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## TS Inter 2nd Year Maths 2B Question Paper March 2019

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.

Write the parametric equations of the circle 2x^{2} + 2y^{2} = 7.

Solution:

Given circle equation is 2x^{2} + 2y^{2} = 7

⇒ x^{2} + y^{2} = \(\frac{7}{2}\)

Here centre (h, k) = (0, 0) and

radius r = \(\sqrt{\frac{7}{2}}\)

Circle Parametric equations are

Question 2.

Find the value of k, if the points (1, 3) and (2, k) are conjugated with respect to the circle x^{2} + y^{2} = 35.

Solution:

Let S = x^{2} + y^{2} – 35= 0

Since the points (1, 3) and (2, k) are conjugated with respect to the circle S = 0

∴ S_{12} = 0

⇒ x_{1}x_{2} + y_{1}y_{2} – 35 = 0

⇒ 2 + 3k – 35 = 0 ⇒ 3k = 33 ⇒ k = 11

Question 3.

Find the equation of radical axis of the circles x^{2} + y^{2} + 4x + 6y – 7 = 0, 4(x^{2} + y^{2})+ 8x + 12y – 9 = 0.

Solution:

Let S ≡ x^{2} + y^{2} + 4x + 6y – 7 = 0,

S^{1} ≡ x^{2} + y^{2} + 2x + 3y – \(\frac{9}{4}\) = 0

The equation of the radical axis of the circles S = 0, S^{1} = 0 is S – S^{1} = 0

⇒ (x^{2} + y^{2} + 4x + 6y – 7) – (x^{2} + y^{2} + 2x + 3y – \(\frac{9}{4}\)) = 0

⇒ x^{2} + y^{2} + 4x + 6y – 7 – x^{2} – y^{2} – 2x – 3y + \(\frac{9}{4}\) = 0

⇒ 8x + 12y – 19 = 0.

Question 4.

Find the equation of the normal to the parabola y^{2} = 4x which is parallel to y – 2x + 5 = 0.

Solution:

Given parabola equation is y^{2} = 4x

Here 4a = 4 ⇒ a = 1

Given line equation is y – 2x + 5 = 0

⇒ y = 2x – 5

∴ Slope m = 2

Since required normal is parallel to y = 2x – 5

∴ Slope of normal = 2

⇒ -t = 2 ⇒ t = -2

Equation of the normal at t is

y + xt = 2at + t^{3}

⇒ y + x(-2) = 2(1) (-2) + (-2)^{3}

⇒ y – 2x = -4 – 8

⇒ 2x – y – 12 = 0.

Question 5.

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

Solution:

If e_{1}, e_{2} are the eccentricities of hyperbola and conjugate hyperbola then \(\frac{1}{e_1^2}+\frac{1}{e_2^2}\) = 1

Question 6.

Evaluate \(\int \frac{1+\cos ^2 x}{1-\cos 2 x} d x\), on I ⊂ R\{nπ : n ∈ z}.

Solution:

Question 7.

Evaluate \(\int \frac{1}{x \log x[\log (\log x)]}\)dx, on (1, ∝).

Solution:

Question 8.

Evaluate \(\int_0^a(\sqrt{a}-\sqrt{x})^2\) dx

Solution:

Question 9.

Find \(\int_0^{\pi / 2}\)cos^{11} xdx.

Solution:

Question 10.

Find the general solution of : \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{2 y}{x}\)

Solution:

Given \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{2 y}{x}\) ……… (1)

⇒ \(\frac{\mathrm{dy}}{\mathrm{y}}\) = 2\(\frac{d x}{x}\)

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

Integrating

Section – B

(5 × 4 = 20)

II. Short Answer type questions.

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

Question 11.

Find the equation of the circle which cuts orthogonally the circle x^{2} + y^{2} – 4x + 2y – 7 = 0 and having the centre at (2, 3).

Solution:

Let x^{2} + y^{2} + 2gx + 2fy + c = 0 be the required circle equation ………. (1)

Given centre (-g, -f) = (2, 3)

∴ -g = 2,-f = 3

⇒ g = -2, f = -3.

Given circle equation in x^{2} + y^{2} – 4x + 2y – 7 = 0 …….. (2)

Since (1), (2) cut orthogonally

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

⇒ 2(-2)(-2)+2(-3)(1) = -7 + c

⇒ 8 – 6 = -7 + c

⇒ 2 = -7 + c ⇒ c = 9

Hence required circle equation is

x^{2} + y^{2} + 2(-2)x + 2(-3)y + 9 = 0

⇒ x^{2} + y^{2} – 4x – 6y + 9 = 0.

Question 12.

The line y = mx + c and the circle x^{2} + y^{2} = a^{2} intersect at A and B. If AB = 2λ, then show that:

c^{2} = (1 + m^{2})(a^{2} – λ^{2}).

Solution:

Given circle equation is x^{2} + y^{2} = a^{2}

Centre c = (0, 0) and radius r = a

d = The ⊥ distance from c to line

y = mx + c

Question 13.

Find the equation of the ellipse with focus at (1, -1), e = \(\frac{2}{3}\) and directrix as x + y + 2 = 0.

Solution:

Given focus s = (1, -1)

Eccentricity e = \(\frac{2}{3}\)

Equation of directrix is x + y + 2 = 0

Let P(x, y) be any point on the ellipse

\(\frac{S P}{P M}\) = e

⇒ SP = e.PM

⇒ SP^{2} = e^{2}.PM^{2}

⇒ 9x + 9y^{2} – 18x + 18y + 18 = 2x^{2} + 2y^{2} + 8 + 4xy + 8x + 8y

⇒ 7x^{2} – 4xy + 7y^{2} – 26x + 10y + 10 = 0

Hence required ellipse equation is

7x^{2} – 4xy + 7y^{2} – 26x + 10y + 10 = 0.

Question 14.

Find the equations of tangents to the ellipse 2x^{2} + 3y^{2} = 11 at the points whose ordinate is 1.

Solution:

Given ellipse equation is 2x^{2} + 3y^{2} = 11

Given ordinate y = 1

⇒ 2x^{2} + 3 = 11

⇒ 2x^{2} = 8

⇒ x^{2} = 8

⇒ x = ±2

∴ Points on the ellipse are P(1, 2) and Q(-2, 1)

Case (i): P(2, 1)

∴ Equation of the tangent at P is S_{1} = 0

⇒ 2x(2) + 3y(1) – 11 = 0

⇒ 4x + 3y – 11 = 0

∴ Equation of the normal at P is

3(x – 2) – 4(y – 1) = 0

⇒ 3x – 6 – 4y + 4 = 0

⇒ 3x – 4y – 2 = 0

Case (ii): Q (-2,1)

Equation of the tangent at Q is S_{1} = 0

⇒ 2x(-2) + 3y(1) – 11 = 0

⇒ – 4x + 3y – 11 = 0

⇒ 4x – 3y + 11 = 0

Equation of the normal at Q is

3(x + 2) + 4(y – 1) = 0

⇒ 3x + 6 + 4y – 4 = 0

⇒ 3x + 4y + 2 = 0.

Question 15.

Find the foci, eccentricity, equations of the directrix, length of latus rectum of the hyperbola x^{2} – y^{2} = 4.

Solution:

Given equation of the hyperbola is x^{2} – 4y^{2} = 4

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

Centre = (0, 0)

Eccentricity e = \(\sqrt{\frac{a^2+b^2}{a^2}}\)

= \(\sqrt{\frac{4+1}{4}}\)

= \(\sqrt{\frac{5}{4}}\)

= \(\frac{\sqrt{5}}{2}\)

Question 16.

Find: \(\int_0^{2 \pi}\) sin^{4} x cos^{6} x dx.

Solution:

Question 17.

Solve the differential equation : cos x\(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y sin x = sec^{2} x.

Solution:

Given cos x\(\frac{d y}{d x}\) + y sin x = sec^{2} x

⇒ \(\frac{d y}{d x}\) + y tanx = sec^{3}x …….. (1)

Here P = tan x, Q = Sec^{3} x

Section – C

(5 × 7 = 35)

III. Long Answer type questions.

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

Question 18.

Show that the points (9, 1), (7, 9), (-2, 12), (6, 10) are concyclic and find the equation of the circle on which they lie.

Solution:

Let A = (9, 1), B = (7, 9), C = (-2, 12), D = (6, 10)

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

(x – x_{1})(x – x_{2}) + (y – y_{1})(y – y_{2}) = 0

⇒ (x – 9)(x – 7) + (y – 1 )(y – 9) = 0

⇒ x^{2} – 7x – 9x + 63 + y^{2} – 9y – y + 9 = 0

⇒ x^{2} + y^{2} – 16x – 10y + 72 = 0 ……… (1)

Slope of \(\overline{\mathrm{AB}}\) = \(\frac{y_2-y_1}{x_2-x_1}\) = \(\frac{9-1}{7-9}\) = \(\frac{8}{-2}\) = -4

Equation of \(\overline{\mathrm{AB}}\) is y – 1 = – 4(x – 9)

⇒ y – 1 = – 4x + 36

⇒ 4x + y – 37 = 0

The equation of circle passing through A, B is

(x^{2} + y^{2} – 16x – 10y + 72) + C(4x + y – 37) = 0

If this circle passes through C(-2, 12) then

4 + 144 + 32 – 120 + 72 + C(-8 + 12 – 37) = 0

⇒ 132 – 33C = 0

⇒ 33C = 132 ⇒ C = 4

∴ Equation of the circle passes through A, B, C is

(x^{2} + y^{2} – 16x – 10y + 72) + 4(4x + y – 37) = 0

⇒ x^{2} + y^{2} – 16x – 10y + 72 + 16x + 4y – 148 = 0

⇒ x^{2} + y^{2} – 6y – 76 = 0 ……….. (2)

x^{2} + y^{2} – 6y – 76 = (6)^{2} + (10)^{2} – 6(10) – 76

= 36 + 100 – 60 – 76 = 0

∴ D(6, 10) lies on (2)

∴ A, B, C, D are concydic and circle equation is x^{2} + y^{2} – 6y – 76 = 0.

Question 19.

Show that, four common tangents can be drawn for the circles given by x^{2} + y^{2} – 14x + 6y + 33 = 0 and x^{2} + y^{2} + 30x – 2y + 1 = 0 and find the internal and external centres of similitude.

Solution:

Given circle equations are x^{2} + y^{2} – 14x + 6y + 33 = 0

x^{2} + y^{2} + 30x – 2y + 1 = 0

Centres A = (7, -3), B = (-15, 1)

r_{1} = \(\sqrt{49+9-33}\) = \(\sqrt{25}\) = 5

∴ Four tangents can be drawn for the given circles.

Let P, Q be the internal and external centres of similitude.

∴ P divides \(\overline{\mathrm{AB}}\) in the ratio 1 : 3 internally

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 cot θ_{1} + cot θ_{2} is a constant ‘d’. Then show that all such P lie on a horizontal line.

Solution:

Equation of a tangent to the Parabola y^{2} = 4ax is a

y = mx + \(\frac{a}{m}\)

This tangent passes through P(x_{1}y,)

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

⇒ my_{1} = m^{2}x_{1} + a

⇒ m^{2}x_{1} – my_{1} + a = 0 which is a quadratic eq. in m

If m_{1}, m_{2} are the slopes of the tangents drawn from P then m_{1}, m_{2} are the roots of the equation m^{2}x_{1} – my_{1} + a = 0

Question 21.

Evaluate : ∫e^{ax} sin(bx + c)dx; (a, b, c ∈ R; b ≠ 0) on R.

Solution:

Question 22.

Obtain 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.

Find : \(\int_0^\pi x \sin ^7 x \cos ^6 x d x .\)

Solution:

Question 24.

Solve the differential equation : \(\frac{d y}{d x}\) = \(\frac{2 y+x+1}{2 x+4 y+3}\)

Solution:

Integrating

\(\int \frac{1}{2} d z\) + \(\frac{1}{2} \int \frac{1}{4 z+5}\)dz = \(\int d x\) + \(\frac{c}{8}\)

⇒ \(\frac{1}{2}\)z + \(\frac{1}{2} \cdot \frac{1}{4}\)log|4z + 5| = x + \(\frac{c}{8}\)

⇒ 4z – log|4z + 5| = 8x + c

⇒ 4(x + 2y) + log|4(x + 2y) + 5| = 8x + c

⇒ 8y – 4x + log|4x + 8y + 5| = c

∴ The general solution of (1) is

8y – 4x + log|4x + 8y + 5| = c.