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
Given radius = 6
⇒ \(\sqrt{g^2+f^2-c}\) = 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)e^x dx on
((2n – \(\frac{1}{2}\))π, (2n + \(\frac{1}{2}\))π), n∈Z
Solution:
∫(Tanx + Log Secx) e^x dx = ∫e^xTanx dx + ∫e^x.log sec x dx
= ∫e^xTanx dx + log sec x. e² – ∫\(\frac{1}{\sec x}\).Secx Tanx. e^x dx
= ∫e^x Tanx dx + ∫e^xlog Secx – ∫e^x Tanx dx
= e^x log Secx + c
∴ ∫(Tanx + Log Secx) e^x dx = e^x 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

II. Short Answer Type questions.

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
= e^log|secx|
= Secx.

Section – C
(5 × 7 = 35)

III. Long Answer typecfoestions,

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 I_n = ∫sinⁿ x dx, then show that

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

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)