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## TS Inter 1st Year Maths 1B Question Paper March 2017

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 value of ‘y’, if the line joining the points (3, y) and (2, 7) is parallel to the line joining the points (-1, 4), (0, 6).

Solution:

Let A = (3, 4)

B = (2, 7)

C = (-1, 4)

D = (0, 6)

Slope of \(\overline{\mathrm{AB}}\) = \(\frac{7-y}{2-3}\) = \(\frac{7-y}{-1}\) = y – 7

Slope of \(\overline{\mathrm{CD}}\) = \(\frac{6-4}{0-(-1)}\) = \(\frac{2}{1}\) = 2

Since \(\overline{\mathrm{AB}}\) is parallel to \(\overline{\mathrm{CD}}\)

∴ y – 7 = 2

⇒ y = 9.

Question 2.

Find the value of ‘p’, if the straight lines x + p = 0, y + 2 = 0 and 3x + 2y + 5 = 0 are concurrent.

Solution:

Given straight line equations are

x + p = 0 ⇒ x = -p ………. (1)

y + 2 = 0 ⇒ y = -2 ………. (2)

3x + 2y + 5 = 0 …….. (3)

The’point of intersection of (1) and (2) is (-p, -2)

since (1), (2), (3) are concurrent

∴ (-p, -2) lies on (3)

∴ 3(-p) + 2(-2) + 5 = 0

⇒ -3p – 4 + 5 = 0

⇒ 3p = 1 ⇒ p = \(\frac{1}{3}\).

Question 3.

Find the fourth vertex of the parallelogram whose consecutive vertices are (2, 4, -1), (3, 6, -1) and (4, 5, 1).

Solution:

Let A = (2, 4, -1)

B = (3, 6, -1)

C = (4, 5, 1)

D = (x, y, z)

Since ABCD is a parallelogram.

∴ Mid point of AC = Mid point of BD

Question 4.

Find the angle between the planes x + 2y + 2z – 5 = 0 and 3x + 3y + 2x – 8 = 0.

Solution:

Given planes equations are x + 2y + 2z – 5 = 0 ……. (1)

3x + 3y + 2z – 8 = 0 …… (2)

Let ‘θ’ be the acute angle between (1) and (2)

Question 5.

Compute \(\lim _{x \rightarrow 0} x^2 \sin \left(\frac{1}{x}\right)\)

Solution:

Question 6.

Compute \(\lim _{x \rightarrow \infty} \frac{8|x|+3 x}{3|x|-2 x}\)

Solution:

Question 7.

If f(x) = \(7^{x^3+3 x}\) (x > 0), then find f'(x).

Solution:

Question 8.

If x = tan(e^{-y}), then show that \(\frac{d y}{d x}\) = \(\frac{-e^y}{1+x^2}\).

Solution:

Given x = tan(e^{-y})

⇒ e^{-y} = tan^{-1} x

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

Question 9.

Find dy and ∆y of y = x^{2} + x at x = 10 when ∆x = 0.1.

Solution:

Let y = f(x) = x^{2} + x

Given 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(10)\(\left(\frac{1}{10}\right)\) + 0.01 + 0.1

Question 10.

Verify Rolle’s theorem for the function f: [-3, 8] → IR be defined by f(x) = x^{2} – 5x + 6.

Solution:

Given f(x) = x^{2} – 5x + 6

Since f is a second degree polynomial

∴ f is continuous on [-3, 8] and f is derivable on (-3, 8)

Also f(-3) = (-3)^{2} – 5(-3) + 6

= 9 + 15 + 6 = 30

f(8) = 8^{2} – 5(8) + 6

= 64 – 40 + 6 = 30

∴ f(- 3) = f(8)

∴ f satisfies all the conditions of Rolle’s theorem.

∴ There exists c ∈ (-3, 8) such that f(c) = 0

f(x) = x^{2} – 5x + 6

⇒ f'(x) = 2x – 5

⇒ f'(c) = 2c – 5

⇒ 0 = 2c – 5

⇒ 2c = 5

⇒ c = \(\frac{5}{2}\) ∈ (-3, 8)

Hence Rolle’s theorem is verified.

Section – B

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 locus of P, so that the area of ∆RAB is 9 sq. units.

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 sq. 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

Here θ = \(\frac{\pi}{4}\) = 45°

Question 13.

x – 3y – 5 = 0 is the perpendicular bisector of the line segment joining the points A, B. If A = (-1, -3), find the coordinates of ‘B’.

Solution:

Given A = (-1, -3)

Let B = (a, b) and L = x – 3y – 5 = 0

OR

Given A = (-1, -3)

Let B = (a, b) and L = x – 3y – 5 = 0

Question 14.

Show that

where a and b are real constants, is continuous at x = 0.

Solution:

Question 15.

If ay^{4} = (x + b)^{5} then 5yy” = (y’)^{2}.

Solution:

Given ay^{4} = (x + b)^{5} ……… (1)

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

a.4y^{3} = 5(x + b)^{4} …… (2)

Question 16.

Find the lengths of subtangent, subnormal at a point t on the curve x = a(cos t + t sin t), y = a(sin t – t cos t).

Solution:

Given curve equations are

x = a(cos t + t sin t) ………… (1)

y = a(sin t – t cos t) ………. (2)

differentiate (1) w.r.to ‘x’ on both sides, we have

Question 17.

The volume of a cube is increasing at a rate of 9 cubic centimetres per second. How fast is the surface area increasing when the length of the edge is 10 centimetres ?

Solution:

Let x, v, s be the length of the edge, volume and surface area or a cube respectively.

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 (5, -2), (-1,2) and (1, 4).

Solution:

Let A = (5, -2)

B = (-1, 2)

C = (1, 4)

Let \(\overline{\mathrm{AD}}\) be the perpendicular drawn from A to \(\overline{\mathrm{BC}}\) and \(\overline{\mathrm{BE}}\) be the perpendicular drawn from B to \(\overline{\mathrm{AC}}\).

Question 19.

Show that the area of the triangle formed by the lines ax^{2} + 2hxy + by^{2} = 0 and the line lx + my + n = 0 is \(\left|\frac{n^2 \sqrt{h^2-a b}}{a m^2-2 h l m+b l^2}\right|\).

Solution:

Let \(\overleftrightarrow{\mathrm{OA}}\) and \(\overleftrightarrow{\mathrm{OB}}\) be the pair of straight lines represented by the equation

ax^{2} + 2hxy + by^{2} = 0 (see figure)

and \(\overleftrightarrow{\mathrm{AB}}\) be the line lx + my + n = 0

Let ax^{2} + 2hxy + by^{2}

≡ (l_{1}x + m_{1}y) (l_{2}x + m_{2}y), and \(\overleftrightarrow{\mathrm{OA}}\) and \(\overleftrightarrow{\mathrm{OB}}\) be the lines.

l_{1}x + m_{1}y = 0 and

l_{2}x + m_{2}y = 0 respectively.

Let A = (x_{1}, y_{1}) and B = (x_{2}, y_{2}).

Then l_{1}x_{1} + m_{1}y_{1} = 0 and

lx_{1} + my_{1} + n = 0.

So, by the rule of cross-multiplication, we obtain

Question 20.

The condition for the line joining the origin to the point of intersection of the circle x^{2} + y^{2} = a^{2} and the line lx + my = 1 to coincide.

Solution:

Given circle equation is x^{2} + y^{2} = a^{2} …… (1)

Given line equation is lx + my = 1 ……… (2)

Homegenising (1) with help of (2)

The combined equation of OA, OB is

Question 21.

Find the direction cosines of two lines which are connected the relation l + m + n = 0 and mn – 2nl – 2lm = 0.

Solution:

Given l + m + n = 0 …….. (1)

mn – 2nl – 2lm = 0 ……….. (2)

From (1), l = -(m + n)

Substituting in (2),

mn + 2n(m + n) + 2m (m + n) = 0

mn + 2mn + 2n^{2} + 2m^{2} + 2mn = 0

2m^{2} + 5mn + 2n^{2} = 0

(2m + n) (m + 2n) = 0

2m = -n or m = -2n

Case (i): 2m_{1} = -n_{1}

From l_{1} = -m_{1} – n_{1}

\(\frac{l_1}{1}\) = \(\frac{m_1}{1}\) = \(\frac{x_1}{-2}\)

D.Rs of the first line are 1, 1, -2

D.Cs of this line are \(\frac{1}{\sqrt{6}}\), \(\frac{1}{\sqrt{6}}\), \(\frac{2}{\sqrt{6}}\)

Case (ii): m_{2} = -2n_{2}

From (1) l_{2} = -m_{2} – n_{2} = +2n_{2} – n_{2} = n_{2}

\(\frac{l_2}{1}\) = \(\frac{m_2}{-2}\) = \(\frac{n_2}{1}\)

d.cs of the second line are 1, -2, 1

d.cs of this line are \(\frac{1}{\sqrt{6}}\), \(\frac{-2}{\sqrt{6}}\), \(\frac{1}{\sqrt{6}}\)

Question 22.

If \(\sqrt{1-x^2}+\sqrt{1-y^2}\) = a(x – y) then prove that \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\sqrt{\frac{1-y^2}{1-x^2}}\).

Solution:

Question 23.

At a point (x_{1}, y_{1}) on the curve x^{3} + y^{3} = 3axy, show that the tangent is (\(x_1^2\) – ay_{1})x + (\(y_1^2\) – ax_{1})y = ax_{1}y_{1}.

Solution:

Given curve equation is x^{3} + y^{3} = 3axy

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

Question 24.

A window is in the shape of a rectangle surmounted by a semi-circle. If the perimeter of the window is 20 ft, find the maximum area.

Solution:

Let length or the rectangle be 2x and breadth be y so that radius of the semi-circle is x.

Given that perimeter of the window 20 ft.

⇒ 2x + 2y + πx = 20

⇒ 2y = 20 – 2x – πx

⇒ y = 10 – x – \(\frac{\pi}{2}\)x

∴ Area A = 2x.y + \(\frac{1}{2} \cdot \pi x^2\)