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Maths: II B Section A I) Very short Answer Type Questions i) Answer all questions ii) Each question carries 2 marks. 1) If the centre of the circle x2 + y2 + ax + by – 12 = 0 is (2, 3), find the values of a, b and the radius of the circle. 2) If the length of the tangent from (2, 5) to the circle x2 + y2 - 5x + 4y + K = 0 is √37 , then find K 3) Find K, if the following pair of circles are orthogonal x2 + y2 - 6x - 8y + 12 = 0 and x2 + y2 - 4x + 6y + K = 0 4) If
, 2 is one extremity of a focal chord of the parabola y2 = 8x. Find the coordinates of the other
extremity. 5) If e, e1 are the eccentricities of a hyperbola and conjugate hyperbola, prove that 6) Evaluate ∫
+
=1
dx [
7) Evaluate ∫
] (
)
dx
8) Evaluate ∫ |1 − |dx 9) Find the area of the region bounded by the curve y = x2 + 2, x – axis and the lines x = 1. X = 2 /
10) Find the order and degree of the differential equation
II
= 1+
Section – B
Short Answer type question i) Answer any 5 questions ii) Each question carries 4 marks. 11) Find the length of the chord intercepted by the circle x2 + y2 - x + 3y - 22 = 0 on the line y = x - 3 12) Find the equation to the circle cutting orthogonally to the circles x2 + y2 - 4x - 6y + 11 = 0, x2 + y2 - 10x - 4y + 21 = 0 and has 2x + 3y = 7 as diameter. 13) Find the length of the Major axis, Minor axis, latusrectum, eccentricity and equation of the directrices of the Ellipse 4x2 + y2 - 8x + 2y + 1 = 0 14) Find the equation of tangent to the ellipse 2x2 + y2 = 8 which is parallel and perpendicular to x – 2y – 4 = 0 15) S.T. Eq of normal at P (0) to the hyperbola S = 0 is + = a2 + b2 16) Find lim
→
+
+ …….+
17) Solve the differential equation xdy =
+
dx
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Section - C III ) Long Answer type questions i) Answer any 5 questions ii) Each question carries seven marks. 18) Find the Value of ‘C’ if the points (2, 0), (0, 1), (4, 5) and (0, C) are Concylic. 19) Find the direct common tangents to the circles x2 + y2 + 22x – 4x - 100= 0, x2 + y2 - 22x +4y + 100= 0 20) Find the equation of the parabola whose axis is parallel to y – axis and which passes through the points (4, 5), (-2, 11) and (-4, 21) 21) Evaluate ∫
dx (
22) Show that ∫
)
dx = log2
23) Show that ∫
dx =
√
log √2 + 1
=
24) Solve
I
SOLUTIONS
1. Given x2 + y2 + ax + by – 12 = 0 ,
Centre = a = -4
b = -6
r=
+
= (2, 3)
+ 12 =
+
+ 12
√25 = 5
2. x2 + y2 - 5x + 4y + K = 0 P (2, 5) L.T =
= √4 + 25 − 10 + 20 +
= √39 + K = -2
= √37
= √37
3. x2 + y2 - 6x - 8y + 12 = 0 x2 + y2 - 4x + 6y + K = 0 Orthogenal 2 ⃓ = 2 ⃓ = C + ⃓ 2 (-3) (-2) + 2(-4) (3) = 12 + K 12 – 24 = 12 + K K = -24
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, 2 = (x1, y1) one extrimity
4.
purcebd y2 = 8x a=2
Other extremity (x2, y2) x 1 x 2 = a2 x2 = 4
y1y2 = -4a2 2 y2 = -4(2)2
x2 = 8
= -16 y2 = -8
(8, -8)
5.
= = +
=
+
=1
6. ∫
dx = ∫ =
7. ∫
[
] (
)
+ +C
dx
=t [ ] dx = dt + (1 + x) dx = dt ∫
dt = ∫ = Tan t + C = Tan ( )+C
8. ∫ |1 − |dx = ∫ |1 − |dx + ∫ |1 − |dx = ∫ (1 − )dx - ∫ (1 − )dx =
− 1−
-
−
- (2 − 2) − 1 −
- −is not = 1 responsible for any inadvertent error that may have crept in the guess paper being manabadi.com published on NET. The guess paper published on net is for the information to the examinees. 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.
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9. A = ∫ (
+ 2)dx + 2
+ 4 -
+ 2
+2=
10.
= 1+ Order = 2 Degree = 3
II 11) x2 + y2 - x + 3y - 22 = 0 C r=
,− +
+ 22 =
=
√
line x – y – 3 = 0 d=
√
=
√
length of the chord = 2 √ =2
− −
= 4 √6
12) x2 + y2 - 4x - 6y + 11 = 0 (1) 2 2 x + y - 10x - 4y + 21 = 0 (2) 2x + 3y – 7 = 0 (3) 2 2 Cle x + y + 2gx + 2fy + 1 = 0 (4) (1) R (4) orthogonal 2g(-2) + 2f(-3) = (+1) (2) R (4) orthogonal 2g(-5) + 2f(-2) = (+2) Sub -6g + 2f = 10 Cention (3) -2g – 3f = 7 Sclug f = -1 g = -2 C = 3 cle x2 + y2 - 4x - 2y + 3 = 0 manabadi.com is not responsible for any inadvertent error that may have crept in the guess paper being published on NET. The guess paper published on net is for the information to the examinees. 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.
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13. Ellipse 4x2 + y2 - 8x + 2y + 1 = 0 4 (x2 - 2x + 1) + (y2 + 2y + 1) = 4 4 ( − 1) + ( − 1) = 4 (
)
+
(
)
=1
(ℎ, ) = (1, −1) a=1 b=2 e=
=
=
√
Major axis 2a = 2 × 2 = 4 Minor axis 2a = 2 × 1 = 2 latusrectun =
=
×
=1
y=K±
Directsicces
y = -1 ± √ y+1=±
√
14) Ellipse 2x2 + y2 = 8 +
=1
(1)
Given line x – 2y – 11 = 0
(2)
Slope of line = Equal Tangent
y = mx ± √
+
Parallel to (1)
y= x ± 4× + 8 y = x ±3
x – 2y ± 3 = 0 Perpendicular to (1) m = -2 y = 2 x ± 4(4) + 8 2x + y = ± √24
15. hyp
-
=1
Slope of Tangent = P(asec , tan ) at P ( ) =
∙ ∙
=
Slope of normal = manabadi.com is not responsible for any inadvertent error that may have crept in the guess paper being ( net ) the information to the examinees. This does Equation of normal - tanpublished = − is secfor published on NET. The guess ypaper on 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.
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16.
lim
+
→
lim →
-
=
+
=
+ +
+ …….+
∑ ∑
lim →
=∫
dx
= |log(1 + )| = log 6 - log 1 = log 6
III
Section - C
18. Given pts (2, 0), (0, 1), (4, 5) and (0, C) Cle x2 + y2 + 2gx + 2fy + 1 = 0 (I) (2, 0) ⇒ (1) ⇒ 4 + 0 + 4g + 1 = 0 (0, 1) ⇒ 0 + 1 + 2g (0) + 2f + 1 = 0 (4, 5) ⇒ 16 + 25 + 8g + 10f + 1 = 0 (1) (2) ⇒ -3 – 4g + 2f = 0 4g – 2f = -3 (4) (2) (3) ⇒ -40 – 8g – 8f = 0 g +f = -5 (5) solve (4) (5) (1) ⇒
g=
(1) (2) (3)
f=
4+4
+C=0 C=
∴ substituting (I) ⇒ x2 + y2 Hisn (0, C) ⇒
x-
y+
=0
3C2 – 17C + 14 = 0 ⇒ (3C – 14) (C – 1) = 0 C=1 : C=
19. line x2 + y2 + 22x – 4x - 100= 0 x2 + y2 - 22x +4y + 100= 0 (1) C1 = (-11, 2) C2 = (11, -2) r1 = 15 r2 = 15 r1 : r2 = 15 : 5 = 3 : 1
(1) (2)
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E.C.S = =(
)
∓
(
,
)
,
(
)
( )
)
= (22, -4) Equation of tangent (22, -4) y + u = m (x – 22) mx – y – (22m + 4) = 0 It is tangent of (1) d=r | (
)
(
)|
= 15
√
m=
,
m=
Equation of tangent 3x + 4y = 50 7x – 24y – 250 = 0
20. Eqn of parabola parallel to y – axis y = ax2 + bx + C (1) Given pts (4, 5), (-2, 11) and (-4, 21) (4, 5) ⇒ 5 = 16a + 4b + C = 0 (2) (-2, 11) ⇒ 11 = 4a - 2b + C = 0 (3) (-4, 21) ⇒ 21 = 16a - 4b + C = 0 (4) (3) (2) ⇒ 6 = -12a – 6b (4) (3) ⇒ 10 = -12a – 2b Substituting b = -2 a=½ C=5 Parabola (1) ⇒
y = x2 - 2x + 5 x2 – 2y – 4x + 10 = 0
21.
dx
∫
2 sin + 3 cos + 4 = A
(3 sin + 4 cos + 5) + B (3 sin + 4 cos + 5) + C
(1)
= A (3 cos − 4 sin ) + B (3 sin + 4 cos + 5) +C Compare 3A + 4B = 3 -4A + 3B = 2 5B + C = 4 Substituting (2) R (3)
(2) (3) (4) B=
From (2) ⇒
3A +
=3 A=
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dx =
∫ =
+
∫
log (3 sin + 4 cos + 5) + Let
∫
x +
∫
dx +
∫
dx
(6)
dx
∫ Tan
=t
Sin x = Cos x = dx = dx = ∫
∫
dx
=2 ∫
dt
= 2 ∫( =
)
dt
=
Substituting in (6)
22.
∫
(
)
dx
x = Tan dx = sec
d (
I=∫
)
∙ sec
= ∫ log(1 + Tan =
→
d
)d
-
= ∫ log 1 + Tan
−
∫ log 1 + ∫ log
d d
∫ log 2 - I = 2 2I = ∫ log 2 d 2I =
log 2
I=
log 2
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23. I = ∫ →
dx
-
I=∫
dx
= ∫
- ∫
I+I=
dx dx
∫
Tan
=t
Sin x = Cos x = dx = ∙
∫
dt
=2∫ =2∫ =2∙
√
dt (
√
log
)
√ √
= √2 log √2 + 1 I=
24.
√
log √2 + 1
=
(1)
x=x+h y=y+K =
( (
) )
(
)
(
)
( (
) )
6ℎ + 5 − 7 = 0 2ℎ + 18 − 14 = 0 Substituting h = =
K= (1)
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(1) ⇒ V+x
= =
⇒ ∫
–
dV
+
∫
= -3 ∫ dv
dx
= -3 ∫
dx
2 log ( 3V – 2) + log ( 2V + 1) = -3 log x + log C
⇒
log 3V – 2
( 2V + 1)
3V – 2
( 2V + 1) = C
3
( 2 + 1) = C
–2 –
+
(
)
= log
=C
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