practice paper – 6 solutions -...

Practice Paper – 667

Senior Inter ✦ Mathematics – IIB

SOLUTIONS

PRACTICE PAPER – 6

SECTION – A

I.I.I.I.I. 1. Find the value of k if the points (4, 2) and (k, –3) are conjugatepoints with respect to the circle x2 + y2 – 5x + 8y + 6 = 0.

Sol. Equation of the circle is x2 + y2 – 5x + 8y + 6 = 0

Polar of P(4, 2) is

x.4 + y.2 – 25

(x + 4) + 4 (y + 2) + 6 = 0

8x + 4y – 5x – 20 + 8y + 16 + 12 = 0

3x + 12y + 8 = 0

P(4, 2), Q(k, –3) are conjugate points

Polar of P passes through Q

∴ 3k – 36 + 8 = 0

3k = 28 ⇒ k = 3

28.

2. If the circle x2 + y2 + ax + by – 12 = 0 has the centre at (2, 3)

then find a, b and the radius of the circle.

Sol. Equation of the circle is x2 + y2 + ax + by – 12 = 0

Centre =

−−

2b

,2a

= (2, 3)

2a− = 2

2b− = 3

a = – 4 b = –6g = –2, f = –3, c = –12

radius = cfg 22 −+ = 1294 ++ = 5



3. x2 + y2 +4x + 6y – 7 = 0, 4(x2 + y2) + 8x + 12y – 9 = 0 find

the equation of the radical axis of the circles.

Sol. S – S' = 0 is radical axis.

(x2 + y2 + 4x + 6y – 7)

–

−+++

49

y3x2yx 22= 0

2x + 3y – 4

19 = 0 ⇒ 8x + 12y – 19 = 0

4. Find the equations of axis and directrix of the parabola

y2 + 6y – 2x + 5 = 0.

Sol. y2 + 6y = 2x – 5

Adding '9' on both sides we get,

y2 + 6y + 9 = 2x – 5 + 9

[y – (–3)]2 = 2x + 4

[y – (–3)]2 = 2[x – (–2)]

Comparing with (y – k)2 = 4a (x – h) we get,

(h, k) = (–2, –3), a = 21

Equation of the axis y – k = 0 i.e. y + 3 = 0

Equation of the directrix x – h + a = 0

i.e., x – (–2) + 21

= 0

2x + 5 = 0.

5. If the angle between the asymptotes is 30º then find itseccentricity.

Sol. Angle between the asymptotes = 2θ = 30º

θ = 15º

tan θ = tan 15º = ab



e2 = 2

22

a

ba += 1 + tan2 15º

= sec2 15º =

2

13

22

+ =

324

324

324

8

−−×

+

= 2)26(3484

)324(8 −=−=−

Eccentricity = e = 26 −

6. Solve ∫ dxxeccosxsec 22 on I ⊂⊂⊂⊂⊂ R \

∈

π+∪∈π Zn:

2)1n2(}Zn:n{ .

Mar., May '07

Sol. ∫ sec2 x. cosec2 x dx

= ∫ dxxsinxcos

122

= ∫ +dx

xsin.xcos

xcosxsin22

22

= ∫ dxxcos

12 + ∫ dx

xsin

12

= ∫ sec2 x dx + ∫ cosec2 x dx

= tan x – cot x + C

7. Solve cot (log x)

dxx∫ , x ∈∈∈∈∈ I ⊂ ⊂ ⊂ ⊂ ⊂ (0, ∞∞∞∞∞) \ {enπ : n ∈ ∈ ∈ ∈ ∈ Z). Mar.

'05

Sol. t = log x ⇒ dt = x

dx

∫ dxx

)x(logcot= ∫ cot t dt = log (sin t) + C

= log (sin (log x) + C



8. Evaluate ∫ −4

0

dx|x2|

Sol. = ∫∫ −+−4

2

2

0

dx|x2|dx|x2|

= dx)2x(dx)x2(4

2

2

0−∫+−∫

=

4

2

22

0

2

x22x

2x

x2

−+

−

= ( )

−−−−

−

24

48824

4

= 2 – 0 + 2 = 4

9. Find the area under the curve f(x) = sin x in [0,2πππππ].

Sol.

f(x) = sin x,

We know that in [0, π], sin x ≥ 0 and [π, 2π], sin x ≤ 0

Required area = dx)xsin(dxxsin2

1∫∫π

π

π

−+

= (– cos x)0π π

π��

= – cos π + cos 0 + cos 2π – cos π

= – (– 1) + 1 + 1 –(–1) = 1 + 1 + 1 + 1

= 4.

Y

X2ππO

–1

π/2

3π/2

1



10. Find the general solution of dydx

= 2yx

.

Sol.dxdy

= xy2

∫ ydy

= ∫ xdx

2

log c + log y = 2 log x

log cy = log x2

Solution is cy = x2 where c is a constant

SECTION – B

II.II.II.II.II. 11. If the abscissae of points A, B are the roots of the equation,x2 + 2ax – b2 = 0 and ordinates of A, B are roots of

y2 + 2py – q2 = 0, then find the equation of a circle for

which AB is a diameter.

Sol. Equation of the circle is

(x – x1) (x – x2) + (y – y1) (y – y2) = 0

x2 – x(x1 + x2) + x1x2 + y2 – y (y1 + y2) + y1y2 = 0

x1, x2 are roots of x2 + 2ax – b2 = 0

y1, y2 are roots of y2 + 2py – q2 = 0

x1 + x2 = – 2a y1 + y2 = – 2p

x1x2 = – b2 y1y2 = – q2

=−==++

c/aproductb/arootsofsum0cbxax2

Equation of circle be

x2 – x (– 2a) – b2 + y2 – y (– 2p) – q2 = 0

x2 + 2xa + y2 + 2py – b2 – q2 = 0



12. If the angle between the circles

x2 + y2 – 12x – 6y + 41 = 0 and

x2 + y2 + kx + 6y – 59 = 0 is 45° find k.

Sol. Suppose θ is the angle between the circles

x2 + y2 – 12x – 6y + 41 = 0

and x2 + y2 + kx + 6y – 59 = 0

g1 = – 6, f1 = –3, c1 = 41,

g2 = 2k

, f2 = 3, c2 = – 59

cos θ = 21

212121

rr2f.f2g.g2cc −−+

cos 45o =

5994k

419362

3).3(22k

)6(25941

2

++−+

−−−−−

684

k.2.2

18k618

2

12

+

++−=

684

k.4

k6

2

12

+

=

Squaring and cross - multiplying

+ 68

4k

42

= 18k2

[ ]4

272k2 2 + = 9k2

k2 + 272 = 18 k2

17k2 = 272



k2 = 17272

= k2 = 16

k = ± 4.

13. Find the equation of the tangent and normal to the ellipse9x2 + 16y2 = 144 at the end of the latus rectum in the firstquadrant. Mar. '07

Sol. Given ellipse is 9x2 + 16y2 = 144

9y

16x 22

+ =1

e = 2

22

a

ba − =

16916 −

= 47

End of the latus rectum in first quadrant

P

a

b,ae

2

=

49

,7

Equation of the tangent at P is

21

21

b

yy

a

xx+ = 1

+

49

.9y

167

.x = 1

4y

16x7 + = 1 or 7 x + 4y = 16

Equation of the normal at P is

1

2

1

2

yyb

xxa − = a2 – b2

−

49y9

7

x16 = 16 – 9

7

x16 – 4y = 7

16x – 4 7y = 7 7 .



14. If e, e1 are the eccentricities of a hyperbola and its conjugate

hyperbola prove that 21

2 e

1

e

1+ = 1.

Sol. Equation of the hyperbola is 2

2

2

2

b

y

a

x − = 1

∴ b2 = a2(e2 – 1) ⇒ e2 – 1= 2

2

a

b

e2 = 1 + 2

22

2

2

a

ba

a

b +=

∴ 22

2

2 ba

a

e

1

+= –––– (1)

Equation of the conjugate hyperbola is

2

2

2

2

b

y

a

x − = – 1 ⇒ 2

2

2

2

a

x

b

y− = 1

⇒ a2 = b2 (e12 – 1) ⇒ 2

1e – 1 = 2

2

b

a

21

e = 1 + 2

2

b

a = 2

22

b

ba +

22

2

21 ba

b

e

1

+= –––– (2)

Adding (1) and (2)

22

2

22

2

21

2 ba

b

ba

a

e

1

e

1

++

+=+ = 22

22

ba

ba

++

= 1.

15. Solvedx

4 + 5 sin x∫ Mar. '05

Sol. t = tan 2x

⇒ dt = sec2 2x

. 21

dx

dx = 22 t1

dt2

2x

sec

dt2

+=



sin x = 22 t1

t2

2x

tan1

2x

tan2

+=

+

I = ∫ ∫ ++=

++

+ t10t44

dt2

t1t2

54

t1

dt2

2

2

2

= ∫ ∫

−

+

=++

222

43

45

t

dt21

12t5

t

dt21

= C

43

45

t

43

45

tlog

43

.2

1.

21 +

++

−+

= C4t21t2

log31

C8t42t4

log31 +

++=+

++

= C2

2x

tan2

12x

tan2log

31 +

+

+

16. Evaluate /4

0

log (1+ tan x) dxπ

∫

Sol. I =π

π + − ∫�

��

=π π −

+ π +

∫�

��

��

=π

−+ + ∫�

��

��



=

π + + − +

∫�

��

��

=π

− +∫�

��

=π π

− +∫ ∫� �

��

= I−π 4/

0)x(2log

2I = 2log4π

I = 2log8π

17. Evaluate dxdy

= tan2 (x + y).

Sol.dxdy

= tan2 (x + y)

put v = x + y

dxdv

= 1 + dxdy

= 1 + tan2 v = sec2 v

∫ vsec

dv2 = ∫ dx

= ∫ cos2 v . dv = x + c

∫ +2

)v2cos1( dv = x + c

∫ (1 + cos 2v) dv = 2x + 2c

v + 2

v2sin = 2x + 2c

2v + sin 2v = 4x + c'

2(x + y) + sin 2(x + y) = 4x + c'

x – y – 21

sin [2(x + y)] = c



SECTION – C

III.III.III.III.III. 18. If (2, 0), (0, 1), (4, 5) and (0, c) are concyclic and then find c.

Sol. x2 + y2 + 2gx + 2fy + c1 = 0

Satisfies (2, 0), (0, 1) (4, 5) we get

4 + 0 + 4g + c1 = 0 –– (i)

0 + 1 + 2g. 0 + 2f + c1 = 0 –– (ii)

16 + 25 + 8g + 10f + c1 = 0 –– (iii)

(ii) – (i) we get

– 3 – 4g + 2f = 0

4g – 2f = – 3 –– (iv)

(ii) – (iii) we get

– 40 – 8g – 8f = 0 (or)

g + f = – 5 –– (v)

Solving(iv) and (v) we get

g = – 613

, f = – 617

Substituting g and f value in equation (i) we get

4 + 4

−

613 + c1 = 0

c1 = 3

14

Now equation x2 + y2 – 3

14y

317

x3

13 +− = 0

Now circle passes through (0, c) then

c2 –3

14c

317 + = 0

3c2 – 17c + 14 = 0

⇒ (3c – 14) (c – 1) = 0 (or)

c = 1 or 3

14 .



19. Show that the locus of the point of intersection of the linesx cos ααααα + y sin ααααα = a, x sin ααααα – y cos ααααα = b (ααααα is a parameter)is a circle.

Sol. Equations of the given lines are

x cos α + y sin α = a

x sin α – y cos α = b

Let p (x1, y1) be the point of intersection

x1 cos α + y1 sin α = a ––– (1)

x1 sin α – y1 cos α = b ––– (2)

Squaring and adding (1) and (2)

(x1cos α + y1sin α)2 + (x1sin α – y1cos α)2 = a2 + b2

1122

122

1 yx2sinycosx +α+αα+α+αα 22

122

1 cosysinxsincos

– 2x1y1 cos α sin α = a2 + b2

)cos(siny)sin(cosx 2221

2221 α+α+α+α = a2 + b2

21

21 yx + = a2 + b2

Locus of p(x1, y1) is the circle

x2 + y2 = a2 + b2

20. Find the equation of the parabola whose latus rectum is theline segment of joining the points (–3, 2) and (–3, 1).

Sol. L (–3, 2) and L' (–3, 1) are the ends of the latus rectum.

S is the midpoint of LL'

Co-ordinates of S are

−

23

,3

LL' = 10)12()33( 22 +=−++− = 1

4|a| = 1, ⇒ |a| = 41

⇒ a = ± 41

Case (i) a = – 41

L(–3, 2)

S

L'(–3, 1)



Co-ordinates of A are

+−

23

,41

3

Equation of the parabola is

−+−=

−

41

3x23

y2

⇒ 4

)112x4(4

)3y2( 2 −+−=−

⇒ (2y – 3)2 = –(4x + 11)

Case (ii) a = 41

Co-ordinates of A are

−−

23

,41

3,

Equation of the parabola is

++=

−

41

3x23

y2

4)112x4(

4)3y2( 2 ++=−

i.e., (2y – 3)2 = 4x + 13.

21. Evaluate1

(x a)(x b)(x c)− − −∫ dx

Sol. Let )cx)(bx)(ax(1

−−− ≡ ax

A−

+ bx

B− +

cxC−

⇒ )cx)(bx)(ax(

1−−−

= )cx)(bx)(ax()bx)(ax(C)cx)(ax(B)cx)(bx(A

−−−−−+−−+−−

⇒ 1 = A(x – b) (x – c) + B(x – a) (x – c) + C(x – a) (x –b) –––(1)

Put x = a, we get

1 = A(a – b) (a – c) ⇒ A = )ca)(ba(1

−−



Put x = b, we get

1 = A(0) + B(b – a) (b – c) + C(0)

⇒ B = )cb)(ab(1

−−

Similarly C = )bc)(ac(1

−−

∴ )cx)(bx)(ax(

1−−−

= bx

)cb)(ab(1

ax)ca)(ba(

1

−−−+

−−−

+ cx

)bc)(ac(1

−−−

∴ ∫ −−− )cx)(bx)(ax(1

dx

= )ca)(ba(

1−−

∫ − ax1

dx + )cb)(ab(1

−− ∫ − bx1

dx

+ )bc)(ac(1

−− ∫ − cx1

dx

= )ca)(ba(1

−− log |x – a| + )cb)(ab(1

−− log |x – b|

+ )bc)(ac(1

−− log |x – c| + k

22. Evaluate cos x + 3sin x + 7cos x + sin x + 1∫ dx.

Sol. Let cosx + 3 sin x + 7

= A (cos x + sin x + 1)' + B(cos x + sin x + 1) + C

Comparing the coefficients

A + B = 1, A – B = 3, B + C = 7

A = –1, B = 2, C = 5



∫ ++++1xsinxcos7x3sinxcos

dx = – ∫ +++−

1xsinxcosxcosxsin

dx

+ 2 ∫ dx + 5 ∫ ++ 1xsinxcos1

dx

= – log |cos x + sin x + 1| + 2x + 5I ...(1)

I = ∫ ++ 1xsinxcos1

dx = ∫ +2x

cos2x

sin22x

cos2

dx2

= 21

∫+

2x

tan1

dx2x

sec2

= ∫ + t1dt

=

2x

tant

= log |1 + t| = log

+

2x

tan1

Substituting in ∫ ++++1xsinxcos7xsin3xcos

dx

= – log |cos x + sin x + 1|+ 2x + 5 log2x

tan1+ + C

23. Solve ∫4

dx

(9 x) (x 4)

9

− −

Sol. Put x = 4 cos2θ + 9 sin2θ

dx = (9 – 4) sin2θ dθ

dx = 5 sin2θ dθ

U.L.

x = 4 cos2θ + 9 sin2θ

9 = 4 cos2θ + 9 sin2θ

5 cos2θ = 0

θ = �

π



L.L

x = 4 cos2θ + 9 sin2θ

4 = 4 cos2θ + 9 sin2θ

5 sin2θ = 0

θ = 0

9 – x = 9 – (4 cos2θ + 9 sin2θ) = (9 – 4) cos2θ = 5 cos2θ

x – 4 = 4 cos2θ + 9 sin2θ – 4 = (9 – 4) sin2θ = 5 sin2θ

Let I = �

��

�� − −∫

I =

�

� �

� ��

π

θ θ∫ 5(2 sinθ cosθ)dθ

= �

��

� ��

πθ θ θθ θ∫ =

��

� � ��

� �

ππθ =∫

=

��

π =

π = π

24. Solve dxdy

= 3y 7x 73x 7y 3

− +− −

Sol. Let x = x + h, y = y + k so that dxdy

dxdy =

dxdy

= 3)ky(7)hx(37)hx(7)ky(3

−+−+++−+

= )3k7h3()y7x3()7h7k3(x7y3

−−+−+−+−

– 7h + 3k – 3 = 0

3h – 7k + 7 = 0

h k I

+3 –3 –7 3

–7 7 3 –7

2121h− =

499k+−

= 9491

−+



h = 0 and k = 1

dxdy

= y7x3x7y3

−−

Put y = vx ⇒ dxdy

= v + x. dxdv

v + x. dxdv

= )v73(x)7v3(x

−−

x. vv737v3

dxdv −

−−= =

v737v7

v73v7v37v3 22

−−=

−+−− =

v737v7 2

−−

xdx

7v7

v732

=−

−

∫ ∫∫ =−

−− x

dx

7v7

dvv7dv

7v7

322

ln x = 143

ln 1v1v

+−

– 21

ln |v2 – 1|

14 log x – log c

x = 3 log 1v1v

+−

– 7 log |v2 – 1|

⇒ 14 ln x – ln c

= 3ln (v – 1) – 3ln (v + 1) – 7ln (v + 1) – 7ln (v – 1)

14ln x – ln c = – 10 ln (v + 1) – 4 ln (v – 1)

ln (v + 1)5 + ln (v – 1)2 + ln x7 = ln c

(v + 1)5. (v – 1)2. x7 = c25

1xy

1xy

−

+ . x7 = c

(y – x)2 (y + x)5 = c

[y – (x – 1)]2 (y + x – 1)5 = c

Solution is [y – x + 1]2 (y + x – 1)5 = c.

❖ ❖ ❖ ❖ ❖

practice paper – 6 solutions -...

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