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## AP Inter 2nd Year Maths 2B Question Paper March 2015

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.

Find the value of ‘a’ if 2x^{2} + ay^{2} – 3x + 2y -1 = 0 represents a circle and also find its radius.

Solution:

Given equation is 2x^{2} + ay^{2} – 3x + 2y – 1 = 0 …….. (1)

Comparing the (1) with ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0

We get a = 2, h = 0, b = a, g = \(\frac{-3}{2}\), f = 1, c = -1

Now, equation (1) represents a circle then

a = b; 2 = a ∴ a = 2

The equation of the circle is 2x^{2} + 2y^{2} – 3x + 2y – 1 = 0

x^{2} + y^{2} – \(\frac{3}{2}\)x + y – \(\frac{1}{2}\) = 0

Comparing the above equation with x^{2} + y^{2} + 2gx + 2fy + c = 0

Question 2.

If the length of a tangent from (5, 4) to the circle x^{2} + y^{2} + 2ky = 0 is ‘1’, then find ‘k’.

Solution:

The given equation of the circle is x^{2} + y^{2} + 2ky = 0

Comparing the given equation with x^{2} + y^{2} + 2gx + 2fy + c = 0

We get g = 0, f = k, c = 0

Let, the given point P(x_{1}, y_{1}) = (5, 4)

Given that, the length of the tangent = 1

\(\sqrt{S_{11}}\) = 1 ⇒ s_{11} = 1

\(x_1^2\) + \(y_1^2\) + 2gx_{1} + 2fy_{1} + c = 1

(5)_{1} + (4)_{1} + 2(0)(5) + 2k(4) + 0 = 1

25 + 16 + 8k = 1

8k = -40

k = \(\frac{-40}{8}\) = -5

Question 3.

Find the equation of the common chord of the circles :

(x – a)^{2} + (y – b)^{2} = c^{2}, (x – b)^{2} + (y – a)^{2} = c^{2}, (a ≠ b)

Solution:

Given equations of the circles are S ≡ (x – a)^{2} + (y – b)^{2} = c^{2} = 0

x^{2} + a^{2} – 2ax + y^{2} + b^{2} – 2by – c^{2} = 0

x^{2} + y^{2} – 2ax – 2by + a^{2} + b^{2} – c^{2} = 0

S’ ≡ (x – b)^{2} + (y – a)^{2} = c^{2}

x^{2} + b^{2} – 2bx + y^{2} + a^{2} – 2ay – c^{2} = 0

x^{2} + y^{2} – 2bx – 2ay + a^{2} + b^{2} – c^{2} = 0

The equation of the common chord (radical axis) of the given circles is S – S’= 0

Question 4.

Find the co-ordinates of the points on the parabola: y^{2} = 2x whose focal distance is \(\frac{5}{2}\).

Solution:

Given equation of the parabola is y^{2} = 2x

Comparing with y^{2} = 4ax we get

4a = 2 ⇒ a = \(\frac{1}{2}\)

Let, P(x_{1}, y_{1}) be a point on the parabola y^{2} = 2x

Given that, focal distance = \(\frac{5}{2}\)

x_{1} + a = \(\frac{5}{2}\) ⇒ x_{1} + \(\frac{1}{2}\) = \(\frac{5}{2}\)

⇒ x_{1} = \(\frac{5}{2}\) – \(\frac{1}{2}\) ⇒ x_{1} = 2

Since P(x_{1}, y_{1}) lies on the parabola y^{2} = 2x then

\(y_1^2\) = 2x_{1} ⇒ \(y_1^2\)2(2) ⇒ \(y_1^2\) = 4 ⇒ y_{1} = ±2

∴ The required points are (2, 2), (2, -2).

Question 5.

Define rectangular hyperbola and find its eccentricity.

Solution:

If in a hyperbola the length of the transverse axis (2a) is equal to the length of the conjugate axis (2b), the hyperbola is called a rectangular hyperbola its equation is x^{2} – y^{2} = a^{2} [∵ a = b]

In this case e^{2} = \(\frac{a^2+b^2}{a^2}\) = \(\frac{2 a^2}{a^2}\) = 2 ⇒ e = \(\sqrt{2}\)

The eccentricity of a rectangular hyperbola is \(\sqrt{2}\).

Question 6.

Find \(\int \frac{e^x(1+x \log x)}{x}\)dx

Solution:

Question 7.

Find \(\int \frac{\sin \left(\tan ^{-1} x\right)}{1+x^2}\)dx, x ∈ R

Solution:

Put tan^{-1}x = t then

\(\frac{1}{1+x^2}\)dx = dt

Now, \(\int \frac{\sin \left(\tan ^{-1} x\right)}{1+x^2}\)dx = ∫sin t dt

= -cos t + c

= -cos(tan^{-1}x) + c

Question 8.

Evaluate : \(\int_0^{\pi / 2}\)sin^{5}xcos^{4}xdx

Solution:

Given \(\int_0^{\pi / 2}\)sin^{5}xcos^{4}xdx

= \(\frac{4+1}{5+4}\).\(\frac{4-3}{5+4-2}\).\(\frac{5-1}{5}\).\(\frac{5-3}{5-2}\)

= \(\frac{5}{9}\).\(\frac{1}{7}\).\(\frac{4}{5}\).\(\frac{2}{3}\) = \(\frac{8}{315}\)

Question 9.

Evaluate \(\int_0^2\)|1-x|dx

Solution:

Question 10.

From the differential equation corresponding to

y = A cos 3x + B sin 3x, where A and B are parameters.

Solution:

Given y = A cos 3x + B sin 3x

then \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = – 3A sin 3x + 3B cos 3x

\(\frac{d^2 y}{d x^2}\) = – 9A cos 3x – 9B sin 3x

= – 9(A cos 3x + B sin 3x) = -9y

∴ \(\frac{d^2 y}{d x^2}\) + 9y = 0 where A, B are eliminated.

Section – B

(5 × 4 = 20)

II. Short Answer type questions.

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

Question 11.

Find the equation of circle whose centre lies on the x-axis and passing through (-2, 3) and (4, 5).

Solution:

Let, the equation of the required circle is

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

Center of (1) c(h, k) = (-g, -f) lies on the x-axis then – k = 0 ⇒ f = 0

Since, (1) passes through the point (-2, 3) then

(-2)^{2} + (3)^{2} + 2g(-2) + 2f(3) + c = 0

4 + 9 + 4g – 6f + c = 0

4g + 6f + c = -13

If f = 0, then

-4g + 6(0) + c + 13 = 0

-4g + c + 13 = 0 ………….. (2)

Since, (1) passes through the point (4, 5) then

(4)^{2} + (5)^{2} + 2g(4) + 2f(5) + c = 0

16 + 25 + 8g + 10f + c = 0

8g + 10f + c +41 = 0

If f = 0 then 8g + 10(0) + c + 41 = 0

8g + c + 41 = 0 ……….. (3)

Solve (2) and (3)

Now, substitute the values of g, f, c in (1)

x^{2} + y^{2} + 2g\(\left(\frac{-7}{3}\right)^x\) + 2(0)y – \(\frac{67}{3}\) = 0

3x^{2} + 3y^{2} – 14x – 67 = 0

∴ The equation of the required circle is 3x^{2} + 3y^{2} – 14x – 67 = 0

Question 12.

If x + y = 3 is the equation of the chord AB of the circle :

x^{2} + y^{2} – 2x + 4y – 8 = 0, find the equation of the circle having AB as diameter.

Solution:

Given equation of the straight line is L ≡ x + y – 3 = 0.

Given equation of the circle is g ≡ x^{2} + y^{2} – 2x + 4y – 8 = 0

Equation of the circle passing through A,B is S + λL = 0

x^{2} + y^{2} – 2x + 4y – 8 + λ, (x + y – 3) = 0 ………. (1)

x^{2} + y^{2} – (2 – λ)x + (4 + λ)y – (3λ + 8) = 0

Here g = \(\frac{(2-\lambda)}{2}\), f = \(\frac{4+\lambda}{2}\)

Centre of (1) s C = (-g, -f) = \(\left(\frac{2-\lambda}{2},-\frac{4+\lambda}{2}\right)\)

If \(\overline{\mathrm{AB}}\) is a diameter of circle (1) then ‘C’ lies on x + y – 3 = 0.

\(\frac{2-\lambda}{2}\) + \(\frac{-4+\lambda}{2}\) – 3 = 0

2 – λ – 4 – λ – 6 = 0 ⇒ 2A = -8 ⇒ λ = -4

∴ The equation of the circle on \(\overline{\mathrm{AB}}\) as a diameter is

x^{2} + y^{2} – 2x + 4y – 8 – 4(x + y – 3) = 0

⇒ x^{2} + y^{2} – 6x + 4 = 0

Question 13.

Find the equation of tangent and normal to the ellipse 9x^{2} + 16y^{2} = 144 at the end of the latus rectum in the first quadrant.

Solution:

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 equation of the ellipse is x^{2} +3y = 3

\(\frac{x^2}{3}+\frac{3 y^2}{3}\) = 1 ⇒ \(\frac{x^2}{3}+\frac{y^2}{1}\) = 1

Here a^{2} = 3, b^{2} = 1

Given equation of the straight line is 4x + y + k = 0 …………. (1)

y = -4x – k

Comparing with y = mx + c we get m = -4, c = -k

Since eq. (1) is a tangent to the given ellipse then

c^{2} = a^{2}m^{2} + b^{2}

(-k)^{2} = 3(-4)^{2} + 1 ⇒ k^{2} = 49 ⇒ k = ±1

Question 15.

