TS Inter 2nd Year Maths 2B Question Paper March 2018

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TS Inter 2nd Year Maths 2B Question Paper March 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.
Find the equation of the circle whose centre is (-1, 2) and which passes through (5, 6).
Solution:
Let C = (-1, 2) and P = (5, 6)
r = CP = \(\sqrt{(5+1)^2+(6-2)^2}\)
= \(\sqrt{36+16}\)
= \(\sqrt{52}\)
∴ The equation of the circle whose centre is (-1, 2) and which passes through (5, 6) is
(x – h)² + (y – k)² = r²
⇒ (x + 1)² + (y – 2)² = \((\sqrt{52})^2\)
⇒ x² + 2x + 1 + y² – 4y + 4 = 52
⇒ x² + y² + 2x – 4y – 47 = 0

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.
Length or the tangent = \(\sqrt{511}\)
⇒ \(\sqrt{37}\) = \(\sqrt{(2)^2+(5)^2-5(2)+4(5)+k}\)
⇒ \(\sqrt{37}\) = \(\sqrt{4+25-10+20+k}\)
⇒ 37 = k + 39
⇒ k = + 37 – 39
⇒ k = -2

Question 3.
If the angle between the circles x² + y² – 12x – 6y + 41 = 0 and x² + y² + kx + 6y – 59 = 0 is 45°, find k.
Solution:
Given circle equations are
x² + y² – 12x – 6y + 41 = 0 ………… (1)
x² + y² + kx + 6y – 59 = 0 ………. (2)
Here 2g = -12 ⇒ g= -6
2f = -6 ⇒ f = -3 and c = 41
2g¹ = k ⇒ g = -6
2f = -6 ⇒ f = -3 and c¹ = -59.

Question 4.
Find the equation of the parabola whose vertex is (3, -2) and focus is (3, 1).
Solution:
Given vertex = (3, -2)
Focus = (3, 1)
Here the abcissae of the vertex and focus are equal to 3.
∴ The axis of the parabola is x = 3, a line parallel to y – axis and focus is above the vertex.
a = distance between vertex and focus
= \(\sqrt{(3-3)^2+(1+2)^2}\)
= 3
Hence equation of the parabola is
(x – h)² = 4a (y k)
⇒ (x – 3)² = 4.3 (y + 2)
⇒ (x – 3)² = 12(y + 2).

Question 5.
If 3x – 4y + k = 0 is a tangent to x² – 4y² = 5, find the value of k.
Solution:
Given line equation is 3x – 4y + k = 0
⇒ 4y = 3x + k
y = \(\frac{3}{4}\)x + \(\frac{k}{4}\) ……… (1)
Hence m = \(\frac{3}{4}\), c = \(\frac{k}{4}\)
Given Hyperbola equation is x² – 4y² = 5
⇒ \(\frac{x^2}{5}\) – \(\frac{y^2}{5 / 4}\) = 1
Here a² = 5, b² = \(\frac{5}{4}\)
If (1) is a tangent to the Hyperbola x² – 4y² = 5 then
c² = a²m² – b²
⇒ \(\frac{k^2}{16}\) = 5.\(\frac{9}{16}\) – \(\frac{5}{4}\)
⇒ \(\frac{k^2}{16}\) = \(\frac{45-20}{16}\)
⇒ k² = 25
⇒ k = ±5

Question 6.
Evaluate : \(\int \frac{e x}{e^x+1}\) dx
Solution:
I = \(\int \frac{e x}{e^x+1}\)dx
Let e^x + 1 = t
e^x dx = dt
= \(\int \frac{1}{t}\)dt
= log |t| + c
= log |e^x + 1| + c
∴ \(\int \frac{e x}{e^x+1}\)dx = log|e^x + 1| + c

Question 7.
Evaluate : ∫cos³x sin x dx
Solution:
I = ∫cos³ x sinx dx
Let cosx = t
-sinx dx = dt
sinx dx = – dt
= ∫t³(-dt)

Question 8.
Evaluate : \(\int_0^2\)|1 – x| dx
Solution:

Question 9.
Evaluate : \(\int_0^{\pi / 2}\)xsin x dx
Solution:

Question 10.
Find the general solution of \(\frac{d y}{d x}\) = e^x+y
Solution:
Given differential equation is
\(\frac{d y}{d x}\) = e^x+y …….. (1)
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = e^x . e^y
⇒ \(\frac{d y}{e^y}\) = e^x d^x
⇒ e^xdx – e^-ydy = 0
Integrating
∫e^xdx – ∫e^-ydy = c
⇒ e^x – \(\frac{e^{-y}}{-1}\) = c
⇒ e^x + e^-y = c
∴ The general solution of (1) is
e^x + e^-y = c

Section – B

II. Short Answer Type questions.

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

Question 11.
Find the area of the triangle formed by the normal at (3, -4) to the circle
x² + y² – 22x – 4y + 25 = 0 with the co-ordinate axes.
Solution:
Given circle equation is
s ≡ x² + y² – 22x – 4y + 25 = 0
Here 2g = – 22 ⇒ g = – 11
2f = – 4 ⇒ f = – 2
(x₁, y₁ = (3, -2)
The equation of the normal at (3, -2) or
the circle s = 0 is (x – x₁) (y₁ + f) – (y – y₁) (x₁ + g) = 0
⇒ (x – 3) (-4 – 2) – (y + 4) (3 – 11) = 0
⇒ (x – 3) (- 6) – (y + 4) (- 8) = 0
– 6x + 18 + 8y + 32 = 0
⇒ 6x – 8y – 50 = 0
⇒ 3x – 4y – 25 = 0
The area of the triangle formed by the normal at (3, -2) to the circle s = 0 with co-ordinate axis is \(\frac{1}{2} \cdot \frac{c^2}{|a b|}\) sq. units
= \(\frac{1}{2} \frac{(-25)^2}{|3.4|}\)
= \(\frac{625}{24}\) sq.units.

Question 12.
Find the equation and length of the common chord of the two circles:
x² + y² + 3x + 5y + 4 = 0 and
x² + y² + 5x + 3y + 4 = 0.
Solution:
Let s ≡ x² + y² + 3x + 5y + 4 = 0
s¹ ≡ x² + y² + 5x + 3y + 4 = 0
The radical axis of the circles s = 0, s¹ = 0 is s – s¹ = 0
⇒ – 2x + 2y = 0
⇒ x – y = 0 ………….. (1)

d = The perpendicular distance from c to the line (1)
= \(\frac{\left|\frac{-3}{2}+\frac{5}{2}\right|}{\sqrt{1+1}}\)
= \(\frac{1}{\sqrt{2}}\)
The length of the common chord of the two circles s = 0, s¹ = 0 is 2\(\sqrt{r^2-d^2}\)
= 2\(\sqrt{\frac{18}{4}-\frac{1}{2}}\)
= 2\(\sqrt{\frac{18-2}{4}}\)
= 2.2
= 4 units.

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:
Let \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1 be the required ellipse equation.
Given length of the latus rectum = 4
⇒ \(\frac{2 b^2}{a}\) = 4
⇒ 2b² = 4a
⇒ b² = 2a
Given distance between foci is 4\(\sqrt{2}\)
⇒ 2ae = 4\(\sqrt{2}\)
⇒ ae = 2\(\sqrt{2}\)
we know b² = a²(1 – e²)
⇒ 2a = a² – a²e²
⇒ 2a = a² – (2\(\sqrt{2}\))²
⇒ 2a = a² – 8
⇒ a² – 2a – 8 = 0
⇒ a² – 4a + 2a – 8 = 0
⇒ a(a – 4) + 2(a – 4) = 0
⇒ (a – 4) (a + 2) = 0
sine a > 0
∴ a = 4
b² = 2a = 2.4 = 8
Required ellipse equation is
\(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1
⇒ \(\frac{x^2}{16}\) + \(\frac{y^2}{8}\) = 1
⇒ x² + 2y² = 16

Question 14.
Find the eccentricity, length of latus rectum, foci 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 α = 2, β = -1
Also a² = 16, b² = 9

Question 15.
Show that angle between the two asymptotes of a hyperbola
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1 is 2 tan^-1 (b/a) (or) 2 sec^-1(e).
Solution:
Given Hyperbola equation is \(\frac{x^2}{a^2}\) – \(\frac{y^2}{b^2}\) = 1
Equations of the asymptotes are
\(\frac{x}{a}\) + \(\frac{y}{b}\) = 0 and \(\frac{x}{a}\) – \(\frac{y}{b}\) = 0
If 2θ is the angle between the asymptotes
then tanθ = \(\frac{b}{a}\) = slope of the asymptotes
∴ θ = tan^-1\(\left(\frac{b}{a}\right)\)
∴ Angle between the asymptotes = 2θ
= 2tan^-1\(\left(\frac{b}{a}\right)\)
we know that sec²θ = 1 + tan²θ
= 1 + \(\frac{b^2}{a^2}\)
= \(\frac{a^2+b^2}{a^2}\)
= e²
⇒ secθ = e
⇒ θ = sec^-1(e).
∴ Angle between the asymptotes = 2 tan^-1\(\left(\frac{b}{a}\right)\) or sec^-1(e).

Question 16.
Find the area bounded between the curves y = x², y = \(\sqrt{x}\).
Solution:
Given curve equations are y = x² ………. (1)
y = \(\sqrt{x}\) ……. (2)
From (1) & (2)

Question 17.
Solve : \(\frac{d y}{d x}\) + 1 = e^x+y.
Solution:
Given differential equation is
\(\frac{d y}{d x}\) + 1 = e^x+y …… (1)
Let x + y = t
then 1 + \(\frac{d y}{d x}\) = \(\frac{d t}{d x}\)
From (1)
\(\frac{\mathrm{dt}}{\mathrm{dx}}\) = e^t
⇒ \(\frac{d t}{e^t}\) = dx
⇒ dx – e^-tdt = 0
Integrating
∫dx – ∫e^-t dt = c
⇒ x – \(\frac{\mathrm{e}^{-\mathrm{t}}}{-1}\) = c
⇒ x + e^-t = c
⇒ x + e^-(x+y) = c
∴ The general solution of (1) is
x + e^-(x+y) = c.

Section – C

II. Long Answer Type questions.

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

Question 18.
Find the equation of a circle which passes through (4, 1), (6, 5) and having the centre on :
4x + 3y – 24 = 0.
Solution:
Let the equation of the required circle
x² + y² + 2gx + 2fy + c = 0 …… (1)
If (1) Passes through (4, 1) then
16 + 1 + 8g + 2f + c = 0
8g + 2f + c = -17 …… (2)
If (1) passes through (6, 5) then
36 + 25 + 12g + 10f + c = 0
12g + 10f + c = -61 ……… (3)
Since centre (-g, -f) lies on 4x + 3y – 24 = 0
⇒ 4 (-g) + 3 (-f) – 24 = 0
⇒ – 4g – 3f – 24 = 0
⇒ 4g + 3f + 24 = 0 …….. (4)
(3) – (2) ⇒ 4g + 8f = – 44
⇒ -4g + 8f + 44 = 0 …….. (5)
(5) – (4) ⇒ 5f + 20 = 0
⇒ f + 4 = 0
⇒ f = -4
from (5)
4g + 8(-4) + 44 = 0
⇒ 4g + 12 = 0
⇒ g + 3 = 0
⇒ g = -3
from (2)
8(-3) + 2(-4) + c = – 17
-24 – 8 + c = -17
⇒ c = 15
∴ Required circle equation is
x² + y² + 2(-3) x + 2(-4)y + 15 = 0
⇒ x² + y² – 6x – 8y + 15 = 0

Question 19.
Show that the circles:
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:
Given circle equations are
s ≡ x² + y² – 6x – 9y + 13 = 0
s¹ ≡ x² + y² – 2x – 16y = 0
centres A = (3, \(\frac{9}{2}\)) and B = (1, 8)

AB = |r₁ – r₂|
∴ The circles touch each other internally
The point of contact p divides AB externally in the ratio

Equation of the common tangent is s – s¹ = 0
⇒ -4x + 7y + 13 = 0
⇒ 4x – 7y – 13 = 0

Question 20.
Derive the equation of parabola in the standard form, that is y² = 4ax.
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: \(\int_a^b \sqrt{(x-a)(b-x)}\)dx
Solution:

Question 22.
Evaluate : \(\int \frac{d x}{(x+1) \sqrt{2 x^2+3 x+1}}\)
Solution:

Question 23.
Evaluate : \(\int \frac{d x}{4 \cos x+3 \sin x}\)
Solution:

Question 24.
Solve : (1 + y²)dx = (tan^-1y – x) dy.
Solution:
Given differential equation is

The general solution of (1) ¡s