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## AP Inter 2nd Year Maths 2B Question Paper March 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.

Obtain the parametric equation of circle (x – 3)^{2} + (y – 4)^{2} = 8^{2}.

Solution:

Given circle equation is (x – 3)^{2} + (y – 4)^{2} = 8^{2}

Here h = 3, k = 4 and r = 8

Parametric equations are x = h + r cos θ

= 3 + 8 cos θ

y = k + r sin θ

= 4 + 8 sin θ where 0 ≤ θ < 2θ

Question 2.

Find the equation of the normal at P(3, 5) of the circle S = x^{2} + y^{2} – 10x – 2y + 6.

Solution:

Given S ≡ x^{2} + y^{2} – 10x – 2y + 6

Here 2g = -10 ⇒ g = -5

2f = -2 ⇒ f = -1

The equation of the normal of p(3, 5) or the circle S = 0 is

(x – x_{1}) (y_{1} + f) – (y – y_{1}) (x_{1} + g) = 0

⇒ (x – 3) (5 – 1) – (y – 5) (3 – 5) = 0

⇒ (x – 3)4 – (y – 5) (-2) = 0

⇒ 4x – 12 + 2y – 10 = 0

⇒ 4x + 2y – 22 = 0

⇒ 2x + y – 11 = 0

Question 3.

If x^{2} + y^{2} – 5x – 14y – 34 = 0, x^{2} + y^{2} + 2x + 4y + k = 0 circles are orthogonal, then find ‘k’.

Solution:

Given circle equations are

x^{2} + y^{2} – 5x – 14y – 34 = 0 …… (1)

x^{2} + y^{2} + 2x + 4y + k = 0 ……. (2)

Here 2g = -5 ⇒ g = \(\frac{-5}{2}\)

2f = -14 ⇒ f = -7 and c = -34

2g’ = 2 ⇒ g‘ = 1

2f’ = 4 ⇒ f = 2 and c’ = k

Since (1), (2) are orthogonal

∴ 2gg’ + 2ff’ + c + c’

⇒ 2\(\left(\frac{-5}{2}\right)\)(1) + 2(-7) (2) = -34 + k

⇒ -5 – 28 = -34 + k

⇒ k = -33 + 34

⇒ k = 1

Question 4.

Find the value of k if the line 2y = 5x + k is a tangent to the parabola y^{2} = 6x.

Solution:

Given line eq. is 2y = 5x + k

⇒ y = \(\left(\frac{-5}{2}\right)\)x + \(\frac{k}{2}\) …….. (1)

Here m = \(\frac{5}{2}\) and c = \(\frac{k}{2}\)

Given parabola equation is y^{2} = 6x

Here 4a = 6 ⇒ a = \(\frac{6}{4}\)

⇒ a = \(\frac{3}{2}\)

Since (1) is tangent to the parabola y^{2} = 6x

∴ c = \(\frac{a}{m}\)

⇒ \(\frac{k}{2}\) = \(\frac{\frac{3}{2}}{\frac{5}{2}}\)

⇒ \(\frac{k}{2}\) = \(\frac{3}{5}\)

⇒ k = \(\frac{6}{5}\)

Question 5.

Find the eccentricity and length of the latus rectum of the hyperbola 16y^{2} – 9x^{2} = 144.

Solution:

Given hyperbola equation is 16y^{2} – 9x^{2} = 144

⇒ \(\frac{y^2}{9}\) – \(\frac{x^2}{16}\) = 1

Here a^{2} = 16, b^{2} = 9

Eccentricity = \(\sqrt{\frac{a^2+b^2}{b^2}}\) = \(\sqrt{\frac{16+9}{9}}\) = \(\frac{5}{3}\)

Length or the latus rectum = \(\frac{2 a^2}{b}\) = \(\frac{2.16}{3}\) = \(\frac{32}{3}\)

Question 6.

Evaluate : \(\int \frac{\sin \left(\tan ^{-1} x\right)}{1+x^2}\)dx, x ∈ R .

Solution:

I = \(\int \frac{\sin \left(\tan ^{-1} x\right)}{1+x^2} d x\)

Put tan^{-1}x = t

Then \(\frac{1}{1+x^2}\)dx = dt

= ∫sin t dt

= -cos t + c

= -cos(tan^{-1}x) + c

∴ \(\int \frac{\sin \left(\tan ^{-1} x\right)}{1+x^2} d x\) = -cos(tan^{-1}x) + c

Question 7.

