# TS Inter 1st Year Maths 1B Question Paper May 2019

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## TS Inter 1st Year Maths 1B 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 Marks)

I. Very short answer type questions :

2. Each question carries two marks.

Question 1.
Find the equation of the straight line which makes an angle 60° with the positive x-axis measured counter-clockwise and passing through the point (1, 2).
Solution:
Slope of the line m = tan 60° = $$\sqrt{3}$$.
Required straight line equation is y-2=V3(x-1)
⇒ y – 2 = $$\sqrt{3}$$(x – 1),
⇒ y – 2 = $$\sqrt{3}$$x – $$\sqrt{3}$$
⇒ $$\sqrt{3}$$x – y + (2 – $$\sqrt{3}$$) = 0

Question 2.
Find the value of P, if the straight lines 3x + 7y – 1 = 0 and 7x – py + 3 = 0 are mutually perpendicular.
Solution:
Given straight line equations are
3x + 7y – 1 = 0 ……… (1)
7x – py + 3 = 0 ……. (2)
Since (1), (2) are mutually perpendicular.
∴ 3(7) + 7(-P) = 0
⇒ 21 – 7P = 0
⇒ 7P = 21
⇒ P = 3

Question 3.
Find the distance between the mid point of the line segment $$\overline{\mathrm{AB}}$$ and the point (3, -1, 2), where A = (6, 3, -4) and B = (-2, -1, 2).
Solution:
Given A = (6, 3, -4)
B = (-2, -1, 2)
Let P be the midpoint of the line segment $$\overline{\mathrm{AB}}$$

Question 4.
Find the angle between the planes x + 2y + 2z – 5 = 0 and 3x + 3y + 2z – 8 = 0.
Solution:
Given plane equations are
x + 2y + 2z – 5 = 0 ….. (1)
3x + 3y + 2z – 8 = 0 …… (2)
Let ‘θ’ be the angle between the planes (1) and (2)

Question 5.
Find the left hand limit and right hand limit of the function, f(x) = $$\frac{|x-2|}{x-2}$$ at x = 2.
Solution:
Given f(x) = $$\frac{|x-2|}{x-2}$$, x ≠ 2

Question 6.
Compute :

Solution:

Question 7.
Find the derivative of the function, f(x) = 5 sinx + ex Log x.
Solution:
Given f(x) = 5 sinx + ex logx.
Differentiating w.r.to. ‘x’ on both sides, we have
f'(x) = 5cosx + ex. $$\frac{1}{x}$$ + logx.ex
= 5 cosx + ex($$\frac{1}{x}$$ + logx).

Question 8.
If x = esin hy, find $$\frac{\mathrm{dy}}{\mathrm{dx}}$$.
Solution:
Given x = esin hy
Taking logarithm on both sides, we have
logx = sin hy
Differentiating w.r.to ‘x’ on both sides, we have
$$\frac{1}{x}$$ = cos hy . $$\frac{d y}{d x}$$
⇒ $$\frac{d y}{d x}$$ = $$\frac{1}{x \cosh y}$$

Question 9.
Find dy and ∆y of y = f(x) = x2 + x at x = 10. when ∆x = 0.1.
Solution:
Given f(x) = x2 + x, x = 10 and ∆x = 0.1
dy = f’(x) ∆x
= (2x + 1) ∆x
= [2(10) + 1] (0.1)
= (21) (0.1)
= 2.1
∆y = f(x + ∆x) – f(x)
= (x + ∆x)2 + (x + ∆x) – (x2 + x)
= x2 + 2x ∆x + (∆x)2 + x + ∆x – x2 – x
= 2x ∆x + (∆x)2 + ∆x
= 2(10) (0.1)+ (0.1)2 +(0.1)
= 2 + 0.01 + 0.1
= 2.11.

Question 10.
Verify Rolle’s theorem for the function y = f(x) = x2 – 1 on [-1, 1].
Solution:
Given y = f(x) = x2 – 1
since f is a second degree polynomial
∴ f is continuous on [-1, 1] and f is derivable on (-1, 1)
Also f (-1) = (-1 )2 – 1 = 1 – 1 = 0
f(1) = 12 – 1 = 1 – 1 = 0
∴ f(-1) = f(1)
∴ f statistics all the conditions of Rolle’s theorem.
∴ There exists c ∈ (-1, 1) such that f'(c) = 0.
f(x) = x2 + x
⇒ f'(x) = 2x + 1
⇒ f'(x) = 2x + 1
⇒ f'(c) = 2c + 1
⇒ 2c = – 1
⇒ c = $$\frac{-1}{2}$$ ∈ (-1, 1)
Hence Roll’s theorem is verified.

Section – B (5 × 4 = 20 Marks)

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

Question 11.
A(5, 3) and B(3, -2) are two fixed points. Find the equation of the locus of a P, so that the area of triangle PAB is 9.
Solution:
Given A = (5, 3)
B = (3, – 2)
Let P(x1, y1) be any point on the locus.
Given geometric condition is area of ∆PAB is 9 units,

Question 12.
When the axes are rotated through an angle $$\frac{\pi}{4}$$, find the transformed equation of 3x2 + 10xy + 3y2 = 9.
Solution:
Given equation is 3x2 + 10xy + 3y2 – 9 = 0

Question 13.
Transform the equation $$\frac{x}{a}+\frac{y}{b}$$ = 1 into the normal form when a > 0 and b > 0. If the perpendicular distance of the straight line from the origin is p, deduce that $$\frac{1}{p^2}$$ = $$\frac{1}{a^2}+\frac{1}{b^2}$$.
Solution:
Given $$\frac{x}{a}+\frac{y}{b}$$ = 1
⇒ $$\frac{b x+a y}{a b}$$ = 1

Question 14.
Check the continuity of the function f given by

at the point 3.
Solution:
Given f(x) = $$\frac{x^2-9}{x^2-2 x-3}$$, 0 < x < 5, x ≠ 3
= 1.5, x = 3

Question 15.
If y = tan-1$$\left(\frac{2 x}{1-x^2}\right)$$(| x |< 1), then find $$\frac{d y}{d x}$$.
Solution:
Given y = tan-1 $$\left(\frac{2 x}{1-x^2}\right)$$(|x| < 1)
Put x = tan θ
⇒ θ = tan-1x.
∴ y = tan-1$$\left[\frac{2 \tan \theta}{1-\tan ^2 \theta}\right]$$
⇒ y = tan-1 (tan 2θ)
⇒ y = 2θ
⇒ y = 2 tan-1x.
Differentiating with respect to ‘x’ on both sides, we have
$$\frac{d y}{d x}$$ = $$\text { 2. } \frac{1}{1+x^2}$$
$$\frac{d y}{d x}$$ = $$\frac{2}{1+x^2}$$

Question 16.
Find the equations of tangent and normal to the curve x = cost, y = sint, at t = $$\frac{d y}{d x}$$.
Solution:

Question 17.
A stone is dropped into a quiet lake and ripples move in circles at the speed of 5cm/sec. At the instant when the radius of circular ripple is 8cm, how fast is the enclosed area increases ?
Solution:
Let r, A be the radius and area of circular ripple.
Given $$\frac{\mathrm{dr}}{\mathrm{dt}}$$ = 5cm/sec and r = 8.
A = πr2
⇒ $$\frac{\mathrm{dA}}{\mathrm{dt}}$$ = $$\pi .2 \mathrm{r} . \frac{\mathrm{dr}}{\mathrm{dt}}$$
when r = 8
$$\frac{\mathrm{dA}}{\mathrm{dt}}$$ = π.2r.$$\frac{\mathrm{dr}}{\mathrm{dt}}$$
= 80 sq.cm/sec.

Section – C

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

Question 18.
Find the orthocentre of the triangle whose vertices are (-2, -1) (6, -1) and (2, 5).
Solution:
Let A = (-2, -1); B = (6, -1); C = (2, 5)

Question 19.
Show that the product of the perpendicular distances from the origin to the pair of straight lines represented by ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 is $$\frac{|c|}{\sqrt{(a-b)^2+4 h^2}}$$.
Solution:
Let S ≡ ax2 + 2hxy + by2 + 2gx + 2fy + c = 0
Let s = 0 represent pain of lines be
l1x + m1y + n1 = 0 …… (1)
l2x + m2y + n2 = 0 ……. (2)
∴ s ≡ (l1x + n1y + n1) (l2x + m2y + n2)
Comparing the co-efficients of like terms on both sides, we have
l1l2 = a; l1m1 = 2h
m1m2 = b; l1n2 + l2n1 = 2g
n1n2 = c; m1n2 + m2n1 = 2f
The perpendicular distance from origin to the line (1) = $$\frac{\left|\mathrm{n}_1\right|}{\sqrt{l_1^2+\mathrm{m}_1^2}}$$
The perpendicular distance from origin to the line (2) = $$\frac{\left|n_2\right|}{\sqrt{l_2^2+m_2^2}}$$
∴ The product of the perpendicular distances from origin to the pair of straigh lines represented by s = 0 is $$\frac{\left|n_1\right|}{\sqrt{l_1^2+m_1^2}} \cdot \frac{\left|n_2\right|}{\sqrt{l_2^2+m_2^2}}$$
= $$\frac{\left|n_1 n_2\right|}{\sqrt{\left(l_1^2+m_1^2\right)\left(l_2^2+m_2^2\right)}}$$

Question 20.
Find the condition for the fines joining the origin to the points of intersection of the circle x2 + y2 = a2 and the line lx + my = 1 to coincide.
Solution:
Equation of the circle is x2 + y2 = a2 ……… (1)
Equation of the line AB is lx + my = 1 ………. 2
Homogenising (1) with the help of (2)
∴ The combined equation of OA, OB is x2 + y2 = a2 . 12
= a2 (lx + my)2
= a2 (l2x2 + 2lmxy + m2y2)
= a2l2x2 + 2a2lmxy + a2m2y2
⇒ (a2l2 – 1) x2 + 2a2 lmxy + (a2m2 – 1) = 0
Since the lines OA, OB coincide.
∴ h2 = ab
⇒ (a2 lm)2 = (a2l2 – 1) (a2m2 – 1)
⇒ a4l2m2 = a4l2m2 – a2l2 – a2m2 + 1
⇒ 0 = – a2l2 – a2m2 + 1
⇒ a2l2 + a2m2 = 1
⇒ a2(l2 + m2) = 1.

Question 21.
Find the direction cosines of two lines which are connected by the relations l – 5m + 3n = 0 and 7l2 + 5m2 – 3n2 = 0.
Solution:
Given l – 5m + 3n = 0
⇒ l = 5m – 3n ……. (1)
7l2 + 5m2 – 3n2 = 0
⇒ 7(5m – 3n)2 + 5m2 – 3n2 = 0
⇒ 7(25m2 – 30mn + 9n2) + 5m2 – 3n2 = 0
⇒ 175m2 – 210mn + 63n2 + 5m2 – 3n2 = 0
⇒ 180m2 – 210mn + 60n2 = 0
⇒ 6m2 – 7mn + 2n2 = 0
⇒ 6m2 – 4mn – 3mn + 2n2 = 0
⇒ 2m(3m – 2n) – n (3m – 2n) = 0
⇒ (2m – n) (3m – 2n) = 0
⇒ 2m – n = 0 ………. (2)
3m – 2n = 0 ….. (3)
Solving (1) and (2)

Question 22.
Find the derivative of the function (sinx)logx + xsinx.
Solution:
Given y = (sin x)log x + xsin x.
Let u = (sinx)logx
Taking logarithms on bothsides, we have
log u = log x. log (sin x)
Differentiating w.r.to ‘x’ on bothsides, we have

Differentiating w.r.to x on bothsides, we have

Question 23.
Find the angle between the curves, y2 = 4x; x2 + y2 = 5.
Solution:
Given curve equations are
y2 = 4x ……… (1)
x2 + y2 = 5 ……… (2)
From (1) and (2)
x2 + 4x = 5 ⇒ x2 + 4x – 5 = 0.
⇒ x2 + 5x – x – 5 = 0
⇒ x (x + 5) – 1 (x + 5) = 0
⇒ (x – 1) (x + 5) = 0
⇒ x = 1, x = – 5
If x = 1 then y2 = 4
⇒ y = ± 2.
∴ The points of intersection of the curves (1) and (2) are (1, 2) and (1, -2).
y2 = 4x
Differentiating w.r.to ‘x’ on bothsides, we have

Differentiating w.r.to. ‘x’ on both sides, we have

Question 24.
A window is in the shape of a rectangle surrounded by a semicircle. If the perimeter of the window is 20 ft., find the maximum area.
Solution:
Let length of the rectangle be 2x and breadth by y and radius of the semi-circle in x.
∴ Perimeter = 20
⇒ y + 2x + y + πx = 20
⇒ 2x + 2y + πx = 20
⇒ 2y = 20 – 2x – πx
⇒ y = 10 – x – $$\frac{\pi}{2}$$x.

Window area = Area of the rectangle + Area of semi-circle