AP Inter 2nd Year Maths 2B Question Paper March 2020

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AP Inter 2nd Year Maths 2B Question Paper March 2020

Time : 3 Hours
Max. Marks : 75

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

Section – A (10 × 2 = 20)

I. Very Short Answer Type Questions.

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

Question 1.
Find the other end of the diameter of the circle x² + y² – 8x – 8y + 27 = 0 if one end of it is (2, 3).
Solution:
A = (2, 3) and AB is the diameter of the circle x² + y² – 8x – 8y + 27 = 0

Question 2.
Define chord of contact and find the chord of contact of (1, 1) to the circle x² + y² = 9.
Solution:
Equation of the circle is x² + y² = 9
Equation of the chord of contact is
x.1 + y.1 = 9
i.e. x + y = 9

Question 3.
Find k if the circles x² + y² – 5x – 14y – 34 = 0 and x² + y² + 2x + 4y + k = 0 are orthogonal.
Solution:
g₁ = \(\frac{-5}{2}\) ;
f₁ = -7 ;
c₁ = -34
g₂ = 1 ; f₂ = 2 ; c₂ = k
Two circles are said to be orthogonal
2g₁g₂ + 2f₁f₂ = c₁ + c₂
2\(\left(\frac{-5}{2}\right)\)(1) + 2(-7)(2) = -34 + k
-5 – 28 = -34 + k
– 33 = – 34 + k
k = 34 – 33 ⇒ k = 1

Question 4.
Find the equation of the parabola whose vertex is (3, -2) and focus is (3, 1).
Solution:
The abcissae of the vertex and focus are equal to 3. Hence the axis of the parabola is x = 3, a line parallel to y-axis. focus is above the vertex.
a = distance between focus and vertex = 3.
∴ Equation of the parabola
(x – 3)² = 4(3) (y + 2)
i.e.,(x – 3)² = 12(y + 2)

Question 5.
If 3x – 4y + k = 0 is a tangent to the hyperbola x² – 4y² = 5, find the value of k.
Solution:
Equation of the hyperbola x² – 4y² = 5
\(\frac{x^2}{5}\) – \(\frac{y^2}{\left(\frac{5}{4}\right)}\) = 1
a² = 5, b² = \(\frac{5}{4}\)
Equation of the given line is 3x – 4y + k = 0
4y = 3x + k
y = \(\frac{3}{4}\)x + \(\frac{k}{4}\)
m = \(\frac{3}{4}\) c = \(\frac{k}{4}\),
condition for tangency is c² = a²m² – b²
\(\frac{k^2}{16}\) = 5.\(\frac{9}{16}\) – \(\frac{5}{4}\)
k² = 45 – 20 = 25
k = ±5

Question 6.
Evaluate : ∫\(\frac{\cos x}{(1+\sin x)^2}\)dx.
Solution:
∫\(\frac{\cos x d x}{(1+\sin x)^2}\)
t = 1 + sin x ⇒ dt = cos x dx
\(\int \frac{\cos x d x}{(1+\sin x)^2}\) = \(\int \frac{d t}{t^2}\) = –\(\frac{1}{t}\) + C
= –\(\frac{1}{1+\sin x}\) + C

Question 7.
Evaluate : ∫x log x dx on (0, ∞).
Solution:
We take the First function U = log x abd
Second Function V = X
From the “By parts rule” we have

Question 8.
Evaluate : \(\lim _{n \rightarrow \infty} \frac{1+2^4+3^4+\ldots \ldots \ldots+n^4}{n^5}\)
Solution:

Question 9.
Find : \(\int_{-\pi / 2}^{\pi / 2}\)sin²xcos⁴x dx
Solution:

Question 10.
Solve: y(1 + x)dx + x(1 + y)dy = 0.
Solution:
The given equation can be written as
\(\frac{(1+x)}{x}\).dx + \(\frac{(1+y)}{y}\).dy = 0
\(\int\left(1+\frac{1}{x}\right)\) dx + \(\int\left(1+\frac{1}{y}\right)\) dy = 0
x + log x + y + log y = c
x + y + log (xy) = c is the required solution.

Section – B
(5 × 4 = 20 Marks)

II. Short Answer Type questions.

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

Question 11.
Find the area of the triangle formed by the tangent at P(x₁, y₁) to the circle \(x_1^2\) + \(y_1^2\) = a² with the co-ordinate axes where x₁y₁ ≠ 0.
Solution:
Equation of the circle is x² + y² = a²
Equation of the tangent at p(x₁, y₁) is
xx₁ + yy₁ = a² ……… (1)

This tangent cuts X – axis at A and Y – axis at B
Changing into intercept form

Question 12.
If the two circles x² + y² + 2gx + 2fy = 0 and x² + y² + 2g’x + 2fy = 0 touch each other then show that f’g = fg’.
Solution:

Question 13.
S and T are the foci of an ellipse and B is one end of the minor axis. If STB is an equilateral triangle, then find die eccentricity of the ellipse.
Solution:

Let \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1 (a > b) be an ellipse whose foci are S and T, B is an end of the minor axis such that STB is equilateral, then SB = ST = TB. We have S(ae,0).
T = (-ae, 0) and B(0, b)
Consider SB = ST ⇒ (SB)² = (ST)² ⇒ (ae)² + b² = 4a2e²
∴ a²e² + a² (1 – e)² = 4a²e²
[∵ b² = a²(1 – e²)]
e² = \(\frac{1}{4}\)
∴ Eccentricity of the ellipse is \(\frac{1}{2}\).

Question 14.
Find the condition for the line x cos α + y sin α = P to be a tangent to the ellipse \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1.
Solution:
Equation of the ellipse is
\(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1 ……. (1)
Equation of the line is x cos α + y sin α = p
y sin α = -x cos α + p

Question 15.
Find the centre, foci, eccentricity, equation of the directrices of the hyperbola x² – 4y² = 4.
Solution:
Equation of the hyperbola is \(\frac{x^2}{4}\) – \(\frac{y^2}{1}\) = 1
a² = 4, b² = 1
Centre is c (0,0)
a²e² = a² + b² = 4 + 1 = 5
ae = \(\sqrt{5}\)
Foci are (±ae, 0) = (±\(\sqrt{5}\), 0)
Eccentricity = \(\frac{\mathrm{ae}}{\mathrm{a}}\) = \(\frac{\sqrt{5}}{2}\)
Equations of directrices are x = ±\(\frac{a}{e}\)
= ±2. \(\frac{2}{\sqrt{5}}\)
⇒ \(\sqrt{5}\)x = ±4
⇒ \(\sqrt{5}\)x ± 4 = 0
Length of the latus rectum = \(\frac{2 b^2}{a}\) = \(\frac{2.1}{2}\)

Question 16.
Find the area of the region bounded by the parabolas y² = 4x and x² = 4y.
Solution:
Equations of the given curves are
y² = 4x ……… (1)
x² = 4y = 4y …….. (2)
\(\left(\frac{x^2}{4}\right)^2\) = 4x
\(\frac{x^4}{16}\) = 4x
x⁴ = 64 ⇒ x⁴ = 0 or x³ = 64, x = 4

Question 17.
Solve : (x² + y²)dx = 2xy dy.
Solution:
Given equation can be written as
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{x^2+y^2}{2 x y}\)
This is a homogeneous function.

Section – C

III. Long Answer Type questions.

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

Question 18.
Find the equation of a circle which passes through (4,1) and (6,5) and having the centre on 4x + 3y – 24 = 0.
Solution:
Equation of circle be
x² + y² + 2gx + 2fy + c = 0 passes through (4, 1) and (6, 5) then
4² + 1² + 2g(4) + 2f(1) + c = 0 …….. (i)
6² + 5² + 2g(6) + 2f(5) + c = 0 …….. (ii)
Centre lie on 4x + 3y – 24 = 0
∴ 4(-g) + 3 (-f) – 24 = 0 ……….. (iii)
(ii) – (i) we get
44 + 4g + 8f = 0 ………. (iv)
Solving (iii) and (iv) we get f = -4, g = -3, c = -15
∴ Required equation of circle is x² + y² – 6x – 8y + 15 = 0

Question 19.
Find the equation of the circle which touches the circle x² + y² – 2x – 4y – 20 = 0 externally at (5, 5) with radius 5.
Solution:
x² + y² – 2x – 4y – 20 = 0
C = (1, 2)

h = 9 k = 8
Equation of circle be
(x – 9)² + (y – 8)² = 25
x² + y² – 18x – 16y + 120 =0
If (h, k) is the centre of the required circle (5, 5) is the mid-point of (1, 2) and (h, k).

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 tan θ₁ + tan θ₂ is a constant b. Then show that P lies on the line y = bx.
Solution:
Let the coordinates of P be (x₁, y₁) and the equation of the parabola y₂ = 4ax. Any tangent to the parabola is y = mx + \(\frac{a}{m}\), if this passes through (x₁, y₁) then
y₁ = mx₁ + \(\frac{\mathrm{a}}{\mathrm{m}}\)
i.e., m²x₁ – my₁ + a = 0 ………….. (1)
Let the roots of (1) be m₁, m₂.

[∴ The tangents make angles θ₁, θ₂ with its axis (x-axis) then their slopes m₁ = tan θ₁, and m₂ = tanθ₂].
∴ b = \(\frac{y_1}{x_1}\) ⇒ y₁ = bx₁
∴ p(x₁, y₁) lies on the line y = bx

Question 21.
Evaluate: ∫\(\frac{1}{1+\sin x+\cos x} d x\)
Solution:

Question 22.
If I_n = ∫cosⁿx dx, then show that I_n = \(\frac{1}{n}\)cos^n-1x sin x + \(\frac{n-1}{n}\)I_n-2 (where n ≥ 2)
Solution:
I_n = ∫cosⁿx dx = ∫cos^n-1x.cos x dx
= cos^n-1x.sin x – ∫sin x. (n – 1)cos^n-2x(-sin x)dx
= cos^n-1x.sin x + (n – 1)∫cos^n-2x(1 – cos²x)dx
= cos^n-1x.sinx + (n – 1)I_n-2 – (n – 1)I_n
∴ I_n(1 + n – 1) = cos^n-1x.sin x + (n – 1)I_n-2
I_n = \(\frac{\cos ^{n-1} x \sin x}{n}\) + \(\frac{n-1}{n}\)I_n-2

Question 23.
Show that : \(\int_0^{\pi / 2} \frac{x}{\sin x+\cos x}\)dx = \(\frac{\pi}{2 \sqrt{2}}\)log (\(\sqrt{2}\) + 1)
Solution:

Question 24.
Solve : x log x\(\frac{d y}{d x}\) + y = 2 log x.
Solution:
\(\frac{d y}{d x}\) + \(\frac{1}{x \log x}\).y = \(\frac{2}{x}\)
I.F. = \(e^{\int \frac{d x}{x \log x}}\) = e^{log(log x)} = log x
y.log x = 2∫\(\frac{\log x}{x}\)dx
= (log x)² + c