
2
−h 1 1 = 2 = 6 2 2
, 2 h2 = 9 h = ±3 (D) e angular bisectors of ax 2 + 2hxy + by 2 + (x 2 + y 2) = 0 is h(x 2 – y 2) – (a – b)xy = 0 which are angular bisectors of ax 2 + 2hxy + by 2 = 0. e two pairs are equally inclined for any . 34. A - r, B - q, C - s, D - p Solving the equation 2m3 – 3m2 – 3m + 2 = 0, we get 2(m3 + 1) – 3 m(m + 1) = 0
(m + 1)(2m2 – 5m + 2) = 0 (m + 1)(2m – 1)(m – 2) = 0 m = –1, 1/2 or 2
37. (4) : Let B(a, b), C (c, b), A(a, d )
Equation of the given lines can be written as mi2 x – mi y = –1 (A) Algebraic sum of the intercepts made by the lines
on x -axis
= −
1 + 1 + 4 = − 21 = − 4 4 mi2 1
(B) Algebraic sum of the intercepts made by the lines
1 1 3 = −1 + 2 + = mi 2 2 (C) Let pi denote the perpendicular distance of the l ine from the origin, then on y -axis
pi
=
1 / mi
=
1 + mi2
1
pi = 1
=
1+1
+
4
2
+
1 + (1 / 4)
1
+
+
1/ 2 1+ 4
1 (5 2 + 9 5 ) 10
=
2 5 2 5 (D) l i = length of the line intercepted between the coordinates 2
−1 1 2 1 1 2 + m li = 1 + 1 + 16 + 4 + 16 + 4 mi i 5 (4 2 + 9 5 ) = 2 +2 5 + = 4
4
35. (2) : tanq1 = x
y (0, 1) D
and tanq2 = 1 – y
q1 + q2 = 45° tan q1 + tan q2 =1 1 − tan q1 tan q2
1
Since,
x + (1 − y ) =1 1 − x(1 − y )
45°
A (0, 0)
B(1, 0)
2 x y = 1 + x
…(i)
Now, perimeter = 1 − x + y + (1 − x ) By using (i), we get Perimeter = 2
+y
2
ar (PAB) = Since, the triangle is an equilateral triangle, 1 3 = 2(d( P, OA) + d( P, OB) + d( P, AB)) 2 For all positions of P d (P , OA) + d (P , OB) + d (P , AB) = 3
A 2
x
O
+ 3
y
= 1
D
C
16 + 9
=4
Area of ABD = AP BP =
=5
24 = 12 2
BP = 3
40. (1) : Solving the altitudes AM , CL
1 , 1 lies on x + y – k = 0 k =1 2 2 41. (2) : x = x cosq – y sin q, y = x sin q + y cos q
Orthocentre =
36. (3) : ar (OAB) = ar(OPA) + ar( OPB) +
3k = 3
x
12 + 6 + 2
AB = AP 2 + BP 2
2
k= 3
AP =
Q(1, y ) P (x , 0)
2 38. (4) : Solving 2x + 3 y = 1 x + 2 y = 1, A = (–1, 1) 1 = y Orthocentre = (0,0) 2 + x slope of altitude AD = –1 Equation of BC is x – y = k B Solving, x – y = k 1 + 2k , 1 − k x + 2 y = 1, we get B = 3 3 1− k , slope of AC = –2/3 Slope of OB = 1 + 2k −1 1− k 3 = k= 1 + 2k 2 8 1 Equation of BC is x − y + = 0 8 –8x + 8 y – 1 = 0 a = –8, b = 8 39. (5) : Let AC and BD intersect at P
C (1, 1)
2
a + c ,b 2 b + d E, (mid point of AB) is a, 2 b + d b− 2 = 1 b − d = 2 Given slope of CE = 1 c−a c −a b − d (b − d ) Slope of AD = =2 =4 a+c c−a −a en D, (mid point of BC ) is
3 4 4
Substitute x and y in the equation a = 2cos2q + 3cosq sinq, b = 2sin2q – 3cosq sinq a + b = 2 42. (2) : Transformed equation is ( y + 2) 2 – 8(x + 1) – 4( y + 2) + 12 = 0 y 2 = 8x y 2 = 4ax a = 2
a + b + c , a + b + c = (0, 0) 3 3 a + b + c = 0 a3 + b3 + c3 = 3abc k = 3
43. (3) : G =
nn MATHEMATICS TODAY | NOVEMBER ‘15
47
INDEFINITE & DEFINITE INTEGRATION *
ALOK KUMAR, B.Tech, IIT Kanpur
This column is aimed at Class XII students so that they can prepare for competitive exams such as JEE Main/Advanced, etc. and be also in command of what is being covered in their school as part of NCERT syllabus. The problems here are a happy blend of the straight and the twisted, the simple and the difficult and the easy and the challenging. INDEFINITE INTEGRATION
DEFINITION
Integration is the inverse process of differentiation. Let F ( x ) be a differentiable function of x such d [F (x)] = f ( x). en F (x ) is called the integral that dx of f ( x ). It is written as f ( x ) dx = F ( x) + C , where f (x ) is to be integrated, is called integrand and F (x ) is called the anti-derivative or primitive of f (x ), here C is known as constant of integration and can take any real value.
METHODS OF INTEGRATION
For finding the integral of complicated functions, generally these methods are used. Integration by substitution Integration by parts Integration by partial fractions Integration by Substitution: Direct Substitution n +1
( f (x )) ( f ( x ))n f ( x) dx = n +1
f (x )
f (x ) dx = 2
+ C
*
f (x) + C
Let f (x ) = t dt = f (x )dx dt = 2 t + C = 2 f (x) + C t
Standard Substitution
In some standard integrand or a part of it, we have standard substitution. List of standard substitution is as follows: Expression
Substitution
x 2 + a2 or
x2
+ a2
x = a tan q or a cot q
x 2 − a2 or
x2
− a2
x = a sec q or a cosecq
a2 − x 2 or
a2
− x2
x = a sin q or x = a cosq
a + x or a − x
Let f (x ) = t f (x )dx = dt n+ 1 n +1 t ( f (x )) n t dt = + C = + C n +1 n +1 f (x ) dx = loge | f (x )| + C f ( x )
Let f (x ) = t f (x )dx = dt 1 dt = log | t | + C = log | f (x )| + C t
x x 2 a2
n
x = a cos 2q
expression inside the bracket = t x = a cos 2 q + bsin 2 q
(x − a) (b − x) 1 1−
( x + a)
1 1 1+ n ( x + b) n
(n N , n 1)
Alok Kumar is a winner of INDIAN NATIONAL MATHEMATICS OLYMPIAD (INMO-91). He trains IIT and Olympiad aspirants.
48
MATHEMATICS TODAY | NOVEMBER ‘15
x+a x +b
= t
Integration by Parts If u and v be two functions of x , then integral of the product of these two functions is given by
A A1 A2 f (x ) 1 = + + + .... ........ + 1 2 g (x ) (x − a1) (x − a1 ) (x − a ) 1
du uv dx = u v dx − v dx dx dx
1.
In applying the above rule care has to be taken in the selection of the first function(u) and the second function (v ). Normally we use the following methods: If in the product of the two functions, one of the function is not directly integrable (e.g . lnx , sin– 1x , cos–1x , tan–1x etc.) then we take it as the first function and the remaining function is taken as the second function. e.g . In the integration of
x tan−1 x dx , tan−1 x
is taken as the first function and x as the second function.
2.
If there is no other function, then unity is taken as the second function. e.g . In the integration of
tan−1 x dx , tan−1 x is taken as the first function and 1 as the second function. 3.
......... +
If both of the functions are directly integrable, then the first function is chosen in such a way that the following preference order for the first function
(Inverse Trigonometrical, Logarithmic, Algebraic, Trigonometric, Exponential) In the above as ILATE e. g . In the integration of
x sin xdx , x is taken as the first function and sin x is taken as the second function. An important result
e x ( f (x) + f (x)) dx = e x f (x ) + C Integration by Partial fractions When integrand is a rational function i.e. of the form f (x ) , where f (x ) and g (x ) are the polynomial functions g (x ) of x , we use the method of partial fraction. For example, we can rewrite
1 1 1 as − (3x − 1)(3x + 2) 3(3x − 1) 3(3x + 2)
then we can put
+
B2 x + C 2
+ ........ (x 2 + b1x + c1 ) (x 2 + b1x + c1)2 B x + C 1 1 + ......... ......... + 1 2 (x + b1x + c1 )
Here A1, A2,........., A , ............, B1, B2,..........., B 1 1 .........,C 1, C 2,......., C 1 ...........are the real constants and these can be calculated by reducing both sides of the above equation as identity in polynomial form and then by comparing the coefficients of like powers. e constants can also be obtained by putting some suitable numerical values of x in both sides of the identity. If degree of f (x ) is more than or equal to degree of g (x ), then divide f (x ) by g (x ) so that the remainder has degree less than that of g (x ). ALGEBRAIC INTEGRALS Using the technique of standard substitution and integration by parts, we can derive the following formulae :
dx
1
x
dx
1
x −a
dx
1
x +a
a2 + x 2 = a tan−1 a + C x 2 − a2 = 2a ln x + a + C a2 − x 2 = 2a ln a − x + C
If degree of f ( x ) is less than degree of g ( x ) and g (x ) = (x − a1 )1 ..........(x 2 + b1x + c1 )1 .........,
B1x + C 1
dx a2 − x 2 dx x 2 + a2 dx x 2 − a2
x a
= sin−1 + C = ln x + x 2 + a 2 + C = ln x + x 2 − a 2 + C
x 2 a2 x a − x 2 + sin−1 + C 2 2 a x 2 2 x 2 + a 2 dx = x +a 2 a2 + ln(x + x 2 + a 2 ) + C 2
a2 − x 2 dx =
x
2
−a
2
dx =
x
2
x
2
2
−a −
a
2
2
2
2
ln(x + x − a ) + C
MATHEMATICS TODAY | NOVEMBER ‘15
49
Integrals of the form dx dx , , 2 ax 2 + bx + c ax + bx + c
ax2 + bx + c dx
2
b 4ac − b2 , bx + c = a x + + 2a 4a b Put x + = t and use the standard formula. 2a ax 2 +
Integrals of the form (ax + b)dx (ax + b)dx , , 2 2 cx ex f + + cx + ex + f
(ax + b)
cx 2 + ex + f dx
Here write ax + b = A(2cx + e) + B Find A and B by comparing the coefficients of x and constant term. Integrals of the form
(ax 2 + bx + c)dx (ax 2 + bx + c)dx , , 2 2 (ex + fx + g ) (ex + fx + g )
(ax2 + bx + c)
(ex 2 + fx + g ) dx
H e r e p u t a x 2 + b x + c = A ( e x 2 + f x + g ) + B(2ex + f ) + C, find the values of A, B and C by comparing the coefficients of x 2 , x and constant term. Integrals of the form 1 dx Hereput ax + b = t (ax + b) ex 2 + fx + g
Integrals of the form (ax + b)dx
(cx + d)
(ax2 + b)
ex 2 + fx + g
Here put (ax + b) = A(cx + e) + B, find the values of A and B by comparing the coefficients of x and constant term. Integrals of the form
(ax 2 + bx + c)dx gx 2 + hx + i
Here put ax 2 + bx + c = A(ex + f ) (2 gx + h) + B(ex + f ) + C , find the values of A, B and C by comparing the coefficients of x 2 , x and constant term. 50
MATHEMATICS TODAY | NOVEMBER ‘15
Integrals of the form dx
(ax2 + b)
(cx 2 + d) 1 Here 1st put x = and then the expression inside the t square root as y 2 . Integrals of the form
xm(a + bxn ) p dx ( p 0) Here we have the following cases : Case I : If p is a natural number, then expand (a + bx n) p by binomial theorem and integrate. Case II : If p is a negative integer and m, n are rational numbers, put x = t k, when k is the LCM of denominators of m and n. m +1 is an integer and p is a rational Case III : If n number, put (a + bx n) = t k, when k is the denominator of p. m +1 a + bx n p + is an integer, put Case IV : If n x n where k is the denominator of p. Integrals of the form
= t k ,
f (sin x , cos x ) g (sin x, cos x ) dx = R(sin x, cos x )dx , where f and g both are polynomials in sin x and cos x . Here we can convert them in algebraic form by putting x tan = t aer writing 2 x x 2 tan 1 − tan2 2 and cos x = 2 sin x = x x 1 + tan2 1 + tan2 2 2
(ex + f )
(cx 2 + d)
Here put cx 2 + d = t 2 .
