# 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)

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

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 In = ∫cosnx dx, then show that In = $$\frac{1}{n}$$cosn-1x sin x + $$\frac{n-1}{n}$$In-2 (where n ≥ 2)
Solution:
In = ∫cosnx dx = ∫cosn-1x.cos x dx
= cosn-1x.sin x – ∫sin x. (n – 1)cosn-2x(-sin x)dx
= cosn-1x.sin x + (n – 1)∫cosn-2x(1 – cos2x)dx
= cosn-1x.sinx + (n – 1)In-2 – (n – 1)In
∴ In(1 + n – 1) = cosn-1x.sin x + (n – 1)In-2
In = $$\frac{\cos ^{n-1} x \sin x}{n}$$ + $$\frac{n-1}{n}$$In-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. = $$e^{\int \frac{d x}{x \log x}}$$ = elog(log x) = log x
y.log x = 2∫$$\frac{\log x}{x}$$dx
= (log x)2 + c