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## AP Inter 2nd Year Maths 2B Question Paper March 2016

Time : 3 Hours

Max. Marks : 75

Note: This question paper contains three sections A, B and C.

Section – A

(10 × 2 = 20)

I. Very Short Answer Type Questions.

- Attempt ALL questions.
- Each question carries TWO marks.

Question 1.

If the circle x^{2} + y^{2} – 4x + 6y + a = 0 has radius 4, then find a.

Solution:

Given circle x^{2} + y^{2} – 4x + 6y + a = 0

Radius ⇒ \(\sqrt{2^2+3^2-a}\) = 4

⇒ \(\sqrt{13-a}\) = 4 ⇒ a = -3.

Question 2.

Obtain the parametric equations of the circle (x – 3)^{2} + (y – 4)^{2} = 8^{2}.

Solution:

From the given equation,

centre C(h, k) = (3, 4) and radius r = 8.

∴ The parametric equations of the circle (x – 3)^{2} + (y – 4)^{2} = 8^{2} are

x = 3 + 8 cos θ.

y = 4 + 8 sinθ.

Question 3.

Find the value of k, if the circles x^{2} + y^{2} + 4x + 8 = 0 and x^{2} + y^{2} – 16y + k = 0 are orthogonal.

Solution:

The circles x^{2} + y^{2} + 4x + 8 = 0 and x^{2} + y^{2} – 16y + k = 0 cut each other orthogonally

∴ 2(gg’ + ff) = e + c’

⇔ 2 × 2 × 0 + 2(0) (-8) = k + 8

⇔ k = -8.

Question 4.

Find the coordinates of the points on the parabola y^{2} = 8x whose focal distance is 10.

Solution:

The focal distance of a point on the parabola y^{2} = 4ax is | x_{1} + a |

Here a = 2

∴ |x_{1} + 2| = 10

⇒ x_{1} = 8

⇒ \(y_1^2\) = 64

⇒ y_{1} = ± 8.

∴ Co-ordinates of the points (8, ±8).

Question 5.

If the eccentricity of a hyperbola is \(\frac{5}{4}\), then find the eccentricity of its conjugate hyperbola.

Solution:

Let e and e_{1} be the eccentricities of the hyperbola and its conjugate respectively

Question 6.

Evaluate \(\int \frac{d x}{\cosh x+\sinh x}\) on R.

Solution:

Question 7.

Evaluate \(\int \frac{x^8}{1+x^{18}}\) on R.

Solution:

Question 8.

Find \(\int_{-\pi / 2}^{\pi / 2}\)sin^{2}xcos^{4}x dx

Solution:

Question 9.

Evaluate \(\int_0^\pi \sqrt{2+2 \cos \theta}\) dθ

Solution:

Question 10.

Find the order and degree of the differential equation

\(\left[\frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^3\right]^{6 / 5}\) = 6y.

Solution:

The given differential equation

∴ The order and degree of the equation are 2 and 1.

Section – B

(5 × 4 = 20)

II. Short Answer Type Questions.

- Attempt ANY FIVE questions.
- Each question carries FOUR marks.

Question 11.

Find the pole of 3x + 4y – 45 = 0 with respect to the circle x^{2} + y^{2} – 6x – 8y + 5 = 0.

Solution:

Let (x_{1}, y_{1}) be the pole of 3x + 4y – 45 = 0 ……… (1)

The equation of polar of (x_{1}, y_{1}) w.r.t the circle

x^{2} + y^{2} – 6x – 8y + 5 = 0 is

xx_{1} + yy_{1} – 3(x + x_{1}) -4 (y + y_{1}) + 5 = 0

⇒ x(x_{1} – 3) + y(y_{1} – 4) – (3x_{1} + 4y_{1} – 5) = 0 ………… (2)

From (1) and (2)

\(\frac{x_1-3}{3}\) = \(\frac{y_1-4}{4}\) = \(\frac{-\left(3 x_1+4 y_1-5\right)}{-45}\)

\(\frac{x_1-3}{3}\) = \(\frac{y_1-4}{4}\)

⇒ 4x_{1} -12 = 3y_{1} -12 ⇒ 4x_{1} – 3y_{1} = 0 …….. (3)

\(\frac{y_1-4}{4}\) = \(\frac{3 x_1+4 y_1-5}{45}\)

⇒ 45y_{1} – 180 = 12x_{1} + 16y_{1} – 20

⇒ 12x_{1} – 29y_{1} + 160 = 0 ………. (4)

Solving (3) and (4) we get 12xx – 9yx = 0

12x_{1} – 29y_{1} = -160

∴ 20y_{1} = 160 ⇒ y_{1} = 8 and from (3) 4x_{1} = 24 ⇒ x_{1} = 6

∴ Pole of 3x + 4y – 45 = 0 w.r.t x^{2} + y^{2} – 6x – 8y + 5 = 0 is (6, 8).

Question 12.

