# TS Inter 2nd Year Maths 2B Question Paper May 2018

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## TS Inter 2nd Year Maths 2B Question Paper May 2018

Time : 3 Hours
Max. Marks : 75

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 – 4x + 6y + c = 0 represents a circle with radius 6, then find the value of C.
Solution:
Given circle equation is x2 + y2 – 4x + 6y + c = 0 .
Here 2g = -4 ⇒ g = -2
2f = 6 ⇒ f = 3 and c = c
⇒ $$\sqrt{g^2+f^2-c}$$ = 6
⇒ $$\sqrt{4+9-c}$$ = 6
⇒ 13 – c = 36
⇒ c = 13 – 36
⇒ c = -23.

Question 2.
If the length of the tangent from (2, 5) to the circle x2 + y2 – 5x + 4y + k = 0 is $$\sqrt{37}$$ then find k.
Solution:
Let S = x2 + y2 – 5x + 4y + k = 0
Given length of the tangent from (2, 5) to the circle S = 0 is $$\sqrt{37}$$
$$\sqrt{S_{11}}$$ = $$\sqrt{37}$$
⇒ S11 = 37
⇒ 4 + 25 – 5(2) + 4(5) + k = 37
⇒ 29 – 10 + 20 + k = 37
⇒ k = 37 – 39
⇒ k = -2

Question 3.
If pair of circles x2 + y2 – 5x – 14y – 34 = 0,
x2 + y2 + 2x + 4y + k = 0 are orthogonal, then find the value of k.
Solution:
Given circle equations are
x2 + y2 – 5x – 14y – 34 = 0 ……. (1)
x2 + y2 + 2x + 4y + k = 0 ……. (2)
Since (1), (2) are orthogonal
∴ 2g1g2 + 2f1f2 = c1 + c2
⇒ (-5) (1) + (-14) (2) = -34 + k
⇒ -5 – 28 = -34 + k
⇒ k = – 33 + 34
⇒ k = 1.

Question 4.
Find the value of k, if the line 2y = 5x + k is a tangent to the parabola y2 = 6x.
Solution:
Given line equation is 2y = 5x + k
⇒ y = $$\frac{5}{2}$$x + $$\frac{k}{2}$$ ….. (1)
Here m = $$\frac{5}{2}$$, c = $$\frac{k}{2}$$
Given parabola equation is y2 = 6x …….. (2)
Here 4a = 6 ⇒ a = $$\frac{6}{4}$$
⇒ a = $$\frac{3}{2}$$
If (1) is a tangent to the Parabola (2) then
c = $$\frac{a}{m}$$
⇒ $$\frac{k}{2}$$ = $$\frac{\frac{3}{2}}{\frac{5}{2}}$$
⇒ $$\frac{k}{2}$$ = $$\frac{3}{5}$$
⇒ k = $$\frac{6}{5}$$

Question 5.
If e„ e1, are the eccentricities of a hyperbola and its conjugate hyperbola, then prove that $$\frac{1}{\mathrm{e}^2}+\frac{1}{\mathrm{e}_1^2}$$ = 1.
Solution:

Question 6.
Evaluate : ∫Sec2x Cosec2 x dx, x ∈ I ⊂ R \
({nπ : n ∈ Z} ∪ {(2n + 1)$$\frac{\pi}{2}$$:n∈Z})
Solution:
∫Sec2x Cosec2 xdx = ∫$$\frac{1}{\cos ^2 x \sin ^2 x}$$dx
= ∫$$\frac{\sin ^2 x+\cos ^2 x}{\sin ^2 x \cos ^2 x}$$dx
= ∫$$\frac{1}{\cos ^2 x}$$ + ∫$$\frac{1}{\sin ^2 x}$$
= ∫sec2x dx + ∫cosec2xdx
= Tanx – Cotx + c
∴ ∫Sec2x Cosec2 x dx = Tanx – cotx + c

Question 7.
Evaluate ∫(Tanx + Log Sec x)ex dx on
((2n – $$\frac{1}{2}$$)π, (2n + $$\frac{1}{2}$$)π), n∈Z
Solution:
∫(Tanx + Log Secx) ex dx = ∫exTanx dx + ∫ex.log sec x dx
= ∫exTanx dx + log sec x. e2 – ∫$$\frac{1}{\sec x}$$.Secx Tanx. ex dx
= ∫ex Tanx dx + ∫exlog Secx – ∫ex Tanx dx
= ex log Secx + c
∴ ∫(Tanx + Log Secx) ex dx = ex log Secx + c

Question 8.
Evaluate $$\int_0^1\left(\frac{x^2}{x^2+1}\right) d x$$
Solution:

Question 9.
Find the area bounded by the parabola y = x2, the X-axis and the lines x = -1, x = 2.
Solution:

Question 10.
Find the order and degree of the differential equation
$$\frac{d^2 y}{d x^2}$$ = $$\left[1+\left(\frac{d y}{d x}\right)^2\right]^{5 / 3}$$
Solution:
Given differential equation is

Section – B

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

Question 11.
Find the value of k, if x + y – 5 = 0 and 2x + ky – 8 = 0 are conjugate with respect to the circle x2 + y2 – 2x – 2y – 1 =0.
Solution:
Given line equations are
x + y – 5 = 0 …….. (1)
2x + ky – 8 = 0 ………. (2)
Here l1 = 1, m1 = 1, n1 = – 5
l2 – 2, m2 = k, n2 = – 8
Given circle equation is S ≡ x2 + y2 – 2x – 2y – 1 = 0 ……. (3)
Here 2g = -2 ⇒ g = -1
2f = -2 ⇒ f = -1 and c = -1
Radius r = $$\sqrt{g^2+f^2-c}$$ = $$\sqrt{1+1-(-1)}$$ = $$\sqrt{3}$$
If (1), (2) are conjugate w.r. to the circle S = 0 then

Question 12.
Find the radical centre of the circles x2 + y2 – 4x – 6y + 5 = 0, x2 + y2 – 2x – 4y – 1 = 0 and x2 + y2 – 6x – 2y = 0.
Solution:
Let S ≡ x2 + y2 – 4x – 6y + 5 = 0
S’ ≡ x2 + y2 – 2x – 4y – 1 = 0
S” ≡ x2 + y2 – 6x – 2y = 0
The radical axis of S = 0, S’ = 0 is S – S’ = 0
⇒ -2x – 2y + 6 = 0
⇒ x + y – 3 = 0 ………. (1)
The radical axis or S = 0, S” = 0 is S – S” = 0
⇒ 2x – 4y + 5 = 0 ……… (2)
Solving (1) and (2)

Question 13.
If a tangent to the ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}$$ = 1 (a > b) meets its major axis and minor axis at M and N respectively, then prove that $$\frac{a^2}{(C M)^2}+\frac{b^2}{(C N)^2}$$ = 1 where C is the centre of the ellipse.
Solution:
Given ellipse equation is $$\frac{x^2}{a^2}$$ + $$\frac{y^2}{b^2}$$ = 1 (a > b)
Let P(θ) (a cosθ, b sinθ) is any point on the ellipse.
Equation of the tangent at P(θ) is S1 = 0

Question 14.
Find the eccentricity, co-ordinates of foci, length of latus rectum and the equations of directrices of the ellipse 9x2 + 16y2 – 36x + 32y – 92 = 0.
Solution:
Given ellipse equation is 9x2 + 16y2 – 36x + 32y – 92 = 0
⇒ 9(x2 – 4x + 4) + 16 (y2 + 2y + 1) = 92 + 36 + 16
⇒ 9(x – 2)2 + 16 (y + 1)2 = 144
⇒ $$\frac{(x-2)^2}{16}$$ + $$\frac{(y+1)^2}{9}$$ = 1
Here a2 = 16, b2 = 9
Also h = 2, k = -1

Question 15.
Find the equations of the tangents to the hyperbola 3x2 – 4y2 = 12 which are
(i) Parallel and
(ii) Perpendicular to the line y = x – 7.
Solution:
Given hyperbola equation is 3x2 – 4y2 = 12
⇒ $$\frac{x^2}{4}-\frac{y^2}{3}$$ = 1
Here a2 = 4, b2 = 3

i) The tangent is parallel to y = x – 7
∴ m = slope of the tangent = 1
Equation of the parallel tangents are
y = mx ± $$\sqrt{a^2 m^2-b^2}$$
⇒ y = x± $$\sqrt{4-3}$$
⇒ y = x ± 1
⇒ y = x + 1, y = x – 1
⇒ x – y + 1 = 0, x – y – 1 = 0

ii) The tangents perpendicular to y = x – 7
∴ m = slope of the tangent = -1
∴ Equation of the perpendicular tangents are
y = (-1)x ± $$\sqrt{4-3}$$
y = -x ± 1
⇒ y = -x + 1, y = -x – 1
⇒ x + y – 1 = 0, x + y + 1 = 0

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

Question 17.
Solve $$\frac{d y}{d x}$$ + yTanx = Cos3x. dx
Solution:
Given differential equation is
$$\frac{\mathrm{dy}}{\mathrm{dx}}$$ + yTanx = Co3x ……. (1)
Here P = Tan x and Q = Cos3x.
I.F = $$\text { e } \int^{P d x}$$ = e∫Tanx dx
= elog|secx|
= Secx.

Section – C
(5 × 7 = 35)

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

Question 18.
Find the equation of a circle passing through the points (1, 2), (3, -A), (5, -6).
Solution:
Let A = (1, 2), B = (3, -4), C = (5, -6) and D = (19, 8)
The equation of the circle which having $$\overline{\mathrm{AB}}$$ as a diameter is
(x – 1) (x – 3) + (y – 2) (y + 4) = 0
⇒ x2 – 4x + 3 + y2 + 2y – 8 = 0
⇒ x2 + y2 – 4x + 2y – 5 = 0 …….. (1)
Equation of $$\overline{\mathrm{AB}}$$ is
y – 2 = $$\frac{-4-2}{3-1}$$(x – 1)
y – 2 = -3(x – 1)
y – 2 = -3x + 3
3x + y – 5 = 0
The equation of circle passing through A, B is (x2 + y2 – 4x + 2y – 5) + λ(3x + y – 5) = 0 If this circle through C (5, – 6) then
(25 + 36 – 20 – 12 – 5) + λ(15 – 6 – 5) = 0
⇒ 24 + 4λ = 0
⇒ 6 + λ = 0
⇒ λ = – 6
Equation to the circle through A, B, C is
(x2 + y2 – 4x + 2y – 5) – 6 (3x + y – 5) = 0
⇒ x2 + y2 – 4x + 2y – 5 – 18x – 6y + 30 = 0
⇒ x2 + y2 – 22x – 4y + 25 = 0 …….. (2)
Substituting D (19, 8)
361 + 64 – 22(19) – 4(8) + 25 = 361 + 64 – 418 – 32 + 2
= 450 – 450 = 0
∴ D lies on (2)
Hence A, B, C, D are concyclic.

Question 19.
Show that x2 + y2 – 6x – 9y + 13 = 0, x2 + y2 – 2x – 16y = 0 touch each other. Find the point of contact and the equation of common tangent at their point of contact.
Solution:
Equations of the circles are

AB = $$\sqrt{(3-1)^2+\left(\frac{9}{2}-8\right)^2}$$ = $$\sqrt{4+\frac{49}{4}}$$ = $$\frac{\sqrt{65}}{2}$$
AB = |r1 – r2|
∴ The circles touch each other internally. The point of contact ‘P’ divides AB
externally in the ratio r1 : r2 = $$\frac{\sqrt{65}}{2}$$ : $$\sqrt{65}$$
= 1 : 2 Co-ordinates of P are

∴ Equation of the common tangent is
S1 – S2 = 0
-4x + 7y + 13 = 0
4x – 7y – 13 = 0

Question 20.
Derive the equation of a parabola in the standard form y2 = 4ax with a diagram.
Solution:
To study the nature of the curve, we prefer its equation in the simplest possible form we proceed us follows to derive such an equation.

Let S be the focus, I be the directrix as shown in fig. Let Z be the projection of ‘S’ on I and ‘A’ be the midpoint of SZ. A lies on the parabola because SA = AZ. A is called the vertex of the parabola. Let YAY be the straight line through A and parallel to the directrix. Now take ZX as the -axis and YY 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 $$\frac{S P}{P M}$$ = 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 ∫(6x + 5) ($$\sqrt{6-2 x^2+x}$$) dx.
Solution:
Let 6x + 5 = A (1 – 4x) + B
Equating the co-efficients of x
6 = -4 A ⇒ A = $$\frac{-3}{2}$$
Equating the constants
A + B = 5
B = 5 – A = 5 + $$\frac{3}{2}$$ = $$\frac{13}{2}$$

Question 22.
If In = ∫sinn x dx, then show that

for n ≥ 2 and deduce the value of ∫sin4x x dx.
Solution:
In = ∫sinn x dx
= ∫sinn-1x sin xdx
= sinn-1 x (-cos x) – ∫(n – 1) sin(n-2) x cos x (-cos x) dx
= -sinn-1 xcosx + (n- 1) ∫sinn-2x cos2 x dx
= – sinn-1 xcosx + (n – 1) ∫sinn-2x(1 – sin2x)dx
= -sinn-1 xcosx + (n- 1) ∫sinn-2xdx – (n – 1) ∫sin,sup>n xdx
= -sinn-1 xcosx + (n – 1) In-2 – (n – 1)In

Question 23.
Evaluate $$\int_0^\pi\left(\frac{x \sin ^3 x}{1+\cos ^2 x}\right)$$dx
Solution:

Question 24.
Solve xy2dy – (x3 + y3) dx = 0.
Solution:
Given differential equation is
xy2dy – (x3 + y3) dx = 0
⇒ xy2dy = (x3 + y3) dx
⇒ $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = $$\frac{x^3+y^3}{x y^2}$$ ……… (1)
This is a homogeneous differential equation
Put y = Vx
$$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = V + x$$\frac{\mathrm{dv}}{\mathrm{dx}}$$
Substituting these values in (1)