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## AP Inter 2nd Year Maths 2B Question Paper May 2016

Time : 3 Hours

Max. Marks : 75

Note:

This question paper consists of THREE sections A, B and C.

Section – A

I. Very Short Answer Type questions.

- Attempt ALL the questions.
- Each question carries TWO marks.

Question 1.

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

Solution:

The length of the tangent from (x_{1}, y_{2}) to the circle x^{2} + y^{2} + 2gx + 2fy + c = 0 is \(\sqrt{\mathrm{s}_{11}}\)

The length of the tangent from (5, 4) to the circle

x^{2} + y^{2} + 2ky = 0 is 1

⇒ \(\sqrt{25+16+8 k}\) = 1

⇒ 8k = -40 k = -5.

Question 2.

Find the pole of ax + by + c = 0, (c ≠ 0) with respect to x^{2} + y^{2} = r^{2}.

Solution:

Let (x_{1}, y_{1}) be the pole

Equation of the polar of (x_{1}, y_{1}) w.r.t. the circle

x^{2} + y^{2} = r^{2} is xx_{1} + yy_{1} = r^{2} ……….. (1)

but the given polar is ax + by + c = 0 …………… (2)

From (1) & (2)

∴ \(\frac{\mathrm{x}_1}{\mathrm{a}}\) = \(\frac{\mathrm{y}_1}{\mathrm{b}}\) = \(\frac{r^2}{-c}\)

Pole = (x_{1}, y_{1}) = \(\left(\frac{-a r^2}{c}, \frac{-b r^2}{c}\right)\)

Question 3.

Find the equation of the radical axis of the circles

x^{2} + y^{2} – 2x – 4y – 1 = 0

x^{2} + y^{2} – 4x – 6y + 5 = 0

Solution:

The equation of the radical axis of the circles

S = x^{2} + y^{2} – 2x – 4y – 1 = 0 and S^{1} = x^{2} + y^{2} – 4x – 6y + 5 = 0 is S – S^{1} = 0

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

⇒ 2x + 2y – 6 = 0

⇒ x + y – 3 = 0

Question 4.

Find the equation of tangent to the parabola y^{2} = 16x inclined at an angle 60° with its axis.

Solution:

Given parabola is y^{2} = 16x ……… (1)

∴ Slope of the tangent is m = tan 60° = \(\sqrt{3}\)

The equation of the tangent to the parabola (1) having slope

\(\sqrt{3}\) is y = mx + \(\frac{a}{m}\)

⇒ y = \(\sqrt{3}\)x + \(\frac{4}{\sqrt{3}}\) = \(\frac{3 x+4}{\sqrt{3}}\) = \(\sqrt{3}\)y = 3x + 4

⇒ 3x – \(\sqrt{3}\)y = 0.

Question 5.

If the eccentricity of the hyperbola is \(\frac{5}{4}\), then find the eccentricity of the conjugate hyperbola.

Solution:

Let e and e^{x} be the eccentricities of the hyperbola and its conjugate respectively

Question 6.

Evaluate the integral

\(\int \frac{(3 x+1)^2}{2 x}\)dx, x ∈ I ⊂ R/{0}

Solution:

Question 7.

Evaluate the integral

∫e^{x} (sec x + sec x tan x) dx

Solution:

We know that

∫e^{x} (f(x) + f'(x))dx = e^{x}f(x) + c

∴ e^{x} = [sec x + sec x tan x dx] = e^{x} sec x + c

Question 8.

Evaluate the definite integral

\(\int_0^\pi \sqrt{2+2 \cos \theta}\) dθ

Solution:

Question 9.

Evaluate the definite integral

\(\int_0^{\pi / 2} \sin ^6 x \cdot \cos ^4 x\) dx

Solution:

Question 10.

Find the general solution of x + y \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0

Solution:

x + y\(\frac{d y}{d x}\) = 0 dx

y dy = -x dx

∫y dy = -∫x dx

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

Section – B

(5 × 4 = 20)

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 – 22z = 0 on the line y = x – 3.

