AP Inter 2nd Year Maths 2B Question Paper March 2015

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AP Inter 2nd Year Maths 2B Question Paper March 2015

Time : 3 Hours
Max. Marks : 75

Note: This question paper contains 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.
Find the value of ‘a’ if 2x² + ay² – 3x + 2y -1 = 0 represents a circle and also find its radius.
Solution:
Given equation is 2x² + ay² – 3x + 2y – 1 = 0 …….. (1)
Comparing the (1) with ax² + 2hxy + by² + 2gx + 2fy + c = 0
We get a = 2, h = 0, b = a, g = \(\frac{-3}{2}\), f = 1, c = -1
Now, equation (1) represents a circle then
a = b; 2 = a ∴ a = 2
The equation of the circle is 2x² + 2y² – 3x + 2y – 1 = 0
x² + y² – \(\frac{3}{2}\)x + y – \(\frac{1}{2}\) = 0
Comparing the above equation with x² + y² + 2gx + 2fy + c = 0

Question 2.
If the length of a tangent from (5, 4) to the circle x² + y² + 2ky = 0 is ‘1’, then find ‘k’.
Solution:
The given equation of the circle is x² + y² + 2ky = 0
Comparing the given equation with x² + y² + 2gx + 2fy + c = 0
We get g = 0, f = k, c = 0
Let, the given point P(x₁, y₁) = (5, 4)
Given that, the length of the tangent = 1
\(\sqrt{S_{11}}\) = 1 ⇒ s₁₁ = 1
\(x_1^2\) + \(y_1^2\) + 2gx₁ + 2fy₁ + c = 1
(5)₁ + (4)₁ + 2(0)(5) + 2k(4) + 0 = 1
25 + 16 + 8k = 1
8k = -40
k = \(\frac{-40}{8}\) = -5

Question 3.
Find the equation of the common chord of the circles :
(x – a)² + (y – b)² = c², (x – b)² + (y – a)² = c², (a ≠ b)
Solution:
Given equations of the circles are S ≡ (x – a)² + (y – b)² = c² = 0
x² + a² – 2ax + y² + b² – 2by – c² = 0
x² + y² – 2ax – 2by + a² + b² – c² = 0
S’ ≡ (x – b)² + (y – a)² = c²
x² + b² – 2bx + y² + a² – 2ay – c² = 0
x² + y² – 2bx – 2ay + a² + b² – c² = 0
The equation of the common chord (radical axis) of the given circles is S – S’= 0

Question 4.
Find the co-ordinates of the points on the parabola: y² = 2x whose focal distance is \(\frac{5}{2}\).
Solution:
Given equation of the parabola is y² = 2x
Comparing with y² = 4ax we get

4a = 2 ⇒ a = \(\frac{1}{2}\)
Let, P(x₁, y₁) be a point on the parabola y² = 2x
Given that, focal distance = \(\frac{5}{2}\)
x₁ + a = \(\frac{5}{2}\) ⇒ x₁ + \(\frac{1}{2}\) = \(\frac{5}{2}\)
⇒ x₁ = \(\frac{5}{2}\) – \(\frac{1}{2}\) ⇒ x₁ = 2
Since P(x₁, y₁) lies on the parabola y² = 2x then
\(y_1^2\) = 2x₁ ⇒ \(y_1^2\)2(2) ⇒ \(y_1^2\) = 4 ⇒ y₁ = ±2
∴ The required points are (2, 2), (2, -2).

Question 5.
Define rectangular hyperbola and find its eccentricity.
Solution:
If in a hyperbola the length of the transverse axis (2a) is equal to the length of the conjugate axis (2b), the hyperbola is called a rectangular hyperbola its equation is x² – y² = a² [∵ a = b]
In this case e² = \(\frac{a^2+b^2}{a^2}\) = \(\frac{2 a^2}{a^2}\) = 2 ⇒ e = \(\sqrt{2}\)
The eccentricity of a rectangular hyperbola is \(\sqrt{2}\).

Question 6.
Find \(\int \frac{e^x(1+x \log x)}{x}\)dx
Solution:

Question 7.
Find \(\int \frac{\sin \left(\tan ^{-1} x\right)}{1+x^2}\)dx, x ∈ R
Solution:
Put tan^-1x = t then
\(\frac{1}{1+x^2}\)dx = dt
Now, \(\int \frac{\sin \left(\tan ^{-1} x\right)}{1+x^2}\)dx = ∫sin t dt
= -cos t + c
= -cos(tan^-1x) + c

Question 8.
Evaluate : \(\int_0^{\pi / 2}\)sin⁵xcos⁴xdx
Solution:
Given \(\int_0^{\pi / 2}\)sin⁵xcos⁴xdx
= \(\frac{4+1}{5+4}\).\(\frac{4-3}{5+4-2}\).\(\frac{5-1}{5}\).\(\frac{5-3}{5-2}\)
= \(\frac{5}{9}\).\(\frac{1}{7}\).\(\frac{4}{5}\).\(\frac{2}{3}\) = \(\frac{8}{315}\)

Question 9.
Evaluate \(\int_0^2\)|1-x|dx
Solution:

Question 10.
From the differential equation corresponding to
y = A cos 3x + B sin 3x, where A and B are parameters.
Solution:
Given y = A cos 3x + B sin 3x
then \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = – 3A sin 3x + 3B cos 3x
\(\frac{d^2 y}{d x^2}\) = – 9A cos 3x – 9B sin 3x
= – 9(A cos 3x + B sin 3x) = -9y
∴ \(\frac{d^2 y}{d x^2}\) + 9y = 0 where A, B are eliminated.

Section – B
(5 × 4 = 20)

II. Short Answer type questions.

1. Attempt ANY FIVE questions.
2. Each question carries FOUR marks.

Question 11.
Find the equation of circle whose centre lies on the x-axis and passing through (-2, 3) and (4, 5).
Solution:
Let, the equation of the required circle is
x² + y² + 2gx + 2fy + c = 0 ……… (1)
Center of (1) c(h, k) = (-g, -f) lies on the x-axis then – k = 0 ⇒ f = 0
Since, (1) passes through the point (-2, 3) then
(-2)² + (3)² + 2g(-2) + 2f(3) + c = 0
4 + 9 + 4g – 6f + c = 0
4g + 6f + c = -13
If f = 0, then
-4g + 6(0) + c + 13 = 0
-4g + c + 13 = 0 ………….. (2)
Since, (1) passes through the point (4, 5) then
(4)² + (5)² + 2g(4) + 2f(5) + c = 0
16 + 25 + 8g + 10f + c = 0
8g + 10f + c +41 = 0
If f = 0 then 8g + 10(0) + c + 41 = 0
8g + c + 41 = 0 ……….. (3)
Solve (2) and (3)

Now, substitute the values of g, f, c in (1)
x² + y² + 2g\(\left(\frac{-7}{3}\right)^x\) + 2(0)y – \(\frac{67}{3}\) = 0
3x² + 3y² – 14x – 67 = 0
∴ The equation of the required circle is 3x² + 3y² – 14x – 67 = 0

Question 12.
If x + y = 3 is the equation of the chord AB of the circle :
x² + y² – 2x + 4y – 8 = 0, find the equation of the circle having AB as diameter.
Solution:
Given equation of the straight line is L ≡ x + y – 3 = 0.
Given equation of the circle is g ≡ x² + y² – 2x + 4y – 8 = 0
Equation of the circle passing through A,B is S + λL = 0
x² + y² – 2x + 4y – 8 + λ, (x + y – 3) = 0 ………. (1)
x² + y² – (2 – λ)x + (4 + λ)y – (3λ + 8) = 0
Here g = \(\frac{(2-\lambda)}{2}\), f = \(\frac{4+\lambda}{2}\)
Centre of (1) s C = (-g, -f) = \(\left(\frac{2-\lambda}{2},-\frac{4+\lambda}{2}\right)\)
If \(\overline{\mathrm{AB}}\) is a diameter of circle (1) then ‘C’ lies on x + y – 3 = 0.
\(\frac{2-\lambda}{2}\) + \(\frac{-4+\lambda}{2}\) – 3 = 0
2 – λ – 4 – λ – 6 = 0 ⇒ 2A = -8 ⇒ λ = -4
∴ The equation of the circle on \(\overline{\mathrm{AB}}\) as a diameter is
x² + y² – 2x + 4y – 8 – 4(x + y – 3) = 0
⇒ x² + y² – 6x + 4 = 0

Question 13.
Find the equation of tangent and normal to the ellipse 9x² + 16y² = 144 at the end of the latus rectum in the first quadrant.
Solution:

Question 14.
Find the value of ‘k’ if: 4x + y + k = 0 is a tangent to the ellipse x² + 3y² = 3.
Solution:
Given equation of the ellipse is x² +3y = 3
\(\frac{x^2}{3}+\frac{3 y^2}{3}\) = 1 ⇒ \(\frac{x^2}{3}+\frac{y^2}{1}\) = 1
Here a² = 3, b² = 1
Given equation of the straight line is 4x + y + k = 0 …………. (1)
y = -4x – k
Comparing with y = mx + c we get m = -4, c = -k
Since eq. (1) is a tangent to the given ellipse then
c² = a²m² + b²
(-k)² = 3(-4)² + 1 ⇒ k² = 49 ⇒ k = ±1

Question 15.
Find the equations of the tangents to the hyperbola :3x² – 4y² = 12 which are :

1. Parallel and
2. Perpendicular to the line : y = x – 7.

Solution:

Question 16.
Find \(\int_0^{\pi / 2} \frac{d x}{4+5 \cos x}\)
Solution:

Question 17.
Solve the differential equation : (xy² + x) dx + (yx² + y) dy = 0.
Solution:

Section – C
(5 × 7 = 35)

III. Long Answer type questions.

1. Attempt ANY FIVE questions.
2. Each question carries SEVEN marks.

Question 18.
If (2, 0), (0,1), (4, 5) and (0, C) are concyclics then find ‘C’.
Solution:
Let, the equation of the required circle is
x² + y² + 2gx + 2fy + c = 0 ……. (1)
Since (1) passes through the point (2, 0) then
(2)² + (0)² + 2g(2) + 2f(0) + 0 = 0
4 + 4g + c = 0 ⇒ 4g + c = -4 ……… (2)
Since (1) passes through the point (0, 1) then
0² + 1² + 2g(0) + 2f(1) + c = 0
1 + 2f + c = 0 ⇒ 2f + c = -1 ……. (3)
Since, (1) passes through the point (4, 5) then
(4)² + (5)² + 2g(4) + 2f(5) + c = 0
16 + 25 + 8g + 10f + c = 0
8g + 10f + c = – 41 ……….. (4)


Substitute the point (0, c) in (7)
3(0)² + 3c² – 13(0) – 17c + 14 = 0
3c² – 17c + 14 = 0
3c² – 3c – 14c + 14 = 0
3c(c – 1) – 14(c – 1) = 0
(c – 1) (3c – 14) = 0
c – 1 = 0 3c – 14 = 0
c = 1 c = \(\frac{14}{3}\)
∴ c = 1 or \(\frac{14}{3}\)

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

∴ Given circle are each circle lies completely outside the other.
Let, A₁ be the internal centre of similitude.
The internal centre of similitude A₁ divides C₁C₂ in the ratio r₁ : r₂ (1 : 3) internally.
∴ The internal centre of similitude

The equation to the pair of transverse common tangents is SS₁₁ = \(S_1^2\)

(x₂ + y₂ – 4x – 10y + 28) = (-2x – y + 7)₂
x₂ + y₂ – 4x – 10y + 28 = 4x₂ + y₂ + 4g + 4xy – 14y – 28x
3x₂ + 4xy – 24x – 4y + 21 = 0
consider 3x₂ + 4xy = 0
x(3x + 4y) = 0
x = 0 and 3x + 4y = 0
∴ 3x² + 4xy – 24x – 4y + 21 = (x + 1)(3x + 4y + k)
Comparing coefficient of x on both sides we get
k + 3l = -24 …… (3)
Comparing coefficient of x on both sides we get
4l = -4
l = -1
Substitute the value of l in (3)
k + 3(-1) = -24
k – 3 = -24
k = -24 + 3
k = -21
∴ The equation of the transverse common tangents are x – 1 = 0 and 3x + 4y – 21 = 0

Question 20.
Evaluate: \(\int \frac{2 \cos x+3 \sin x}{4 \cos x+5 \sin x}\)dx
Solution:
Let,
2cosx + 3sinx = λ\(\frac{\mathrm{d}}{\mathrm{dx}}\)(4cosx + 5sinx) + μ(4cosx + 5sinx x) + γ
= λ(-4sinx + 5cosx) + μ(4cos x + 5sin x) + γ …….. (1)
= 4λ.sinx + 5λcosx + 4μcosx + 5μsinx + γ
Comparing the coefficients of cos x on both sides, we get
5λ + 4μ = 2 ………. (2)
Comparing the coefficients of sinx on both sides, we get
-4λ + 5μ = 3 ………. (3)

Question 21.
Obtain reduction formula : ∫tanⁿx dx for integer n ≥ 2 and evaluate: ∫tan⁶xdx.
Solution:

Question 22.
Derive the standard form of the parabola.
Solution:
The equation of a parabola in the standard form is y² = 4ax.
Let, ‘S’ be the focus and L = 0 be the directrix of the parabola.
Let, ‘P’ be a point on the parabola.
Let, M, Z be the projections (Foot of the ⊥rs) of P, S on the directrix,
L = 0 respectively. .
Let, N be the projection of P and SZ.
Let, A be the mid point of SZ.
Since, SA = AZ, A lies on the parabola.
Let, AS = a. Take AS, the principle axis of the parabola as X-axis and AY perpendicular to SZ as y-axis. ,
Then, S = (a, 0) and the parabola is in the standard form.
Let, P = (x, y)
Now, PM = NZ = AN + AZ = x + a
The equation to the parabola is

Question 23.
Evaluate \(\int_0^\pi \frac{x \sin x}{1+\sin x}\)dx
Solution:

Question 24.
Solve : (1 + y²)dx = (tan^-1y – x)dy.
Solution: