ANSWERS
I.
1) Given circle x2 + y2 +2gx +2fy -12 = 0
Given centre (-g, -f) = (2, 3)
g = -2
f = -3
r=
+
−
= √4 + 9 + 12 = 5
2) Given centre x2 + y2 = 17
Points (4,k) (2,3) are conjugate
Then S12 = 0
x1x2 + y1y2 -17 = 0
4(2)+k(3) -17 = 0
3k – 9 = 0
k=3
3) Given circles x2+y2-12x-6y+41
C1 (6,3)
r1 = √36 + 9 − 41 = 2
x2+y2+4x+6y-59 = 0
c2 (-2, -3) r2 = √4 + 9 + 59 = √72 = 6√2
d = c1c2 = √64 + 36 = 10
cosθ =
=
θ=
√
=
√
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4) Equation of parabola y2 = 8x
One entremets (x,y) = (
Other entremets (x2y2)
Since x1x2 = a2,
a=2
,2 )
x2 = 4
y1y2 = 4-a2
2y2 = -4x4
=8
ℎ = (8, -8)
y2 = -8
5) Foci (h+ae, k) = (8,2)
(h-ae, k) = (4,2)
e=2
h+ae = 8
h-ae = 4
2ae = 4
ae = 2
a(2) = 1
a=1
2h = 12
h=6
k=2
b2 = a2 (e2-1)
= 1(4-1)
=3
(
) (
(
)
)
-
−
=1
(
)
=1
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(1 +
∫
6)
(
( ( )+
Since ∫
Here f(x) = Tanx
Implies ∫
7) ∫
=t
(
dx = ∫
+
+
( )] =
f1(x) =
+
]
)
)
( )
=
( )
9x8dx = dt
x8dx = dt
∫
( )
=∫
dt
= Tan-1 t+c
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= Tan-1(x9) + c
8) ∫
Here m = 5
n =4
.
Im,n =
m is odd
……….
. .
=
. .
=
9) Given parabola y = x2
=
x = -1
Area = ∫
=[ ]
=
= −
(
x=2
)
=3
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10)
= [1 + ( ) ]
Given D.E
Cubing both sides
(
) = [1 + (
) ]
Order = 2
Degree = 3
II.
Short answers
11) Given circle x2+y2-3x+7y+14 = 0
Centre c ( ,
r=
+
)
− 14 =
=
Given x+y+1 = 0
 (1)
tr distance from ( ,
) to line (1)
d=
|
√
|
=
=
|
√
√
|
√
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should NOT follow the same. While all efforts have been made to make the guess paper available on this website as authentic as
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r = d line touches the circle.
Point at contact is foot at perpendicular
from c ( , ) to line (1)
ℎ−
=
=
3
=
2
+
h= +
=2
,
7
1
= +
2
2
k=
point at centre (2, -3)
12. Given circles S = x2+y2-8x-6y+21 = 0
= −3

(1)

(2)
S1 = x2+y2-2x-15 = 0
Equation of circle passing through P of I of (1) and (2)
S+λS1 = 0
x2+y2-8x-6y+21+λ (x2+y2-2x-15) = 0
and it is passes through (1,2)
(3)  1+4-8-12+21+λ (1+4-2-15) = 0
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6+λ (-12) = 0
λ=
=
(3)  x2+y2-8x-6y+21+ (x2+y2-2x-15) = 0
x2+y2-6x-4y+a = 0
13. Given Ellipse 9x2+16y2 = 144
+
=1
a=4
b=3
e=
e=
=
√
c (0,c)
Foci = (± ae , c) = ± (√7 , )
Length of major axis = 2a = 2x4 = 8
Minor axis = 2b = 2x3 = 6
Latasrectum =
=
=
=
Equation of directories x = ±
x= ±
√

