AP Inter 2nd Year Maths 2B Question Paper March 2018

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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.

1. Attempt all questions.
2. Each question carries two marks.

Question 1.
Obtain the parametric equation of circle (x – 3)² + (y – 4)² = 8².
Solution:
Given circle equation is (x – 3)² + (y – 4)² = 8²
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 = x² + y² – 10x – 2y + 6.
Solution:
Given S ≡ x² + y² – 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 – x₁) (y₁ + f) – (y – y₁) (x₁ + 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 x² + y² – 5x – 14y – 34 = 0, x² + y² + 2x + 4y + k = 0 circles are orthogonal, then find ‘k’.
Solution:
Given circle equations are
x² + y² – 5x – 14y – 34 = 0 …… (1)
x² + y² + 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 y² = 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 y² = 6x
Here 4a = 6 ⇒ a = \(\frac{6}{4}\)
⇒ a = \(\frac{3}{2}\)
Since (1) is tangent to the parabola y² = 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 16y² – 9x² = 144.
Solution:
Given hyperbola equation is 16y² – 9x² = 144
⇒ \(\frac{y^2}{9}\) – \(\frac{x^2}{16}\) = 1
Here a² = 16, b² = 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^-1x = t
Then \(\frac{1}{1+x^2}\)dx = dt
= ∫sin t dt
= -cos t + c
= -cos(tan^-1x) + c
∴ \(\int \frac{\sin \left(\tan ^{-1} x\right)}{1+x^2} d x\) = -cos(tan^-1x) + 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 = \(\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\)sin²xcos⁴x dx
since sin²xcos⁴x is an even function
= 2\(\int_0^{\frac{\pi}{2}}\)sin²xcos⁴x dx
= 2.\(\frac{2-1}{2+4}\).\(\frac{4-1}{4}\).\(\frac{1}{2}\).\(\frac{\pi}{2}\)
= 2.\(\frac{1}{6}\).\(\frac{3}{4}\).\(\frac{1}{2}\).\(\frac{\pi}{2}\)
= \(\frac{\pi}{16}\)
∴ \(\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\)sin²xcos⁴x = \(\frac{\pi}{16}\)

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.

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

Question 11.
Find the length of the chord intercepted by the circle x² + y² – x + 3y – 22 = 0 on the line y = x – 3.
Solution:
Given circle equation is x² + y² – 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) x² + y² – 4x – 6y + 5 = 0
ii) x² + y² – 2x – 4y – 1 = 0
iii) x² + y² – 6x- 2y = 0
Solution:
Let S ≡ x² + y² – 4x – 6y + 5 = 0
S’ ≡ x² + y² – 2x – 4y – 1 = 0
S” ≡ x² + y² – 6x – 2y = 0
The radical axis of S = 0, S’ = 0 is S – S’ = 0
⇒ x² + y² – 4x – 6y + 5 – x² – y² + 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
⇒ x² + y² – 4x – 6y + 5 – x² – y² + 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 + 4a² = 15a
⇒ 4a² – 15a – 4 = 0
⇒ 4a² – 16a + a – 4 = 0
⇒ 4a (a – 4) + 1 (a – 4) = 0
⇒ (a – 4) (4a + 1) = 0
Since a > 0
∴ a = 4
from (1) e = \(\frac{1}{4}\)
b² = a² (1 – e²)
= 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 x² – 4y² = 4.
Solution:
Given hyperbola equation is x² – 4y² = 4
⇒ \(\frac{x^2}{4}\) – \(\frac{y^2}{1}\) = 1
Here a² = 4, b² = 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 + x²)\(\frac{d y}{d x}\) + y = \(e^{\tan ^{-1} x}\).
Solution:
Given (1 + x²)\(\frac{d y}{d x}\) + 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.

1. Attempt any five questions.
2. 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 x² + y² + 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
x² + y² + 2(-2)x + 2(-3)y + 11 = 0
⇒ x² + y² – 4x – 6y + 11 = 0.

Question 19.
Find the direct common tangents of the circles x² + y² + 22x – 4y – 100 = 0 and x² + y² – 22x + 4y + 100 = 0.
Solution:
Given circle equations are
S ≡ x² + y² + 22x – 4y – 100 = 0
S’ ≡ x² + y² – 22x + 4y + 100 = 0
C₁ = (-11, 2) r₁ = \(\sqrt{121+4+100}\) = 15
C₂ = (-11, 2) r₂ = \(\sqrt{121+4-100}\) = 5
d = C₁C₂ = \(\sqrt{(11+11)^2+(-2-2)^2}\)
= \(\sqrt{484+16}\)
= \(\sqrt{500}\) = 10\(\sqrt{5}\)
d > r₁ + r₂
∴ The two circles do not intersect.
Let P be the external centre of similitude.
∴ P divides C₁C₂ in ratio r₁ : r₂ = 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 \(S_1^2\) = SS₁₁
⇒ [x(22) + y(- 4) + 11 (x + 22) – 2(y – 4) – 100]²
= (x² + y² + 22x – 4y – 100) (484 + 16 + 484 + 16 – 100)
⇒ [22x – 4y + 11x + 242 – 2y + 8 – 100]² = (x² + y² + 22x – 4y – 100)(900)
⇒ [33x – 6y + 150]² = (x² + y² + 22x – 4y – 100)(900)
⇒ [11x – 2y + 50]² = (x² + y² + 22x – 4y – 100)(100)
⇒ 121x² + 4y² + 2500 – 44xy -200y + 1100x = 100x² + 100y² + 2200x – 400y – 10000
⇒ 21x² – 96y² – 44xy – 1100x + 200y + 12500 = 0
21x² – 44xy – 96y² = 21x² – 72xy + 28xy – 96y²
= 3x(7x – 24y) + 4y(7x – 24y)
= (3x + 4y)(7x – 24y)
21x² – 44xy – 96y² – 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 (x₁, y₁), (x₂, y₂) and (x₃, y₃) to the parabola y² = 4ax (a > 0) is \(\frac{1}{16 a}\)|(y₁ – yy₂) (y₂ – y₃) (y₃ – y₁)| 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 – ∫tanⁿx dx, n being a positive integer n ≥ 2 and deduce the value of ∫tan⁶ x dx.
Solution:
Let I_n = ∫Tanⁿx dx
= ∫Tan^n-2x Tan²x dx
= ∫Tan^n-2x(sec²x – 1)dx
= ∫Tan^n-2xsec²x dx – ∫Tan^n-2xdx

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 : (x²y – 2xy²)dx = (x³ – 3x²y)dy.
Solution:
Solution:
Given differential equation is
(x²y – 2xy²)dx = (x³ – 3x²y)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