Find the equation of the circle which cuts the circles

x^{2} + y^{2} – 4x – 6y +11 = 0 and x^{2} + y^{2} – 10x – 4y + 21 = 0 orthogo nally and has the diameter along the straight line 2x + 3y = 7.

Solution:

Equations of given circle are

x^{2} + y^{2} – 4x – 6y + 11 = 0 ……….. (1)

x^{2} + y^{2} – 10x – 4y + 21 = 0 ………… (2)

Let the equation of the required circle be

x^{2} + y^{2} + 2gx + 2fy + c = 0

If this cuts (1) and (2) orthogonally then

2g(-2) + 2f (-3) = c + 11 ⇒ 4g + 6g = -c – 11 ………. (3)

and if (1) and (3) are orthogonal then

2g(-5) + 2f (-2) = c + 21

⇒ -10g – 4f = c + 21

⇒ 10g + 4f = -c – 21 ………(4)

from (3) and (4)

-6g + 2f = 10

⇒ 6g – 2f = -10 ⇒ 3g – f = -5 ……. (5)

Since the centre of circle (1) (-g, – f) lies on the diameter

2x + 2y = 7, we have

2(-g) + 3 (-f) = 7 ⇒ 2g + 3f = -7 ……. (6)

Solving (5) and (6)

Question 13.

Show that the points of intersection of the perpendicular tan gents to an ellipse lie on a circle.

Solution:

The equation of the ellipse \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1 (a > b). Any tangent to it in the slope intercept form is

y = mx ± \(\sqrt{a^2 m^2+b^2}\)

Let the perpendicular tangents intersect at P(x_{1}, y_{1})

P lies on (1) for some real m,

⇒ y_{1} = mx_{1} ± \(\sqrt{a^2 m^2+b^2}\)

∴ (y_{1} – mx_{1})^{2} = a^{2}m^{2} + b^{2} (or)

(\(x_1^2\) – a^{2}) m^{2} – 2x_{1}y_{1}m + (\(y_1^2\) – b^{2}) =0 being a quadratic equation in ‘m’ has two roots say m_{1} and m_{2} then m_{1}, m_{2} are the slopes of tangents from P to the ellipse.

If, however, one of the perpendicular tangents in Vertical, then such pair of perpendicular tangents intersect at one of the points (±a, ±b) and any of these points satisfies x^{2} + y^{2} = a^{2} + b^{2}

∴ The point of intersection of perpendicular tangents to the ellipse S = 0 lies on the circle x^{2} + y^{2} = a^{2} + b^{2}.

Question 14.

Find the value of k if 4x + y + k = 0 is a tangent to the ellipse x^{2} + 3y^{2} = 3.

Solution:

Given ellipse is x^{2} + 3y^{2} = 3

Hence a^{2} = 3 and b^{2} = 1

Equation of the given line is 4x + y + k = 0

⇒ y = -4x – k where m = -4 and c = -k the condition for tangency is c^{2} = a^{2}m^{2} + b^{2}

⇒ k^{2} = 3 (16) + 1

⇒ k^{2} = 49

⇒ k = ±7.

Question 15.

Find the center, foci, eccentricity, equation of directrices to the hyperbola x^{2} – 4y^{2} = 4.

Solution:

x^{2} – 4y^{2} = 4

this can be written as \(\frac{x^2}{4}\) – \(\frac{y^2}{1}\) = 1

⇒ a^{2} = 4 and b^{2} = 1

⇒ a = 2 and b = 1

Question 16.

Evaluate \(\int_0^{\pi / 2} \frac{d x}{4+5 \cos x}\)

Solution:

Question 17.

Solve the differential equation

Solution:

The equation can be written as

Section – C

III. Long Answer Type Questions.

- Attempt ANY FIVE questions.
- Each question carries SEVEN marks.

Question 18.

Find the equation of a circle which passes through (4, 1), (6, 5) and having the center on 4x + 3y – 24 = 0.

Solution:

Let the equation of the required circle be

x^{2} + y^{2} + 2gx + 2fy + c = 0 …….. (1)

The centre of the circle is (-g, -f)

since (1) passes through (4, 1) we have

4^{2} + 1^{2} + 2g(4) + 2f(1) + c = 0

⇒ 8g + 2f + c = -17 ………. (2)

Also since (1) passes through the point (6, 5) we have

6^{2} + 5^{2} + 2g(6) + 2f(5) + c = 0

⇒ 12g + 10f + c = -61 ……. (3)

Given that the centre (-g, -f) lies on the line 4x + 3y – 24 = 0

we have -4g – 3f – 24 = 0

⇒ 4g + 3f = -24 ………….. (4)

from (2) and (3)

8g + 2f + c = – 17

12g + 10f + c = – 61

We have -4g – 8f – 44 = 0

⇒ g + 2f + 11 = 0 …………. (5)

Solving (4) and (5)

substituting in (2) we get

8(-3) + 2(-4) – c = -17

⇒ -24 – 8 + c = -17

⇒ c = -17 + 32 = 15

∴ from (1) the equation of the required circle is

x^{2} + y^{2} + 2 (-3) x + 2 (-4) y + 15 = 0

⇒ x^{2} + y^{2} – 6x – 8y + 15 = 0.

