AP Inter 1st Year Maths 1B Question Paper March 2020

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AP Inter 1st Year Maths 1B Question Paper March 2020

Time : 3 Hours
Max. Marks : 75

Section – A
(10 × 2 = 20)

I. Very Short Answer Type Questions.

1. Answer all questions.
2. Each question carries two marks.

Question 1.
Find the equation of the straight lirje passing through (-4, 5) and cutting off equal and non zero intercepts on the co-ordinate axes.
Solution:
Equation of the line in the intercept form is \(\frac{x}{a}+\frac{y}{b}\) = 1
Given a = b
Equation of the line is \(\frac{x}{a}+\frac{y}{b}\) = 1
⇒ x + y = a
This line passes through P (-4, 5)
_ 4 5 = a ⇒ a = 1
Equation of the required line is x + y = 1 or x + y = – 1 = 0.

Question 2.
Find the area of the triangle formed by the straight line x – 4y + 2 = 0 and the co-ordinate axes.
Solution:
Equation of AB is x – 4y + 2 = 0
-x + 4 = 2 ⇒ \(\frac{x}{-2}+\frac{y}{\left(\frac{1}{2}\right)}\) = 1
a = -2, b = \(\frac{1}{2}\)
Area of ∆OAB = \(\frac{1}{2}\) |ab|
= \(\frac{1}{2}\)|-2 × \(\frac{1}{2}\)| = \(\frac{1}{2}\) sq. units

Question 3.
Find the coordinates of the vertex ‘C of ∆ABC if its centroid is origin and the vertices A, B are (1, 1, 1) and (-2, 4, 1) respectively.
Solution:
A (1, 1, 1) B (-2, 4, 1) and (x, y, z) are the vertices of ∆ABC.
G is the centroid of ∆ABC.
Co – ordinates of G are
\(\left(\frac{1-2+x}{3} \cdot \frac{1+4+y}{3} \cdot \frac{1+1+z}{3}\right)\) = (0,0, 0)
\(\frac{x-1}{3}\) = 0. \(\frac{y+5}{3}\) = 0. \(\frac{x+2}{3}\) = 0
x – 1 = 0, y + 5 = 0, z + 2 = 0
x = 1, y = -5, z = -2.
∴ Co – ordinates of c are (1, -5, -2).

Question 4.
Find the equation of the plane passing through (-2, 1, 3) and having (3, -5, 4) as d.r.’s of its normal.
Solution:
d.r.’s of the normal are (3, -5, 4) and the plane passes through (-2, 1, 3) Equation of the plane is 3(x + 2) – 5(y – 1) + 4(z – 3) = 0
3x + 6 – 5y + 5 + 4z – 12 = 0
3x – 5y + 4z – 1 = 0.

Question 5.
Compute the limit \(\lim _{x \rightarrow 0} \frac{\sin a x}{x \cos x}\)
Solution:

Question 6.
Compute the limit \(\lim _{x \rightarrow \infty} \frac{11 x^3-3 x+4}{13 x^3-5 x^2-7}\)
Solution:

Question 7.
Find the derivative of the function \(\frac{1-\cos 2 x}{1+\cos 2 x}\).
Solution:

Question 8.
Define the derivative of a function.
Solution:
definition : Let f be a function defined on a neighbourhood of a real number a. Then f in said to be differentiable or derivable at a if \(\operatorname{Lt}_{x \rightarrow a} \frac{f(x)-f(a)}{x-a}\) exists finitely. The limit is called the derivative of differential coefficient of f at a. It is denoted by f(a).

Question 9.
Find ∆y and dy for the function y = x² + 3x + 6 for the value of x = 10 and ∆x = 0.01.
Solution:
∆y = f(x + ∆x) – f(x)
= f(1o.o1) – f(1o)
= [(10.0)² + 3 (10.01) +6] – [10² + 3 (10) + 6]
= 100.2001 + 30.03 + 6 – 100 – 30 – 6
= 0.2001 + 0.03
= 0.2301
y = x² + 3x + 6
dy = (2x + 3) dx
= (2.10 + 3) (0.01) = 0.23

Question 10.
State Legrange’s Mean value theorem.
Solution:
(i) If a function f(x) ie., (i) continuous on [a, b].
(ii) 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.

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.
P(x, y) is any point on the locus.
Given condition is | PA – PB | = 8
PA – PB = 8 ……. (1)
PA² – PB² = [(x – 5)² + (y – 0)²] – [(x + 5)² + (y – 0)²]
= x² – 10x + 25 + y² – x² – 10x – 25 – y²
= -20x
(PA + PB) (PA – PB) = – 20x
(PA + PB) 8 = -20x
PA + PB = –\(\frac{5}{2}\) x …….. (2)
Adding (1) and (2),
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 – 1 + y² = 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 α, find the transformed equation of x cos α + y sin α = p.
Solution:
The given equation is x cox α + y sin α = p
∴ The axes are rotated through an angle α
x = x’ cos α – y’ sin α
y = x’ sin α + y’ cos α
The given equation transformed to
(x’cos α – y’ sin α) cos α + (x’ sin α + y’ cos α) sin α = p
⇒ x’ (cos² α + sin² α) = p ⇒ x’ = p
The equation transformed to x = p.

Question 13.
Transform the equation 4x – 3y + 12 = 0 into
i) intercept form and
ii) normal form
Solution:
Slope – intercept form :
3y = 4x + 12
y = \(\left(\frac{4}{3}\right)\)x + 4
Intercept form :
4x – 3y + 12 = 0
-4x + 3y = 12

Question 14.
Check the continuity of f given by

Solution:

Question 15.
Find the derivative of f(x) = sin 2x using the First Principle.
Solution:

Question 16.
Find the equations of tangent and normal to the curve y = x³ + 4x² at (-1, 3)
Solution:
Equatión of the curve is y = x³ + 4x²
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 3x² + 8x
At P (-1, 3),
Slope of the tangent 3(-1)² + 8(-1)
= 3 – 8 = -5
Equation of the tangent at P(-1, 3) is
y – Y₁ = (x – x₁)
y – 3 = -5(x + 1) = -5x – 5
5x + y + 2 = 0
Equation of the normal at P is
y – y₁ = –\(\frac{1}{f^{\prime}\left(x_1\right)}\)(x – x₁)
y – 3 = \(\frac{1}{5}\)(x + 1)
5y – 15 = x + 1
x – 3y + 16 = 0.

Question 17.
A container is in the shape of an inverted cone has height 8 m and radius 6 m at the top. If it is filled with water at the rate 2m³/minute, how fast is the height of water changing when the level is 4 m ?
Solution:
h = 8m = OC
r = 6m = AB
\(\frac{\mathrm{dv}}{\mathrm{dt}}\) = 2m³/minute

∆ OAB and OCD
are similar angle then
\(\frac{C D}{A B}\) = \(\frac{O C}{O A}\)
\(\frac{r}{6}\) = \(\frac{h}{8}\)
r = h \(\frac{3}{4}\)
Volume of cone v = \(\frac{1}{3}\) πr²h
v = \(\frac{1}{3}\)πr²\(\frac{9}{16}\)h

Section – C

III. Long Answer Type Questions

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

Question 18.
a) If Q (h, k) is the foot of the perpendicular from p(x₁, y₁) on the straight line ax + by + c = 0 then prove that
\(\frac{h-x_1}{a}\) = \(\frac{k-y_1}{b}\) = \(\frac{-\left(a x_1+b y_1+c\right)}{a^2+b^2} .\)
b) Find the foot of the perpendicular drawn from (4, 1) upon the straight line 3x – 4y +12 = 0.
Solution:
Equation of \(\overleftrightarrow{\mathrm{PQ}}\) which is normal to the given straight line

b) Find the foot of the perpendicular drawn from (4, 1) upon the straight line 3x – 4y + 12 = 0.
Sol.
Equation of the line is 3x – 4y + 12 = 0
If (x², y²) is the foot of the perpendicular from (x₁, y₁) on the line

Question 19.
a) Let the equation ax² +2hxy + by² = 0 represents a pair of straight lines and the angle between them is e then show that
cos θ = \(\frac{|a+b|}{\sqrt{(a-b)^2+4 h^2}}\)
b) Find the angle between the pair of lines represented by the equation x² – 7xy + 12y² = 0.
Solution:
It is Obvious that (a – b)² + 4h² > o.
Let H ≡ ax² + 2hxy + by² = (l₁x + m₁y) (l₂x + m₂y). Then the lines represented by the given equation are l₁x + m₁y = 0 and l₂x + m₂y = 0. Further, l₁l₂ = a, m₁m₂ = b and l₁m₂ + l₂m₂1 = 2h.
Therefore, the angle 0 between these lines is given by

b)

Question 20.
Show that the lines joining foe origin to foe points of intersection of the curve x² – xy + y² + 3x + 3y – 2 = 0 and the straight line x – y – \(\sqrt{2}\) = 0 are mutually perpendicular.
Solution:

Equation of the curve is
x² – xy + y² + 3x + 3y – 2 = 0
Equation of AB is x – y – \(\sqrt{2}\) = 0
x – y = \(\sqrt{2}\)
\(\frac{x-y}{\sqrt{2}}\) = 1 …..(2)
Homogenising. (1) with the help of (2)
combined equation of OA, OB is

Question 21.
Find the angle between two diagonals of a cube.
Solution:
Let ‘O’ one of the vertices of the cube taken as origin and the three sides OA, OB, OC are taken as co-ordinate axes.
Let OA = OB = OC are taken as co-ordinae axes. Let OA = OB + OC = a the four diagonals are, and
The co-ordinates of the vertices of the cube are
O(0, 0, 0), A(a, 0, 0, B(0, a, o), C(0, 0, a)
F(a, a, 0), D(a, a, 0), E(a, 0, a), G(0, a, a)
D.Rs of AG are (a, – 9, a – 0, a – 0 – = (-a, a, a). If O is the angle between the diagonals OF and AG.

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.
Find the angle between the curves y² = 4x; x² + y² = 5.
Solution:
Eliminating y; we get x² + 4x = 5
x² + 4x – 5 = 0
(x – 1) (x + 5) = 0
x – 1 = 0 or x + 5 = 0
x = 1 or -5
Now, y² = 4x
x = 1 ⇒ y² = 4
y = ±2
x = – 5 ⇒ y is not real.
∴ Points of intersection of P(1, 2) and
Q (1, 2) equation of the first curve is y² = 4x

Question 24.
The profit function p(x) of a company, selling x items per day is given p(x) = (150 – x) x – 1600. Find the number of items that the company should sell for maximum profit. Also find the maximum profit.
Solution:
Given that the profit function is
P(x) = (150 – x) x – 1600 ……. (1)
For maxima or minima \(\frac{d p(x)}{d x}\) = 0.
∴ (150 – x) (1) + x (-1) ≠ 0
i.e., x = 75
Now, \(\frac{d^2 p(x)}{d x^2}\) = -2 and \(\left[\frac{d^2 p(x)}{d x^2}\right]_{x=75}\) < 0
∴ The profit P(x) is maximum for x = 75.
∴ The company should sell 75 items a day to make maximum profit.
The maximum profit will be P(75) = 4025.