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

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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.

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

Question 1.
Write the parametric equations of the circle 2x² + 2y² = 7.
Solution:
Given circle equation is 2x² + 2y² = 7
⇒ x² + y² = \(\frac{7}{2}\)
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 x² + y² = 35.
Solution:
Let S = x² + y² – 35= 0
Since the points (1, 3) and (2, k) are conjugated with respect to the circle S = 0
∴ S₁₂ = 0
⇒ x₁x₂ + y₁y₂ – 35 = 0
⇒ 2 + 3k – 35 = 0 ⇒ 3k = 33 ⇒ k = 11

Question 3.
Find the equation of radical axis of the circles x² + y² + 4x + 6y – 7 = 0, 4(x² + y²)+ 8x + 12y – 9 = 0.
Solution:
Let S ≡ x² + y² + 4x + 6y – 7 = 0,
S¹ ≡ x² + y² + 2x + 3y – \(\frac{9}{4}\) = 0
The equation of the radical axis of the circles S = 0, S¹ = 0 is S – S¹ = 0
⇒ (x² + y² + 4x + 6y – 7) – (x² + y² + 2x + 3y – \(\frac{9}{4}\)) = 0
⇒ x² + y² + 4x + 6y – 7 – x² – y² – 2x – 3y + \(\frac{9}{4}\) = 0
⇒ 8x + 12y – 19 = 0.

Question 4.
Find the equation of the normal to the parabola y² = 4x which is parallel to y – 2x + 5 = 0.
Solution:
Given parabola equation is y² = 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 + t³
⇒ 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 e₁, e₂ 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 \(\int_0^{\pi / 2}\)cos¹¹ 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.

1. Answer ANY FIVE questions.
2. Each question carries FOUR marks.

Question 11.
Find the equation of the circle which cuts orthogonally the circle x² + y² – 4x + 2y – 7 = 0 and having the centre at (2, 3).
Solution:
Let x² + y² + 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 x² + y² – 4x + 2y – 7 = 0 …….. (2)
Since (1), (2) cut orthogonally
∴ 2g₁g₂ + 2f₁f₂ = c₁ + c₂
⇒ 2(-2)(-2)+2(-3)(1) = -7 + c
⇒ 8 – 6 = -7 + c
⇒ 2 = -7 + c ⇒ c = 9
Hence required circle equation is
x² + y² + 2(-2)x + 2(-3)y + 9 = 0
⇒ x² + y² – 4x – 6y + 9 = 0.

Question 12.
The line y = mx + c and the circle x² + y² = a² intersect at A and B. If AB = 2λ, then show that:
c² = (1 + m²)(a² – λ²).
Solution:
Given circle equation is x² + y² = a²
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
⇒ SP² = e².PM²

⇒ 9x + 9y² – 18x + 18y + 18 = 2x² + 2y² + 8 + 4xy + 8x + 8y
⇒ 7x² – 4xy + 7y² – 26x + 10y + 10 = 0
Hence required ellipse equation is
7x² – 4xy + 7y² – 26x + 10y + 10 = 0.

Question 14.
Find the equations of tangents to the ellipse 2x² + 3y² = 11 at the points whose ordinate is 1.
Solution:
Given ellipse equation is 2x² + 3y² = 11
Given ordinate y = 1
⇒ 2x² + 3 = 11
⇒ 2x² = 8
⇒ x² = 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 S₁ = 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 S₁ = 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 x² – y² = 4.
Solution:
Given equation of the hyperbola is x² – 4y² = 4
Here a² = 4, b² = 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: \(\int_0^{2 \pi}\) sin⁴ x cos⁶ x dx.
Solution:

Question 17.
Solve the differential equation : cos x\(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y sin x = sec² x.
Solution:
Given cos x\(\frac{d y}{d x}\) + y sin x = sec² x
⇒ \(\frac{d y}{d x}\) + y tanx = sec³x …….. (1)
Here P = tan x, Q = Sec³ x

Section – C
(5 × 7 = 35)

III. Long Answer type questions.

1. Answer ANY FIVE questions.
2. 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 – x₁)(x – x₂) + (y – y₁)(y – y₂) = 0
⇒ (x – 9)(x – 7) + (y – 1 )(y – 9) = 0
⇒ x² – 7x – 9x + 63 + y² – 9y – y + 9 = 0
⇒ x² + y² – 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
(x² + y² – 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
(x² + y² – 16x – 10y + 72) + 4(4x + y – 37) = 0
⇒ x² + y² – 16x – 10y + 72 + 16x + 4y – 148 = 0
⇒ x² + y² – 6y – 76 = 0 ……….. (2)
x² + y² – 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 x² + y² – 6y – 76 = 0.

Question 19.
Show that, four common tangents can be drawn for the circles given by x² + y² – 14x + 6y + 33 = 0 and x² + y² + 30x – 2y + 1 = 0 and find the internal and external centres of similitude.
Solution:
Given circle equations are x² + y² – 14x + 6y + 33 = 0
x² + y² + 30x – 2y + 1 = 0
Centres A = (7, -3), B = (-15, 1)
r₁ = \(\sqrt{49+9-33}\) = \(\sqrt{25}\) = 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 y² = 4ax and these tangents make angles θ₁, θ₂ with its axis, such that cot θ₁ + cot θ₂ is a constant ‘d’. Then show that all such P lie on a horizontal line.
Solution:
Equation of a tangent to the Parabola y² = 4ax is a
y = mx + \(\frac{a}{m}\)
This tangent passes through P(x₁y,)
∴ y₁ = mx₁ + \(\frac{\mathrm{a}}{\mathrm{m}}\)
⇒ my₁ = m²x₁ + a
⇒ m²x₁ – my₁ + a = 0 which is a quadratic eq. in m
If m₁, m₂ are the slopes of the tangents drawn from P then m₁, m₂ are the roots of the equation m²x₁ – my₁ + 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 = ∫cotⁿ x dx; n being a positive integer; n ≥ 2 and deduce the value of ∫cot⁴ 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.