AP Inter 2nd Year Maths 2B Question Paper March 2016

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AP Inter 2nd Year Maths 2B Question Paper March 2016

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.
If the circle x² + y² – 4x + 6y + a = 0 has radius 4, then find a.
Solution:
Given circle x² + y² – 4x + 6y + a = 0
Radius ⇒ \(\sqrt{2^2+3^2-a}\) = 4
⇒ \(\sqrt{13-a}\) = 4 ⇒ a = -3.

Question 2.
Obtain the parametric equations of the circle (x – 3)² + (y – 4)² = 8².
Solution:
From the given equation,
centre C(h, k) = (3, 4) and radius r = 8.
∴ The parametric equations of the circle (x – 3)² + (y – 4)² = 8² are
x = 3 + 8 cos θ.
y = 4 + 8 sinθ.

Question 3.
Find the value of k, if the circles x² + y² + 4x + 8 = 0 and x² + y² – 16y + k = 0 are orthogonal.
Solution:
The circles x² + y² + 4x + 8 = 0 and x² + y² – 16y + k = 0 cut each other orthogonally
∴ 2(gg’ + ff) = e + c’
⇔ 2 × 2 × 0 + 2(0) (-8) = k + 8
⇔ k = -8.

Question 4.
Find the coordinates of the points on the parabola y² = 8x whose focal distance is 10.
Solution:
The focal distance of a point on the parabola y² = 4ax is | x₁ + a |
Here a = 2
∴ |x₁ + 2| = 10
⇒ x₁ = 8
⇒ \(y_1^2\) = 64
⇒ y₁ = ± 8.
∴ Co-ordinates of the points (8, ±8).

Question 5.
If the eccentricity of a hyperbola is \(\frac{5}{4}\), then find the eccentricity of its conjugate hyperbola.
Solution:
Let e and e₁ be the eccentricities of the hyperbola and its conjugate respectively

Question 6.
Evaluate \(\int \frac{d x}{\cosh x+\sinh x}\) on R.
Solution:

Question 7.
Evaluate \(\int \frac{x^8}{1+x^{18}}\) on R.
Solution:

Question 8.
Find \(\int_{-\pi / 2}^{\pi / 2}\)sin²xcos⁴x dx
Solution:

Question 9.
Evaluate \(\int_0^\pi \sqrt{2+2 \cos \theta}\) dθ
Solution:

Question 10.
Find the order and degree of the differential equation
\(\left[\frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^3\right]^{6 / 5}\) = 6y.
Solution:
The given differential equation

∴ The order and degree of the equation are 2 and 1.

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 pole of 3x + 4y – 45 = 0 with respect to the circle x² + y² – 6x – 8y + 5 = 0.
Solution:
Let (x₁, y₁) be the pole of 3x + 4y – 45 = 0 ……… (1)
The equation of polar of (x₁, y₁) w.r.t the circle
x² + y² – 6x – 8y + 5 = 0 is
xx₁ + yy₁ – 3(x + x₁) -4 (y + y₁) + 5 = 0
⇒ x(x₁ – 3) + y(y₁ – 4) – (3x₁ + 4y₁ – 5) = 0 ………… (2)
From (1) and (2)
\(\frac{x_1-3}{3}\) = \(\frac{y_1-4}{4}\) = \(\frac{-\left(3 x_1+4 y_1-5\right)}{-45}\)
\(\frac{x_1-3}{3}\) = \(\frac{y_1-4}{4}\)
⇒ 4x₁ -12 = 3y₁ -12 ⇒ 4x₁ – 3y₁ = 0 …….. (3)
\(\frac{y_1-4}{4}\) = \(\frac{3 x_1+4 y_1-5}{45}\)
⇒ 45y₁ – 180 = 12x₁ + 16y₁ – 20
⇒ 12x₁ – 29y₁ + 160 = 0 ………. (4)
Solving (3) and (4) we get 12xx – 9yx = 0
12x₁ – 29y₁ = -160
∴ 20y₁ = 160 ⇒ y₁ = 8 and from (3) 4x₁ = 24 ⇒ x₁ = 6
∴ Pole of 3x + 4y – 45 = 0 w.r.t x² + y² – 6x – 8y + 5 = 0 is (6, 8).

