AP Inter 2nd Year Maths 2B Question Paper May 2018

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AP Inter 2nd Year Maths 2B Question Paper May 2018

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 equation of the circle passing through (3, 4) and having the centre at (-3, 4).
Solution:
Let P = (3, 4) and d = (-3, 4)
r = dp
= \(\sqrt{(-3-3)^2+(4-4)^2}\)
= \(\sqrt{36}\)
= 6
∴ The equation of the circle passing through (3, 4) and having the centre at (-3, 4) is
(x + 3)² + (y – 4)² = 6²
⇒ x² + 6x + 9 + y² – 8y + 16 = 36
⇒ x² + y² + 6x – 8y – 11 = 0

Question 2.
If the length of the tangent from (5, 4) to the circle x² + y² + 2ky = 0 is ‘1’ then find ‘k’.
Solution:
Let S = x² + y² + 2ky = 0
Given \(\sqrt{S_{11}}\) = 1
⇒ \(\sqrt{25+16+2 . k .4}\) = 1
⇒ 41 + 8k = 1
⇒ 8k = -40
⇒ k = -5

Question 3.
Find the angle between the circles x² + y² – 12x – 6y + 41 = 0, x² + y² + 4x + 6y – 59 = 0.
Solution:
Given circle equations are x² + y² – 12x – 6y + 41 = 0
x² + y² + 4x + 6y – 59 = 0
d₁ = (6, 3) and r₁ = \(\sqrt{36+9-41}\) = 2
d₂ = (-2, -3) and r₂ = \(\sqrt{4+9+59}\) = \(\sqrt{72}\) = 0
d = d₁d₂ = \(\sqrt{(-2-6)^2+(-3-3)^2}\) = \(\sqrt{64+36}\) = 10
Let ‘θ’ be the angle between the circles
∴ cos θ = \(\frac{d^2-r_1^2-r_2^2}{2 r_1 r_2}\)
= \(\frac{100-4-72}{2.2 .6 \sqrt{2}}\)
= \(\frac{24}{24 \sqrt{2}}\)
= \(\frac{1}{\sqrt{2}}\)
∴ θ = 45°

Question 4.
If (\(\frac{1}{2}\), 2) is one extremity of a focal chord of the parabola y² = 8x, find the co-ordinates of the other extremity.
Solution:
Given parabola equation is y² = 8x
Here 4a = 8 ⇒ a = 2
Given that one extremity of focal chord
(at², 2at) = (\(\frac{1}{2}\), 2)
⇒ 2at = 2
⇒ at = 1
⇒ 2.t = 1
⇒ t = \(\frac{1}{2}\)
∴ The other extremity of focal chord = \(\left(\frac{a}{t^2}, \frac{-2 a}{t}\right)\)
= \(\left(\frac{2}{\left(\frac{1}{2}\right)^2}, \frac{-2.2}{\frac{1}{2}}\right)\)
= (8, -8)

Question 5.
Find the equation of the hyperbola whose foci are (±5, 0), the transverse axis is of length 8.
Solution:
Given Foci are (±5, 0)
∴ ae = 5
Length of transverse axis = 8
⇒ 2a = 8
⇒ a = 4
ae = 5 ⇒ 4e = 5 ⇒ e = \(\frac{5}{4}\)
We know b² = a²(e² – 1)
= 16(\(\frac{25}{16}\) – 1)
= 25 – 16
= 9
∴ Equation of the hyperbola is \(\frac{x^2}{a^2}\) – \(\frac{y^2}{b^2}\) = 1
⇒ \(\frac{x^2}{16}\) – \(\frac{y^2}{9}\) = 1
⇒ 9x² – 16y² = 144

Question 6.
Evaluate: \(\int \frac{(3 x+1)^2}{2 x}\)dx x ∈ I ⊂ R – {0}
Solution:

Question 7.
Evaluate : \(\int \frac{\log x}{x^2}\)dx on (0, ∞)
Solution:

Question 8.
Evaluate : \(\lim _{n \rightarrow \infty} \frac{2^k+4^k+6^k+\ldots \ldots+(2 n)^k}{n^{k+1}}\) by using the method of finding definite integral of the limit of a sum.
Solution:

