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## AP Inter 2nd Year Maths 2B Question Paper March 2020

Time : 3 Hours

Max. Marks : 75

Note : This question paper consists of 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 other end of the diameter of the circle x^{2} + y^{2} – 8x – 8y + 27 = 0 if one end of it is (2, 3).

Solution:

A = (2, 3) and AB is the diameter of the circle x^{2} + y^{2} – 8x – 8y + 27 = 0

Question 2.

Define chord of contact and find the chord of contact of (1, 1) to the circle x^{2} + y^{2} = 9.

Solution:

Equation of the circle is x^{2} + y^{2} = 9

Equation of the chord of contact is

x.1 + y.1 = 9

i.e. x + y = 9

Question 3.

Find k if the circles x^{2} + y^{2} – 5x – 14y – 34 = 0 and x^{2} + y^{2} + 2x + 4y + k = 0 are orthogonal.

Solution:

g_{1} = \(\frac{-5}{2}\) ;

f_{1} = -7 ;

c_{1} = -34

g_{2} = 1 ; f_{2} = 2 ; c_{2} = k

Two circles are said to be orthogonal

2g_{1}g_{2} + 2f_{1}f_{2} = c_{1} + c_{2}

2\(\left(\frac{-5}{2}\right)\)(1) + 2(-7)(2) = -34 + k

-5 – 28 = -34 + k

– 33 = – 34 + k

k = 34 – 33 ⇒ k = 1

Question 4.

Find the equation of the parabola whose vertex is (3, -2) and focus is (3, 1).

Solution:

The abcissae of the vertex and focus are equal to 3. Hence the axis of the parabola is x = 3, a line parallel to y-axis. focus is above the vertex.

a = distance between focus and vertex = 3.

∴ Equation of the parabola

(x – 3)^{2} = 4(3) (y + 2)

i.e.,(x – 3)^{2} = 12(y + 2)

Question 5.

If 3x – 4y + k = 0 is a tangent to the hyperbola x^{2} – 4y^{2} = 5, find the value of k.

Solution:

Equation of the hyperbola x^{2} – 4y^{2} = 5

\(\frac{x^2}{5}\) – \(\frac{y^2}{\left(\frac{5}{4}\right)}\) = 1

a^{2} = 5, b^{2} = \(\frac{5}{4}\)

Equation of the given line is 3x – 4y + k = 0

4y = 3x + k

y = \(\frac{3}{4}\)x + \(\frac{k}{4}\)

m = \(\frac{3}{4}\) c = \(\frac{k}{4}\),

condition for tangency is c^{2} = a^{2}m^{2} – b^{2}

\(\frac{k^2}{16}\) = 5.\(\frac{9}{16}\) – \(\frac{5}{4}\)

k^{2} = 45 – 20 = 25

k = ±5

Question 6.

Evaluate : ∫\(\frac{\cos x}{(1+\sin x)^2}\)dx.

Solution:

∫\(\frac{\cos x d x}{(1+\sin x)^2}\)

t = 1 + sin x ⇒ dt = cos x dx

\(\int \frac{\cos x d x}{(1+\sin x)^2}\) = \(\int \frac{d t}{t^2}\) = –\(\frac{1}{t}\) + C

= –\(\frac{1}{1+\sin x}\) + C

Question 7.

Evaluate : ∫x log x dx on (0, ∞).

Solution:

We take the First function U = log x abd

Second Function V = X

From the “By parts rule” we have

Question 8.

Evaluate : \(\lim _{n \rightarrow \infty} \frac{1+2^4+3^4+\ldots \ldots \ldots+n^4}{n^5}\)

Solution:

Question 9.

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

Solution:

Question 10.

Solve: y(1 + x)dx + x(1 + y)dy = 0.

Solution:

The given equation can be written as

\(\frac{(1+x)}{x}\).dx + \(\frac{(1+y)}{y}\).dy = 0

\(\int\left(1+\frac{1}{x}\right)\) dx + \(\int\left(1+\frac{1}{y}\right)\) dy = 0

x + log x + y + log y = c

x + y + log (xy) = c is the required solution.

Section – B

(5 × 4 = 20 Marks)

II. Short Answer Type questions.

- Attempt any five questions.
- Each question carries four marks.

Question 11.

Find the area of the triangle formed by the tangent at P(x_{1}, y_{1}) to the circle \(x_1^2\) + \(y_1^2\) = a^{2} with the co-ordinate axes where x_{1}y_{1} ≠ 0.

Solution:

Equation of the circle is x^{2} + y^{2} = a^{2}

Equation of the tangent at p(x_{1}, y_{1}) is

xx_{1} + yy_{1} = a^{2} ……… (1)

This tangent cuts X – axis at A and Y – axis at B

Changing into intercept form

Question 12.

If the two circles x^{2} + y^{2} + 2gx + 2fy = 0 and x^{2} + y^{2} + 2g’x + 2fy = 0 touch each other then show that f’g = fg’.

Solution:

Question 13.

S and T are the foci of an ellipse and B is one end of the minor axis. If STB is an equilateral triangle, then find die eccentricity of the ellipse.

Solution:

Let \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1 (a > b) be an ellipse whose foci are S and T, B is an end of the minor axis such that STB is equilateral, then SB = ST = TB. We have S(ae,0).

T = (-ae, 0) and B(0, b)

Consider SB = ST ⇒ (SB)^{2} = (ST)^{2} ⇒ (ae)^{2} + b^{2} = 4a2e^{2}

∴ a^{2}e^{2} + a^{2} (1 – e)^{2} = 4a^{2}e^{2}

[∵ b^{2} = a^{2}(1 – e^{2})]

e^{2} = \(\frac{1}{4}\)

∴ Eccentricity of the ellipse is \(\frac{1}{2}\).

Question 14.

Find the condition for the line x cos α + y sin α = P to be a tangent to the ellipse \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1.

Solution:

Equation of the ellipse is

\(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1 ……. (1)

Equation of the line is x cos α + y sin α = p

y sin α = -x cos α + p

Question 15.

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

Solution:

Equation of the hyperbola is \(\frac{x^2}{4}\) – \(\frac{y^2}{1}\) = 1

a^{2} = 4, b^{2} = 1

Centre is c (0,0)

a^{2}e^{2} = a^{2} + b^{2} = 4 + 1 = 5

ae = \(\sqrt{5}\)

Foci are (±ae, 0) = (±\(\sqrt{5}\), 0)

Eccentricity = \(\frac{\mathrm{ae}}{\mathrm{a}}\) = \(\frac{\sqrt{5}}{2}\)

Equations of directrices are x = ±\(\frac{a}{e}\)

= ±2. \(\frac{2}{\sqrt{5}}\)

⇒ \(\sqrt{5}\)x = ±4

⇒ \(\sqrt{5}\)x ± 4 = 0

Length of the latus rectum = \(\frac{2 b^2}{a}\) = \(\frac{2.1}{2}\)

Question 16.

Find the area of the region bounded by the parabolas y^{2} = 4x and x^{2} = 4y.

Solution:

Equations of the given curves are

y^{2} = 4x ……… (1)

x^{2} = 4y = 4y …….. (2)

\(\left(\frac{x^2}{4}\right)^2\) = 4x

\(\frac{x^4}{16}\) = 4x

x^{4} = 64 ⇒ x^{4} = 0 or x^{3} = 64, x = 4

Question 17.

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

Solution:

Given equation can be written as

\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{x^2+y^2}{2 x y}\)

This is a homogeneous function.

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) and (6,5) and having the centre on 4x + 3y – 24 = 0.

Solution:

Equation of circle be

x^{2} + y^{2} + 2gx + 2fy + c = 0 passes through (4, 1) and (6, 5) then

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

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

Centre lie on 4x + 3y – 24 = 0

∴ 4(-g) + 3 (-f) – 24 = 0 ……….. (iii)

(ii) – (i) we get

44 + 4g + 8f = 0 ………. (iv)

Solving (iii) and (iv) we get f = -4, g = -3, c = -15

∴ Required equation of circle is x^{2} + y^{2} – 6x – 8y + 15 = 0

Question 19.

Find the equation of the circle which touches the circle x^{2} + y^{2} – 2x – 4y – 20 = 0 externally at (5, 5) with radius 5.

Solution:

x^{2} + y^{2} – 2x – 4y – 20 = 0

C = (1, 2)

h = 9 k = 8

Equation of circle be

(x – 9)^{2} + (y – 8)^{2} = 25

x^{2} + y^{2} – 18x – 16y + 120 =0

If (h, k) is the centre of the required circle (5, 5) is the mid-point of (1, 2) and (h, k).

Question 20.

From an external point’P tangents are drawn to the parabola y^{2} = 4ax and these tangents make angles θ_{1}, θ_{2} with its axis, such that tan θ_{1} + tan θ_{2} is a constant b. Then show that P lies on the line y = bx.

Solution:

Let the coordinates of P be (x_{1}, y_{1}) and the equation of the parabola y_{2} = 4ax. Any tangent to the parabola is y = mx + \(\frac{a}{m}\), if this passes through (x_{1}, y_{1}) then

y_{1} = mx_{1} + \(\frac{\mathrm{a}}{\mathrm{m}}\)

i.e., m^{2}x_{1} – my_{1} + a = 0 ………….. (1)

Let the roots of (1) be m_{1}, m_{2}.

[∴ The tangents make angles θ_{1}, θ_{2} with its axis (x-axis) then their slopes m_{1} = tan θ_{1}, and m_{2} = tanθ_{2}].

∴ b = \(\frac{y_1}{x_1}\) ⇒ y_{1} = bx_{1}

∴ p(x_{1}, y_{1}) lies on the line y = bx

Question 21.

Evaluate: ∫\(\frac{1}{1+\sin x+\cos x} d x\)

Solution:

Question 22.

If I_{n} = ∫cos^{n}x dx, then show that I_{n} = \(\frac{1}{n}\)cos^{n-1}x sin x + \(\frac{n-1}{n}\)I_{n-2} (where n ≥ 2)

Solution:

I_{n} = ∫cos^{n}x dx = ∫cos^{n-1}x.cos x dx

= cos^{n-1}x.sin x – ∫sin x. (n – 1)cos^{n-2}x(-sin x)dx

= cos^{n-1}x.sin x + (n – 1)∫cos^{n-2}x(1 – cos^{2}x)dx

= cos^{n-1}x.sinx + (n – 1)I_{n-2} – (n – 1)I_{n}

∴ I_{n}(1 + n – 1) = cos^{n-1}x.sin x + (n – 1)I_{n-2}

I_{n} = \(\frac{\cos ^{n-1} x \sin x}{n}\) + \(\frac{n-1}{n}\)I_{n-2}

Question 23.

Show that : \(\int_0^{\pi / 2} \frac{x}{\sin x+\cos x}\)dx = \(\frac{\pi}{2 \sqrt{2}}\)log (\(\sqrt{2}\) + 1)

Solution:

Question 24.

Solve : x log x\(\frac{d y}{d x}\) + y = 2 log x.

Solution:

\(\frac{d y}{d x}\) + \(\frac{1}{x \log x}\).y = \(\frac{2}{x}\)

I.F. = \(e^{\int \frac{d x}{x \log x}}\) = e^{log(log x)} = log x

y.log x = 2∫\(\frac{\log x}{x}\)dx

= (log x)^{2} + c