mathematicstoday nov 2015.pdf

MATHEMATICS TODAY | NOVEMBER ‘15 7

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8 Maths Musing Problem Set - 155

10 Jee Work Outs

16 You Ask, We Answer

18 Math Musing Solutions

19 Olympiad Corner

23 Ace Your Way CBSE XI(Series 6)

34 Concept Boosters (XI)

48 Concept Boosters (XII)

63 Mock Test PaperWB JEE 2016

72 Ace Your Way CBSE XII(Series 6)

82 Math Archives

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Vol. XXXIII No. 11 November 2015Corporate Office:Plot 99, Sector 44 Institutional Area, Gurgaon -122 003 (HR), Tel : 0124-4951200e-mail : [email protected] website : www.mtg.inRegd. Office:406, Taj Apartment, Near Safdarjung Hospital, Ring Road, New Delhi - 110029.Managing Editor : Mahabir SinghEditor : Anil Ahlawat

CONTENTS

162372

82

63

34

19

1048

8

MATHEMATICS TODAY | NOVEMBER ‘158

JEE MAIN1. 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

2. �e area bounded by the curves y = | 1 – x2| and y = 5 – |x| is

(a) 18 (b) 19 (c) 20 (d) 213. Let {a, b} be a subset of {1,2,3,…,100}. �e probability

of a2– b2 is divisible by 5 is

(a) 3199

(b) 3299

(c) 3499

(d) 3599

4. In a triangle ABC, if A : B : C = 1 : 2 : 4, then tan B tan C + tan C tan A + tan A tan B =

(a) –4 (b) –5 (c) –6 (d) –75. P is a point on the line r i j k

= + −5 7 2 + s i j k( )3 − + and Q is a point on the line r i j k

= − + +3 3 6 + t i j k( ).− + +3 2 4 If PQ

is parallel to the vector

2 7 5i j k + − , then | |PQ

=(a) 78 (b) 2 78 (c) 4 78 (d) 3 78 CED

JEE ADVANCED6. �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

COMPREHENSIONLet Z be a complex number such that Z = reiq and

ZZ

+ =1 2

7. If q p=6

, then r =

(a) 7 32+ (b)

3 12+

(c) 5 12+ (d) 5 1

2−

Maths Musing was started in January 2003 issue of Mathematics Today with the suggestion of Shri Mahabir Singh. The aim of Maths Musing is to augment the chances of bright students seeking admission into IITs with additional study material.

During the last 10 years there have been several changes in JEE pattern. To suit these changes Maths Musing also adopted the new pattern by changing the style of problems. Some of the Maths Musing problems have been adapted in JEE benefitting thousand of our readers. It is heartening that we receive solutions of Maths Musing problems from all over India.Maths Musing has been receiving tremendous response from candidates preparing for JEE and teachers coaching them. We do hope that students will continue to use Maths Musing to boost up their ranks in JEE Main and Advanced. Prof. Dr. Ramanaiah Gundala, Former Dean of Science and Humanities, Anna University, Chennai

Set 155

8. If q p=3

, then r =

(a) 7 32+ (b) 5 1

2−

(c) 3 1

2−

(d) 7 32−

INTEGER MATCH

9. In a quadrilateral ABCD, BC = 8, CD = 12, AD = 10,

∠A= ∠B = p3

. If AB m n= + , then the sum of the

digits of (m + n) isMATCHING LIST

10. Column-I Column-II

P.If lim ( ) ,

/

x

xx f x

xe

→+ +

=0

131

then lim ( )x

f xx→

=0 2

1. 0

Q. �e number of roots of the equation tan x + sec x = 2 cos x in [0, 2p] is

2. 2

R. Let x 2+ y2 – 4x – 2y – 11 = 0 be a circle. �e area of the quadrilateral formed by the tangents from the point (4, 5) with a pair of radii is

3. 4

S. Let S and S1 be the foci of the ellipse 9x2+ 5y2 = 30y and a point P be (3, 3). �e area of triangle PSS1 is

4. 6

5. 8 P Q R S(a) 2 2 5 4(b) 1 1 3 4(c) 5 4 3 2(d) 4 2 3 5

See Solution set of Maths Musing 154 on page no. 18

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


SECTION-1This 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.

1. �e area bounded by y = 2 – |2 – x| and yx

= 3| |

is k e− 3 3

2log ( ) , where k is

2. Let f (x) = [x] + {x}2, f : R → R, then area of �gure bounded by y = f –1(x), y = 0 between the ordinates x = 1/2 and x = 5 is ([·] = G.I.F) and {} = fractional

function) 40 2 12−

k, where k is

3. 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

4. If y = 2sin(x) + sin(2x), 0 x p, then the area enclosed by the curve and the x-axis is

5. �e value of p psin( ln )x

xdx

e

1

37

is

6. If Ax

x xdx B x

x xdx=

−=

− cos

cos sinsin

cos sin

/ /

0

3 2

0

3 2p p and

then the value of A + B is

7. �e area bounded by y = xe|x| and lines |x| = 1, y = 0 is

8. If ( ) ln( ),11

12

2

2 40

1 −+ +

=x dx

x xb then �nd the value of b.

SECTION-2This 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.

9. If xe

edx f x e g x C

x

xx

( )( ) ( ) ln ( ) ,

11 2

+= + − +

then

(a) f (x) = x – 1 (b) g x e

e

x

x( ) = + −

+ +

1 1

1 1

(c) g x e

e

x

x( ) = + +

+ −

1 1

1 1 (d) f (x) = 2(x – 2)

10. If l x xdx= sec2 4 cosec= + + +A x B x C x Dcot tan cot ,3 then

(a) A = –1/3 (b) B = 2(c) C = –2 (d) none of these

11. A curve g x x x x x x dx( ) ( ) ( )= + + + + 27 2 6 21 6 5 4 is passing through origin, then

(a) g( )1 37

7= (b) g( )1 2

7

7=

(c) g( )− =1 17

(d) g( )− =1 314

7

12. If f x xx x

dx( ) = +− +

8

4 24

2 2 and f (0) = 0, then

(a) f(x) is an odd function

PAPER-1

By : V i d yal a n k ar I n s t i t u t e , P ear l C en t r e, S en ap at i B ap at M ar g , D ad ar ( W) , M u m b ai - 28. T el .: ( 022) 2 4306367


(b) f(x) has range R(c) f(x) has atleast one real root(d) f(x) is a monotonic function

13. Let = f x ex

xx

( ) , .sin

0

If 3

3

1

4 ex

dx f k f lxsin( )

( ) ( )= − , then

(a) k = 64 (b) k = 32(c) l = 2 (d) l = 1

14. If f (x) = ae2x + bex + cx satis�es the conditions

f (0) = –1, f (loge2) = 31, ( ( ) ) ,log ( )

f x cx dxe

− =3920

4

then(a) a = 5 (b) b = –6 (c) c = 2 (d) a = 3

15. A function y = f (x) satisfying the di�erential

equation dydx

x y x xx

− + =sin cos sin ( )2

2 0 is such

that y → 0 as x → then

(a) lim ( )x

f x→

=0

1 (b) f x dx( )/

pp

20

2

(c) f x dx( )/

10

2p (d) f (x) is even

16. �e function f x t dtx

( ) = − 1 4

0 is such that

(a) It is de�ned on the interval [–1, 1](b) It is an increasing function(c) It is an odd function(d) �e point (0, 0) is the point of in�ection

17. If I dtt

I dtt

x

x1 2 2

1

11

1 1=

+=

+ and 2

/ for x > 0, then

(a) I1 = I2 (b) I1 > I2

(c) I2 > I1 (d) I x21

4= −−cot p

18. Which of the following de�nite integrals when simpli�ed reduces to zero?

(a) ln sincos/

/ 4 34 35

3 10 ++

xx

dxp

p(b) sin cos4 5

0

2

2 2x x dx

p

(c)

sinsin

4

0

xx

dxp

(d) e x x dxx−

− (cos sin )0

SECTION-3This section contains TWO questions. Each question contains two columns, Column I and Column II. Column I has four entries (A), (B), (C) and (D), Column 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.

19. Match the following:

Column I Column II

(A) If 2

1 4

x

xdx

−

= ksin–1(f (x)) + C,

then k is greater than

p. 0

(B) If

( )( )

xx x

dx5

7 6+

=+

+a xx

ck

kln ,1

then ak is less than

q. 1

(C) If x

x xdx

4

2 211

++

( )

= ++

+k x mx

nln | | ,1 2

where n is the constant of integration, then mk is greater than

r. 3

(D) If dxx5 4+ cos

=

+

−k m x Ctan tan12

then

k/m is greater than

s. 4

(A)

(B)

(C)

(D)

P Q R S T

P Q R S T

P Q R S T

P Q R S T


PAPER-2SECTION-1

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. �e order of the di�erential equation of all circles,

having centre on y-axis and passing through the origin is

2. Area bounded by y = x sin x and x-axis between x = 0 and x = 2p is kp, where k is

3. �e area enclosed between the curve y = loge(x + e) and the coordinate axes is

4. Evaluate 1 1

1 2

2 101x

xx

dx/

cos −

ec

5. If x x x

xdx k

2

0

22

2

sin sin cosp

p p

p

−= ,

then k is

6. Sum of the two integral e dx e dxx x( ) ( / )

/

/+ −

−

−+ 5 9 2 3

1 3

2 3

4

5 2 23 is

is

7. If f t dt x tf t dtx

x( ) ( ) ,

0

1 = + then �nd the value of 2f(1).

8. If lim(tan )

,x

xt dt

x k→

−

+=

1 2

02

2

1

p then k is

SECTION-2This 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.

9. If sin (sec ) ( ) ( ) ,x d x f x g x c= − + then (a) f (x) = secx (b) f (x) = tanx(c) g(x) = 2x (d) g(x) = x

10. If xx

dx f x g x C4

61 11

123

++

= − + − −tan ( ( )) tan ( ( )) , then

(a) both f (x) and g(x) are odd functions(b) f (x) is monotonic function(c) f (x) = g(x) has no real roots

(d)

f xg x

dxx x

c( )( )

= − + +1 3

3

11. If x x

x

e dx e f x cx x2

232

1

1

− +

+

= +

( )

( ) , then

(a) f(x) is an even function(b) f(x) is a bounded function(c) �e range of f(x) is (0, 1](d) f(x) has two points of extrema

12. �e solution of di�erential equation

( ) ( )tan1 02 1+ + − =

−y x e

dydx

y is

(a) 21 12xe e ky ytan tan− −

= +

(b) xe e ky y2 21 1tan tan− −= +

(c) xe y kytan tan−

= +−1 1

(d) ( ) tanx ke y− =−

21

13. Which of the following is the integrating factor of

x xdydx

y xlog log ?+ = 2

(a) x (b) ex

(c) loge(x) (d) loge(logex)

20. Match the following:Column I Column II

(A) x x dxln(sin ) =0

pp. − p

2

22ln( )

(B)ln( tan )1

0

4+ = q q

p

dq.

p4

12

2+ ln

(C)coscos sin

/ x dxx x10

2

+ +=

pr.

p8

2ln

(D)sin

( ) /

/ −

−=

1

2 3 20

1 2

1

x dx

xs.

p4

12

2− ln


14. �e value of min( [ ], [ ])x x x x dx− − − −−2

2is ([·]

denotes the greatest integer function) (a) 0 (b) 1(c) 2 (d) none of these

15. Let f(a) > 0, and let f(x) be a non decreasing

continuous function in [a, b]. �en 1

b af x dx

a

b

− ( ) is (a) less than equal to f (b) (b) greater than equal to f (a)(c) maximum value bf (b)

(d) minimum value f a

b a( )−

16. �e values of which satisfy sin sin/

xdx = 22

p

( [0, 2p]) are equal to (a) p/2 (b) 3p/2 (c) 7p/6 (d) 11p/6

SECTION-3This 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

Paragraph for Q. No. 17 & 18An even function f is de�ned and integrable everywhere and is periodic with period 2.

Also function g x f t dt g Ax

( ) ( ) ( ) .= = and 10

17. Function g(x) is (a) odd (b) even(c) neither even nor odd(d) can’t be determined

18. Value of g(2) in terms of A is (a) 2A (b) A/2 (c) 4A (d) A/4

Paragraph for Q. No. 19 & 20Sometimes equations which are not linear can be reduced to the linear form by suitable transformations.

For example an equation of the form dydx

Py Qyn+ = ,

where P and Q are functions of x only or constants and n ( 0 and 1) is a constant.

dydx

Py Qyn+ =

y

dydx

Py Qn n− −+ =1

... (i)

[Dividing by yn on both sides]

Let y1 – n = z − =−( )1 n ydydx

dzdx

n

Put the value of ydydx

n− in (i), we get

dzdx

n Pz n Q+ − = −( ) ( )1 1 which is linear in z.

So, solution is

z e C n Q e dxn Pdx n Pdx = + − − −( ) ( )( )1 11

i.e. y e C n Q e dxn n Pdx n Pdx1 1 11− − − = + − ( ) ( )( )

�is form is called Bernoulli’s equation.

19. Solution of dydx

y x yx

= −( )

( )

2, where (x) is a

function of x is (a) (x) = x + C (arbitrary constant)

(b) ( )xy

x C= + (arbitrary constant)

(c) y (x) = –x + C (arbitrary constant)(d) y (x) = x + C (arbitrary constant)

20. Solution of the di�erential equation

dydx xy x y

=+

112 2[ sin ]

is

(a) e y yx

Cy2 12

12 22(cos sin )− −

=

(b) e y yx

Cy2 12

12 22(cos sin )+ −

=

(c) e y yx

Cy2 12

12 22(cos sin )− +

=

(d) e y yx

Cy2 12

12 22(cos sin )+ +

=

ANSWERS KEYPAPER-1

1. (4) 2. (3) 3. (7) 4. (4) 5. (2) 6. (0) 7. (2) 8. (3) 9. (b, d) 10. (a, c)11. (a, c) 12. (a, b, c, d) 13. (a, d) 14. (a, b) 15. (a, b, c, d) 16. (a, b, c, d) 17. (a, d) 18. (a, b, c, d) 19. (A) → (p, q); (B) → (r, s); (C) → (p); (D) → (p, q) 20. (A) → (p), (B) → (r), (C) → (s), (D) → (s)

PAPER-21. (1) 2. (4) 3. (1) 4. (0) 5. (8) 6. (0) 7. (1) 8. (4) 9. (b, d) 10. (a, c, d)11. (a, b, c) 12. (a) 13. (c) 14. (b) 15. (a, b) 16. (a, b , c, d) 17. (a) 18. (a)19. (b) 20. (a)

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

nn


1. Integrate tan( ) tan( ) tanx x xdx− + 2– Rakesh Avasthi, Patna

Ans. tan tan( )2x x x= + + −

= + + −

− + −tan( ) tan( )

tan( )tan( )x x

x x 1

or tan 2x – tan 2xtan(x + )tan(x – ) = tan (x + ) + tan (x – ) tan(x – )tan(x + )tan2x = tan 2x – tan(x + ) – tan (x – )

Let I x x x dx= − + tan( )tan( )tan 2

I x x x dx= − + − − tan tan( ) tan( )2

= + + + − +12

2log sec log cos( ) log cos( )x x x c

= + − +log sec cos( ) cos( ) .2x x x c

2. If two distinct chords, drawn from the point (p, q) on the circle x2 + y2 = px + qy (where pq 0) are bisected by the x-axis, then prove that p2 > 8q2

– Nalini Sharma, Bangalore

Ans. Let PQ be a chord of the given circle passing through P (p, q) and the coordinates of Q be (x, y). Since PQ is bisected by the x-axis, the mid-point of PQ lies on the x-axis which gives y = –q.Now Q lies on the circle x2 + y2 – px – qy = 0so x2 + q2 – px + q2 = 0 x2 – px + 2q2 = 0 …(i)

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.

which gives two values of x and hence the

coordinates of two points Q and R (say), so that

the chords PQ and PR are bisected by x-axis. If the

chords PQ and PR are distinct, the roots of (i) are

real and distinct.

P p q( , )

QR

y

xO

the discriminant p2 – 8q2 > 0

p2 > 8q2.

3. If SC

n nrr

n=

= 1

0 and t r

Cn n

rr

n=

= ,

0 then prove that

tS

nn

n=

2– Ramesh Ahuja, Ludhiana

Ans. Given, SC

n nrr

n=

= 1

0 and t r

Cn n

rr

n=

=

0

SC

n nn rr

n=

−= 1

0

nS nC

n nn rr

n=

−=

0

nS n rC

rC

n nn r

nn rr

n= − +

− −=

0

nS n rC

rC

n nn rr

n

nrr

n= −

+

−= =

0 0

nS nC

nC C

rC

n nn

nn

n nrr

n= + − + +

+

− =1 1

1 1 0...

nSn = tn + tn

tS

nn

n=

2 nn

Y UASKWE ANSWER


SOLUTION SET 154

1. (c) : Let . stand for H or T. �e 4 consecutive heads start with 1st, 2nd, 3rd, ... 7th tossHHHH.........., THHHH ...........THHHH .........., ..THHHH.................... ..........THHHH

Prob. = + =12

62

144 5

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 + sin21

Minimum value = f p2

= cos21

�e di�erence = 1 + sin21 – cos21 = 1 – cos23. (b) : Additional digits 0, 4 or 1, 3 or 2, 2

00124 → 3 42

36 =!!

numbers

001123 → 4 2 42

48! !!

+ = numbers

01222 → 43

42

16!!

!!

+ = numbers

Total numbers = 36 + 48 + 16 = 1004. (c) : Let c = kb, 20, b, kl are in H.P.

b = 40 20−k

, k = 2, 4, 5, 10, 20 →

b = 30, 35, 36, 38, 39 Number of value of b = 5

5. (b) : Let x + y = z → dx = 1 22

−+

z

dzx + c1 = z – 2ln(z + 2) x + y + z = cey/2

6. (a,b) : p = (x1, y1) → t12 = x1

2 + y21 – a2, t2

2 = x21 + y1

2– 2ax1, t3

2 = x12 + y1

2 – 2ay1, t14 = t2

2t32 + a4 → locus of p is

(x + y) (x2 + y2 – ax – ay) = 0, straight line and circle.7. (b) : f(x + y) = ex f(y) + ey f(x)

x = y = 0 → f(0) = 0

+ =

+x

f x yy

f x y( ) ( )

−

= −

=f x f x

e

f y f y

ekx y

( ) ( ) ( ) ( ), constant

f (x) – f(x) = kex → f(x) = (kx + c)ex

f(0) = 0 → c = 0, f (0) = 1 → k = 1, f(x) = xex

It has mininum at x = –1.

8. (a) : Area = −− xe dxx0

= 1

1

y

xO

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)16

9

2 = – 81 + 30b2 – b4 = 144 – (b2 – 15)2 144

9, maximum area = 910. (b) : P → 4, Q → 5, R → 3, S → 2

zn z0

z1 z2

zn z1 z2 = q

z z z zz zz z0 1 2 1

0 1

2 1 2− = −

−−

( ) cis q

= +

−1

21

2 2 1i z ztan ( )q

z Az Az A i0 1 212

12

= + = −

, tan q

A 2 1

2 1=

+( cos )q

n = 6, |A|2 = 1, n = 8, |A|2 = 1 12

+

n = 10, |A|2 = 3 5

2+

, n = 12, |A|2 = 2 3+nn

Solution Sender of Maths MusingSET-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

1. Jayanthi (Hyderabad)

2. Khokon Kumar Nandi (W.B.)

3. Gouri Sankar Adhikari (W.B.)


1. Prove that

a b c b c c a a bab

+ + + + +3

14

2 2 23

( ) ( ) ( )c

,

where a, b, c > 0. Equality holds if a = b = c.

2. 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 2r3 – 7r – 3 = 0.

3. Prove that sin2(q + ) + sin2(q + ) – 2cos( – ) sin(q + )

sin(q + ) = sin2( – ). …(1)

4. 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 + DF2 = EF2.Prove that BP2 + DQ2 = PQ2

5. �e numbers 2, 4, 8, 16, ..., 2n 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 Tn denote the maximum and minimum possible value of this �nal number, respectively. Determine a formula for Sn and Tn in terms of n.

SOLUTIONS

1. By the arithmetic-geometric mean inequality, we have

a2b + ab2 + b2c + bc2 + c2a + ca2 6 a b c abc6 6 66 6= , which implies 9(a2b + ab2 + b2c + bc2 + c2a + ca2 + 2abc) 8(a2b + ab2 + b2c + bc2 + c2a + ca2 + 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)).

Using the arithmetic-geometric mean inequality again, we then have

34

( )( )( )a b b c c a+ + +

+ + + + + + +( ). ( ) ( ) ( )a b c a b c b c a c a b3

+ + + + +( ) ( )( )( )a b c abc a b b c c a …(i)

From (i), it follows that

14 3

2 2 23

( ) ( ) ( ) .a b b c c aabc

a b c+ + + + +

2. 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 1 2 3

2 290

2+ = − ,

and cos 1 2 3 3

2 290

2 2+

= −

=cos sin .

Next we apply the addition formula for the cosine:

cos cos sin sin sin , 1 2 1 2 3

2 2 2 2 2− = …(i)

where (see �gures)

sin / , cos , 1 12

21 2

24 1

2= = −

rr

r

sin , cos ; 2 2

2

21

21= = −

rr

r

sin / ;3

23 2=

r


α1

r1

α

r1

α

r

We substitute these expressions into (i) and obtain, a�er multiplying both sides by 2r2,

4 1 1 1 32 2r r r− − − = Now write it in the form

( )( ) ,4 1 1 3 12 2r r r− − = + and square, obtaining

(4r2 – 1)(r2 – 1) = 9r2 + 6r + 1, which is equivalent to r(2r3 – 7r – 3) = 0. Since r 0, we have 2r3 – 7r – 3 = 0,

which was to be shown.

3. 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 (2q + + ) cos ( – ) – cos2( – ) +

cos ( – ) cos(2q + + )) = 2 (1 – cos2( – )) = 2 sin2( – ).

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 x2 + y2 = EF2 to

x2 + y2 = (1 – x)2 + (1 – y)2 – 2. 12

.(1 – x) (1 – y),

which simpli�es to 1 – x – y – xy = 0,

and so, y = 11−+

xx

. …(i)

Furthermore, BP is the angle-bisector of ∠EBC, in EBC so, by a well-known formula,

BP = 2 60

1EB BC

EB BCx

x

. .cos.

+

=+

Similarly, DQ = yy +1

.Using (i), DQ = 12− x .

Now, PQ = DB – BP – DQ

= 1 – xx

x xx+

− − = ++1

12

12 1

2

( ).

Finally,

BP2 + DQ2 = xx

x+

+ −

1

12

2 2

= + ++

= ++

=1 24 1

14 1

2 4

2

2 2

22x x

xx

xPQ

( )( )

( ),

which completes the proof.

5. We shall prove by induction that if we perform the operation on the numbers a1, a2, ..., an–1, an, where a1 < a2 < ...< an, then

Ta a a a a a

nnn

nn

nn= + + + + + +−

−−− −

1 2 3 22

11 12 4 8 2 2 2

... .

For n = 2, we have two numbers, a1 and a2 and thus T2 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 o� 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 o� 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, ..., bp 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

xb b b b bp

ppp

pp + + + + +−

−−− −

1 2 22

11 12 4 2 2 2

.... ,


and similarly,

yc c c c ck p

k pk pk p

k pk p + + + + +− −

− −− −− −

−− −

1 2 22

11 12 4 2 2 2

.... .

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

x y c b b b b b

a a a

kk

kk

kk

+ + + + + + +

+ +

−−

−−

−−2 2 4 8 2 2 2

2 4

1 1 2 32

21

11

1 2 3

...

88 2 2 222

11 1+ + + +−

−−− −... ,

a a akk

kk

kk

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

a a a a

a a a a a a a

r r r r

r r r rr

2 4 8 16

4 8 16 2 4 8 161 2 3

= + + +

+ + + + + + +

...

... ....+ −arr1

2

Adding

a a a a a a a ar

rrr

rr

kk

k1 2 3 1 11

22

114 8 16 2 2 2 2 2

+ + + + + + + + +− ++

++

−−... ... kk−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

Tn

n

n

n

n

n

n= + + + + + +−

−

−

− −

22

44

88

22

22

22

2

2

1

1 1...

= (n – 1). 1 + 2 = n + 1 �us, our formula for Tn 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 = 22

24

28

22

22

22

1 2 3

2

2

1

1

1

n n n

n n n+ + + + + +− −

− − −...

= 2n–1 + 2n–3 + 2n–5 + ... + 25–n + 23 –n + 22–n

= 23–n . (1 + 22 + 24 + ... + 22n–4) + 22 – n

= 23–n. 4 13

21

2n

n−

−− +

= 8 4 13 2

42

2 43 2

1 2 1( )n

n n

n

n

− +−

+ = +

nn


STRAIGHT LINES

SLOPE OR GRADIENT OF A LINEIf q is the inclination of a non-vertical line, then 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.Note :

�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 de�ned.

Slope of a Line Passing through Two Given PointsSlope m of a non-vertical line passing through the points

(x1,y1) and(x2, y2) is given by m

y yx x

=−−

( )( )

.2 1

2 1

Condition for Parallelism and PerpendicularityLet L1 and L2 be two lines whose slopes are m1 and m2 respectively. �en,(i) L1||L2 m1 = m2(ii) L1 L2 m1m2 = –1

ANGLE BETWEEN TWO LINES If is the acute angle between two non-vertical lines L1 and L2 with slopes m1 and m2, then

tan , =−

+m m

m m2 1

1 21 where 1 + m1m2

DEFINITION

A path traced by a point in a constant direction and endlessly in its opposite direction is known as a straight line.

Distance FormulaDistance between the points A(x1, y1) and B(x2, y2) is given by

AB x x y y= − + −( ) ( )2 12

2 12

Section Formula Let a point M divides a line segment joining the points (x1, y1) and (x2, y2) in the ratio m : n, then

Types of division Coordinates of MInternal mx nx

m nmy ny

m n2 1 2 1++

++

,

Externalmx nx

m nmy ny

m n2 1 2 1−−

−−

,

Mid-point x x y y1 2 1 22 2+ +

,

Area of TriangleArea of ABC whose vertices are A(x1, y1), B(x2, y2) and C(x3, y3) is given by

= − + − + −12 1 2 3 2 3 1 3 1 2

. | ( ) ( ) ( ) |x y y x y y x y y sq. unit.

Note : The points A, B, C are collinear area of ABC = 0.


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.

VARIOUS FORMS OF THE EQUATION OF A LINEEquation of the x-axis : Since ordinate of any point on the x-axis is zero. Hence, the equation of the x-axis is y = 0Equation 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 (x1, y1) is y – y1 = m(x – x1).Two-point form : �e equation of the straight line which passes through the point (x1, y1) and (x2, y2)

is

y yy yx x

x x− =−−

−12 1

2 11( ).

Intercept form : �e equation of the straight line making intercepts a and b on x- axis and y-axis

respectively is x

ayb

+ =1

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 (x1, y1) and making an angle q with the positive direction of x-axis is

x x y y

r−

=−

=1 1cos sinq 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

y AB

x CB

= − − , which is of the form y = mx + c

whose slope, m A

B= − and y- intercept,

c C

B= −

(ii) Reduction to intercept form : If C 0, then the equation Ax + By + C = 0 can be written as

xC A

yC B−

+−

=/ /

1 (provided A 0, B 0)

which is of the form xa

yb

+ =1

where x-intercept, a CA

= − and y-intercept, b CB

= −

(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


xcos + ysin – p = 0 ...(i) Comparing both equations, we get

cos

( ), sin

( ) =

+=

+

A

A B

B

A B2 2 2 2

and p C

A B=

+( )2 2

Hence, equation (i) can be written as

+

+=

+

A

A Bx B

A By C

A B( ) ( ) ( )2 2 2 2 2 2

which is the required form. Sign of the equation is taken as so that p should be

positive.

DISTANCE OF A POINT FROM A LINE Length of perpendicular from a given point P(x1,y1) on a line Ax + By + C = 0, is given by

d

Ax By C

A B=

+ +

+

| |.1 1

2 2

Note : Length of the perpendicular from the origin on

the straight line Ax + By + C = 0 is | |

( ).

C

A B2 2+

DISTANCE BETWEEN TWO PARALLEL LINES Distance between the parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0 is given by

dC C

A B=

−

+

| |.2 1

2 2

CONIC SECTIONS

DEFINITION A conic section or conic is thelocus of a point which moves in a plane so that its distance from a f ixed point is in a constant ratio to its distance from a �xed straight line.�e �xed point is called focus, the �xed straight line is called directrix, and the constant ratio is called the eccentricity, which is denoted by e.

From �gure, we havePSPK

e= =constant

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 called 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.

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

x2 + y2 = a2 General Form : Equation x2 + y2 + 2gx + 2fy + c = 0 represents general form of circle with centre at

(–g, –f) and radius = + − + g f c g f c2 2 2 2( ).


Standard equation of parabola y2 = 4ax y2 = – 4ax x2 = 4ay x2 = – 4ay

Graph

Eccentricity e = 1 e = 1 e = 1 e = 1Focus S(a, 0) S(–a, 0) S(0, a) S(0, –a)

Equation of directrix x + a = 0 x – a = 0 y + a = 0 y – a = 0

Equation of axis y = 0 y = 0 x = 0 x = 0

Vertex A(0, 0) A(0, 0) A(0, 0) A(0, 0)Extremities of latus rectum (a, ±2a) (–a, ±2a) (±2a, a) (±2a, –a)Length of latus rectum 4a 4a 4a 4a

Equation of latus rectum x – a = 0 x + a = 0 y – a = 0 y + a = 0

Note : Parabola is symmetric with respect to the axis of the parabola. If the equation has y2 term, then the axis of symmetry is along the x-axis and if the equation has x2 term, then the axis of symmetry is along the y-axis.

ELLIPSE

DEFINITION An ellipse is the set of all points in a plane, the sum of whose distances from two �xed points in the plane is constant.

PARABOLA

DEFINITION A Parabola is the set of all points in a plane that are equidistant from a �xed line and a �xed point.


Fundamental terms Ellipse (Horizontal ellipse) Conjugate Ellipse (Vertical ellipse)

Standard equationxa

yb

a b2

2

2

2 1+ = , xa

yb

a b2

2

2

2 1+ = ,

Graph

Centre (0, 0) (0, 0)

Vertices (±a, 0) (0, ±b)

Length of major axis 2a 2b

Length of minor axis 2b 2a

Foci (±ae, 0) (0, ±be)

Equation of directrices x ae= y b

e=

Eccentricity e ba

= −12

2 e ab

= −12

2

Length of latus rectum 2 2ba

2 2ab

Ends of latus rectum

ae b

a,

2

ab

be2

,

Focal distance or radii |SP| = (a – ex1) and |SP| = (a + ex1) |SP| = (b – ey1) and |SP| = (b + ey1)

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 2c, then

a2 = b2 + c2

Ellipse is symmetric w.r.t. both the coordinate axes.

�e foci always lie on the major axis. i.e, major axes is along the x-axis if the coe�cient of x2 has the

larger denominator and it is along the y-axis if the coe�cient of y2 has the larger denominator.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.


HYPERBOLA

DEFINITION A hyperbola is the set of all points in a plane, the di�erence of whose distance from two �xed points is constant.

Fundamental terms Hyperbola Conjugate Hyperbola

Standard equation xa

yb

2

2

2

2 1− = yb

xa

2

2

2

2 1− =

Graph

Centre (0, 0) (0, 0)Vertices (±a, 0) (0, ±b)Length of transverse axis 2a 2bLength of conjugate axis 2b 2aFoci (±ae, 0) (0, ±be)

Equation of directrices x ae= y b

e=

Eccentricity e ba

= +12

2 e ab

= +12

2

Length of latus rectum 2 2ba

2 2ab

Ends of latus rectum

ae b

a,

2

ab

be2

,

Di�erence of focal radii = |SP| – |SP| 2a 2bDistance between foci 2ae 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


COORDINATE AXESLet XOX , YOY , ZOZ be the three mutual ly 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 co-ordinate 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 planes, called the coordinate planes, divides the space into eight parts known as octants. �ese octants could be named as OXYZ, XOYZ, XOYZ, XOYZ, XOYZXOYZ, XOYZ and XOYZ.

COORDINATES OF A POINT IN SPACELet P be any point in space. Draw PL, PM, PN perpendicular to the yz, zx and xy planes respectively then

(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.

�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, XOYZ, XOYZ, XOYZ, XOYZ, XOYZ, XOYZ and XOYZ are denoted by I, II, III, ..., VIII respectively.

I II III IV V VI VII VIIIx + – – + + – – +y + + – – + + – –z + + + + – – – –

DISTANCE BETWEEN TWO POINTSThe distance between the points P(x1, y1, z1) and Q(x2, y2, z2) is given by

PQ x x y y z z= − + − + −( ) ( ) ( )2 12

2 12

2 12

SECTION FORMULALet M be a point which divides the line joining the points A(x1, y1, z1) and B(x2, y2, z2) in the ratio m : n, then

Types of division

Coordinates of M

Internalmx nx

m nmy ny

m nmz nz

m n2 1 2 1 2 1++

++

++

, ,

Externalmx nx

m nmy ny

m nmz nz

m n2 1 2 1 2 1−−

−−

−−

, ,

Mid-pointx x y y z z1 2 1 2 1 2

2 2 2+ + +

, , .

INTRODUCTION TO THREE DIMENSIONAL GEOMETRY


COORDINATES OF CENTROID OF A TRIANGLE�e coordinates of the centroid of a triangle whose vertices are (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) are

x x x y y y z z z1 2 3 1 2 3 1 2 33 3 3

+ + + + + +

, ,

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 x2 = – 16y

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. Find the equation of a circle with centre (3, –2) and radius 5.

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.

Short Answer Type

6. If A(–2, 1), B(2, 3) and C(–2, –4) are three points, �nd the angle between BA and BC.

7. (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 o� equal and positive intercepts from the axes and passes through the point (, ).

8. Reduce the lines 3x – 4y + 4 = 0 and 4x – 3y + 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, �nd the coordinates of the point C.

Long Answer Type

11. �e equation of a straight line which passes through the point (acos3q, asin3q) and perpendicular to x sec q + y cosec q = a is

12. Find the length of axes of the ellipse whose eccentricity is 4/5 and whose foci coincide with those of the hyperbola 9x2 – 16y2 + 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 – 4y = 1.

14. (i) If the point (h, 0), (a, b) and (0, k) lie on a line,

show that ah

bk

+ =1.

(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.

SOLUTIONS

1. We have, m = − 3 = −tanq 3

tanq = tan (180 – 60°) q= 120°

2. We have, equation of parabola, x2 = –16y

Comparing with x2 = –4ay, we get a = 4

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 the x-axis, when the coe�cient 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 = r2. Here, h = 3, k = –2 and r = 5. �e required equation of the circle is (x – 3)2 + (y + 2)2 = 52

x2 + y2 – 6x + 4y – 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 y z= − −

= − −−

= − −

( ) ( ) , ( ) ( ) , ( ) ( )2 4 3 22 3

2 3 3 12 3

2 2 3 42 3

x = –2, y = –9, z = 8.So, required point is R (–2, –9, 8).


6. 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,

m13 1

2 224

12

= −− −

= =( )

, and m2

4 32 2

74

= − −− −

=

Let q be the angle between BA and BC. �en,

tanq =−

+=

−

+ = =

m mm m

2 1

1 21

74

12

1 74

12

54

158

23

=

−q tan 1 23

7. (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 m2 34 2

1 25

Also, the required line has y-intercept 4. So, c = 4Substituting the value of m and c in y = mx + c, we get

y x x y= − + + − =25

4 5 20 0 w h i ch i s t he required equation of line

(ii) Let the equation of the line be xa

yb

+ =1 which

cuts o� intercepts a and b with the coordinate axes.It is given that a = b. �erefore the equation of the line is

xa

ya

x y a+ = + =1

... (i)

But, it passes through (,). So, + = a.Putting the value of a in (i), we get

x + y = + which the required equation of line

8. We have, 3x – 4y + 4 = 0 –3x + 4y = 4

−− +

+− +

=− +

3

3 4

4

3 4

4

3 42 2 2 2 2 2

x y

( ) ( ) ( )

− + =35

45

45

x y

�is is the normal form of 3x – 4y + 4 = 0 and the length of the perpendicular from the origin to it is given by p1 = 4/5Now, 4x –3y + 12 = 0 –4x + 3y = 12

−− +

+− +

=− +

4

4 3

3

4 3

12

4 32 2 2 2 2 2

x y

( ) ( ) ( )

− + =45

35

125

x y

�is is the normal form of 4x – 3y + 12 = 0 and the length of the perpendicular from the origin to it is

given by p2125

= .

Clearly, p2 > p1. �erefore, the line 3x – 4y + 4 = 0 is nearer to the origin.

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 ya

xb

2

2

2

2 1− = ,

...(i)

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 ya

xa

2

2

2

2101−

−=

…(ii)

Since (2, 3) lies on hyperbola (ii),

−−

− =9 410

12 2a a 90 – 9a2 – 4a2 = 10a2 – a4 a4 – 23a2 + 90 = 0 a2 = 18, 5a2 = 18 b2 = 10 – 18 = – 8 not possible.a2 = 5 b2 = 10 – 5 = 5

From (i), the equation of hyperbola is y x2 2

5 51− =

or y2 – x2 = 5.10. Let the coordinates of C be (x, y, z). Coordinates of

the centroid G = (1, 1, 1). �en

x + −3 13

= 1, y − +5 73

= 1, z + −7 63

= 1

x = 1, y = 1, z = 2.Hence, coordinates of C are (1, 1, 2).

11. We have x secq + y cosecq = a y cosecq = a – x secq

y x a= − +seccosec cosec

qq q

y = (– tanq)x + a sinq


slope = –tanq Slope of line to x secq + y cosecq = a is

1tan

cotq

q=

Equation of line through (a cos3q, a sin3q) and slope cot q is y – a sin3q = cotq (x– a cos3q)

y x a a− = −cossin

sin cos cotqq

q q q3 3

x y a acos sin sin sin cossin

q q q q qq

− = − +

3

4

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 asy x x y2 2 2 2

9 161

16 91− = − + =or Here a = 3, b = 4

�e eccentricity of the hyperbola is given by

e b

a= +1

2

2 = + =1 169

53

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

xb

ya

2

2

2

2 1+ =

... (1)

Since 2ae = SS = 10 =a . 45

5 =a 25

4Now, b2 = a2(1 – e2)

= −

=b2 625

161 16

2522516

Hence, equation of the required ellipse is 16225

16625

12 2x y+ =

Length of major axis of the ellipse = =2 252

a and

Length of minor axis = =2 152

b .

13. Let the equation of the required circle bex2 + y2 + 2gx + 2fy + c = 0 ... (i)It passes through the point (3, 7) 9 + 49 + 6g + 14f + c = 0 or 6g + 14f + c + 58 = 0 ... (ii)Also point (5, 5) lies on (i) 25 + 25 + 10g + 10f + c = 0 or 10g + 10f + c + 50 = 0 ... (iii)Centre of (i), i.e., (–g, –f) lies on line x – 4y = 1 –g + 4f = 1 or g – 4f + 1 = 0 ... (iv)Subtracting (iii) from(ii), we get–4g + 4f + 8 = 0 or –g + f + 2 = 0 ... (v)Adding (iv) and (v), we get–3f + 3 = 0 f = 1Putting f = 1 in (iv), we have g – 4 + 1 = 0 g = 3Putting g = 3, f = 1 in (ii), we get6(3) + 14(1) + c + 58 = 0or c + 18 +14 + 58 = 0 or c + 90 = 0 c = – 90Substituting the values of g = 3, f = 1 and c = – 90 in (i), we getx2 + y2 + 6x + 2y – 90 = 0which 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 haveslope of AB = slope of BC

−−

= −−

−

= −ba h

k ba

ba h

b ka

00 ( )

( )

ab = (a – h)(b – k) ak + hb = hk

+ =ah

bk

1 [on dividing both sides by hk]


(ii) Let A(5, 1), B(1, –1) and C(11, 4) be the given points. �en,

Slope of AB = − −−

= −−

=( )( )

1 11 5

24

12

and slope of BC = − −−

= =4 111 1

510

12

( )

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.

15. We know that the equation of the line AB is given by

yx−−

= −−

sincos

sin sincos cos

qq

q q

−−

=

+

−

− +

yx

sincos

cos sin

sin si

qq

q q

q

22 2

22

nn q−

2

−−

=

+

− +

yx

sincos

cos

sin

qq

q

q2

2

+

+

+

x ycos sin q q

2 2

− +

+ +

=cos .cos sin sinq q q q

2 20

+

+

+

−

+ −

=x ycos sin cos q q q q

2 2 20

+

+

+

−

−

=x ycos sin cos q q q

2 2 20

... (i)Let d be the perpendicular distance from the origin to line (i). �en,

d = +

+ +

−

−

+

02

02 2

22

cos sin cos

cos

q q q

q +

+

sin2

2 q

= −

cos q

2

Hence, the required distance is cos . q−

2

nn

MTG o f f e r s “ C l a s s r o o m S t u d y M a t e r i a l ” f o r

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MATHEMATICS TODAY | NOVEMBER ‘1534 MATHEMATICS TODAY | NOVEMBER ‘1534

* Al ok K umar is a w inner of I N DI AN N AT I O N AL MAT HE MAT I CS O L Y MP I AD ( I N MO - 9 1 ) . He trains I I T and O l ymp iad asp irants.

CARTESIAN & POLAR COORDINATESY

O XX

Y

II quadrant

(–, +)

I quadrant

(+, +)

III quadrant

(–, –)IV quadrant

(+, –)

X O x X

yr

Y

Y

P x y( , )

P(x, y) - Cartesian coordinates P(r, q) - Polar coordinates

r x yyx

= + =2 2 , tanq

DISTANCE FORMULA, SECTION FORMULA & AREA

Distance formula : Distance between the points P(x1, y1) and Q(x2, y2) is

PQ x x y y= − + −( ) ( )2 12

2 12

Section formula : The coordinates of a point dividing the line segment joining the points (x1, y1) and (x2, y2) internally, in the ratio m : n are

mx nxm n

my nym n

2 1 2 1++

++

,

and externally in the ratio m : n are

mx nx

m nmy ny

m n2 1 2 1−−

−−

,

Mid-Point formula : In particular, if m = n, the coordinates of the mid-point of the line segment joining the points (x1, y1) and (x2, y2) is

x x y y1 2 1 22 2+ +

, .

Area of a Triangle : Let (x1, y1), (x2, y2) and (x3, y3) respectively be the coordinates of the vertices A, B, C of a ABC. �en the area of triangle ABC, is

12 1 2 3 2 3 1 3 1 2[ ( ) ( ) ( )]x y y x y y x y y− + − + −

= 12

111

1 1

2 2

3 3

x yx yx y

While using formulas, order of the points (x1, y1), (x2, y2) and (x3, y3) has not been taken into account. If we plot the points A(x1, y1), B(x2, y2) and C(x3, y3), 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 �nding the area of triangle ABC, we take modulus.

Remark : In case of polygon with vertices (x1, y1), (x2, y2), ...., (xn, yn) in order, then area of polygon is given by 12 1 2 1 2 2 3 2 3|( ) ( ) ....x y y x x y y x− + − +

+ (xn – 1 yn – yn – 1xn) + (xn y1 – ynx1)|

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.

*ALOK KUMAR, B.Tech, IIT Kanpur

MATHEMATICS TODAY | NOVEMBER ‘15 35MATHEMATICS TODAY | NOVEMBER ‘15 35

SLOPE & EQUATION OF LINEIf q is the inclination of a line L, (0 q< p) then tanqis called the slope or gradient of the line L.�e slope of a line whose inclination is 90° is not de�ned.�e slope of a line is denoted by m.�us, m = tanq,q 90°It may be observed that the slope of x-axis is zero and slope of y-axis is not de�ned.Slope of a line when coordinates of any two points

on the line are given :

m

y yx x

dydxPQ =

−−

= =2 1

2 1tanq

P Q

( , )x y1 1 ( , )x y2 2

Conditions for parallelism and perpendicularity

of lines in terms of their slopes :If lines L1 || L2, then m1 = m2 If L1 L2, then m1·m2 = –1

VARIOUS FORMS OF THE EQUATION OF A LINE:

Line Parallel to x-axis • Line Parallel to y-axis

O x

y = a

y a= –

x

x b= – x = by

O

Point slope form

Equation of line L having slope m and passing through (x1, y1) isy – y1 = m(x – x1)Two point form : Equation of line L passing through P(x1, y1) and Q(x2, y2) is

y yy yx x

x x− =−−

−12 1

2 11( )

or y y

y yx x

x x− =−−

−22 1

2 12( )

Slope-intercept form

If line L having slope m and intercept on y - a x i s i s c , then equation of l i ne L i s y = mx + c

Intercept form

Suppose a line L makes intercept ‘a’ on x-axis a n d ‘ b ’ on y - a x i s . Then its equation is xa

yb

+ =1

Normal form

L e t l e n g t h o f perpendicular from orig in to l ine L i s OP = p and is the angle which is made by OP with +ve x-axis. �e equation of line L is xcos + ysin = pGeneral Equation of a Line

Ax + By + C = 0

DIFFERENT FORMS OF Ax + By + C = 0Slope Intercept Form

If B 0, then Ax + By + C = 0 ....(i)can be written as

y A

Bx C

B= − −

....(ii)

Compare (ii) with y = mx + c, we get

Slope of line = m AB

= −

Intercept on y-axis = c CB

= −

Intercept form

If C 0, then Ax + By + C = 0 ....(i)can be written as

xC

A

yCB

−

+−

=1

...(ii)

Compare (ii) with xa

yb

+ =1 , we get ....(iii)

a = Intercept on x-axis = −CA

b = Intercept on y-axis = −CB

Normal form

Let Ax + By + C = 0 ....(i)can be written asAx + By = –C ....(ii)Compare (ii) with xcos + ysin = p

y

O x

x

Ly

O

(0, )c

a

b

y

O x

O

P

x

L

y

p


so A B Cpcos sin

= = −

cos =−pA

C and sin =

−pBC

Using sin2 + cos2 = 1

−

+

−

=pAC

pBC

2 21

=+

= +

p CA B

p C

A B2

2

2 2 2 2

Proper choice of signs is made so that p should be +ve.

So, cos sin =

+=

+

A

A B

B

A B2 2 2 2and

PARAMETRIC EQUATIONS OF A STRAIGHT LINE

In �gure given below, let BAP be a straight line through a given point A (x1, y1), the angle of slope being q. �e positive direction of the line is in the sense BAP. (Direction of increasing ordinate is called the positive direction of the line).For the points P (x, y) and Q (X, Y) shown in the �gure, AP is regarded as a positive vector and AQ as a negative vector, as indicated by the arrows.From the general de�nitions of cosq and sinq, we havecos , sinq q=

−=

−x xAP

y yAP

1 1

or x – x1 = AP cosq, y – y1 = AP sinq

or x x y y

r−

=−

= 1 1cos sinq 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

tanq =−

+m m

m m1 2

1 21, as m1m2 –1

�us obtuse angle = 180° – qCollinearity of three points

If A, B, C are collinear, then

• mAB = mBC = mAC• |AB| + |BC| = |AC| or |BC| + |CA| = |BA| or |CA| + |AB| = |CB|

LENGTH OF PERPENDICULAR AND REFLECTIONLength of perpendicular from ( x1, y1) to a straight line ax + by + c = 0 is given by

ax by c

a b1 1

2 2

+ +

+If two lines ax + by + c1 = 0 and ax + by + c2 = 0

are parallel, then dc c

a b=

−

+1 22 2

Re�ection of a point about a line

Let the image of a point (x1, y1) about the line ax + by + c = 0 is (x2, y2), then

x xa

y yb

ax by c

a b2 1 2 1 1 1

2 22−

=−

= −+ +

+

and let the foot of perpendicular from a point (x1, y1) on the line ax + by + c = 0 is (x2, y2) then

x xa

y yb

ax by c

a b2 1 2 1 1 1

2 2−

=−

= −+ +

+

CENTRES CONNECTED WITH A TRIANGLE (w.r.t. ABC, where A (x1, y1), B (x2, y2), C (x3, y3), BC = a, CA = b & AB = c).Centroid : �e point of concurrency of the medians of a triangle is called the centroid of the triangle. The centroid of a triangle divides each median in the ratio 2 : 1. �e coordinates of centroid

are given by Gx x x y y y

+ + + +

1 2 3 1 2 33 3

,

Orthocentre : The point of concurrency of the altitudes of a triangle is called the orthocentre of the triangle. T h e c o o r d i n a t e s o f t h e orthocentre are given by

Hx A x B x C

A B C

+ ++ +

1 2 3tan tan tantan tan tan

,

y A y B y CA B C

1 2 3tan tan tantan tan tan

+ ++ +

A

y

( , )x yP

( , )X Y

O Bx

( , )x y1 1

–

+

Q

F

A

B D C

EG

2 : 1

1 : 1

F

B D C

H E

A


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 of the angles of a triangle is called the incentre of the triangle. The coordinates of the incentre are given by

I

ax bx cxa b c

ay by cya b c

+ ++ +

+ ++ +

1 2 3 1 2 3,

Excentre : Coordinate of excentre opposite to

A is given by BLLC

cb

= ,

I1

B LC

bcA

AlsoAII L

b ca

1

1= − +

Iax bx cx

a b cay by cy

a b c

11 2 3

1 2 3

− + +

− + +

− + +− + +

,

and similarly for excentres ( I 2 & I 3) o p p o s i t e t o ΑB and ΑC are given by

Iax bx cx

a b cay by cy

a b c21 2 3 1 2 3− +− +

− +− +

,

Iax bx cx

a b cay by cy

a b c31 2 3 1 2 3+ −+ −

+ −+ −

,

Circumcentre : The point of concurrency of the perpendicular bisectors of the sides of a triangle is called circumcentre of the triangle. The coordinates of the circumcentre are given by

Ox A x B x C

A B C

+ ++ +

1 2 32 2 22 2 2

sin sin sinsin sin sin

,

y A y B y CA B C

1 2 32 2 22 2 2

sin sin sinsin sin sin

+ ++ +

Remarks :• CircumcentreO, centroid G and orthocentre

H of a ABC are collinear. G divides OH in the ratio 1 : 2, i.e. OG : GH = 1 : 2

• Inanisoscelestrianglecentroid,orthocentre,incentre and circumcentre lie on the same line and in an equilateral triangle all these four points coincide.

POSITION OF A GIVEN POINT RELATIVE TO A GIVEN LINE & CONCURRENCY

�e �gure shows a point P(x1, y1) 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 x1. Putting x = x1 in the equation ax + by + c = 0

gives y-coordinate of N = −+( )ax c

b1

If P(x1, y1) lies above the line, then we have

yax c

bi e y

ax cb1

11

1 0 −+

++

( )

. .( )

i eax by c

b. .

( )1 1 0+ +

i eL x y

b. .

( , )1 1 0 . . . . ( i )

Hence, if P(x1, y1) satis�es equation (i), it would mean that P lies above the line ax + by + c = 0,

and if L x y

b( , )1 1 0 , it would mean that P lies

below the line ax + by + c = 0.Remark : If (ax1 + by1 + c) and (ax2 + by2 + c) have same signs, it implies that (x1, y1) and (x2, y2) both lie on the same side of the line ax + by + c = 0. If the quantities ax1 + by1 + c and ax2 + by2 + c have opposite signs, then they lie on the opposite sides of the line.Concurrency of straight lines : �e condition for three lines a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, a3x + b3y + c3 = 0 to be concurrent is

(i) a b ca b ca b c

1 1 1

2 2 2

3 3 3

0=

B

N

A

M

CL

I

c b:

B/2B/2

ab

c:

+

A/2A/2

C/2C/2

� 2 C

B L

O

A

Csin2 : sin2C B

sin2A

:sin2

+ sin2

B

C

� 2 B


(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 Angle bisector is the locus of a point which moves in such a way so that i t s d i s t a n c e f r o m t w o i n t e r s e c t i n g lines remains same.The equations of the two bisectors of the angles between the lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are a x b y c

a b

a x b y c

a b1 1 1

12

12

2 2 2

22

22

+ +

+=

+ +

+

If the two given lines are not perpendicular i .e . , a1a2 + b1b2 0 and not parallel i .e . , a1b2 a2b1, 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 m1 and take one of the bisectors and let its slope be m2. If q be the acute angle between them, then

�nd tanq =−

+m m

m m1 2

1 21

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 + b1y + c1 = 0 and a2x + b2y + c2 = 0, then

a x b y c

a b

a x b y c

a b1 1 1

12

12

2 2 2

22

22

+ +

+=

+ +

+

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

a x b y c

a b

a x b y c

a b1 1 1

12

12

2 2 2

22

22

+ +

+=

+ +

+

ora x b y c

a b

a x b y c

a b1 1 1

12

12

2 2 2

22

22

+ +

+= −

+ +

+

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

a x b y c

a b

a x b y c

a b1 1 1

12

12

2 2 2

22

22

+ +

+= +

+ +

+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(x1, y1) and equally inclined with the lines

a1x + b1y + c1= 0 and a2x + b2y + 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

Q MPD

N

O a xb y

c

22

2

++

= 0

ax

by

c

11

1

++

= 0


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 �nding the equation of the locus

of a point :(i) If we are �nding the equation of the locus of

a point P, assign coordinates (h, k) or (x1, y1) 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 AXESChanges of Axes (Shifting of origin without

rotation of axes) :Let P (x, y) with respect to axes OX and OY.Let O (a , b ) with respect to axes OX and OY and let P (x , y) with respect to axes OX 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 – bThus 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 a r e t h e t w o p e r p e n d i c u l ar l ines obt a ine d 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 = xcosq – ysinq y = xsinq + ycosq

PAIR OF STRAIGHT LINES The genera l equat ion of second degree

ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents a

pair of straight lines if a h gh b fg f c

= 0

abc + 2fgh – af 2 – bg2 – ch2 = 0 The homogeneous second degree equation ax2 + 2hxy + by2 = 0 represents a pair of straight lines through the origin. If lines through the origin whose joint equation is ax 2 + 2hxy + by 2 = 0 , are y = m 1x and y = m2x, then y2 – (m1 + m2)xy + m1m2x2 = 0 and

y hb

xy ab

x2 22 0+ + = are identical. If θ is the angle

between the two lines, then

tan( )

q = + −+

= −+

m m m mm m

h aba b

1 22

1 2

1 2

241

2

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 ax2 + 2hxy + by2 = 0, then

m m h

bm m a

b1 2 1 22+ = − =&

.(iii) �e equation to the straight lines bisecting

t h e ang l e b e t we e n t h e s t r a i g ht l i n e s

ax hxy by

x ya b

xyh

2 22 2

2 0+ + =−−

= is


(iv) A homogeneous equat ion of degree n represents n straight lines (in general) passing through origin.

Joint equation

of pair of lines joining the origin and the points of intersection of a line and a curve If the line lx + my + n = 0, ((n 0) i.e. the line not passing through origin) cuts the curve ax2 + 2hxy + by2 + 2gx + 2fy + 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.

ax2 + 2hxy + by2 + (2gx + 2fy) lx myn

clx my

n+−

+

+−

=

20

lx myn

clx my

n+−

+

+−

=

20

is the equation of the lines OA and OB.

PROBLEMS

SECTION-I

Single Correct Answer Type

1. �e line x + y = 1 meets x-axis at A and y-axis at B. P is the mid-point of AB. P1 is the foot of the perpendicular from P to OA; M1 is that from P1 to OP; P2 is that from M1 to OA; M2 is that from P2 to OP; P3 is that from M2 to OA and so on. If Pn denotes the nth foot of the perpendicular on OA from Mn – 1, then OPn =

(a) 12

(b) 12n (c) 1

2 2n/ (d) 12

2. �e orthocentre of the triangle formed by the lines x + y = 1, 2x + 3y = 6 and 4x – y + 4 = 0 lies in(a) I quadrant (b) II quadrant(c) III quadrant (d) IV quadrant3. 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

(a) 3 (b) − 3 (c) 13

(d) 1

4. A straight line passes through the point of intersection of x – 2y – 2 = 0 and 2x – 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

(a) (– , 4) (7, )(b) (– , 5) (7, )(c) (– , 4) (5, 7) (7, )(d) (– , 4) (4, 5) (7, )5. 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) a ba b+−

(b) a ba b−+

(c) ab

(d) ba

6. �e values of m for which the system of equations 3x + my = m and 2x – 5y = 20 has a solution satisfy the conditions x > 0, y > 0 are given by the set

(a)

m m: −

132

(b)

m m:

172

(c)

m m m: −

132

172

or

(d)

m m m: −

30 152

or

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) f (4) (b) 12

6f ( ) (c) 13

6f ( ) (d) 13

5f ( )

8. If 9x2 + 2hxy + 4y2 + 6x + 2fy – 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 these9. 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 = m10. 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.

(a) a2 3 13 1−+

(b) 32

3 13 1

2a −+

(c) 3 3 13 1

2a −+

(d) a2

11. �e area of the triangle formed by the line x + y = 3 and the angular bisectors of pair of straight lines x2 – y2 + 2y = 1 is (a) 8 sq.units (b) 6 sq.units (c) 4 sq.units (d) none of these

Y

A

B

XO


12. 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 l2 + m2 = 1, then the locus of its circumcentre is (a) (x2 – y2)2 = x2 + y2 (b) (x2 + y2)2 = x2 – y2 (c) (x2 + y2)2 = 4x2y2 (d) (x2 – y2)2 = (x2 + y2)2

13. 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 14. 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) y2 = 2x (b) y2 = 4x(c) y = 2x (d) x + y = 2xy 15. 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) x2 + y2 = 3l2 (b) x2 + y2 = l2(c) x2 + y2 = 4l2 (d) x2 + y2 = 2l2

16. In a ABC, the coordinates of B are (0, 0)

AB = 2, =ABC p3

and the mid point of BC is (2,0).

�e centroid of triangle is

(a)

12

32

,

(b)

53

13

,

(c)

4 33

13

+

,

(d) 4 33

13

−

,

17. A ray travelling along the line 3x – 4y = 5 a�er being re�ected from a line 'l ' travels along the line 5x + 12y = 13. �en the equation of the line 'l ' is (a) x + 8y = 0 (b) x – 8y = 0 (c) 32x + 4y + 65 = 0 (d) 32x – 4y + 65 = 018. All points inside the triangle formed by A(1, 3), B(5, 6), C(–1, 2) will satisfy (a) 2x + 2y 0 (b) 2x + y + 1 0(c) 2x + 3y – 12 0 (d) –2x + 11 019. 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

(a)

32

2 1( )− sq. units (b) 3

2sq. units

(c)

32

sq. units

(d)

34

sq. units

SECTION-II

Multiple Correct Answer Type

20. P is a point inside a ABC of area K (K > 0). �e lengths of perpendiculars drawn to the sides BC, CA, AB

of lengths a, b, c are respectively P1, P2, P3. aP

bP

cP1 2 3

+ +

is minimum when(a) P is incentre of ABC(b) P is equidistant to all the 3 sides

(c) P P P Ka b c1 2 3

2= = =+ +

(d) P is orthocentre of ABC

21. Equations (b – c)x + (c – a)y + (a – b) = 0 and (b3 – 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

22. �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 – 4y + 2 = 0 (b) 3x + 4y = 18(c) x = 2 (d) y = 3

23. �e equation of the diagonal of the square formed by the pairs of lines xy + 4x – 3y – 12 = 0 and xy – 3x + 4y – 12 = 0 is (a) x – y = 0 (b) x + y + 1 = 0(c) x + y = 0 (d) x – y + 1 = 0

24. Under rotation of axes through q,x cos + y sin = P changes to xcos + ysin = P, then (a) cos = cos( – q) (b) cos = cos( – q) (c) sin = sin( – q) (d) sin = sin( – q)

25. 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

(a) 85

165

,

(b) 715

1415

,

(c) 65

125

,

(d) 4

158

15,

26. 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


(a) x = 1 (b) x + y = 1 (c) y = 9 (d) x – 2y = 3

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 x2 + 2hxy + y2 = 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

(a) sec 2 = h (b) cos = +12

hh

(c) m1 + m2 = –2sec2 (d) cot = +−

hh

11

SECTION-III

Comprehension Type

Paragraph for Question No. 29 to 31In a PQR, with PQ = r, QR = p, PR = q, the cosine values of the angles are given by

cos ; cosP

q r pqr

Qp r q

pr=

+ −=

+ −2 2 2 2 2 2

2 2

cos R

p q rpq

=+ −2 2 2

2 and the area of PQR is

= = =1

212

12

pq R qr P pr Qsin sin sin

Let ABCD be a parallelogram whose diagonal equations are AC x + 2y – 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

(a) 103

(b) 2 (c) 203

(d) 11 2

3

30. �e length of side AB is equal to

(a) 2 583

(b) 4 58

3 (c) 58

3 (d) 58

31. �e length of BC is equal to

(a) 4 103

(b) 2 103

(c) 8 103

(d) 103

SECTION-IVMatrix-Match Type

32. Match the following Column I with Column II

Column I Column II(A) The distance between the lines

( ) ( )x y x y+ + + − =7 4 2 7 42 02 is(p) 2

(B) If the sum of the distance of a point from two perpendicular lines in a plane is 1, then its locus is |x| + |y| = k, where k is equal to

(q) 7

(C) If 6x + 6y + m = 0 is acute angle bisector of lines x + 2y + 4 = 0 and 4x + 2y – 1 = 0, then m is equal to

(r) 3

(D) Area of the triangle formed by the lines y2 – 9xy + 18x2 = 0 and y = 6 is

(s) 1

33. Match the following:

Column I Column II(A) T h e v a l u e o f k f o r w h i c h

4x2 + 8xy + ky2 = 9 is the equation of a pair of straight lines, is

(p) 3

(B) If the sum of the slopes of the lines given by x2 – 2Cxy – 7y2 = 0 is four times their product, then the value of C is

(q) –3

(C) If the gradient of one of the lines x2 + hxy + 2y2 = 0 is twice that of the other, then h =

(r) 2

(D) If the lines ax2 + 2hxy + by2 = 0 are equally inclined to the lines ax2 + 2hxy + by2 + (x2 + y2) = 0, then the value of can be

(s) 4

34. If y m xm

iii

= + =1 1 2 3( , , ) represent three straight

lines whose slopes are the roots of the equation 2m3 – 3m2 – 3m + 2 = 0, thenMatch the following:

Column I Column II(A) A l g e b r a i c s u m o f t h e

intercepts made by the lines on x-axis is

(p) ( )4 2 9 54+

(B) A l g e b r a i c s u m o f t h e intercepts made by the lines on y-axis is

(q) 3/2


(C) Sum of the distances of the lines from the origin is

(r) –21/4

(D) Sum of the lengths of the lines intercepted between the coordinate axes is

(s) ( )5 2 9 510+

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. 36. Consider 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 =

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 is38. If the orthocentre of the triangle formed by 2x + 3y – 1 = 0, x + 2y – 1 = 0, ax + by – 1 = 0 is at the

origin then b a− =

4

39. �e area of the rhombus ABCD is 24. �e equation of the diagonal BD is 4x + 3y + 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 + 5y – 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 2x2 + 3xy – 6x + 2y – 4 = 0 changes to aX2 + 2hXY + bY2 + 2gX + 2fY + c = 0, then (a + b) is equal to 42. Origin is shi�ed to (1, 2) then the equation y2 – 8x – 4y + 12 = 0 changes as y2 = 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

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 PP1 is perpendicular to OA.

OP1 = P1P = 1/2 By

O A x

PM1M

P P P1

Equation of line OP is y = xWe have, (OMn – 1)2 = (OPn)2 + (PnMn – 1)2 = 2(OPn)2 = 2pn

2 (say)Also, (OPn – 1)2 = (OMn – 1)2 + (Pn – 1Mn – 1)2

(OPn – 1)2 = (OMn – 1)2 + (Pn – 1Mn – 1)2

= + −2 12

21

2p pn n = =− −p p p pn n n n

21

21

14

12

= = = = = =− − −Op p p p pn n n n n n12

12

12

12

1 2 2 1 1.....

2. (a) : Coordinates of A and

B are (–3, 4) and −

35

85

, .

Let orthocentre be P(h, k)

�en, (slope of PA) × (slope of BC) = –1

kh−+

= −43

4 1

4k – 16 = –h –3 h + 4k = 13 ... (i)and slope of PB × slope of AC = – 1

−

+ −

= −

k

h

8535

23

1

−+

=5 85 3

23

1kh

10k – 16 = 15h + 9 15h – 10k + 25 = 0 3h – 2k + 5 = 0 ... (ii)

Solving eqs. (i) and (ii), we get h k= =37

227

,

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 x2 – y2 – 2x + 1 = 0 ... (i)Let equation of line passes through (2, 0) be y = m(x – 2) ... (ii)By homogenizing equation (i) with the help of line (ii),

we get x y xmx y

mmx y

m2 2

22

2 20− −

−

+

−

=

Q coe�cient of x2 + coe�cient of y2 = 0

= m 13

4. (d) : As line passes through the point of intersection of x – 2y – 2 = 0 and 2x – by – 6 = 0

y

A C

B

O x

2+ 3

= 6

xy

4–

+ 4

= 0

xy

xy+

= 1


It can be represented as (x – 2y – 2) + (2x – by – 6) = 0As it passes through the origin –2 – 6 = 0 = –3 Equation of the line is –x + (6 – b)y = 0

Its slope is 16−b

As its angle with y = 0 is less than p/4

− −

1 16

1b

6 – b > 1 or b – b < –1 b < 5 or b > 7 But b 4 (as the lines intersect) b (– , 4) (4, 5) (7, )

5. (b) : Equation of the line is

xa

yb

+ =1

Transformed equation is

1 1 1a

x yb

x y( cos sin ) ( sin cos ) − + + =

Since, intercepts are equal x-coe�cient y-coe�cient

= −+

tan a ba b

6. (d) : Solving the equations, we get

x mm

y mm

=+

= −+

252 15

2 602 15

,

But x > 0, y > 0 25m > 0, 2m + 15 > 0, 2m – 60 > 0or 25 m < 0, 2m + 15 < 0, 2m – 60 < 0

−m m30 152

or

7. (c) : Area =

= =

4 12

83

4

643

23

6

f (x) = 2x

y

(0, 4)

(0, –4)

(–8/3, 0) (8/3, 0) xO

8. (a) : Since the given equation represents a pair of parallel lines, we have h2 = ab h = ±6

Condition for pair of lines 9 3

43 3

0h

h ff −

=

108 ± 36f – 9f 2 – 144 = 0 f = 2 and h = 6 f = –2, h = –69. (b) : By solving the sides of the rhombus, the vertices are

0 0 0 0, , , , , , ,−

−

nm

nl

nm

nl

Area = 12

2 2 2 2nm

nl

n lm

= =

10. (b) : Shaded area is the region traced by P= ABC – 3ABD

= − 3

432 2

152a a a tan

= −

+

32

3 13 1

2a

A

B C

D I

11. (d) 12. (a)13. (b) : a = 2,

b = 8 a + b = 10 mouse14. (a) : a = x-intercept, b = y-intercept 2h = a + b, k2 = ab

xa

yb

+ =1, substitute (1, 1)

1 1 1a b+ = a + b = ab

2h = k2 y2 = 2x15. (b) : Let OA = a, OB = b, AB = 3l A = (a, 0), B = (0, b)

Let G(x, y) = a b3 3

,

, a = 3x, b = 3y

a2 + b2 = 9l2 x2 + y2 = l2

16. (b) : Let A(h, k) then cos602

1 = =h h

sin602

3 = =k k

A( , )1 3 and Centroid = 53

33

,

17. (b) : �e line ‘l’ can be any one of the bisectors of the angles between the lines 3x – 4y = 5 and 5x + 12y = 13

Angular bisectors, 3 4 5

55 12 13

13x y x y− −

= + −

x – 8y = 0, 32x + 4y – 65 = 018. (b) : L1 2x + 2y = 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


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

= =32

1 32

sq. units

B C G

A

O F x

y

ExD

y x= – + 2

y x= – 1

(3/2, 1/2)

y = 1

y x= –

yx

=+

1

20. (a, b, c, d) : Given, 12 1 2 3( )aP bP cP K+ + =

= + +y aP

bP

cP1 2 3

is minimum.

When yK

aP bP cP aP

bP

cP

= + + + +

12 1 2 3

1 2 3( ) is

minimum.

but yK

a b c abPP

PP

bcPP

PP

caPP

P= + + + +

+ +

+ +

22 2 2 1

2

2

1

2

3

3

2

1

3

33

1P

yK

a b c abPP

PP

bcPP

PP

caPP

P= + + + +

+ +

+ +

22 2 2 1

2

2

1

2

3

3

2

1

3

33

1P

+ + + + +12

2 2 22 2 2K

a b c ab bc ac( )

+ +y a b cK

( )2

2

y is minimum whenPP

PP

PP

PP

PP

PP

1

2

2

1

2

3

3

2

1

3

3

11= = = = = =

i.e., when P1 = P2 = P3 P is incentre of ABC.21. (a, b, c, d) : �e two lines will be identical if there 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

22. (b, c) : tanq = 12

Let slope of required line be m 2

52

Slope of given lines = –2

m

mm+

−= = − 2

1 212

34

or

�e lines are 3x + 4y – 18 = 0, x – 2 = 0

23. (a, b) : (x – 3)(y + 4) = 0, (x + 4)(y – 3) = 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 = 024. (a, c) : xcos + ysin = P Axis rotated through an angle ‘q’Transformed equation is cos(xcosq – ysinq) + sin(xsinq + ycosq) = Por xcos( – q) + ysin( – q) = P xcos + ysin = 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

x y x y+ −=

− −

12

7 55 2

2x – 6y = 0 and 6x + 2y = 5�erefore, the equations of diagonals are x – 3y + 8 = 0 and 3x + y – 6 = 0. �us the required vertex will be the point where these lines meet the line y = 2x. Solving

these lines, we get possible coordinates as 85

165

,

and 65

125

,

.

26. (a, c) : �e intersection point of the given diagonals

is P

32

5,

Equation of angularbisectors of the diagonalsare

D

A B

C

P(3/2, 5)

y x+ 8 – 17 = 0

y x– 8 + 7 = 0

y x y x+ −=

− +8 1765

8 765

= =x y32

5 and

Let length of BC be a and that of CD be b

�en tan //

q = = =ab

ab

22

8

Also ab = 8 a = 8, b = 1So, equations of sides are y = 1, y = 9, x = 1 and x = 2.

27. (c, d) : =

− −

− − −−

=

m m

m m

2 2 5 0

1 7 51 1 1

02

(m – 3)(m2 – m + 2) = 0 For m = 3, the lines become parallel.


28. (a, b, c, d) : Let lines of x2 + 2hxy + y2 = 0 be given by y = m1x and y = m2x m1 + m2 = –2h Slope of y + x is –1

tan , tan =

+−

=+

−m

mm

m1

1

2

2

11

11

tan tan =

+−

= −+

−

mm

mm

1

1

2

2

11

11

and

(for +ve signs, both gives the same value but m1 m2)

= −+

= +−

m m1 211

11

tantan

, tantan

m1 + m2 = –2sec2 h = sec2

cos2 1 =h

2 1 12cos − =h

= +

= +−

cos cot/

12

11

1 2hh

hh

29. (c) 30. (a) 31. (b)Given, P be the point of intersection

tan sinq q=

− +

+= =

12

2

1 134

35

Area of CPB

= =12

2PC PB sinq

= =PB BD103

203

D

A B

C

P

cos( )p q− = − =+ −

=4

5

4 1009

2 2 103

2 583

2ABAB

Again from CPB, BC = 2 103

32. A - p; B - s; C - q; D - r(A) ( ) ( )x y x y+ + + − =7 4 2 7 42 02 + + + − + − =( )[ ] ( )x y x y x y7 7 7 2 3 2 7 42 0

+ + + − + + =( )[ ] ( )x y x y x y7 7 7 2 3 2 7 7 2 0 + + + − =( )( )x y x y7 7 2 7 3 2 0

x y x y+ + = + − =7 7 2 0 7 3 2 0 and

= ++

= =d 7 2 3 21 49

10 250

2

(B) Let two perpendicular linesare coordinate axes.�en, PM + PN = 1 h + k = 1 Hence, the locus is x + y =1But if the point lies in otherquadrants also, then |x| + |y| = 1 Hence, value of k is 1.(C) Angle bisector between the lines x + 2y +4 = 0 and 4x + 2y – 1 = 0 is

x y x y+ ++

= − − ++

2 41 4

4 2 116 4

( )

+ + = − − +x y x y2 4 4 2 12

( )

+ + = − − +2 2 4 4 2 1( ) ( )x y x y Since aa1 + bb1 < 0, so +ve sign gives acute angle bisector.Hence, 2x + 4y + 8 = –4x – 2y + 1 6x + 6y + 7 = 0 m = 7(D) We have, y2 – 9xy + 18x2 = 0or y2 – 6xy – 3xy + 18x2 = 0 y(y – 6x) – 3x(y – 6x) = 0 y – 3x = 0 and y – 6x = 0 �e third line is y = 6. �erefore, area of the triangle formed by these lines,

= = − =12

0 0 11 6 12 6 1

12

6 12 3| | sq. units

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 = 4m1m2

− = − −−

= −

=2 4 27

4 17

2hb

ab

C C( )

(C) 2

22 1

22

6122 2 2

22

m m h m h+ = − = −

=,

h2 = 9 h = ±3 (D) �e angular bisectors of ax2 + 2hxy + by2 + (x2 + y2) = 0 is h(x2 – y2) – (a – b)xy = 0which are angular bisectors of ax2 + 2hxy + by2 = 0. �e two pairs are equally inclined for any . 34. A - r, B - q, C - s, D - pSolving the equation 2m3 – 3m2 – 3m + 2 = 0, we get 2(m3 + 1) – 3m(m + 1) = 0


(m + 1)(2m2 – 5m + 2) = 0 (m + 1)(2m – 1)(m – 2) = 0 m = –1, 1/2 or 2 Equation of the given lines can be written as mi

2x – mi y = –1(A) Algebraic sum of the intercepts made by the lines

on x-axis = − = − + +

= −1 1 1

44 21

42mi

(B) Algebraic sum of the intercepts made by the lines

on y-axis = = − + + =1 1 2 12

32mi

(C) Let pi denote the perpendicular distance of the line from the origin, then

pm

mpi

i

ii=

+ =

++

++

+1

1

11 1

21 1 4

1 21 42

/( / )

/

= + + = +1

245

12 5

110

5 2 9 5( )

(D) li = length of the line intercepted between the coordinates

= −

+

1 12

2 2

m mi i li = + + + + +1 1 16 4 1

1614

= + + = +2 2 5 5

44 2 9 5

4( )

35. (2) : tanq1 = x

and tanq2 = 1 – y Since, q1 + q2 = 45°

+

−=

tan tantan tanq qq q

1 2

1 211

D(0, 1)

A(0, 0)

P( , 0)x

B(1, 0)

Q y(1, )

C(1, 1)45°

1

2

y

x

+ −− −

= =+

x yx y

y xx

( )( )1

1 11 2

1 …(i)

Now, perimeter = 1 1 2 2− + + − +x y x y( ) By using (i), we get Perimeter = 2

36. (3) : ar OAB ar OPA ar OPB ar PAB( ) ( ) ( ) ( ) = + + = 34

4

ar OAB ar OPA ar OPB ar PAB( ) ( ) ( ) ( ) = + + = 3

44

Since, the triangle is an equilateral triangle,

3 12

2= + +( ( , ) ( , ) ( , ))d P OA d P OB d P AB

For all positions of P d(P, OA) + d(P, OB) + d(P, AB) = 3

= =k k3 3 3

37. (4) : Let B(a, b), C(c, b), A(a, d)

�en D, (mid point of BC) is a c b+

2

,

E, (mid point of AB) is a b d, +

2

Given slope of CE = 1− +

−= −

−=

b b d

c ab dc a

2 1 2

Slope of AD = b da c a

b dc a

−+ −

= −−

=

2

2 4( )

38. (4) : Solving 2x + 3y = 1x + 2y = 1, A = (–1, 1) Orthocentre = (0,0) slope of altitude AD = –1Equation of BC is x – y = k

D CB

O

2+ 3

= 1

xy

xy

+2

=1

A

Solving, x – y = k

x + 2y = 1, we get B k k= + −

1 23

13

,

Slope of OB =−+

11 2

kk

, slope of AC = –2/3

−+

= = −11 2

32

18

kk

k

Equation of BC is x y− + =18

0

–8x + 8y – 1 = 0 a = –8, b = 8

39. (5) : Let AC and BD intersect at P

AP = + ++

=12 6 216 9

4

Area of ABD AP BP= = =242

12 BP = 3

AB AP BP= + =2 2 5

40. (1) : Solving the altitudes AM, CL

Orthocentre = 12

12

,

lies on x + y – k = 0 k =1

41. (2) : x = xcosq – ysinq, y = xsinq + ycosq Substitute x and y in the equation a = 2cos2q + 3cosqsinq, b = 2sin2q – 3cosqsinq a + b = 242. (2) : Transformed equation is (y + 2)2 – 8(x + 1) – 4(y + 2) + 12 = 0 y2 = 8x y2 = 4ax a = 2

43. (3) : G a b c a b c= + + + +

=

3 30 0, ( , )

a + b + c = 0 a3 + b3 + c3 = 3abc k = 3

nn


INDEFINITE INTEGRATION

DEFINITIONIntegration is the inverse process of di�erentiation.Let F(x) be a differentiable function of x such

that d

dxF x f x[ ( )] ( ).= �en F(x) is called the integral

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 INTEGRATIONFor finding the integral of complicated functions, generally these methods are used.Integration by substitutionIntegration by partsIntegration by partial fractionsIntegration by Substitution: Direct Substitution

( ( )) ( ) ( ( ))f x f x dx f xn

Cnn

=+

++1

1

Let f(x) = t f (x)dx = dt

=+

+ =+

++ +

t dt tn

C f xn

Cnn n1 1

1 1( ( ))

= +

f xf x

dx f x Ce( )( )

log | ( )|

Let f(x) = t f (x)dx = dt

= + = +1t

dt t C f x Clog | | log | ( ) |

= +f xf x

dx f x C( )( )

( )2

Let f(x) = t dt = f (x)dx

= + = +dt

tt C f x C2 2 ( )

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 a x a x a a2 2 2 2+ + =or ortan cotq q

x a x a x a a2 2 2 2− − =or or secsec coq q

a x a x x a x a2 2 2 2− − = =or or ssin coq q

a x a x x a+ − =or cos 2q

x x a tn

=2 2 expression inside the bracket

( ) ( ) cos sinx a b x x a b− − = +2 2q q

1 11 1 1 1

( ) ( )

( , )

x a x b

n N nx ax b

t

n n+ +

++

=− +

* Alok Kumar is a winner of INDIAN NAtIoNAl MAtheMAtIcs olyMpIAD (INMo-91).

he trains IIt and olympiad aspirants.

INDEFINITE & DEFINITE INTEGRATION

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.

*ALOK KUMAR, B.Tech, IIT Kanpur


Integration by PartsIf u and v be two functions of x, then integral of the product of these two functions is given by

uv dx u v dx dudx

v dx dx= −

In applying the above rule care has to be taken

in the selection of the �rst function(u) and the second function (v). Normally we use the following methods:

1. 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 �rst function and the remaining function is taken as the second function. e.g. In the integration of

x x dx xtan , tan− − 1 1 is taken as the �rst 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 , tan− − 1 1x dx x is taken as the �rst function and 1 as the second function.

3. If both of the functions are directly integrable, then the �rst function is chosen in such a way that the following preference order for the �rst function

(Inverse Trigonometrical, Logarithmic, Algebraic, Trigonometric, Exponential)In the above as ILATE e.g. In the integration of

x xdx xsin , is taken as the first function and sin x is taken as the second function. An important result

e f x f x dx e f x Cx x( ( ) ( )) ( )+ = +Integration by Partial fractionsWhen integrand is a rational function i.e. of the form f xg x

( )( )

, where f(x) and g(x) are the polynomial functions

of x, we use the method of partial fraction. For example, we can rewrite

13 1 3 2

13 3 1

13 3 2( )( ) ( ) ( )x x x x− + −

−+

as

If degree of f(x) is less than degree of g(x) and g x x a x b x c( ) ( ) ..........( ) .........,= − + +1

21 1

1 1

then we can put

f xg x

Ax a

Ax a

A

x a( )( ) ( ) ( )

........( )

....=−

+−

+ +−

+1

1

2

12

1

11

.........( ) ( )

........

...

++

+ ++

++ +

+B x C

x b x cB x C

x b x c1 1

21 1

2 22

1 12

.......( )

.........++

+ ++

B x C

x b x c

1 1

121 1

Here A1, A2,........., A1, ............, B1, B2,..........., B1

.........,C1, C2,......., C1 ...........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 coe�cients 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 INTEGRALSUsing the technique of standard substitution and integration by parts, we can derive the following formulae :

dxa x a

xa

C2 211

+= +− tan

dxx a a

x ax a

C2 21

2−= −

++ ln

dxa x a

x aa x

C2 21

2−= +

−+ ln

dx

a x

xa

C2 2

1

−= +− sin

dx

x ax x a C

2 22 2

+= + + + ln

dx

x ax x a C

2 22 2

−= + − + ln

a x dx x a x a xa

C2 2 2 22

12 2

− = − + + −sin

x a dx x x a

a x x a C

2 2 2 2

22 2

2

2

+ = +

+ + + +

ln( )

x a dx x x a a x x a C2 2 2 22

2 22 2

− = − − + − + ln( )


Integrals of the form

dxax bx c

dx

ax bx cax bx c dx2 2

2

+ + + ++ + , ,

Here in each case, write

ax2 + bx + c = +

+ −a x b

aac b

a24

4

2 2,

Put x ba

t+

=

2and use the standard formula.


( ) , ( ) ,ax b dx

cx ex f

ax b dxcx ex f

+

+ +

++ + 2 2

( )ax b cx ex f dx+ + + 2

Here write ax + b = A(2cx + e) + BFind A and B by comparing the coe�cients of x and constant term.


( )

( ), ( )

( ),ax bx c dx

ex fx g

ax bx c dxex fx g

2

2

2

2+ +

+ +

+ ++ +

( ) ( )ax bx c ex fx g dx2 2+ + + +Here put ax 2 + bx + c = A (ex 2 + fx + g ) + B(2ex + f) + C, find the values of A, B and C by comparing the coefficients of x2, x and constant term.


dx

ax b ex fx gax b

t( )+ + ++ = 2

1Hereput


( )

( )

ax b dx

cx d ex fx g

+

+ + + 2

Here put (ax + b) = A(cx + e) + B, �nd the values of A and B by comparing the coe�cients of x and constant term.


( )

( )

ax bx c dx

ex f gx hx i

2

2

+ +

+ + +

Here put ax2 + bx + c = A(ex + f) (2gx + h) + B(ex + f) + C, �nd the values of A, B and C by comparing the coe�cients of x2, x and constant term.


x dx

ax b cx d( ) ( )2 2+ +Here put cx2 + d = t2.


dx

ax b cx d( ) ( )2 2+ +

Here 1st put x = 1t

and then the expression inside the

square root as y2. Integrals of the form

x a bx dx pm n p( ) ( )+ 0

Here we have the following cases :Case I : If p is a natural number, then expand (a + bxn)p by binomial theorem and integrate.Case II : If p is a negative integer and m, n are rational numbers, put x = tk, when k is the LCM of denominators of m and n.

Case III : If mn+1 is an integer and p is a rational

number, put (a + bxn) = tk, when k is the denominator of p.

Case IV : If m

np+ +1

is an integer, put a bxx

tn

nk+ = ,

where k is the denominator of p. Integrals of the form

f x xg x x

dx R x x dx(sin , cos )(sin ,cos )

(sin , cos ) ,

=

where f and g both are polynomials in sin x and cos x.Here we can convert them in algebraic form by putting

tan x t2= a�er writing

sintan

tancos

tan

tanx

x

x x

x

x=+

=−

+

22

12

12

12

2

2

2and


p x q x ra x b x c

dxsin cossin cos

+ ++ +

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 coe�cients of sinx, cosx


and constant term by this technique. �e given integral becomes sum of 3 integrals in which �rst two are very

easy and in 3rd, we can put tan x t2=


(sin cos )m nx x dx• henm, n N(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 m n+ −

22

is a negative integer, then substitute cotx = p or tanx = p which ever is found suitable.

DEFINITE INTEGRATION

DEFINITION

f x dxa

b( ) is the integration of f(x) w.r.t. x with

x = a as lower limit and x = b as upper limit.GEOMETRICAL INTERPRETATION OF DEFINITE INTEGRALLet f(x) be a function de�ned on a closed interval

[a, b]. �en f x dxa

b( ) represents the algebraic sum

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 de�nite integral may be positive, zero or negative.

y f x= ( )

x b=xx = a

+

–

+

––

y

FUNDAMENTAL THEOREM OF CALCULUS If f(x) is a continuous function on [a, b], then

ddx

f t dt f x x a ba

x( ) ( ) ( [ , ]) =

Now if we take F(x) = f(x), then by the above theorem

ddx

F x f t dt F x f xa

x( ) ( ) ( ) ( )−

= − = 0

− = =F x f t dt c

a

x( ) ( ) (constant say)

= +F x f t dt ca

x( ) ( )

Now F a f t dt c c ca

a( ) ( )= + = + = 0

So, F x f t dt F aa

x( ) ( ) ( )= +

= +

− =

F b f t dt F a

F b F a f t dt

a

b

a

b

( ) ( ) ( )

( ) ( ) ( )

Hence, if f x dx g x c( ) ( ) ,= +

then f x dx g x g b g aa

b

a

b( ) ( ) ( ) ( )= = −

GENERAL PROPERTY OF DEFINITE INTEGRAL

If f x dx F x C( ) ( ) ,= + then f x dx F b F aa

b( ) ( ) ( ) = −

P E R I O D I C P R O P E RT I E S O F D E F I N I T E INTEGRAL

If f(x) is a periodic function with period p then,

f x dx n f x dx n Ip

np( ) ( )=

00

Proof : Let I f x dx f x dx f x dx

f x dx

np p

p

p= = +

+ + +

( ) ( ) ( )

( ) .......

0 0

2

ff x dxn p

np

p

p( )

− 12

3

Let I f x dxk p

kp

11

= −

( )

Put x = (k – 1)p + then,x → (k – 1)p, t → 0

and x → kp, t → p

I f k p t dt f t dt f x dxppp

1000

1= − + = = ( ) ( ) ( )

If f(x) is a periodic function with period p, then

f x dx n m f x dx n m Imp

np p( ) ( ) ( ) , = −

0

DIFFERENTIATION OF DEFINITE INTEGRAL

If F x g t dtf x

f x( ) ( ) ,

( )

( )=

1

2

then F(x) = g(f2(x))f 2(x) – g(f1(x))f 1(x)


APPROXIMATION IN DEFINITE INTEGRALIf f1(x) f(x) f2 (x) x [a, b],

then f x dx f x dx f x dxa

b

a

b

a

b

1 2( ) ( ) ( )

If absolute maximum and minimum values of f(x), when x [a, b] is M and m respectively,

then m b a f x dx M b aa

b( ) ( ) ( )− −

f x dx f x dxa

b

a

b( ) ( )

From �gure it is clear.

D E F I N I T E I N T E G R A L O F P I E C E W I S E CONTINUOUS FUNCTIONS

Suppose we have to evaluate f x dxa

b( ) , but either

f(x) is not continuous at x = c1, c2, ...., cn or it is not de�ned at these points. In both cases we have to break the limit at c1, c2, ...., cn . DEFINITE INTEGRAL AS THE LIMIT OF A SUM

Consider f x dxa

b( ) , for simplicity, we can take

f(x) 0 x [a, b].

x = a + r b a n(( – )/ )

x = a x = b

x

f x( )

�en f x dxa

b( ) 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.

Area of the rth part can be assumed a rectangle,

with width equal to b an−

and height equal to

f a r b an

+ −

So that area =−

+ −

=

b an

f a r b anr

n

1 but this

is only approximated area. To get the actual area, take rectangle with width tends to zero, hence

f x dx b an

f a b an

rn r

n

a

b( ) = −

+ −

→ =

lim1

�is is used in both ways i.e. to evaluate the de�nite integral as a limit of sum and also used in �nding the sum of in�nite terms of some series.

GAMMA FUNCTION If n is a positive number, then the improper integral

e x dxx n− −

1

0is defined as Gamma function and

is denoted by n.

i e n e x dxx n. ., = − − 1

0

Properties of Gamma Function ( )i 1 1= ( ) ( )ii n n n+ =1

( ) , ( ) ( )!iii If then n N n n + = 1

( ) ( / )iv 1 2 = p

Use of gamma function to �nd

sin cos/

m nx x dx0

2p

sin cosm nx x dx

m n

m n =

+

+

+ +

0

21

21

2

2 22

p/

Di�erentiation of function inside integral sign

If I f x dxa

b( ) ( , ) , = then this function can be

di�erentiated w.r.t. , i.e;

dId

f xdx

a

b( ) ( )

=

,

ROBLEMS


PROBLEMSSECTION-I

Single Correct Answer Type

1. Let F x x t dt t dt x xx x

( ) sin cos cos .= + + − 0

2 2

02

�en area bounded by xF(x) and ordinate x = 0 and x = 5 with x-axis is

(a) 16 (b) 252

(c) 352

(d) 25

2. Area bounded by the circle which is concentric

with the ellipse x y2 2

25 91+ = and which passes through

4 95

, ,−

the vertical chord common to both circle and

ellipse on the positive side of x-axis is

(a) 48125

920

365

1tan−

− (b) 2 9

201tan−

(c) 48125

920

1tan−

(d) none of these

3. Area bounded between the curves y x= −4 2 and y2 = 3|x| is

(a) p −13

(b) 2 13 3p −

(c) 2 33

p − (d) 2 33 3p −

4. �e area of the smaller portion enclosed by the curves x2 + y2 = 9 and y2 = 8x is

(a) 23

94

92

13

1+ −

−p sin

(b) 2 23

94

92

13

1+ −

−p sin

(c) 2 23

94

92

13

1+ +

−p sin

(d) 23

94

92

13

1+ +

−p sin

5. If sec

sin( )

sin,

2

2010 20102010x

xdx P x

xC

−= + then value

of P p3

is

(a) 0 (b) 13

(c) 3 (d) 1

6. Let I x f x x dxz

z

1

2

32

2

= −−

( ( ))sec

tan

and I f x x dxz

z

2

2

32

2

= −−

( ( ))sec

tan

where 'f ' is a continuous function and 'z' is any real

number, then II

1

2=

(a) 32

(b) 12

(c) 1 (d) 23

7. I x xdxnn= − tan .1

0

1 If anIn+2 + bnIn = cn n N,

n 1, then(a) al, a2, a3, ........... are in A.P. (b) b1, b2, b3, ........... are in G.P.(c) c1, c2, c3, ........... are in H.P. (d) a1, a2, a3, ........... are in H.P.

8. Let bt t a tt

dt a xx

x cos sin sin ,4 4 42

0

−

= then a

and b are given by(a) 1/4, 1 (b) 2, 2 (c) –1, 4 (d) 2, 4

9. e xdxe

x

xsec

/

/ 2

24

4

1−−p

p is equal to

(a) 0 (b) 2 (c) e (d) 2e

10. Value of x x x x dx+ − + − −

2 1 2 1

1

5

is

(a) 83

(b) 163

(c) 323

(d) 343

11. If f x t dtx

( ) ( ) /= + − 1 3 1 2

0 and g (x) is the inverse of

f, then the value of g x

g x

( )

( )2 is

(a) 3/2 (b) 2/3 (c) 1/3 (d) 1/2

12. 2 1

1 2x x

xdx

+ +

=−

sin

cosp

p

(a) p2

4 (b) p2 (c) 0 (d)

p2


13. If t f t dt x xx

21

1 02

= −

sin , , ,

sin

p then the

value of f 13

is

(a) 13

(b) 3 (c) 13

(d) 3

14. Let , I e x dx n N nnx n= −

(sin ) , 10

then II2008

2006 equals

(a) 2007 20062008 12

+

(b) 2008 20072008 12

+

(c) 2006 20042008 12

−

(d) 2008 2007

2008 12−

15. If secA t dt

tB

t tdt=

+=

+ 11

12 211

,( )

cosin qq

then

A A B

e B

A B

A B

2

2

2 2

1

1 1

+ −

+ −

=

(a) sin q (b) cosec q (c) 0 (d) 1

16. �e area of the region whose boundaries are de�ned by the curves y = 2 cos x, y = 3 tanx and the y-axis, is

(a) 1 3 23

+

ln (b) 1 32

3 3 2+ −ln ln

(c) 1 32

3 2+ −ln ln (d) ln 3 – ln 2

SECTION-II

Multiple Correct Answer Type

17. If u dxx x

v x dxx x

=+ +

=+ + 4 2

0

2

4 207 1 7 1

and , then

(a) u > v (b) u < v (c) u = v (d) u = p6

18. If f(x) is an odd and periodic function with period T, then

(a) f x dxT

( )0

0 = (b) f x dxT

( )0

30 =

(c) f x dx f x dx a Ra

a T T( ) ( )

+

= for any 0

(d) f x dxT

( )/

0

20 =

19. If the area bounded by y = (sinx)cosecx,y = (cosecx)sinx,y = (sinx)sinx and y = (cosecx)cosecx, between the ordinates

x = 0, x = p2

denoted by A1, A2, A3 and A4 then

(a) A4 is the greatest (b) A1 is the least (c) A2 > A3 (d) A2 < A3

20. 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)(c) f t dt g t dt( ) ( )= − 0

5

0

55

(d) f t dt g t dt( ) ( )= − − 0

5

0

55

21. If 3 4

2 43x

x xdx+

− − = log |x – 2| + k log f(x) + c, then (a) k = –1/2 (b) f(x) = x2 + 2x + 2 (c) f(x) = |x2 + 2x + 2| (d) k = 1/4

22. xx

xx

dx+−

+ −

+

−

−

11

11

22 2

1 2

1 2

/

/is

(a) 4 43

ln

(b) 4 3

4ln

(c) −

ln 81256

(d) ln 25681

23. �e value of t dt

tdt

t te

x

e

x

1 121

21+

++

/

tan

/

cot

( ) is

(a) 12 2+ tan x

(b) 1

(c) p4

(d) 21 2

1

1

pdt

t+−

24. If x x dx f x g xx

− −

= 12

0( ) ( ), where [x] and {x}

are integral and fractional parts of x, respectively then

(a) f x x( ) { }=2

(b) g(x) = {x} – 1

(c) f x x( ) [ ]=2

(d) g(x) = [x] – 1


25. If I x xx x

dx= ++ +

5 71 2

8 6

2 7 2( ), then I is equal to

(a) xx x

C7

7 22 1+ ++ (b) x

x xC

5

2 71 2+ ++

(c) −+ +

+12 17 2x x

C

(d) p xq x

p x q x( )( )

, deg ( ) deg ( )= = 7

SECTION-III

Comprehension Type

Paragraph for Question No. 26 to 28Two real valued di�erentiable functions f(x) and g(x)satisfy the following conditions:

(i) =−

f xf x g x

( )( ) ( )

3 (ii) =

−g x

g x f x( )

( ( ) ( ))23

(iii) f(0) = 5 (iv) g(0) = –1 then

26. 1

f xdx

( )=

(a) x – ln (f(x)) + c (b) 12

( ln ( ))x f x c− +

(c) 13

( ln ( ))x f x c− + (d) 1

x f xc

−+

ln ( ( ))

27. 1g x

dx( )

=

(a) x – ln (g(x)) + c (b) 12

( ln ( ) )x g x c− +

(c) 13

( ln ( ) )x g x c− + (d) 1

x g xc

−+

ln ( )

28. g xf x

dx( )( )

=

(a) x + 3ln (f(x)) + c (b) x – 3ln (f(x)) + c (c) 3ln (f(x)) – x + c (d) 3(x – ln (f(x)) + c

Paragraph for Question No. 29 to 31

If , thendx

x xA f x c

( )sin { ( )}

1 12 21

+ −= +−

29. �e value of A is

(a) 1 (b) 12

(c) − 12

(d) 0

30. �e f(x) is

(a) 1

1

2

2+

−

x

x (b)

1

1

2

2−

+

x

x (c) 1

1

2

2−

+

x

x (d) 1 2− x

31. f x dx( ) is

(a) 2 tan–1 x – x + c (b) 2tan–1x + x + c(c) tan–1x + 2x + c (d) none of these

Paragraph for Question No. 32 to 34Let f(x) be a differentiable function satisfying (x – y) f(x + y) – (x + y)f(x – y) = 4xy(x2 – y2) for all x, y R. If f(1) = 1, then

32. �e function f at x = 0 attains(a) Local maximum (b) Local minimum(c) Point of in�exion (d) none of these

33. �e value of f x dx( )−1

2 is

(a) 0 (b) 14

(c) 114

(d) 154

34. �e area of the region bounded by the curves y = f(x) and y = x2 is

(a) 14

sq.units (b) 1

12sq.units

(c) 712

sq.units (d) 1112

sq.units

SECTION-IV

Matrix-Match Type

35. Match the following :

Column I Column II

(A)cosec x dx

x

+

1

02

,

, p(p) 1

2 21cos cos−

+x c

(B)

sincos

,

,

xx

dx

x

3 2

04

−

p

(q) sin–1(sinx cosx) + c

(C)

coscos

,

,

xx

dx

x

2

04

p(r) − +−1

221sin ( cos )x c

(D)cos

cos sin

22 4

x

x xdx

+ (s) sin–1(2sinx – 1) + c


36. Match the de�nite integrals in column I with their values in column II.

Column I Column II

(A) x x x dx(sin (sin ) cos (cos ))2 2

0+

p(p) p2

16

(B) ( sin cos )/

20

42

x x x dx+p

(q) p2

2

(C) p

p

p

lnln ( sin )

/

/

21 2

4

4+

− x dx (r) p2

4

(D)8

3 3

3 4 2

2 20

x x x

x xdx

cos sin

p p

p

− + (s) p2

8

SECTION-IV

Integer Answer Type

37. Let f(x) be a di�erentiable function such that

f x x e f x t dttx

( ) ( ) ,= + −−2

0then 6f(1) =

38. Let , be roots of the quadratic equation 18x2 – 9px + p2 = 0, where < . Also f(x) = x2 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 �nd the sum of the digits in

39. Given lnsin ln/

x dx =pp

212

0

2

and then xx

dx k ksin

ln , .........../

= =2

0

2

22pp

40. Find the value of 29 1

4 1

4 7

0

1

4 6

0

1

( )

( )

.

−

−

x dx

x dx

41. If f x a x b f( ) cos( ) ,= +

=p p1

2

and f x dx( )/

/= +

2 11 2

3 2

p , then �nd the value of,

− +− −12 1 1p

(sin cos ).a b

42. Let F(x) be a non-negative continuous function

de�ned on R such that F x F x( )+ +

=1

23 and the

value of F x dx( ) .is 9000

0

1500

�en the numerical value

of isSOLUTIONS

1. (b) : F x x t dt tdt x xxx

( ) sin cos cos= + − + 2 2 2

00

= +

− +sin (sin ) cosx t t x xx

x

0

2

0

2 222

= sin2x + x2 – x2 + cos2x = 1

A xF x dx x dx x= = =

= ( ) ( )( )1

2252

0

5

0

5 2

0

5

2. (a) : Eccentricity of ellipse = 45

Co-ordinate of foci = (4, 0), (-4, 0)

−

4 95

, is one of

the end point of latusrectum

x

y

P(4, –9/5)

O

Q(4, 9/5)

Required area is

12

4 95

2 920

22

2

1

pp +

−−tan area of POQ

=

−

=

−− −481

259

2012

4 185

48125

920

361 1tan tan55

3. (c) : Required area = 2 4 32

0

1− −

x x dx

= − +

−

−2

24 4

2 23 2

32 1

3 2

0

1x x x x

sin/

= −2 33

p

(1, 3)(–1, 3)

y x= 4 – 2

–2 2x

y x2 = 3| |

y

O


4. (b) : x2 + y2 = 9, x2 + 8x – 9 = 0

= − +x 8 64 362 O

(0, 3)

(3, 0)(–3, 0)

(0, –3)

2 29x y

28y x

x

y

= − = −x 8 102

9 1,

x = 1

Area enclosed = + −

= + −

2 2 2 9

2 2 2 9

0

12

1

3

2

1

3

xdx x dx

xdx x dx00

1

On simplifying, we get

Area = + −

−2 2

394

92

13

1p sin

5. (c) : sec

sin

2

20102010x

xdx

−

= − = −− sec (sin )(sin )

2 20102010 1 22010 1x x

xdx I I

Applying by parts on I1, we get

I xx

x x

xdx1 2010 20112010= +

tan(sin )

tan cos

(sin )

= +

tan(sin ) (sin )

xx

dxx2010 20102010

= − = + = +I I I xx

C P xx

C1 2 2010 2010tan

(sin )( )

(sin )

= =P p p

3 33tan

6. (a) : I x f x x dxz

z

1

2

32

2

= −−

( ( ))sec

tan

I f x x dxz

z

2

2

32

2

= −−

( ( ))sec

tan

I x f x x dxz

z

1

2

3 32

2

= − −−

( ) (( ) )sec

tan

= −−

2 3 31

2

2

2

I f x x dxz

z

( ( ))sec

tan

= −−

3 32

22f x x dx

z

z( ( ))

sec

tan = =2 3 3

21 21

2I I

II

7. (a) : I xn

x xn x

dxn

n n=

+

−

++

+−

+

1

1

0

1 1

20

1

1 11

1tan

( )n I xx

dxn

n+ = −

+

+

14 1

1

20

1p

( )n I xx

dxn

n+ = −

++

+

34 1

2

3

20

1p

+ + + = −++( ) ( )n I n I

nn n1 32

122

p

an = (n + 3) a1, a2, a3, ......are in A.P. bn = (n + 1) b1, b2, ......are in A.P.

c

nn = −+

p2

12

is not in any progression.

8. (a) : Since, bt t a tt

dt a xx

x cos sin sin4 4 42

0

−

=Di�erentiating both sides w.r.t. x

− =− bx x a x

x

a x x x

xcos sin cos sin4 4 4 4 4

2 2

On comparing, we get b = 4a a = 1/4 and b = 1

9. (a) : Let Ie x dx

e

x

x=−−

sec

/

/ 2

24

4

1p

p

If f x e xe

x

x( ) sec=−

2

2 1

− =−

−

−f x e xe

x

x( ) sec2

2 1=

−e x

e

x

xsec2

21= −

−= −e x

ef x

x

xsec ( )

2

2 1 I = 0 (f(x) is an odd function)

10. (d) : x x+ −( ( ))2 1

= − + + − = − +( ) ( ) ( )x x x1 1 2 1 1 12 2

And x x x x− − = − + − −2 1 1 1 2 12 2( ) ( ( )) ( )

= − −x 1 1

Then x x x x dx+ − + − −

2 1 2 1

1

5


= − + + − − ( ) ( )x dx x dx1 1 1 11

5

1

5

= − + + − − + − − ( ) ( ) ( )x dx x dx x dx1 1 1 1 1 11

2

1

5

2

5

= + + − + − ( ) ( ) ( )x dx x dx x dx1 1 11

4

0

1

0

4

= +

+ −

+ −

23

23

23

3 2

0

43 2

0

13 2

1

4x x x x x x/ / /

= +

+ −

+ −

− −

=16

34 1 2

3163

4 23

1 343

11. (a) : f x t dtx

( ) ( ) /= + − 1 3 1 2

0

i e f g x t dtg x

. . [ ( )] ( ) /= + − 1 3 1 2

0

i e x t dtg x

. . ( ) /= + − 1 3 1 2

0[ g is inverse of f f [g(x)] = x] Di�erentiating with respect to x, we have 1 = (1 + g3(x))–1/2.g(x)i.e. (g(x))2 = 1 + g(x)3

Di�erentiating again with respect to x, we have

2 3 2 = g x g x g x g x( ) ( ) ( ( )) ( )

=g x

g x

( )

( ( ))232

12. (b) : I x xx

dxx x

xdx=

+=

− + 4

14

120

20

sincos

sin

cos

p p p

=+

=+

= =

2 41

81

2 2

20

20

2

2 2

I xx

dx xx

dx

I I

p p

p p

p psincos

sincos

/

13. (d) : t f t dt xx

21

1 = − ( ) sinsin

Di�erentiating both sides with respect to ‘x’ 0 – sin2x.f (sinx).cosx = –cosx cosx [1 – sin2x.f(sinx)] = 0But cosx 0

So, (sin )sin

f xx

= 12 f 1

33

=

14. (b) : I e x dxnx n= −

(sin )0

= − +− − −

[sin ( )] sin cosn x n xx e n x x e dx01

0

= + − −

0 1

0n x x e dxn x(sin cos )

= −

− − + − −

− −

−

n x x e

n x n x x

n x

n n

[(sin cos )( )]

{ sin ( )sin cos }(

10

2 2

01 ee dxx− )

= + − + − −− −

0 1 12 2

0n e x n x x dxx n n{ sin ( ) sin ( sin )}

= − −− −

n e n x n x dxx n n{( ) sin sin }1 2

0= n(n – 1)In – 2 – n2In (1 + n2) In = n(n – 1)In–2

orI

In nn

n

n−= −

+22

11

( )

15. (c) : Bt t

dt= 1

1 ( )1+ 2

cosecq

Let 11

021

tu B udu

uA B A B= = −

+ + = = −

sinq

−

−

−

=

A A A

e A

A

2

0 2

2

1

1 2 1

0

16. (b) : Solving 2 cos x = 3 tan x, we get 2 – 2 sin2 x = 3 sin x

= =sin x x12 6

p

Required area = ( cos tan )

sin ln sec |

/

/

2 3

2 3 1

0

6

06

x x dx

x x

−

= − = −

p

p 33 2 32

3ln ln+

17. (c, d) : ux x

dx=+ +

17 14 2

0

Put xt

dxt

dt= = −1 12,


u

t t

dtt

t dtt t

v=+ +

−

=

+ +=

1

1 7 1 7 14 2

2

0 2

4 20

u = v

So, u v ux

x xdx+ = =

+

+ +

21

7 1

2

4 20

=+

−

+

=+

=

−

21 1

1 99 3

2

20

2u xdx

xx

dtt

p

18. (a, b, c) : Since f x dx f T x dxTT

( ) ( )= −00

= is periodic with periodic f x dx f x TT

( ) ( ( ) )−

0

= is odd− f x dx fT

( ) ( )

0 =2 f x dx

T( ) 0

0

(a) is correct

Since f x dx f x dxTT

( ) ( )= 300

3 (b) is also correct.

(c) is a known result.

19. (a, b, c) : For x

02

, p 0 < sin x < 1 < cosec x < So, sinxcosecx < sinxsinx < cosecxsinx < cosecxcosecx

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

I f t dt= ( )0

5

Indeed I g t dt= − ( )50

5

( f(t) = g(t – 5) on replacing x by t – 5 in (ii))

= − g t dt g( ) ( )50

5 is even

Choice (c) is correct and choice (d) is false.

21. (a, b, c) : By using partial fraction, we get 3 4

2 4 2 2 23 2x

x xA

xBx C

x x+

− −=

−+ +

+ + Solving this equation, we get A = 1, B = –1 and C = –1

+− −

=−

− ++ +

3 42 4

12

12 23 2

xx x

dxx

dx xx x

dx

= − − + + +log | | log | |x x x c2 1

22 22

Hence k = –1/2, so that f(x) = |x2 + 2x + 2|

22. (a, c, d) : Let I xx

xx

dx= +−

− −+

−

11

11

2

1 2

1 2

/

/

=−

= −−

−

41

2 412 2

0

1 2

1 2

1 2 xx

dx xx

dx( )

/

/

/

= − − = −

4 1 4 34

20

1 2ln ln

/x

=

=

= −

4 43

25681

81256

ln ln ln

23. (b, d) : Let I dtt te

x=

+ ( )/

cot

1 21

Put tz

= 1 = −dt

zdz1

2

=−

+

=+I z

dz

z z

z dz

zx

e

e

x1

1 1 1 1

2

2

2( )tan

tan

=

+t dt

tx

e

( )tan 1 2

+

++

12 2

11 1/

tan

/

cot

( ) ( )e

x

e

xt dt

tdt

t t

=

++

+ tt

dt tt

dte

x

x

e

( ) ( )/

tan

tan1 121

2

=

+= +

t dt

tt

e

e

e

e

( )ln

/ /112

121

21

= + − +

12

1 1 122ln( ) lne

e = =1

212(ln )e

Also, 21

41

4 121

1

20

11

p p pdt

tdt

t+=

+=

−

− . tan = =44

1p

p.


24. (a, b) : I x dx x dx

x

x x x= −

= −

+

−

+

{ } { }

{ }

[ ] { } [ ]12

12

12

0 0

[[ ]

[ ] { }

x

x x

dx+

= −

+ −

[ ] { } { }

{ }x x dx x dx

x12

12

0

1

0

= −

+ −

[ ]

{ }x x dx x dx

x12

12

0

1

0

= −

+ −

[ ]

{ }

x x x x xx2

0

1 2

02 2 2 2

= +−

= − 01

2 21

{ } { } { } { }x x x x

25. (a, d) : We can write

I x x

xx x

dx= +

+ +

5 7

1 1 2

8 6

145 7

2 =+

+ +

5 7

1 1 2

6 8

5 7

2x x

x x

dx

Put so thattx x

= + +1 1 25 7 ,

I dtt t

C= − = + 21 =

+ ++x

x xC

7

2 71 2

(26 - 28) :

26. (c) :

1 12 3f x

dxe

dxx( )=

+

=+

= − + +−

−−

ee

dx e cx

xx

2 31

32 3ln ( )

= − +13

( ln ( ))x f x c

27. (c) : 1 1

3 4g xdx

edxx( )

=−

=−

= − − +−

−−

ee

dx e cx

xx

3 41

33 4ln | |

= − +13

( ln | ( )|)x g x c

28. (b) :3 4

3 21 3 2

3 23

−

+= −

+

= − + e

edx e

edx x f x c

x

x

x

x( ) ln ( ( ))

(29 - 31) :

29. (c) : dx

x x( )( )1 12 2+ − =

+

−

dx

xx x

32 21 1 1 1

Put 1 1 2 222

3xt

xdx t dt− = − =

−+

=− + =− − +− −tdt

t tt C x

xC

( )tan tan2

1 12

212 2

12

12

= − −+

+−12

11

12

2sin xx

C = −A 12

30. (b) : f x xx

( ) = −+

11

2

2

31. (a) : f x dx xx

dx( ) = −+

11

2

2

=+

−

21

12xdx = 2tan–1x – x + c

(32 - 34) :32. (c) 33. (d) 34. (b)(x – y)f(x + y) – (x + y)f(x – y) = 4xy(x2 – y2)

++

−−−

=f x y

x yf x y

x yxy

( ) ( )4

So,f x h

x hf x h

x hxh

( ) ( )++

−−−

= 4

+ − +

+

−− − +

−

=

f x h f x f xx h

f x h f x f xx h

xh

( ) ( ) ( )

( ) ( ) ( )4

+ −+

−

− −−

= +−

−

f x h f xx h

f x h f xx h

xh f xx h x

( ) ( ) ( ) ( )

( )4 1 1++

h

+ −+

−

− −−

= +

−

f x h f xx h

f x h f xx h

xhhf x

x h

( ) ( ) ( ) ( ) ( )4

22 2

+ −

+

+− −−

−

f x h f xh x hf x h f x

h x h

( ) ( ).

( ) ( )

1

1= +

−4

22 2xf x

x h

( )

Taking limit on both sides as h →0 ;

+

= +

= +f x

xf x

xx

f x

x

f xx

xf x

x

( ) ( ) ( ) ( ) ( )4

222 2


= +f x xf x

x( )

( )2 2

+−

=f x

f xx

x( )( )

,2 2

which is a linear di�erential equation.

Integrating factor = =−e

xx

dx11

Its solution is f xx

xx

c( ) = +1 2 12

= + = +f x

xx c f x x cx

( )( )2 3

But f(1) = 1 c = 0So, f(x) = x3 f (x) = 3x2 and f (x) = 6x Here, f (x) = 0 x = 0 So, at x = 0, f(x) has point of in�exion.

x dx x dx x dx x3 3 3

1

24

1

2

1

1

1

20 1

4154

= + = +

=

−− �e area of the region bounded by the curves y = f(x) and y = x2 is

( ) .x x dx x x2 3

0

1 3 4

0

1

3 413

14

112

− = −

= − = sq units

35. A → s; B → p; C → r; D → q

(A) cosecx dxx

xdx+ =

+ 1

1 sin

sin

=−

=−

=cos

sin ( sin ) ( )[ sin ]x

x xdx dt

t tt x

1 1Put

= sin–1(2sin x – 1) + c

(B) sincos

sin

cos

xx

dxx dx

x3 212 2 2−

=−

=

+−1

2 21cos cos x c

(C) coscos

sincos cos

xx

dx xx x

dx2

12

21 2 2

=−

= −

−=

12 1

22

dt

tt x( cos )Put

= − +−12

21sin ( cos )x c

(D)

cos

cos sin

cos

sinsin sin2 2 2

4 2

222 4 2

1x dx

x x

x dx

x

x c+

=−

=

+ −

36. A → q; B → q; C → r; D → r

(A) I x x x dx= + (sin (sin ) cos (cos ))2 2

0

p

= − +I x x x dx( )(sin (sin ) cos (cos ))pp

2 2

0

Adding 2 2 2

0I x x dx= +p

p(sin (sin ) cos (cos ))

2 2 2 2

0

2I x x dx= +p

p(sin (sin ) cos (cos ))

/

I x x dx= +pp

(sin (sin ) cos (cos ))/

2 2

0

2

Also I x x dx= +pp

[sin (cos ) cos (sin )]/

2 2

0

2

Adding 2 2 2 2 2

0

2I dx= = =p p p p

p( / )

/

I = p2/2

(B) Let f x x( ) sin= 2

Then x f x x xx

x x = =( ) . cos cos22

I f x x f x dx x f x= + = ( ( ) ( )) ( )/

/

04

0

4 22

pp

= =2 4 2 22 2( / ) sin( / ) /p p p

(C) ln sin ln sin cos/

/

/

/1 2

4

4

4

4+ = +

−− x dx x x dxp

p

p

p

= +− ln sin( / )/

/2 4

4

4x dxp

p

p= ln sin

/2

0

2t dt

p

= + ln lnsin//

20

2

0

2dt t dt

pp

= − =−( / )ln ( / ) ln ( / ) lnp p p4 2 2 2 4 2

+ =

−

= −

−

p p p p

p

p

lnln sin

lnln

/

/

21 2

2 42

44

4 2x dx

(D) Ix x x

x xdx=

− +3 4 2

2 20 3 3

cos sin

p p

p

Ix x x

x xdx=

−

− +( ) cos sinp

p p

p 3 4 2

2 20 3 3


Adding 2 4 2

0I x x dx= p

pcos sin

I x x dx x x dx= = p p

pp

22 4 2 4 2

0

2

0

2. cos sin cos sin

//

=

=p p p( ) ( / ) /3 1 16 4 2

2 322

37. (8) : f x x e f x t dt x e e f t dttx

x tx

( ) ( ) ( )= + − = −− − 2

0

2

0 = − − + − −f x x e e f x x e e f xx x x x( ) ( ( )) ( )2 2

= + = + +f x x x f x x x k23

23

2( )

But f(0) = 0 k = 0 =f 1 43

38. (3) : (fog)(x) = f(cosx) = cos2xLet , be roots of the equation

18 9 06 3

2 2x x− + = = =p p p p,

Area = = cos/

/2

6

3

12x dx

p

p p

39. (2) :

I x x dx x x x xdx= = − + 2 2 2

0

2

0

2

0

22cosec cot cot

/ // p pp

= + + −→ + 0 2 2

0

2

02

0

2lim

tanlnsin lnsin/

/

x

xx

x x x dxpp

= − − =→ +

0 2 22

12

20

limx

x xlnsin ln lnp p

lim limx x

xx

xx

x→ →+ +

=−

=

0 0

21

1

1 0lnsin/

sin.cos

40. (7) : I x dx= − ( )1 14 7

0

1

= −

+ −x x x x x dx1 7 4 14 7

0

14 6 3

0

1( )

= − + − = −28 28 1 29 28 14 6 4 6

0

1

0

1I x dx I x dx( ) ( )

−

−

=

29 1

4 1

7

4 7

0

1

4 6

0

1

( )

( )

x dx

x dx

41. (6) : f (x) = –apsin(px)

= − = − = = −f a a a1

2 21p p p psin

a x b dx a x bx

a b a

cos sin

/

/

/

/p p

p

p

+ = +

= − +

−

1 2

3 2

1 2

3 2

32 pp

+

b2

− + = + =2 2 1 1a b bp p

So, − − + = − − +

=− −12 1 1 12

20 61 1

p pp(sin ( ) cos )

42. (4) : We have, 1)F x F x( ) ....(+ +

=1

23

Replace by in (1), we getx x + 12

F x F x+

+ + =1

21 3 2...( )

From (1) and (2), we get F(x) = F(x + 1) .....(3) F(x) is periodic function.

Now consider I F x dx F x dx= = ( ) ( )0

1500

0

11500

= + 15000

12

12

1F x dx F x dx( )

Using property ofperiodic functioon

Put in integral, we getndx y= + 12

2

I F x dx F y dy

F x F x

= + +

= + +

1500 12

1500 12

0

12

0

12

= dx dx1500 3 10

12

0

12

Using( )

Hence I =

= =1500 3 1

2750 3 2250( )

nn


CATEGORY IFor each correct answer one mark will be awarded, whereas, for each wrong answer, 25% of total marks (1/4) will be deducted. If candidates mark more than one answer, negative marking will be done.

1. �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

(a) 23

(b) 143

(c) 163

(d) 53

2. Let

a b c, and be three non-zero vectors, no two of which are collinear. If the vector

a b+ 2 is collinear with c and

b c+ 3 is collinear with a , then

a b c+ +2 6 is equal to (a) a (b)

b (c) c (d)

0 3. If ABCD is a rhombus whose diagonals cut at the

origin O, then OA OB OC OD

+ + + equals

(a) AB AC

+ (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 = 3AB, then the position vector of C is (a) 3 2

a b− (b) 3 2

b a−(c) 3 2

b a+ (d) 2 3

a b−

5. If the vectors

a b c a b c a b c+ + − + − − + 3 2 3 4 3 5, and and

c a b c a b c− + − − +3 2 3 4 3 5 and 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 − + + + +2 3 5 2 3

a b c a b c, and 7 a c− are

(a) collinear (b) non-coplanar(c) non-collinear (d) none of these

7. If ABCDEF is a regular hexagon with AB a

=and BC b

= , then CE

= (a)

b a− (b) −

b(c)

b a− 2 (d) none of these

8. If a i j k= + +^ ^ ^2 2 and

b i j k= + +3 6 2^ ^ ^ , then the vector in the direction of a and having magnitude | |

b is

(a) 7 2 2( )^ ^ î j k+ + (b) 79

2 2( )^ ^ î j k+ +

(c) 73

2 2( )^ ^ î j k+ + (d) none of these

9. Let ABCD be the parallelogram whose sides AB

and AD are represented by the vectors 2 4 5i j k^ ^ ^+ −and i j k^ ^ ^+ +2 3 respectively. If a is a unit vector

parallel to AC

, then a is equal to

(a) 13

2 3( )^ ^ î j k+ + (b) 13

3 6 2( )^ ^ î j k+ +

(c) 17

3 6 3( )^ ^ î j k− − (d) 17

3 6 2( )^ ^ î j k+ −

10. �e value of , for which the four points 2 3i j k^ ^ ^,+ −i j k^ ^ ^,+ +2 3 3 4 2i j k^ ^ ^+ − and i j k^ ^ ^− + 6 are coplanar, is (a) –2 (b) 8 (c) 6 (d) 0

11. Consider points A, B, C, D with position vectors

7 4 7i j k^ ^ ^− + , i j k^ ^ ^− +6 10 , − − +i j k^ ^ ^3 4 and

5 5i j k^ ^ ^− + respectively. �en ABCD is a

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 published in the subsequent issue.

UNIT-V : VECTORS

By : Sankar Ghosh, HOD(Math), Takshyashila


(a) square (b) rhombus (c) rectangle (d) none of these

12. �e angle between the two vectors i j k^ ^ ^+ + and

2 i j k^ ^ ^− +2 2 is equal to

(a) cos−

1 23

(b) cos−

1 16

(c) cos−

1 56 (d) cos−

1 13

13. If

p q and are non-collinear unit vectors and | | ,p q+ = 3 then ( ) ( )2 3 3

p q p q− + is equal to

(a) 0 (b) 1/3 (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

(a) 172

(b) 142

(c) 72

(d) 41

15. If

a b c, , are unit vectors such that

a b c+ + = 0then

a b b c c a + + = (a) 3/2 (b) –3/2 (c) 2/3 (d) 1/2

16. If

a b a b a mb m + + = and then ( ) ( ), (a) –1 (b) 1

(c) − | |

| |

a

b

2

2 (d) 1

2

17. If the sum of two unit vectors is also a unit vector, then the angle between the two unit vectors is

(a) p3

(b) p2

(c) p4

(d) 23p

18. 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

19. Let a i j k= − +^ ^ ^2 3 . If

b is a vector such that

a b b a b = − =| | |2 7 and | , then | |

b =

(a) 7 (b) 3 (c) 7 (d) 3

20. If

a b c, and 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 (a) | | | | | |

a b c2 2 2+ +

(b) | | | | | |

a b c+ +

(c) 2 2 2 2| | | | | |

a b c+ +

(d) 12

2 2 2| | | | | |

a b c+ + 21. If = + +2 i j k^ ^ ,̂

= − +i j k^ ^ ^2 2 and = − +3 4 2i j k^ ^ ^,

then the projection of ( )

+ in the direction of

is

(a) 173

(b) 53

(c) 1743

(d) − 18

22. If

a b c, and are three vectors such that

a b c+ + = 0 and | | , | | , | | ,

a b c= = =3 5 7 the angle between a and

b is(a) 60° (b) 30° (c) 45° (d) 90°

23. If

a i j k b= + + =^ ^ ^, | |2 2 5 and angle between

a b and

a b and is p/6, then the area of the triangle formed by these two vectors as two sides is

(a) 154

(b) 152

(c) 15 (d) 15 32

24. If

a b and 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

25. Let

a b c, , be three vectors such that

a b c =and

c a b = , then

(a)

a b c = | |2 (b)

c a b = | |2

(c)

b c a = | |2 (d)

a b c||( )

26. Let

a b and be two vectors of equal magnitude

inclined at an angle q, then a sin q2

is equal to

(a) | |

a b−2

(b) | |

a b+2

(c) | |

a b− (d) | |

a b+

27. a is perpendicular to

b c and , | | , | | ,

a b= =2 3

| |c = 4 and the angle between

b c and is 23p , then

[ ]

a b c is equal to

(a) 4 3 (b) 6 3 (c) 12 3 (d) 18 3

28. �e unit vector perpendicular to i j^ ^− and coplanar

with i j^ ^+ 2 and 2 3i j

^ ^+ is


(a) 2 5

29

i j^ ^−

(b) 2 5i j^ ^+

(c) i j^ ^+

2 (d) i j

^ ^+

29. If

a i k b x i j x k= − = + + −^ ^ ^ ^ ^

, ( )1 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

30. If | | | |

a b a b + =2 2 144 and | |a = 4 , then | |

b is equal to (a) 12 (b) 3 (c) 8 (d) 4

CATEGORY IIEvery 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.

31. �e vector of magnitude 12, which is perpendicular

to both the vectors 4 3i j k^ ^ ^− + and − + −2 2i j k

^ ^ ^

is (a) − + +4 8 8i j k

^ ^ ^ (b) − + +2 4 4i j k^ ^ ^

(c) − + +6 12 12i j k^ ^ ^ (d) none of these

32.

a b and are such that | | , | |

a b= =1 4 ,

a b = 2. If

c a b b= −2 3 , then the angle between

b c and is

(a) p6

(b) 56p (c) p

3 (d) 2

3p

33. Let OA a OB a b OC b

= = + =, ,10 2 , 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

34. �e value of (constant) when 2 i j k^ ^ ^,− +

i j k^ ^ ^+ −2 3 and 3 5i j k^ ^ ^+ + be coplanar, is (a) 2 (b) –4(c) 4 (d) none of these

35. If [ ] [ ]

+ + + = , then is equal to (a) 1 (b) 2 (c) 3 (d) 4

CATEGORY III

In this section more than 1 answer can be correct. Candidates will have to mark all the correct answers, for which 2 marks will be awarded. If, candidates 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)

36. If a is any vector, then

(a) ( ) ( ) ( )^ ^ ^ ^ ^ ^ a i i a j j a k k a + + =

(b) ( ) ( ) ( ) | |^ ^ ^ a i a j a k a + + =2 2 2 2

(c) i a i j a j k a k a^ ^ ^ ^ ^ ^( ) ( ) ( ) + + = 2(d) only (a) and (b)

37. �e vectors a i j k= + + ^ ^ ^2 ,

b i j k= + −^ ^ ^ and c i j k= − +2 ^ ^ ^ are coplanar, if = (a) –2 (b) 3 1+

(c) 1 3− (d) 2

38. If

a b c, and are any three vectors, then which of the following is equal to

a b b c c a + + ? (a) ( ) ( )

a b b c− −

(b) ( ) ( )

c b a c− −

(c) ( ) ( )

a b a c− −

(d) 12

a b c b c a c a b − + − + − ( ) ( ) ( )

39. [ ]

a b c is equal to

(a) [ ][ ] [ ][ ]

a b c a b c−

(b) [ ][ ] [ ][ ]

a 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 k^ ^ ^2 (d) − + −13

43

13

i j k^ ^ ^


SOLUTIONS

1. (b) : Given that, the line PQ makes equal angles with the axes. Let the angle be . cos2 + cos2 + cos2 = 1 3cos2 = 1

cos = 13

�us, direction cosines are 13

13

13

, ,

Now, PM = projection of AP on PQ

= − + + − + − ( ) ( ) ( )2 3 13

3 5 13

1 2 13

= 23

and AP = − + + − + − =( ) ( ) ( )2 3 3 5 1 2 62 2 2

AM AP PM= − = − =( ) ( )2 2 6 43

143

2. (d) : Given that

a b+ 2 is collinear with c + =

a b xc x R2 ,Also,

b c a+ 3 is collinear with + =

b c ya y R3 , Now,

a b c x c+ + = +2 6 6( ) ... (i)

Also,

a b c y a+ + = +2 6 1 2( ) ... (ii)From (i) and (ii), we get ( ) ( )x c y a+ = +6 1 2

+ − + =( ) ( )1 2 6 0y a x c

1 + 2y = 0 and x + 6 = 0 y = –1/2 , x = –6 + + =

a b c2 6 0

3. (b) : We know that diagonalsof a rhombus bisect each other. B D

A

C

O = − −OA OC OB OD

and =So, OA OB OC OD O

+ + = +

4. (b) : Given that AC = 3AB B divides AC internally in the ratio 1 : 2Let c be the position vector of C

A B C1

= + +

= −

b c a c b a1 21 2

3 2

5. (d) : Given that

a b c+ + 3 , − + −2 3 4

a b c and

a b c− +3 5 are coplanar. It is possible to express one of them as a linear combinaton of the other two.

Let us write

a b c x a b c y a b c+ + = − + − + − + 3 2 3 4 3 5( ) ( )where x, y are scalars.Now, comparing the coe�cients of

a b c, and from both the sides, we get –2x + y = 1, –4x + 5y = 3 and 3x – 3y = Solving �rst two equations, we get x = –1/3 and y = 1/3. �ese values of x and y satisfy the last equation as well.

− −

= = −3 1

33 1

32

6. (a) : Let O be the origin of reference.�en, OA a b c

= − + +2 3 5 OB a b c

= + +2 3 OC a c

= −7 = − = + + − − + +AB OB OA a b c a b c

( ) ( )2 3 2 3 5

= − −3 2

a b cAlso, AC OC OA a c a b c

= − = − − − + +( ) ( )7 2 3 5

= − −9 3 6

a b c

AC a b c

= − −3 3 2( ) =AC AB

3

�e above equation shows that the vectors AC AB

and have the same or parallel supports.But these vectors have a common initial point A. So, AC AB

and have the same support. A, B, C are collinear.

7. (c) : �e sides AB BC

and of a regular hexagon ABCDEF represents the vectors

a b and respectively.�en AC AB BC a b

= + = +

Now, AC CD AD

+ =

= −CD AD AC

= − +2

b a b( ) ( || )

AD BC

= −CD b a

Now in CDE, CE CD DE b a a b a

= + = − + − = −( ) 2

8. (c) : Here | |

b = + + =3 6 2 72 2 2

Now, any vector along a with magnitude | |

b = 7 is

7 7

1 2 22 2

2 2 2 =

+ ++ +

aa

i j k| |

( )^ ^ ^ = + +73

2 2( )^ ^ ^i j k

QMP(–3, 5, 2)

A(–2, 3, 1)

A

B

C

D

E

F

a b


9. (d) : Let

R i j k1 2 4 5= + −^ ^ ^

and

R i j k2 2 3= + +^ ^ ^

R = +

R R1 2

A B

CD

RR2

R1

= + − + + +2 4 5 2 3i j k i j k^ ^ ^ ^ ^ ^

= + −3 6 2i j k^ ^ ^

Unit vector along AC RR

i j k

= =+ −

+ +| |

^ ^ ^3 6 2

9 36 4

= + −1

73 6 2( )^ ^ ^i j k

10. (a) : Let the four points are

A i j k= + −2 3^ ^ ^, B i j k= + +^ ^ ^,2 3

C i j k= + −3 4 2^ ^ ^ and

D i j k= − +^ ^ ^ 6

Now, AB i j k i j k

= + + − + −( ) ( )^ ^ ^ ^ ^ ^2 3 2 3

= − − +AB i j k ^ ^ ^4 ,

AC i j k i j k

= + − − + −( ) ( )^ ^ ^ ^ ^ ^3 4 2 2 3

= + −AC i j k ^ ^ ^

and AD i j k i j k

= − + − + −( ) ( )^ ^ ^ ^ ^ ^ 6 2 3

= − − + +AD i j k ^ ^ ^( ) 3 7

Now, AB AC AD

, and will be coplanar, if

AD x AB y AC

= + [x, y are scalars]

− − + + = − − + + + −i j k x i j k y i j k^ ^ ^ ^ ^ ^ ^ ^ ^( ) ( ) ( ) 3 7 4

Comparing the coe�cients of i j k^ ^ ^, and , we get –x + y = –1, 4x – y = 7 and –x + y = –( + 3)Solving the �rst two equations, we get x = 2 and y = 1.�ese values of x and y satisfy the last equation as well. –2 + 1 = – – 3 = –211. (d) : Given that,

OA i j k OB i j k

= − + = − +7 4 7 6 10^ ^ ^ ^ ^ ^,

OC i j k OD i j k

= − − + = − +^ ^ ^ ^ ^ ^3 4 5 5 and

Now, | | ( ) ( ) ( )AB

= − + − + + −7 1 4 6 7 102 2 2

= + + =36 4 9 7

| | ( ) ( ) ( )BC

= + + − + + −1 1 6 3 10 42 2 2

= + + =4 9 36 7

| | ( ) ( ) ( )CD

= − − + − + + −1 5 3 1 4 52 2 2

= + + =36 4 1 41

| | ( ) ( ) ( )DA

= − + − + + −5 7 1 4 5 72 2 2

= + + =4 9 4 17which does not satisfy the conditions of square, rhombus and rectangle.

12. (d) : Let a i j k= + +^ ^ ^ ,

b i j k= − +2 2 2^ ^ ^

If q is the angle between the two vectors

a b and , then

cos

| || |q = = − +

a ba b

2 2 23 12

=

=23 2 3

13

=

−q cos 1 13

13. (d) : Given that, | |p q+ = 3

+ + =

p q p q2 2 2 3

Since

p q and are unit vectors

+ + = =1 1 2 3 12

p q p q

Now, ( ) ( )2 3 3

p q p q− + = + − −6 2 9 32 2

p p q q p q

= −

− = − = −6 7 1

23 3 7

212

14. (d) : Given that

= − = =3 5 3i k^ ^, | | and

= =

q| || | cos 3

= = + =10 5 3 3 1 102 2cos [ | | ]q

=cosq 350

Now, area of parallelogram =| |

= = − =| || | sin

q 50 1 950

41

15. (b) : Given that

a b c a b c+ + = + + =0 0| | + + + + + =| | | | | | ( )

a b c a b b c c a2 2 2 2 0

+ + = −

a b b c c a 32

(

a b c, , are unit vectors)

16. (c) : Since

a b a b =, 0

Now, ( ) ( )

a b a mb+ + = + + + =

a m a b b a m b2 20

( ( ) ( ))

a b a mb+ +

+ + + =| | ( ) | |

a m a b m b2 21 0

+ = = −| | | | | || |

a m b m a

b2 2

2

20


17. (d) : Let a b^ ând be two unit vectors.

We have, | |^ â b+ 2 1=

+ + =| | ( ) | |^ ^ ^ â a b b2 22 1 = −2 1a b^ ^

= − = −a b a b^ ^ ^ ^| || | cos12

12

q

= −

cos cosq p p3

=q p23

18. (d) : AC i k i j k j

= + − + = −( ) ( )^ ^ ^ ^ ^ ^

AB j k i j k i

= + − + = −( ) ( )^ ^ ^ ^ ^ ^

A i j( + )^ ^ B j k( + )^ ^

C i k( + )^ ^

= − − = =cosq q p1 0 02 2

12 3

19. (a) : Here, a i j k= − +^ ^ ^2 3

= + + =| |a 1 4 9 14

Also,

a b b a b = − =| | | |2 7and

− =| |

a b 2 7

− + =| | | |

a a b b2 22 7

− + =14 2 72 2| | | |

b b =| |

b 7

20. (a) : | | | | | | | |

a b c a b c+ + = + +2 2 2 2

+ + + 2( )

a b b c c a ... (i)

a b c a b c + + =( ) ( ) 0 ... (ii)

b c a b c a + + =( ) ( ) 0 ... (iii)

and

c a b c a b + + =( ) ( ) 0 ... (iv)

Adding (ii), (iii) and (iv), we get

2 0( )

a b b c c a + + = (i) becomes

| | | | | | | |

a b c a b c+ + = + +2 2 2 2

21. (a) : Here

= + + = − +2 2 2i j k i j k^ ^ ^ ^ ^ ^,

and

= − +3 4 2i j k^ ^ ^

+ = − +

5 3 3i j k^ ^ ^

Now, projection of ( )

+ on is

( )| |

( )( )

+ = + − − +

+ +=1 2 3 2 3

1 2 2

1732 2 2

22. (a) : Given that

a b c+ + = 0

+ = − + = −

a b c a b c| | | |2 2

+ + =| | | | | |

a b a b c2 2 22

9 + 25 + 2 · 3 · 5cosq = 49

= =30 15 12

cos cosq q

q= 60°

23. (a) : Here a i j k= + +^ ^ ^2 2

= + + =| |a 1 2 2 32 2 2

Required area = 12

12

| | | || | sin

a b a b = q

= =1

23 5

6154

sin p

24. (b) : Since

a b and are two unit vectors, = =| | | |

a b 1

Now,

a b a b = = =| || | cos cosq p1 13

12

| | | | | | ( )

a b a b a b+ = + + =2 2 2 2 3

+ = | |

a b 3 1

25. (d) : We have,

a b c =

a c b c and ... (i)Also,

c a b =

c b a b and ... (ii)

From (i) and (ii), we get

a c a b a b c and ||( )

26. (a) : Here, | | | |

a b=

− = + −| | | | | | | || | cos

a b a b a b2 2 2 q

= + −| | | | | | cos a a a2 2 22 q

= − =2 1 4

22 2 2a a( cos ) sinq q

− =| | sin

a b a22q

= −a a bsin | |q

212

27. (c) : Given that | | , | | , | |

a b c= = =2 3 4

= [ ] ( )

a b c a b c =

a b c n| || | sin ^23p


= = =| || || |sin [ | || | cos | |]^ ^

a b c a n a n a23

0p

= 2 3 4 32

=12 3

28. (c) : Given two vectors lie in xy-plane. �erefore, a vector coplanar with them is

a a i a j= +1 2

^ ^

Since a i j i j= + + +( ) ( )^ ^ ^ ^2 2 3

+ = + + +a i a j i j1 2 1 2 2 3^ ^ ^ ^( ) ( ) Comparing coe�cients of i j^ ^ and , we get a1 = 1 + 2, a2 = 2 + 3

= + + +a i j( ) ( )^ ^1 2 2 3

Now, a i j is perpendicular to ^ ^−

− =a i j( )^ ^ 0 + − + =( ) ( )1 2 2 3 0

–1 – = 0 = –1

= − −a i j^ ^.

= = − −

+=

+a a

ai j i j^^ ^ ^ ^

| |( )

1 1 2

29. (a) : Here, we have

a i k b x i j x k= − = + + −^ ^ ^ ^ ^, ( )1and c y i x j x y k= + + + −^ ^ ^( )1

Now, [ ]

a b c x xy x x y

=−−

+ −

1 0 11 1

1 Applying C3 → C3 + C1 , we get

[ ] [( ) ]

a b c xy x x

x x=+

= + − =1 0 0

1 11

1 1 1

�us, [ ]

a b c depends neither on x nor on y.

30. (b) : Given that | | | |

a b a b + =2 2 144 and | |a = 4

But | | | | | | | |

a b a b a b + =2 2 2 2

= 144 16 2| |

b

= =| | | |

b b2 9 3

31. (a) : Let

= − +4 3i j k^ ^ ^ ,

= − + −2 2i j k^ ^ ^ and

= + +x i y j z k^ ^ ^ be the required vector.

Given that | |

=12

x2 + y2 + z2 = 144 ... (i)Also given that

= + + − + =0 4 3 0( ) ( )^ ^ ^ ^ ^ ^x i y j z k i j k

4x – y + 3z = 0 ... (ii)

and

= + + − + − =0 2 2 0( ) ( )^ ^ ^ ^ ^ ^x i y j z k i j k

–2x + y – 2z = 0 ... (iii)From (ii) and (iii), we get

x y z

2 3 6 8 4 2−=− +

=−

i e x y z. ., (−

= = =1 2 2

say)

x = –, y = 2 and z = 2Putting these values in (i), we get

2 + 42 + 42 = 144

92 = 144 2 = 16 = 4 x = –4, y = z = 8 or x = 4, y = z = –8�erefore, the required vectors are

= − + +4 8 8i j k^ ^ ^ or

= − −4 8 8i j k^ ^ ^

32. (b) : By Lagrange’s identity, we have

( ) | | | | ( )

a b a b a b = − 2 2 2 2

= − | | | | | | ( )

a b a b a b2 2 2 2

= 16 – 4 = 12

Again, | | ( ) ( )

c a b b a b b2 2 3 2 3= − −

| | | | ( ) ( ) | |

c a b b a b b a b b2 2 24 6 6 9= − − +

= + = =4 9 02 2| | | | ( ) [ ] )

a b b b a b b a b

= (4 × 12) + (9 × 16) = 192

= =| |c 192 8 3

Now,

b c b a b b = −( )2 3

= − = − = −0 3 3 16 48

2b

Let the angle between

b c and beq


= = −

= −cos| || |

q

b cb c

484 8 3

32

=cos cosq p56

=q p56

33. (c) : Here p = area of the quadrilateral OABC = ar(BOA) + ar(COB)

= + 1

212

( ) ( )OA OB OB OC

= + 1

2| |OA OB OB OC

C B

AO a

b

= + + + 1

210 2 10 2| ( ) ( ) |

a a b a b b

= + + + 1

210 2 10 2| ( ) ( ) ( ) ( ) |

a a a b a b b b

= + + = =1

20 12 0 0| ( ) | [ ]

a b a a b b

= 6 | |

a b

p = 6q k = 6

34. (b) : We know, three vectors will be coplanar, if their scalar triple product will be zero.

Now, ( ) ( )^ ^ ^ ^ ^ ^

^ ^ ^

2 2 3 2 1 11 2 3

i j k i j ki j k

− + + − = −−

= + +i j k^ ^ ^7 5

+ + − + + −( ) {( ) ( )}^ ^ ^ ^ ^ ^ ^ ^ ^3 5 2 2 3i j k i j k i j k

= + + + +( ) ( )^ ^ ^ ^ ^ ^3 5 7 5i j k i j k

= 3 + 7 + 25 = 7 + 28 By the condition of coplanarity, we have 7 + 28 = 0 = –4

35. (b) : We have [ ]

+ + +

= + + +( ) {( ) ( )}

= + + + + ( ) ( )

= + + + =( ) ( ) [ ]

0

= + + +

( ) ( ) ( ) ( )

+ +

( ) ( )

= + + + + +[ ] [ ]

0 0 0 0

= + =[ ] [ ] [ ]

2

= 236. (a, b, c) : Let a x i y j z k= + +^ ^ ^

(a) ( ) ( ) ( )^ ^ ^ ^ ^ ^ a i i a j j a k k + +

[ ( ) ]^ ^ ^ ^ ^

a i x i y j z k i x = + + =

= + + =x i y j z k a^ ^ ^

(b) ( ) ( ) ( ) | |^ ^ ^ a i a j a k x y z a + + = + + =2 2 2 2 2 2 2

(c) i a i j a j k a k^ ^ ^ ^ ^ ^( ) ( ) ( ) + +

= − + − + − ( ) ( ) ( ) ( ) ( ) ( )^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^i i a i a i j j a j a j k k a k a kk^

= − + − + − a i a i a j a j a k a k( ) ( ) ( )^ ^ ^ ^ ^ ^

= − + + = − =3 3 2 a x i y j z k a a a( )^ ^ ^

37. (a, b, c) : Here,

a i j k b i j k= + + = + − ^ ^ ^ ^ ^ ^,2 and

c i j k= − +2 ^ ^ ^ are coplanar.

= −−

=

a b c( ) 01 2

1 12 1

0

(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 = –2or 2 – 2 – 2 = 0

= − −

= = 2 4 4 1 22 1

2 2 32

1 3( )

�e values of are –2, 1 3 3+ − and 1 .

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

a b b c− −

= − − +

a b a c b b b c

= + + = = −

a b b c c a b b a c c a[ ]0 and

− − = + + ( ) ( )

a b b c a b b c c a


Similarly, it can be shown that

( ) ( ) ) ( )

c b a c a b a c− − − − and (

each is equal to

a b b c c a + +

Now, 12

{ ( ) ( ) ( )}

a b c b c a c a b − + − + −

= − + − + − 1

2( )

a b a c b c b a c a c b

= + + − + + 1

2[( ) ( )]

a b b c c a a c b a c b

= + + 1

22 2 2[ ( ) ( ) ( )]

a b b c c a

= + +

a b b c c a

− + − + −12

{ ( ) ( ) ( )}

a b c b c a c a b

= + +

a b b c c a

39. (a, b, c) : [ ]

a b c

Let

A a B b C c= = = , ,

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

40. (a, d) : Let c x i y j z k= + +^ ^ ^

Given that | | | | | |

a b c= = = 2

x2 + y2 + z2 = 2 ... (i)

Also, cosq = = =

a b b c c a2 2 2 2 2 2

=+

=+1

2 2 2y z x y

x + y = 1, y + z = 1Subtracting, x – z = 0 x = z = 1 – yPutting the values of x and z in (i), we get (1 – y)2 + y2 + (1 – y)2 = 2 3y2 – 4y = 0 y(3y – 4) = 0 y = 0 or 4/3

When y = 0, x = z = 1 = +c i k^ ^

When

y x z= = = −43

13

,

= − + −c i j k13

4( )^ ^ ^

nn


INDEFINITE INTEGRALS

DEFINITIONLet f(x) be a function. Then the family of all its primitives (or antiderivatives) is called the inde�nite integral of f(x) and is denoted by ∫ f(x)dx.�e symbol f x dx( ) is read as the inde�nite integral of f(x) with respect to x.

�us, d

dxx C f x f x dx x C( ( ) ) ( ) ( ) ( ) , + = = +

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 di�erential of x.

G E O M E T R I C A L I N T E R P R E TAT I O N O F INDEFINITE INTEGRAL�e statement f x dx F x C y( ) ( ) = + = (say)represents a family of curves. �e di�erent values of C will correspond to di�erent members of this family. Hence, the inde�nite 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.

YOUR WAY CBSE XII

Series6

I , A I &NTEGRALS PPLICATION OF NTEGRALSD EIFFERENTIAL QUATIONS

SOME FUNDAMENTAL INTEGRATION FORMULAS

(i) x dxxn

C nnn

=+

+ −+1

11,

(ii) 1x

dx x C= + log | |

(iii) e dx e Cx x= +(iv) a dx

aa

Cxx

= + log(v) sin cosx dx x C = − +

(vi) cos sinx dx x C = +(vii) sec tan2 x dx x C = +(viii) cosec 2x dx x C = − +cot

(ix) sec tan secx x dx x C = +(x) cosec cosec x x dx x Ccot = − +(xi)

12 2

1

a xdx

xa

C−

=

+ −sin

(xii) −−

=

+ −1

2 21

a xdx x

aCcos

(xiii) 1 12 2

1

a xdx

axa

C+

=

+ −tan

(xiv) −+

=

+ −1 1

2 21

a xdx

axa

Ccot

(xv) 1 12 2

1

x x adx

axa

C−

=

+ − sec

(xvi) −−

=

+ −1 1

2 2

1

x x adx

axa

Ccosec


SOME PROPERTIES OF INDEFINITE INTEGRAL(a) The process of differentiation and integration

are inverse of each other i.e., ddx

f x dx f x( ) ( )=

and = + f x dx f x C( ) ( ) , where C is any arbitrary constant.

(b) Two inde�nite 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

that d

dxf x dx d

dxg x dx( ) ( ) ,=

then f x dx( )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.

(d) �e integral of the sum or the di�erence of two functions is equal to the sum or di�erence of their integrals i.e.,

[f1(x) ± f2(x)]dx = f1(x)dx ± f2(x)dx Similarly, [ f1(x) ± f2(x) ± ... ± fn(x)]dx = f1(x)dx ± f2(x)dx

± ... ± fn(x)dx

METHODS OF INTEGRATION(a) Integration by substitution(b) Integration using partial fractions(c) Integration by parts(a) Integration by SubstitutionBy 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.

Some Important SubstitutionsIntegrals Substitutions

(i) f ax b dx( )+ Put ax + b = t

(ii) x f x dxn n− 1 ( ) Put xn = t

(iii) f x f x dxn( ) ( ) or

f xf x

dx

)

( )Put f(x) = t

(iv)dx

ax bx cdx

ax bx c2 2+ + + +

or Transformed into standard form by expressing

ax bx c a xba

ca

ba

22 2

22 4+ + = +

+ −

then put

xba

t+ =2

(v) ( )ax b cx d dx+ + or ax bcx d

dx++ Put ax + b = A(cx + d) + B

(vi)dx e

ax bx cdx

++ + 2

or dx e

ax bx c

dx+

+ +

2

or ( )dx e ax bx c dx+ + + 2

Put ( ) ( )dx e Ad

dxax bx c B+ = + + +2

(vii) a x b xc x d x

dxsin cossin cos

++ Put a sin x + b cos x = + +

+Ad

dxc x d x

B c x d x

( s i n c o s )

( s i n c o s )

(b) Integration by Partial FractionsIf 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 xq x

dx( )( )

by decomposing

p xq x

( )( )

into partial fractions.

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:


Factor in the denominator

Corresponding partial fraction

(i) (x – a)(x – b) Ax a

Bx b−

+−

(ii) (x – b)2A

x bB

x b( ) ( )−+

− 2

(iii) (x – c)3 Ax c

Bx c

Cx c( ) ( ) ( )−

+−

+−2 3

Ax c

Bx c

Cx c( ) ( ) ( )−

+−

+−2 3

(iv) (x – a)2(x – b)A

x aB

x aC

x b−+

−+

−( )2

(v) (ax2 + bx + c)Ax B

ax bx c+

+ +( )2

(vi) (x – a)(bx2 + cx + d)where bx2 + cx + d cannot be factorised further

Ax a

Bx Cbx cx d−

+ ++ +2

(c) Integration by PartsThe 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

uvdx u vdx dudx

vdx dx= − ( )

In words, integral of the product of two functions = �rst function × integral of the second – integral of (di�erential coe�cient of the �rst function × integral of the second function).Note : We choose the �rst function as the function which comes �rst in the word ILATE, whereI–stands for the inverse trigonometric function (sin–1x, cos–1x, tan–1x etc.)L – stands for the logarithmic functionsA – stands for the algebraic functionsT – stands for the trigonometric functionsE – stands for the exponential functionsIf there is no other function, then unity is taken as the second function.

Some Particular Integrals

(i) cot log |sin | log cosx dx x C x C ec= + = − +(ii) tan log |cos | log |sec |x dx x C x C= − + = +(iii) sec log |sec tan |x dx x x C = + +

(iv) cosec cosec x dx x x C= − + log | cot |

(v)

dx

x a a

x a

x aC

2 2

1

2−= −

++ l o g

(vi)

dx

a x a

a x

a xC

2 2

1

2−= +

−+ l o g

(vii)

dx

x a

x x a C2 2

2 2

+= + + + l o g

(viii) dx

x a

x x a C2 2

2 2

−= + − + l o g

(ix) a x dxxa x

ax a x C

2 2 2 2

22 2

2

2

+ = +

+ + + +

l o g

(x) x a dxxx a

ax x a C

2 2 2 2

22 2

2

2

− = −

− + − +

l o g

(xi) a x dxxa x

a x

aC

2 2 2 22

1

2 2− = − +

+ −s i n

Special Integrals(i) ex[f(x) + f (x)]dx = exf(x) + C(ii) ekx[kf(x) + f (x)]dx = ekxf(x) + C

DEFINITE INTEGRALS

DEFINITIONLet F(x) be antiderivative of f(x), then for any two values of the independent variable x, say a and b, the di�erence F(b) – F(a) is called the de�nite integral of

f(x) from a to b and is denoted by f x dxa

b( ) . �us

f x dx F b F a

a

b( ) ( ) ( ) = −

where numbers a and b are called the limits of integration; a is the lower limit and b is the upper limit.

DEFINITE INTEGRAL AS THE LIMIT OF A SUMLet f be a continuous function defined in a closed

interval [a, b], then the de�nite integral f x dxa

b( ) is the

area bounded by the curve y = f(x), the ordinates x = a, x = b and the x-axis.


Also,

f x dx h f a f a h f a n ha

b

h( ) lim ( ) ( ) ... { ( ) } = + + + + + −

→01

where, h b an

h= − →, 0 as n →

FUNDAMENTAL THEOREM OF INTEGRAL CALCULUS(i) Area function : �e function A(x) denotes the

area function and is given by A(x) = f x dxa

x( ) .

(ii) First Fundamental Theorem : Let f (x) be 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 de�ned in the closed interval [a, b] and f (x) dx = F(x), then

f x dx F x F b F aa

b

a

b( ) [ ( )] ( ) ( ).= = −

PROPERTIES OF DEFINITE INTEGRALS

(i)

f x dx f y dy

a

b

a

b( ) ( ) =

(ii)

f x dx f x dxa

b

b

a( ) ( ) = −

(iii)

f x dxa

a( ) = 0

(iv)

f x dx f x dx f x dx a c ba

b

a

c

c

b( ) ( ) ( ) , = +

(v)

f x dx f a x dxa a

( ) ( )0 0 = −

(vi)

f x dx f a b x dxa

b

a

b( ) ( ) = + −

(vii) f x dx

f x

f x dx f xa

aa( )

, ( )

( ) , (− =

0

20

if is an odd function

if )) is an even function

(viii) f x dx

f a x f x

f x dx f a x f x

aa( )

, ( ) ( )

( ) , ( ) ( )0

2

0

0 2

2 2 =

− = −

− =

if

if

APPLICATION OF INTEGRALS

1. Area of a curve between two ordinates : Let y = f(x) be a continuous and �nite function in [a, b].

Case I : If the curve y = f(x) lies above the x-axis, then the area bounded b y t h e c u r v e y = f(x), x-axis and the ordinates x = a and x = b is given

by y dxa

b

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

( )− y dx

a

b

2. Area of a curve between two abscissas : C a s e I : I f t h e

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

x dy

c

d

Case II : If the

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

( )− x dy

c

d


Note : If some portion of the curve is above the x-axis and some is below the x-axis as shown in the �gure, 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.

ab

Y

X

3. Area between two curves : Case I : The area

O a b

y = f x( )

y = g x( )

Y

Xx a= x b=

Y

X

bounded by two

curves y = f(x) and y = g ( x ) , w h i ch intersects at the ordinates x = a and x = b, is given by

f x g x dx

a

b( ) ( )−

Case II : The area

O

y = c

y = dY

X

d

c

Y

X

xgy

= xfy

=

b o u n d e d b y t w o c u r v e s x = f ( y ) and x = g(y) which i nt e r s e c t s a t t h e abscissas, y = c and y = d is given by

| ( ) ( ) |f y g y dy

c

d−

Note : 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,

[ ( ) ( )] [ ( ) ( )]f x g x dx g x f x dxe

d

c

e− + −

y f x=

Y

y g x=

deO XX

Y

c

y x= g y x= f

DIFFERENTIAL EQUATION

DEFINITIONAn equation involving an independent variable, dependent variable and the derivatives of the dependent variable, is called a di�erential equation.

A differential equation involving derivatives

of the dependent variable with respect to only one independent variable is called an ordinary di�erential equation and a di�erential equation involving derivatives with respect to more than one independent variables is called a partial di�erential equation.

ORDER AND DEGREE OF A DIFFERENTIAL EQUATION�e order of highest order derivative appearing in a di�erential equation is called the order of the di�erential equation. �e power of the highest order derivative appearing in a di�erential equation, a�er it is made free from radicals and fraction, is called the degree of the di�erential equation.Note :

�e degree of a di�erential equation which is not a

polynomial equation in derivatives is not de�ned.Order and degree (if defined) of a differential

equation are always positive integers.

SOLUTION OF A DIFFERENTIAL EQUATIONAny relation between the dependent and independent variables (not involving the derivatives) which, satisfy the given di�erential equation is called a solution of the di�erential equation.General Solution : �e solution which contains arbitrary constants is called the general solution (primitive) of the di�erential 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 GIVENSuppose an equation of a family of curves contains n arbitrary constants (called parameters).Then, we obtain its differential equation, as given below.Step I : Di�erentiate 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(i) If the equation is

dydx

f x= ( ) , then y = f(x)dx + C 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

is

dydx

f ax by c= + +( ), then put ax + by + c = z.

(iv) Homogeneous equation : If a first order, first degree di�erential equation is expressible in the form

dydx

f x yg 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

(v) Linear Equation : If the equation is dydx

Py Q+ = , where P and Q are functions of x, then

y e Q e dx CPdx Pdx = + ,

where e Pdx is the integrating factor (I.F.).

OR

If the equation is dxdy

Px Q+ = , where P and Q are

functions of y, then

x e Q e dy CPdy Pdy = + ,

where e Pdy is the integrating factor.

VERY SHORT ANSWER TYPE

1. Evaluate : 2(x+3)dx

2. Evaluate : x x dxcos/

0

2p

3. Find the di�erential 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 di�erential

equation dydx

dydx

+

=sin 0

5. Solve : dydx

ex y= +

6. Evaluate : cos3x sin x dx

SHORT ANSWER TYPE7. For what values of c and a is the following equation

satis�ed?

(sin cos ) sin( )2 2 12

2x x dx x c a+ = − +

8. Evaluate : log

(log )

x

xdx

−

12

9. If the area enclosed between the curves y = ax2 and x = ay2 (a > 0) is 1 square unit, then �nd 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), �nd its equation.

11. Evaluate : sin( sin cos )

/ 2

0

2

1x

x xdx

+p

12. Solve : xdydx

y x− = log

LONG ANSWER TYPE

13. Evaluate :

x

x xdx

8 2+ −

14. Evaluate

( )x x dx2

1

3+ as limit of sum.

15. Evaluate : x

x xdx

4

21 1( )( )− +16. Solve :

xyx

y dx x dy yyx

x dy y dxcos ( ) sin ( )

+ =

−

17. Find the area of the region {(x, y) : x2 + y2 2ax, y2 ax, x 0, y 0}.

18. Evaluate the following integrals :

(i) x

xdx

2

2

1

4

+

+

(ii)

xx x

dx2

4 29

2 81+

− +

SOLUTIONS

1. 2(x+3)dx = 2x 23dx = 82xdx

= 8 22

+x

Clog = +

+22

3( )

log

xC

2. x x dx x x x dxcos [ sin ] sin/

//

0

2

02

0

21

pp

p

= −

[Integrating by parts]

= + = −p pp

2 210

2[cos ] /x


3. Given equation is y = Aex + Be–x ...(i)

Now, dydx

Ae Bex x= − −

d y

dxAe Be yx x

2

2 = + =− [From (i)]

d y

dxy

2

2 0− = is required di�erential equation.

4. �e highest order derivative present in the

di�erential equation is dydx

. So, it is of order 1.

Also, L.H.S. of the di�erential equation cannot be

expressed as a polynomial in dydx

. So, its degree is not de�ned.

5. We have,

dydx

ex y= +

dydx

e ex y= e–y dy = ex dx

e–y dy = ex dx –e–y = ex + C, which is the required solution.

6. We have, cos sin3 x x dx Put cos x = t sin x dx = –dt

cos sin3 34

4x x dx t dt t C = − = − +

= − +14

4cos x C

7. (sin cos ) cos sin2 2 22

22

x x dx x x C+ = − + + ,

where C is an arbitrary constant.

= − +1

22 2(sin cos )x x C

= −

+1

212

2 12

2sin cosx x C

= −

+

12 4

24

2cos sin sin cosp px x C

= −

+

12

24

sin x Cp

But (sin cos ) sin( )2 2 12

2x x dx x c a+ = − + −

+ = − +1

22

412

2sin sin( )x C x c ap

c = p4

and a = C an arbitrary constant.

8. Let Ix

xdx=

−

log

(log )

12

Put log x = t x = et dx = et dt

I tt

e dt et t

dtt t= − = −

1 1 12 2

= − 1 1t

e dtt

e dtt t

Integrating �rst integral by parts, we get

It

et

e dtt

e dtt t t= − − − 1 1 1

2 2

= + = +1t

e C xx

Ctlog

9. Solving, y = ax2 and x = ay2, we get point of

intersection O(0, 0) and Aa a1 1,

.

Y

Y

X XO

y ax= 2Q x,y( )2

P x,y( )1

(0, 0)

A1a

1a

,

Since, area enclosed between the curves = 1

xa

ax dxa

−

= 2

0

11

/

23 3

13 2 3

0

1

ax a x

a/

/−

=

2

31

31 13 2 3a a

aa

− =/ 2

3 312 2a a

− =

1

312a

= a2 13

= a = 13 [ a > 0]

10. �e equation of the normal to a curve at a point (x, y) is

( ) ( )Y ydydx

X x− + − = 0

Since it passes through the point (3, 0), we have

( ) ( ) ( )0 3 0 3− + − = = −ydydx

x ydydx

x

= − y dy x dx( )3 = − +y

x x C2 2

23

2 x2 + y2 – 6x – 2C = 0,Which is the equation of the curve.


I F e e e xx

xdx x x. . log log( )=

= = = = − − −−

111

So, the required solution is y × I .F. = {Q × (I.F.)}dx + C

yx

xx x

dx C =

+

1 1log

= +yx

xx

dx C(log ) 12

= − − −

+

yx

xx x x

dx C(log ) 1 1 1

[Integrating by parts]

= − + +yx

xx x

dx Clog 1

2=−

− +log xx x

C1

y = Cx – (log x + 1)

13. Let I x

x xdx=

+ −

8 2

Let x = A(1 – 2x) + B or x = –2Ax + (B + A)Equating the coe�cient of x and constant term, we get

–2A = 1 A = – 12

and 0 = B + A B = –A = 12

x x= − − +12

1 2 12

( )

Ix

x xdx=

− − +

+ −

12

1 2 12

8 2

( )

= − −

+ −+

+ −

12

1 2

8

12 82 2

x

x xdx dx

x x

Let

I I I= − +12

121 2 ,

where

I x

x xI dx

x x1 2 2 2

1 2

8 8= −

+ −=

+ − and

Now, I x

x xdx1 2

1 2

8= −

+ −

Put 8 + x – x2 = t (1 – 2x)dx = dt

I dt

tt dt t x x1 1 2

1 2 22 2 8= = = = + − −/

/

Also, I dx

x x

dx

x x2 2 28 8=

+ −=

− −

( )

Since the curve passes through (3, 4), we have 9 + 16 – 18 – 2C = 0 C = 7/2 x2 + y2 – 6x – 7 = 0 is the required equation of the curve.

11. Let I xx x

dx=+

sin( sin cos )

/ 2

0

2

1

p

...(i)

�en, I xx x

dx= −+ − −

sin [( ) ]sin[( ) ]cos[( ) ]

/ 2

0

2 21 2 2

pp p

p // /

f x dx f a x dxa a

( ) ( )0 0 = −

or I xx x

dx=+

cos( sin cos )

/ 2

0

2

1

p

...(ii)

Adding (i) and (ii), we get

21 1

2 2

0

2

0

2I x x

x xdx dx

x x= +

+=

+ (sin cos )( sin cos ) ( sin cos )

/ /p p

=

+sec

(sec tan )

/ 2

20

2 xx x

dxp

(By dividing numerator and denominator by cos2x)

=

+ +=

+ + sec

( tan tan ) ( ),

/ 2

20

2

201 1

xx x

dx dtt t

p

[Put tanx = t sec2x dx = dt,

When x = 0 t = 0 and x = p2 t → ]

=

+

+

= +

−

dt

t

t

12

32

23

2 132 2

0

1

0tan

= −

= −

=

− −23

13

23 2 6

23 3

1 1tan ( ) tan

p p p

12. �e given equation can be written asdydx x

yx

x− =1 log

�is is of the linear form dydx

Py Q+ = , where

Px

= −1 and

Q

xx

=log .


=− − +

+

=

− −

dx

x x

dx

x8 2 12

14

14

334

12

2 2

=

− −

= −

−dx

x

x

332

12

1 233 22 2

1sin //

= −

−sin 1 2 133

x

I x x x C= − + − + −

+−1

22 8 1

22 1

332 1sin

= − + − + −

+−8 1

22 1

332 1x x x Csin

14. Let

I x x dx= + ( )2

1

3

By de�nition, we know

f x dx h f a f a h f a h

f a n ha

b

h( ) lim [ ( ) ( ) ( )

... ( ( ) )],

= + + + +

+ + + −→0

2

1

where h b an

= −

Here, a = 1, b = 3, f(x) = x2 + x and hn n

= − =3 1 2

I h f f h f h f h

f n hh

= + + + + + +

+ + + −→

lim [ ( ) ( ) ( ) ( )

... ( ( ) )]0

1 1 1 2 1 3

1 1

I h h h

h hh

= + + + + +

+ + + + + + +→

lim [{ } {( ) ( )}

{( ) ( )} ... {( (0

2 2

2

1 1 1 1

1 2 1 2 1 nn hn h−

+ + −1

1 1

2) )( ( ) )}]

I h h h

n h hh

= + + + + +

+ + − + + + +→

lim [{ ( ) ( ) ...

... ( ( ) ) } { ( )0

2 2 2

2

1 1 1 2

1 1 1 1 (( )... ( ( ) )}]

1 21 1

++ + + −

hn h

I h n h n

h nn

h= + + + + + −

+ + + + −+ +

→lim [{ ( ... ( ))

( ... ( ) )}{

02 2 2 2

2 1 2 3 1

1 2 1hh n( ... ( ))}]1 2 3 1+ + + + −

I h n h n n

h n n n n h

h= + −

+ − −

+ +

→lim ( )

( )( )

0

2

2 12

1 2 16

nn n( )−

12

I h n h n n

h n n nh

= + −

+ − −

→lim ( )

( )( )0

2

2 3 12

1 2 16

In

nn

n n

nn n n

n= + −

+ − −

→lim ( )

( )( )

2 2 6 12

4 1 2 162

h

n=

2

I nn

n nnn

= + −

+

− −

→

lim ( )( )4 6 1 43

1 2 12

= + −

+ −

−

→

limn n n n

4 6 1 1 43

1 1 2 1

I = + − + − − = + + =4 6 1 0 43

1 0 2 0 4 63

383

( ) ( )( )

15. We have,

xx x

dx4

21 1( )( )− +

xx x

xx x x

4

2

4

3 21 1 1( )( ) ( )− +=

− + −

= + +

− + −( )

( )x

x x x1 1

13 2

x

x xx

x x

4

2 21 11 1

1 1( )( )( )

( )( )− += + +

− + ...(i)

Let 1

1 1 1 12 2( )( ) ( ) ( )x xA

xBx Cx− +

=−

+ ++

�en, 1 = A(x2 + 1) + (Bx + C)(x – 1) ...(ii)

Putting x = 1 in (ii), we get A = 12

Comparing coe�cients of x2 on both sides of (ii), we get

A + B = 0 B = –A = −12

Comparing the constant terms on both sides of (ii), we get

A – C = 1 C A= − = −1 12

11 1

12 1

12

12

12 2( )( ) ( ) ( )x x x

x

x− +=

−+− −

+ …(iii)

xx x

xx

xx

4

2 21 11 1

2 112

11( )( )

( )( )

( )( )− +

= + +−

− ++

[By (i) & (iii)]


x

x xdx x dx dx

xx

xdx dx

x

4

2

2 2

1 11 1

2 114

21

12

( )( )( )

( )

( ) (

− += + +

−

− +

−

++ 1)

= + + − − + − +−x x x x x C2

2 12

12

1 14

1 12

log log( ) tan

16. �e given equation may be written as

xyx

yyx

y yyx

xyx

xdydx

cos sin sin cos+

− −

= 0

dydx

x y x y y x yy y x x y x x

=+−

{ cos( ) sin( )}{ sin( ) cos( )}

/ // /

dydx

y x y x y x y xy x y x y x

=+

−{cos( ) ( )sin( )}( )

{( )sin( ) cos( )}/ / / /

/ / /,,

which is clearly homogeneous, being a function of (y/x).

Putting, y = vx and dydx

v x dvdx

= +

in it, we get

v x dvdx

v v v vv v v

+ = +−

(cos sin )( sin cos )

x dvdx

v vv v v

=−

2 cos( sin cos )

( sin cos )cos

v v vv v

dvx

dx− =

tanv dv dvv x

dx− = 2

–log|cos v| – log |v| + log C = 2log |x| log |cos v| + log |v| + 2log |x| = log C log |x2 v cos v| = log C |x2 v cos v| = C x2 v cos v = ± C = C1 (say)

xyyx

Ccos = 1 , which is the required solution.

17. �e equations of the given curves are x2 + y2 = 2ax ... (i) and y2 = ax ... (ii) Now, clearly x2 + y2 = 2ax is a circle with its centre B(a, 0) and radius = a units.Since y2 = ax is a parabola with O(0, 0) as its vertex and the x-axis as its axis, thensolving (i) and (ii), we get the points of intersection O(0, 0) and A(a, a).

Required area = − − 2 2

00

ax x dx ax dxaa

= − − − a x a dx a x dxaa

2 2

00

( )

= − − − + −

−

−( ) ( ) sin

/

x a a x a a x aa

a x

a2 2 2

1

0

3 2

0

2 2

23

aa

= − − −

− −a a a2

12

1 22

02

1 23

sin ( ) sin ( )

= −

pa a2

24

23

sq. units

18. (i) x

xdx x

xdx

2

2

2

2

1

4

4 3

4

+

+= + −

+

= +

+−

+

x

xdx dx

x

2

2 2 2

4

43

2

= + −+

x dx dx

x2 2

2 22 3

2

= + + + +

− + + +

x x x x

x x C2

2 22

2

3 4

2 22

2 2

2

log

log

= + + + +

− + + +

x x x x

x x C2

4 2 4

3 4

2 2

2

log

log

= + − + + +x x x x C2

4 42 2log

(ii) Let I x

x xdx= +

− +2

4 29

2 81

=+

− +=

+

+ −

1 9

2 81

1 9

81 2

2

22

2

22

x

xx

dx x

xx

dx

=+

−

+ −

1 9

9 18 2

2

2x

xx

dx

Put xx

t− =9 1 2+

=x

dx dt

I dtt

t C=+

=

+ −

21

1614 4

tan

=

−

+ = −

+

− −14

9

414

94

1 12

tan tanx

x C xx

C

nn


1. If [.] stands for the greatest integer function,

31

2x dx is equal to

(a) 3 (b) 4 (c) 5 (d) 6

2. If du

eu

x

−=

1 6p ,

ln 2 then the value of x is

(a) 4 (b) ln 8(c) ln 4 (d) None of these3. �e equation of a curve is y = f(x). �e tangents at

(1, f(1)), (2, f(2)) and (3, f(3)) make angles p p6 3

, and p4

respectively with positive direction of x-axis, then the

value of f x f x dx f x dx ( ) ( ) ( )2

3

1

3 + is equal to

(a) − 13

(b) 13

(c) 0 (d) None of these4. Let f (x) be a function de�ned by

f x t t t dtx

( ) ( ) ,= − + 2

13 2 1 ≤ x ≤ 3.

�en the range of f(x) is

(a) [0, 2] (b) −

14

4,

(c) −

14

2, (d) None of these

5. All the values of 'a' for which

( ( ) )a a x x2 3

1

24 4 4+ − + dx 12 are given by

(a) a = 3 (b) a ≤ 4(c) 0 a 3 (d) None of these

6. Evaluate ( )sin( )cos ( )sin

12 1 22 2

++ − +

x x

x x x x xdx

7. Show that the value of sin

sin,

n x

xdx

+

12

20

p n N, is

independent of n. Find the value of the integral.8. Find the orthogonal trajectories of the family of circles having their centres on the y-axis and touching the x-axis.9. Find the area bounded by x = cos–1y, x-axis and the lines |x| = 1.10. Show that the solution of the di�erential equation yp2–p = (p + 2) (py – 1), p dy/dx, represents two families,one of which is the family of curves having subnormal unity and the other is the family of parallel lines.

SOLUTIONS

1. (b) : [ ]31

2x dx = [ ] [ ] [ ]

/

/

/

/3 3 3

1

4 3

4 3

5 3

5 3

2x dx x dx x dx + +

= + + = 3 4 5 41

4 3

4 3

5 3

5 3

2dx dx dx

/

/

/

/

2. (c) : Put eu – 1 = t2 eudu = 2t dt

=+

=

−− −

p6

21

22

1

11

1

1tdtt t

te e

x x

( )tan

= − −−2 12

1tan ex p

− = − = =−tan tan1 13

13

3e ex xp p

=ex 4 x = ln 4

3. (a) : Here f f ( ) tan , ( ) tan ,16

13

23

3= = = =p p

f ( ) tan34

1= =p . Now, f x f x dx f x ( ) ( ) { ( )}2

32

2

312 =

Math Archives, as the t it le itself suggests, is a col lection of various challenging problems related to the topics of JEE (Main & Advanced) Syllabus. This section is basically aimed at providing an extra insight and knowledge to the candidates preparing for JEE (Main & Advanced). In every issue of MT, challenging problems are offered with detailed solution. The readers' comments and suggestions regarding the problems and solutions offered are always welcome.

thrchives

M10 Best Problems10 Best Problems

Prof. Shyam Bhushan*10 Best Problems

B y : Prof. Shyam Bhushan, D i r ec t o r , N ar ay an a I I T A c ad em y , J am s h ed p u r . M o b . : 09334870021


= −

12

3 22 2{ ( )} { ( )}f f

and f x dx f x f f ( ) [ ( )] ( ) ( )1

3

13 3 1 = = −

Value = −

+ − = −1

21 3 1 1

313

2 2( )

4. (c) : f (x) = x(x2– 3x + 2) = x(x – 1) (x – 2)�e sign scheme for f (x) is as below

+0 1 2

+– –

f '(x) 0 in 1 x 2 and f '(x) 0 in2 ≤ x 3. f(x) is decreasing in [1, 2] and increasing in [2, 3].

min f(x) = f(2) = x x x dx( )2

1

23 2− +

= − +

= −x x x

43 2

1

2

414

max. f(x) = the greatest among {f(1), f(3)}.

f x x x dx( ) ( ) ,1 3 2 02

1

1= − + = f x x x dx( ) ( )3 3 2 22

1

3= − + =

max f(x) = 2, so the range = −

14

2,

5. (a) : a a x x dx2 3

1

24 4 4+ − + ( )

= + − +a x a x x212 2

12 4

122 2[ ] ( )[ ] [ ] = a2 + (2 – 2a)(3) + 15

Given, a2 – 6a + 21 ≤ 12 a2 – 6a + 9 ≤ 0 (a – 3)2 ≤ 0

(a – 3)2 = 0 a = 36. Now ( )sin sin ( )cos1 1+ = − + + x x dx x x x C

I x xx x x

dx= +− + −

( )sin

[sin ( )cos ]11 12

= − + −− + +

+12

1 11 1

ln sin ( )cossin ( )cos

x x xx x x

C

7. Let In x

xdxn =

+

sin

sin

12

20

p

�en I In x n x

xdxn n+ − =

+

− +

1

0

32

12

2

sin sin

sin

p

= + =+

+ =2 1 21

1 00

0cos( ) .[sin( ) ]n x dxn

n xp

p

Hence I0 = I1 = I2 = I3 = ……

�us In = I0 = dx0

pp =

8. �e family of circles is given byx2 + y2 – ay = 0, a R …(1)Di�erentiating w.r.t. x, we have

2 2 0 2x y dydx

a dydx

a x dxdy

y+ − = = +

Putting this value in (1), we have

x y x dxdy

y y x y xy dxdy

2 2 2 22 0 2 0+ − +

= − − =

For the orthogonal trajectories, replace dydx

by −

dxdy

,

to get x y xy dydx

dydx

y xxy

2 22 2

2 02

− + = = −

�is is a homogeneous di�erential equation in x and y. Putting y = vx, we have

v x dvdx

v x xx vx

vv

+ = − = −2 2 2 2

21

2

x dvdx

v vv

vv

= − − = − +2 2 21 22

12

dxx

vv

dv= −+2

1 2

log |x| + log |1 + v2| = constant |x| (1 + y2/x2) = constant = b (say)Which is required orthogonal trajectories.9. x = cos–1y y = cos x, x [0, p]

x y= cos–1

1

1

y

xO

x = 1

The required area (shaded portion) is shown in the adjacent �gure.

Required area = cos |sin | sinx dx x0

1

01 1 = =

cos |sin | sinx dx x0

1

01 1 = = sq. units.

10. �e given di�erential equation can be re-written as

dydx

y dydx

dydx

y dydx

−

= +

−

1 2 1

dydx

dydx

y dydx

= + − =2 1 0or dxdy

y dydx

= =0 1or

Now dxdy

x c= =0 , a family of parallel lines (lines

parallel to y-axis).

y dydx

= 1 the solution curve has subnormal unity.

�e curves are y2 = 2x + k, k is a constant.nn


Despite the many engineering colleges that India boasts of, talented engineers who

can really build things are a scarce commodity. But a five-year dual-degree course introduced by IIT-Madras is slowly changing and challenging the status quo. Called Engineering Design, the course aims to ingrain basic engineering concepts much beyond the duration of the course. As a result of this decade-long experiment, the graduates are making successful forays into difficult areas like biomedicine, automotives and medical technology. “We wanted to bring engineering close to practice. Traditionally , there is a lot more theory than domain knowledge, that’s not always good,” said R Krishna Kumar, professor, department of engineering design at IIT-Madras.Kumar said over the past few years, he has seen several graduates in a batch opting to set up their own ventures. For instance, there is Tarun Mehta and Swapnil Jain of Ather Energy, who took a leap to build India’s first smart electric scooter. Alumni Chinmay Deodhar is building minimally invasive surgical tools, while Kartik Mehta is building the country’s first sanitary napkin making machine. For a country that has seen a large number of startups in the software space, IIT-Madras is filling a crucial gap by training students with skill sets required to build hardware products. But more importantly, the course is allowing engineers to think beyond the obvious, potentially serving as a blueprint for other colleges.“By the time we started Ather, we

just knew where to look for what. We had built vehicles several times over from scratch,” said Tarun Mehta, CEO of Ather, whose firm raised ` 75 crore from Tiger Global recently. When the department opened shop for applications in 2005, Chinmay Deodhar’s well wishers advised him against opting for it.“But the curriculum design was too good to refuse,” said the 28-year old Deodhar. And for a valid reason: besides concepts of mechanical and electronics engineering, the course included a smattering of a foreign language and legal know-how. More importantly, they were also taught to appreciate various forms of art: from clay modeling, to studying prehistoric, Egyptian and Islamic art, among others. The decision probably served Deodhar well, as he now heads Croleon Innovation Labs, which is building tools to perform minimally invasive surgeries. The three-year-old firm’s latest product is expected to reduce the incision size in surgeries to half a millimeter from 5 mm. While several IITs in India have a design course in their curriculum, few colleges have set up a separate department, IIT-Guwahati being one of them.Since students are required to come up with a project every year for

five years, there is a very large component of hands on learning.Balaji Teegala re-called his course on Vehicle Dynamics,

which he expected would have a lot of mathematical modeling to deal within the four walls of the classroom.Instead, the professor put a car in front of them, attached with simulators, allowing them to experiment with different controls to understand what effect it had on dynamics.

“Most of the work revolved around thinking about and coming up with new innovative things. As a student, I always thought, what is it that will create a wow-factor?” said Teegala, who is developing an electronic labour monitoring tool for expecting mothers as cofounder of Brun Healthcare.Buoyed by the success, Krishna Kumar has decided to flip around the entire model of the course: Practicals first, to spark curiosity , followed by theory. “Instead of teaching them machine tools, we allow them to play and experiment with it. So, the next time they read, they will not lose interest,” said Krishna Kumar. The course has also been able to arrest the brain drain to a certain extent as well. “The best part is, almost everyone of us have stayed back in India. Very few have gone out of the country to study or work,” said Deodhar.

Courtesy : The Economic Times

Engineering a new class

Designed to be bigCompanies that have come out of IIT-Madras’ Engineering Design Course

ATHER ENERGY- Makes smart electric bike

BRUN HEALTHCARE- Makes labour monitoring tool for expecting mothers

EMBRYYO TECHNOLOGIES - Makes medical devices

CROLEON INNOVATION LABS- Makes minimally invasive surgical tools

SARAL DESIGNS- Makes manufacturing machines for sanitary pads

number of startups in the software space,

up with a project every year for five years,very large component of hands on learning.Balaji Teegala re-called his course on Vehicle Dynamics,

which he expected

Instead of teaching them machine tools,

we allow them to play and experiment

with it. So, The next time they read, they

will not lose interest.R Krishna Kumar , Professor,

Dept. of Engineering Design, IIT-Madras

mathematicstoday nov 2015.pdf

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