Here in each case, write
Integrals of the form x dx
Integrals of the form
p sin x + q cos x + r dx a sin x + b cos x + c Here put p sin x + q cosx + r = A(a sinx + b cosx + c) + B(a cosx – b sinx ) + C, values of A, B and C can be obtained by comparing the coefficients of sinx , cosx
and constant term by this technique. e given integral becomes sum of 3 integrals in which first two are very x easy and in 3rd , we can put tan = t 2 Integrals of the form
(sinm x cosn x )dx
a
Now F (a) = f (t )dt + c = 0 + c = c a
So, F (x) = f (t )dt + F(a)
(i) If one of them is odd, then substitute for term of even power. (ii) If both are odd, substitute either of them. (iii) If m, n are both even, use trigonometric identities only. If m and n are rational numbers and
•
a
x
hen m, n N
•
x
F(x) = f (t )dt + c
m + n − 2 2
is a negative integer, then substitute cotx = p or tanx = p which ever is found suitable.
a b
F (b) = f (t )dt + F(a) a
b
F (b) − F(a) = f (t)dt a
Hence, if f ( x )dx = g ( x) + c, b
then f (x )dx = g ( x) a
GENERAL PROPERTY OF DEFINITE INTEGRAL a
b
f (x)dx is the integration of f ( x ) w.r.t. x with a
PERIODIC PROPERTIES OF DEFINITE INTEGRAL If f (x ) is a periodic function with period p then,
x = a as lower limit and x = b as upper limit. GEOMETRICAL INTERPRETATION OF DEFINITE INTEGRAL Let f (x ) be a function defined on a closed interval b
np
0
p
0
f ( x ) dx = n f ( x)dx n I
Proof :Let I =
f (x)dx represents the algebraic sum a
of the areas of the region bounded by the curve y = f (x ), x –axis and the lines x = a, x = b. e value of the definite integral may be positive, zero or negative. y +
y = f(x ) +
– –
–
x x=b
FUNDAMENTAL THEOREM OF CALCULUS If f ( x ) is a continuous function on [a, b], then d x f (t )dt = f (x) ( x [a, b]) dx a
Now if we take F (x ) = f ( x ) , then by the above theorem x d F ( x) − f (t )dt = F ( x) − f ( x) = 0 dx a x
F(x) − f (t )dt = constant = c (say) a
= g (b) − g (a)
If f ( x ) dx = F ( x) + C, then f (x ) dx = F (b) − F(a)
DEFINITION
x=a
a
b
DEFINITE INTEGRATION
[a, b]. en
b
np
p
2 p
0
0
p
f (x)dx = f ( x)dx + f ( x)dx
3 p
np
2 p
n−1 p
+ f (x)dx + ....... + f f (x)dx Let I1 =
kp
f (x)dx
k −1 p
Put x = (k – 1) p + t hen, x → (k – 1) p, t → 0 and x → kp, t → p p
p
p
0
0
0
I1 = f (k − 1) p + t dt = f (t )dt = f (x)dx
If f (x ) is a periodic function with period p, then
np
p
mp
0
f (x)dx = (n − m) f (x)dx n, m I
DIFFERENTIATION OF DEFINITE INTEGRAL
If F (x) =
f2 (x )
g (t )dt ,
f1(x )
then F (x ) = g ( f 2(x )) f 2(x ) – g ( f 1(x )) f 1(x ) MATHEMATICS TODAY | NOVEMBER ‘15
51
APPROXIMATION IN DEFINITE INTEGRAL If f 1(x ) f (x ) f 2 (x ) x [a, b], b
b
b
a
a
a
then f1( x) dx f ( x) dx f2 ( x)dx If absolute maximum and minimum values of f ( x ) , when x [a, b] is M and m respectively, b
Area of the r th part can be assumed a rectangle, b−a with width equal to and height equal to n b − a f a + r n So that area
n b−a b − a f a + r = but this n n r =1
then m (b − a) f ( x) dx M (b − a)
is only approximated area. To get the actual area, take rectangle with width tends to zero, hence
b
b
b
a
a
a
n
b − a f a + b − a r n n
f (x)dx = nlim → r =1
f (x)dx f (x) dx
a
is is used in both ways i.e. to evaluate the definite integral as a limit of sum and also used in finding the sum of infinite terms of some series.
From figure it is clear. DEFINITE INTEGRAL OF PIECEWISE CONTINUOUS FUNCTIONS b
Suppose we have to evaluate f ( x )dx , but either a
f (x ) is not continuous at x = c1, c2, ...., cn or it is not defined at these points. In both cases we have to break the limit at c1, c2, ...., c n . DEFINITE INTEGRAL AS THE LIMIT OF A SUM b
Consider
f (x)dx , for simplicity, we can take a
f (x ) 0
x [a, b].
GAMMA FUNCTION If n is a positive number, then the improper integral e − x x n−1dx is defined as Gamma function and
0
is denoted by n. − x n−1 i.e., n = e x dx
0
Properties of Gamma Function (i) 1 = 1
(n + 1) = nn (iii) If n N , then (n + 1) = (n)! (ii)
(iv )
(1 / 2) = p
Use of gamma function to find p/2 sinm x cosn x dx
f (x )
0
m +1 n +1 2 2 m n sin x cos x dx = m + n + 2 0 2 2
p /2
x=b
x=a
x x = a + r((b – a)/n)
Differentiation of function inside integral sign b
b
en
a
f ( x )dx represents the area bounded by the
curve y = f (x ), x –axis and the lines x = a and x = b i.e. the above shaded area. Now this area can be divided into n parts. 52
MATHEMATICS TODAY | NOVEMBER ‘15
If I () = f (x, )dx , then this function can be a
differentiated w.r.t. , i.e; b
f (x , ) dx a ROBLEMS
dI () = d
PROBLEMS
2 − tan2 z
SECTION-I
6.
Let I1 =
Single Correct Answer Type x
1.
x
Let F ( x) = sin x cos t dt + 2 t dt + cos 2 x − x 2 .
0
0
en area bounded by xF (x ) and ordinate x = 0 and x = 5 with x -axis is 25 35 (a) 16 (b) (c) (d) 25 2 2 Area bounded by the circle which is concentric 2. x 2 y 2 with the ellipse + = 1 and which passes through 25 9 4, − 9 , 5 the vertical chord common to both circle and ellipse on the positive side of x -axis is
− 36 (b) 2 tan−1 9 20 20 5 25 (c) 481 tan−1 9 (d) none of these 20 25 (a)
481
tan
−1
9
Area bounded between the curves y = 4 − x 2 and y 2 = 3|x | is 2p − 1 p −1 (a) (b) 3 3 3 3.
2p − 3 2p − 3 (c) (d) 3 3 3 e area of the smaller portion enclosed by the 4. curves x 2 + y 2 = 9 and y 2 = 8x is 2 9p 9 −1 1 (a) + − sin 3 3 4 2
5.
p of P is 3 (a) 0
sin2010 x
(b)
1 3
dx =
(c)
f (x(3 − x))dx
sec2 z
where 'f ' is a continuous function and ' z ' is any real I number, then 1 = I 2 (a) 3 2
(b)
1 2
(c) 1
(d) 2 3
1
7.
In
= xn tan−1 xdx. If anI n+2 + bnI n = cn n N , 0
n 1, then
(a) (b) (c) (d)
al, a2, a3, ........... are in A.P. b1, b2, b3, ........... are in G.P. c1, c2, c3, ........... are in H.P. a1, a2, a3, ........... are in H.P. x
8.
bt cos 4t − a sin 4t a sin 4 x dt = , then a 2 x t 0
Let
and b are given by (a) 1/4, 1 (b) 2, 2 (c) –1, 4 p/ 4 x 2 e sec xdx 9. is equal to 2 x e −1 − p/ 4 (a) 0 (b) 2 (c) e
(d) 2, 4
5
10.
Value of x + 2 x − 1 + x − 2
8 3
(b) 16 3
(c)
32 3
(d) 2e
x − 1 dx is (d)
34 3
x
11.
3 −1/2 If f (x ) = (1 + t ) dt and g (x ) is the inverse of
0
2 9p 9 −1 1 + + sin 3 3 4 2 If
and I2 =
(a)
2 9p 9 −1 1 + + sin (c) 2 3 3 4 2
sec2 x − 2010
2 − tan2 z
1
(b) 2 2 + 9p − 9 sin−1 1 3 4 2 3
(d)
x f ( x(3 −x))dx
sec2 z
f , then the value of P( x )
sin2010 x
3
(a) 3/2
+ C , then value 12.
(d) 1
g (x ) g 2 ( x )
(b) 2/3
p 2x 1 + sin x
2
− p 1 + cos x
2 p (a)
4
(b) p2
is
(c) 1/3
(d) 1/2
(c) 0
(d)
dx =
p 2
MATHEMATICS TODAY | NOVEMBER ‘15
53
1
13.
t
If
p f t dt = 1 − sin x, x 0, , then the 2
2
sin x
1 3 is
14.
1 3
(b)
(d)
3
(c)
Let In
1 3
(d) 3
= e − x (sin x)n dx, n N , n 1 0
I then 2008 equals I 2006
(a) (c)
2007 2006 20082 + 1 2006 2004 20082 − 1 sin q
15.
(b) 2008 2007 20082 + 1 2008 2007 (d) 20082 − 1
cosecq
t dt
1 + t 2 , B =
If A =
1
A then e A+ B
(a) sin (c) 0
1
A2
B
B2
−1 = −1
A2 + B2
1
q
1 t (1 + t 2 )
dt
q
SECTION-II Multiple Correct Answer Type
dx
2
x dx
x 4 + 7x 2 + 1 and v = x 4 + 7 x 2 + 1, then 0
p (a) u > v (b) u < v (c) u = v (d) u =
6 18. If f (x ) is an odd and periodic function with period T , then (a)
3T
f (x) dx = 0 0
54
0
If the area bounded by y = (sinx )cosecx ,y = (cosecx )sinx , y = (sinx )sinx and y = (cosecx )cosecx , between the ordinates 19.
p
x = 0, x =
denoted by A1, A2, A3 and A4 then 2 (a) A4 is the greatest (b) A1 is the least (c) A2 > A3 (d) A2 < A3 Given two functions f and g which are integrable on every interval and satisfy (i) f is odd, g is even (ii) g (x ) = f (x + 5), then (a) f (x – 5) = g (x ) (b) f (x – 5) = – g (x ) 20.
(b)
5
5
0 0 g (5 − t)dt 5 5 (d) f (t )dt = − g (5 − t )dt 0 0 3x + 4 dx = log |x – 2| + k log f (x ) + c, If 3 x − 2 x − 4 (c)
f (t )dt =
x + 1 2 x − 1 2 22. + x + 1 − 2dx is x − 1 −1/2 4 3 (a) 4 ln (b) 4 ln 3 4 256 81 ln (c) − ln (d) 81 256 tan x 23.
(a) (c)
f (x) dx = 0 0
MATHEMATICS TODAY | NOVEMBER ‘15
e value of 1 2 + tan2 x
24.
t dt 2
cot x
+
(d)
4
dt
t(1 + t 2 )
1/e 1 + t 1/e (b) 1
p x
0
T
f (x) dx = 0
1/2
e area of the region whose boundaries are defined by the curves y = 2 cos x, y = 3 tanx and the y -axis, is 2 3 (a) 1 + 3 ln (b) 1 + ln 3 − 3 ln 2 3 2 3 (c) 1 + ln 3 − ln 2 (d) ln 3 – ln 2 2
If u =
0
then (a) k = –1/2 (b) f (x ) = x 2 + 2x + 2 (c) f (x ) = | x 2 + 2x + 2| (d) k = 1/4
16.
17.
a
f (x) dx = f (x) dx for any a R
21.
(b) cosec (d) 1
T
T /2
value of f (a)
(c)
a +T
1
2
1
is
dt
p 1 + t 2 −1
If x − x − dx = f (x ) g (x ), where [x ] and {x } 2 0
are integral and fractional parts of x , respectively then {x } 2 [ x ] (c) f (x ) = 2 (a) f (x ) =
(b) g (x ) = { x } – 1 (d) g (x ) = [x ] – 1
5x 8 + 7 x 6
(x2 + 1 + 2x 7 )2
25. If I =
x 7
(a)
2 x 7 + x 2 + 1
+ C
31.
dx , then I is equal to
(b)
x 5 x 2 + 1 + 2 x 7
(a) 2 tan–1 x – x + c (c) tan–1x + 2x + c
+ C
(d) p( x ) , deg p(x ) = deg q(x ) = 7 q(x )
32. e function f at x = 0 attains (a) Local maximum (b) Local minimum (c) Point of inflexion (d) none of these
SECTION-III
2
Paragraph for Question No. 26 to 28 Two real valued differentiable functions f (x ) and g (x ) satisfy the following conditions: f (x ) − g (x ) 2( g (x ) − f (x )) (i) f (x ) = (ii) g (x ) = 3 3
26.
(a) (c) 27.
(a) (c) 28.