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√7
14. Equation of Ellipse 2x2 + y2 = 8
=
= ±16
= 1  (1)
Given line x-2y-4 = 0  (2)
i.
Tangent is parallel to (2)
Slope m =
Equation Tangent y = mx ± √
y= x± 4 ×
y= x±3
+
+ 8
2y = x ± 6
x-2y ± 6 = 0
ii.
Tangent is perpendicular to (2)
Slope m = -2
Equation of Tangent y = mx ± √
y = -2x ± √16 + 8
+
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15. Given hyperbola
−
2x+y = ± √24
=1
 (1)
 (2)
Line lx+my+n = 0
Let line (2) touches the hyperbola at P (x,y)
Since equation Tangent at P (x,y) to (1) S1 = 0
−
=1
 (3)
=
Compare (2) and (3)
=
=
x1 =
=
y1 =
it is a point on line (2)
(2) l (
)+(
-a2l2+b2m2+n2 = 0
)+
=0
a2l2 – b2m2 = n2
16.
I = ∫ log(1 +
Put x = - x
)
 (1)
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I = ∫ log
1+
∫ log
∫ log
(
−
1+
) dx
dx
dx
2I = ∫ log 2dx − ∫ log(1 +
)
2I =∫ log 1
= log 2 ( - c)
I
= log 2
–
17.
=
It is linear D.E
Here P = - Tanx
I.F =
∫
Solution
y cosx = ∫
=
∫
∫
Q=
=
=
.
.
∫
∫
.
=
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guess paper published on net is for the information to the examines. this does not constitute to be a Main Question paper and
should NOT follow the same. While all efforts have been made to make the guess paper available on this website as authentic as
possible. manabadi or any staff persons will not be responsible for any loss to persons caused by any shortcoming, defect or
inaccuracy in the Guess papers provided by Manabadi.com website.
y cosx = ∫
y cosx =
III.
+
Long answer questions
18.
Given points (2,0)(0,1)(4,5)(0.C)
Let equal circle x2+y2+2gx+2fg+c = 0
 (1)
(1) passing through (2,0)
(1)  4 + 4g + c = 0
 (2)
(1)passing through (0,1)
(1)  1 + 2f +c = 0
 (3)
(1)passing through (4,5)
(1)  16 + 25 + 8g + 10f + c = 0
8g + 10f + c +41 = 0
 (4)
Solving (3) and (4)
(2) & (3)
(4) - (3) 
g+ f+5=0
(2) - (3) 
4g - 2f + 3 = 0
Solving
1
g
5
f
1
1
1
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inaccuracy in the Guess papers provided by Manabadi.com website.
-2
3
4
=
-2
=
g=
f=
put in (2)
c=
since Equation of circle (1) 
x2 + y2 -
x-
y+
3x2 + 3y2 -13x-17y +14
=0
=0
It is through (0, c)
0 + 3c2 -17c +14 = 0
(3c -14) (c-1) = 0
c=
19.
c≠1
Given circles
x2 + y2 -4x – 6y -12 = 0
 (1)
x2 + y2 + 6x + 18y + 26 = 0  (2)
(1)  c1 (2,3)
(2)  c2 (-3, -9)
r1 = √4 + 9 + 12 = 5
r2 = √9 + 81 − 26 = 8
c1c2 = √25 + 144 = 13
r1r2 = 5 + 8
= 13
c1c2 = r1r2
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implies Two circles touch each other extremely
point at contact is internal centre of similitude it divides c1c2 in the ratio r1-r2 internally
c1 (2,3)c2 (-3, -9)
r1r2 = 5 : 8
P=
(
)
( )( )
( )
Point of contact = ,
Each of Tangent at P is S1 = 0
x( )+y(
) -2 (x+ ) -3 (y-
) -12 = 0
5x + 12y +19 = 0
20.
let S is the focus and ‘l’ is the direction of parabola ‘z’ is the foot at perpendicular
from S to direction.
A is midpoint at sz such that SA = AZ = 9
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Let SA2 is x-axis and A is origin then line through A and +r to S2 is y-axis
p
M
Z
A
S (a o)
x-axis
y - axis
Focus ‘S’ on x-axis and SA = 9
Focus S = (9,c)
And direction is parallel to y-axis then its each x = -9
x+a=0
let P (x, y) is a point on parabola
by def
= 1
SP = PM
( − )+
( − )+
+
−2
= 4ax1
=
|
√
|
= (x1+a)2
+
=
+
+2
Loss of P (x,y)  y2 = 4ax
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21.
∫
Rule Nr = A
Dr + B (Dr)
2cosx + 3 sinx = A
(4cosx+5sinx) + B (4cosx+5sinx)
2cosx + 3sinx = A (-4sinx + 5 cosx) + B (4cosxy + 5sinx)
= (5A + 4B) cosx + (-4A+5B) sinx
Comparing cosx1 sinx co-efficient
5A + 4B = 2  (1)
-4A + 5B = 3  (2)
Solving
4 x (1) + 5 x (2)
20A + 16B = 8
-20A + 25 B = 15
41B = 23
B =
(1)  A =
∫
=
∫
+
∫
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inaccuracy in the Guess papers provided by Manabadi.com website.
log (4cosx+5sinx) +
22.
∫(
x+c
 (1)
√)
 x=
Put 1+x =
dx =
)-(
3 + 2x –x2 = 3 + 2 (
=
)
=
(1)  I = ∫
= -∫(4
=
=
=
(
[4
− 1)
)
=
(4
− 1) +
− 1] +
+
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23.
∫
Put sinx-cosx = t
(cosx+sinx)dx = dt
(sinx-cosx)2 = t2
1-2sinxcosx = t2
1-t2 = 2sinxcosx
= sin2x
x =  sin - cos = 0
V.L
x = 0  sin0 – cos0 = 0-1 = -1
I =
=∫
=∫
=
=
=
∫
( )
(
[
)
| |
]
(−1)(−2)
log3
3
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24.
=
(
Put x-y = V
1−
)

=
=1−
(1)  1 −
1−
(1)
=
=
=
=
Such that ∫
2
=∫
+5
+2
2+
2V + log(V+2) = x+c
1
+2
=
=
2(x-y) + log (x-y+2) = x+c
x-2y + log (x-y+2) = c
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