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

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

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.
If x² + y² + 2gx + 2fy – 12 = 0 represents a circle with centre (2, 3) find g, f and its radius.
Solution:
Given circle equation is x² + y² + 2gx + 2fy – 12 = 0.
centre = (- g, – f)
⇒ (2,3) = (-g,-f)
⇒ – g = 2, – f = 3
⇒ g = -2, f = -3 and c = – 12
Radius = \(\sqrt{g^2+f^2-c}\)
= \(\sqrt{4+9-(-12)}\)
= \(\sqrt{4+9+12}\) = \(\sqrt{25}\) = 5.

Question 2.
Find the value of k if the points (1, 3) and (2, k) are conjugate with respect to the circle x² + y² = 35.
Solution:
Let S ≡ x² + y² – 35 = 0
Since the points (1, 3) and (2, k) an conjugate with respect to the circle s = 0
∴ s₁₂ = 0
⇒ x₁x₂ + y₁y₂ – 35 = 0
⇒ 1(2) + 3(k) – 35 = 0
⇒ 3k = 33 ⇒ k = 11.

Question 3.
Find the angle between the circles
x² + y² – 12x – 6y + 41 = 0
x² + y² + 4x + 6y – 59 = 0.
Solution:
Given circle equations are
x² + y² – 12x – 6y + 41 = 0 ……. (1)
x² + y² + 4x + 6y – 59 = 0 …… (2)

Question 4.
Find the co-ordinates of the points on the parabola y² = 8x whose focal distance is 10.
Solution:
Given parabola equation is y² = 8x
Here 4a = 8 ⇒ a = 2
∴ Focus = (a, 0) = (2, 0)
Let p(x₁, y₁) be a point on the parabola.
Given sp = 10

⇒ (x₁ + 12) (x₁ – 8) = 0
⇒ x₁ = – 12 (or) x₁ = 8
If x₁ = -12 then \(y_1{ }^2\) = – 96 < 0
If x₁ = 8 then \(y_1^2\) = 64 ⇒ y₁ = ± 8
∴ Co-ordinates on the parabola are (8, 8) and (8, -8).

Question 5.
If the eccentricity of the hyperbola is \(\frac{5}{4}\), then find the eccentricity of its conjugate hyperbola.
Solution:
If e₁, e₂ are the eccentricities of hyperbola and conjugate hyperbola then \(\frac{1}{\mathrm{e}_1^2}\) + \(\frac{1}{\mathrm{e}_2^2}\) = 1

Question 6.
Evaluate : ∫\(\frac{d x}{1+e^x}\), x ∈ R
Solution:

Question 7.
Evaluate: ∫\(\frac{1}{\cos h x+\sinh x}\)dx on R.
Solution:

Question 8.
Evaluate \(\int_0^2|1-x| d x\)
Solution:

Question 9.
Evaluate \(\int_0^{\pi / 2} \frac{\sin ^5 x}{\sin ^5 x+\cos ^5 x} d x .\)
Solution:

Question 10.
Find the differential equation corresponding to xy = ae^x + be^-x, a and b are parameters.
Solution:
Give xy = ae^x + be^-x
diffefentiating w.r.to ‘x’ on both sides, we have

Section-B

II. Short Answer Type Questions.

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

Question 11.
Find the equation of tangent and normal at (3, 2) of the circle x² + y² – x – 3y – 4 = 0.
Solution:
Let S = x² + y² – x – 3y – 4 = 0.
The equation of tangent at (3, 2) of the circle s = 0 is s₁ = 0
⇒ x(3) + y(2) – \(\frac{1}{2}\)(x + 3) – \(\frac{3}{2}\)(y + 2) – 4 = 0.
⇒ 3x + 2y – \(\frac{1}{2}\)x – \(\frac{3}{2}\) – \(\frac{3}{2}\)y – 3 – 4 = 0
⇒ \(\frac{5}{2}\)x + \(\frac{1}{2}\)y – \(\frac{17}{2}\) = 0
⇒ 5x + y – 17 = 0.
The equation of normal at (3, 2) of the circle s = 0 in
1(x – 3) – 5(y – 2) = 0
⇒ x – 3 – 5y + 10 = 0
⇒ x – 5y + 7 = 0.

Question 12.
Find the radical centre of the circles x² + y² – 4x – 6y + 5 = 0 x² + y² – 2x – 4y – 1 = 0, x² + y² – 6x – 2y = 0.
Solution:
Lets s ≡ x² + y² – 4x – 6y + 5 = 0
s¹ = x² + y² – 2x – 4y – 1 = 0
s¹¹ = x² + y² – 6x – 2y = 0
The radical axis of s = 0, s¹ = 0 in s – s¹ = 0
⇒ -2x – 2y + 6 = 0
⇒ x + y – 3 = 0 ……….. (1)
The radical axis of s = 0, s¹¹ = 0 is s – s¹¹ = 0
⇒ 2x – 4y + 5 = 0 ……… (2)
solving (1) and (2), we have

Question 13.
Find the equation of the ellipse referred to its major and minor axes as the co-ordinate axes x,y – respectively with latus rectum of length 4 and distance between foci 4\(\sqrt{2}\).
Solution:

Question 14.
Find the equation of the tangents 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

ii)
Perpendicular to x + y + 2 = 0
Slope of tangent be ‘1’ as it is perpendicular to above line
y = mx ± \(\sqrt{a^2 m^2+b^2}\)
y = x ± \(\sqrt{4+8}\)
y = x ± 2\(\sqrt{3}\).
⇒ x – y ± 2\(\sqrt{3}\) = 0.

Question 15.
Find the centre, eccentricity, foci, directrices and the length of the latus rectum of the hyperbola 4(y + 3)² – 9(x – 2)² = 1.
Solution:
Given hyperbola equation is 4(y + 3)² – 9 (x – 2)² = 1

Question 16.
Evaluate : \(\int_0^4\left(16-x^2\right)^{\frac{5}{2}}\)
Solution:
I = \(\int_0^4\left(16-x^2\right)^{\frac{5}{2}} d x\)
put x = 4 sin θ
dx = 4 cos θ dθ

Question 17.
Solve : \(\frac{d y}{d x}\) + y tan x = cos³ x.
Solution:

Section – C

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 concyclic, then find C.
Solution:
x² + y² + 2gx + 2fy + c₁ = 0
Satisfies (2, 0), (0, 1) (4, 5) we get
4 + 0 + 4g + c₁ = 0 ……. (i)
0 + 1 + 2g. 0 + 2f + c₁ = 0 ……. (ii)
16 + 25 + 8g + 10f + c₁ = 0 …….. (iii)
(ii) – (i) we get
– 3 – 4g + 2f = 0
4g – 2f = – 3 …….. (iv)
(ii) – (iii) we get
– 40 – 8g – 8f = 0 (or)
g + f = – 5 …… (v)
Solving(iv) and (v) we get
g = –\(\frac{13}{6}\) = –\(\frac{17}{6}\)
Substituting g and f values in equation (i) we get
4 + 4(-\(\frac{13}{6}\)) + c₁ = 0
c₁ = \(\frac{14}{3}\)
Now equation x² + y² – \(\frac{13}{3}\)x – \(\frac{17}{3}\)y + \(\frac{14}{3}\) = 0
Now circle passes through (0, c) then
c² – \(\frac{17}{3}\)c + \(\frac{14}{3}\) = 0
3c² – 17c + 14 = 0
⇒ (3c – 14) (c – 1) = 0
(or)
c = 1 or \(\frac{14}{3}\)

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:
Given circle equations are
S₁ ≡ x² + y² – 6x – 2y + 1 = 0
S₂ ≡ x² + y² + 2x – 8y + 13 = 0
Centre A = (3, 1), r₁ = \(\sqrt{9+1-1}\) = 3
Centre B = (-1, 4), r₂ = \(\sqrt{1+16-13}\) = 2
AB = \(\sqrt{(-1-3)^2+(4-1)^2}\)
= \(\sqrt{16+9}\) = 5
= r₁ + r₂
AB = r₁ + r₂
∴ The given circles touch each other externally.
Let ‘P’ be the common point of the circles S₁ = 0, S₂ = 0
P divides AB in the ratio 3:2 internally
∴ P = \(\left(\frac{3(-1)+2(3)}{3+2}, \frac{3(4)+2(1)}{3+2}\right)\)
= \(\left(\frac{3}{5}, \frac{14}{5}\right)\)
Common tangent equation is S₁ – S₂ = 0
x² + y² – 6x – 2y + 1 – (x² + y² + 2x – 8y + 13) = 0
⇒ x² + y² – 6x – 2y + 1 – x² – y² – 2x + 8y – 13 = 0
⇒ – 8x + 6y – 12 = 0
⇒ 4x – 3y + 6 = 0.

Question 20.
Derive the equation of a parabola in the standard form y² = 4ax with diagram.
Solution:
Equation of a parabola in standard form :
To study the nature of the curve, we prefer its equation in the simplest possible form. We proceed as follows to derive such an equation.
Let S be the focus, l be the directrix as shown in fig. Let z be the projection of ‘S’ on l and ‘A’ be the midpoint of \(\overline{S Z}\). A lies on the parabola because SA = AZ. A is called the vertex of the parabola. Let \(\overline{Y A Y^{\prime}}\) be the straight line through A and parallel to the directrix. Now take \(\overline{\mathrm{zx}}\) as the X-axis and \(\overline{Y Y^{\prime}}\) as the Y- axis.
Then A is the origin (0, 0). Let S = (a, 0), (a>0). Then z = (-a, 0) and the equation of the directrix is x + a = 0.

If P (x, y) is a point on the parabola and PM is the perpendicular distance from P to the directrix l, then \(\frac{S P}{S M}\) = e = 1
∴ (SP)² = (PM)²
⇒ (x – a)² + y² = (x + a)²
∴ y² = 4ax
Conversely if P(x, y) is any point such that y² = 4ax then
SP = \(\sqrt{(x-a)^2+y^2}\) = \(\sqrt{x^2+a^2-2 a x+4 a x}\) = \(\sqrt{(x+a)^2}\)
= |x + a| = PM
Hence P(x, y) is on the locus. In other words, a necessary and sufficient condition for the point P(x, y) to be on the parabola is that y² = 4ax.
Thus the equation of the Parabola is y² = 4ax.

Question 21.
Evaluate : ∫\(\sqrt{\frac{5-x}{x-2}}\) on (2, 5).
Solution:
I = \(\int \sqrt{\frac{5-x}{x-2}} d x\)

Question 22.
Obtain reduction formula I_n = ∫sin⁴ x.dx, n be a positive integer, n ≥ 2 and deduce the value of ∫sin⁴ x.dx.
Solution:

Question 23.
Find the area bounded between the curves y² = 4ax, x² = 4by (a > 0, b > 0).
Solution:
Given curve equations are y² = 4ax, x² = 4by (a > 0, b > 0).
⇒ \(\left(\frac{x^2}{4 b}\right)^2\) = 4ax
⇒ \(\frac{x^4}{166^2}\) = 4ax
⇒ x⁴ = 64 ab²x
⇒ x⁴ – 64ab²x = 0
⇒ x(x³ – 64ab²) = 0
⇒ x = 0 (or) x = 4a^1/3b^2/3

Question 24.
Find the equation of a curve whose gradients is \(\frac{d y}{d x}\) = \(\frac{y}{x}\) – cos²\(\frac{y}{x}\); where x > 0, y > 0 and which passes through the point
Solution: