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## TS Inter 2nd Year Maths 2B Question Paper May 2017

Section – A

(10 × 2 = 20)

Time : 3 Hours

Max. Marks : 75

Note : This question paper consists of three Sections – A, B and C.

Section – A

I. Very Short Answer Type Questions.

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

Question 1.

Find the equation of a circle which is concentric with x^{2} + y^{2} – 6x – 4y – 12 = 0 and passing through (-2, 14).

Solution:

Let s ≡ x^{2} + y^{2} – 6x – 4y – 12 = 0

The equation of a circle which is concentric with s = 0 is x^{2} + y^{2} – 6x – 4y + k = 0 …….. (1)

If (1) passes through (-2, 14) then

4 + 196 + 12 – 56 + k = 0

⇒ 156 + k = 0

⇒ k = -156

Hence required circle equation is x^{2} + y^{2} – 6x – 4y – 156 = 0

Question 2.

Find the value of ‘k’, if the points (1, 3) and (2, k) are conjugate 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 conjugate with respect to the circle s = 0

∴ S_{12} = 0

⇒ x_{1} x_{2} + y_{1} y_{2} – 35 = 0

⇒ 1(2) + 3(k) – 35 = 0

⇒ 3k – 33 = 0

⇒ 3k = 33

⇒ k = 11

Question 3.

Find the equation of the radical axis of the circles x^{2} + y^{2} – 3x – 4y + 5 = 0 and 3(x^{2} + y^{2}) – 7x + 8y – 11 = 0.

Solution:

Let s ≡ x^{2} + y^{2} – 3x – 4y + 5 = 0

S^{1} = x^{2} + y^{2} – \(\frac{7}{3}\)x + \(\frac{8}{3}\)y – \(\frac{11}{3}\) = 0

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

Question 4.

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

Solution:

Given parabola equation is y^{2} = 8x

Hence 4a = 8 ⇒ a = 2

∴ Focus s = (2, 0)

Let p(x_{1}, y_{1}) be a point on the parabola

Given sp = 10

If x_{1} = 8 then \(y_1^2\) = 64 ⇒ y_{1} = ±8

∴ co-ordinates on the parabola are (8, 8), (8, -8)

Question 5.

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

Solution:

Given hyperbola equation is x^{2} – 4y^{2} = 5

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

⇒ \(\frac{x^2}{5}\) – \(\frac{y^2}{5 / 4}\) = 1 …… (1)

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

Given line equation is 3x – 4y + k = 0

⇒ 4y = 3x + k

⇒ y = \(\frac{3}{4}\)k + \(\frac{k}{4}\) …….. (2)

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

If (2) is a tangent to the hyperbola (1) then

Question 6.

Evaluate \(\int \sqrt{1-\sin 2 x} d x\) on I ⊂ \(\left[2 n \pi-\frac{3 \pi}{4}, 2 n \pi+\frac{\pi}{4}\right]\), n ∈ Z

Solution:

Question 7.

Evaluate \(\int \cos \sqrt{x}\) dx on R

Solution:

Let I = ∫cos\(\sqrt{x}\) dx

Let \(\sqrt{x}\) = t

Then \(\frac{1}{2 \sqrt{x}}\)dx = dt

⇒ dx = 2\(\sqrt{x}\) dt

⇒ dx = 2t dt

Question 8.

Evaluate \(\int_{-\pi / 2}^{\pi / 2} \frac{\cos \mathrm{x}}{1+\mathrm{e}^{\mathrm{x}}}\)dx

Solution:

Let I = \(\int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^x} d x\) ………. (1)

Question 9.

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

Solution:

Area = \(\int_{-1}^2 y d x\)

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:

Order = 2

Degree = 1

Section – B

II. Short Answer Type Questions.

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

Question 11.

Find the equation of the tangent to x^{2} + y^{2} – 2x + 4y = 0 at (3, -1). Also find the equation of tangent parallel to it.

Solution:

Let s = x^{2} + y^{2} – 2x + 4y = 0

Center = (1, -2) = (-g, -f)

Radius r = \(\sqrt{1+4}\) = \(\sqrt{5}\)

The equation of the tangent to the circle s = 0 at (3, -1) is s_{1} = 0

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

⇒ x(3) + y(-1) – (x + 3) + 2(y – 1) = 0

⇒ 3x – y – x – 3 + 2y – 2 = 0

⇒ 2x + y – 5 = 0 ………. (1)

∴ Slope of the tangent = \(\frac{-2}{+1}\) = -2 = m

The equation on the tangent parallel to (1) is (y – f) = m(x – g) ±

r\(\sqrt{1+\mathrm{m}^2}\)

⇒ y+ 2 = -2(x – 1) ± \(\sqrt{5} \sqrt{1+4}\)

⇒ y + 2 = -2x + 2 ± 5

⇒ 2x + y = ± 5 ⇒ 2x + y – 5 = 0 and 2x + y + 5 = 0

Question 12.

Find the equation of the circle passing through the points of intersection of the circles x^{2} + y^{2} – 8x – 6y + 21 = 0, x^{2} + y^{2} – 2x – 15 = 0 and (1, 2).

Solution:

Let S ≡ x^{2} + y^{2} – 8y – 6y + 21 = 0

S^{1} ≡ x^{2} + y^{2} – 2x – 15 = 0

S – S^{1} = x^{2} + y^{2} – 8x – 6y + 21 – x^{2} – y^{2} + 2x + 15

= -6x – 6y + 36

The equation of the circle passing through the points of intersection of the circles s = 0, s^{1} = 0 is

s + λ(s – s^{1}) = 0

⇒ (x^{2} + y^{2} – 8x – 6y + 21) + λ(-6x – 6y + 36) = 0

If this circle passing (1, 2) then

(1 + 4 – 8 – 12 + 21) + λ(-6 – 12 + 36) = 0

⇒ 6 + 18λ = 0

⇒ 18λ = -6

⇒ λ = \(\frac{-1}{3}\)

Hence required circle equation is

(x^{2} + y^{2} – 8x – 6y + 21) – \(\frac{1}{3}\)(-6x – 6y + 36) = 0

⇒ 3x^{2} + 3y^{2} – 24x – 18y + 63 + 6x + 6y – 36 = 0

⇒ 3x^{2} + 3y^{2} – 18x – 12y + 27 = 0

⇒ x^{2} + y^{2} – 6x – 4y + 9 = 0

Question 13.

If the normal at one end of a latus rectum of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1 passes through one end of the minor axis, then show that e^{4} + e^{2} = 1 (e = eccentricity of the ellipse).

Solution:

Given ellipse equation is \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1

Let be the one end of the latus rectum.

∴ The equation of the normal at L is

\(\frac{a^2 x}{a e}\) – \(\frac{b^2 y}{b^2 / a}\) = a^{2} – y^{2}

⇒ \(\frac{a x}{e}\) – ay = a^{2} – b^{2} is a line

Passes through the one end B’ = (0, -b)

⇒ \(\frac{a(0)}{e}\)a(-b) = a^{2} – b^{2}

⇒ ab = a^{2} – b^{2}

⇒ ab = a^{2} – a^{2}(1 – e^{2})

⇒ ab = a^{2}e^{2}

⇒ e^{2} = \(\frac{b}{a}\)

⇒ e^{4} = \(\frac{b^2}{a^2}\)

⇒ e^{4} = \(\frac{a^2\left(1-e^2\right)}{a^2}\)

⇒ e^{4} = 1 – e^{2}

⇒ e^{4} + e^{2} = 1

Question 14.

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

Solution:

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

Question 15.

Prove that the point of intersection of two perpendicular tangents to the hyperbola \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1 lies on the circle x^{2} + y^{2} = a^{2} – b^{2}.

Solution:

Let p(x_{1}, y_{1}) be the point or intersection of two perpendicular tangents to the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1

Equation or the tangent can be taken as y = mx ± \(\sqrt{a^2 m^2-b^2}\)

This tangent passes through (x_{1}, y_{1})

This is a quadratic equation in m giving the values set m_{1}, m_{2} which are the slopes of the tangents passing through p.

Since tangents an perpendicular

∴ m_{1}m_{2} = -1

⇒ \(\frac{y_1^2+b^2}{x_1^2-a^2}\) = -1

⇒ \(y_1^2\) + b^{2} – \(\mathrm{x}_1^2\) + a^{2}

⇒ \(\mathrm{x}_1^2\) + \(y_1^2\) = a^{2} – b^{2}

∴ The locus of P is x^{2} + y^{2} = a^{2} – b^{2}.

Question 16.

Evaluate \(\int_{-a}^a x^2\left(a^2-x^2\right)^{3 / 2}\) dx.

Solution:

Let I = \(\int_{-a}^a x^2\left(a^2-x^2\right)^{3 / 2}\) dx

Since x^{2}(a^{2} – x^{2})^{3/2} is an even function

\(2 \int_0^a x^2\left(a^2-x^2\right)^{3 / 2}\) dx

Put x = a sin θ

⇒ dx = a cosθ dθ

Equation or the tangent can be taken as

y = mx ± \(\sqrt{a^2 m^2-b^2}\)

This tangent passes through (x_{1}, y_{1})

This is a quadratic equation in m giving the values set m_{1}, m_{2} which are the slopes of the tangents passing through p.

Since tangents an perpendicular = -1

m_{1}m_{2} = -1

⇒ \(\frac{y_1^2+b^2}{x_1^2-a^2}\) = 1

⇒ \(\mathrm{y}_1^2\) + b^{2} – \(x_1^2\) + a^{2}

⇒ \(\mathrm{x}_1^2\) + \(\mathrm{y}_1^2\) = a^{2} – b^{2}

∴ The locus of P is x^{2} + y^{2} = a^{2} – b^{2}

Or

Let I = \(\int_{-a}^a x^2\left(a^2-x^2\right)^{3 / 2} d x\)

Since x^{2}(a^{2} – x^{2})^{3/2} is an even function

\(2 \int_0^a x^2\left(a^2-x^2\right)^{3 / 2}\) dx

Put x = a sin θ

⇒ dx = a cosθ dθ

Question 17.

Solve the differential equation \(\frac{d y}{d x}\) = e^{x-y} + x^{2}e^{-y}.

Solution:

Given differential equation is

\(\frac{d y}{d x}\) = e^{x-y} + x^{2}e^{-y}

= e^{x}.e^{-y} + x^{2}e^{-y}

⇒ \(\frac{d y}{e^{-y}}\) = (e^{x} + x^{2})dx

⇒ e^{y} dy = (e^{x} + x^{2})dx

integrating

∫e^{y}dy = ∫(e^{x} + x^{2})dx + c

⇒ e^{y} = e^{x} + \(\frac{x^3}{3}\) + c

Hence the general solution of (1) is

⇒ e^{y} = e^{x} + \(\frac{x^3}{3}\) + c

Section – C

(5 × 7 = 35)

III. Long Answer Type Questions.

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

Question 18.

Find the equation of a circle passing through (2, -3) and (-4, 5) and having a centre on 4x + 3y + 1 =0.

Solution:

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

be the required circle.

If (1) passes through (2, -3) then

4 + 9 + 4g – 6f + c = 0

4g – 6f + c = -13 …….. (2)

If (1) passes through (-4, 5) then

16 + 25 – 8g + 10f + c = 0

⇒ 8g – 10f – c = 41 …….. (3)

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

⇒ 4(-g) + 3(-f) + 1 = 0

⇒ -4y – 3f + 1 =0

⇒ 4g + 3f – 1 = 0 ……….. (4)

(2) + (3) ⇒ 12g – 16f = 28

⇒ 3g – 4f = 7

⇒ 3g – 4f – 7 = 0 …….. (5)

Solving (4) & (5)

g = -1, f = 1

From (2) + 4 + 6 + c = – 13

⇒ c = -23

∴ g = -1, f = 1, c = -23

Hence required circle equation is

x^{2} + y^{2} – 2x + 2y – 23 = 0

Question 19.

Show that the circles x^{2} + y^{2} – 4x – 6y – 12 = 0 and 5(x^{2} + y^{2}) – 8x – 14y – 32 = 0 touch each other. Also find the point of contact and common tangent at this point of contact.

Solution:

= \(\sqrt{\frac{36}{25}+\frac{64}{25}}\) = \(\sqrt{\frac{100}{25}}\) = \(\sqrt{4}\) = 2

∴ AB = r_{1} – r_{2}

∴ The circles touch internally.

p divides AB externally in the ratio 5 : 3

p = \(\left(\frac{5 . \frac{4}{5}-3.2}{5-3}, \frac{5 . \frac{7}{5}-3.3}{5-3}\right)\)

= (-1, -1)

Equation of the tangent to the circle s = 0 at p is s_{1} = 0.

⇒ x(x_{1}) + y(y_{1}) – 2(x + x_{1}) – 3(y + y_{1}) – 12 = 0

⇒ x(-1) + y(-1) – 2(x – 1) -3 (y – 1) – 12 = 0

⇒ -x – y – 2x + 2 – 3y + 3 – 12 = 0

⇒ -3x – 4y – 7 = 0

⇒ 3x + 4y + 7 = 0

Question 20.

Find the equation of the parabola whose axis is parallel to x – axis and which passes through the points (-2, 1), (1, 2) and (-1, 3).

Solution:

Since axis is parallel to x-axis

∴ Equation of the parabola is

x = ay^{2} + by + c ………. (1)

since (1) passes through the points (-2, 1), (1,2), (-1,3)

-2 = a + b + c ……… (2)

1 = 4a + 2b + c ……… (3)

-1 = 9a + 3b + c …….. (4)

Question 21.

Evaluate \(\int \sqrt{\frac{5-x}{x-2}}\) dx on (2, 5).

Solution:

Question 22.

Obtain the reduction formula for ln = ∫Tan^{n}x dx, n being a positive integer n ≥ 2 and deduce the value of ∫Tan^{6}xdx.

Solution:

Question 23.

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

Solution:

Put x = tan θ ⇒ dx = sec^{2}θdθ

x = 0 ⇒ θ = 0

x = 1 ⇒ θ = \(\frac{\pi}{4}\)

Question 24.

Find the solution of the equation

x(x – 2)\(\frac{\mathrm{dy}}{\mathrm{dx}}\) – 2(x – 1 )y = x^{3} (x – 2), which satisfies the condition that y = 9 when x = 3.

Solution:

Given differential equation is

\(\frac{\mathrm{dy}}{\mathrm{dx}}\) – \(\frac{2(x-1)}{x(x-2)}\)y = x^{2}

This is a linear differential equation of first order in y.

Here p = \(\frac{-2(x-1)}{x(x-2)}\) and Q = x^{2}

The general solution of (1) is