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 \(\frac{d y}{d x}\) = e^x-y + x²e^-y.
Solution:
Given differential equation is
\(\frac{d y}{d x}\) = e^x-y + x²e^-y
= e^x.e^-y + x²e^-y
⇒ \(\frac{d y}{e^{-y}}\) = (e^x + x²)dx
⇒ e^y dy = (e^x + x²)dx
integrating
∫e^ydy = ∫(e^x + x²)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.

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