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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.
- Attempt ALL questions.
- 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.
- Attempt ANY FIVE questions.
- 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 :
- Parallel and
- 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.
- Attempt ANY FIVE questions.
- 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: