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

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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 :

2. 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) = xex 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 = x2
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)

1. Attempt any five questions.
2. 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
PA2 – PB2 = [(x – 5)2 + (y – 0)2] – [(x + 5)2 + (y – 0)2] ……… (1)
= x2 – 10x + 25 + y2 – x2 – 10x – 25 – y2
= – 20x
(PA + PB) (PA – PB) = -20x
(PA + PB) 8 = – 20x
(PA + PB) = – $$\frac{5}{2}$$x
2PA = $$\frac{5 x}{2}$$ + 8 = $$\frac{-5 x+16}{2}$$
4PA = – 5x + 16
16PA2 = (-5x + 16)2
16 [x – 5)2 + y2] = (-5x + 16)2
16 [x2 – 10x + 25 + y2] = [-5x + 16]2
16x2 + 16y2 – 160x + 400 = 25x2 + 256 – 160x
9x2 – 16y2 = 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 17x2 – 16xy + 17y2 = 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 (x1 ± r cos θ, y1 ± 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) = k2(1) – k = k2 – k
From (1)
k2 – k = 2
⇒ k2 – k – 2 =0
⇒ k2 – 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 cm3/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 cm3/sec.
We know v = x3
$$\frac{\mathrm{dv}}{\mathrm{dt}}$$ = $$3 x^2 \frac{d x}{d t}$$
$$\frac{d x}{d t}$$ = $$\frac{1}{3 x^2} \frac{d v}{d t}$$
∴ Surface area increasing at the rate of $$\frac{8}{3}$$ cm2/sec

Section – C

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

1. Attempt any five questions.
2. 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 ax2 + 2hxy + by2 = 0 is

Solution:
let ax2 + 2hxy + by2 ≡ (l1x + m1y) (l2x + m2y)
Then the separate equations of the lines represented by the equation
ax2 + 2hxy + by2 = 0 are
L1 : l1x + m1y = 0 and L2 : l1x + m2y = 0
Also, we have l1l2 = a; m1m2 = b and
l1m2 + l2m1 = 2h
d1 = length of the perpendicular

Question 20.
Find the angle between the lines joining the origin to the points of intersection of the curve x2 + 2xy + y2 + 2x + 2y – 5 = 0 and the line 3x – y + 1 = 0.
Solution:

Equation of the curve is
x2 + 2xy + y2 + 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
x2 + 2xy + y2 + 2x.1 + 2y.1 – 5.12 = 0
x2 + 2xy + y2 + 2x (y – 3x) + 2y (y – 3x) – 5 (y – 3x)2 = 0
x2 + 2xy + y2 + 2xy – 6x2 + 2y2 – 6xy – 5(y2 + 9x2 – 6xy) = 0
-5x2 – 2xy + 3y2 – 5y2 – 45x2 + 30 xy = 0
-50x2 + 28xy – 2y2 = 0
i.e., 25x2 – 14xy + y2 = 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 cos2α + cos2β + cos2γ + cos2δ.
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 xm.yn = am+n (mn ≠ 0) meets the co-ordinate axes in A, B, then show that AP : BP is a constant.
Solution:
Given curve equation is xm. yn = am+n
Taking logarithms on both sides, we have
log xm. yn = log am+n
⇒ log xm + log yn = (m + n) log a
⇒ m log x + n log y = (m + n) log a ……… (1)
Let P (x1, y1) 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 = x2 + nπ2

∴ 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 $$\frac{4 l}{\pi+4}$$