Evaluate \(\int e^x\left[\frac{1+x \log x}{x}\right] d x\) on (0, ∞)

Solution:

Question 8.

Evaluate : \(\int_0^\pi \sqrt{2+2 \cos \theta d \theta}\)

Solution:

\(\int_0^\pi \sqrt{2+2 \cos \theta} d \theta\) = \(\int_0^\pi \sqrt{2(1+2 \cos \theta)} d \theta\)

= \(\int_0^\pi \sqrt{2.2 \cos ^2 \frac{\theta}{2}} d \theta\)

Question 9.

Find : \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^2 x \cos ^4 x d x\).

Solution:

I = \(\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\)sin^{2}xcos^{4}x dx

since sin^{2}xcos^{4}x is an even function

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

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

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

= \(\frac{\pi}{16}\)

∴ \(\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\)sin^{2}xcos^{4}x = \(\frac{\pi}{16}\)

Question 10.

Find the order and degree of : x^{1/2}\(\left[\frac{d^2 y}{d x^2}\right]^{1 / 3}\) + x\(\frac{d y}{d x}\) + y = 0.

Solution:

Given differential equation is

Section – B

II. Short Answer Type Questions.

- Attempt any five questions.
- Each question carries four marks.

Question 11.

Find the length of the chord intercepted by the circle x^{2} + y^{2} – x + 3y – 22 = 0 on the line y = x – 3.

Solution:

Given circle equation is x^{2} + y^{2} – x + 3y – 22 = 0

Centre C = (\(\frac{1}{2}\), \(\frac{-3}{2}\))

Radius r = \(\sqrt{\frac{1}{4}+\frac{9}{4}-(-22)}\)

= \(\sqrt{\frac{1+9+88}{4}}\)

= \(\sqrt{\frac{98}{4}}\)

d = The ⊥las distance from c to the line x – y – 3 = 0

Question 12.

Find the radical centre of the three circles :

i) x^{2} + y^{2} – 4x – 6y + 5 = 0

ii) x^{2} + y^{2} – 2x – 4y – 1 = 0

iii) x^{2} + y^{2} – 6x- 2y = 0

Solution:

Let S ≡ x^{2} + y^{2} – 4x – 6y + 5 = 0

S’ ≡ x^{2} + y^{2} – 2x – 4y – 1 = 0

S” ≡ x^{2} + y^{2} – 6x – 2y = 0

The radical axis of S = 0, S’ = 0 is S – S’ = 0

⇒ x^{2} + y^{2} – 4x – 6y + 5 – x^{2} – y^{2} + 2x + 4y + 1 = 0

⇒ -2x – 2y + 6 = 0

⇒ 2x + 2y – 6 = 0

⇒ x + y – 3 = 0 ……… (1)

The radical axis of S = 0, S” = 0 is S – S” = 0

⇒ x^{2} + y^{2} – 4x – 6y + 5 – x^{2} – y^{2} + 6x + 2y = 0

⇒ 2x – 4y + 5 = 0 …….. (2)

Solving (1) and (2)

Question 13.

Find the equation of the ellipse in the standard form whose distance between foci is 2 and the length of latus rectum is \(\frac{15}{2}\).

Solution:

Given distance between foci is 2

⇒ 2ae = 2

⇒ ae = 1 ……. (1)

Also length of the latus rectum = \(\frac{15}{2}\)

⇒ \(\frac{2 b^2}{a}\) = \(\frac{15}{2}\)

⇒ \(\frac{-a^2\left(e^2-1\right)}{a^1}\) = \(\frac{15}{4}\)

⇒ \(\frac{-a^2 e^2+a^2}{a}\) = \(\frac{15}{4}\)

⇒ \(\frac{-1+a^2}{a}\) = \(\frac{15}{4}\)

⇒ – 4 + 4a^{2} = 15a

⇒ 4a^{2} – 15a – 4 = 0

⇒ 4a^{2} – 16a + a – 4 = 0

⇒ 4a (a – 4) + 1 (a – 4) = 0

⇒ (a – 4) (4a + 1) = 0

Since a > 0

∴ a = 4

from (1) e = \(\frac{1}{4}\)

b^{2} = a^{2} (1 – e^{2})

= 16(1 – \(\frac{1}{16}\))

= 16(\(\frac{15}{16}\))

= 15

Hence required ellipse equation is

\(\frac{x^2}{16}\) + \(\frac{y^2}{15}\) = 1

Question 14.

If a tangent to the ellipse \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1 (a > b) meets its major axis and minor axis at M and N respectively, then prove that \(\frac{a^2}{(\mathrm{CM})^2}\) + \(\frac{b^2}{(C N)^2}\) = 1 where C is the centre of the ellipse.

Solution:

Given Ellipse equation is

\(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1(a > b)

Let P(a cos θ, b sin θ) be any point on the ellipse.

Equation of the tangent at p(θ) is

Question 15.

Find the centre, foci, eccentricity, equation of the directrices, length of the latus rectum of the hyperbola x^{2} – 4y^{2} = 4.

Solution:

Given hyperbola equation is x^{2} – 4y^{2} = 4

⇒ \(\frac{x^2}{4}\) – \(\frac{y^2}{1}\) = 1

Here a^{2} = 4, b^{2} = 1

Centre = (0, 0)

Eccentricity = \(\sqrt{\frac{a^2+b^2}{a^2}}\) = \(\sqrt{\frac{4+1}{4}}\) = \(\frac{\sqrt{5}}{2}\)

Foci = (± a e, 0) = (± 2.\(\frac{\sqrt{5}}{2}\), 0)

= (±\(\sqrt{5}\), 0)

Equations of divertrices are x = ± \(\frac{\mathrm{a}}{\mathrm{e}}\)

⇒ x = ±\(\frac{2}{\frac{\sqrt{5}}{2}}\)

⇒ x = ±\(\frac{4}{\sqrt{5}}\)

Length or the latus rectum = \(\frac{2 b^2}{a}\) = \(\frac{2.1}{2}\) = 1

Question 16.

Find the area bounded by the curves y = sin x and y = cos x between any two consecutive points of intersection.

Solution:

Two points or intersection of the curves y = sin x and y = cos x are \(\left(\frac{\pi}{4}, \frac{1}{\sqrt{2}}\right)\) and \(\left(\frac{5 \pi}{4}, \frac{-1}{\sqrt{2}}\right)\)

Two points or intersection of the curves y = sin x and y = cos x are \(\left(\frac{\pi}{4}, \frac{1}{\sqrt{2}}\right)\) and \(\left(\frac{5 \pi}{4}, \frac{-1}{\sqrt{2}}\right)\)

= –\(\left[\frac{-2}{\sqrt{2}}-\frac{2}{\sqrt{2}}\right]\)

= \(\sqrt{2}\) + \(\sqrt{2}\)

= 2\(\sqrt{2}\) sq. units.

Question 17.

Solve : (1 + x^{2})\(\frac{d y}{d x}\) + y = \(e^{\tan ^{-1} x}\).

Solution:

Given (1 + x^{2})\(\frac{d y}{d x}\) + y = tan^{-1} x

⇒ \(\frac{d y}{d x}\) + \(\frac{1}{1+x^2}\)y = \(\frac{\tan ^{-1} x}{1+x^2}\) ……. (1)

This is a linear differential equation of the first order in y

Section – C

III. Long Answer Type Questions.

- Attempt any five questions.
- Each question carries seven marks.

Question 18.

Find the equation of circle passing through the three points (3, 4), (3, 2), (1, 4).

Solution:

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

If (1) passes through (3, 4) then

9 + 16 + 6g + 8f + c = 0

⇒ 6g + 8f + c = -25 …… (2)

If (1) passes through (3, 2) then

9 + 4 + 6g + 4f + c = 0

⇒ 6g + 4f + c = -13 ……… (3)

If (1) passes through (1, 4) then

1 + 16 + 2g + 8f + c = 0

⇒ 2g + 8f + c = -17 ……. (4)

(2) – (3) ⇒ 4f = – 12 ⇒ f = -3

(2) – (4) ⇒ 4g = – 8 ⇒ g = -2

From (4)

2(-2) + 8(-3) + c = – 17

⇒ c = -17 + 4 + 24

⇒ c = 11

Hence required circle equation is

x^{2} + y^{2} + 2(-2)x + 2(-3)y + 11 = 0

⇒ x^{2} + y^{2} – 4x – 6y + 11 = 0.

Question 19.

Find the direct common tangents of the circles x^{2} + y^{2} + 22x – 4y – 100 = 0 and x^{2} + y^{2} – 22x + 4y + 100 = 0.

Solution:

Given circle equations are

S ≡ x^{2} + y^{2} + 22x – 4y – 100 = 0

S’ ≡ x^{2} + y^{2} – 22x + 4y + 100 = 0

C_{1} = (-11, 2) r_{1} = \(\sqrt{121+4+100}\) = 15

C_{2} = (-11, 2) r_{2} = \(\sqrt{121+4-100}\) = 5

d = C_{1}C_{2} = \(\sqrt{(11+11)^2+(-2-2)^2}\)

= \(\sqrt{484+16}\)

= \(\sqrt{500}\) = 10\(\sqrt{5}\)

d > r_{1} + r_{2}

∴ The two circles do not intersect.

Let P be the external centre of similitude.

∴ P divides C_{1}C_{2} in ratio r_{1} : r_{2} = 15 : 5 = 3 : 1 externally.

P = \(\left(\frac{3(11)-1(-11)}{3-1}, \frac{3(-2)-1(2)}{3-1}\right)\)

= \(\left(\frac{33+11}{2}, \frac{-6-2}{2}\right)\) = (22, -4)

∴ Equation of pair of direct common tangents to the circle

S = 0 is \(S_1^2\) = SS_{11}

⇒ [x(22) + y(- 4) + 11 (x + 22) – 2(y – 4) – 100]^{2}

= (x^{2} + y^{2} + 22x – 4y – 100) (484 + 16 + 484 + 16 – 100)

⇒ [22x – 4y + 11x + 242 – 2y + 8 – 100]^{2} = (x^{2} + y^{2} + 22x – 4y – 100)(900)

⇒ [33x – 6y + 150]^{2} = (x^{2} + y^{2} + 22x – 4y – 100)(900)

⇒ [11x – 2y + 50]^{2} = (x^{2} + y^{2} + 22x – 4y – 100)(100)

⇒ 121x^{2} + 4y^{2} + 2500 – 44xy -200y + 1100x = 100x^{2} + 100y^{2} + 2200x – 400y – 10000

⇒ 21x^{2} – 96y^{2} – 44xy – 1100x + 200y + 12500 = 0

21x^{2} – 44xy – 96y^{2} = 21x^{2} – 72xy + 28xy – 96y^{2}

= 3x(7x – 24y) + 4y(7x – 24y)

= (3x + 4y)(7x – 24y)

21x^{2} – 44xy – 96y^{2} – 1100x + 200y + 12500 = (3λ + 4y + l) (7x – 24y + m)

7l + 3m = – 1100

-24l + 4m = 200

Solving l = – 50, m = – 250

∴ Required direct common tangents are

3x + 4y – 50 = 0

7x – 24y – 250 = 0

Question 20.

Prove that the area of the triangle formed, by the tangents at (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) to the parabola y^{2} = 4ax (a > 0) is \(\frac{1}{16 a}\)|(y_{1} – yy_{2}) (y_{2} – y_{3}) (y_{3} – y_{1})| sq-units.

Solution:

Question 21.

Evaluate : \(\int \frac{2 \cos x+3 \sin x}{4 \cos x+5 \sin x} d x\).

Solution:

Let 2 cos x + 3 sin x = A\(\frac{d}{d x}\) (4 cosx + 5 sin x) + B(4 cosx + 5 sinx)

= A(-4 sin x + 5 cos x) + B(4 cosx + 5 sinx)

now equating like terms on both sides, we have

2 = 5A + 4B ⇒ 5A + 4B – 2 = 0 …….. (1)

3 = -4A + 5B ⇒ 4A + 5B + 3 = 0 …….. (2)

Solving (1) and (2) ”

Question 22.

Obtain reduction formula for I_{n} – ∫tan^{n}x dx, n being a positive integer n ≥ 2 and deduce the value of ∫tan^{6} x dx.

Solution:

Let I_{n} = ∫Tan^{n}x dx

= ∫Tan^{n-2}x Tan^{2}x dx

= ∫Tan^{n-2}x(sec^{2}x – 1)dx

= ∫Tan^{n-2}xsec^{2}x dx – ∫Tan^{n-2}xdx

Question 23.

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

Solution:

I = \(\int_0^{\pi / 2} \frac{x}{\sin x+\cos x} d x\) …. (1)

Question 24.

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

Solution:

Solution:

Given differential equation is

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

⇒ \(\frac{d y}{d x}\) = \(\frac{x^2 y-2 x y^2}{x^3-3 x^2 y}\) …….. (1)

Put y = Vx

\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = V + x\(\frac{d V}{d x}\)

Substituting these values in (1), we get