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

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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.

1. Attempt all questions.
2. 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 – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0
⇒ (x + 4) (x – 3) + (y – 3) (y + 4) = 0
⇒ x₂ – 3x + 4x – 12 + y₂ + 4y – 3y – 12 = 0
⇒ x₂ + x + y – 24 = 0.

Question 2.
Find the Polar of (3, -1) with respect to 2x² + 2y² = 11.
Solution:
Let S ≡ x² + y² – \(\frac{11}{2}\) = 0
The polar of (3, -1) with respect to s = 0 is s₁ =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 x² + y² + 4x + 6y – 7 = 0, 4(x² + y²) + 8x + 12y – 9 = 0.
Solution:
Let s ≡ x² + y² + 4x + 6y – 7 = 0
s¹ ≡ x² + y² + 2x + 3y – \(\frac{9}{4}\) = 0
The radical axis of s = 0, s¹ = 0 is s – s¹ = 0
⇒ (x² + y² + 4x + 6y – 7) – (x² + y² + 2x + 3y – \(\frac{9}{4}\)) = 0
⇒ x² + y² + 4x + 6y – 7 – x² – y² – 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 a² = 16 and b² = 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 = x², y = x³
Solution:
Given curve equations are y = x², y = x³

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^-1x + tan^-1y = c
∴ The general solution of (1) is
tan^-1x – tan^-1y = c.

Section – B

II. Short Answer Type Questions.

1. Attempt any five questions.
2. Each question carries four marks.

Question 11.
Find the equation of the tangent at the point 30° (parametric value of θ) of the circle x² + y²+ 4x + 6y – 39 = 0.
Solution:
Given circle equation is s ≡ x² + y² + 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 2x² + 2y² + 5x – 6y + 4 = 0.
Solution:
Let x² + y² + 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 x² + y² + 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 9x² + 16y² = 144.
Solution:
Given 9x² + 16y² = 144
⇒ \(\frac{x^2}{16}+\frac{y^2}{9}\) = 1
Here a² = 16 ⇒ a = 4
b² = 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 2x² + y² = 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 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 tanθ₁ + tanθ₂ = 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(x₁, y₁) be the point of intersection of the tangents.

Question 16.
Find the area bounded between the curves y² = 4x, y² = 4(4 – x).
Solution:
Given curve equations are y² = 4x ………. (1)
y² = 4(4 – x) …………… (2)
From (1) & (2)
4x = 4 (4 – x)
⇒ 4x = 16 – 4x
⇒ 8x = 16
⇒ x = 2

From (1) y² = 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^xsecx cosx dx + c
⇒ y cosx = ∫e^xdx + c
⇒ y cosx = e^x + c

Section – C
(5 × 7 = 35)

III. Long Answer Type Questions.

1. Attempt any five questions.
2. 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 x² + y² + 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
x² + y² – 2x + 2y – 23 = 0.

Question 19.
Find the transverse common tangents of the circles x² + y² – 4x – 10 y + 28 = 0 and x² + y² + 4x – 6y + 4 = 0.
Solution:


(-2x – y + 7)² = (x² + y² – 4x – 10y + 28)
4x² + y² + 4xy – 28x – 14y + 4 = x² + y² – 4x – 10y + 28
3x² + 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 x² + y² = 2a² and the parabola y² = 8ax are y = ± (x + 2a).
Solution:
The equation of tangent to parabola y² = 8ax is y = mx + 2 a/m
⇒ my = m²x + 2a
⇒ m²x – my + 2a = 0 …….. (1)
If (1) touches the circle x² + y² = 2a² then
r = d
⇒ \(\sqrt{2} a\) = \(\frac{|2 a|}{\sqrt{m^4+m^2}}\)
⇒ 2a² = \(\frac{4 a^2}{m^4+m^2}\)
⇒ 1 = \(\frac{2}{m^4+m^2}\)
⇒ m⁴ + m² = 2
⇒ m⁴ + m² – 2 = 0
⇒ m⁴ + 2m² – m² – 2 = 0
⇒ m² (m² + 2) – 1 (m² + 2) = 0
⇒ (m² + 2) (m² – 1) = 0
⇒ m² – 1 = 0
⇒ m² = 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 = ∫cosⁿx dx, n being a positive integer, n ≥ 2, and deduce the value of ∫cos³x 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.