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 = aex + be-x, a and b are parameters.
Solution:
Give xy = aex + be-x
diffefentiating w.r.to ‘x’ on both sides, we have

Section-B

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

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 In = ∫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 = 4a1/3b2/3

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: