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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 :

- Answer all the questions.
- 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 + e^{x} Log x.

Solution:

Given f(x) = 5 sinx + e^{x} logx.

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

f'(x) = 5cosx + e^{x}. \(\frac{1}{x}\) + logx.e^{x}

= 5 cosx + e^{x}(\(\frac{1}{x}\) + logx).

Question 8.

If x = e^{sin hy}, find \(\frac{\mathrm{dy}}{\mathrm{dx}}\).

Solution:

Given x = e^{sin 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) = x^{2} + x at x = 10. when ∆x = 0.1.

Solution:

Given f(x) = x^{2} + 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) – (x^{2} + x)

= x^{2} + 2x ∆x + (∆x)^{2} + x + ∆x – x^{2} – 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) = x^{2} – 1 on [-1, 1].

Solution:

Given y = f(x) = x^{2} – 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) = 1^{2} – 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) = x^{2} + 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)

II. Short Answer Type Questions.

- Attempt any five questions.
- 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(x_{1}, y_{1}) 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 3x^{2} + 10xy + 3y^{2} = 9.

Solution:

Given equation is 3x^{2} + 10xy + 3y^{2} – 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^{-1}x.

∴ y = tan^{-1}\(\left[\frac{2 \tan \theta}{1-\tan ^2 \theta}\right]\)

⇒ y = tan^{-1} (tan 2θ)

⇒ y = 2θ

⇒ y = 2 tan^{-1}x.

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 = πr^{2}

⇒ \(\frac{\mathrm{dA}}{\mathrm{dt}}\) = \(\pi .2 \mathrm{r} . \frac{\mathrm{dr}}{\mathrm{dt}}\)

when r = 8

\(\frac{\mathrm{dA}}{\mathrm{dt}}\) = π.2r.\(\frac{\mathrm{dr}}{\mathrm{dt}}\)

= 80 sq.cm/sec.

Section – C

III. Long Answer Type Questions.

- Attempt any five questions.
- 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 ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 is \(\frac{|c|}{\sqrt{(a-b)^2+4 h^2}}\).

Solution:

Let S ≡ ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0

Let s = 0 represent pain of lines be

l_{1}x + m_{1}y + n_{1} = 0 …… (1)

l_{2}x + m_{2}y + n_{2} = 0 ……. (2)

∴ s ≡ (l_{1}x + n_{1}y + n_{1}) (l_{2}x + m_{2}y + n_{2})

Comparing the co-efficients of like terms on both sides, we have

l_{1}l_{2} = a; l_{1}m_{1} = 2h

m_{1}m_{2} = b; l_{1}n_{2} + l_{2}n_{1} = 2g

n_{1}n_{2} = c; m_{1}n_{2} + m_{2}n_{1} = 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 x^{2} + y^{2} = a^{2} and the line lx + my = 1 to coincide.

Solution:

Equation of the circle is x^{2} + y^{2} = a^{2} ……… (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 x^{2} + y^{2} = a^{2} . 1^{2}

= a^{2} (lx + my)^{2}

= a^{2} (l^{2}x^{2} + 2lmxy + m^{2}y^{2})

= a^{2}l^{2}x^{2} + 2a^{2}lmxy + a^{2}m^{2}y^{2}

⇒ (a^{2}l^{2} – 1) x^{2} + 2a^{2} lmxy + (a^{2}m^{2} – 1) = 0

Since the lines OA, OB coincide.

∴ h^{2} = ab

⇒ (a^{2} lm)^{2} = (a^{2}l^{2} – 1) (a^{2}m^{2} – 1)

⇒ a^{4}l^{2}m^{2} = a^{4}l^{2}m^{2} – a^{2}l^{2} – a^{2}m^{2} + 1

⇒ 0 = – a^{2}l^{2} – a^{2}m^{2} + 1

⇒ a^{2}l^{2} + a^{2}m^{2} = 1

⇒ a^{2}(l^{2} + m^{2}) = 1.

Question 21.

Find the direction cosines of two lines which are connected by the relations l – 5m + 3n = 0 and 7l^{2} + 5m^{2} – 3n^{2} = 0.

Solution:

Given l – 5m + 3n = 0

⇒ l = 5m – 3n ……. (1)

7l^{2} + 5m^{2} – 3n^{2} = 0

⇒ 7(5m – 3n)^{2} + 5m^{2} – 3n^{2} = 0

⇒ 7(25m^{2} – 30mn + 9n^{2}) + 5m^{2} – 3n^{2} = 0

⇒ 175m^{2} – 210mn + 63n^{2} + 5m^{2} – 3n^{2} = 0

⇒ 180m^{2} – 210mn + 60n^{2} = 0

⇒ 6m^{2} – 7mn + 2n^{2} = 0

⇒ 6m^{2} – 4mn – 3mn + 2n^{2} = 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} + x^{sinx}.

Solution:

Given y = (sin x)^{log x} + x^{sin 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, y^{2} = 4x; x^{2} + y^{2} = 5.

Solution:

Given curve equations are

y^{2} = 4x ……… (1)

x^{2} + y^{2} = 5 ……… (2)

From (1) and (2)

x^{2} + 4x = 5 ⇒ x^{2} + 4x – 5 = 0.

⇒ x^{2} + 5x – x – 5 = 0

⇒ x (x + 5) – 1 (x + 5) = 0

⇒ (x – 1) (x + 5) = 0

⇒ x = 1, x = – 5

If x = 1 then y^{2} = 4

⇒ y = ± 2.

∴ The points of intersection of the curves (1) and (2) are (1, 2) and (1, -2).

y^{2} = 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