I P E (QUESTIONS) - IIT JEE Mathematicskishoremaths.in/.../2016/02/Important-Questions_Part_2.pdfFind the area of the triangle formed by the coordinate axes and the line 2x – 4y

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MATHEMATICS–II

2 MARKS

1. Find the equation to the locus of a point P from which the distance to (2, 0) is equal to the distance from P to the y–axis.

Ans. Let A (2, 0). Let P(x, y) be a point in the locus. The distance from P to y–axis is |x|

Given condition is PA =|x| PA2 = x

2 (x – 2)

2 + (y – 0)

2 = x

2

x2 – 4x + 4 + y

2 = x

2 y

2 – 4x + 4 = 0.

2. If the distances from P to the points (5, –4), (7, 6) are in the ratio 2 : 3, then find the locus of P. Ans. Let P = (x, y) and A(5, –4), B(7, 6) be the given points

given condition is PA : PB = 2 : 3 3PA = 2PB 9PA2 = 4PB

2

9[(x – 5)2 + (y + 4)

2] = 4[(x – 7)

2 + (y – 6)

2]

9(x2 –10x + 25 + y

2 + 8y + 16) = 4(x

2 – 14x + 49 + y

2 – 12y + 36)

5x2

+ 5y2 – 34x + 120y + 29 = 0

The locus of P is 5x2 + 5y

2 – 34x + 120y + 29 = 0

3. Find the locus of the point P such that PA2 + PB

2 = 2c

2 where A(a, 0), B(–a, 0).

Ans. Let P = (x, y)

given condition is PA2 + PB

2 = 2c

2 (x – a)

2 + (y – 0)

2 + (x + a)

2 + (y – 0)

2 = 2c

2

x2 + a

2 – 2xa + y

2 + x

2 + a

2 + 2xa + y

2 = 2c

2

2x2 + 2y

2 + 2a

2 = 2c

2 x

2 + y

2 = c

2 – a

2.

The locus of P is x2 + y

2 = c

2 – a

2.

4. Find the locus of a point P if the line joining the points (2, 3) and (–1, 5) subtends a right angle at P. Ans. Let P = (x, y) and A(2, 3), B(–1, 5) be the given points.

Given condition is APB = 90 PA2 + PB

2 = AB

2

(x – 2)2 + (y – 3)

2 + (x + 1)

2 + (y – 5)

2 = (–1 – 2)

2 + (5 – 3)

2

x2 – 4x + 4 + y

2 – 6y + 9 + x

2 + 2x + 1 + y

2 – 10y + 25 = 9 + 4

2x2 + 2y

2 – 2x – 16y + 26 = 0 x

2 + y

2 – x – 8y + 13 = 0

The locus of P is x2 + y

2 – x – 8y + 13 = 0.

5. Find the transformed equation of 2x2 + 4xy + 5y

2 = 0 when the origin is shifted to (3, 4) by translation of axes.

Ans. Let (X, Y) be the new coordinates of the point (x, y).

x = X + 3, y = Y + 4 The transformed equation is 2(X + 3)

2 + 4(X + 3)(Y + 4) + 5(Y + 4)

2 = 0

2(X2 + 6X + 9) + 4(XY + 4X + 3Y +12) + 5(Y

2 + 8y + 16) = 0

2X2 + 12X + 18 + 4XY + 16X + 12Y + 48 + 5Y

2 + 40Y + 80 = 0

2X2 + 4XY + 5Y

2 + 28X + 52Y + 146 = 0

6. Find the transformed equation of 4x2 + 9y

2 – 8x + 36y + 4 = 0 when the axes are translated to the point (1, –2).

Ans. Let (X, Y) be the new coordinates of the point (x, y).

x = X + 1, y = Y – 2.

The transformed equation is 4(X + 1)2 + 9(Y – 2)

2 – 8(X + 1) + 36(Y – 2) + 4 = 0.

4(X2 + 2X + 1) + 9(Y

2 – 4Y + 4) – 8X – 8 + 36Y – 72 + 4 = 0 4X

2 + 9Y

2 = 36.

7. Choose a new origin so that the equation 2x2 + 7y

2 + 8x – 14y + 15 = 0 may be translated to the form in which

the first degree terms be missing. Find the transformed equation also. Ans. Comparing the given equation with ax

2 + 2hxy + by

2 + 2gx + 2fy + c = 0.

We get a = 2, b = 7, g = 4, f = –7.

Required point = g f 4 ( 7)

, ,a b 2 7

= (–2, 1).

The transformed equation is 2(X – 2)2 + 7(Y + 1)

2 + 8(X – 2) – 14(Y + 1) + 15 = 0.

2(X2 + 4 – 4X) + 7(Y

2 + 2Y + 1) + 8X – 16 – 14Y – 14 + 15 = 0 2X

2 + 7Y

2 = 0.

8. If the axes are rotated through an angle 30, then find the coordinates of (1, 2) in the new system. Ans. Let (X, Y) be the new coordinates of the given point. The transformation table is

30 X Y

1 3

2 –1/2

2 1/2

3

2

Now X =3 1 3 2 1 3 1 2 3

1 2 , Y 1 22 2 2 2 2 2

.

The new coordinates are3 2 1 2 3

,2 2

.

9. If the coordinates of a point P are transformed to (4, –3) when the axes are rotated through an angle of 135, then find P.


Ans. Let (x, y) be the coordinates of P. The transformation table is

135 4 –3

x 1

2

1

2

Y 1

2

1

2

x = 1 1 4 3 1

4 32 2 2 2 2

y = 1 1 4 3 7

4 32 2 2 2 2

P = 1 7,

2 2

.

10. The line x y

a b= 1 meets x–axis at P. Find the equation of the line perpendicular to this line at P.

Ans. The line meets x-axis at P(a, 0). Slope of the given line is b/a. Slope of the required line is –a/b.

Equation of the required line is y – 0 =a y x a x y a

(x a)b a b b b a b

.

11. If the portion of the straight line intercepted between the axes is bisected at (2p, 2q), write the equation of the straight line.

Ans. Let a, b be the intercepts of the line.

The line cuts x-axis at A(a, 0), y-axis at B(0, b)

Midpoint of AB is a b a b a b

, , (2p,2q) 2p, 2q2 2 2 2 2 2

a = 4p, b = 4q

Equation of the line is x y x y

1 44p 4q p q

.

12. Find the equation of the straight lien of inclination 135 with x–axis and an intercept – 3 on the y–axis.

Ans. Inclination is 135 Slope = tan 135 = –1 Equation of the line having slope –1 and y-intercept –3 is y = (–1)(x – 3)

x + y + 3 = 0.

13. Find the area of the triangle formed by the coordinate axes and the line 2x – 4y – 7 = 0.

Ans. Area of the triangle formed =2( 7) 49

sq.unit.2 2( 4) 16

14. Find the area of the triangle formed by the co–ordinate axes and the line x cos + y sin = p.

Ans. Area of the triangle =2 2 2( p) p p

sq. unit.2 | cos sin | | 2sin cos | | sin2 |

15. Find the equation of the straight line passing through (–4, 5) and cutting off equal intercepts on the coordinate axes.

Ans. Let a, b be the intercepts of the line

Intercepts are equal a = b

Equation of the line is x y

1a a x + y = a

The line passes through (–4, 5) –4 + 5 = a a = 1

Equation of the required line is x + y = 1.

16. Find the equation of the straight line passing through (–2, 4) and making intercepts whose sum is zero. Ans. Let a, b be the intercepts of the line.

a + b = 0 b = –a

Equation of the line is x y

1 x y aa a

The line passes through (–2, 4) –2 – 4 = a a = –6. Equation of the required line is x – y + 6 = 0. 17. Reduce the equation into perpendicular form x + y + 1 = 0. Ans. The given equation is x + y + 1 = 0

x y 1 ( 1)x ( 1)y 1 1

cos x sin y4 42 2 2 2 2 2 2

1 5 5 1

x cos x y sin x cos y sin4 4 4 42 2

.


18. Find the distance between the parallel lines 5x – 3y – 4 = 0, 10x – 6y – 9 = 0. Ans. Given lines are 5x – 3y – 4 = 0, 10x – 6y – 9 = 0

10x – 6y – 8 = 0, 10x – 6y – 9 = 0

Distance between the parallel lines =| 8 9 | 1 1

100 36 136 2 34

.

19. Find the ratio in which the line joining the points A(–1, –1) and B(2, 1) divides the line joining C(3, 4) and D(1, 2).

Ans. Slope of 1 1 2

AB is2 1 3

Equation to AB is y + 1 = 2

3(x + 1) 3y + 3 = 2x + 2 2x – 3y – 1 = 0

The ratio in which AB divides CD is –L11 : L22 = 2[2(3) – 3(4) – 1] : [2(1) – 3(2) – 1]

= –(6 – 12 – 1) : (2 – 6 – 1) = –(–7) : (–5) = –7 : 5, i.e., 7 : 5 externally.

20. Find the acute angle between the pair of lines 2x2 – 3xy – 6y

2 = 0.

Ans. If is the acute angle between the pair of lines 2x2 – 3xy – 6y

2 = 0, then

cos = 1

2 2

| 2 6 | 4 4 4cos

64 9 73 73(2 6) ( 3)

.

21. Find k if the equation x2 – y

2 + 2x + 2y + k = 0 represents a pair of lines.

Ans. Comparing the given equation with ax2 + 2hxy + by

2 + 2gx + 2fy + c = 0 we get

a = 1, b = –1, c = k, f = 1, g = 1, h = 0. Since the given equation represents a pair of lines. abc + 2fgh – af

2 – bg

2 – ch

2 = 0.

1(–1)k + 2(1)(1)(0) – (1)2 – (–1)(1)

2 – k(0)

2 = 0 k = 0.

22. Show that the points (–2, 3, 5), (1, 2, 3), (7, 0, –1) are collinear. Ans. Let A (–2, 3, 5), B (1, 2, 3), C (7, 0, –1).

AB = 2 2 2( 2 1) (3 2) (5 3) 9 1 4 14

BC = 2 2 2(1 7) (2 0) (3 1) 36 4 16 56 2 14

AC = 2 2 2( 2 7) (3 0) (5 1) 81 9 36 126 3 14

AB + BC = 14 2 14 3 14 AC.

Given points are collinear.

23. Find the coordinates of the point which divides the line joining the points (2,–3, 1) and (3,4,–5) in the ratio 1 : 3. Ans. Let A(2, –3, 1), B(3, 4, –5)

The point which divides AB in the ratio 1 : 3 is

1(3) 3(2) 1(4) 3( 3) 1( 5) 3(1) 3 6 4 9 5 3 9 5 2 9 5 1

, , , , , , , ,1 3 1 3 1 3 4 4 4 4 4 4 4 4 2

.

24. Find the ratio in which the xy–plane divides the line segment joining the points (2, 4, 5), (3, 5, –4).

Ans. Let A(2, 4, 5), B(3, 5, –4). xy-plane divides AB in the ratio = 5 : –(–4) = 5 : 4.

25. Find the co–ordinates of the point at which yz–plane intersects the line segment joining the points (–2, 3, 7) and (6, –1, 2).

Ans. Let A(–2, 3, 7), B(6, –1, 2).

yz-plane divides AB in the ratio –(–2) : 6 = 1 : 3

The point at which yz-plane intersects 6 6 1 9 2 21 23

AB is , , 0,2,1 3 1 3 1 3 4

.

26. If a ray makes an angles

,3 3

with x–axis and y-axis, respectively, find the angle made by the ray with z–axis.

Ans. Let , , be the angles made by the ray with the positive x, y, z axes respectively.