Question 19.

Show that the circles x^{2} + y^{2} – 6x – 2y + 1 = 0, x^{2} + y^{2} + 2x – 8y + 13 = 0 touch each other. Find the point of contact and the equation of common tangent at their point of contact.

Solution:

Let S ≡ x^{2} + y^{2} – 6x – 2y + 1 = 0 and

S’ ≡ x^{2} + y^{2} + 2x – 8y + 13 = 0 be the given circles then centre of circles are C_{1} = (3, 1) and C_{2} = (-1, 4)

radius of circles are r_{1} = \(\sqrt{9+1-1}\) = 3

and r_{2} = \(\sqrt{1+16-13}\) = 2

Distance between centres

C_{1}C_{2} = \(\sqrt{(3+1)^2+(1-4)^2}\) = \(\sqrt{16+9}\) = 5 and r_{1} + r_{2} = 3 + 2 = 5

Since C_{1}C_{2} = r_{1} + r_{2} the two circles touch externally at a point.

Let P be the point of contact of circles such that r_{1} : r_{2} = 3 : 2

Since P divides C_{1}, C_{2} internally in the ratio 3 : 2 coordinates of point of contact

The equation of common tangent is S – S’ = 0

⇒ x^{2} + y^{2} – 6x – 2y + 1 – x^{2} – y^{2} – 2x + 8y + 13 = 0

⇒ -8x + 6y – 12 = 0

⇒ 4x – 3y + 6 = 0

∴ The equation of the common tangent at the point of contact is 4x – 3y + 6 = 0.

Question 20.

Show that the common tangent to the parabola y^{2} = 4ax and x^{2} = 4by is x\(a^{\frac{1}{3}}\) + y\(b^{\frac{1}{3}}\) + \(a^{\frac{2}{3}}\)\(a^{\frac{2}{3}}\) =0.

Solution:

The equations of the parabola are

y^{2} = 4ax …….. (1)

x^{2} = 4by ………… (2)

Equation of any tangent to (1) is of the form y = mx + \(\frac{a}{m}\) …. (3)

If the line (3) is a tangent to (2) also, the points of intersection of (2) and (3) coincide.

Substituting the value of y from (3) in (2) we get

x^{2} = 4b(mx + \(\frac{a}{m}\))

⇒ mx^{2} – 4bm^{2}x – 4ab = 0

Which should have equal roots.

∴ Its discriminant is zero. Hence

16b^{2}m^{4} – 4m(-4ab) = 0

16b (bm^{4} + am) = 0

m(bm^{3} + a) = 0. But m ≠ 0

∴ m = \(\frac{-a^{1 / 3}}{b^{1 / 3}}\) substituting in (3) the equation of the common tangent becomes y = \(\left(\frac{a}{b}\right)^{1 / 3}\) + \(\frac{a}{\left(-\frac{a}{b}\right)^{\frac{1}{3}}}\) (or)

a^{1/3}x + b^{1/3}y + a^{2/3}b^{2/3} = 0.

Question 21.

Evaluate \(\int \frac{2 \sin x+3 \cos x+4}{3 \sin x+4 \cos x+5}\)dx.

Solution:

\(\int \frac{2 \sin x+3 \cos x+4}{3 \sin x+4 \cos x+5} d x\)

Since there exists constants in both numerator and denominator, we determine constant A, B and C such that

2 sin x + 3 cos x + 4

= A\(\frac{d}{d x}\)(3 sin x + 4 cos x + 5) + g(3 sin x + 4 cos x + 5) + c

= A(3 cos x – 4 sin x) + B(3 sin x + 4 cos x + 5) + c …… (1)

Comparing both sides the coefficients of sin x, cos x and constants

-4A + 3B = 2

⇒ 4A – 3B + 2 = 0 ………. (2)

⇒ 3A + 4B – 3 = 0 ……….. (3)

⇒ 5B + c – 4 = 0 ……… (4)

Solving (2) and (3)

Question 22.

Obtain the reduction formula for I_{n} = ∫cot^{n} x dx, n being a positive integer, n ≥ 2 and deduce the value of ∫cot^{4}x dx.

Solution:

Question 23.

Evaluate \(\int_0^{\pi / 4}\)log(1 + tan x)dx.

Solution:

Question 24.

Solve the differential equation (x^{2} + y^{2}) dx = 2xy dy.

Solution:

(x^{2} + y^{2}) dx = 2xy dy

The given equation can be written as

\(\frac{d y}{d x}\) = \(\frac{x^2+y^2}{2 x y}\) ……… (1)

Which is a homogeneous equation, since the numerator and denominator on the right are homogeneous functions each of degree 2.