Find the equations of the tangents to the hyperbola :3x^{2} – 4y^{2} = 12 which are :

- Parallel and
- Perpendicular to the line : y = x – 7.

Solution:

Question 16.

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

Solution:

Question 17.

Solve the differential equation : (xy^{2} + x) dx + (yx^{2} + y) dy = 0.

Solution:

Section – C

(5 × 7 = 35)

III. Long Answer type questions.

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

Question 18.

If (2, 0), (0,1), (4, 5) and (0, C) are concyclics then find ‘C’.

Solution:

Let, the equation of the required circle is

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

Since (1) passes through the point (2, 0) then

(2)^{2} + (0)^{2} + 2g(2) + 2f(0) + 0 = 0

4 + 4g + c = 0 ⇒ 4g + c = -4 ……… (2)

Since (1) passes through the point (0, 1) then

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

1 + 2f + c = 0 ⇒ 2f + c = -1 ……. (3)

Since, (1) passes through the point (4, 5) then

(4)^{2} + (5)^{2} + 2g(4) + 2f(5) + c = 0

16 + 25 + 8g + 10f + c = 0

8g + 10f + c = – 41 ……….. (4)

Substitute the point (0, c) in (7)

3(0)^{2} + 3c^{2} – 13(0) – 17c + 14 = 0

3c^{2} – 17c + 14 = 0

3c^{2} – 3c – 14c + 14 = 0

3c(c – 1) – 14(c – 1) = 0

(c – 1) (3c – 14) = 0

c – 1 = 0 3c – 14 = 0

c = 1 c = \(\frac{14}{3}\)

∴ c = 1 or \(\frac{14}{3}\)

Question 19.

Find the transverse common tangents of the circles : x^{2} + y^{2} – 4x – 10y + 28 = 0 and x^{2} + y^{2} + 4x – 6y + 4 = 0.

Solution:

∴ Given circle are each circle lies completely outside the other.

Let, A_{1} be the internal centre of similitude.

The internal centre of similitude A_{1} divides C_{1}C_{2} in the ratio r_{1} : r_{2} (1 : 3) internally.

∴ The internal centre of similitude

The equation to the pair of transverse common tangents is SS_{11} = \(S_1^2\)

(x_{2} + y_{2} – 4x – 10y + 28) = (-2x – y + 7)_{2}

x_{2} + y_{2} – 4x – 10y + 28 = 4x_{2} + y_{2} + 4g + 4xy – 14y – 28x

3x_{2} + 4xy – 24x – 4y + 21 = 0

consider 3x_{2} + 4xy = 0

x(3x + 4y) = 0

x = 0 and 3x + 4y = 0

∴ 3x^{2} + 4xy – 24x – 4y + 21 = (x + 1)(3x + 4y + k)

Comparing coefficient of x on both sides we get

k + 3l = -24 …… (3)

Comparing coefficient of x on both sides we get

4l = -4

l = -1

Substitute the value of l in (3)

k + 3(-1) = -24

k – 3 = -24

k = -24 + 3

k = -21

∴ The equation of the transverse common tangents are x – 1 = 0 and 3x + 4y – 21 = 0

Question 20.

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

Solution:

Let,

2cosx + 3sinx = λ\(\frac{\mathrm{d}}{\mathrm{dx}}\)(4cosx + 5sinx) + μ(4cosx + 5sinx x) + γ

= λ(-4sinx + 5cosx) + μ(4cos x + 5sin x) + γ …….. (1)

= 4λ.sinx + 5λcosx + 4μcosx + 5μsinx + γ

Comparing the coefficients of cos x on both sides, we get

5λ + 4μ = 2 ………. (2)

Comparing the coefficients of sinx on both sides, we get

-4λ + 5μ = 3 ………. (3)

Question 21.

Obtain reduction formula : ∫tan^{n}x dx for integer n ≥ 2 and evaluate: ∫tan^{6}xdx.

Solution:

Question 22.

Derive the standard form of the parabola.

Solution:

The equation of a parabola in the standard form is y^{2} = 4ax.

Let, ‘S’ be the focus and L = 0 be the directrix of the parabola.

Let, ‘P’ be a point on the parabola.

Let, M, Z be the projections (Foot of the ⊥rs) of P, S on the directrix,

L = 0 respectively. .

Let, N be the projection of P and SZ.

Let, A be the mid point of SZ.

Since, SA = AZ, A lies on the parabola.

Let, AS = a. Take AS, the principle axis of the parabola as X-axis and AY perpendicular to SZ as y-axis. ,

Then, S = (a, 0) and the parabola is in the standard form.

Let, P = (x, y)

Now, PM = NZ = AN + AZ = x + a

The equation to the parabola is

Question 23.

Evaluate \(\int_0^\pi \frac{x \sin x}{1+\sin x}\)dx

Solution:

Question 24.

Solve : (1 + y^{2})dx = (tan^{-1}y – x)dy.

Solution: