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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.

- Attempt all questions.
- 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} = 6^{2}

⇒ x^{2} + 6x + 9 + y^{2} – 8y + 16 = 36

⇒ x^{2} + y^{2} + 6x – 8y – 11 = 0

Question 2.

If the length of the tangent from (5, 4) to the circle x^{2} + y^{2} + 2ky = 0 is ‘1’ then find ‘k’.

Solution:

Let S = x^{2} + y^{2} + 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 x^{2} + y^{2} – 12x – 6y + 41 = 0, x^{2} + y^{2} + 4x + 6y – 59 = 0.

Solution:

Given circle equations are x^{2} + y^{2} – 12x – 6y + 41 = 0

x^{2} + y^{2} + 4x + 6y – 59 = 0

d_{1} = (6, 3) and r_{1} = \(\sqrt{36+9-41}\) = 2

d_{2} = (-2, -3) and r_{2} = \(\sqrt{4+9+59}\) = \(\sqrt{72}\) = 0

d = d_{1}d_{2} = \(\sqrt{(-2-6)^2+(-3-3)^2}\) = \(\sqrt{64+36}\) = 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 y^{2} = 8x, find the co-ordinates of the other extremity.

Solution:

Given parabola equation is y^{2} = 8x

Here 4a = 8 ⇒ a = 2

Given that one extremity of focal chord

(at^{2}, 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 b^{2} = a^{2}(e^{2} – 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

⇒ 9x^{2} – 16y^{2} = 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 \(\int_0^{2 \pi}\)sin^{2}x cos^{4}x dx .

Solution:

\(\int_0^{2 \pi}\)sin^{2}cos^{4}x dx = 4\(\int_0^{\frac{\pi}{2}}\)sin^{2}x.cos^{4}x dx

= 4.\(\frac{4-1}{2+4}\).\(\frac{4-3}{2+4-2}\).\(\frac{1}{2}\).\(\frac{\pi}{2}\)

= 4.\(\frac{3}{6}\).\(\frac{1}{4}\).\(\frac{1}{2}\).\(\frac{\pi}{2}\)

= \(\frac{\pi}{8}\)

∴ \(\int_0^{2 \pi}\)sin^{2}xcos^{4}x dx = \(\frac{\pi}{8}\)

Question 10.

Solve \(\frac{d y}{d x}\) = e^{x-y} + x^{2}e^{-y}.

Solution:

\(\frac{d y}{d x}\) = e^{x – y} + x^{2}e^{-y}

⇒ \(\frac{d y}{d x}\) = e^{x} . e^{-y} + x^{2}e^{-y}

⇒ \(\frac{d y}{e^{-y}}\) = ∫(e^{x} + x^{2}) dx

⇒ e^{y} dy = ( e^{x} + x^{2}) dx

Integrating

∫e^{y}dy = ∫(e^{x} + x^{2}) dx + c

⇒ e^{y} = e^{x} + \(\frac{x^3}{3}\) + c

Section – B

II. Short Answer Type Questions.

- Answer ANY FIVE questions.
- 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

⇒ r^{2} – d^{2} = 1

⇒ r^{2} – 4 = 1

⇒ r^{2} = 5

⇒ r = \(\sqrt{5}\)

Required circle equation is (x – h)^{2} + (y – k)^{2} = r^{2}

⇒ (x + 2)^{2} + (y – 3)^{2} = (\(\sqrt{5}\))^{2}

⇒ x^{2} + 4x + 4 + y^{2} – 6y + 9 = 5

⇒ x^{2} + y^{2} + 4x – 6y + 8 = 0

Question 12.

Find the equation of the circle which passes through origin and intersects the circles x^{2} + y^{2} – 4x + 6y + 10 = 0, x^{2} + y^{2} + 12y + 6 = 0 orthogonally.

Solution:

Let x^{2} + y^{2} + 2gx + 2fy + c = 0 …….. (1) be the required circle.

If (1) passes through origin then c = 0

If (1) is orthogonal to x^{2} + y^{2} – 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 x^{2} + y^{2} + 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 x^{2} + y^{2} + 2(\(\frac{-7}{4}\))x + 2(\(\frac{1}{2}\))y + 0 = 0

⇒ x^{2} + y^{2} – \(\frac{7}{2}\)x + y = 0

⇒ 2x^{2} + 2y^{2} – 7x + 2y = 0

Question 13.

Find the length of the latus rectum, eccentricity, centre and foci of the ellipse 4x^{2} + y^{2} – 8x + 2y + 1 = 0.

Solution:

Given ellipse equation is 4x^{2} + y^{2} – 8x + 2y + 1 = 0

⇒ 4(x^{2} – 2x + 1 – 1) + (y^{2} + 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 a^{2} = 1 ⇒ a = 1

b^{2} = 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 3x^{2} + 13y^{2} = 78. Prove that a common tangent is inclined to the major axis at an angle \(\frac{\pi}{4}\).

Solution:

Given ellipse equation is 3x^{2} + 13y^{2} = 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} = 4^{2}

⇒ x^{2} + y^{2} = 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

c^{2} = a^{2}m^{2} + b^{2}

⇒ \(\left(\frac{4}{\sin \theta}\right)^2\left(\frac{-\cos \theta}{\sin \theta}\right)^2\) = 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 cos^{2}θ + 6(1 – cos^{2}θ)

⇒ 16 = 26 cos^{2}θ + 6 – 6 cos^{2}θ

⇒ 20 cos^{2}θ = 10

⇒ cos^{2}θ = \(\frac{1}{2}\)

⇒ 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 θ_{1}, θ_{2} with transverse axis of a hyperbola.

Show that the point of intersection of these tangents lies on the curve 2xy = k(x^{2} – a^{2}) when tan θ_{1} + tan θ_{2} = 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(x_{1}, y_{1}) be the point of intersection of the tangents.

⇒ 2x_{1}y_{1} = K(\(x_1^2\) – a^{2})

∴ The point P(x_{1}, y_{1}) lies on the curve

2xy = K(x^{2} – a^{2})

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.

- Answer ANY FIVE questions.
- Each question carries SEVEN marks.

Question 18.

Find the equation of the circle passing through (2, 1), (5, 5) and (-6, 7).

Solution:

Let x^{2} + y^{2} + 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

x^{2} + y^{2} + 2\(\left(\frac{1}{2}\right)\)x + 2(-6)y + 5 = 0

⇒ x^{2} + y^{2} + x – 12y + 5 = 0.

Question 19.

Show that x^{2} + y^{2} – 6x – 9y + 13 = 0, x^{2} + y^{2} – 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 S_{1} = x^{2} + y^{2} – 6x – 9y + 13 = 0

s_{2} = x^{2} + y^{2} – 2x – 16y = 0

d_{1} = (3, \(\frac{9}{2}\)), d_{2} = (1, 8)

Equation of the common tangent is S_{1} – S_{2} = 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 = ay^{2} + 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 = –\(\frac{-5}{2}\)y^{2} + \(\frac{21}{2}\)y – 10

⇒ 2x = -5y^{2} + 21y – 20

⇒ 5y^{2} + 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} = ∫cos^{n} x dx then show that I_{n} = \(\frac{1}{n}\)cos^{n-1}x sin x + \(\frac{n-1}{n}\)I_{n-2}, n being a positive integer n ≥ 2 and deduce the value of ∫cos^{5}x dx.

Solution:

Question 23.

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

Solution:

Question 24.

Solve (x^{3} – 3xy^{2}) dx + (3x^{2}y – y^{3}) dy = 0.

Solution:

Given (x^{3} – 3xy^{2}) dx + (3x^{2}y – y^{3}) dy = 0

⇒ (x^{3} – 3xy^{2}) dx = -(3x^{2}y – y^{3}) 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