Question 12.
Find the equation of the circle which cuts the circles
x² + y² – 4x – 6y +11 = 0 and x² + y² – 10x – 4y + 21 = 0 orthogo nally and has the diameter along the straight line 2x + 3y = 7.
Solution:
Equations of given circle are
x² + y² – 4x – 6y + 11 = 0 ……….. (1)
x² + y² – 10x – 4y + 21 = 0 ………… (2)
Let the equation of the required circle be
x² + y² + 2gx + 2fy + c = 0
If this cuts (1) and (2) orthogonally then
2g(-2) + 2f (-3) = c + 11 ⇒ 4g + 6g = -c – 11 ………. (3)
and if (1) and (3) are orthogonal then
2g(-5) + 2f (-2) = c + 21
⇒ -10g – 4f = c + 21
⇒ 10g + 4f = -c – 21 ………(4)
from (3) and (4)
-6g + 2f = 10
⇒ 6g – 2f = -10 ⇒ 3g – f = -5 ……. (5)
Since the centre of circle (1) (-g, – f) lies on the diameter
2x + 2y = 7, we have
2(-g) + 3 (-f) = 7 ⇒ 2g + 3f = -7 ……. (6)
Solving (5) and (6)

Question 13.
Show that the points of intersection of the perpendicular tan gents to an ellipse lie on a circle.
Solution:
The equation of the ellipse \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1 (a > b). Any tangent to it in the slope intercept form is
y = mx ± \(\sqrt{a^2 m^2+b^2}\)
Let the perpendicular tangents intersect at P(x₁, y₁)
P lies on (1) for some real m,
⇒ y₁ = mx₁ ± \(\sqrt{a^2 m^2+b^2}\)
∴ (y₁ – mx₁)² = a²m² + b² (or)
(\(x_1^2\) – a²) m² – 2x₁y₁m + (\(y_1^2\) – b²) =0 being a quadratic equation in ‘m’ has two roots say m₁ and m₂ then m₁, m₂ are the slopes of tangents from P to the ellipse.

If, however, one of the perpendicular tangents in Vertical, then such pair of perpendicular tangents intersect at one of the points (±a, ±b) and any of these points satisfies x² + y² = a² + b²
∴ The point of intersection of perpendicular tangents to the ellipse S = 0 lies on the circle x² + y² = a² + b².

Question 14.
Find the value of k if 4x + y + k = 0 is a tangent to the ellipse x² + 3y² = 3.
Solution:
Given ellipse is x² + 3y² = 3
Hence a² = 3 and b² = 1
Equation of the given line is 4x + y + k = 0
⇒ y = -4x – k where m = -4 and c = -k the condition for tangency is c² = a²m² + b²
⇒ k² = 3 (16) + 1
⇒ k² = 49
⇒ k = ±7.

Question 15.
Find the center, foci, eccentricity, equation of directrices to the hyperbola x² – 4y² = 4.
Solution:
x² – 4y² = 4
this can be written as \(\frac{x^2}{4}\) – \(\frac{y^2}{1}\) = 1
⇒ a² = 4 and b² = 1
⇒ a = 2 and b = 1

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

Question 17.
Solve the differential equation

Solution:
The equation can be written as

Section – C

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 (4, 1), (6, 5) and having the center on 4x + 3y – 24 = 0.
Solution:
Let the equation of the required circle be
x² + y² + 2gx + 2fy + c = 0 …….. (1)
The centre of the circle is (-g, -f)
since (1) passes through (4, 1) we have
4² + 1² + 2g(4) + 2f(1) + c = 0
⇒ 8g + 2f + c = -17 ………. (2)
Also since (1) passes through the point (6, 5) we have
6² + 5² + 2g(6) + 2f(5) + c = 0
⇒ 12g + 10f + c = -61 ……. (3)
Given that the centre (-g, -f) lies on the line 4x + 3y – 24 = 0
we have -4g – 3f – 24 = 0
⇒ 4g + 3f = -24 ………….. (4)
from (2) and (3)
8g + 2f + c = – 17
12g + 10f + c = – 61
We have -4g – 8f – 44 = 0
⇒ g + 2f + 11 = 0 …………. (5)
Solving (4) and (5)

substituting in (2) we get
8(-3) + 2(-4) – c = -17
⇒ -24 – 8 + c = -17
⇒ c = -17 + 32 = 15
∴ from (1) the equation of the required circle is
x² + y² + 2 (-3) x + 2 (-4) y + 15 = 0
⇒ x² + y² – 6x – 8y + 15 = 0.

Question 19.
Show that the circles x² + y² – 6x – 2y + 1 = 0, x² + y² + 2x – 8y + 13 = 0 touch each other. Find the point of contact and the equation of common tangent at their point of contact.
Solution:
Let S ≡ x² + y² – 6x – 2y + 1 = 0 and
S’ ≡ x² + y² + 2x – 8y + 13 = 0 be the given circles then centre of circles are C₁ = (3, 1) and C₂ = (-1, 4)
radius of circles are r₁ = \(\sqrt{9+1-1}\) = 3
and r₂ = \(\sqrt{1+16-13}\) = 2
Distance between centres
C₁C₂ = \(\sqrt{(3+1)^2+(1-4)^2}\) = \(\sqrt{16+9}\) = 5 and r₁ + r₂ = 3 + 2 = 5
Since C₁C₂ = r₁ + r₂ the two circles touch externally at a point.
Let P be the point of contact of circles such that r₁ : r₂ = 3 : 2
Since P divides C₁, C₂ internally in the ratio 3 : 2 coordinates of point of contact

The equation of common tangent is S – S’ = 0
⇒ x² + y² – 6x – 2y + 1 – x² – y² – 2x + 8y + 13 = 0
⇒ -8x + 6y – 12 = 0
⇒ 4x – 3y + 6 = 0
∴ The equation of the common tangent at the point of contact is 4x – 3y + 6 = 0.

Question 20.
Show that the common tangent to the parabola y² = 4ax and x² = 4by is x\(a^{\frac{1}{3}}\) + y\(b^{\frac{1}{3}}\) + \(a^{\frac{2}{3}}\)\(a^{\frac{2}{3}}\) =0.
Solution:
The equations of the parabola are
y² = 4ax …….. (1)
x² = 4by ………… (2)
Equation of any tangent to (1) is of the form y = mx + \(\frac{a}{m}\) …. (3)
If the line (3) is a tangent to (2) also, the points of intersection of (2) and (3) coincide.
Substituting the value of y from (3) in (2) we get
x² = 4b(mx + \(\frac{a}{m}\))
⇒ mx² – 4bm²x – 4ab = 0
Which should have equal roots.
∴ Its discriminant is zero. Hence
16b²m⁴ – 4m(-4ab) = 0
16b (bm⁴ + am) = 0
m(bm³ + a) = 0. But m ≠ 0
∴ m = \(\frac{-a^{1 / 3}}{b^{1 / 3}}\) substituting in (3) the equation of the common tangent becomes y = \(\left(\frac{a}{b}\right)^{1 / 3}\) + \(\frac{a}{\left(-\frac{a}{b}\right)^{\frac{1}{3}}}\) (or)
a^1/3x + b^1/3y + a^2/3b^2/3 = 0.

Question 21.
Evaluate \(\int \frac{2 \sin x+3 \cos x+4}{3 \sin x+4 \cos x+5}\)dx.
Solution:
\(\int \frac{2 \sin x+3 \cos x+4}{3 \sin x+4 \cos x+5} d x\)
Since there exists constants in both numerator and denominator, we determine constant A, B and C such that
2 sin x + 3 cos x + 4
= A\(\frac{d}{d x}\)(3 sin x + 4 cos x + 5) + g(3 sin x + 4 cos x + 5) + c
= A(3 cos x – 4 sin x) + B(3 sin x + 4 cos x + 5) + c …… (1)
Comparing both sides the coefficients of sin x, cos x and constants
-4A + 3B = 2
⇒ 4A – 3B + 2 = 0 ………. (2)
⇒ 3A + 4B – 3 = 0 ……….. (3)
⇒ 5B + c – 4 = 0 ……… (4)
Solving (2) and (3)

Question 22.
Obtain the reduction formula for I_n = ∫cotⁿ x dx, n being a positive integer, n ≥ 2 and deduce the value of ∫cot⁴x dx.
Solution:

Question 23.
Evaluate \(\int_0^{\pi / 4}\)log(1 + tan x)dx.
Solution:

Question 24.
Solve the differential equation (x² + y²) dx = 2xy dy.
Solution:
(x² + y²) dx = 2xy dy
The given equation can be written as
\(\frac{d y}{d x}\) = \(\frac{x^2+y^2}{2 x y}\) ……… (1)
Which is a homogeneous equation, since the numerator and denominator on the right are homogeneous functions each of degree 2.