Question 9.
Find the value of \(\int_0^{2 \pi}\)sin²x cos⁴x dx .
Solution:
\(\int_0^{2 \pi}\)sin²cos⁴x dx = 4\(\int_0^{\frac{\pi}{2}}\)sin²x.cos⁴x dx
= 4.\(\frac{4-1}{2+4}\).\(\frac{4-3}{2+4-2}\).\(\frac{1}{2}\).\(\frac{\pi}{2}\)
= 4.\(\frac{3}{6}\).\(\frac{1}{4}\).\(\frac{1}{2}\).\(\frac{\pi}{2}\)
= \(\frac{\pi}{8}\)
∴ \(\int_0^{2 \pi}\)sin²xcos⁴x dx = \(\frac{\pi}{8}\)

Question 10.
Solve \(\frac{d y}{d x}\) = e^x-y + x²e^-y.
Solution:
\(\frac{d y}{d x}\) = e^{x – y} + x²e^-y
⇒ \(\frac{d y}{d x}\) = 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

Section – B

II. Short Answer Type Questions.

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

Question 11.
Find the equation of the circle with centre (-2, 3) cutting a chord length 2 units on 3x + 4y + 4 = 0.
Solution:
Given chord equation is 3x + 4y + 4 = 0
Centre d = (-2, 3)
d = Length of the perpendicular from d to the line \(\overline{\mathrm{AB}}\)
= \(\frac{|3(-2)+4(3)+4|}{\sqrt{3^2+4^2}}\)
= \(\frac{|-6+12+4|}{\sqrt{25}}\)
= \(\frac{10}{5}\)
Given length of the chord = 2
⇒ 2\(\sqrt{r^2-d^2}\) = 2
⇒ \(\sqrt{r^2-d^2}\) = 1
⇒ r² – d² = 1
⇒ r² – 4 = 1
⇒ r² = 5
⇒ r = \(\sqrt{5}\)
Required circle equation is (x – h)² + (y – k)² = r²
⇒ (x + 2)² + (y – 3)² = (\(\sqrt{5}\))²
⇒ x² + 4x + 4 + y² – 6y + 9 = 5
⇒ x² + y² + 4x – 6y + 8 = 0

Question 12.
Find the equation of the circle which passes through origin and intersects the circles x² + y² – 4x + 6y + 10 = 0, x² + y² + 12y + 6 = 0 orthogonally.
Solution:
Let x² + y² + 2gx + 2fy + c = 0 …….. (1) be the required circle.
If (1) passes through origin then c = 0
If (1) is orthogonal to x² + y² – 4x + 6y + 10 = 0 then 2g(-2) + 2f(3) = c + 10
⇒ -4g + 6f = 0 + 10
⇒ -2g + 3f = 5
⇒ 2g – 3f + 5 = 0 ………… (2)
If (1) is orthogonal to x² + y² + 12y + 6 = 0 then 2g(0) + 2f (6) = c + 6
⇒ 12f = 6
⇒ f = \(\frac{1}{2}\)
From (2)
2g – 3(\(\frac{1}{2}\)) + 5 = 0
⇒ 4g – 3 + 10 = 0
⇒ 4g + 7 = 0
⇒ g = \(\frac{-7}{4}\)
Hence required circle equation is x² + y² + 2(\(\frac{-7}{4}\))x + 2(\(\frac{1}{2}\))y + 0 = 0
⇒ x² + y² – \(\frac{7}{2}\)x + y = 0
⇒ 2x² + 2y² – 7x + 2y = 0

Question 13.
Find the length of the latus rectum, eccentricity, centre and foci of the ellipse 4x² + y² – 8x + 2y + 1 = 0.
Solution:
Given ellipse equation is 4x² + y² – 8x + 2y + 1 = 0
⇒ 4(x² – 2x + 1 – 1) + (y² + 2y + 1 – 1) + 1 = 0
⇒ 4 (x – 1 )² – 4 + (y + 1 )² – 1 + 1 = 0
⇒ 4(x – 1)² + (y + 1)² = 4
⇒ \(\frac{(x-1)^2}{1}\) + \(\frac{(y+1)^2}{4}\) = 1
Here a² = 1 ⇒ a = 1
b² = 4 ⇒ b = 2
∴ a < b
∴ Length of the latus rectum = \(\frac{2 a^2}{b}\) = \(\frac{2.1}{2}\) = 1
eccenticity = \(\sqrt{\frac{b^2-a^2}{b^2}}\) = \(\sqrt{\frac{4-1}{4}}\) = \(\frac{\sqrt{3}}{2}\)
centre = (-1, 1)
Foci = (-1, 1±2.\(\frac{\sqrt{3}}{2}\))
= (-1, 1±\(\sqrt{3}\))

Question 14.
A circle of radius 4 is concentric with the ellipse 3x² + 13y² = 78. Prove that a common tangent is inclined to the major axis at an angle \(\frac{\pi}{4}\).
Solution:
Given ellipse equation is 3x² + 13y² = 78
⇒ \(\frac{3 x^2}{78}\) + \(\frac{13 y^2}{78}\) = 1
⇒ \(\frac{x^2}{26}\) + \(\frac{y^2}{6}\) = 1
∴ Centre of the ellipse is (0, 0)
∴ Equation of the circle is (x – 0)² + (y – 0)² = 4²
⇒ x² + y² = 16
Equation of tangent at p(θ) to the circle is x(4 cos θ) + y (4 sin θ) = 16
x cos θ + y sin θ = 4
⇒ y sin θ = – x cos θ + 4
⇒ y = \(\left(\frac{-\cos \theta}{\sin \theta}\right)\) + \(\frac{4}{\sin \theta}\) ………. (2)
If (2) s a tangent to the ellipse then
c² = a²m² + b²
⇒ \(\left(\frac{4}{\sin \theta}\right)^2\left(\frac{-\cos \theta}{\sin \theta}\right)^2\) = 26\(\left(\frac{-\cos \theta}{\sin \theta}\right)^2\) + 6
⇒ \(\frac{16}{\sin ^2 \theta}\) = 26.\(\frac{\cos ^2 \theta}{\sin ^2 \theta}\) + 6
⇒ 16 = 26 cos²θ + 6(1 – cos²θ)
⇒ 16 = 26 cos²θ + 6 – 6 cos²θ
⇒ 20 cos²θ = 10
⇒ cos²θ = \(\frac{1}{2}\)
⇒ cos θ = \(\frac{1}{\sqrt{2}}\)
∴ θ = \(\frac{\pi}{4}\)

Question 15.
Tangents to the hyperbola \(\frac{x^2}{a^2}\) – \(\frac{y^2}{b^2}\) = 1 angles θ₁, θ₂ with transverse axis of a hyperbola.
Show that the point of intersection of these tangents lies on the curve 2xy = k(x² – a²) when tan θ₁ + tan θ₂ = k.
Solution:
Given hyperbola equation is \(\frac{x^2}{a^2}\) – \(\frac{y^2}{b^2}\) = 1
Equations of the tangents to the hyperbola can be taken as y = mx ± \(\sqrt{a^2 m^2-b^2}\)
Let P(x₁, y₁) be the point of intersection of the tangents.

⇒ 2x₁y₁ = K(\(x_1^2\) – a²)
∴ The point P(x₁, y₁) lies on the curve
2xy = K(x² – a²)

Question 16.
Evaluate : \(\int_0^{\frac{\pi}{4}}\)log(1 + tanx)dx.
Solution:

= \(\frac{\pi}{4}\)log 2
∴ I = \(\frac{\pi}{8}\)log 8
Hence \(\int_0^{\pi / 4}\)log(1 + tan x)dx = \(\frac{\pi}{8}\) log 2

Question 17.
Solve : \(\frac{1}{x} \frac{d y}{d x}\) + y e^x = \(e^{(1-x) e^x}\)
Solution:
Given \(\frac{1}{x} \cdot \frac{d y}{d x}\) + y.e^x = x.\(e^{-(x-1) e^x}\) …….. (1)
This is a linear differential equation of first order is y.

Section – C

III. Long Answer Type Questions.

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

Question 18.
Find the equation of the circle passing through (2, 1), (5, 5) and (-6, 7).
Solution:
Let x² + y² + 2gx + 2fy + c = 0 ………. (1) be the required circle.
If (1) passes through (2, 1) then
4 + 1 + 2g(2) + 2f(1) + c = 0
⇒ 4g + 2f + c = -5 …….. (2)
If (1) passes through (5, 5) then
25 + 25 + 2g(5) + 2f(5) + c = 0
⇒ 10g + 10f + c = -50 ……. (3)
If (1) passes through (-6, 7) then
36 + 49 + 2g(-6) + 2f(7) + c = 0
⇒ 85 – 12g + 14f + c = 0
⇒ 12g – 14f – c = 85 ……….. (4)
(3) – (2) ⇒ 6g + 8f = – 45 ⇒ 6g + 8f + 45 = 0 ………. (5)
(4) + (3) ⇒ 22g – 4f = 35 ⇒ 22g – 4f – 35 = 0 ………. (6)
Solving (5) and (6), we get

\(\frac{g}{-280+180}\) = \(\frac{f}{990+210}\) = \(\frac{1}{-24-176}\)
\(\frac{g}{-100}\) = \(\frac{f}{1200}\) = \(\frac{1}{-200}\)
g = \(\frac{1}{2}\), f = -6
From (2)
4(\(\frac{1}{2}\)) + 2(-6) + c = -5
2 – 12 + c = -5
c = 5
∴ g = \(\frac{1}{2}\), f = -6, c = 5
Hence required circle equation is
x² + y² + 2\(\left(\frac{1}{2}\right)\)x + 2(-6)y + 5 = 0
⇒ x² + y² + x – 12y + 5 = 0.

Question 19.
Show that x² + y² – 6x – 9y + 13 = 0, x² + y² – 2x – 16y =0, circles touch each other. Find the point of contact and the equation of common tangent at their point of contact.
Solution:
Let S₁ = x² + y² – 6x – 9y + 13 = 0
s₂ = x² + y² – 2x – 16y = 0
d₁ = (3, \(\frac{9}{2}\)), d₂ = (1, 8)

Equation of the common tangent is S₁ – S₂ = 0
⇒ -4x + 7y + 13 = 0
4x – 7y – 13 = 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:
Axis is parallel to X-axis, general equation be x = ay² + by + c ……… (1)
(1) Passes through (-2, 1)
⇒ 2 = a + b + c ……………. (2)
(1) Passes through (1, 2)
⇒ 1 = 4a + 2b + c ………. (3)
(1) Passes through (-1, 3)
⇒ – 1 = 9a + 3b + c …….. (4)
(3) – (2) ⇒ 3a + b = 3 ……… (5)
(4) – (3) ⇒ 5a + b = – 2 …… (6)

Required parabola equation is
x = –\(\frac{-5}{2}\)y² + \(\frac{21}{2}\)y – 10
⇒ 2x = -5y² + 21y – 20
⇒ 5y² + 2x – 21y + 20 = 0.

Question 21.
Evaluate : ∫(6x + 5)\(\sqrt{6-2 x^2+x}\) dx
Solution:
Let 6x + 5 = A(1 – 4x) + B
Equating the coefficients of x
6 = -4A ⇒ A = \(\frac{-3}{2}\)
Equating the constants
A + B = 5
B = 5 – A = 5 + \(\frac{3}{2}\) = \(\frac{13}{2}\)

Question 22.
If I_n = ∫cosⁿ x dx then show that I_n = \(\frac{1}{n}\)cos^n-1x sin x + \(\frac{n-1}{n}\)I_n-2, n being a positive integer n ≥ 2 and deduce the value of ∫cos⁵x dx.
Solution:

Question 23.
Evaluate: \(\int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x\)
Solution:

Question 24.
Solve (x³ – 3xy²) dx + (3x²y – y³) dy = 0.
Solution:
Given (x³ – 3xy²) dx + (3x²y – y³) dy = 0
⇒ (x³ – 3xy²) dx = -(3x²y – y³) dy
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{x^3-3 x y^2}{-3 x^2 y+y^3}\)
which is a homogeneous equation.
Put y = vx