(iv) g (0) = –1 then
1
f (x ) dx = x – ln ( f (x )) + c (b) 1 (x − ln f (x )) + c 2 1 1 +c (x − ln f (x )) + c (d) x − ln ( f (x )) 3 1 dx = g (x ) x – ln ( g (x )) + c (b) 1 (x − ln g (x ) ) + c 2 1 1 +c (x − ln g (x ) ) + c (d) x − ln g (x ) 3 g (x ) dx = f (x )
−1
(a) 0
2
1 + x
1 − x 2
(b)
1 − x 2
(A)
1 + x 2
(c)
11 sq.units 12
Column II
cos x 1 cos −1 + c p (p) 2 x 0, 2 2
sin x
(B)
3 − cos 2x dx , p 4
(q) sin–1(sinx cosx ) + c
x 0, cos x
2
1 − x
15 4
cosec x + 1 dx ,
cos 2x dx , p x 0, 4
(d) 0
2
(d)
Column I
(C)
1
(d)
35. Match the following :
−
11 4
Matrix-Match Type
(b) x – 3ln ( f (x )) + c (d) 3(x – ln ( f (x )) + c
(c)
(c)
SECTION-IV
Paragraph for Question No. 29 to 31 dx If = A sin−1{ f (x )} + c, then (1 + x 2 ) 1 − x 2
(a)
7 sq.units 12
29. e value of A is 1 (a) 1 (b) 2 30. e f (x ) is
1 4
(b)
34. e area of the region bounded by the curves y = f (x ) and y = x 2 is 1 1 sq.units sq.units (a) (b) 12 4
(c)
(a) x + 3ln ( f (x )) + c (c) 3ln ( f (x )) – x + c
f (x)dx is
33. e value of
Comprehension Type
(iii) f (0) = 5
(b) 2tan–1x + x + c (d) none of these
Paragraph for Question No. 32 to 34 Let f ( x ) be a differentiable function satisfying (x – y ) f (x + y ) – (x + y ) f (x – y ) = 4xy (x 2 – y 2) for all x, y R. If f (1) = 1, then
−1 + C 2 x 7 + x 2 + 1
(c)
f (x)dx is
2
(d) 1 − x 2 1 + x
(D)
cos 2x 2
4
cos x + sin x
dx
(r)
1
−
2
1
sin − ( cos 2 x ) + c
(s) sin–1(2sinx – 1) + c
MATHEMATICS TODAY | NOVEMBER ‘15
55
36. Match the definite integrals in column I with their
values in column II. Column I
Column II
p
(A)
2
x(sin2 (sin x ) + cos2 (cos x ))dx
(p)
0
p
16
/4
(2 sin x
x cos x )dx (q)
+
p2 2
0
(D)
ln 2
p/4
ln ( 1 + sin 2 x ) dx
(r)
p2
p 8 x 3 cos 4 x sin2 x
p2 − 3px + 3x 2
dx
0
. en the numerical value
f (x ) = x
(s)
1. (b) : F (x ) = sin x cos t dt
2
x
+ e −t f (x − t )dt , then 6 f (1) =
, be roots of the quadratic equation – 9 px + p2 = 0, where < . Also f (x ) = x 2 and
g (x ) = cosx . If the area bounded by the curve y = ( fog ) (x ), the vertical lines x = , x = and x -axis is
p then find the sum of the digits in , p /2 p 1 39. Given lnsin x dx = ln 2
0
0
x
2 0
5
2
− x 2 + cos2 x
5
5
4 5 Co-ordinate of foci = (4, 0), (-4, 0) 9 y 4, − is one of 5 the end point of latus rectum Required area is O 2. (a) : Eccentricity of ellipse =
2 92 1 p 4 + 2 2p 5
=
481 −1 9 1 18 481 tan−1 9 − 36 tan − 4 = 20 5 20 2 5 25 25 1
1
4 7
29 (1 − x ) dx 40. Find the value of
0 1
.
4 (1 − x 4 )6 dx 0
x P ( 4, –9/5)
2
0
Q(4, 9/5)
9 2 tan−1 − area of POQ 20
2
x dx = k p ln 2, then k = ...........
sin x
+ 2 tdt − x 2 + cos2 x
x 2 25 A = xF (x )dx = (x )(1)dx = = 2 0 2 0 0
38. Let
3. (c) : Required area = 2
4 − x2
0
− 3x dx 1
x 3 2 x 3/2 2p − 3 2 4 1 x − = = 2 4 − x + sin − 3 2 2 3 2 0 y
1 = p 2
41. If f (x ) = a cos( px ) + b, f 3/2
x
8
0
p /2
x
= sin2 x + x 2 – x 2 + cos2 x = 1
37. Let f (x ) be a differentiable function such that
(–1, 3)
(1, 3)
y2 = 3|x |
2
f (x)dx = p + 1, then find the value of, 1/2
12
− (sin−1 a + cos−1 b). p 56
is
p2
Integer Answer Type
and
of
9000
F (x )dx is
4
− p/ 4
SECTION-IV
and
value of
1 = 3 and the 2
t 2 x = sin x (sin t )0 + 2
0
18x 2
1500
0
p
(C)
defined on R such that F (x ) + F x +
SOLUTIONS 2
(B)
p
42. Let F (x ) be a non-negative continuous function
MATHEMATICS TODAY | NOVEMBER ‘15
y = 4 – x 2
–2
O
2
x
y
4. (b) : x 2 + y 2 = 9,
x 2 +
8x – 9 = 0
x =
2− tan2 z
(0, 3)
=3
−8 64 + 36
x
2 −8 10 x = = −9,1 2
O
(–3, 0)
2
2 y
(3, 0)
(0, –3)
y
x = 1
2
On simplifying, we get
2 9p 9 −1 1 + − sin 3 3 4 2 sec2 x − 2010
Area = 2
1
+ 2010 2010
dx
(sin x )2010
(sin x )
tan x
+ C = 2010
(sin x )
P( x )
+ C 2010
(sin x )
p p P = tan = 3 3 3 2 − tan2 z
6. (a) : I = 1
x f ( x(3 − x)) dx
sec2 z
2
2 − tan z
I2
=
f ( x(3 − x)) dx
2 − tan2 z
2 0 1 + x
x n+3
p
1
4
2 0 1 + x
= −
dx
p 1 (n + 1)In + (n + 3)In+ 2 = − 2 n+2 an = (n + 3) a1, a2, a3, ......are in A.P. bn = (n + 1) b1, b2, ......are in A.P. p 1 cn = − is not in any progression. 2 n+2 a sin 4 x bt cos 4t − a sin 4t dt = x t 2 0
Differentiating both sides w.r.t. x bx cos 4 x − a sin 4 x a 4 x cos 4 x − sin 4 x
=
2
x
x 2
On comparing, we get b = 4a a = 1/4 and b = 1 p/ 4 e x sec2 x dx 9. (a) : Let I = 2 x − p/4 e − 1
If f ( x ) =
e x sec2 x
e2 x − 1 e − x sec2 x e x sec2 x e x sec2 x = = − = − f (x ) f (− x ) = e −2 x − 1 e 2 x − 1 1 − e 2 x
I = 0 ( f (x ) is an odd function) 10. (d) :
x + (2 ( x − 1))
(3 − x ) f ((3 − x)x) dx
And x − 2 (x − 1)
2
2 − tan2 z
2
sec z
3 f ( x(3 − x)) dx
= ( ( x − 1) )2 + 12 − 2 ( x − 1) = x − 1 − 1
sec z
2I1 =
dx
= ( x − 1)2 + 12 + 2 ( x − 1) = ( x − 1) + 1
sec2 z
I1 =
4
dx = I1 − I2
tan x cos x + I 1 = 2010 dx 2010 2011 (sin x ) (sin x )
I = I1− I 2 =
1
= −
(n + 3)I n+2
x n+1
p
8. (a) : Since,
(sin x )2010 Applying by parts on I 1, we get
tan x
3 2
x
= sec2 x (sin x )−2010 − 2010
=
=
1
dx
sin2010 x
tan x
7. (a) :
(n + 1)I n
I 1 I 2
x n+1 −1 1 x n+1 1 I n = tan x − dx n +1 0 0 n + 1 1 + x 2
x
8 x
2I1 = 3I 2
f (x(3 − x)) dx
sec2 z
9
3 1 Area enclosed = 2 2 2 xdx + 9 − x 2 dx 0 1 3 1 2 = 2 2 2 xdx + 9 − x dx 0 1
5. (c) :
5
Then
1
x+2
x − 1 +
x−2
x − 1 dx
MATHEMATICS TODAY | NOVEMBER ‘15
57
5
5
= (x − 1) + 1 dx + ( x − 1) − 1 dx 1
14. (b) : In
1
5
2
5
1 4
1
2
0
= ( x − 1 + 1)dx + (1 − x − 1)dx + ( x − 1 −1)dx 1
4
0 4
1 1
4
16 2 16 2 34 = + 4 + 1 − + − 4 − − 1 = 3 3 3 3 3
− n {− sinn x + (n − 1)sinn−2 x cos 2 x}( −e − x )dx 0
= 0 + n e − x {− sinn x + (n − 1) sinn−2 x (1 − sin 2 x)}dx
(1 + t 3 )−1/2 dt
= n(n – 1) I n – 2 – n2I n (1 + n2) I n = n(n – 1)I n–2
or
0
g x
15. (c) : B =
[ g is inverse of f f [ g (x )] = x ] Differentiating with respect to x , we have 1 = (1 + g 3(x ))–1/2. g (x ) i.e. ( g (x ))2 = 1 + g (x )3 Differentiating again with respect to x , we have g (x ) 3 = 2 g ( x ) g (x) = 3( g ( x))2 g ( x ) 2 2 ( g ( x )) p p p − x sin x x sin x 12. (b) : I = 4 dx = 4 dx 2 2 x x 1 cos 1 cos + + 0 0
2I = 4 p
p /2
sin x
dx = 8p dx 2 2 + + x x 1 cos 1 cos 0 0 2 2 2I = 2p I = p 1
13. (d) :
t
2
f (t )dt = 1 − sin x
sin x
Differentiating both sides with respect to ‘ x ’ 0 – sin2x . f (sinx ).cosx = –cosx cosx [1 – sin 2x . f (sinx )] = 0 But cosx 0 1 1 So, f (sin x ) = f 3 = 3 sin2 x 58
I n n(n − 1) = I n−2 n2 + 1 cosecq
0
sin x
0
0
g x
p
= n e − x {(n − 1) sinn−2 x − nsinn x}dx
0
x cos x)(−e − x )]0
11. (a) : f ( x ) = (1 + t 3 )−1/2 dt
(1 + t 3 )−1/2 dt
0
= n[(sin
x
i.e. x =
+ n sin n −1x cos x e −x dx
0 n−1
2 2 2 = x 3/2 + x + x − x 3/2 + x 3/2 − x 3 0 3 0 3 1
i.e. f [ g (x )] =
= [sin x(−e − x )]0 n
= 0 + n (sinn−1 x cos x )e − x dx
= ( x + 1)dx + (1 − x )dx + ( x − 1)dx 0
= e − x (sin x)ndx
MATHEMATICS TODAY | NOVEMBER ‘15
1
1 Let = u B = t A
A2
e0
A2
1
2 A2
1 t (1+ t 2 )
sin q
dt
−udu
1 + u2 A + B = 0
A = −B
1
−A −1 = 0 −1
16. (b) : Solving 2 cos x = 3 tan x, we get
2 – 2 sin 2 x = 3 sin x 1 p x = 2 6 p/6 Required area = (2 cos x −3 tan x) dx
sin x =
0
3 2
= 2 sin x − 3 ln sec x |0p/6 = 1 − 3 ln 2 + ln 3
1
x 4 + 7x 2 + 1 dx
17. (c, d) : u =
0
−1 1 Put x = , dx = 2 dt t t
0
u=
1
+
1 7
t 4 t 2 u = v
−dt t 2dt = v 2 = 4 2 t t t + + 7 1 +1 0
1 + x 2
x 4 + 7x 2 + 1dx
So, u + v = 2u =
0
1 1 + 2 dx x dt p = = 2u = 2 2 3 0 x − 1 + 9 − t + 9 x
T
21. (a, b, c) : By using partial fraction, we get
3x + 4
x
3x + 4 3
− 2x − 4
0
0
T
1 2 2 Hence k = –1/2, so that f (x ) = |x + 2x + 2|
= log | x − 2 | − log | x 2 + 2x + 2 |+ c
1/2
−1/2 1/2
1/2
4 x
=
2 −1/2 x − 1
0
T
= − f (x ) dx ( f is odd) 0
2 f (x )d x = 0 0
(a) is correct
4 x
(x 2 − 1) dx
dx = −2
= −4 ln x 2 − 1
T
x + 1 − x − 1 2 dx x − 1 x + 1
22. (a, c, d) : Let I =
= f (− x ) dx ( f (x ) is periodic with periodic T ) 0
x + 1
1
x − 2 dx − x2 + 2x + 2 dx
dx =
T
f (x) dx = f (T − x )dx
1/2
3 = −4 ln 4
0
4 256 81 = 4 ln = ln = − ln 3 81 256 cot x
3T
T
23. (b, d) : Let I =
f (x) dx = 3 f (x ) dx (b) is also correct. 0
0
(c) is a known result.
p 2 0 < sin x < 1 < cosec x < 19. (a, b, c) : For x 0,
1 z
Put t =
dt = −
tan x
tan x
1/e
5
I = f (t )dt
e
=
5
Indeed I = g (t − 5)dt 0
( f (t ) = g (t – 5) on replacing x by t – 5 in (ii))
1/e
=
+
2 tan x (z
(1 + t 2 )
+ 1)
e
=
t dt
2 tan x (1 + t )
dt
t(1 + t 2 )
1/e
e
t
z dz
dt +
t
(1 + t 2 ) dt
tan x
e 1 2 = + t ln 1 1/e (1 + t 2 ) 2
t dt
1 1 = ln(1 + e2 ) − ln 1 + 2 = 1 (ln e2 ) = 1 e 2 2
5
= 0 g (5 − t )dt ( g is even) Choice (c) is correct and choice (d) is false.
e
cot x
(1 + t )
dz
z 2
1 1+ 2 z z
t dt
tan x
1
dz
z 2
2
1/e
=
1
−
1
I =
e
20. (b, c) : To test choice (a) and (b), we begin with
computing g (x ). Indeed g (x ) = f (x + 5) g (– x ) = f (– x + 5) g (x ) = – f ( x – 5) Choice (b) is true and choice (a) is ruled out. To test the choices (c) and (d), we compute
dt
t(1 + t 2 ) 1/e
So, sinx cosecx < sinx sin x < cosecx sin x < cosecx cosec x
0
A Bx + C + 2 x − 2 x + 2x + 2
x − 2x − 4 Solving this equation, we get A = 1, B = –1 and C = –1
18. (a, b, c) : Since
Since
=
3
Also,
2
1
dt
p −1 1 + t
= 2
1
4 −1 1 = 4 . p = 1 = .tan 2 p p 4 p 0 1 + t 4
dt
MATHEMATICS TODAY | NOVEMBER ‘15
59
[ x ] + {x }
24. (a, b) : I =
0
[ x ]
{x} − 1 dx = 2
{x} − 1 dx + 2
0 [ x ] + {x }
[ x ]
1
1 ] {x} − dx + = [x 2
1 {x} − 2 dx
0 { x }
0
0
1 x − 2 dx
1 x 14 5 x
+
1
1
x
x 7
Put t =
+ 5
+ 2 x 7 1
t
1 x
dx =
6
+
7 x
1 5 + 7 + 2 x x
2
dx
x 7 x 2 + 1 + 2 x 7
+ C
1 3
= (x − ln f (x )) + c 27. (c) :
1 1 dx = dx x g ( x ) 3 − 4e
e − x −1 = − x dx = ln | 3e − x − 4 | + c 3 3e − 4
1 3
= (x − ln | g (x )|) + c 28. (b) :
3 − 4e x
3 + 2e x 60
dx = 1 −
dx = 2t dt
t 1 1 1 − x 2 tan −1 + C = − tan−1 2 2 2 2x
+ C
32. (c)
3(2e x )
dx = x − 3 ln ( f (x )) + c
3 + 2e x
MATHEMATICS TODAY | NOVEMBER ‘15
33. (d)
34. (b)
(x – y ) f (x + y ) – ( x + y ) f (x – y ) = 4 xy (x 2 – y 2) f (x + y ) f ( x − y ) − = 4 xy x + y x − y So,
1 1 dx = dx x f ( x ) 2e + 3 x − e −1 dx = = ln (2 + 3e − x ) + c x − 3 2 + 3e
=−
x 3
−1
(32 - 34) :
8
(26 - 28) : 26. (c) :
tdt
2
1 1 x 3 1 + x 2 x 2
+ 2, so that
−dt 1 I = = + C = 2 t
2
x
5 5x + 7 x
− 1 = t 2 −
2
1 − x 2 )
dx
=
2 1 1 − 1 1 − x A = − C = − sin + 2 2 1 + x 2 1 − x 2 30. (b) : f (x ) = 1 + x 2 2 31. (a) : f ( x ) dx = 1 − x dx 1 + x 2 2 = − 1 dx = 2tan–1x – x + c 2 1 + x
2 2 25. (a, d) : We can write 6
1
dx
(1 + x2 )(
(t 2 + 2)t
x
I =
Put
−
x2 x 1 x 2 x { } = [x ] − + − 2 2 0 2 2 0 {x} {x } − 1 {x } =0+ = {x } − 1
8
29. (c) :
{x }
0 1
1 = [x ] x − dx + 2
{x } − 1 dx 2
(29 - 31) :
f (x + h) x +h
−
f (x − h) = 4 xh x −h
f ( x + h) − f ( x) + f ( x ) x +h f ( x − h) − f ( x) + f ( x ) − = 4 xh x −h f (x + h) − f (x ) f (x − h) − f ( x ) − x +h x −h 1 1 = 4 xh + f (x ) − x − h x + h f ( x + h) − f (x ) f (x − h ) − f (x ) xh 2hf (x ) − = 4 + 2 2 x +h x −h x −h f ( x + h) − f ( x ) 1 . x + h h 2 f (x ) f ( x − h) − f ( x ) 1 + = 4 x + 2 2 −h x − h x −h Taking limit on both sides as h → 0 ; 2 f (x ) f (x ) f (x ) f ( x ) f ( x ) + = 4 x + 2 = 2x + 2 x
x
x
x
x
36. A → q; B → q; C → r; D → r
f (x ) − f (x ) 2x 2 , f (x ) + = x x which is a linear differential equation. f ( x ) = 2 x 2 +
p
(A) I
0
p
1 dx x =
−
1 Integrating factor = e x 1 1 Its solution is f (x ) = 2 x 2 + c x x f (x ) = x 2 + c f (x) = x 3 + cx x But f (1) = 1 c = 0 So, f (x ) = x 3 f (x ) = 3x 2 and f (x ) = 6x Here, f (x ) = 0 x = 0 So, at x = 0, f (x ) has point of inflexion.
I = (p − x)(sin2 (sin x ) + cos2 (cos x ))dx 0
2
x 3dx =
1
2
2 1 x 3dx = 0 + x 4 4 1
x 3dx +
0
2I = 2p
=
cos x sin x(1 − sin x )
1 + sin x
sin x
I =p
(C)
cos x cos 2 x
dx =
Also I = p
1
2 cos x Then x f ( x ) = x. = x cos x 2 x p2 / 4 p2 / 4 I= ( f ( x ) + x f ( x))dx = x f ( x)
= 2 ( p2 / 4 ) sin(p / 2) = p2 / 2
2 − cos2 x
dt
1 − t 2
2
ln
2
cos x
+ sin
4
x
=
=
cos 2 x
− sin2 2x
0
2
0
p /2
dt + lnsint dt
= ln 2
dx
p /2
sin 2x = sin−1 + c
ln sin x + cos x dx
− p/4
sin−1 ( cos 2 x ) + c
4
1 + sin 2 x dx =
− p/ 4 p /2 ln 2 sin(x + p / 4 ) dx = ln 2 sin t dt
p/ 4
(Put t = cos 2 x )
2 cos 2x dx
p/ 4
− p/4
0
= (p / 4) ln 2 − (p / 2) ln 2 =− (p / 4) ln 2 p p/ 4
p − p − p2 ln 1 + sin 2 x dx = ln 2 = 4 ln 2 ln 2 4 − p/ 4 p x 3 cos 4 x sin2 x
p2 − 3px + 3x 2
(D) I = cos 2 x dx
0
0
sin x dx
2
(D)
2dx = 2p(p / 2) = p2 0
(C)
sin 2 x
=−
Adding 2I = p
p /2
(B) Let f ( x) = 2 sin x
dx
2 1 − cos 2 x 1
0
I = p2/2
cos x + c cos −1 2 2
=−
[sin2 (cos x) + cos2 (sin x)]dx
p/ 4
1
p /2
15 = 4
1
=
(sin2 (sin x) + cos2 (cos x ))dx 0
= sin–1(2sin x – 1) + c sin x 1 (B) dx = 3 − cos 2 x 2
p/ 2
dt [Put t = sin x ] t (1 − t )
dx =
(sin2 (sin x) + cos2 (cos x ))dx 0
1
cosecx + 1 dx =
p /2
x3 x 4 1 1 1 2 3 ( x x ) dx − = − = − = sq. units 3 4 0 3 4 12 0 35. A → s; B → p; C → r; D → q (A)
p
Adding 2I = p (sin2 (sin x ) + cos2 (cos x ))dx
1 −1 −1 e area of the region bounded by the curves y = f (x ) and y = x 2 is 1
= x(sin2 (sin x ) + cos2 (cos x ))dx
dx
0
p (p − x )3 cos 4 x sin2 x
I =
0
p2 − 3px + 3x 2
dx
MATHEMATICS TODAY | NOVEMBER ‘15
61
41. (6) : f (x ) = –ap sin(px )
p
4
2
Adding 2 I = p cos x sin x dx
p
I=
2
0
p /2
.2
cos
4
p 1 f = −ap sin = −ap = p a = −1 2 2
p /2
2
x sin x dx
0
= p cos 4 x sin2 x dx 0
(3 1)1 (p / 2) = p2 / 32 6 4 2 x 2 37. (8) : f (x ) = x + e −t f (x − t )dt
=p
−a 3b a b = + − + p 2 p 2
x
0
f (x ) = 2x − e − x (e x ( f x − x 2 )) + e − x e x f (x )
−
x 3 f x = 2x + x f (x ) = + x 2 + k 3 4 But f (0) = 0 k = 0 f 1 = 3 2 38. (3) : ( fog )(x ) = f (cosx ) = cos x Let , be roots of the equation 2
18 x
2
2
p
p
6
3
− 9px + p = 0 = , = p/3
Area =
cos2 x dx =
p /6
2a
−12 −1 −12 p (sin (−1) + cos −1 1) = − + 0 = 6 p p 2
42. (4) : We have, F (x ) + F x +
p
F x +
12
1 + F x + 1 = 3 2
2
p /2
/2
2
p/2
2
0
x = 0 + lim + 2 x lnsin x 0p/2 − 2 lnsin x dx x →0+ tan x 0 p 1 = 0 − 2 lim x lnsin x − 2 ln = p ln 2 2 2 x →0+ 1 .cos x lnsin x = 0 lim+ 1 / x = lim+ sin x 1 x x → → 0 0 − 2 x 1 40. (7) : I
1500
Now consider I =
1
1
= −28I + 28 (1 − x ) dx 29I = 28 (1 − x 4 )6 dx 4 6
0
0
29 (1 − x 4 )7 dx 0 1
=7
4 (1 − x 4 )6 dx 0
62
MATHEMATICS TODAY | NOVEMBER ‘15
1
F (x )dx = 1500 F (x )dx
0
1 2
0
1
Using property of = 1500 F x dx + F (x )dx periodic functioon 1 0 2
Put x = y + 1 2
0
1
.....(3)
F (x ) is periodic function.
= (1 − x 4 )7 1dx
1 4 7 = x 1 − x + 7 4 x (1 − x 4 )6 x 3dx 0 0
....(1)
...(2)
From (1) and (2), we get F (x ) = F (x + 1)
0
1 = 3 2
1 Replace x by x + in (1), we get 2
p I = x cosec x dx = − x 2 cot x 0 + 2x cot xdx
1
2
+b = +1 b= 1 p p
So,
39. (2) :
p /2
3/2
a sin px a cos px + bdx = p + bx 1/2 1/2
= x 2 − e − x e t f (t )dt
0
3/2
1 in 2nd integral, we get 2 1 2
I = 1500 F x dx + F y + 0
0
1 2
1 dy 2 1 2
1 = 1500 F x + F x + dx = 1500 3dx Using(1) 2 0 0 1 = 750 3 = 2250 2
Hence I = 1500(3)
nn
The entire syllabus of Mathematics of WB-JEE is being divided into six units, on each unit there will be a Mock Test Paper (MTP) which will be publi shed in the subsequent issue.
UNIT-V : VECTORS CATEGORY I
For each correct answer one mark will be awarded, whereas, for each wrong answer, 25% of total marks (1/4) will be deducted. If candidates mark more than one answer, negative marking will be done.
(a) collinear (c) non-collinear
(b) non-coplanar (d) none of these
7. If ABCDEF is a regular hexagon with AB = a and BC = b , then CE = (a) b − a (b) −b
1. e distance of the point A(–2, 3, 1) from the line PQ through P (–3, 5, 2) which makes equal angles
with the axes is 2 (a) (b) 3
(c) b − 2a
(d) none of these
^
^
^
14 16 5 (c) (d) 3 3 3
^
^
7 ^ ^ ^ ( i + 2 j + 2 k) 9 (d) none of these
^
(a) 7(i + 2 j + 2 k)
(b)
(c)
3. If ABCD is a rhombus whose diagonals cut at the origin O, then OA + OB + OC + OD equals
^
^
2. Let a , b and c be three non-zero vectors, no two of which are collinear. If the vector a + 2b is collinear with c and b + 3c is collinear with a , then a + 2b + 6c is equal to (a) a (b) b (c) c (d) 0
^
8. If a = i + 2 j + 2 k and b = 3 i + 6 j + 2 k , then the vector in the direction of a and having magnitude | b | is
7 ^ ^ ^ ( i + 2 j + 2 k) 3
9. Let ABCD be the parallelogram whose sides AB
and AD are represented by the vectors 2 ^i + 4 ^j − 5 k^ ^ ^ ^ and i + 2 j + 3 k respectively. If a is a unit vector
(a) AB + AC
parallel to AC , then a is equal to
(b) O
(c) 2( AB + BC )
(d) AC + BD
4. If a , b are the position vectors of A, B respectively and C is a point on AB produced such that AC = 3 AB, then the position vector of C is (a) 3a − 2b (b) 3b − 2a (c) 3b + 2a (d) 2a − 3b
5. If the vectors
a + b + 3c , − 2a + 3b − 4 c and a − 3b + 5c are coplanar, then the value of is (a) 2 (b) –1 (c) 1 (d) –2
6. e three points A, B, C with position vectors −2a + 3b + 5c , a + 2b + 3 c and 7a − c are
1 ^ ^ ^ 1 (i + 2 j + 3 k) (b) (3 ^i + 6 ^j + 2 k^) 3 3 1 1 (c) (3 ^i − 6 ^j − 3 k^) (d) (3 ^i + 6 ^j − 2 k^) 7 7 10. e value of , for which the four points 2 ^i + 3 ^j − k^, ^ ^ ^ ^ ^ ^ ^ ^ ^ i + 2 j + 3 k , 3i + 4 j − 2 k and i − j + 6 k are coplanar, is (a) –2 (b) 8 (c) 6 (d) 0 (a)
11. Consider points A, B, C , D with position vectors ^
^
^
7i −4 j +7k , ^ ^
^
^
^
i − 6 j + 10 k ,
^
^
^
−i −3 j + 4k
and
^
5 i − j + 5 k respectively. en ABCD is a
By : Sankar Ghosh, HOD(Math), Takshyashila MATHEMATICS TODAY | NOVEMBER ‘15
63
(a) square (c) rectangle 12.
(b) rhombus (d) none of these ^
^
^
^
^
21.
^
^
(b) 1/3
^
^
^
^
^
^
22.
If
a , b and c
are three vectors such that a + b + c = 0 and | a | = 3, | b | = 5, | c | = 7 , the angle between a and b is (a) 60° (b) 30° (c) 45° (d) 90°
(a) 0
^
If = 2 i + j + k, = i − 2 j + 2 k and = 3 i − 4 j + 2 k , then the projection of ( + ) in the direction of is 17 5 1 17 (a) (b) (c) (d) − 3 8 43 3
If p and q are non-collinear unit vectors and | p + q | = 3 , then (2 p − 3q) (3 p + q) is equal to
^
e angle between the two vectors i + j + k and 2 i − 2 j + 2 k is equal to 2 1 (a) cos −1 (b) cos −1 3 6 −1 5 −1 1 (c) cos (d) cos 6 3
13.
(c) 2 | a |2 + | b |2 + | c |2 1 (d) | a |2 + | b |2 + | c |2 2
(c) –1/3 (d) –1/2
14.
^
^
If = 3 i − k , | | = 5 and = 3 , then the area of the parallelogram for which and are adjacent sides is 17 14 7 (a) (b) (c) (d) 41 2 2 2
15.
If a, b , c are unit vectors such that a + b + c = 0 then a b + b c + c a = (a) 3/2 (b) –3/2 (c) 2/3 (d) 1/2
19.
^
^
Let a , b , c be three vectors such that a b = c and c a = b , then
Let a = i − 2 j + 3 k . If b is a vector such that a b = | b |2 and |a − b | = 7 , then | b | =
(a) a b = | c |2
(b) c a = | b |2
(c) b c = | a |2
(d) a ||(b c )
26.
Let a and b be two vectors of equal magnitude
q
^ ^ ^
^
and b is p/6, then the area of the triangle formed by these two vectors as two sides is 15 15 15 3 (a) (b) (c) 15 (d) 4 2 2 If a and b are two unit vectors inclined at an angle p/3, then the value of |a + b | is (a) equal to 1 (b) greater than 1 (c) equal to 0 (d) less than 1
inclined at an angle q, then a sin is equal to 2
If i + j, j + k, i + k are the position vectors of the vertices of a triangle ABC taken in order, then angle A is equal to (a) p/2 (b) p/5 (c) p/6 (d) p/3
25.
If the sum of two unit vectors is also a unit vector, then the angle between the two unit vectors is p p p 2p (a) (b) (c) (d) 3 2 4 3 ^ ^ ^
^
18.
^
17.
24.
If a b and (a + b ) (a + mb), then m = (a) –1 (b) 1 1 − | a |2 (c) (d) 2 | b |2
^
If a = i + 2 j + 2 k, | b | = 5 and angle between a
16.
23.
|a + b | (b) 2
|a − b | (a) 2
(c) | a − b |
(d) | a + b |
(a)
7
(b)
3
(c) 7
(d) 3
(a) | a |2 + | b |2 + | c |2
(b) | a | + | b | + | c |
MATHEMATICS TODAY | NOVEMBER ‘15
(a) 4 3
64
If a , b and c are three non-zero vectors such that each one of them being perpendicular to the sum of the other two vectors, then the value of | a + b + c | is
20.
a is perpendicular to b and c , | a | = 2, | b | = 3, 2p | c | = 4 and the angle between b and c is , then 3 [a b c ] is equal to
27.
28.
(b) 6 3
(c) 12 3 (d) 18 3 ^
^
e unit vector perpendicular to i − j and coplanar ^
^
^
^
with i + 2 j and 2 i + 3 j is
^
(a)
^
^
i+j
(c)
2
^
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 mark one correct and one incorrect answer, then no marks will be awarded. But if, candidate makes only correct, without marking any incorrect, formula below will be used to allot marks. 2 × (no. of correct response/total no. of correct options)
(b) 2 i + 5 j
29 ^
29.
CATEGORY III
^
2i −5 j
^
^
(d) i + j
^ ^
^
^
^
If a = i − k, b = x i + j + (1 − x ) k and
^
^
^
c = y i + x j + (1 + x − y) k , then [a b c ] depends on (a) neither x nor y (b) both x and y (c) only x (d) only y
36.
31.
^
^
^
^
^
^
^
(c) −6 i + 12 j + 12 k
^
p
6
(b)
33.
34.
(c)
p
3
(d)
^
^
^
^
(a) –2
(b)
(c) 1 − 3
(d) 2
3 +1
If a, b and c are any three vectors, then which of the following is equal to a b + b c + c a ? (a) (a − b ) (b − c )
^
^
(a) 2 (c) 4
2p 3
(c) (a − b ) (a − c ) 1 (d) a (b − c ) + b (c − a) + c (a − b) 2
39.
[ a b c ] is equal to
^
when
^
^
^
2 i − j + k,
^
(b) –4 (d) none of these
(a) [ a b][ c ] − [ a ][b
(b) [ a
][c
c ]
b ] − [ a c ][ b]
(c) [ b ][a
c] − [ b
a][
c ]
(d) [ ][a b c ]
40.
e vectors a , b , c are of same magnitude and taken pairwise, they contain equal angles. If
^ ^ ^ ^ a = i + j, b = j + k , then vector c =
^
^
(a) ^i + k^
(b) j + k
(c) − ^i + ^j + 2 k^
1 4 1 (d) − ^i + ^j − k^ 3 3 3
If [ + + + ] = [ ] , then is equal to (a) 1 (b) 2 (c) 3 (d) 4
i + 2 j − 3 k and 3 i + j + 5 k be coplanar, is
35.
e value of (constant) ^
^
Let OA = a, OB = 10a + 2b , OC = b , where O, A, C are non-collinear points. Let p denotes the area of the quadrilateral OABC and q denotes the area of the parallelogram with OA and OC as adjacent sides. If p = kq, then k = (a) 2 (b) 3 (c) 6 (d) 4
^
5p 6
^
38.
(a)
^
e vectors a = ^i + ^j + 2 k^ , b = ^i + ^j − k^ and
^
(b) (c − b ) ( a − c )
c = 2 i − j + k are coplanar, if =
a and b are such that | a | = 1, | b | = 4 , a b = 2. If c = 2a b − 3b , then the angle between b and c is
37.
(d) none of these
32.
^ ^
(c) i (a i ) + j (a j) + k (a k) = 2a (d) only (a) and (b)
^
to both the vectors 4 i − j + 3 k and −2 i + j − 2 k is ^ ^ ^ ^ ^ ^ (a) −4 i + 8 j + 8 k (b) −2 i + 4 j + 4 k
^ ^
e vector of magnitude 12, which is perpendicular ^
^ ^
(b) (a ^i )2 + ( a ^j)2 + (a k^)2 = | a |2
CATEGORY II
Every correct answer will yield 2 marks. For incorrect response, 25% of full mark (1/2) would be deducted. If candidates mark more than one answer, negative marking will be done.
If a is any vector, then (a) (a i ) i + (a j) j + (a k) k = a
If | a b |2 + | a b |2 = 144 and | a | = 4 , then | b | is equal to (a) 12 (b) 3 (c) 8 (d) 4
30.
MATHEMATICS TODAY | NOVEMBER ‘15
65
SOLUTIONS
1. (b) : Given that, the line
A(–2, 3, 1)
PQ makes equal angles with the axes. Let the angle be . cos2 + cos2 + cos2 = 1 M P (–3, 5, 2) 3cos2 = 1 cos = 1 3 1 1 1 , , us, direction cosines are 3 3 3 Now, PM = projection of AP on PQ 1 1 1 2 = (−2 + 3) + (3 − 5) + (1 − 2) = 3 3 3 3
Q
2 2 2 and AP = (−2 + 3) + (3 − 5) + (1 − 2) = 6
2
Now, a + 2b + 6 c = (x + 6)c
Also, a + 2b + 6c = (1 + 2 y )a From (i) and (ii), we get (x + 6) c = (1 + 2 y )a (1 + 2 y)a − ( x + 6) c = 0 1 + 2 y = 0 and x + 6 = 0 , x = –6 y = –1/2 a + 2b + 6c = 0
... (ii)
B
D
O
5. (d) : Given that a + b + 3c ,
E F
D
A
C
C
B
B
b
CD = b − a Now in CDE, CE = CD + DE = b − a + (−a) = b − 2a
8. (c): Here | b | = 32
−2a + 3b − 4c and
a − 3b + 5c are coplanar. It is possible to express one of them as a linear combinaton of the other two. MATHEMATICS TODAY | NOVEMBER ‘15
a
( AD || BC )
1
A
Now, AC + CD = AD
C
Let c be the position vector of C 1 c + 2 a b= c = 3b − 2a 1+ 2
CD = AD − AC = 2b − (a + b )
66
en AC = AB + BC = a + b
B divides AC internally in the ratio 1 : 2
ABCDEF represents the vectors a and b respectively.
A
4. (b) : Given that AC = 3 AB
of a rhombus bisect each other. OA = −OC and OB = − OD So, OA + OB + OC + OD = O
7. (c): e sides AB and BC of a regular hexagon
3. (b) : We know that diagonals
e above equation shows that the vectors AC and AB have the same or parallel supports. But these vectors have a common initial point A. So, AC and AB have the same support. A, B, C are collinear.
... (i)
OC = 7a − c
AB = OB − OA = (a + 2b + 3c ) − (−2a + 3b + 5c ) = 3a − b − 2c Also, AC = OC − OA = (7a − c ) − (−2a + 3b + 5c ) = 9a − 3b − 6c AC = 3(3a − b − 2c ) AC = 3 AB
Also, b + 3c is collinear with a b + 3c = ya, y R
OB = a + 2b + 3c
4 14 AM = ( AP) − (PM ) = 6 − = 3 3 2. (d) : Given that a + 2b is collinear with c a + 2b = xc , x R 2
Let us write a + b + 3c = x(−2a + 3b − 4c ) + y(a − 3b + 5c ) where x , y are scalars. Now, comparing the coefficients of a , b and c from both the sides, we get –2x + y = 1, –4x + 5 y = 3 and 3x – 3 y = Solving first two equations, we get x = –1/3 and y = 1/3. ese values of x and y satisfy the last equation as well. 1 1 3 − − 3 = = −2 3 3 6. (a) : Let O be the origin of reference. en, OA = −2a + 3b + 5c
+ 62 + 22 = 7
Now, any vector along a with magnitude | b | = 7 is
a 7 ^ ^ ^ 7 = ( i + 2 j + 2 k) |a | 12 + 22 + 22
7 3
= (^i + 2 ^j + 2 k^)
^
9. (d) : Let R1 = 2 i
^
^
+ 4 ^j − 5 k^
D
^
C
and R2 = i + 2 j + 3 k
= 4 + 9 + 4 = 17
R = R1 + R2 = 2 ^i + 4 ^j − 5 k^ + ^i + 2 ^j + 3 k^ ^
^
R
R2
A
^
= 3i + 6 j − 2k
which does not satisfy the conditions of square, rhombus and rectangle.
^
^
^
^
^
^
^
A = 2 i + 3 j − k, B = i
^
cos q =
^
^
^
^
^
^
| a || b |
=
^
^
2−2+2 3 12
^
^
^
^
^
^
^
^ ^
^
^ ^
^
− i − ( + 3) j + 7 k = x(− i − j+ 4 k) + y( i+ j − k) ^ ^
^
Comparing the coefficients of i , j and k , we get –x + y = –1, 4x – y = 7 and –x + y = –( + 3) Solving the first two equations, we get x = 2 and y = 1. ese values of x and y satisfy the last equation as well. –2 + 1 = – – 3 = –2 11. (d) : Given that,
^
^
^
^
^
^
^
^
^
OA = 7 i − 4 j + 7 k, OB = i − 6 j + 10 k
^ ^
^
OC = − i − 3 j + 4 k and OD = 5 i − j + 5 k
2 2 2 Now, | AB | = (7 − 1) + (−4 + 6) + (7 − 10)
= 36 + 4 + 9 = 7
2
2
2
= 4 + 9 + 36 = 7 | CD | = (−1 − 5)2 + (−3 + 1)2 + (4 − 5)2
= 36 + 4 + 1 = 41
2
2
^
2
^
2
= 3 i − k, | | = 5 and = 3 = | || | cos q = 3
10 5 cos q = 3 [ | | = 32 + 12 = 10 ] 3 cos q = 50 Now, area of parallelogram =| |
= | || |sin q = 50 1 −
9 = 41 50
15. (b) : Given that a + b + c
= 0 |a + b + c | = 0 2 | a | + | b |2 + | c |2 + 2(a b + b c + c a) = 0
a b +b c + c a = −
16. (c): Since a b ,
Now,
| BC | = (1 + 1) + (−6 + 3) + (10 − 4)
1 2
2
14. (d) : Given that
[x , y are scalars]
^
1 7 1 = 6 − 7 − 3 = 3 − = −
^
= 3
^
Now, AB, AC and AD will be coplanar, if
1 3
Now, (2 p − 3q ) (3 p + q ) = 6 p + 2 p q − 9 q p − 3 q
^
AD = x AB + y AC
3 2 3
=
AD = − i − ( + 3) j + 7 k
AC = ^i + ^j − k^ and AD = (i − j + 6 k ) − (2 i + 3 j − k )
2
=
1+ 1 + 2 p q = 3 p q =
AC = (3 i + 4 j − 2 k) − (2 i + 3 j − k)
^
Since p and q are unit vectors
^
^
^
p 2 + q 2 + 2 p q = 3
^
AB = − i − j + 4 k ,
− j + 6k ^ ^ ^ ^ ^ ^ Now, AB = ( i + 2 j + 3 k) − (2 i + 3 j − k) ^
13. (d) : Given that, | p + q |
^
^
a b
^
^
1 q = cos−1 3
C = 3 i + 4 j − 2 k and D = i
^
= (3 ^i + 6 ^j − 2 k^)
+ 2 j + 3 k,
^
^
+ j + k , b = 2i −2 j +2k
If q is the angle between the two vectors a and b , then
1 7 10. (a) : Let the four points are ^
^
12. (d) : Let a = i
B
R1
R 3i +6 j −2k Unit vector along AC = = |R| 9 + 36 + 4
| DA | = (5 − 7)2 + (−1 + 4)2 + (5 − 7)2
3 2
( a , b , c are unit vectors)
ab = 0 2 2 (a + b ) (a + mb ) = a + m a b + b a + m b = 0 ( (a + b ) (a + mb))
| a |2 +(m + 1)a b + m | b |2 = 0
2
2
|a | + m|b | = 0 m = −
| a |2
| b |2
MATHEMATICS TODAY | NOVEMBER ‘15
67
^ ^ 17. (d) : Let a and b be two unit vectors.
22. (a) : Given that a + b + c
^ ^
^2
| a | +2(a b) + | b | = 1 2 a b = −1 1 2
^
^
a^ b = − | a^ || b | cos q = −
1 2
^
^
^ ^
^
^
^ ^
1− 0 − 0
=
2 2
^
^ ^ ^
^ ^
1 2
p 1 q = A(i^+ ^j ) 2 3
^
^
+ 2 j + 2k
| a | = 12 + 22 + 22 = 3
^
C ( i + k ) ^
^
1 2
1 2
Required area = | a b | = | a || b | sin q
B ( j + k)
p 15 1 = 3 5 sin =
− 2 ^j + 3 k^
19. (a) : Here, a = i
^
23. (a) : Here a = i
AB = ( j + k ) − (i + j ) = k − i
cos q =
q = 60°
18. (d) : AC = ( i + k ) − (i + j ) = k − j
30 cos q = 15 cos q =
2p p cos q = cos p − q = 3 3
| a |2 + | b |2 + 2 a b = | c |2 9 + 25 + 2 · 3 · 5cosq = 49
^ ^
We have, | a + b | = 1 ^ 2
=0
a + b = −c | a + b |2 =| −c |2
^2
^
2
6
4
24. (b) : Since a and b are two unit vectors,
| a | = 1 + 4 + 9 = 14
|a | = |b | = 1
Also, a b = | b |2 and | a − b | = 7
| a − b |2 = 7
| a | −2 a b + | b | = 7
2
Now, a b = | a || b |cos q = 1 1 cos
2
20. (a) : | a + b + c |2 = | a |2 + |b |2
25. (d) : We have, a b = c
+ |c |2 + 2(a b + b c + c a )
a (b + c ) = 0 b (c + a ) b (c + a ) = 0 and c (a + b ) c (a + b ) = 0
a c and b c Also, c a = b c b and a b
Adding (ii), (iii) and (iv), we get 2(a b + b c + c a ) = 0
(i) becomes | a + b + c |2 = | a |2 + |b |2 + | c |2
... (i)
... (ii)
... (ii)
... (iii) ... (iv)
From (i) and (ii), we get a c and a b a ||(b c )
26. (a) : Here, | a | = | b |
| a − b | = | a |2 + | b |2 −2 | a || b | cos q
... (i)
a (b + c )
2
|a + b | = 3 1
3
| a + b |2 = | a |2 + |b |2 + 2(a b ) = 3
14 − 2 | b |2 + | b |2 = 7 | b | = 7
p 1 =
= | a |2 + | a |2 −2 | a |2 cos q
= 2 ^i + ^j + k^, = i^ − 2 ^j + 2 k^
21. (a) : Here ^
^
^
and = 3 i − 4 j + 2 k
^
^
^
q
2 q 1 q | a − b | = 2a sin a sin = | a − b | 2 2 2
+ = 5 i − 3 j + 3 k Now, projection of ( + ) on is ( + ) 1 + (−2)(−3) + 2 3 17 = = 2 2 2 3 | | 1 +2 +2
= 2a2 (1 − cos q) = 4a2 sin2
68
MATHEMATICS TODAY | NOVEMBER ‘15
= 2, | b | = 3, | c | = 4 [a b c ] = a (b c ) = a | b || c |sin 2p n^ 3
27. (c): Given that | a |
2p ^ ^ = | a || b || c |sin [ a n = | a || n |cos 0 = | a |] 3
3 = 12 3 2 28. (c): Given two vectors lie in xy -plane. erefore, a vector coplanar with them is =234
^
31. (a) : Let
^
^
^
x 2 + y 2 + z 2 = 144
^
Also given that
(1 + 2 ) − (2 + 3) = 0 –1 – = 0 = –1
^
^ ^
^ ^
^
^ ^
^
2
2
2
erefore, the required vectors are
and c = y i + x j + (1 + x − y ) k −1 1 0 Now, [a b c] = x 1 1− x y x 1 + x − y
^
(a b )2 = | a |2 | b |2 −( a b)2 | a b |2 = | a |2 | b |2 −(a b)2
= 16 – 4 = 12
1 0 0 [a b c ] = x 1 1 = 1[(1 + x) − x ] = 1 y x 1 + x
us, [a b c ] depends neither on x nor on y .
30. (b) : Given that | a b |2 + | a b |2 = 144
^
32. (b) : By Lagrange’s identity, we have
Applying C 3 → C 3 + C 1 , we get
^
^ ^ ^ = −4 i + 8 j + 8 k or = 4 i − 8 j − 8 k
^
^
9 = 144 = 16 = 4 x = –4, y = z = 8 or x = 4, y = z = –8
^ ^
^
^
... (iii)
2
^ ^
29. (a) : Here, we have a = i − k, b = x i + j + (1 − x) k
Again, | c |2 = (2a b − 3b) (2a b − 3 b)
| c |2 = 4 | a b |2 −6b (a b ) − 6b (a b) + 9 | b |2
= 4 | a b |2 + 9 | b |2
( b a b) = [b a b] = 0)
= (4 × 12) + (9 × 16) = 192
| c | = 192 = 8 3
and | a | = 4
But | a b |2 + | a b |2 = | a |2 | b |2
Now, b c = b (2a b − 3b) 2
144 = 16 | b |2
^ ^
... (ii) ^
2
i+ j a (− i − j) = = |a | 1+1 2
^
+ 4 + 4 = 144
^
a = − i − j.
^
x y z = = = (say) −1 2 2 x = –, y = 2 and z = 2 Putting these values in (i), we get
^ ^
^
^
i.e.,
a (i − j ) = 0
... (i)
From (ii) and (iii), we get y x z = = 2 − 3 −6 + 8 4 − 2
Now, a is perpendicular to ^i − ^j
^
^
^
–2x + y – 2z = 0
a = (1 + 2 ) i + (2 + 3 ) j
^
and = 0 (x i + y j + zk) (−2 i + j − 2 k) = 0
^
Comparing coefficients of ^i and ^j , we get a1 = 1 + 2, a2 = 2 + 3
a=
^
a1 i + a2 j = (1 + 2 ) i + (2 + 3) j
^
^
4x – y + 3z = 0
^
= 0 (x i + y j + z k) (4 i − j + 3 k) = 0
Since a = (^i + 2 ^j ) + (2 ^i + 3 ^j )
^
Given that | | = 12
^
^
= x i + y j + z k be the required vector.
a = a1 i + a2 j
^
^
= 4 i − j + 3 k , = −2 i + j − 2 k and
= 0 − 3 b = −3 16 = −48
| b |2 = 9 | b | = 3
Let the angle between b and c be q MATHEMATICS TODAY | NOVEMBER ‘15
69
3 cos q = = =− 2 | b || c | 4 8 3 cos q = cos
5p 6
q=
5p 6
^
a
^
^
[ a a = b b = 0]
^^
^
^
^^
^
^
^
^
^
^
^
i
^
^
^
^
^
^
^
^
^
^
^
^
^
^ ^
^
^
^
^
^
^
^
^
^
^
^
^
1
a (b c ) = 0 1
^
^
2 −1 = 0
2 −1
(2 – 1) – 1( + 2) + 2(–1 – 2 ) = 0 3 – – – 2 – 2 – 4 = 0 3 – 6 – 4 = 0 3 + 22 – 22 – 4 – 2 – 4 = 0 2( + 2) – 2( + 2) – 2( + 2) = 0 ( + 2)(2 – 2 – 2) = 0 either + 2 = 0 = –2 or 2 – 2 – 2 = 0
=
2 4 − 4 1(−2) 2 2 3 = =1 3 2 1 2
= a b− a c − b b + b c
= ( + ) ( + + + )
= ( + ) ( + + ) [ = 0]
MATHEMATICS TODAY | NOVEMBER ‘15
= a b + b c + c a [ b b = 0 and a c = − c a]
(a − b ) (b − c ) = a b + b c + c a 70
^
^
38. (a, b, c, d) : (a − b ) (b − c )
= ( + ) {( + ) ( + )}
^
e values of are –2, 1 + 3 and 1 − 3 .
35. (b) : We have [ + + + ]
^
^
k
= 3 + 7 + 25 = 7 + 28 By the condition of coplanarity, we have 7 + 28 = 0 = –4
^
^
j
= (3 i + j + 5 k) ( i + 7 j + 5 k )
= 3a − (x i + y j+ z k) = 3a − a = 2 a
(3 i + j + 5 k) {(2 i − j + k) ( i + 2 j − 3 k)}
^ ^
= a − ( i a) i + a − ( j a) j + a − ( k a) k
= i + 7 j + 5k
^
c = 2 i − j + k are coplanar.
Now, (2 i − j + k) ( i + 2 j− 3 k) = 2 −1 1 1 2 −3
^
^
^
37. (a, b, c) : Here, a = i + j + 2 k, b = i + j − k and
their scalar triple product will be zero.
^
34. (b) : We know, three vectors will be coplanar, if
^
^
= (i i )a − (i a ) i + (j j )a − (j a ) j + (k k )a − ( k a) k
p = 6q k = 6
^
^
(c) i^ (a i^) + ^j ( a ^j) + k^ ( a k^)
= 6| a b |
^
^
^ ^ ^ (b) (a i )2 + (a j)2 + (a k)2 = x2 + y2 + z2 = | a |2
^ ^
A
1 2
^ ^
= x i + y j+ z k = a
= | 0 + 12(a b ) + 0 |
= |10(a a) + 2(a b) + 10(a b) + 2( b b)|
^
[ a i = ( x i + y j+ z k) i = x]
b
^ ^
B
C
1 = | a (10a + 2b ) + (10a + 2b) b | 2
^
O
^
(a) (a i) i + (a j) j+ ( a k) k
= | OA OB + OB OC |
1 2
36. (a, b, c) : Let a = x i + y j+ z k
1 1 (OA OB) + (OB OC) 2 2
= 2
Here p = area of the quadrilateral OABC = ar(BOA) + ar(COB)
= [ ] + 0 + 0 + 0 + 0 + [ ]
33. (c):
1 2
+ ( ) + ( )
= [ ] + [ ] = 2[ ]
=
= ( ) + ( ) + ( ) + ( )
−48
b c
Similarly, it can be shown that
^
(c − b ) (a − c ) and (a − b ) (a − c )
Given that | a | = | b | = | c | = 2
each is equal to a b + b c + c a
x 2 + y 2 + z 2 = 2 a b b c c a = = Also, cos q =
1 2
1 2
1 2
= [2(a b ) + 2( b c ) + 2( c a)]
= a b+ b c + c a
2 2
1 {a (b − c ) + b (c − a) + c (a − b )} 2
39. (a, b, c) : [ a
c ]
b
^
^
^
nn
1 3
c = (− i + 4 j − k )
Let A = a, B = b , C = c
^
4 1 When y = , x = z = − 3 3
= a b+ b c +c a
^
2 2
When y = 0, x = z = 1 c = i + k
= [(a b + b c + c a) − (a c + b a + c b)]
1 y + z x + y = = 2 2 2 x + y = 1, y + z = 1 Subtracting, x – z = 0 x = z = 1 – y Putting the values of x and z in (i), we get (1 – y )2 + y 2 + (1 – y )2 = 2 3 y 2 – 4 y = 0 y (3 y – 4) = 0 y = 0 or 4/3
2 2
= (a b − a c + b c − b a + c a − c b )
... (i)
1 {a (b − c ) + b (c − a) + c (a − b)} 2
Now,
^
^
40. (a, d) : Let c = x i + y j + z k
So, [ A B C] = A ( B C) = B (C A) = C ( A B) Now,
[ A B C] = A (B C) = ( a) [( b) ( c )]
= ( a ) [{( b ) c } − {( b ) }c ]
[ A B C] = [ b c ][ a ] − [ b ][ a c ]
[ a b c ] = [ a ][c b]
− [ a c ][ b]
Again, [ A B C] = B (C A) = ( b ) [( c ) ( a)]
= ( b ) [{( c ) a} − {( c ) }a]
[ A B C] = [ c a][ b ] − [ c ][ b a]
[ a b c ] = [ c a][ b ]
− [ c ][ b a]
Similarly, [ A B C] = C ( A B)
[ a b c ] = [ a b ][ c ]
− [ a ][b c ]
MATHEMATICS TODAY | NOVEMBER ‘15
71
Similarly, it can be shown that
^
(c − b ) (a − c ) and (a − b ) (a − c )
Given that | a | = | b | = | c | = 2
each is equal to a b + b c + c a
x 2 + y 2 + z 2 = 2 a b b c c a = = Also, cos q =
1 2
1 2
1 2
= [2(a b ) + 2( b c ) + 2( c a)]
= a b+ b c + c a
2 2
1 {a (b − c ) + b (c − a) + c (a − b )} 2
39. (a, b, c) : [ a
c ]
b
^
^
^
nn
1 3
c = (− i + 4 j − k )
Let A = a, B = b , C = c
^
4 1 When y = , x = z = − 3 3
= a b+ b c +c a
^
2 2
When y = 0, x = z = 1 c = i + k
= [(a b + b c + c a) − (a c + b a + c b)]
1 y + z x + y = = 2 2 2 x + y = 1, y + z = 1 Subtracting, x – z = 0 x = z = 1 – y Putting the values of x and z in (i), we get (1 – y )2 + y 2 + (1 – y )2 = 2 3 y 2 – 4 y = 0 y (3 y – 4) = 0 y = 0 or 4/3
2 2
= (a b − a c + b c − b a + c a − c b )
... (i)
1 {a (b − c ) + b (c − a) + c (a − b)} 2
Now,
^
^
40. (a, d) : Let c = x i + y j + z k
So, [ A B C] = A ( B C) = B (C A) = C ( A B) Now,
[ A B C] = A (B C) = ( a) [( b) ( c )]
= ( a ) [{( b ) c } − {( b ) }c ]
[ A B C] = [ b c ][ a ] − [ b ][ a c ]
[ a b c ] = [ a ][c b]
− [ a c ][ b]
Again, [ A B C] = B (C A) = ( b ) [( c ) ( a)]
= ( b ) [{( c ) a} − {( c ) }a]
[ A B C] = [ c a][ b ] − [ c ][ b a]
[ a b c ] = [ c a][ b ]
− [ c ][ b a]
Similarly, [ A B C] = C ( A B)
[ a b c ] = [ a b ][ c ]
− [ a ][b c ]
MATHEMATICS TODAY | NOVEMBER ‘15
71
Series 6
YOUR WAY CBSE XII INDEFINITE INTEGRALS
DEFINITION
Let f ( x ) be a function. Then the family of all its primitives (or antiderivatives) is called the indefinite integral of f (x ) and is denoted by ∫ f (x )dx . e symbol f (x )dx is read as the indefinite integral of f (x ) with respect to x .
d (( x ) + C ) = f (x ) f (x )dx = (x ) +C , us, dx where (x ) is primitive of f (x ) and C is an arbitrary
constant known as the constant of integration. Here, ∫ is integral sign, f (x ) is the integrand, x is the variable of integration and dx is the element of integration or differential of x . GEOMETRICAL INTERPRETATION OF INDEFINITE INTEGRAL
e statement f (x )dx = F (x ) + C = y (say) represents a family of curves. e different values of C will correspond to different members of this family. Hence, the indefinite integral of a function represents geometrically, a family of curves placed parallel to each other having parallel tangents at the points of intersection of the curves of the family with the lines perpendicular to the axis representing the element of integration.
72
MATHEMATICS TODAY | NOVEMBER ‘15
INTEGRALS, APPLICATION OF INTEGRALS & DIFFERENTIAL EQUATIONS
SOME FUNDAMENTAL INTEGRATION FORMULAS
x n+1 x dx = + C, n −1 n +1
1 (ii) dx = log | x | + C x (iii) e dx = e + C a (iv) a dx = + C log a (v) sin x dx = − cos x + C (vi) cos x dx = sin x + C (vii) sec x dx = tan x + C (viii) cosec x dx = − cot x + C (ix) sec x tan x dx = sec x + C (x) cosec x cot x dx = −cosec x + C (i)
n
x
x
x
x
2
2
(xi)
(xii)
x + C 2 a − x 2 1 x − a2 − x 2 dx = cos−1 a + C
a
1
dx = sin−1
1 −1 x dx tan + C = a a a2 + x 2 1 1 x − 2 2 dx = cot −1 + C a a a + x
(xiii)
(xiv)
(xv)
x
(xvi)
1
1 x dx = sec−1 + C a a x 2 − a2 1 1 x − dx = cosec−1 + C a a x x 2 − a2 1
SOME PROPERTIES OF INDEFINITE INTEGRAL (a) The process of differentiation and integration d are inverse of each other i.e., f ( x )dx = f ( x ) dx and f ( x )dx = f ( x) + C, where C is any arbitrary constant. (b) Two indefinite integrals with the same derivative lead to the same family of curves so they are equivalent. So, if f and g are two functions such d d f ( x ) dx = g ( x) dx , then f ( x ) dx that dx dx
(d) e integral of the sum or the difference of two functions is equal to the sum or difference of their integrals i.e., [ f 1(x ) ± f 2(x )]dx = f 1(x )dx ± f 2(x )dx Similarly, [ f 1(x ) ± f 2(x ) ± ... ± f n(x )]dx = f 1(x )dx ± f 2(x )dx ± ... ± f n(x )dx METHODS OF INTEGRATION (a) Integration by substitution (b) Integration using partial fractions (c) Integration by parts (a) Integration by Substitution By suitable substitution, the variable x in f (x ) dx is changed into another variable t so that the integrand f (x ) is changed in to F (t ), which is some integral or algebraic sum of standard integrals.
and g ( x ) dx are equivalent. (c) If c is any constant and f (x ) is a function of x , then c f (x )dx = c f (x )dx . Some Important Substitutions Integrals (i) (ii) (iii)
(iv)
Substitutions
f (ax + b)dx xn−1 f (xn )dx n
f (x )
Put ax + b = t Put x n = t
f (x )dx or
dx
ax2 + bx + c
or
f x )
f (x ) dx
Put f (x ) = t
dx
Transformed into standard form by expressing
b 2 c b2 ax + bx + c = a x + + − 2a a 4a2 b then put x + = t
ax 2 + bx + c
2
2a
(v) (vi)
(ax + b)
cx + d dx or
dx + e dx or 2 ax + bx + c
or (dx + e) (vii)
ax
2
ax + b
cx + d dx dx + e ax
2
dx
+ bx + c
Put (dx + e) =
A
d dx
(ax 2 + bx + c) + B
+ bx + c dx
a sin x + b cos x dx c sin x + d cos x
(b) Integration by Partial Fractions If p(x ) and q(x ) are two polynomials such that the degree of p(x ) is less than the degree of q(x ), then we can evaluate p( x ) p( x ) into partial fractions. dx by decomposing q(x ) q(x )
Put ax + b = A(cx + d ) + B
Put a sin x + b cos x = A
d dx
(c sin x + d cos x) + B(c sin x + d cos x)
Method : First resolve the denominator of the given fraction into simplest factors. On the basis of these factors, we obtain the corresponding partial fraction as per rules given below:
MATHEMATICS TODAY | NOVEMBER ‘15
73
Factor in the denominator
A
(x – a)(x – b)
(i)
+
(iii) (x – c)3
B
A B + (x − c) (x − c)2
+
(iv)
(x – a)2(x – b)
(v)
(ax 2 + bx + c)
(iv)
x −a x −b A B + (x − b) (x − b)2
(x – b)2
(ii)
cosec x dx = log | cosec x − cot x | + C 1 dx x−a (v) = +C log 2 2 x+a 2a x −a
Corresponding partial fraction
(vi)
dx
(vii)
C 3
( x − c) A B C + + x − a (x − a)2 x − b
x
Ax + B
2
+ a2
dx
(viii)
(ix)
x
2
− a2
x+
x
2
+ a2 + C
= log
x+
x
2
− a2 + C
x
a2 + x2 dx = 2 +
2
(ax + bx + c)
(x – a)(bx + cx + d ) where bx 2 + cx + d A + Bx + C cannot be factorised x − a bx 2 + cx + d further
(x)
x
2
− a2 dx =
(c) Integration by Parts
2
a
a
2 x
2
x
a
2
+ x2
2
x
− 2
+C
a−x
= log
2
(vi)
a+x
1
dx
a2 − x2 = 2a log
log 2
log 2
x+
a
2
+ x2 + C
x
2
− a2 + C
− a2
2
2
2
x+ a
2
x +C
sin −1
The process of integration of the product of two functions is known as integration by parts. If u and v are two functions of x , then du uvdx = u ( vdx ) − vdx dx dx In words, integral of the product of two functions = first function × integral of the second – integral of (differential coefficient of the first function × integral of the second function). Note : We choose the first function as the function which comes first in the word ILATE, where I–stands for the inverse trigonometric function (sin–1x , cos–1x , tan–1x etc.) L – stands for the logarithmic functions A – stands for the algebraic functions T – stands for the trigonometric functions E – stands for the exponential functions If there is no other function, then unity is taken as the second function.
DEFINITION Let F (x ) be antiderivative of f (x ), then for any two values of the independent variable x , say a and b, the difference F (b) – F (a) is called the definite integral of
Some Particular Integrals
DEFINITE INTEGRAL AS THE LIMIT OF A SUM Let f be a continuous function defined in a closed
(ii) tan x dx = − log | cos x | +C = log |sec x | + C (iii) sec x dx = log |sec x + tan x | + C (i)
74
cot x dx = log |sin x | + C = − log cos ec x + C
MATHEMATICS TODAY | NOVEMBER ‘15
(xi)
a
−x
dx =
2
a
−x +
2
a
Special Integrals
ex [ f (x ) + f (x )]dx = ex f (x ) + C (ii) ekx [kf (x ) + f (x )]dx = ekx f (x ) + C (i)
DEFINITE INTEGRALS
b
f (x ) from a to b and is denoted by b
f (x)dx . us a
f (x)dx = F(b) − F(a) a
where numbers a and b are called the limits of integration; a is the lower limit and b is the upper limit.
b
interval [a, b], then the definite integral f ( x )dx is the a
area bounded by the curve y = f (x ), the ordinates x = a, x = b and the x -axis.
Also,
APPLICATION OF INTEGRALS
b
h f (a) + f (a + h) + ... + f {a + (n − 1)h} f (x)dx = hlim →0 a
where, h =
1.
y = f (x ) be a continuous and finite function in [a, b]. Case I : If the curve y = f (x ) lies above the x - axis, then the area bounded by the cur ve y = f (x ), x -axis and the ordinates x = a and x = b is given
b−a , h → 0 as n → n
FUNDAMENTAL THEOREM OF INTEGRAL CALCULUS (i) Area function : e function A(x ) denotes the x
area function and is given by A(x ) = f (x )dx . a
(ii) First Fundamental Theorem : Let f (x ) be
b
a continuous function defined in the closed interval [a, b] and A(x ) be the area function, then A(x ) = f (x ), for all x [a, b]. (iii) Second Fundamental Theorem : Let f (x ) be a continuous function defined in the closed interval [a, b] and f (x ) dx = F (x ), then b
by y dx a
Case II : If the curve
y = f (x ) lies below the x - axis, then the area between the curve y = f (x ), x - a x i s a n d t h e ordinates x = a and x = b is given by
f (x)dx = [F(x)]a = F(b) − F(a). b
a
PROPERTIES OF DEFINITE INTEGRALS (i)
(ii)
b
b
a
a
(− y) dx
a
a
b
a
2.
f (x)dx = − f (x)dx a
(iii)
f (x)dx = 0 a
(iv)
(v)
(vi)
b
f (x)dx = f ( y)dy b
Area of a curve between two ordinates : Let
Area of a curve between two abscissas : Case I : If the
curve x = f ( y ) lies to the right of the y - axis, then the area bounded by the curve x = f ( y ), y - a x i s a n d t h e abscissa y = c and y = d is given by
b
c
b
a a
a a
c
0
0
c
b
b
C a s e I I : If the
f (x)dx = f (x)dx + f (x)dx, a c b f (x)dx = f (a − x)dx f (x)dx = f (a + b − x)dx a
a
if f (x ) is an odd function 0, a (vii) f (x )dx = 2 f ( x )dx, if f (x ) is an even function −a 0 if f (2a − x ) = − f ( x ) 0, 2a a (viii) f (x )dx = 2 f ( x )dx , if f (2a − x ) = f (x ) 0 0 a
d
x dy curve x = f ( y ) lies to the left of the y - axis, then the area bounded by the curve x = f ( y ), y - a x i s a n d t h e abscissa y = c and y = d is given by d
(− x) dy c MATHEMATICS TODAY | NOVEMBER ‘15
75
If some portion of the curve is above the x -axis and some is below the x -axis as shown in the figure, then A1 < 0 and A2 > 0. erefore, the area A bounded by the curve y = f (x ), x -axis and the ordinates x = a and x = b is given by A = | A1| + A2.
Not e :
DIFFERENTIAL EQUATION
DEFINITION
An equation involving an independent variable, dependent variable and the derivatives of the dependent variable, is called a differential equation. A differential equation involving derivatives of the dependent variable with respect to only one independent variable is called an ordinary differential equation and a differential equation involving derivatives with respect to more than one independent variables is called a partial differential
a
X
b
equation.
Y
3.
ORDER AND DEGREE OF A DIFFERENTIAL EQUATION
Area between two curves : Case I : The area Y
bounded by two curves y = f (x ) and y = g ( x ), which intersects at the ordinates x = a and x = b, is given by
y = f(x )
X
y = g(x ) x = b x=a X a b
O
Y
e order of highest order derivative appearing in a differential equation is called the order of the differential equation. e power of the highest order derivative appearing in a differential equation, aer it is made free from radicals and fraction, is called the degree of the differential equation. Note :
b
f (x) − g (x) dx
a
Case II : The area
Y
bounded by two y = d x d c u r v e s x = f ( y ) = g and x = g ( y ) which y intersects at the y = c c abscissas, y = c and X O y = d is given by
SOLUTION OF A DIFFERENTIAL EQUATION
x =
f y
X
Y
d
| f ( y ) − g ( y)| dy c
If f (x ) ≥ g (x ) in [ c, e] and f (x ) ≤ g (x ) in [ e, d ] where c < e < d , then area bounded by the curves is,
Note :
e
d
c
e
Y
y = fx
y = gx
y = gx
y = f x
c
O
e
d
Y 76
Any relation between the dependent and independent variables (not involving the derivatives) which, satisfy the given differential equation is called a solution of the differential equation. General Solution : e solution which contains arbitrary constants is called the general solution (primitive) of the differential equation. Particular Solution : e solution obtained from the general solution by giving particular value to arbitrary constants is called a particular solution. FORMATION OF A DIFFERENTIAL EQUATION WHOSE GENERAL SOLUTION IS GIVEN
[ f (x) − g(x)] dx + [ g(x) − f (x)] dx
X
e degree of a differential equation which is not a polynomial equation in derivatives is not defined. Order and degree (if defined) of a differential equation are always positive integers.
MATHEMATICS TODAY | NOVEMBER ‘15
X
Suppose an equation of a family of curves contains n arbitrary constants (called parameters). Then, we obtain its differential equation, as given below. Step I : Differentiate the equation of the given family of curves n times to get n more equations. Step II : Eliminate n constants, using these ( n + 1) equations. This gives us the required differential equation of order n.
METHODS OF SOLVING FIRST ORDER, FIRST DEGREE DIFFERENTIAL EQUATIONS dy (i) If the equation is = f (x ) , then y = f (x )dx + C dx is the solution. (ii) Variable separable : If the given differential equation can be expressed in the form f (x )dx = g ( y )dy , then f (x )dx = g ( y )dy + C is the solution. (iii) Reducible to variable separable : If the equation dy is = f (ax + by + c), then put ax + by + c = z . dx (iv) Homogeneous equation : If a first order, first degree differential equation is expressible in the form dy f (x , y ) , = dx g (x , y )
where f (x , y ) and g (x , y ) are homogeneous functions of the same degree in x and y , then put y = vx dy (v) Linear Equation : If the equation is + Py = Q, dx where P and Q are functions of x , then y e
Pdx
= Q e
where e
Pdx
Pdx
dx + C ,
SHORT ANSWER TYPE
7. For what values of c and a is the following equation satisfied? 1 (sin 2 x + cos 2 x )dx = sin(2 x − c ) + a 2 log x − 1 dx 8. Evaluate : (log x )2
9. If the area enclosed between the curves y = ax 2 and x = ay 2 (a > 0) is 1 square unit, then find the value of a. 10. e line normal to a given curve at each point (x , y ) on the curve passes through the point (3, 0). If the curve contains the point (3, 4), find its equation. p /2 sin2 x 11. Evaluate : dx (1 + sin x cos x )
0
dy 12. Solve : x − y = log x dx LONG ANSWER TYPE
13. Evaluate :
is the integrating factor (I.F.). OR
dx + Px = Q, where P and Q are dy functions of y , then If the equation is x e
= Q e Pdy dy + C, Pdy where e is the integrating factor. Pdy
VERY SHORT ANSWER TYPE
1. Evaluate : 2(x +3)dx p /2 2. Evaluate : x cos x dx
x
8 + x − x 2 dx
3
14. Evaluate
(x2 + x)dx as limit of sum. 1
15. Evaluate :
x 4
(x − 1)(x 2 + 1) dx
16. Solve :
x cos y ( y dx + x dy) = y sin y (x dy − y dx ) x x 17. Find the area of the region {(x , y ) : x 2 + y 2 2ax , y 2 ax , x 0, y 0}. 18. Evaluate the following integrals : x 2 + 1 x 2 + 9 (i) (ii) dx dx 4 2 2 x − 2 x + 81 x + 4
0
3. Find the differential equation of the family of curves y = Aex + Be–x , where A and B are arbitrary constants. 4. Determine the order and degree of the differential dy dy + sin = 0 equation dx dx dy x + y 5. Solve : =e dx 6. Evaluate : cos3x sin x dx
SOLUTIONS
1.
2(x +3)dx = 2x 23dx = 8 2x dx
2( x +3) = + C log 2 p /2 p /2 /2 p x cos x dx = [ x sin x]0 − 1 sin x dx 2 x + C = 8 log 2
2.
0
0
[Integrating by parts]
p p = + [cos x ]0p/2 = − 1 2
2
MATHEMATICS TODAY | NOVEMBER ‘15
77
3.
Given equation is y = Aex + Be–x dy Now, = Ae x − Be− x dx
4.
5.
dx 2 d 2 y
= Ae x + Be− x = y
log x − 1
(log x )2 dx
Let I =
Put log x = t x = et dx = et dt I = t −2 1 et dt = et 1 − 12 dt t t t 1 1 = et dt − et dt t t Integrating first integral by parts, we get 1 1 1 t I = et − − et dt − e dt 2 t t t 2 x 1 = et + C = + C t log x
− y = 0 is required differential equation. 2
We have,
8.
[From (i)]
dx e highest order derivative present in the dy differential equation is . So, it is of order 1. dx Also, L.H.S. of the differential equation cannot be dy expressed as a polynomial in . So, its degree is dx not defined.
6.
d 2 y
...(i)
9.
Solving, y = ax 2 and x = ay 2, we get point of 1 1 intersection O(0, 0) and A , . a a
dy x + y =e dx
Y
Q(x,y 2)
dy x y = e e e–y dy = e x dx dx x – y e –y dy =x e dx –e = e + C , which is the required solution.
y =
ax 2
X
O (0, 0)
We have, cos3 x sin x dx t 4 cos x sin x dx = − t dt = − + C 4 1 = − cos4 x + C 4 cos 2 x sin 2 x (sin 2 x + cos 2 x )dx = − + + C , 2 2 where C is an arbitrary constant. 1 = (sin 2x − cos 2 x ) + C 2 1 1 1 = sin 2x − cos 2x + C 2 2 2 1 p p = cos sin 2x − sin cos 2x + C 4 4 2 p 1 = sin 2x − + C 4 2 1 sin(2 x − c ) + a But (sin 2 x + cos 2 x )dx = 2 1 1 p sin 2 x − + C = sin(2 x − c) + a 4 2 2
3
Y
3
Since, area enclosed between the curves = 1 1/a
x 2 a − ax dx = 1 0 1/a 2 3/2 a 3 x − x = 1 3 0 3 a
78
1 1 , a a
P(x,y 1)
X
Put cos x = t sin x dx = –dt
7.
A
c =
p and a = C an arbitrary constant. 4
MATHEMATICS TODAY | NOVEMBER ‘15
10.
2 3 a 1 3a
2
1
a
a 3
− 3/2
=1
1
a3 1 a2 = 3
=1
2
3a 1
3a2
a =
− 2
=1
[ a > 0]
3
e equation of the normal to a curve at a point (x , y ) is dy (Y − y ) + ( X − x ) = 0 dx Since it passes through the point (3, 0), we have dy dy (0 − y ) + (3 − x ) = 0 y = (3 − x ) dx dx 2 2 y = 3x − x + C y dy = (3 − x) dx 2 2 x 2 + y 2 – 6x – 2C = 0, Which is the equation of the curve.
Since the curve passes through (3, 4), we have 9 + 16 – 18 – 2C = 0 C = 7/2 2 x + y 2 – 6x – 7 = 0 is the required equation of the curve.
p /2
11.
Let I =
0
sin2 x dx (1 + sin x cos x )
p /2
en, I =
0
1
I .F . = e
...(i)
I =
or
0
2 I =
0 p /2
2
2
(sin x + cos x ) dx = (1 + sin x cos x )
=
sec2 x
2 0 (sec x + tan x )
dx
(1 + sin x cos x ) 0
dx
(By dividing numerator and denominator by cos2x ) p /2 sec2 x dt = dx = , 2 2 0 (1 + tan x + tan x ) 0 (t + t + 1)
sec2x dx = dt , p When x = 0 t = 0 and x = t → ] 2 2 dt 1 2t + 1 − tan = = 2 3 2 3 0 1 3 0 t + + 2 2 2 1 = tan−1 () − tan−1 3 3 p p 2 2p = − = 3 2 6 3 3 [Put tanx = t
12.
e given equation can be written as dy 1 log x − y = dx x x dy + Py = Q , where is is of the linear form dx −1 log x . P = and Q = x x
y 1 1 1 = (log x ) − − − dx + C x x x x [Integrating by parts]
...(ii)
p /2
x
y 1 = (log x ) 2 dx + C x x
Adding (i) and (ii), we get
p /2
1 log x 1 y x = x x dx + C
sin2 [(p/2) − x ] dx 1 + sin[(p/2) − x]cos[(p/2) − x ]
cos2 x dx (1 + sin x cos x )
−1
= e− log x = e log(x ) = x −1 =
So, the required solution is y × I .F . = {Q × (I.F.)}dx + C
a a f (x)dx = f (a − x)dx 0 0
p /2
− x dx
13.
− log x 1 y log x 1 − + C =− + 2 dx + C = x x x x x y = Cx – (log x + 1)
x
8 + x − x 2 dx
Let I =
Let x = A(1 – 2x ) + B or x = –2 Ax + (B + A) Equating the coefficient of x and constant term, we get 1 1 –2 A = 1 A = – and 0 = B + A B = – A = 2 2 1 1 x = − (1 − 2 x ) + 2 2 1 1 − (1 − 2 x ) + 2 dx I = 2
=− Let
8 + x − x 2
1 2
1 − 2 x 8 + x − x 2
dx +
1 1 I = − I1 + I 2 , where 2 2 1 − 2 x and I 2 = 8 + x − x 2
I 1 =
1 2
dx
8 + x − x 2 dx
8 + x − x 2
1 − 2 x
8 + x − x 2 dx
Now, I 1 =
Put 8 + x – x 2 = t (1 – 2x )dx = dt I 1 = dt = t −1/2dt = 2 t = 2 8 + x − x 2 t 1/2 dx dx Also, I 2 = = 8 + x − x 2 8 − (x 2 − x )
MATHEMATICS TODAY | NOVEMBER ‘15
79
=
=
dx
8 − x 2 − 2 x
1 1 1 + + 2 4 4
dx 2
33 1 2 − x − 2 2
=
I = lim h 2n + 3h n(n − 1) h →0 2 + h2 n(n − 1)(2n − 1) 6 I = lim 2 2n + 6 n(n − 1) n n→ n 2 4 n(n − 1)(2n − 1) + 2 n 6 h = 2 n I = lim 4 + 6 n − 1 + 4 (n − 1)(22n − 1) n 3 n → n 1 4 = lim 4 + 6 1 − + 1 − 1 2 − 1 n 3 n n n→ 4 38 I = 4 + 6(1 − 0) + (1 − 0)(2 − 0) = 4 + 6 + =
dx 33 1 − x − 4 2
2
x − 1/2 = sin−1 33 /2
2 x − 1 = sin−1 33
I =−
1 2 8 + x − x 2 2
1 2 x − 1 + sin−1 + C 33 2
1 2x − 1 = − 8 + x − x 2 + sin−1 + C 33 2 3
14.
Let I = (x 2 + x ) dx
4
1
15.
By definition, we know b
h [ f (a) + f (a + h) + f (a + 2h) f (x)dx = hlim →0 a
x 4
I = lim h [{12 + 1} + {(1 + h)2 + (1 + h)} h →0
+ {(1 + 2h)2 + (1 + 2h)} + ... + {(1 + (n − 1)h)2 + (1 + (n − 1)h)}]
I = lim h [{12 + (1 + h)2 + (1 + 2h)2 + ... h →0
... + (1 + (n − 1)h)2 } + {1 + (1 + h) + (1 + 2h) +... + (1 + (n − 1)h )}]
I = lim h [{n + 2h(1 + 2 + 3 + ... + (n − 1)) h →0
+ h2 (12 + 22 + ... + (n − 1)2 )} +{n + h(1 + 2 + 3 + ... + (n − 1))}]
I = lim h n + 2h n(n − 1) h →0 2 + h2 n(n − 1)(2n − 1) + n + h n(n − 1) 6 2 80
MATHEMATICS TODAY | NOVEMBER ‘15
Let
3
x
(x − 1)(x 2 + 1) dx
where h =
I = lim h [ f (1) + f (1 + h) + f (1 + 2h) + f (1 + 3h) h →0 +... + f (1 + (n −1)h)]
3
We have,
+ ... + f (a + (n − 1)h)],
b−a n 3 −1 2 Here, a = 1, b = 3, f (x ) = x 2 + x and h = = n n
3
(x − 1)(x 2 + 1)
=
x 4
(x 3 − x 2 + x − 1) 1 = (x + 1) + 3 2 (x − x + x − 1) 4 1 x (x + 1) + = ...(i) (x − 1)(x 2 + 1) (x − 1)( x 2 + 1) 1
(x − 1)( x 2 + 1)
=
A Bx + C + 2 (x − 1) (x + 1)
en, 1 = A(x 2 + 1) + (Bx + C )(x – 1) ...(ii) 1 Putting x = 1 in (ii), we get A = 2 2 Comparing coefficients of x on both sides of (ii), we get 1 A + B = 0 B = – A = − 2 Comparing the constant terms on both sides of (ii), we get −1 A – C = 1 C = A − 1 = 2 1 1 x − − 1 1 = + 22 2 …(iii) 2 (x − 1)(x + 1) 2( x − 1) (x + 1) 1 1 (x + 1) x 4 = (x + 1) + − 2 2 2(x − 1) 2 (x + 1) (x − 1)(x + 1) [By (i) & (iii)]
