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## TS Inter 2nd Year Maths 2B Question Paper May 2019

Time : 3 Hours

Max. Marks : 75

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 whose extremities of a diameter are (-4, 3), (3, -4).

Solution:

The equation of the circle whose extremities of a diameter are (-4, 3), (3, -4) is

⇒ (x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

⇒ (x + 4) (x – 3) + (y – 3) (y + 4) = 0

⇒ x_{2} – 3x + 4x – 12 + y_{2} + 4y – 3y – 12 = 0

⇒ x_{2} + x + y – 24 = 0.

Question 2.

Find the Polar of (3, -1) with respect to 2x^{2} + 2y^{2} = 11.

Solution:

Let S ≡ x^{2} + y^{2} – \(\frac{11}{2}\) = 0

The polar of (3, -1) with respect to s = 0 is s_{1} =0

⇒ x(3) + y(-1) – \(\frac{11}{2}\) =0

⇒ 3x – y – \(\frac{11}{2}\) = 0

⇒ 6x – 2y – 11 = 0.

Question 3.

Find the equation of the radical axis of the circles x^{2} + y^{2} + 4x + 6y – 7 = 0, 4(x^{2} + y^{2}) + 8x + 12y – 9 = 0.

Solution:

Let s ≡ x^{2} + y^{2} + 4x + 6y – 7 = 0

s^{1} ≡ x^{2} + y^{2} + 2x + 3y – \(\frac{9}{4}\) = 0

The radical axis of s = 0, s^{1} = 0 is s – s^{1} = 0

⇒ (x^{2} + y^{2} + 4x + 6y – 7) – (x^{2} + y^{2} + 2x + 3y – \(\frac{9}{4}\)) = 0

⇒ x^{2} + y^{2} + 4x + 6y – 7 – x^{2} – y^{2} – 2x – 3y + \(\frac{9}{4}\) = 0

⇒ 2x + 3y + (\(\frac{9}{4}\) – 7) = 0

⇒ 2x + 3y + \(\left(\frac{9-28}{4}\right)\) = 0

⇒ 2x + 3y – \(\frac{19}{4}\) = 0

⇒ 8x + 12y – 19 = 0

Question 4.

Find the equation of the parabola whose vertex is (3, -2) and focus is (3, 1).

Solution:

The abecissae of the vertex and focus are equal to 3.

∴ The axis of the parabola is x = 3, a line parallel to y – axis, focus is above the vertex.

∴ a = \(\sqrt{(3-3)^2+(1+2)^2}\) = 3.

Hence equation of the required parabola is

(x – 3)^{2} = 4(3) (y + 2)

⇒ (x – 3)^{2} = 12 (y + 2).

Question 5.

Find the product of lengths of the perpendiculars from any point on the hyperbola \(\frac{x^2}{16}-\frac{y^2}{9}\) = 1 to its asymptotes.

Solution:

Equation of the hyperbola is \(\frac{x^2}{16}-\frac{y^2}{9}\) = 1

Here a^{2} = 16 and b^{2} = 9.

∴ Product of the perpendiculars from any point on the hyperbola to its asymptotes = \(\frac{a^2 b^2}{a^2+\bar{b}^2}\) = \(\frac{16(9)}{16+9}\) = \(\frac{144}{25}\)

Question 6.

Evaluate \(\int \frac{x^8}{1+x^{18}}\)dx on R.

Solution:

Question 7.

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

Solution:

Question 8.

Evaluate \(\int_0^{\infty} \frac{d x}{x^2+a^2}\)

Solution:

Question 9.

Find the area bounded between the curves y = x^{2}, y = x^{3}

Solution:

Given curve equations are y = x^{2}, y = x^{3}

Question 10.

Solve \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{1+y^2}{1+x^2}\)

Solution:

\(\int \frac{d x}{1+x^2}\) – \(\int \frac{d y}{1+y^2}\) = c

⇒ tan^{-1}x + tan^{-1}y = c

∴ The general solution of (1) is

tan^{-1}x – tan^{-1}y = c.

Section – B

II. Short Answer Type Questions.

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

Question 11.

Find the equation of the tangent at the point 30° (parametric value of θ) of the circle x^{2} + y^{2}+ 4x + 6y – 39 = 0.

Solution:

Given circle equation is s ≡ x^{2} + y^{2} + 4x + 6y – 39 = 0

Here 2g = 4 ⇒ g = 2

2f = 6 ⇒ f = 3 and c = – 39

Radius r = \(\sqrt{g^2+f^2-c}\)

= \(\sqrt{4+9-(-39)}\)

= \(\sqrt{4+9+39}\)

= \(\sqrt{52}\) = 2\(\sqrt{13}\)

The equation of the tangent at 0 of the circle.

s = 0 is (x + g) cos θ + (y + f) sin θ = r

Question 12.

Find the equation of the circle which passes through the point (2, 0), (0, 2) and orthogonal to the circle 2x^{2} + 2y^{2} + 5x – 6y + 4 = 0.

Solution:

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

If (1) passes through the point (2, 0) then

4 + 0 + 4g + 0 + c = 0

⇒ 4g + c = – 4 ……… (2)

If (1) passes through the point (0, 2) then

0 + 4 + 0 + 4f + c = 0

⇒ 4f + c = – 4 ………. (3)

(2) – (3)

⇒ 4g – 4f = 0

⇒ g – f = 0

⇒ g = f ………… (4)

If (1) is orthogonal to x^{2} + y^{2} + 5/2x – 3y + 2 = 0 then

29\(\left(\frac{5}{4}\right)\) + 2f\(\left(\frac{-3}{2}\right)\) = c + 2

⇒ \(\left(\frac{5}{2}\right)\)g – 3f = c + 2

⇒ \(\left(\frac{5}{2}\right)\)g – 3g = c + 2 [∴ from (4)]

⇒ \(\frac{-1}{2}\)g = c + 2

⇒ – g = 2c + 4

⇒ g = – 2c – 4 ………. (5)

From (2)

4(-2c – 4) + c = -4

⇒ -8c – 16 + c = -4

⇒ -7c = 12

⇒ c = -12/7

From (5)

Question 13.

Find the length of the major axis, minor axis, latus rectum, eccentricity of the ellipse 9x^{2} + 16y^{2} = 144.

Solution:

Given 9x^{2} + 16y^{2} = 144

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

Here a^{2} = 16 ⇒ a = 4

b^{2} = 9 ⇒ b = 3

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 = \(\sqrt{\frac{a^2-b^2}{a^2}}\) = \(\sqrt{\frac{16-9}{16}}\) = \(\frac{\sqrt{7}}{4}\)

Question 14.

Find the equation of the tangent to the ellipse 2x^{2} + y^{2} = 8 which are

i) parallel to x – 2y – 4 = 0

ii) perpendicular to x + y + 2 = 0.

Solution:

i) Parallel to x – 2y – 4 = 0

Slope will be : \(\frac{\sqrt{1}}{2}\)

ii) Perpendicular to x + y + 2 = 0 ,

Slope of tangent be T as it is perpendicular to above line y = mx ± +Ja2m2 + b2 => y = x ± ^4 + 8

y = mx ± \(\sqrt{a^2 m^2+b^2}\)

y = x ± \(2 \sqrt{3}\)

⇒ y = x ± \(\sqrt{4+8}\)

⇒ x – y ± \(2 \sqrt{3}\) = 0

Question 15.

Tangents to the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1 makes 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 tanθ_{1} + tanθ_{2} = k.

Solution:

Equation of the hyperbola is \(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1

Equation of the tangent 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.

Question 16.

Find the area bounded between the curves y^{2} = 4x, y^{2} = 4(4 – x).

Solution:

Given curve equations are y^{2} = 4x ………. (1)

y^{2} = 4(4 – x) …………… (2)

From (1) & (2)

4x = 4 (4 – x)

⇒ 4x = 16 – 4x

⇒ 8x = 16

⇒ x = 2

From (1) y^{2} = 8

⇒ y = ± 2\(\sqrt{2}\)

∴ Points of intersection are A(2, 2\(\sqrt{2}\)) and B(2, -2\(\sqrt{2}\))

From fig required area is symmetric about x – axis

Question 17.

Solve \(\frac{\mathrm{dy}}{\mathrm{dx}}\) – y tan x = e^{x} sec x.

Solution:

Given differential equation is

\(\frac{\mathrm{dy}}{\mathrm{dx}}\) – y tanx = e^{x} sec x ….. (1)

Here P = – tanx and Q = e^{x} secx

I.F. = \(e^{\int P d x}\)

= \(e^{-\int \tan x d p}\)

= e^{-log|sec x|}

= e^{log cosx} = cos x.

∴ The general solution of (1) is

y(I.F.) = ∫Q(I.F)dx + c

⇒ y cosx = ∫e^{x}secx cosx dx + c

⇒ y cosx = ∫e^{x}dx + c

⇒ y cosx = e^{x} + c

Section – C

(5 × 7 = 35)

III. Long Answer Type Questions.

- Attempt any five questions.
- 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:

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

be the required circle.

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

4 + 9 + 4g – 6f + c = 0

4g – 6f + c = -13 ……….. (2)

If (1) passes through (-4, 5) then

16 + 25 – 8g + 10f + c = 0

⇒ 8g – 10f – c = 41 ………. (3)

center (-g, -f) lies on 4x + 3y + 1 = 0

⇒ 4(-g) + 3(-f ) +1 = 0

⇒ -4y – 3f + 1 = 0

⇒ 4g + 3f – 1 = 0 ……… (4)

(2) + (3) ⇒ 12g – 16f = 28

⇒ 3g – 4f = 7

⇒ 3g – 4f – 7 = 0 ……… (5)

Solving (4) & (5)

g = -1, f = 1

From(2) + 4 + 6 + c = -13

⇒ c = -23

∴ g = -1, f = 1, c = -23

Hence required circle equation is

x^{2} + y^{2} – 2x + 2y – 23 = 0.

Question 19.

Find the transverse common tangents of the circles x^{2} + y^{2} – 4x – 10 y + 28 = 0 and x^{2} + y^{2} + 4x – 6y + 4 = 0.

Solution:

(-2x – y + 7)^{2} = (x^{2} + y^{2} – 4x – 10y + 28)

4x^{2} + y^{2} + 4xy – 28x – 14y + 4 = x^{2} + y^{2} – 4x – 10y + 28

3x^{2} + 4xy – 24x – 4y + 21 = 0

(3x + 4y – 21); (x – 1) = 0

3x + 4y – 21 = 0; x – 1 = 0.

Question 20.

Show that the equation of common tangents to the circle x^{2} + y^{2} = 2a^{2} and the parabola y^{2} = 8ax are y = ± (x + 2a).

Solution:

The equation of tangent to parabola y^{2} = 8ax is y = mx + 2 a/m

⇒ my = m^{2}x + 2a

⇒ m^{2}x – my + 2a = 0 …….. (1)

If (1) touches the circle x^{2} + y^{2} = 2a^{2} then

r = d

⇒ \(\sqrt{2} a\) = \(\frac{|2 a|}{\sqrt{m^4+m^2}}\)

⇒ 2a^{2} = \(\frac{4 a^2}{m^4+m^2}\)

⇒ 1 = \(\frac{2}{m^4+m^2}\)

⇒ m^{4} + m^{2} = 2

⇒ m^{4} + m^{2} – 2 = 0

⇒ m^{4} + 2m^{2} – m^{2} – 2 = 0

⇒ m^{2} (m^{2} + 2) – 1 (m^{2} + 2) = 0

⇒ (m^{2} + 2) (m^{2} – 1) = 0

⇒ m^{2} – 1 = 0

⇒ m^{2} = 1

⇒ m = ± 1.

∴ Required tangents are

⇒ y = 1 (x) + \(\frac{2 a}{1}\), y = – 1 (x) + \(\frac{2 a}{-1}\)

⇒ y = x + 2a, y = -x – 2a

⇒ y = x + 2a, y = – (x + 2a)

⇒ y = ± (x + 2a).

Question 21.

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

Solution:

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

= A(4 cosx + 5 sinx) + B (- 4 sinx + 5 cosx)

Equating the co-efficient of sinx and cosx, we get

3 = 5A – 4B ⇒ 5A – 4B – 3 = 0 …… (1)

2 = 4A + 5B ⇒ 4A + 5B – 2 = 0 ……… (2)

Solving (1) and (2)

Question 22.

Obtain the reduction formula for I_{n} = ∫cos^{n}x dx, n being a positive integer, n ≥ 2, and deduce the value of ∫cos^{3}x dx.

Solution:

Question 23.

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

Solution:

Question 24.

Solve \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{4 x+6 y+5}{3 y+2 x+4}\)

Solution:

Integrating

⇒ 8z + 9 log |8z + 23| = 64x + c

⇒ 8(2x + 3y) + 9 log |8(2x + 3y) + 23| = 64x + c

⇒ 24y – 48x + 9log|16x + 24y + 23| = c

⇒ y – 2x + 3/8 log |16x + 24y + 23| = k where k = c/24

∴ The general solution of (1) is

y – 2x + 3/8 log |16x + 24y + 23| = k.