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

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

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.
Find the centre and radius of the circle : x² + y² – 4x – 8y – 41 =0.
Solution:
x² + y² – 4x – 8y – 41 = 0 …… (i)
x² + y² + 2gx + 2fy – 41 = 0 …… (ii)
Comparing (i) and (ii) we get
g = -2, f = -4, c = -41
Radius = \(\sqrt{g^2+f^2-c}\)
= \(\sqrt{4+16+41}\)
= \(\sqrt{61}\) units
Centre = (-g, -f) = (2, 4)

Question 2.
If the length of the tangent from (5, 4) to the circle x² + y² + 2ky = 0 is 1, then find k.
Solution:
Length of tangent
= \(\sqrt{S_{11}}\) = \(\sqrt{(5)^2+(4)^2+8 k}\)
But, length of tangent
∴ 1 = \(\sqrt{25+16+8 k}\)
Squaring both sides we get 1 = 41 + 8k
k = – 5 units.

Question 3.
Find the equation of the common chord of the circles : x² + y² – 4x – 4y + 3 = 0 and x² + y² – 5x – 6y + 4 = 0.
Solution:
(x² + y² – 4x – 4y + 3) – (x² + y² – 5x – 6y + 4) = 0
x + 2y – 1 = 0 Equation of common chord.

Question 4.
Find the equation of the tangent to the parabola y² = 6x at the positive end of the latus rectum.
Solution:
(a, 2a) Here, 4a = 6 ⇒ a = \(\frac{3}{2}\)
(\(\frac{3}{2}\), 3)
equation of tangent yy₁ = 2a (x + x₁)
= 3(x + x₁)
3y = 3(x + \(\frac{3}{2}\))
2y – 2x – 3 = 0 is the equation of tangent
Slope of tangent is 1
Slope of normal is -1
Equation of normal is y – 3 = – 1(x – \(\frac{3}{2}\))
2x + 2y – 9 = 0

Question 5.
If the angle between the asymptotes of the hyperbola is 30°, then find its eccentricity.
Solution:
Angle between the asymptotes = 2θ = 30°

Question 6.
Evaluate the integral \(\int \frac{\sin ^2 x}{1+\cos 2 x}\) dx on I ⊂ R\{(2n ± 1)π: n ∈ Z}.
Solution:

Question 7.
Evaluate the integral : \(\int \frac{2 x+1}{x^2+x+1}\) dx, x ∈ R.
Solution:
t = x² + x + 1
dt = (2x + 1)dx
\(\int \frac{2 x+1}{x^2+x+1}\) dx = \(\int \frac{d t}{t}\)
= log|t| + C
= log|x² + x + 1| + C

Question 8.
Evaluate the definite integral: \(\int_2^3 \frac{2 x}{1+x^2} d x\)
Solution:

Question 9.
Find the area of the region enclosed by y = x³ + 3, y = 0, x = -1, x = 2.
Solution:
Required are PABQ

Question 10.
Form the differential equation corresponding to: y = A cos 3x + 6 sin 3x, where A and B are parameters.
Solution:
We have y = A cos 3x + B sin 3x
Differentiating w.r.to x dy
\(\frac{d y}{d x}\) = -3A sin 3x + 3B cos 3x dx
Differentiating again w.r.to. x
\(\frac{d^2 y}{d x^2}\) = – 9(A cos 3x + B sin 3x)
= -9(A cos 3x + B sin 3x)
= -9y
is \(\frac{d^2 y}{d x^2}\) + 9y = 0.

Alternate method :

Eliminating A, B from the equation
y = A cos 3x + B sin 3x
\(\frac{d y}{d x}\) = -3A sin 3x + 3B sin cos 3x

y(27 sin² 3x + 27 cos² 3x) – (-9 sin 3x.cos3x + 9cos3x.sin3x)\(\frac{d y}{d x}\) + (3 cos² 3x + 3 sin² 3x)\(\frac{d^2 y}{d x^2}\) = o
27y + 3.\(\frac{d^2 y}{d x^2}\) = 0 or \(\frac{d^2 y}{d x^2}\) + 9y = 0
This is the required differential equation.

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 length of the chord intercepted by the circle : x² + y² – x + 3y – 22 = 0 on the line y = x – 3.
Solution:
Equation of the circle is
S ≡ x² + y² – x + 3y – 22 = 0

Equation of the line is y = x – 3 ⇒ x – y – 3 = 0
P = distance from the centre

Question 12.
Show that the angle between the circles x² + y² = a², x² + y² = ax + ay is \(\frac{3 \pi}{4}\).
Solution:

Question 13.
Find the length of latus rectum, eccentricity, foci and the equations of directrices of the ellipse : 9x² + 16y² = 144.
Solution:
Given equation is 9x² + 16y² = 144
\(\frac{x^2}{16}+\frac{y^2}{9}\) = 1
Length of major axis = 2a = 2. 4 = 8
Length of minor axis = 2b = 2. 3 = 6
Length of latus rectum = \(\frac{2 b^2}{a}\) = \(\frac{2.9}{4}\) = \(\frac{9}{2}\)
Eccentricity = \(\sqrt{\frac{a^2-b^2}{a^2}}\) = \(\sqrt{\frac{16-9}{16}}\) = \(\sqrt{\frac{7}{4}}\)
Centre is c (0,0)
Foci are (± ae, 0) = ( ± \(\sqrt{7}\), 0)
Equations of the directrices are x = ±\(\frac{\mathrm{a}}{\mathrm{e}}\)
x = ±4.\(\frac{4}{\sqrt{7}}\) = ±\(\frac{16}{\sqrt{7}} \sqrt{7} x\) = ±16

Question 14.
Find the equation of tangent and normal to the ellipse x² + 8y² = 33 at (-1, 2).
Solution:
Equation of the tangent is
\(\frac{x x_1}{a^2}\) + \(\frac{y y_1}{b^2}\) = 1
x(-1) + 8y(2) = 33
⇒ -x + 16y = 33
⇒ x – 16y + 33 = 0
Equation of the normal is
-16x + 2 + k = 0
It passes through P(-1, 2)
-16 + 2 + k = 0 ⇒ k = 14
Equation of the normal is
16x + y + 14 = 0

Question 15.
Find the equations of the tangents to the hyperbola : 3x² – 4y² = 12 which are :
i) Parallel and
ii) Perpendicular to the line y = x – 7.
Solution:
i) Equation of the hyperbola is 3x² – 4y² = 12
\(\frac{x^2}{4}-\frac{y^2}{3}\) = 1
a² = 4, b² = 3
The tangent is parallel to y = x – 7
m = slope of the tangent = 1
Equation of the parallel tangents are
y = mx ± \(\sqrt{a^2 m^2-b^2}\)
y = x ± \(\sqrt{4-3}\)
y = x ± 1

ii) The tangent is perpendicular to y – x = 7
m – slope of the tangent = (-1)
Equation of the perpendicular tangents are
y = (-1)x ± \(\sqrt{4(-1)^2-3}\)
= -x ± 1
x + y = ± 1

Question 16.
Evaluate: \(\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x\)
Solution:

Question 17.
Solve the differential equation: (xy² + x)dx + (yx² + y)dy = 0
Solution:
(xy² + x) dx + (yx² + y) dy = 0
x(y² + 1) dx + y (x² + 1) dy = 0
Dividing with (1 + x²) (1 + y²)
\(\frac{x d x}{1+x^2}\) + \(\frac{y d y}{1+y^2}\) = 0
Integrating
\(\int \frac{x d x}{1+x^2}\) + \(\int \frac{y d y}{1+y^2}\) = 0
\(\frac{1}{2}\)[(log (1 + x²) + log (1 + y²)] = log c
log (1 + x²) (1 + y²) = 2 log c = log c²
Solution is (1 + x²) (1 + y²) = k when k = c².

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 (2, -3) and (- 4, 5) and having the centre on 4x + 3y + 1 = 0.
Solution:
x² + y² +2gx + 2fy + c = 0 ……. (i)
Equation (i) passes through (2, – 3), (- 4, 5)
∴ 4 + 9 + 4g – 6f + c = 0 …… (ii)
16 + 25 – 8g + 10f + c = 0 ……….. (iii)
Equation (iii) – (ii) we get
28 – 12g + 16f = 0
(or) 3g – 4f = 7
Centre lies on (- g, – f) lies on 4x + 3y + 1 = 0
then 4(- g) + 3(- f) + 1 = 0
3g – 4f – 7 = 0
Solving we get f = – 1
g = 1
Now, substituting f, g values in equation (ii) we get
4 + 9 + 4(1) – 6(-1) + c = 0, c = -23
x² + y² + 2x – 2y – 23 = 0 is required equation of circle.

Question 19.
Show that: x² + y² – 6x – 9y + 13 = 0, x² + y² – 2x – 16y = 0 touch each other. Find the point of contact and the equation of common tangent at their point of contact.
Solution:
Equations of the circles are

∴ The circles touch each other internally. The point of contact ‘p’ divides AB externally in the ratio r₁ : r₂ = \(\frac{\sqrt{65}}{2}\) : \(\sqrt{65}\) = 1 : 2 co-ordinates of p are

∴ Equation of the common tangent is S₁ – S₂ = 0
-4x + 7y + 13 = 0
4x – 7y – 13 = 0

Question 20.
Derive the equation of a parabola in standard form.
Solution:
Equation of Parabola : General form – Let S(α, β) be the focus and ax + by + c = 0 be directrix then by definition the equation of parabola be

Question 21.
Evaluate : \(\int \frac{d x}{(1+x) \sqrt{3+2 x-x^2}}\) on (1, 3).
Solution:

Question 22.
Obtain the reduction formula for ∫sinⁿ x dx for an integer n ≥ 2 and deduce the value of : ∫sin⁴x dx
Solution:

Question 23.
Evaluate : \(\int_0^1 \frac{\log (1+x)}{1+x^2}\) dx
Solution:

Question 24.
Solve the differential equation : sin^-1\(\left(\frac{d y}{d x}\right)\) = x + y.
Solution: