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

Time : 3 Hours

Max. Marks : 75

Section – A

(10 × 2 = 20)

I. Very Short Answer type questions.

- Attempt all questions.
- Each question carries two marks.

Question 1.

Find the centre and radius of the circle : x^{2} + y^{2} – 4x – 8y – 41 =0.

Solution:

x^{2} + y^{2} – 4x – 8y – 41 = 0 …… (i)

x^{2} + y^{2} + 2gx + 2fy – 41 = 0 …… (ii)

Comparing (i) and (ii) we get

g = -2, f = -4, c = -41

Radius = \(\sqrt{g^2+f^2-c}\)

= \(\sqrt{4+16+41}\)

= \(\sqrt{61}\) units

Centre = (-g, -f) = (2, 4)

Question 2.

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

Solution:

Length of tangent

= \(\sqrt{S_{11}}\) = \(\sqrt{(5)^2+(4)^2+8 k}\)

But, length of tangent

∴ 1 = \(\sqrt{25+16+8 k}\)

Squaring both sides we get 1 = 41 + 8k

k = – 5 units.

Question 3.

Find the equation of the common chord of the circles : x^{2} + y^{2} – 4x – 4y + 3 = 0 and x^{2} + y^{2} – 5x – 6y + 4 = 0.

Solution:

(x^{2} + y^{2} – 4x – 4y + 3) – (x^{2} + y^{2} – 5x – 6y + 4) = 0

x + 2y – 1 = 0 Equation of common chord.

Question 4.

Find the equation of the tangent to the parabola y^{2} = 6x at the positive end of the latus rectum.

Solution:

(a, 2a) Here, 4a = 6 ⇒ a = \(\frac{3}{2}\)

(\(\frac{3}{2}\), 3)

equation of tangent yy_{1} = 2a (x + x_{1})

= 3(x + x_{1})

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

2y – 2x – 3 = 0 is the equation of tangent

Slope of tangent is 1

Slope of normal is -1

Equation of normal is y – 3 = – 1(x – \(\frac{3}{2}\))

2x + 2y – 9 = 0

Question 5.

If the angle between the asymptotes of the hyperbola is 30°, then find its eccentricity.

Solution:

Angle between the asymptotes = 2θ = 30°

Question 6.

Evaluate the integral \(\int \frac{\sin ^2 x}{1+\cos 2 x}\) dx on I ⊂ R\{(2n ± 1)π: n ∈ Z}.

Solution:

Question 7.

Evaluate the integral : \(\int \frac{2 x+1}{x^2+x+1}\) dx, x ∈ R.

Solution:

t = x^{2} + x + 1

dt = (2x + 1)dx

\(\int \frac{2 x+1}{x^2+x+1}\) dx = \(\int \frac{d t}{t}\)

= log|t| + C

= log|x^{2} + x + 1| + C

Question 8.

Evaluate the definite integral: \(\int_2^3 \frac{2 x}{1+x^2} d x\)

Solution:

Question 9.

Find the area of the region enclosed by y = x^{3} + 3, y = 0, x = -1, x = 2.

Solution:

Required are PABQ

Question 10.

Form the differential equation corresponding to: y = A cos 3x + 6 sin 3x, where A and B are parameters.

Solution:

We have y = A cos 3x + B sin 3x

Differentiating w.r.to x dy

\(\frac{d y}{d x}\) = -3A sin 3x + 3B cos 3x dx

Differentiating again w.r.to. x

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

= -9(A cos 3x + B sin 3x)

= -9y

is \(\frac{d^2 y}{d x^2}\) + 9y = 0.

Alternate method :

Eliminating A, B from the equation

y = A cos 3x + B sin 3x

\(\frac{d y}{d x}\) = -3A sin 3x + 3B sin cos 3x

y(27 sin^{2} 3x + 27 cos^{2} 3x) – (-9 sin 3x.cos3x + 9cos3x.sin3x)\(\frac{d y}{d x}\) + (3 cos^{2} 3x + 3 sin^{2} 3x)\(\frac{d^2 y}{d x^2}\) = o

27y + 3.\(\frac{d^2 y}{d x^2}\) = 0 or \(\frac{d^2 y}{d x^2}\) + 9y = 0

This is the required differential equation.

Section – B

(5 × 4 = 20 marks)

II. Short Answer Type questions.

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

Question 11.

Find the length of the chord intercepted by the circle : x^{2} + y^{2} – x + 3y – 22 = 0 on the line y = x – 3.

Solution:

Equation of the circle is

S ≡ x^{2} + y^{2} – x + 3y – 22 = 0

Equation of the line is y = x – 3 ⇒ x – y – 3 = 0

P = distance from the centre

Question 12.

Show that the angle between the circles x^{2} + y^{2} = a^{2}, x^{2} + y^{2} = ax + ay is \(\frac{3 \pi}{4}\).

Solution:

Question 13.

Find the length of latus rectum, eccentricity, foci and the equations of directrices of the ellipse : 9x^{2} + 16y^{2} = 144.

Solution:

Given equation is 9x^{2} + 16y^{2} = 144

\(\frac{x^2}{16}+\frac{y^2}{9}\) = 1

Length of major axis = 2a = 2. 4 = 8

Length of minor axis = 2b = 2. 3 = 6

Length of latus rectum = \(\frac{2 b^2}{a}\) = \(\frac{2.9}{4}\) = \(\frac{9}{2}\)

Eccentricity = \(\sqrt{\frac{a^2-b^2}{a^2}}\) = \(\sqrt{\frac{16-9}{16}}\) = \(\sqrt{\frac{7}{4}}\)

Centre is c (0,0)

Foci are (± ae, 0) = ( ± \(\sqrt{7}\), 0)

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

x = ±4.\(\frac{4}{\sqrt{7}}\) = ±\(\frac{16}{\sqrt{7}} \sqrt{7} x\) = ±16

Question 14.

Find the equation of tangent and normal to the ellipse x^{2} + 8y^{2} = 33 at (-1, 2).

Solution:

Equation of the tangent is

\(\frac{x x_1}{a^2}\) + \(\frac{y y_1}{b^2}\) = 1

x(-1) + 8y(2) = 33

⇒ -x + 16y = 33

⇒ x – 16y + 33 = 0

Equation of the normal is

-16x + 2 + k = 0

It passes through P(-1, 2)

-16 + 2 + k = 0 ⇒ k = 14

Equation of the normal is

16x + y + 14 = 0

Question 15.

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

i) Parallel and

ii) Perpendicular to the line y = x – 7.

Solution:

i) Equation of the hyperbola is 3x^{2} – 4y^{2} = 12

\(\frac{x^2}{4}-\frac{y^2}{3}\) = 1

a^{2} = 4, b^{2} = 3

The tangent is parallel to y = x – 7

m = slope of the tangent = 1

Equation of the parallel tangents are

y = mx ± \(\sqrt{a^2 m^2-b^2}\)

y = x ± \(\sqrt{4-3}\)

y = x ± 1

ii) The tangent is perpendicular to y – x = 7

m – slope of the tangent = (-1)

Equation of the perpendicular tangents are

y = (-1)x ± \(\sqrt{4(-1)^2-3}\)

= -x ± 1

x + y = ± 1

Question 16.

Evaluate: \(\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x\)

Solution:

Question 17.

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

Solution:

(xy^{2} + x) dx + (yx^{2} + y) dy = 0

x(y^{2} + 1) dx + y (x^{2} + 1) dy = 0

Dividing with (1 + x^{2}) (1 + y^{2})

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

Integrating

\(\int \frac{x d x}{1+x^2}\) + \(\int \frac{y d y}{1+y^2}\) = 0

\(\frac{1}{2}\)[(log (1 + x^{2}) + log (1 + y^{2})] = log c

log (1 + x^{2}) (1 + y^{2}) = 2 log c = log c^{2}

Solution is (1 + x^{2}) (1 + y^{2}) = k when k = c^{2}.

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 (2, -3) and (- 4, 5) and having the centre on 4x + 3y + 1 = 0.

Solution:

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

Equation (i) passes through (2, – 3), (- 4, 5)

∴ 4 + 9 + 4g – 6f + c = 0 …… (ii)

16 + 25 – 8g + 10f + c = 0 ……….. (iii)

Equation (iii) – (ii) we get

28 – 12g + 16f = 0

(or) 3g – 4f = 7

Centre lies on (- g, – f) lies on 4x + 3y + 1 = 0

then 4(- g) + 3(- f) + 1 = 0

3g – 4f – 7 = 0

Solving we get f = – 1

g = 1

Now, substituting f, g values in equation (ii) we get

4 + 9 + 4(1) – 6(-1) + c = 0, c = -23

x^{2} + y^{2} + 2x – 2y – 23 = 0 is required equation of circle.

Question 19.

Show that: x^{2} + y^{2} – 6x – 9y + 13 = 0, x^{2} + y^{2} – 2x – 16y = 0 touch each other. Find the point of contact and the equation of common tangent at their point of contact.

Solution:

Equations of the circles are

∴ The circles touch each other internally. The point of contact ‘p’ divides AB externally in the ratio r_{1} : r_{2} = \(\frac{\sqrt{65}}{2}\) : \(\sqrt{65}\) = 1 : 2 co-ordinates of p are

∴ Equation of the common tangent is S_{1} – S_{2} = 0

-4x + 7y + 13 = 0

4x – 7y – 13 = 0

Question 20.

Derive the equation of a parabola in standard form.

Solution:

Equation of Parabola : General form – Let S(α, β) be the focus and ax + by + c = 0 be directrix then by definition the equation of parabola be

Question 21.

Evaluate : \(\int \frac{d x}{(1+x) \sqrt{3+2 x-x^2}}\) on (1, 3).

Solution:

Question 22.

Obtain the reduction formula for ∫sin^{n} x dx for an integer n ≥ 2 and deduce the value of : ∫sin^{4}x dx

Solution:

Question 23.

Evaluate : \(\int_0^1 \frac{\log (1+x)}{1+x^2}\) dx

Solution:

Question 24.

Solve the differential equation : sin^{-1}\(\left(\frac{d y}{d x}\right)\) = x + y.

Solution: