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
Practice Paper – 668
Senior Inter ✦ Mathematics – IIB
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
Practice Paper – 669
Senior Inter ✦ Mathematics – IIB
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
Practice Paper – 670
Senior Inter ✦ Mathematics – IIB
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
�����
= – 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
Practice Paper – 671
Senior Inter ✦ Mathematics – IIB
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
Practice Paper – 672
Senior Inter ✦ Mathematics – IIB
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
Practice Paper – 673
Senior Inter ✦ Mathematics – IIB
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 .
Practice Paper – 674
Senior Inter ✦ Mathematics – IIB
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
+=
Practice Paper – 675
Senior Inter ✦ Mathematics – IIB
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
−+ + ∫�
��� ���� ��
��� �
Practice Paper – 676
Senior Inter ✦ Mathematics – IIB
=
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
Practice Paper – 677
Senior Inter ✦ Mathematics – IIB
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 .
Practice Paper – 678
Senior Inter ✦ Mathematics – IIB
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)
Practice Paper – 679
Senior Inter ✦ Mathematics – IIB
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
−−
Practice Paper – 680
Senior Inter ✦ Mathematics – IIB
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
Practice Paper – 681
Senior Inter ✦ Mathematics – IIB
∫ ++++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
Practice Paper – 682
Senior Inter ✦ Mathematics – IIB
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
−+
Practice Paper – 683
Senior Inter ✦ Mathematics – IIB
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.
❖ ❖ ❖ ❖ ❖