Solution:

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

Comparing with general equation S = 0 we get

given line is y = x – 3 and L = x – y – 3 = 0

AB is the chord intercepted by the line L = 0 on the circle.

CM = Perpendicular distance from C\(\left(\frac{1}{2},-\frac{3}{2}\right)\) to x – y – 3 = 0

Question 12.

If x + y = 3 is the equation of the chord AB of the circle x^{2} + y^{2} – 2x + 4y – 8 = 0, find the equation of the circle having \(\overline{\mathrm{AB}}\) as diameter.

Solution:

Let S ≡ x^{2} + y^{2} – 2x + 4y – 8 = 0 be the given circle

Let L ≡ x + y – 3 = 0 be the given line

Given that L = 0 intersects the circle S = 0 at AB

Equation of the circle passing through the point of intersection of S = 0 and L = 0 is S + λL = 0 where λ is a constant

(x^{2} + y^{2} – 2x + 4y – 8) + λ(x + y – 3) = 0

⇒ x^{2} + y^{2} + (λ – 2) x + y (λ + 4) – (8 + 3λ) = 0 …….. (1)

Centre of the circle = \(\left[\left(\frac{\lambda-2}{2}\right),-\left(\frac{\lambda+4}{2}\right)\right]\)

Given that the line L = 0 is a diameter of the circle (1)

⇒ Centre \(\left(\frac{2-\lambda}{2},-\frac{\lambda+4}{2}\right)\) lies on the line x + y – 3 = 0

∴ \(\frac{2-\lambda}{2}\) + \(\left[-\left(\frac{\lambda+4}{2}\right)\right]\) – 3 = 0

⇒ -λ + 2 – λ – 4 – 6 = 0

⇒ -2λ – 8 = 0

⇒ 2λ = -8

⇒ λ = -4

∴ The equation of the required circle from (1) is

x^{2} + y^{2} + (-4 – 2) x + (-4 + 4) y – 8 – 3(-4) = 0

⇒ x^{2} + y^{2} – 6x + 4 = 0.

Question 13.

Find the equation of the ellipse with focus at (1, -1), e = \(\frac{2}{3}\) and directrix as x + y + 2 = 0.

Solution:

Given that focus S = (1, -1) and e = \(\frac{2}{3}\) and equation of directrix is L = x + y + 2 = 0

Let (x_{1}, y_{1}) be any point on the ellipse and PM is the perpendicular distance from P to the directrix L = 0.

Question 14.

The tangent and normal to the ellipse x^{2} + 4y^{2} = 4 at a point P(θ) on its meets the major axis in Q and R respectively. If 0 < θ < \(\frac{\pi}{2}\) and QR = 2, then show that θ = cos^{-1}\(\left(\frac{2}{3}\right)\)

Solution:

The equation of the tangent at P(θ) on the ellipse S = 0 is

⇒ -3 cos^{2}θ + 4 = 4 cos θ

⇒ 3cos^{2}θ + 4 cos θ – 4 = 0

⇒ 3 cos θ(cos θ + 2) – 2(cos θ + 2) = 0

⇒ (cos θ + 2)(3cosθ – 2) = 0

∴ cos θ + 2 = 0: solution is not admissive (∵ -1 ≤ cos θ ≤ 1) and 3 cosθ – 2 = 0

⇒ cos θ = \(\frac{2}{3}\)

⇒ θ = cos^{-1} \(\left(\frac{2}{3}\right)\)

Question 15.

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

Solution:

x^{2} – 4y^{2} = 4

This can be written as \(\frac{x^2}{4}\) – \(\frac{y^2}{1}\) = 1

⇒ a^{2} = 4 and b^{2} = 1

⇒ a = 2 and b = 1

i) centre – (0, 0)

Question 16.

Find the area enclosed by the curves y = x^{2} + 1, y = 2x – 2, x = -1, x = 2

Solution:

Question 17.

Solve the differential equation

Solution:

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 each of the three points (3, 4), (3, 2) and (1, 4).

Solution:

Let the equation of the required circle is

x^{2} + y^{2} + 2gx + 2fy + c = 0 …… (1)

Since (1) passes through the point (3, 4) then

(3)^{2} + (4)^{2} + 2g(3) + 2f(4) + c = 0

∴ 9 + 16 + 6g + 8f + c = 0 ⇒ 6g + 8f + c = -25 ……… (2)

Since (1) passes through the point (3, 2)

∴ 9 + 4 + 6g + 4f + c = 0 ⇒ 6g + 4f + c = -13 …….. (3)

Since (1) passes through the point (1,4)

∴ 1 + 16 + 2g + 8f + c = 0 ⇒ 2g + 8f + c = -17 …….. (4)

from (2) & (3)

Substitute the value of g, f in (2)

6(-2) + 8(-3) + c = -25

-12 – 24 + c = -25 ⇒ c = -25 + 36

∴ c = 11

Substitute the values of g, f, c in (1)

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

= x^{2} + y^{2} – 4x – 6y + 11 = 0

∴ The equation of the required circle is

x^{2} + y^{2} – 4x – 6y + 11 = 0

Question 19.

Show that the circles x^{2} + y^{2} – 6x – 2y + 1 = 0, x^{2} + y^{2} + 2x – 8y + 13 = 0, touch each other. Find the point of contact and the equation of common tangent at their point of contact.

Solution:

Let S ≡ x^{2} + y^{2} – 6x – 2y +1 = 0 and

S’ ≡ x^{2} + y^{2} + 2x – 8y + 13 = 0 be the given circles then centre of circles are C_{1} = (3, 1) and C_{2} = (-1, 4)

Since C_{1}C_{2} = r_{1} + r_{2} the two circles touch externally at a point at P be the point of contact of circles such that r_{1}: r_{2} = 3 : 2

Since P divides C_{1}, C_{2} internally in the ratio 3 : 2 coordinates of point of contact

The equation of common tangent is S – S’ = 0

⇒ x^{2} + y^{2} – 6x – 2y + 1 – x^{2} – y^{2} – 2x + 8y + 13 = 0

⇒ -8x + 6y – 12 = 0

⇒ 4x – 3y + 6 = 0

∴ The equation of the common tangent at the point of contact is 4x – 3y + 6 = 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:

Let A = (-2, 1), B (1, 2) and C = (-1, 3) be the given points.

Let the equation of the parabola whose axis is parallel to X – axis and passing through the points A, B, C be

x = ly^{2} + my + n ……. (1)

If parabola (1) passes through A(-2, 1) then

-2 = l + m + n ………. (2)

If (1) passes through B(1, 2) then 1 = 4l + 2m + n …….. (3)

If (1) passes through C (-1, 3) then -1 = 9l + 3m + n …….. (4)

From (2) and (3)

-3l – m = -3 ⇒ 3l + m = 3 …….. (5)

From (3) and (4)

-5l – m = 2 = 5l + m = -2 ………. (6)

Solving (5) and (6) we get

Question 21.

Evaluate the integral

\(\int_0^\pi \frac{x \sin x}{1+\cos ^2 x}\)dx

Solution:

We write x + 1 = A\(\frac{d}{d x}\)(x^{2} + 3x + 12) + B = A (2x + 3) + B

On comparing the coefficients of like powers of x on both sides of the above equation, we get

2A = 1 ⇒ A = \(\frac{1}{2}\) and 3A + B = 1

⇒ B = 1 – 3A = 1 – \(\frac{3}{2}\) = –\(\frac{1}{2}\)

A = \(\frac{1}{2}\) and B = –\(\frac{1}{2}\)

Question 22.

Obtain the reduction formula for ∫ tan^{n}x dx for an integer n ≥ 2 and deduce the value of ∫ tan^{6}x dx.

Solution:

Question 23.

Evaluate the integral

\(\int_0^\pi \frac{x \sin x}{1+\cos ^2 x}\)dx

Solution:

Question 24.

Find the equation ofa curve whose gradient is \(\frac{d y}{d x}\) = \(\frac{y}{x}\) – cos^{2} \(\left(\frac{y}{x}\right)\) where x >0, y > 0 and which passes through the point (1, \(\frac{\pi}{4}\))

Solution: