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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.

- Attempt all questions.
- Each question carries two marks.

Question 1.

If x^{2} + y^{2} – 4x + 6y + c = 0 represents a circle with radius 6, then find the value of C.

Solution:

Given circle equation is x^{2} + y^{2} – 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 x^{2} + y^{2} – 5x + 4y + k = 0 is \(\sqrt{37}\) then find k.

Solution:

Let S = x^{2} + y^{2} – 5x + 4y + k = 0

Given length of the tangent from (2, 5) to the circle S = 0 is \(\sqrt{37}\)

\(\sqrt{S_{11}}\) = \(\sqrt{37}\)

⇒ S_{11} = 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 x^{2} + y^{2} – 5x – 14y – 34 = 0,

x^{2} + y^{2} + 2x + 4y + k = 0 are orthogonal, then find the value of k.

Solution:

Given circle equations are

x^{2} + y^{2} – 5x – 14y – 34 = 0 ……. (1)

x^{2} + y^{2} + 2x + 4y + k = 0 ……. (2)

Since (1), (2) are orthogonal

∴ 2g_{1}g_{2} + 2f_{1}f_{2} = c_{1} + c_{2}

⇒ (-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 y^{2} = 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 y^{2} = 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„ e_{1}, 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 : ∫Sec^{2}x Cosec^{2} x dx, x ∈ I ⊂ R \

({nπ : n ∈ Z} ∪ {(2n + 1)\(\frac{\pi}{2}\):n∈Z})

Solution:

∫Sec^{2}x Cosec^{2} xdx = ∫\(\frac{1}{\cos ^2 x \sin ^2 x}\)dx

= ∫\(\frac{\sin ^2 x+\cos ^2 x}{\sin ^2 x \cos ^2 x}\)dx

= ∫\(\frac{1}{\cos ^2 x}\) + ∫\(\frac{1}{\sin ^2 x}\)

= ∫sec^{2}x dx + ∫cosec^{2}xdx

= Tanx – Cotx + c

∴ ∫Sec^{2}x Cosec^{2} 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^{x}Tanx dx + ∫e^{x}.log sec x dx

= ∫e^{x}Tanx dx + log sec x. e^{2} – ∫\(\frac{1}{\sec x}\).Secx Tanx. e^{x} dx

= ∫e^{x} Tanx dx + ∫e^{x}log 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 = x^{2}, 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.

- Attempt any five questions.
- 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 x^{2} + y^{2} – 2x – 2y – 1 =0.

Solution:

Given line equations are

x + y – 5 = 0 …….. (1)

2x + ky – 8 = 0 ………. (2)

Here l_{1} = 1, m_{1} = 1, n_{1} = – 5

l_{2} – 2, m_{2} = k, n_{2} = – 8

Given circle equation is S ≡ x^{2} + y^{2} – 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 x^{2} + y^{2} – 4x – 6y + 5 = 0, x^{2} + y^{2} – 2x – 4y – 1 = 0 and x^{2} + y^{2} – 6x – 2y = 0.

Solution:

Let S ≡ x^{2} + y^{2} – 4x – 6y + 5 = 0

S’ ≡ x^{2} + y^{2} – 2x – 4y – 1 = 0

S” ≡ x^{2} + y^{2} – 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 S_{1} = 0

Question 14.

Find the eccentricity, co-ordinates of foci, length of latus rectum and the equations of directrices of the ellipse 9x^{2} + 16y^{2} – 36x + 32y – 92 = 0.

Solution:

Given ellipse equation is 9x^{2} + 16y^{2} – 36x + 32y – 92 = 0

⇒ 9(x^{2} – 4x + 4) + 16 (y^{2} + 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 a^{2} = 16, b^{2} = 9

Also h = 2, k = -1

Question 15.

Find the equations of the tangents to the hyperbola 3x^{2} – 4y^{2} = 12 which are

(i) Parallel and

(ii) Perpendicular to the line y = x – 7.

Solution:

Given hyperbola equation is 3x^{2} – 4y^{2} = 12

⇒ \(\frac{x^2}{4}-\frac{y^2}{3}\) = 1

Here a^{2} = 4, b^{2} = 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 = Cos^{3}x. dx

Solution:

Given differential equation is

\(\frac{\mathrm{dy}}{\mathrm{dx}}\) + yTanx = Co^{3}x ……. (1)

Here P = Tan x and Q = Cos^{3}x.

I.F = \(\text { e } \int^{P d x}\) = e∫Tanx dx

= e^{log|secx|}

= Secx.

Section – C

(5 × 7 = 35)

III. Long Answer typecfoestions,

- Attempt any five questions.
- 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

⇒ x^{2} – 4x + 3 + y^{2} + 2y – 8 = 0

⇒ x^{2} + y^{2} – 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 (x^{2} + y^{2} – 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

(x^{2} + y^{2} – 4x + 2y – 5) – 6 (3x + y – 5) = 0

⇒ x^{2} + y^{2} – 4x + 2y – 5 – 18x – 6y + 30 = 0

⇒ x^{2} + y^{2} – 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 x^{2} + y^{2} – 6x – 9y + 13 = 0, x^{2} + y^{2} – 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 = |r_{1} – r_{2}|

∴ The circles touch each other internally. The point of contact ‘P’ divides AB

externally in the ratio r_{1} : r_{2} = \(\frac{\sqrt{65}}{2}\) : \(\sqrt{65}\)

= 1 : 2 Co-ordinates of P are

∴ Equation of the common tangent is

S_{1} – S_{2} = 0

-4x + 7y + 13 = 0

4x – 7y – 13 = 0

Question 20.

Derive the equation of a parabola in the standard form y^{2} = 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} + y^{2} = (x + a)^{2}

∴ y^{2} = 4ax.

Conversely if P (x, y) is any point such that y^{2} = 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 y^{2} = 4ax.

Thus the equation of the parabola is y^{2} = 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} = ∫sin^{n} x dx, then show that

for n ≥ 2 and deduce the value of ∫sin^{4}x x dx.

Solution:

I_{n} = ∫sin^{n} x dx

= ∫sin^{n-1}x 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-2}x cos^{2} x dx

= – sin^{n-1} xcosx + (n – 1) ∫sin^{n-2}x(1 – sin^{2}x)dx

= -sin^{n-1} xcosx + (n- 1) ∫sin^{n-2}xdx – (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 xy^{2}dy – (x^{3} + y^{3}) dx = 0.

Solution:

Given differential equation is

xy^{2}dy – (x^{3} + y^{3}) dx = 0

⇒ xy^{2}dy = (x^{3} + y^{3}) 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)