= 60, = 60. But cos2 + cos

2 + cos

2 = 1 cos

260 + cos

260 + cos

2 = 1

2 21 1 1 1cos 1 cos cos

4 4 2 2

= 45 or 135.

27. Find the direction ratios and direction cosines of the line joining the points (–4, 1, 7), (2, –3, 2). Ans. Let A(–4, 1, 7), B(2, –3, 2)

Direction ratio’s of AB are (2 + 4, –3 – 1, 2 – 7) = (6, –4, –5) = (–6, 4, 5)

Direction cosines of AB are

2 2 2

6 4 5 6 4 5, , , ,

36 16 25 36 16 25 77 77 77( 6) 4 5

.

28. Find the angle between the lines whose direction ratios are (1, –2, 1) and (–1, 1, 0).

Ans. If is an angle between the lines whose d.r’s are (1, –2, 1) and (–1, 1, 0) then


cos = 1 2 1 2 1 2

2 2 2 2 2 2

1 1 1 2 2 2

a a b b c c 1( 1) ( 2) 1(0) 3 3

21 4 1. 1 1 0 6. 2a b c . a b c

= 150. Angles between the given lines are 30, 150.

29. If P(2, 3, –6) and Q(3, –4, 5) are two points, then find the angle POQ where O(0, 0, 0) is origin.

Ans. D.r’s of OP are (2 – 0, 3 – 0, –6 – 0) = (2, 3, –6)

D.r’s of OQ are (3 – 0, –4 – 0, 5 – 0) = (3, –4, 5)

cosPOQ =2 2 2 2 2 2

(2)(3) (3)( 4) ( 6)(5) 6 12 30

4 9 36 9 16 25(2) (3) ( 6) (3) ( 4) (5)

= 1 136 36 18 2 18 2 18 2. POQ cos cos .

35 35 3549 50 25 2

30. Evaluate:

x

xx 0

a 1Lt

b 1 (a > 0, b > 0, b 1).

Ans. x x

bx xx 0 x 0

a 1 (a 1) / x logaLt Lt log a

logbb 1 (b 1) / x

31. Show that

x 0

x +1 1 1Lt =

x 2.

Ans.

x 0 x 0 x 0

x 1 1 x 1 1x 1 1 x 1 1Lt Lt Lt

x x x 1 1 x x 1 1

=

x 0 x 0

x 1 1 1Lt Lt

2x 1 1 0 1 1x x 1 1

32. Show that x 0

xLt = 1

1+ x 1 x.

Ans. x 0 x 0

x 1 x 1 xxLt Lt

1 x 1 x1 x 1 x

= x 0

1 1Lt 1 x 1 x [1 1] 1.

2 2

33. Show that

x 4

3 5 + x 1Lt =

x 4 6.

Ans.

x 4 x 4 x 4

3 5 x 3 5 x3 5 x 9 (5 x)Lt Lt Lt

x 4 (x 4) 3 5 x (x 4) 3 5 x

= x 4

(x 4) 1 1Lt

6(x 4)(3 5 x) 3 5 4

.

34. Show that:

3 3

x 0

1+ x 1 x 2Lt =

x 3.

Ans. Put 1 + x = y, 1 – x = z as x 0, y 1,z 1

3 3 3 3

x 0 x 0

1 x 1 x 1 x 1 1 x 1Lt Lt

x x x

= 1 11/ 3 1/ 3 1/ 3 1/ 333 1 13 3

y 1 z 1 y 1 z 1

y 1 z 1 y 1 z 1 1 1 1 1 2Lt Lt Lt Lt 1 1

y 1 (z 1) y 1 z 1 3 3 3 3 3

.

35. Show that:x 0

sinaxLt = a

xcosx.

Ans. x 0 x 0

sinax sinax a aLt Lt 1 a

xcosx ax cosx 1

36. Show that:

2

2 2x 0

1 cos2mx 2mLt =

sin nx n.

Ans.

2 22 2 2 2

2 2 2 2 2x 0 x 0 x 0

1 cos2mx 2sin mx sinmx nx m x 2mLt Lt Lt 2

mx sinnxsin nx sin nx n x n

37. Show that

2

2 2 2x a

sin(x a)tan (x a)Lt = 0

(x a ).


Ans. 2 2

2 2 2 2 2x a x a

sin(x a)tan (x a) sin(x a)tan (x a)Lt Lt

(x a ) (x a) (x a)

=

22

2 2x a x a x a

tan(x a) 1 1Lt sin(x a). Lt Lt 0 1 0.

x a (x a) 4a

38. Show that

2

0

1 cosmθ mLt =

1 cosnθ n.

Ans.

22

2 2 22 00

2 2 2 2 20 0

20 0

sin(m / 2)sin (m / 2) LtLt1 cosm 2sin (m / 2) m / 4 m

Lt Lt1 cosn 2sin (n / 2 sin (n / 2) n / 4 nsin(n / 2)Lt Lt

.

39. Show that:

3

2x /2

1 sin x 3Lt =

2cos x.

Ans. 3 2

2x / 2 x / 2

1 sin x (1 sinx)(1 sinx sin x)Lt Lt

(1 sinx)(1 sinx)cos x

= 2 2

x / 2

1 sinx sin x 1 sin / 2 sin / 2 1 1 1 3Lt

1 sinx 1 sin / 2 1 1 2

40. Evaluate:

3x 1

sin(x 1)Lt

x 1.

Ans. 3 2x 1 x 0

sin(x 1) sin(x 1)Lt Lt

x 1 (x 1)(x x 1)

= 2x 0 x 0

sin(x 1) 1 1 1Lt Lt 1.

(x 1) 1 1 1 3x x 1

41. Show that: 2n

1+ 2 + 3 + ....+ n 1Lt =

2n.

Ans. 2 2n n n

1 2 ....n n(n 1) 1 1 1 1Lt Lt Lt 1 (1 0)

2 n 2 2n 2n

42. Show that:

3 3 3

2 2n

1 + 2 + ....+ n 1Lt =

4n (n +1)

Ans. 3 3 3 2 2 2 2

2 2 2 2 2n n n

1 2 ... n n (n 1) 1 (1 1/n) 1 (1 0) 1Lt Lt Lt .

4 4 (1 0) 4n (n 1) 4n (n 1) (1 1/n )

43. Show that:

x

x 0

3 1Lt = 2log3

1+ x 1

Ans.

x

x x

x 0 x 0 x 0 x 0

(3 1) 1 x 13 1 3 1Lt Lt Lt Lt 1 x 1 log3 1 0 1 2log3.

1 x 1 x1 x 1

44. Show that (i)

x

x 0

e 1Lt = 2

1+ x 1. (ii)

2x 3

x 3Lt = 0

| x 9 |

Ans. (i) x x x

x 0 x 0 x 0

e 1 e 1 1 x 1 e 1Lt Lt Lt 1 x 1

x1 x 1 1 x 1 1 x 1

= x

x 0 x 0

e 1Lt Lt 1 x 1 2

x

(ii) x 3 x 3 x 32

x 3 x 3 x 3Lt Lt Lt

x 3x 3 x 3x 9

= x 3 x 3

1 0Lt x 3 Lt 0

x 3 6

45. Find ‘k’ if f(x) =3

5

x + 8

x + 32, (for x –2) and f(x) = k(for x = –2) is continuous at x = –2.

Ans. f(x) is continuous at x = –2 x 2Lt f(x) f( 2)

k = 3 3

5 5x 2 x 2 x 2

x 8 x 8 x 2Lt f(x) Lt Lt

x 2x 32 x 32


=

3 3

3 1x 2

5 5 5 1

x 2

x ( 2)Lt

3( 2) 3 4 3x ( 2)

5 16 20x ( 2) 5( 2)Lt

x ( 2)

46. If f(x) =

2

2

x 9

x 2x 3 for x 3 and f(x) is continuous at x = 3 then find f(3).

Ans. f(x) is continuous at x = 3 f(3) =x 3Lt f(x)

f(3) =2

2x 3 x 3 x 3

x 9 (x 3)(x 3) x 3 6 3Lt Lt Lt

(x 1)(x 3) x 1 4 2x 2x 3

47. Find the derivative of cosx

sinx + cosx.

Ans. 2

d d(sin x cos x) (cos x) cos x (sin x cos x)

d cos x dx dx

dx sin x cos x (sin x cos x)

= 2 2sin x sinxcosx cos x sinxcosx 1

1 sin2x 1 sin2x

48. Find the derivative of log(x + 2x +1 )

Ans. 2 2

2

d 1 dlog x x 1 x x 1

dx dxx x 1

= 2

2 2 2 2

1 1 d 1 11 (x 1) 1 2x

dxx x 1 2 x 1 x x 1 2 x 1

= 2

2 2 2

1 x 1 x 1

x x 1 x 1 x 1

.

49. Find the derivative of cos x .

Ans. d 1 d 1 dcos x cos x sin x x

dx dx dx2 cos x 2 cos x

= sin x 1 sin x

2 x2 cos x 4 x cos x

.

50. Find the derivative of sin–1

2

2x

1+ x

Ans. Put x = tan

1 1 1

2 2

d 2x d 2 tan d dsin sin sin (sin2 ) {2 }

dx dx dx dx1 x 1 tan

= 1

2

d 22tan x

dx 1 x

51. Find the derivative of cos–1

2

2x

1 x

Ans. Put x = tan

2 2

1 1 1

2 2

d 1 x d 1 tan d dcos cos cos (cos2 ) (2 )

dx dx dx dx1 x 1 tan

= 1

2

d 22tan x

dx 1 x

52. If 2x2

– 3xy + y2 + x + 2y – 8 = 0, then find

dy

dx.

Ans. d

dx(2x

2 – 3xy + y

2 + x + 2y – 8) = 0 4x – 3

dy dy dyx y 2y 1 2 0

dx dx dx

dy dy 3y 4x 1

(2y 3x 2) (4x 1 3y)dx dx 2y 3x 2

.

53. If x = a(t – sin t), y = a(1 + cos t) then find y2.

Ans. dx dy

a(1 cos t), asin tdt dt

2

dy dy dt asin t 2sin t / 2.cos t / 2 t. cot

dx dt dx a(1 cos t) 22sin t / 2

.


2 22

2

22 4

1 t tcosec cosec

d y d t 1 t dt 1 1 12 2 2cot .cosec . . .t tdx 2 2 2 dx a(1 cos t) 2a 4adx

2sin sin2 2

.

54. If y = a enx

+ be–nx

then show that y2 = n2y.

Ans. y = aenx

+ be–nx

y1 = naenx

– nbe–nx

y2 = n2 ae

nx + n

2be

–nx = n

2[ae

nx + be

–nx] = n

2y.

55. Find the intervals in which the function x3 – 3x

2 is increasing or decreasing.

Ans. Let f(x) = x3 – 3x

2 f(x) = 3x

2 – 6x

f(x) > 0 3x2 – 6x > 0 x

2 – 2x > 0 x (0, 2)

f(x) is increasing on (–, 0) (2, ) and decreasing on (0, 2).

56. Find the intervals in which the function logx

x is increasing or decreasing.

Ans. f(x) = log x

x f(x) =

2 2

x(1/ x) logx(1) 1 logx

x x

f(x) > 0 2

1 logx0

x

1 – log x > 0 log x < 1 x < e and x > 0.

f(x) is increasing in (0, e)

f(x) < 0 2

1 logx0

x

1 – log x < 0 log x > 1 x > e.

f(x) is decreasing in (e, ).

57. If u = sin(x – y), find

u u,

x y.

Ans. u

x x

[sin(x – y)] = cos(x – y),

u

y y

[sin(x – y)] = –cos(x – y)

58. If z =cosx

siny, then find

z z,

x y.

Ans. z cosx sinx

x x siny siny

z cosx

y y siny y

(cos x . cosec y) = –cosec y . cot y . cos x.

59. If u = tan–1

y

x, then find

2 2

2 2

u u,

x y

Ans. u = 2

1 1

2 2 2 2 2 2

y u y 1 y x y ytan ; tan

x x x x x x1 (y / x) x y x x y

2

2 2 2 2 22 2 2 2

u u y 1 2xyy .2x

x x xx x y x y x y

2

1

2 2 2 2 2

u y 1 y x 1 xtan

y y x y x x1 (y / x) x y x y

2

2 2 2 2 2 2 2 2 2

u u x 1 2xyx. .2y

y y yy x y (x y ) (x y )

60. If z = ex + y

+ f(x) + g(y), prove that zxy = ex + y

.

Ans. z = ex + y

+ f(x) + g(y) zx = z

x

= e

x + y + f(x) zxy = x(z )

y

= e

x + y.

61. If u = log(tan x + tan y), then show that sin 2x

u u+ sin2y = 2

x y.

Ans. u = log [tan x + tan y]

2

2 2u 1 1 sec x cos x cos y sec x cos ysec x sec x

sin x sin yx tan x tan y sin x cos y cos x sin y sin(x y)

cos x cos y

2

2 2u 1 1 sec ycosxcosy cosxsec ysec y sec y

sinx sinyy tanx tany sinxcosy cosxsiny sin(x y)

cosx cosy


u u sec x cos y cos x sec y

sin2x sin2y 2sin x cos x 2sin y cos yx y sin(x y) sin(x y)

= 2[sinxcosy cosxsiny] 2sin(x y)

2sin(x y) sin(x y)

.

62. If z = f(x2 + y

2) then prove that

z zx y = 0

y x.

Ans. z = f(x2 + y

2)

2 2 2 2zf (x y )2x 2x.f (x y )

x

z

y

f(x

2 + y

2) 2y = 2y . f(x

2 + y

2)

z z

x yy x

= x[2y f(x

2 + y

2)] – y[2x . f(x

2 + y

2)]

= 2xy f(x2 + y

2) – 2xy f(x

2 + y

2) = 0.

4 MARKS

1. A(2, 3), B(–3, 4) are two points. If a point P moves such that the area of PAB is 8.5 square unit, then find the locus of P.

Ans. Let P = (x, y). Given condition is, area of PAB = 8.5 sq. unit.

2 3 3 4 5 11 17

172 x 3 y 2 x 3 y2 2

.

|5(3 – y) + 1(2 – x)| = 17 |–(x + 5y – 17)| = 17 (x + 5y – 17)2 = 17

2

x2 + 10xy + 25y

2 – 34x – 170y = 0.

The locus of P is x2 + 10xy + 25y

2 – 34x – 170y = 0.

2. A(2, 3), B(2, –3) are two points. If a point P moves such that PA + PB = 8, then find the locus of P. Ans. Let P = (x, y). Given condition is PA + PB = 8

PA = 8 – PB PA2 = 64 – 16PB – PB

2 PA

2 – PB

2 – 64 = –16PB

(x – 2)2 + (y – 3)

2 – [(x – 2)

2 + (y + 3)

2] – 64 = 2 216 (x 2) (y 3)

y2 – 6y + 9 – y

2 – 6y – 9 – 64 = 2 216 (x 2) (y 3)

–4(3y + 16) = 2 216 (x 2) (y 3) 3y + 16 = 2 24 x 4x 4 y 6y 9

9y2 + 96y + 256 = 16x

2 + 16y

2 – 64x + 96y + 208 16x

2 + 7y

2 – 64x – 48 = 0

The locus of P is 16x2 + 7y

2 – 64x – 48 = 12.

3. A(4, 0), B(–4, 0) are two points. If a point P moves such that |PA – PB| = 4, then show that the locus of P is 3x

2 – y

2 = 12.

Ans. Let P = (x, y)

Given condition is PA – PB = 4 PA = 4 + PB PA2 = 16 + PB

2 + 8PB

PA2 – PB

2 – 16 = 8PB (x – 4)

2 + y

2 – [(x + 4)

2 + y

2] – 16 = 2 28 (x 4) y

x2 – 8x + 16 – x

2 – 8x – 16 – 16 = 2 28 (x 4) y –16(x + 1) = 2 28 (x 4) y

–2(x + 1) = 2 2(x 4) y 4(x2 + 2x + 1) = x

2 + 8x + 16 + y

2 3x

2 – y

2 = 12

The locus of P is 3x2 – y

2 = 12.

4. Find the transformed equation of 17x2 – 16xy + 17y

2 = 225 when the axes are rotated through an angle 45.

Ans. Let (X, Y) be the new coordinates of (x, y)

Angle of rotation = 45.

Now x = X cos 45 – Y sin 45 = X Y

2

, y = X sin 45 + Y cos 45 = X Y

2

The transformed equation is

2 2X Y X Y X Y X Y

17 16 17 2252 2 2 2

2 2 2 2 2 2 2 217 16 17X 2XY Y X Y X 2XY Y 225 9X 25Y 225

2 2 2 .

5. Find the transformed equation of x2 + 2 3 xy – y

2 = 2a

2 when the axes are rotated through an angle 30.

Ans. Let (X, Y) be the new coordinates of (x, y).

The transformation table is

30 X Y

x 3

2 –1/2


y 1/2

3

2

x =3X Y X 3Y

, y2 2

The transformed equation is

2 2

23X Y 3X Y X 3Y X 3Y2 3 a

2 2 2 2

3X2 – 2 3 XY + Y

2 + 2 3 ( 3 X

2 + 3XY – XY – 3 Y

2) – X

2 – 3Y

2 – 2 3 XY = 8a

2

3X2 – 2 3 XY + Y

2 + 6X

2 + 6 3 XY – 2 3 XY – 6Y

2 – X

2 – 3Y

2 – 2 3 XY = 8a

2

8X2 – 8Y

2 = 8a

2 X

2 – Y

2 = a

2.

6. Find the transformed equation of 4xy – 3x2 = a

2 when the axes are rotated through an angle tan

–1 2.

Ans. Let (X, Y) be the new coordinates of (x, y)

Angle of rotation = tan–1

(2) tan = 2 sin = 2/ 5 , cos = 1/ 5

Now x = X cos – Y sin =X 2Y

5

, y = X sin + Y cos =

2X Y

5

The transformed equation is 2

2X 2Y 2X Y X 2Y4 3 a

5 5 5

4

5(2X

2 + XY – 4XY – 2Y

2) –

3

5(X

2 – 4XY + 4Y

2) = a

2 X

2 – 4Y

2 = a

2.

7. Find the transformed equation of x cos + y sin = p when the axes are rotated through an angle ‘’. Ans. Let (X, Y) be the new coordinates of (x, y)

x = X cos – Y sin , y = X sin + Y cos

The transformed equation is (X cos – Y sin ) cos + (X sin + Y cos ) sin = p

X cos2 – Y sin cos + Y sin

2 + Y cos sin = p X(cos

2 + sin

2) = p

X = p.

8. If the transformed equation of a curve is X2 + 2 3 XY – Y

2 = 2a

2 when the axes are rotated through angle 60,

then find the original equation of the curve. Ans. Let (x, y) be the original coordinates of (X, Y).

Angle of rotation = 60.

Now X = x cos 60 + y sin 60 =x 3y

2

, Y = y cos 60 – x sin 60 = y 3x

3

The original equation of the curve is

2 2

2x 3y x 3y y 3x y 3x2 3 2a

2 2 2 2

2 2 2 2 2 2 21 2 3 1(x 2 3xy 3y ) 3y 2xy 3x (y 2 3xy 3x ) 2a

4 4 4

–2x2 + 2y

2 = 2a

2 x

2 – y

2 + a

2 = 0.

9. Show that the angle of rotation of the axes to eliminate xy term in the equation ax

2 + 2hxy + by

2 = 0 is

1

2Tan

–1

2h

a b where (a b).

Ans. Let be the required angle of rotation. Let (X, Y) be the new coordinates of (x, y),

x = X cos – Y sin , y = X sin + Y cos

The transformed equation is a(X cos – Y sin )2 + 2h(X cos – Y sin )(X sin + Y cos ) + b(X sin + Y cos )

2 =

0

a(X2 cos

2 + Y

2 sin

2 – 2XY cos sin ) + 2h (X

2 cos sin + XY cos

2 – XY sin

2 – Y

2 sin cos ) + b(X

2 sin

2 +

Y2 cos

2 + 2XY sin cos ) = 0

X2(a cos

2 + 2h cos sin + b sin

2) + 2XY(–a cos sin + h cos

2 – h sin

2 + b sin cos ) + Y

2(a sin

2 – 2h sin

cos + b cos2) = 0

Since xy term is eliminated,

–a cos sin + h cos2 – h sin

2 + b sin cos = 0

h(cos2 – sin

2) = a cos sin – b sin cos

2h cos 2 = (a – b) sin 2

tan 2 = 1 12h 2h 1 2h2 tan tan

a b a b 2 a b

The required angle of rotation is 11 2htan

2 a b

.

10. Transform the equation 5x – 2y – 7 = 0 into (i) slope–intercept form (ii) intercept form and (iii) normal form.


Ans. (i) Given equation is 5x – 2y – 7 = 0 2y = 5x – 7 y = 5 7

x2 2

.

slope intercept form is y = 5 7

x2 2

.

(ii) Given equation is 5x – 2y – 7 = 0 5x – 2y = 7

5x 2y

7 7 = 1

x y1

(7 /5) ( 7 / 2)

.

intercepts form is x y1

(7 /5) ( 7 / 2)

.

(iii) Given equation is 5x – 2y – 7 = 0 5x – 2y = 7

5x 2y 7

25 4 25 4

5 2 7x y

29 29 29

normal form is x cos + y sin = p where cos 5

29 , sin

2

29 , p =

7

29

11. The portion of a line intercepted between the coordinate axes is bisected by the point (x1, y1). Show that the

equation of the line is 1 1

x y+

x y= 2.

Ans. Let a, b be the intercepts of the line

the line cuts x–axis at A(a, 0), y–axis at B(0, b)

midpoint of AB is Pa b

,2 2

a b

,2 2

= (x1, y1) a

2= x1,

b

2= y1 a = 2x1, b = 2y1

The equation of the line is

1 1 1 2

x y x y1

2x 2y x y = 2.

12. Transform the equation x y

+a b

= 1 into normal form. If the perpendicular distance of the straight line from the

origin is p, deduce that 2 2 2

1 1 1= +

p a b.

Ans. Given equation is x y

a b = 1

2 2 2 2 2 2

x /a y /b 1

1 1 1 1 1 1

a b a b a b

2 2 2 2 2 2

x y 1

1 1 1 1 1 1a b

a b a b a b

which is the normal form.

p =

2 2

1

1 1

a b

2 2 2

1 1 1

p a b .

13. The equation to the base of an equilateral triangle is 3x – 4y + 15 = 0 and one vertex is (1, 2). Find the side and area of the triangle.

Ans. p = Length of the altitude from (1, 2) to {3x – 4y + 15 = 0} =2 2

3 1 4 2 15 102

53 ( 4)

Length of the side =2p 4

3 3 . Area of the triangle =

2p 4

3 3 .

14. If p, q are the perpendiculars from the origin to the lines

x sec + y cosec = a and x cos – y sin – a cos 2, then show that 4p2 + q

2 = a

2.

Ans. p = 2 2

| a |

sec cosec

, q =

2 2

| acos2 || acos2 |

cos sin

4p2 + q

2 =

2 22 2 2 2

2 2

2 2 2 2

4a 4aa cos 2 a cos 2

1 1 sin cos

cos sin cos sin

= 4a2 sin

2 cos

2 + a

2 cos

22 = a

2 sin

22 + a

2cos

22 = a

2(sin

22 + cos

22) = a

2

15. Find the equation of the line passing through the point of intersection of the lines. 2x – 5y + 1 = 0, 3x + 2y = 8 and making equal intercepts on the axes.

Ans. Let P(, ) be the point of intersection of 2x – 5y + 1 = 0, 3x + 2y = 8

2 – 5 + 1 = 0 (1) and 3 + 2 – 8 = 0 (2)


From (1) and (2) we get

1 –5 1 2 –5 2 –8 3 2 By the method of cross multiplication

1 1

2, 1,40 2 3 16 4 15 38 19 19

P = (2, 1)

Equation of the line with equal intercepts is x y1

a a x + y = a

This passes through P(2, 1). 2 + 1 = a a = 3.

The required line equation is x + y = 3.

16. Find the equation of straight line which passes through the intersection point of 2x + 3y = 1, 3x + 4y = 6 and perpendicular to the line 5x – 2y = 7.

Ans. Let P (, ) be the point of intersection of 2x + 3y = 1, 3x + 4y = 6

2 + 3 – 1 = 0 (1) and 3 + 4 – 6 = 0 (2) Solving (1) and (2) we get

1 3 –1 2 3 4 –6 3 4 By the method of cross multiplication

1 1

14, 9.18 4 3 12 8 9 14 9 1

P = (14, –9)

Equation of the line perpendicular to 5x – 2y = 7 and Passing through P is

2(x – 14) + 5(y + 9) = 0 2x – 28 + 5y + 45 = 0 2x + 5y + 17 = 0

17. Find the length of the perpendicular drawn from the point of intersection of the lines 3x + 2y + 4 = 0, 2x + 5y – 1 = 0 to the line 7x + 24y = 15.

Ans. Let P(, ) be the point of intersection of the lines 3x + 2y + 4 = 0, 2x + 5y – 1 = 0

3 + 2 + 4 = 0 (1) and 2 + 5 – 1 = 0 (2) Solving (1) and (2) we get

1 2 4 3 2 5 –1 2 5

1 1

2, 12 20 8 3 15 4 22 11 11

P = (–2, 1) Perpendicular distance from P(–2, 1) to 7x + 24y = 15 is

2 2

| 7( 2) 24(1) 15 | | 14 24 15 | 5 1

25 549 5767 24

.

18. Find the value of k, if the angle between the straight lines 4x – y + 7 = 0 and kx – 5y – 9 = 0 is 45.

Ans. The acute angle between the given lines is 45

cos 45 = 2 2 2 2 2

3(k) ( 1)(1) 1 3k 1

23 ( 1) k 1 10 k 1

10(k2 + 1) = 2(3k – 1)

2 10k

2 + 10 = 2(9k

2 – 6k + 1) 8k

2 – 12k – 8 = 0

2k2 – 3k – 2 = 0 (2k + 1)(k – 2) = 0 k = 2 or –1/2.

19. The base of an equilateral triangle is x + y – 2 = 0 and the opposite vertex is (2, –1). Find the equations of the remaining sides.

Ans. Slope of the base x + y – 2 = 0 of the triangle is –1. Let m be the slope of the side of the triangle.

Now tan 60 = m 1 m 1 m 1

3 31 m m 1 m 1

m 1 m 1 3 1 1 3

m 2 3m 1 m 1 3 1 1 3

.

Equations of the sides of the triangle are y + 1 = (2 3 )(x – 2).

60

(2, –1)

x + y – 2 = 0

20. Find the condition that the lines ax + hy + g = 0, hx + by + f = 0, gx + fy + c = 0 to be concurrent.

Ans. Let P(, ) be the point of concurrence.

a + h + g = 0 (1), h + b + f = 0 (2) and g + f + c = 0 (3) From (1) and (2) we get

1 h g a h


b f h b

2 2 2

1 hf bg gh af,

hf bg gh af ab h ab h ab h

Equation (3) 2 2

hf bg gh afg f c

ab h ab h

= 0 g(hf – bg) + f(gh – af) + c(ab – h

2) = 0

fgh – bg2 + fgh – af

2 + abc – ch

2 = 0 abc + 2fgh – af

2 – bg

2 – ch

2 = 0

The condition is abc + 2fgh – af2 – bg

2 – ch

2 = 0.

21. Show that the line joining the points (2, 3, –4), (–3, 3, –2) is not perpendicular to the line joining the points (–1, 4, 2), (3, 5, 1). Ans. Let A(2, 3, –4), B(–3, 3, –2), C(–1, 4, 2), D(3, 5, 1)

D.r.’s of AB are (–3 – 2, 3 – 3, –2 + 4) = (–5, 0, 2)

D.r.’s of CD are (3 + 1, 5 – 4, 1 – 2) = (4, 1, –1)

a1 a2 + b1b2 + c1c2 = (–5)(4) + (0)(1) + (2)(–1) = –20 – 2 = –22

AB and CD are not perpendicular.

22. Show that

3 2

2x 3

x 8x + 45 7Lt =

32x 3x 9.

Ans. x – 3 is a factor of x3

– 8x2 + 45

x = 3 1 –8 0 45 0 3 –15 –45

1 –5 –15 0

x3 – 8x

2 + 45 = (x – 3)(x

2 – 5x – 15)

2x2 – 3x – 9 = (x – 3)(2x + 3)

3 2 2

2x 3 x 3

x 8x 45 (x 3)(x 5x 15)Lt Lt

(x 3)(2x 3)2x 3x 9

= 2 2

x 3

x 5x 15 3 5(3) 15 21 7Lt

2x 3 2(3) 3 9 3

23. Show that:

1/8 1/8

x 0

(1+ x) (1 x) 1Lt =

x 4.

Ans. 1/ 8 1/ 8 1/ 8 1/ 8

x x 0

(1 x) (1 x) (1 x) 1 (1 x) 1Lt Lt

x x x

= 1 11/ 8 1/ 8

1 18 8

x 0 x 0

(1 x) 1 (1 x) 1 1 1 1 1 1Lt Lt (1) (1) .

(1 x) 1 (1 x) 1 8 8 8 8 4

24. Show that:

x a

a + 2x 3x 2Lt =

3a + x 2 x 3 3

Ans. x a x a

a 2x 3x a 2x 3x 3a x 2 xa 2x 3xLt Lt

3a x 2 x 3a x 2 x 3a x 2 x a 2x 3x

=

3a a 2 a 4 a 2

3 33 a 2a 3a 3 2 3a

.

25. Show that:

2

2x 0

1+ x 1+ xLt = 1

1 x 1 x

Ans.

2 2 2 2

2 2 2 2x 0 x 0

1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 xLt Lt

1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x

=

2 2

2x 0 2

(1 x 1 x ) 1 x 1 xLt

(1 x 1 x) 1 x 1 x

=

2

x 0 2

x(1 x) 1 x 1 x 1 0 1 0 1 1 2Lt 1

1 1 21 0 1 0x(1 x) 1 x 1 x

26. Show that:

x 0

cosecx cotx 1Lt =

x 2.

Ans. x 0 x 0 x 0

cosecx cot x 1 1 cosx 1 cosxLt Lt Lt

x x sinx sinx xsinx


=

2

2 2x 0

2x 0

x 0

sin(x / 2)2 Lt

2sin x / 2 x 2(1/ 2) 1xLt

sinxsinx 1 2xLt

x

.

27. Show that:

3x 0

tanx sinx 1Lt =

2x

Ans. 3 2 2x 0 x 0 x 0

tanx sinx tanx (1 cosx) 1 cosxLt Lt Lt

xx x x

2 22

2x 0 x 0

2sin x / 2 sin(x / 2) 1 1Lt 2 Lt 2

x 2 2x

.

28. Show that f(x) =xtan2x

sin3xsin5x for x 0, f(0) =

2

17 is discontinuous at x = 0.

Ans. x 0 x 0

x tan2x 2 2Lt f(x) Lt f(0)

sin3xsin5x 3 5 15

. f(x) is discontinuous at x = 0.

27. Show that f(x) = cos3x cos4x

xsin2x for x 0, f(0) =

7

4 is continuous at x = 0.

Ans. x 0 x 0 x 0

cos3x cos4x 2sin(7x / 2)sin(x / 2)Lt f(x) Lt Lt

xcos2x xsin2x

= x 0 x 0

x 0

x 0

sin(7x / 2) sin(x / 2)2 Lt Lt

sin(7x / 2) sin(x / 2) x x x2 Ltsin2xx x sin2x

Ltx

= 2(7 / 2)(1/ 2) 7

f(0).2 4

f(x) is continuous at x = 0.

28. If y = x 2 2a + x + a2

log (x + 2 2a + x ) then show that 2 2dy= 2 a + x

dx.

Ans. 2 2 2 2 2 2 2 2 2 2dy d d dx a x a log x a x x a x a log x a x

dx dx dx dx

= 2

2 2 2 2 2 2

2 2

d d a dx a x a x {x} x a x

dx dx dxx a x

=

22 2

2 2 2 2 2 2

2x a 2xx a x 1

2 a x x a x 2 a x

=

2 2 2 2 2 22 2 2 2 2 2

2 2 2 2 2 2 2 2

x a a x x x aa x a x 2 a x

a x x a x a x a x

.

29. Find the derivative of tan–1 21+ x 1

x.

Ans. Put x = tan

2 21 1 1d 1 x 1 d 1 tan 1 d sec 1

tan tan tandx x dx tan dx tan

=

21 1 1d 1 cos d 2sin / 2 d

tan tan tan tandx sin dx 2sin / 2cos / 2 dx 2

= 1

2

d d 1 1tan x

dx 2 dx 2 2(1 x )

30. Find dy

dx if x = a(cos + sin ), y = a(sin – cos ).

Ans.

dy d[a(sin cos )]

dy a(cos sin cos ) sind d tandx ddx a( sin cos sin ) cos

[a(cos sin )]d d


31. If f(x) = loge x then show that f is differentiable at x (0, ) and 1

f'(x) =x

.

Ans. h 0 h 0 h 0

f(x h) f(x) log(x h) logx 1 x hLt Lt Lt log

h h h x

=

x /h

h 0 h 0

1 x h 1 hLt . log 1 Lt log 1

x h x x x

=

1/(h / x)

h / x 0

1 h 1 1log Lt 1 .loge

x x x x

f is differentiable at x (0, ) and f(x) = 1/x.

32. Find the derivative of the function tan (2x + 3) from the definition (first principles). Ans. Let f(x) = tan(2x + 3)

f(x) = h 0

tan(2x 2h 3) tan(2x 3)Lt

h

= h 0

1 sin(2x 2h 3) sin(2x 3)Lt

h cos(2x 2h 3) cos(2x 3)

= h 0

sin(2x 2h 3)cos(2x 3) cos(2x 2h 3)sin(2x 3)Lt

h.cos(2x 2h 3)cos(2x 3)

= h 0 h 0

sin(2x 2h 3 2x 3) 1 sin2hLt Lt 2

h.cos(2x 2h 3)cos(2x 3) cos(2x 2h 3)cos(2x 3) 2h

= 2

1 21 2

cos(2x 3)cos(2x 3) cos (2x 3)

= 2sec

2(2x + 3)

33. Find the derivative of cos x with respect to x from first principle.

Ans. f(x) = cos x

f(x) = h 0 h 0

cos x h cos x 2 x h x x h xLt Lt sin sin

h h 2 2

= h 0

x h xsin

2x h x x h x 2Lt sin

2 2 hx h x

2

=

1/ 2 1/ 2

h 0 h 0 x h x

x h xsin

2x x x (x h) xLt Lt Lt ( 1)

2 (x h) xx h x / 2

= 1/ 2 11 sin x

sin x(1) x2 2 x

34. If xy = e

x–y, then shown that:

2

dy logx=

dx (1+ logx)

Ans. log(xy) = log(e

x – y) y log x = (x – y) log e = x – y y(1 + log x) = x

y = x

1 logx

2 2 2

dy d x (1 logx) . 1 x(1/ x) 1 logx 1 logx

dx dx 1 logx (1 logx) (1 logx) (1 logx)

35. If y = xy, then shown that

2 2dy y y= =

dx x(1 logy) x(1 ylogx).

Ans. log y = log xy = y log x

d d 1 dy 1 dy(logy) ylogx y logx

dx dx y dx x dx

2 2dy 1 y dy 1 ylogx y dy y ylogx

dx y x dx y x dx x(1 ylogx) x(1 logy)

.

36. Find the approximate value of 4 624 .


Ans. Let f(x) = 4 x , x = 625, x = –1

f df =df

dx x =

1 3 31

4 4 41 1 1 1

.x . x .x . x (625) ( 1)4 4 4 4(125)

= –0.002

f(x) = 4 625 5

4 625 f(x x) = f(x) + f = 5 – 0.002 = 4.998

37. Find approximately the value of tan 46 correct to four decimal places, given that 1 = 0.01745 radian.

Ans. Let f(x) = tan x, x = 45, x = 1 = 0.01745

f = df

dx. x = sec

2x. x = sec

2 45. (0.01745) = 2(0.01745) = 0.0349

f(x) = tan 45 = 1

tan 46 = f(x + x) = f(x) + f = 1 + 0.0349 = 1.0349

38. Find the approximate value of sin 46, given that 1 = 0.01745 radius

Ans. Let f(x) = sin x, x = 45, x = 1 = 0.01745

df = cos x. x = 1

. x2

f(x) = sin 45 = 1

2

sin 46 = f(x + x) f(x) + df = 1 1 1 1.01745

x (1 0.01745) 0.71932 2 2 2

39. A circular plate expands when heated from a radius of 5 cm to 5.06 cm. Find the approximate increase, percentage increase in area.

Ans. Let r be the radius, A be the area of the circular plate.

Given r = 5, r = 5.06 – 5 = 0.06. Now A = r2.

A dA

drr = 2rr

The increase in A is A = 2rr = 2(5)(0.06) = 0.6 sq. cm The percentage increase in area is

2

A 2. .r. r r 2(0.06)100 100 2. 100

A r 5r

100 = 2.4

40. The diameter of a sphere is measured to be 20 cm. If an error of 0.02 cm occurs in this, find the errors in volume and surface area of the sphere.

Ans. Let x be the diameter r be the radius, V be the volume and S be the surface area of the sphere.

Given that x = 20cm, x = 0.02 cm

Volume of the sphere V =

3 33 34 4 x 4 x

r .x3 3 2 3 8 6

.

V =6

. 3x

2 . x =

2(20) .(0.02) (400)(0.02)2 2

= 4 cubic cm.

Surface area of the sphere S = 4r2 = 4(x/2)

2 = x

2

S = 2x . x = 2(20)(0.02) = 0.8 sq. cm.

41. If the length of a simple pendulum is decreased by 2%, find the percentage error in its period T. Ans. Let l be the length, T be the period of simple pendulum.

Given 100

= –2. Now T = 2 / g .

T dT 2 1

. .d g 2 g

The percentage error in period = . / g.T

100 100T 2 / g

= 1 1

. 100 ( 2) 1.2 2

42. Show that y = sin3 x cos x has a maximum value at x =

3

and find its value.

Ans. y = sin3x cos x y1 = 3 sin

2x cos x cos x + sin

3x (–sin x) = 3 sin

2x cos

2x – sin

4x

y2 = 3[2 sin x cos x cos2x + sin

2x 2cos x (–sin x)] – 4 sin

3x cos x

= 6 sin x cos3x – 6 sin

3x cos x – 4sin

3x cos x = 6 sin x cos

3x – 10 sin

3x cos x

y1 = 0 3 sin2x cos

2x – sin

4x = 0 sin

2x [3 cos

2x – sin

2x] = 0

sin2x = 0 (or) 3 cos

2x – sin

2x = 0 sin x = 0 (or) tan x = 3 = tan(/3)


x = /3

(y2)x = /3 = 6 sin /3 . (cos /3)3 – 10 (sin /3)

3 cos /3

=

333 1 3 1 3 3 15 3 12 3

6 10 02 2 2 2 8 8 8

y has maximum value at x = /3

The maximum value of y is

333 1 3 3

sin cos3 3 2 2 16

.

43. Find the greatest and least values of 2 sin x + sin 2x over [0, 2].

Ans. Let f(x) = 2 sin x + sin 2x f(x) = 2cos x + 2 cos 2x, f(x) = –2 sin x – 4 sin 2x

f(x) = 0 f(x) = 2 cos x + 2 cos 2x = 0 cos x + 2 cos2x – 1 = 0

2 cos2x + cos x – 1 = 0 (2 cos x – 1)(cos x + 1) = 0 cos x = –1 or 1/2

x = , /3, 5/3.

f() = 0, f(/3) = –2. 3 3

4. 3 3 02 2

At x = /3, f(x) has maximum

Maximum value = f(/3) = 2 . 3 3 3

32 2 2

3 3

f 2 4 3 3 02 2

At x = 5/3, f(x) has minimum.

Minimum value = 5 3 3 3 3

f 23 2 2 2

.

44. The sum of two numbers is 12. Find the numbers so that the sum of squares is minimum. Ans. Let x + y = 12

f(x) = x2 + y

2 = x

2 + (12 – x)

2 f(x) = 2x – 2(12 – x) f(x) = 4

f(x) = 0 2x – 2 (12 – x) = 0 x = 6

f(6) = 4 > 0

When x = 6, f(x) has min. value. x = 6 y = 6.

45. If u = exy

show that u(uxx + uyy) = (2 2x yu u )

Ans. We have from u = exy

that ux = y exy

, uxx = y2 e

xy, uy = xe

xy, uyy = x

2 e

xy.

u(uxx + uyy) = exy

(y2 e

xy + x

2 e

xy) = (y e

xy)2 + (x e

xy)2 = (

2 2x yu u )

46. If u2 =

2 2 2

1

x + y + z then show that

2 2 2

2 2 2

u u u+ +

x y z = 0.

Ans. u2 =

2 2 2

1

x y z u

–2 = x

2 + y

2 + z

2

2 2 2 2(u ) (x y z )x x

=2u–3

3 3u u u

2x u x u xx x x

23 2 3 2 3 2 3 5 2 3

2

u u u[ u x] 3u x u 1 [3u ( u x ) u ] 3u x u

x x x xx

Similarly

2 25 2 3 5 2 3

2 2

u u3u y u , 3u z u

y z

2 2 2

2 2

u u u

zx y

= 3u

5x

2 – u

3 + 3u

5y

2 – u

3 + 3u

5 z

2 – u

3 = 3u

5 (x

2 + y

2 + z

2) – 3u

3

= 3u5(u

–2) – 3u

3 = 3u

3 – 3u

3 = 0

47. If u = x2

tan–1

y

x– y

2 tan

–1

x

y, prove that

2 2 2

2 2

u x y=

y x x + y.

Ans. u = x2tan

–1(y/x) – y

2tan

–1(x/y)

1 2 2

2 2 2

u 1 y 1 12x tan (y / x) x y

x y1 (y / x) x 1 (x / y)

= 2x tan

–1(y/x) –

2 3

2 2 2 2

x y y

x y x y


2 2 31

2 2 2 2

u u y x y y2x tan

y x y x y x x y x y

=

2 2 2 2 2 2 2 3

2 2 2 2 2 2

1 1 x (x y ) x y(2y) 3y (x y ) y (2y)2x

x1 (y / x) (x y ) (x y )

=

2 4 2 2 2 2 2 2 4 4 2 4 4 2 2

2 2 2 2 2 2 2 2 2

2x x x y 2x y 3x y 3y 2y 2x x y 2x y

x y (x y ) x y (x y )

=

2 2 2 2 2 2 2 2 2 2

2 2 2 2 2 2 2 2 2 2 2

2x (x y ) 2x 2x x y x y1

x y (x y ) x y x y x y

.

48. If u = tan–1

2 2x + y

x y, prove that

u u 1x + y =

x y 2sin 2u.

Ans. u =

2 2 2 2 2 2 21 x y x y x [1 y / x ] y

tan tanu xfx y x y x[1 y / x] x

If tan u = z, then z is a homogeneous function in x, y of degree ‘1’.

By Euler’s theorem, z z

x y 1x y

. z x (tanu) y (tanu) tanu

x y

x sec2u

u

x

+ y sec

2u

u

y

= tan u

2

u u tanu 1 1x y [2sinucosu] sin2u

x y 2 2sec u

49. If u = tan–1

3 3x + y

x + y, prove that

u ux + y =

x ysin 2u.

Ans. u =

33

33 3 3 31 2

yx 1

xx y x y ytan tanu x f

yx y x y xx 1

x

If tan u = z then z is a homogeneous function in x, y of degree ‘2’.

By Euler’s theorem, z z

x y 2x y

. z x (tanu) y (tanu) 2tanu

x x

x sec2 u

u

x

+ y sec

2u

u

y

= 2 tan u

2

u u 2tanux y

x y sec u

= [2 sin u cos u] = sin 2u.

50. If u = sin–1

x + y

x + y, prove that

u u 1x + y =

x y 2tan u.

Ans. u = 1

y yx 1 1

x y x y x xsin sinu x

x y x y y y1x 1

xx

If sin u = z, then z is a homogeneous function of degree ½.

By Euler’s Theorem, z z 1 1

x y z x (sinu) y (sinu) sinux y 2 x y 2

u u 1

xcosu ycosu sinu.x y 2

u u 1 sinu 1

x y tanux y 2 cosu 2

.

51. If u = sin–1

x

y + tan

–1

y

x, show that

u ux + y =

x y0.

Ans. u = 1 1 1 1x y 1 y y y

sin tan sin tan f x fy x y / x x x x


u is a homogeneous function in x, y of degree ‘0’.

By Euler’s theorem, u u

x y 0x y

. u = 0.

52. If u = sin–1

( x + y ), then prove that

u u 1x + y =

x y 2tan u.

Ans. u = 1 ysin x y sinu x y x 1

x

If sin u = z, then z is a homogeneous function of degree ½.

By Euler’s Theorem, z z 1 1

x y z x (sinu) y (sinu) sinuy y 2 x y 2

x cos uu u 1

ycosu sinux y 2

.

u u 1 sinu 1

x y tanux y 2 cosu 2

.

7 MARKS

1. Find the circumcentre of the triangle formed by (–2, 3), (2, –1), (4, 0).

Ans. Let A(–2, 3), B(2, –1), C(4, 0) and S(, ) be the circumcentre.

SA = SB = SC SA2 = SB

2 = SC

2

( + 2)2 + ( – 3)

2 = ( – 2)

2 + ( + 1)

2 = ( – 4)

2 +

2

2 + 4 + 4 +

2 – 6 + 9 =

2 – 4 + 4 +

2 + 2 + 1 =

2 – 8 + 16 +

2

4 – 6 + 13 = –4 + 2 + 5 = –8 + 16 8 + 16 8 – 8 + 8 = 0, 4 + 2 – 11 = 0

– + 1 = 0, 4 + 2 – 11 = 0 = 3/2, = 5/2

Circumcentre S (3/2, 5/2). 2. Find the circumcentre of the triangle formed by the lines 3x – y – 5 = 0, x + 2y – 4 = 0,

5x + 3y + 1 = 0.

Ans. Given lines are 3x – y – 5 = 0 (1), x + 2y – 4 = 0 (2), 5x + 3y + 1 = 0 (3) Let A be the point of intersection of (1) and (2) x y 1 –1 –5 3 –1 2 –4 1 2

x y 1 x y 1

4 10 5 12 6 1 14 7 7

x = 2, y = 1. A (2, 1)

Let B be the point of intersection of (3) and (1) x y 1 2 –1 1 2 3 1 5 3

x y 1 x y 1

2 12 20 1 3 10 14 21 7

x = –2, y = 3. B (–2, 3)

Let C be the point of intersection of (3) and (1) x y 1 3 1 5 3 –1 –5 3 –1

x y 1 x y 1

15 1 3 25 5 9 14 28 14

x = 1, y = –2, C (1, –2)

Let S(, ) be the circumcentre of ABC

SA = SB = SC SA2 = SB

2 = SC

2

( – 2)2 + ( – 1)

2 = ( + 2)

2 + ( – 3)

2 = ( – 1)

2 + ( + 2)

2

2 – 4 + 4 +

2 – 2 + 1 =

2 + 4 + 4 +

2 – 6 + 9 =

2 – 2 + 1 +

2 + 4 + 4

–4 – 2 + 5 = 4 – 6 + 13 = –2 + 4 + 5 8 – 4 + 8 = 0, 2 + 6 = 0

2 – + 2 = 0, + 3 = 0 = –6/7, = 2/7.

Circumcentre of (–6/7, 2/7)

3. Find the orthocenter of the triangle formed by the points (–5, –7), (13, 2), (–5, 6). Ans. Let A(–5, –7), B(13, 2) and C(–5, 6)

Slope of BC is 6 2 4 2

5 13 18 9

, Slope of the altitude through A is 9/2.

Equation of the altitude through A is y + 7 = 9/2 (x + 5) 2y + 14 = 9x + 45

9x – 2y + 31 = 0 (1)

Since x-coordinates of C and A are equal, CA is a vertical line.


Equation of the altitude through B is y = 2 (2) Solving (1) and (2), orthocenter is P(–3, 2).

4. Find the orthocenter of the triangle formed by the lines 7x + y – 10 = 0, x – 2y + 5 = 0, x + y + 2 = 0.

Ans. Given lines are 7x + y – 10 = 0 (1), x – 2y + 5 = 0 (2), x + y + 2 = 0 (3) Point of intersection of (2) and (3) is B (–3, 1).

Equation of the altitude through B is (x + 3) – 7(y – 1) = 0 x – 7y + 10 = 0 (4) Point of intersection of (3) and (1) is C(2, –4)

Equation of the altitude through C is 2(x – 2) + (y + 4) = 0 2x + y = 0 (5)

Solving (4) and (5), orthocenter is2 4

,3 3

.

5. If (h, k) is the foot of the perpendicular from (x1, y1) to the line ax + by + c = 0, then show that 1 1 1 1

2 2

h x k y (ax + by + c)= =

a b a + b.

Ans. Let A(x1, y1), P(h, k)

Since P lies in ax + by + c = 0, we get ah + bk + c = 0 ah + bk = –c

Slope of AP is 1

1

k y

h x

Slope of given line is –a/b

Since of AP is perpendicular to the given line 1 1 1

1

k y k y h xa1

h x b b a

1 1 1 1 1 1 1 1 1 1

2 2 2 2 2 2 2 2

h x k y a(h x ) b(k y ) ah bk ax by ax by c (ax by c)

a b a b a b a b a b

6. If (h, k) is the image of (x1, y1) w. r. t the line ax + by + c = 0, then show that 1 1 1 1

2 2

h x k y 2(ax + by + c)= =

a b a + b.

Ans. Let A (x1, y1), B (h, k)

Midpoint of 1 1x h y kAB is P ,

2 2

Since B is the image of A, midpoint P lies on ax + by + c = 0

1 1x h y ka b c 0

2 2

ax1 + by1 + ah + bk + 2c = 0 ah + bk = –ax1 – by1 – 2c

Slope of 1

1

k yAB is

h x

Slope of the given line is a

b

Since AB is perpendicular to the given line,

1 1 1

1

k y h x k ya1

h x b a b

By the law of multipliers in ratio and proportion,

1 1 1 1

2 2

h x k y a(h x ) b(k y )

a b a b

1 1 1 1 1 1 1 1

2 2 2 2 2 2

ah bk ax by ax by 2c ax by 2(ax by c)

a b a b a b

.

7. Show that the equation to the pair of bisectors of angles between the pair of lines ax2 + 2hxy + by

2 = 0 is

h(x2 – y

2) = (a – b)xy.

Ans. Let ax2 + 2hxy + by

2 = 0 represent the lines l1x + m1y = 0 (1), l2x + m2y = 0 (2)

Thus (l1x + m1y) (l2x + m2y) ax2 + 2hxy + by

2 l1l2 = a, m1 m2 = b, l1m2 + l2m1 = 2h

The equations to the bisectors of angles between (1) and (2) are

1 1 2 2

2 2 2 21 1 2 2

x m y x m y0

m m

l l

l l

The combined equation of the bisectors is

2 2

1 1 2 2 1 1 2 2 1 1 2 2

2 2 2 2 2 2 2 2 2 2 2 21 1 2 2 1 1 2 2 1 1 2 2

x m y x m y x m y x m y x m y x m y0 0

m m m m m m

l l l l l l

l l l l l l

2 22 2 2 2

2 2 1 1 1 1 2 2m x m y m x m y 0 l l l l

2 2 2 2 2 2 2 2 2 2 2 2 2 21 2 2 2 1 1 2 1 1 1 2 2x m m y m m m m

l l l l l l

2 2 2 22 2 1 1 1 1 2 22xy m m m m 0

l l l l

2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

1 2 1 2 2 1 2 1 2 1 2 1 1 2 1 2x m m y m m m m m m l l l l l l l l


2 2 2 22 2 1 1 1 1 2 22xy m m m m 0

l l l l

2 2 2 2 2 2 2 2 2 21 2 2 1 1 2 2 1 1 2 1 2 2 1 1 2 1 2 2 1x m m y m m 2xy m m m m m m l l l l l l l l l l

(x2 – y

2) (l1m2 + l2m1) (l1m2 – l2m1) = 2xy (l1m2 – l2m1)(l1l2 – m1m2)

(x2 – y

2) (l1m2 + l2m1) = 2xy(l1l2 – m1m2) (x

2 – y

2) 2h = 2xy(a – b)

h(x2 – y

2) = (a – b)xy

8. Show that the product of the perpendiculars from (,) to the pair of lines ax2 + 2hxy + by

2 = 0 is

2 2

2 2

| a + 2h + b |

(a b) + 4h

.



(l1x + m1y)(l2x + m2y) = ax2 + 2hxy + by

2 l1l2 = a, m1m2 = b, l1m2 + l2m1 = 2h

The length of the perpendicular from (, ) to line (1) is 1 1

2 2

1 1

| m |

m

The length of the perpendicular from (, ) to line (2) is 2 2

2 2

2 2

| m |

m

The product of the perpendiculars is 1 1 2 2

2 2 2 2

1 1 2 2

| m | | m |

m m

=

2 2

1 1 2 2 1 2 1 2 2 1 1 2

2 2 2 2 2 2 2 22 2 2 21 2 1 2 1 2 2 11 1 2 2

( m )( m ) | m m m m |

m m m mm m

= 2 2 2 2

1 2 1 2 2 1 1 2

2 2 2 2

1 2 1 2 1 2 1 2 1 2 2 1 1 2 2 1

| ( m m ) m m | | a 2h b |

( m m ) 2 m m ( m m ) 2 m m (a b) 4h

.

9. Show that the area of the triangle formed by ax2 + 2hxy + by

2 = 0 and x + my + n = 0 is

2 2

2 2

n h ab

| am 2hlm + bl |.



(l1x + m1y)(l2x + m2y) = ax2 + 2hxy + by

2 l1l2 = a, m1m2 = b, l1m2 + l2m1 = 2h

Let the given line be lx + my + n = 0 (3)

Clearly the origin O is the point of intersection of (1) & (2). Let A be the point of intersection of (1) and (3) x y 1 m1 0 l1 m1

m n l m

1 1

1 1 1 1 1 1 1 1

m n nx y 1x ,y

m n 0 0 n m m m m m m

A = 1 1

1 1 1 1

m n n,

m m m m

.

Similarly the point of intersection of (2) and (3) is 2 2

2 2 2 2

m n nB ,

m m m m

The area of OAB = 1 2 2 1

1 1 2 2 2 2 1 1

m n n m n n1

2 m m m m m m m m

= 2 2 2

1 2 2 1 1 2 2 1

2 2

1 1 2 2 1 2 1 2 2 1 1 2

m n m n n ( m m )1 1

2 ( m m )( m m ) 2 m m m m m m m

= 2 2 2 2 2 2

1 2 2 1 1 2 2 1

2 2 2 2 2 2

1 2 1 2 2 1 1 2

n ( m m ) 4 m m1 1 n 4h 4ab n h ab

2 2m m( m m ) m m am 2h m b am 2h m b

.

10. Show that the line ax + by + c = 0 and the pair of lines (ax + by)2 – 3(bx – ay)

2 = 0 form an equilateral triangle

and its area is 2

2 2

c

3 (a + b ) sq. unit.

Ans. (ax + by)2 – 3(bx – ay)

2 = 0 [ax + by + 3 (bx – ay)] [ax + by – 3 (bx – ay)] = 0

[(a + 3 b)x + (– 3 a + b)y] [(a – 3 b)x + ( 3 a + b)y] = 0

This equation represent the lines

(a + 3 b)x + (– 3 a + b)y = 0 (1), (a – 3 b)x + ( 3 a + b)y = 0 (2),

Let the given line be ax + by + c = 0 (3)


If A is an angle between (1) and (3), then

2 2

2 2

a a 3b b 3a bcosA

a b a 3b 3a b

= 2 2 2 2

2 2 2 22 2 2 2 2 2

a 3ab 3ab b a b 1

2a b 4a 4ba b a 3b 2 3ab 3a b 2 3ab

A = 60

The angles between (1) and (3) are 60, 120

If B is an angle between (2) and (3), then

2 2

2 2

a a 3b b 3a bcosB

a b a 3b 3a b

= 2 2 2 2

2 2 2 22 2 2 2 2 2

a 3ab 3ab b a b 1

2a b 4a 4ba b a 3b 2 3ab 3a b 2 3ab

.

B = 60. The angles between (2) and (3) are 60, 120. Since (1) and (2) are different lines it follows that the lines (1), (2), (3) from an equilateral triangle.

The length of the altitude of the triangle, p = The perpendicular distance from the origin O to the line (3)= 2 2

| c |

a b.

Area of the triangle = 2 2

2 2

p c

3 3(a b )

.

11. Prove that the lines represented by x2 – 4xy + y

2 = 0 and x + y = 3 form an equilateral triangle and find its area.

Ans. x2 – 4xy + y

2 = 0 x

2 – 2(x)(2y) + (2y)

2 – 3y

2 = 0 (x – 2y)

2 –

2

3y

(x – 2y + 3 y)(x – 2y – 3 y) = 0 [x + ( 3 – 2)y][x + (– 3 – 2)y] = 0

x2 – 4xy + y

2 = 0 represents the lines

x + (–2 + 3 )y = 0 (1), x + (–2 – 3 )y = 0 (2)

Given line is x + y = 3 (3) If A is an angle between (1) and (3) then

cos A =2

(1)(1) ( 2 3)1 1 2 3 3 1

1 ( 2 3) 1 1 1 4 3 4 3 2 2 8 2 12

=

2

3 1 3 1 3 1 1

22 6 2 2 3 12 6 2

A = 60 or 120 If B is an angle between (2) and (3) then

cos B =

2

3 1(1)(1) ( 2 3)1 1 2 3

1 ( 2 3) 1 1 1 4 3 4 3 2 8 2 12 2

=

2

3 1 3 1 3 1 1

22 6 2 2 3 12 6 2

B = 60 or 120

A = 120, B = 120 is not possible

If A = 120, B = 60 or A = 60, B = 120 then the lines (1) and (2) coincide, which is not true.

A = 60, B = 60

If C is the third angle then clearly C = 60

The triangle ABC is equilateral. The length of the altitude of the triangle, p = The perpendicular distance from the origin to line (3)

= 3 3

1 1 2

Area of the triangle = 22p 1 3 3 3

23 3 2

.

12. Find the centroid of the triangle formed by the lines 12x2 – 20xy + 7y

2 = 0, 2x – 3y + 4 = 0.

Ans. 12x2 – 20xy + 7y

2 = 0 12x

2 – 14xy – 6xy + 7y

2 = 0

2x(6x – 7y) – y(6x – 7y) = 0 (6x – 7y)(2x – y) = 0 The lines represented by this equation are 6x – 7y = 0 ....... (1) and 2x – y = 0 ....... (2) The third side equation is 2x – 3y + 4 = 0 ....... (3) Point of intersection of the lines (1) and (2) is O(0, 0) Point of intersection of the lines (2) and (3) is A(1, 2) Point of intersection of the lines (3) and (1) is B (7, 6)


Centroid of OAB is 0 1 7 0 2 6 8 8

, ,3 3 3 3

13. If S ax2 + 2hxy + by

2 + 2gx + 2fy + c = 0 represents a pair of parallel lines then show that h

2 = ab, af

2 = bg

2 and

the distance between the parallel lines =

2 2g ac f bc2 = 2

a(a + b) b(a + b).

Ans. Let S = 0 represent the lines lx + my + n1 = 0 (1), lx + my + n2 = 0 (2)

(lx + my + n1)(lx + my + n2) = ax2 + 2hxy + by

2 + 2gx + 2fy + c

Comparing both sides, we get l

2 = a, m

2 = b, n1 n2 = c, 2lm = 2h, l(n1 + n2) = 2g, m(n1 + n2) = 2f

h2 = (lm)

2 = l

2 m

2 = ab

2 2 2

2 21 2

2 2 2

1 2

(n n )2g g g g aaf bg

2f m(n n ) f m bf m f

The distance between the parallel lines =

2

1 21 2

2 2

n nn n

a bm

= 2

2 2 21 2 1 2n n 4n n (2g/ ) 4c 4g /a 4c g ac

2a b a b a b a(a b)

Similarly the distance between the parallel lines = 2f bc

2b(a b)

.

14. Show that the pairs of lines 6x2 – 5xy – 6y

2 = 0, 6x

2 – 5xy – 6y

2 + x + 5y – 1 = 0 form a square.

Ans. Given pairs of lines are 6x2 – 5xy – 6y

2 = 0 ........ (1)

6x2 – 5xy – 6y

2 + x + 5y – 1 = 0 ........ (2)

Equation (1) represents a pair of perpendicular lines passing through the origin and Eq. (2) represents a pair of perpendicular lines parallel to (1)

(1) and (2) form a rectangle.

C 2 B

2

A 1 O

1

Let OABC be the rectangle such that (1) represents OA,OC and (2) represents AB,BC .

Equation to AC is x + 5y – 1 = 0

Slope of AC is –1/5

Comparing Eq. (2) with ax2 + 2hxy + by

2 + 2gx + 2fy + c = 0

We get a = 6, b = –6, c = –1, f = 5/2, g = ½, h = –5/2

B = Point of intersection of (2) = 2 2

hf bg gh af,

ab h ab h

= ( 5 / 2)(5 / 2) ( 6)(1/ 2) (1/ 2)( 5 / 2) (6)(5 / 2)

,6( 6) (25 / 4) 6( 6) (25 / 4)

= ( 25 / 4) 3 ( 5 / 4) 15 25 12 5 60 13 65 1 5

, , , ,36 (25 / 4) 36 (25 / 4) 144 25 144 25 169 169 13 13

Slope of (5 /13) 0

OB is 5(1/13) 0

(Slope of AC )(Slope of OB ) = 1(5) 1 AC OB

5

Diagonals of the rectangle OABC are perpendicular OABC is a square.

15. Show that the lines joining the origin to the points of intersection of the line y – 3x + 2 = 0, 7x

2 – 4xy + 8y

2 + 2x – 4y – 8 = 0 are at right angles.

Ans. The given line equation is y – 3x + 2 = 0 3x y1

2

Let A, B, be the points of intersection of the given line and the given curve.

The combined equation of OA and OB is

7x2 – 4xy + 8y

2 + 2(x – 2y)

2

3x y3x y8 0

2 4

7x2 – 4xy + 8y

2 + 3x

2 – xy – 6xy + 2y

2 – 18x

2 + 12xy – 2y

2 = 0

–8x2 + xy + 8y

2 = 0

Coefficient of x2 + Coefficient of y

2 = (–8) + 8 = 0 AOB =

2

.

16. Find the condition that the lines joining the origin to the points of intersection of the line 2x + 3y = k, and the curve 3x

2 – xy + 3y

2 + 2x – 3y – 4 = 0 are at right angles.

Ans. The given line equation is 2x + 3y = k 2x 3y

1k

Let A, B be the points of intersection of the given line and the given curve.



3x2 – xy + 3y

2 + (2x – 3y)

22x 3y 2x 3y

4 0k k

k2(3x

2 – xy + 3y

2) + k(4x

2 – 9y

2) – 4(4x

2 + 12xy + 19y

2) = 0

(3k2 + 4k – 16)x

2 – (k

2 + 48)xy + (3k

2 – 9k – 36)y

2 = 0

Since AOB = /2, we have coefficient of x2 + coefficient of y

2 = 0

3k2 + 4k – 16 + 3k

2 – 9k – 36 = 0 6k

2 – 5k – 52 = 0.

17. If the lines joining the origin to the points of intersection of the line x + 2y = k with the curve 2x

2 – 2xy + 3y

2 + 2x – y – 1 = 0 are at right angles, then find k.

Ans. The given line equation is x + 2y = k x 2y

1k

Let A, B be the points of intersection of the given line and the given curve.


2x2 – 2xy + 3y

2 + (2x – y)

2

2

x 2y (x 2y)0

k k

k2(2x

2 – 2xy + 3y

2) + k(2x

2 + 3xy – 2y

2) – (x

2 + 4xy + 4y

2) = 0

(2k2 + 2k – 1)x

2 – (2k

2 – 3k + 4)xy + (3k

2 – 2k – 4)y

2 = 0

Since AOB = /2 we have coefficient of x2 + coefficient of y

2 = 0

2k2 + 2k – 1 + 3k

2 – 2k – 4 = 0 5k

2 – 5 = 0 k

2 = 1 k = 1

18. Find the condition for the lines joining the origin to the points of intersection of the circle x

2 + y

2 = a

2 and the line lx + my = 1, to coincide.

Ans. Given circle is x2 + y

2 = a

2

Given line is lx + my = 1 Let A, B be the point of intersection of given line and circle.

Combined equation of OA and OB is x2 + y

2 – a

2(lx + my)

2 = 0

x2 + y

2 – a

2l2x

2 – 2a

2lmxy – a

2m

2y

2 = 0

x2(1 – a

2l2) – 2a

2lmxy + (1 – a

2m

2)y

2 = 0 (1)

(1) represents coincident lines (a2lm)

2 = (1 – a

2l2)(1 – a

2m

2)

a4l4m

2 = 1 – a

2l2 – a

2m

2 + a

4l2m

2 a

2(l

2 + m

2) = 1.

19. Find the angle between two lines which are nonparallel and whose direction ratios satisfy the equations 6bc + 5ab – 2ca = 0 and 3a + b + 5c = 0.

Ans. Let 3a + b + 5c = 0 (1), 6bc + 5ab – 2ca = 0 (2)

(1) a = b 5c

3

.

(2) 6bc b 5c

3

(5b – 2c) = 0

b2 + bc – 2c

2 = 0 (b – c)(b + 2c) = 0 b = c or b = –2c.

If b = c, then a = –2c and hence a : b : c = –2c : c : c = –2 : 1 : 1. If b = –2c, then a = –c and hence a : b : c = –c : –2c : c = –1 : –2 : 1

The direction ratios of the lines represented by the given equation are (–2, 1, 1), (–1, –2, 1).

If is the angle between the lines then

cos =2 2 2 2 2 2

| ( 2)( 1) 1( 2) 1(1) | | 1| 1

66 6( 2) 1 1 ( 1) ( 2) 1

.

The angle between the two given lines is 1 1cos

6

.

20. Find the angle between two lines which are nonparallel and whose direction cosines are related by the equations l + m + n = 0, 2lm + 2nl – mn = 0.

Ans. l + m + n = 0 (1), 2lm + 2nl – mn = 0 (2)

(1) n = –l – m

(2) 2lm + 2l(–l – m) – m(–l – m) = 0 –2l2 + lm + m

2 = 0 2l

2 – lm – m

2 = 0

(2l + m)(l – m) = 0 2l + m = 0, l – m = 0 m = –2l or l.

If m = l, then n = –l – l = –2l l : m : n = l : l : –2l = 1 : 1 : –2

If m = –2l, then n = –1 + 2l = l l : m : n = l : –2l : l = 1 : –2 : 1

D.r.’s of the lines are (1, 1, –2) and (1, –2, 1).

If is the angle between the lines then

cos = 2 2 2 2 2 2

| 1(1) 1( 2) ( 2)1| | 1| 1

2 21 1 ( 2) 1 ( 2) 1

.

The angle between the given lines is /3.

21. Find the derivative of tan–1

2 2

2 2

1+ x + 1 x

1+ x 1 x

.

Ans. Put x2 = cos


2 2

1 1

2 2

d 1 x 1 x d 1 cos 1 costan tan

dx dx 1 cos 1 cos1 x 1 x

= 1 1d 2 cos / 2 2 sin / 2 d cos / 2 sin / 2tan tan

dx dx cos / 2 sin / 22 cos / 2 2 sin / 2

= 1 1 2

2 4

d d d 1 1 1 xtan tan cos (x ) 0 2x

dx 4 2 dx 4 2 dx 4 2 2 1 x 1 x

22. If sin y = x sin (a + y) then show that2dy sin (a + y)

=dx sina

.

Ans. sin y = x sin (a + y) x = siny

sin(a y)

2 2 2

dx d siny sin(a y)cosy sinycos(a y) sin(a y y sina

dy dy sin(a y) sin (a y) sin (a y) sin (a y)

2dy sin (a y)

dx sina

.

23. If 2 21 x + 1 y = a(x – y), then show that

2

2

1 ydy=

dx 1 x

.

Ans. Put x = sin , y = sin

2 2 2 21 x 1 y a(x y) 1 sin 1 sin = a(sin – sin )

cos + cos = a(sin – sin )

( )cos

22cos cos a2cos sin a

( )2 2 2 2sin

2

1cot a cot a2 2

– = 2 cot

–1a

sin–1

x – sin–1

y = 2 cot–1

a 1 1 1d d(sin x sin y) (2cot a)

dx dx

2

2 2 2

1 y1 1 dy dy0

dx dx1 x 1 y 1 x

.

24. Show that the sum of the intercepts on the coordinate axes of any tangent to x + y = a is constant.

Ans. Let P (x1, y1) be a point on the given curve.

Given curve is x y a .

1 1

1

x ,y 1

yyd d 1 1 dy dy dyx y a 0

dx dx dx dx dx2 x 2 y x x

Equation of tangent at (x1, y1) is y – y1 = 1 1 11

1 1 1 1 1

y y xy x(x x )

x y y x x

1 1

1 1 1 1

x y x yx y a 1

x y a x a y .

x-intercept = 1a. x , y-intercept = 1a. y

Sum of the intercept = 1 1 1 1a. x a. y a x y a. a = a = constant.

25. Show that the curve x2/3

+ y2/3

= a2/3

(a > 0), the portion of tangent drawn at any point (not on the coordinate for axes) included between the coordinate axes is equal to a.

Ans. The parametric equations are x = a cos3, y = a sin

2;

2

2 2

2

dx dy dy 3asin cos3acos t( sin ), 3asin cos tan

d d dx 3acos sin

.

Equation of the tangent is y – a sin3 = –

sin

cos

(x – a cos

3)

2 2x y x ya(cos sin ) a 1.

cos sin acos asin

If the tangent cuts the axes at A, B respectively then A = (a cos , 0), B = (0, a sin )

AB = 2 2 2 2 2 2(acos 0) (0 asin ) a (cos sin a a.


26. The tangent at any point P on the curve xm

yn = a

m+n cuts the coordinate axes in A and B. Show that AP : PB is

constant.

Ans. xm

yn = a

m + n log(x

m y

n) = log a

m + n m log x + n log y = log a

m + n

m nd d m n dy dy my{mlogx nlogy} {loga } 0

dx dx x y dx dx nx

Slope of the tangent of P (x1, y1) is = 1

P 1

mydy

dx nx

Equation of the tangent at P (x1, y1) is y – y1 = 11

1

my(x x )

nx

nx1y – nx1y1 = –mxy1 + mx1y1 mxy1 + nx1y = x1y1 (m + n)

1 1

1 11 1 1 1

mxy nx y x y1 1

x (m n) y (m n)x y (m n) x y (m n)

m n

Let the tangent cuts x-axis at A and y-axis at B.

OA = 1 1x (m n) y (m n), OB

m n

AP =

2 2 2 2 2 2

1 12 21 1 1 11 1 1

x n m yx (m n) x m x n mxx [y 0] y

m m m

PB = 2 2 2 2 2 2

1 12 21 1 1 11 1 1

n x y my (m n) y n y m y n(x 0) y x

n n n

AP : PB = 2 2 2 2 2 2 2 2

1 1 1 1x n m y n x y m: n : m

m n

27. If p and q are the lengths of the perpendiculars from the origin on the tangent and the normal to the curve x2/3

+ y

2/3 = a

2/3, then show that 4p

2 + q

2 = a

2.

Ans. The parametric equations are x = a cos3, y = a sin

3.

2

2

dy 3asin cos sin

dx cos3acos sin

.

Equation of the tangent is y – a sin3 =

sin

cos

(x – a cos

3)

x sin + y cos = a cos sin (cos2 + sin

2) = (a/2) sin 2.

P = Length of the perpendicular from origin to the tangent

= 2 2

| (a / 2)sin2 | asin2 .

2sin cos

Equation of the normal is y – a sin3 = 3cos

(x acos )sin

x cos – y sin = a(cos4 – sin

4 = a(cos

2 – sin

2)(cos

2 + sin

2) = a . cos 2

q = length of the perpendicular from origin to the normal = 2 2

a.cos2

cos sin

= a cos 2.

4p2 + q

2 =

2a

4 sin22

+ (a cos 2)

2 = a

2(sin

22 + cos

22) = a

2.

28. If the curves x = y2, xy = k cut orthogonally, then show that 8k

2 = 1.

Ans. 2

1 2

d d dy 1 d d(k) dy y(x) (y ) m . (xy) m

dx dx dx 2y dx dx dx x

Since the two curves cut orthogonally. m1 m2 = –1

2 21 y 1 1 1 1 11 x ,y ,xy k . k k 8k 1.

2y x 2 2 82 2

29. Prove that the curves y2 = 4ax and xy = c

2 intersect each other orthogonally if c

4 = 32a

4.

Ans. Given curves are y2 = 4ax (1), xy = c

2 (2)

(2) x2 y

2 = c

4 x

2 . 4ax = c

4 x

3 =

4 4 /3

2/3 1/3

c cx

4a 2 a

y2 =

4 / 3

2 / 3 1/ 3

4a.c

2 .a= 2

4/3 . a

2/3 . c

4/3 y = 2

2/3 . a

1/3 . c

2/3.

Point of intersection of (1) and (2) is P = (c4/3

2–2/3

a–⅓

, 22/3

a1/3

c2/3

)

Equation of the first curve is y2 = 4ax 2d d

{y } {4ax}dx dx

2y dy dy 2a

4adx dx y


Slope of the tangent is m1 = 1/ 3 2 / 3 2 / 3

2 / 3 1/ 3 2 / 3

P

dy 2a2 a c

dx 2 a c

Equation of the second curve is xy = c2

2d d dy dy y{xy} {c } x y 0

dx dx dx dx x

Slope of the tangent is m2 = 2/ 3 1/ 3 2 / 3

4 / 3 2 / 3 2 / 3

4 / 3 2 / 3 1/ 3

p

dy 2 a c2 a c

dx c 2 a

The curves intersect orthogonally m1 m2 = –1

(21/3

a2/3

c–2/3

) (–24/3

a2/3

c–2/3

) = –1 25/3

a4/3

c–4/3

= 1 32a4 = c

4

30. A point P is moving with uniform velocity v along a line AB. O is a point on the perpendicular to AB at A and at

a distance l from it. Show that the angular velocity about O is l2

V

OP.

Ans. Let AOP = at the time t

Now cos = OP

and tan = x

sec2

22

2

d 1 dx v d v v vcos .

dt dt dt OP OP

The angular velocity of P about O is 2

v

OP

O

l

A x P B

31. A man 180 cm high, walks at a uniform rate of 12 km per hour away from a long post of 450 cm high. Find the rate at which the length of his shadow increases.

Ans. AB = lamp post = 450cm PQ = person = 180 cm x = distance of the person from the lamp post y = length of shadow

dx

dt= 12 km/hour

A

450

B Q C y x

180

P

Now ABC ||| PQC x y 450 x y 5

y 180 y 2

2x + 2y = 5y 2x = 3y

dx dy dy dy 2(12)

2 3 2(12) 3 8km/hour.dt dt dt dt 3

32. Gas is leaking out of a spherical ballon at the rate of 900 cc/sec. When the radius of the ballon is 360 cms, find the rate at which the surface area is shrinking.

Ans. Let r be the radius, S be the surface area and V be the volume of the spherical balloon at the time t.

Given r = 360cm. dV

900cc / secdt

V = 3 2

2

4 dV dr dr 900r 4 r

3 dt dt dt 4 (360)

S = 4r2

2

dS dr 9008 r 8 (360) 5

dt dt 4 (360)

The surface area of the balloon is shrinking at the rate of 5 sq. cm/sec.

33. A conical vessel whose vertical angle is 90 is placed with its axis vertical and the vertex downwards. If water flows into it at the rate of 1 c.ft per minute, find the rate at which the level of water is rising when the height of the water in it is 2 ft.

Ans. Let h be the height, r be the radius and V be the volume of the water in the vessel at the time t.

Vertical angle = 90 Semi vertical angle = 45.

tan 45 = r r

1 r hh h

Given that dV

1c.ft /minute, h 2ft.dt

V = 3

21 hr h

3 3

2

2dV 3 h dh dh1 (2)

dt 3 dt dt

dh 1

dt 4

ft. /minute.

The level of water is rising at the rate of 1/4 ft./minute.

34. Sand is being poured on the ground from the orifice of an elevated pipe and forms a pile which has always the shape of a right circular cone whose height is equal to the radius of the base. If the sand is falling at the rate of 6 cubic feet per second, find the rate at which the height of the pile is rising when the height is 5 ft.

Ans. Let h be the height, r be the radius, V be the volume of the conical pile at the time t.


Given h = r, dV

dt= 6 cubic ft./sec.

h = 5 ft.

Now V = 2 31 1r h h

3 3

r

h

2 2dV dh dh dh 6h 6 (5)

dt dt dt dt 25

The height of the pile is rising at the rate of 6ft. / sec.

25

35. Show that the maximum rectangle that can be inscribed in a circle is a square. Ans. Let r be the radius of the circle and x, y be the length, breadth of the rectangle.

x2 + y

2 = 4r

2 y

2 = 4r

2 – x

2 y = 2 24r x

If A is the area of the rectangle, then A = xy

A = 2 2 2 2 4x 4r x 4r x x

Let S = 4r2x

2 – x

4

O r

x

y

2

2 3 2 2

2

dS d S8r x 4x 8r 12x

dx dx

dS0

dx 8r

2x – 4x

3 = 0 4x[2r

2 – x

2] = 0 x = 0 or x = 2 r

2

2

x 2r

d S

dx

= 8r2 – 12(2r

2) = –16r

2 < 0

S has maximum, when x = 2 r

A has maximum, when x = 2 r

If A is maximum, then x = 2 r, y = 2 24r 2r 2r .

The rectangle is a square.

36. Show that the height of the cylinder of the maximum volume that can be inscribed in a sphere of radius a

is2a

3. Also find its volume.

Ans. Let r, h be the radius and height of the cylinder which can be inscribed in a sphere of radius a.

a2 = r

2 +

2 22 2h h

r a4 4

Volume of the cylinder, V = r2h = h

22 h

a4

Let f(h) = 2

2 ha h

4

f(h) = 2

2 3ha

4

f(h) = 3 h

2

f(h) = 0 2

2 3ha

4

= 0 3h2 = 4a

2 h = 2a

3

a

O

h/2 r

2a 3 2a

f . 023 3

V has maximum, when h = 2a/ 3

Maximum volume = 3 3 3 3

22a 2a 1 8a 2a 2a 4 af a . .

43 3 3 3 3 3 3 3 3

cubic unit.

37. Show that when curved surface of a cylinder inscribed in a sphere of radius R is a maximum, then height of

the cylinder is 2 R.

Ans. Let r be the radius and h be the height of the cylinder inscribed in a sphere of radius R. The r2 +

22h

R4

Curve surface area

= 2rh = 2 h.2 4

2 2 2h hR 2 R h

4 4 = f(h)

f(h) = 2 2 4

2

2 R h h / 4

. (2R2h – h

2) = 0

2R2h – h

3 = 0 h

2 = 2R

2 h = 2 R

R

O

h/2 r

When surface are is max. h = 2 R

38. Show that the semi–vertical angle of the cone of maximum volume and of given slant height is tan–1

( 2 ).

Ans. Let r be the base radius, h be the height, l be the slant height and be the semi-vertical angle of the cone.


From the figure, r = l sec , h = l cos

Volume of the cone V = 21 1r h

3 3 (l sec )

2 (l cos ) =

3 2sin cos

3

3 3

2 2 2dV[sin ( sin ) cos 2sin cos ] sin [cos sin ]

d 3 3

r

h

l

2 3

2 2

2

d V[sin { 4cos sin 2sin cos } cos {2cos sin }]

3d

3dV

0d 3

sin (2 cos

2 – sin

2) = 0 2 cos

2 = sin

2 tan

2 = 2

tan = 2 = tan–1

2

= tan–1

2 then2

2

d V

d< 0.

If = tan–1

2 , then V has maximum.

If V is maximum then the semi-vertical angle is tan–1

2 .

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I P E (QUESTIONS) - IIT JEE Mathematicskishoremaths.in/.../2016/02/Important-Questions_Part_2.pdfFind the area of the triangle formed by the coordinate axes and the line 2x – 4y