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## TS Inter 1st Year Maths 1B Question Paper May 2018

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 passing through (-4, 5) and cutting off equal and non-zero intercepts on the co-ordinate axes.

Solution:

Let x – intercept = a

y – intercept = a

Intercept form : \(\frac{x}{a}+\frac{y}{a}\) = 1

⇒ x + y = a

If this straight line passing through (-4, 5) then -4 + 5 = a

⇒ a = 1

∴ Required straight line equation is x + y = 1

⇒ x + y – 1 = 0

Question 2.

Find the distance between the parallel lines 5x – 3y – 4 = 0, 10x – 6y – 9 = 0.

Solution:

Given line equations are

(5x – 3y – 4 = 0) × 2 ⇒ 10x – 6y – 8 = 0 ……. (1)

10x – 6y – 9 = 0 …………………… (2)

The distance between the parallel lines (1) and (2) = \(\frac{\left|c_2-c_1\right|}{\sqrt{a^2+b^2}}\)

= \(\frac{|-9+8|}{\sqrt{10^2+6^2}}\)

= \(\frac{1}{\sqrt{136}}\)

Question 3.

Find the co-ordinates of the vertex C of ∆ABC, if its centroid is the origin and the vertices A, B are (1, 1, 1) and (-2, 4, 1) respectively.

Solution:

Given A = (1, 1, 1)

B = (-2, 4, 1)

Let C = (x, y, z)

Since origin is the centroid of ∆ABC.

Question 4.

Find the intercepts of the plane 4x + 3y – 2z + 2 = 0 on the co-ordinate axes.

Solution:

Given plane equation is 4x + 3y – 2z + 2 = 0.

⇒ 4x + 3y – 2z = – 2

∴ X – intercept = \(\frac{-1}{2}\)

y — intercept = \(\frac{-2}{3}\)

z — intercept = 1

Question 5.

Compute

Solution:

Question 6.

Evaluate

Solution:

Question 7.

If f(x) = xe^{x} sin x then find f (x).

Solution:

Question 8.

If y = \(e^{a \sin ^{-1} x}\) then prove that \(\frac{d y}{d x}\) = \(\frac{a y}{\sqrt{1-x^2}}\).

Solution:

Question 9.

If the increase in the side of a square is 4% then find the approximate percentage of increase in the area of the square.

Solution:

Let x, A be the side, area of a square respectively.

Given \(\frac{\delta x}{x}\) × 100 = 4.

A = x^{2}

δA = 2x δx

\(\frac{\delta A}{A}\) × 100 = \(\frac{2 x \delta x}{x^2}\) × 100

= 2.\(\frac{\delta x}{x}\) × 100

= 2.4

= 8.

∴ The approximate percentage of increase in the area of the square is 8.

Question 10.

Define Lagrange’s mean value theorem.

Solution:

Lagrange’s mean value theorem:

Let a, b ∈ R (a < b). If f : [a, b] → R be a function Such that

(i) f is continuous on [a, b] and

(ii) f is derivable on (a, b) then there exists

c ∈ (a, b) such that f'(c) = \(\frac{f(b)-f(a)}{b-a}\)

Section – B

II. Short Answer Type Questions. (5 × 4 = 20)

- Attempt any five questions.
- Each question carries four marks.

Question 11.

Find the equation of the locus of a point, the difference of whose distances from (-5, 0) and (5, 0) is 8.

Solution:

A(5, 0), B(-5, 0) are the given points.

P(x, y) is any point on the locus.

Given condition is [PA – PB] = 8

PA – PB = 8

PA^{2} – PB^{2} = [(x – 5)^{2} + (y – 0)^{2}] – [(x + 5)^{2} + (y – 0)^{2}] ……… (1)

= x^{2} – 10x + 25 + y^{2} – x^{2} – 10x – 25 – y^{2}

= – 20x

(PA + PB) (PA – PB) = -20x

(PA + PB) 8 = – 20x

(PA + PB) = – \(\frac{5}{2}\)x

Adding (1) and (2),

2PA = \(\frac{5 x}{2}\) + 8 = \(\frac{-5 x+16}{2}\)

4PA = – 5x + 16

16PA^{2} = (-5x + 16)^{2}

16 [x – 5)^{2} + y^{2}] = (-5x + 16)^{2}

16 [x^{2} – 10x + 25 + y^{2}] = [-5x + 16]^{2}

16x^{2} + 16y^{2} – 160x + 400 = 25x^{2} + 256 – 160x

9x^{2} – 16y^{2} = 144

Dividing with 144, locus of P is

\(\frac{9 x^2}{144}-\frac{16 y^2}{144}\) = 1 i.e., \(\frac{x^2}{16}-\frac{y^2}{9}\) = 1

Question 12.

When the axes are rotated through an angle 45°, the transformed equation of a curve is 17x^{2} – 16xy + 17y^{2} = 225. Find the original equation of the curve.

Solution:

Question 13.

A straight line through P(3, 4) makes an angle of 60° with the positive direction of the X – axis. Find the co-ordinates of the points on the line which are 5 units away from P.”

Solution:

Given P = (3, 4) at θ = 60°

The co-ordinates of the points on the line which are 5 units away from P is (x_{1} ± r cos θ, y_{1} ± r sin θ)

= (3 ± 5 cos 60°, 4 + 5 sin 60°)

= \(\left(3 \pm 5\left(\frac{1}{2}\right), 4 \pm 5 \cdot \frac{\sqrt{3}}{2}\right)\)

Question 14.

If f, given by

is a continuous function on R, then find the values of k.

Solution:

Given

Since f is a continuous function on R

Also f(1) = k^{2}(1) – k = k^{2} – k

From (1)

k^{2} – k = 2

⇒ k^{2} – k – 2 =0

⇒ k^{2} – 2k + k – 2 = 0

⇒ k(k – 2) + 1 (k – 2) = 0

∴ k = 2 (or) k = -1

Question 15.

Find the derivative of the function cos (ax) from the first principle.

Solution:

Let f(x) = cos (ax)

f(x + h) = cos a (x + h) ⇒ cos (ax + ah)

Question 16.

Find the equation of tangent and normal to the curve xy = 10 at (2, 5).

Solution:

Given curve equation is xy = 10

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

Question 17.

The volume of a cube is increasing at the rate of 8 cm^{3}/sec. How fast is the surface area increasing, when the length of an edge is 12 cm?

Solution:

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

Given x = 12 cm.

\(\frac{\mathrm{dv}}{\mathrm{dt}}\) = 8 cm^{3}/sec.

We know v = x^{3}

\(\frac{\mathrm{dv}}{\mathrm{dt}}\) = \(3 x^2 \frac{d x}{d t}\)

\(\frac{d x}{d t}\) = \(\frac{1}{3 x^2} \frac{d v}{d t}\)

∴ Surface area increasing at the rate of \(\frac{8}{3}\) cm^{2}/sec

Section – C

III. Long Answer Type Questions. (5 × 7 = 35)

- Attempt any five questions.
- Each question carries seven marks.

Question 18.

Find the circumcentre of the triangle whose vertices are (1, 3), (0, -2) and (-3, 1).

Solution:

Let A = (1, 3)

B = (0, -2)

C = (-3, 1)

Let S (α, β) be the circumcentre of the triangle ABC.

Question 19.

Show that the product of the perpendicular distance from a point (α, β) to the pair of straight lines ax^{2} + 2hxy + by^{2} = 0 is

Solution:

let ax^{2} + 2hxy + by^{2} ≡ (l_{1}x + m_{1}y) (l_{2}x + m_{2}y)

Then the separate equations of the lines represented by the equation

ax^{2} + 2hxy + by^{2} = 0 are

L_{1} : l_{1}x + m_{1}y = 0 and L_{2} : l_{1}x + m_{2}y = 0

Also, we have l_{1}l_{2} = a; m_{1}m_{2} = b and

l_{1}m_{2} + l_{2}m_{1} = 2h

d_{1} = length of the perpendicular

Question 20.

Find the angle between the lines joining the origin to the points of intersection of the curve x^{2} + 2xy + y^{2} + 2x + 2y – 5 = 0 and the line 3x – y + 1 = 0.

Solution:

Equation of the curve is

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

Equation of AB is 3x – y + 1 = 0

y – 3x = 1 ……. (2)

Homogenising (1) with the help of (2), combined equatio of OA, OB is

x^{2} + 2xy + y^{2} + 2x.1 + 2y.1 – 5.1^{2} = 0

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

x^{2} + 2xy + y^{2} + 2xy – 6x^{2} + 2y^{2} – 6xy – 5(y^{2} + 9x^{2} – 6xy) = 0

-5x^{2} – 2xy + 3y^{2} – 5y^{2} – 45x^{2} + 30 xy = 0

-50x^{2} + 28xy – 2y^{2} = 0

i.e., 25x^{2} – 14xy + y^{2} = 0

Suppose θ is the angle between OA and OB

Question 21.

If a ray makes angles α, β and γ with four diagonals of a cube, then find cos^{2}α + cos^{2}β + cos^{2}γ + cos^{2}δ.

Solution:

Let each side of the cube be of length a. Let one of the vertices of the cube be the origin O and the co-ordinate axes be along the three edges \(\overline{\mathrm{OA}}\), \(\overline{\mathrm{OB}}\) and \(\overline{\mathrm{OC}}\) passing through the origin. The co-ordinates of the vertices of the cube with respect to the frame of reference OABC are as shown in figure. The diagonals of he cube are \(\overline{\mathrm{OP}}\), \(\overline{\mathrm{CD}}\), \(\overline{\mathrm{AE}}\) and \(\overline{\mathrm{BF}}\). (a, a, a), (a, a, -a), (-a, a, a) and (a, -a, a) are direction ratios of these diagonals respectively.

Let the direction cosines of the given ray be (l, m, n): If this ray makes the angles α, β, γ and δ with the four diagonals of the cube, then

Question 22.

If y = tan^{-1}\(\left[\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right]\) for 0 < |x| < 1, find \(\frac{d y}{d x}\).

Solution:

Question 23.

If the tangent at any point P on the curve x^{m}.y^{n} = a^{m+n} (mn ≠ 0) meets the co-ordinate axes in A, B, then show that AP : BP is a constant.

Solution:

Given curve equation is x^{m}. y^{n} = a^{m+n}

Taking logarithms on both sides, we have

log x^{m}. y^{n} = log a^{m+n}

⇒ log x^{m} + log y^{n} = (m + n) log a

⇒ m log x + n log y = (m + n) log a ……… (1)

Let P (x_{1}, y_{1}) be any point on the curve.

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

Since the tangent meets the co-ordinate axes at A, B.

Question 24.

A wire of length λ is cut into two parts which are bent respectively in the form of a square and a circle. What are the lengths of the pieces of the wire respectively, so that the sum of the areas is the least ?

Solution:

Suppose x is the side of the square and r is the radius of the circle

Given 4x + 2πr = l

4x = l – 2πr

x = \(\frac{l-2 \pi r}{4}\)

Sum of the area = x^{2} + nπ^{2}

∴ f(r) is least when r = \(\frac{l}{2(\pi+4)}\)

Sum of the area is least when the wire is cut into pieces of length

\(\frac{\pi l}{\pi+4}\) and \(\frac{4 l}{\pi+4}\)