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

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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.

1. Attempt all questions.
2. Each question carries two marks.

Question 1.
Find the equation of a circle which is concentric with x2 + y2 – 6x – 4y – 12 = 0 and passing through (-2, 14).
Solution:
Let s ≡ x2 + y2 – 6x – 4y – 12 = 0
The equation of a circle which is concentric with s = 0 is x2 + y2 – 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 x2 + y2 – 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 x2 + y2 = 35.
Solution:
Let s ≡ x2 + y2 – 35 = 0
Since the points (1, 3) and (2, k) are conjugate with respect to the circle s = 0
∴ S12 = 0
⇒ x1 x2 + y1 y2 – 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 x2 + y2 – 3x – 4y + 5 = 0 and 3(x2 + y2) – 7x + 8y – 11 = 0.
Solution:
Let s ≡ x2 + y2 – 3x – 4y + 5 = 0
S1 = x2 + y2 – $$\frac{7}{3}$$x + $$\frac{8}{3}$$y – $$\frac{11}{3}$$ = 0
The radical axis of the circles s = 0, s1 = 0 is s – s1 = 0

Question 4.
Find the co-ordinates of the points on the parabola y2 = 8x whose focal distance is 10.
Solution:
Given parabola equation is y2 = 8x
Hence 4a = 8 ⇒ a = 2
∴ Focus s = (2, 0)
Let p(x1, y1) be a point on the parabola
Given sp = 10

If x1 = 8 then $$y_1^2$$ = 64 ⇒ y1 = ±8
∴ co-ordinates on the parabola are (8, 8), (8, -8)

Question 5.
If 3x – 4y + k = 0 is a tangent to the hyperbola x2 – 4y2 = 5, then find the value of ‘k’.
Solution:
Given hyperbola equation is x2 – 4y2 = 5
⇒ $$\frac{x^2}{5}$$ – $$\frac{4 y^2}{5}$$ = 1
⇒ $$\frac{x^2}{5}$$ – $$\frac{y^2}{5 / 4}$$ = 1 …… (1)
Here a2 = 5, b2 = $$\frac{5}{4}$$
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 = x3 + 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.

1. Attempt any five questions.
2. Each question carries four marks.

Question 11.
Find the equation of the tangent to x2 + y2 – 2x + 4y = 0 at (3, -1). Also find the equation of tangent parallel to it.
Solution:
Let s = x2 + y2 – 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 s1 = 0
⇒ xx1 + yy1 – (x + x1) + 2(y + y1) = 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 x2 + y2 – 8x – 6y + 21 = 0, x2 + y2 – 2x – 15 = 0 and (1, 2).
Solution:
Let S ≡ x2 + y2 – 8y – 6y + 21 = 0
S1 ≡ x2 + y2 – 2x – 15 = 0
S – S1 = x2 + y2 – 8x – 6y + 21 – x2 – y2 + 2x + 15
= -6x – 6y + 36
The equation of the circle passing through the points of intersection of the circles s = 0, s1 = 0 is
s + λ(s – s1) = 0
⇒ (x2 + y2 – 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
(x2 + y2 – 8x – 6y + 21) – $$\frac{1}{3}$$(-6x – 6y + 36) = 0
⇒ 3x2 + 3y2 – 24x – 18y + 63 + 6x + 6y – 36 = 0
⇒ 3x2 + 3y2 – 18x – 12y + 27 = 0
⇒ x2 + y2 – 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 e4 + e2 = 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
$$\frac{a^2 x}{a e}$$ – $$\frac{b^2 y}{b^2 / a}$$ = a2 – y2
⇒ $$\frac{a x}{e}$$ – ay = a2 – b2 is a line
Passes through the one end B’ = (0, -b)
⇒ $$\frac{a(0)}{e}$$a(-b) = a2 – b2
⇒ ab = a2 – b2
⇒ ab = a2 – a2(1 – e2)
⇒ ab = a2e2
⇒ e2 = $$\frac{b}{a}$$
⇒ e4 = $$\frac{b^2}{a^2}$$
⇒ e4 = $$\frac{a^2\left(1-e^2\right)}{a^2}$$
⇒ e4 = 1 – e2
⇒ e4 + e2 = 1

Question 14.
Find the eccentricity, length of latus rectum, foci and the equations of directrices of the ellipse 9x2 + 16y2 = 144.
Solution:
Given ellipse equation is 9x2 + 16y2 = 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 x2 + y2 = a2 – b2.
Solution:
Let p(x1, y1) 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 (x1, y1)

This is a quadratic equation in m giving the values set m1, m2 which are the slopes of the tangents passing through p.
Since tangents an perpendicular
∴ m1m2 = -1
⇒ $$\frac{y_1^2+b^2}{x_1^2-a^2}$$ = -1
⇒ $$y_1^2$$ + b2 – $$\mathrm{x}_1^2$$ + a2
⇒ $$\mathrm{x}_1^2$$ + $$y_1^2$$ = a2 – b2
∴ The locus of P is x2 + y2 = a2 – b2.

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 x2(a2 – x2)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 (x1, y1)

This is a quadratic equation in m giving the values set m1, m2 which are the slopes of the tangents passing through p.
Since tangents an perpendicular = -1
m1m2 = -1
⇒ $$\frac{y_1^2+b^2}{x_1^2-a^2}$$ = 1
⇒ $$\mathrm{y}_1^2$$ + b2 – $$x_1^2$$ + a2
⇒ $$\mathrm{x}_1^2$$ + $$\mathrm{y}_1^2$$ = a2 – b2
∴ The locus of P is x2 + y2 = a2 – b2

Or

Let I = $$\int_{-a}^a x^2\left(a^2-x^2\right)^{3 / 2} d x$$
Since x2(a2 – x2)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 $$\frac{d y}{d x}$$ = ex-y + x2e-y.
Solution:
Given differential equation is
$$\frac{d y}{d x}$$ = ex-y + x2e-y
= ex.e-y + x2e-y
⇒ $$\frac{d y}{e^{-y}}$$ = (ex + x2)dx
⇒ ey dy = (ex + x2)dx
integrating
∫eydy = ∫(ex + x2)dx + c
⇒ ey = ex + $$\frac{x^3}{3}$$ + c
Hence the general solution of (1) is
⇒ ey = ex + $$\frac{x^3}{3}$$ + c

Section – C
(5 × 7 = 35)

III. Long Answer Type Questions.

1. Attempt any five questions.
2. 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 x2 + y2 + 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
x2 + y2 – 2x + 2y – 23 = 0

Question 19.
Show that the circles x2 + y2 – 4x – 6y – 12 = 0 and 5(x2 + y2) – 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 = r1 – r2
∴ 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 s1 = 0.
⇒ x(x1) + y(y1) – 2(x + x1) – 3(y + y1) – 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 = ay2 + 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 = ∫Tannx dx, n being a positive integer n ≥ 2 and deduce the value of ∫Tan6xdx.
Solution:

Question 23.
Evaluate $$\int_0^1 \frac{\log (1+x)}{1+x^2}$$dx.
Solution:
Put x = tan θ ⇒ dx = sec2θdθ
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 = x3 (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 = x2
This is a linear differential equation of first order in y.
Here p = $$\frac{-2(x-1)}{x(x-2)}$$ and Q = x2

The general solution of (1) is