AP Inter 1st Year Maths 1B Question Paper May 2018

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AP 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 :

1. Answer all the 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.
If the area of the triangle formed by the straight lines x = 0, y = 0 and 3x + 4y = a(a > 0) is 6, find the value of a.
Solution:
Given straight line equation is 3x + 4y = a
⇒ \(\frac{3 x}{a}+\frac{4 y}{a}\) = 1
⇒ \(\frac{x}{\frac{a}{3}}+\frac{y}{\frac{a}{4}}\) = 1
∴ X-intercept = \(\frac{a}{3}\) ;
Y-intercept = \(\frac{a}{4}\)
Given area of the triangle formed by the lines x = 0, y = 0 and 3x + 4y = a is 6 sq. units.
⇒ \(\frac{1}{2}\) (X-intercept) (Y-intercept) | = 6
⇒ \(\frac{1}{2}\) . \(\frac{a}{3}\) . \(\frac{a}{4}\) = 6
⇒ a² = 144
⇒ a = ± 12
since a > 0
∴ a = 12

Question 3.
Show that the points (1, 2, 3) (2, 3, 1) and (3, 1, 2) form an equilateral triangle.
Solution:
Let A = (1, 2, 3)
B = (2, 3, 1)
C = (3, 1, 2)

Question 4.
Find the equation of the plane passing through (1, 1, 1) and parallel to the plane x + 2y + 3z – 7 = 0.
Solution:
The equation of the plane passing through (1, 1,1) and parallel to the plane x + 2y + 3z – 7 = 0 is
a(x – x₁) + b(y – y₁) + c(z – z₁) = 0.
⇒ 1(x – 1) + 2(y – 1) + 3(z – 1) = 0
⇒ x – 1 + 2y – 2 + 3z – 3 = 0
⇒ x + 2y + 3t – 6 = 0

Question 5.
Compute : \(\lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x}\), b ≠ 0, a ≠ b.
Solution:

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

Question 7.
Find the derivative of \(\mathrm{e}^{\sin ^{-1}} x\).
Solution:

Question 8.
Find the derivative of tan^-1\(\sqrt{\frac{1-\cos x}{1+\cos x}}\)
Solution:

Question 9.
Find the approximate value of \(\sqrt[3]{65}\).
Solution:

Question 10.
Verify Rolle’s theorem for sin x – sin 2x on [0, π].
Solution:
Let f(x) = sin x – sin 2x defined over [0, π]
Clearly f is contenuous on [0, π] and f is derivable on (0, π)
Also f(0) = sin 0 – sin 0 = 0
f(π) = sin π – sin 2π = 0 – 0 = 0
∴ f(0) = f(π)
∴ f satisfies all the conditions of Rolle’s theorem.
∴ Ǝ C ∈ (0, π) such that f'(c) = 0
f(x) = sin x – sin 2x
f'(x) = cos x – 2 cos 2x
f (c) = cos c – 2cos 2c
0 = cos c – 2(2cos²c – 1)
0 = cos c – 4 cos²c + 2
⇒ 4 cos²c – cos c – 2 = 0

Section – B
(5 × 4 = 20 Marks)

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) in 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 ……… (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 – 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 origin is shifted to (-1, 2) by the translation of axes find the transformed equation of x² + y² + 2x – 4y + 1 = 0.
Solution:
Here (h, k) = (-1, 2)
x = X + h ⇒ x = X – 1
y = Y + k ⇒ y = Y + 2
Given equation is x² + y² + 2x – 4y + 1 = 0
Now transformed equation is
(X – 1)² + (Y + 2)² + 2(X – 1) -4(Y + 2) + 1 = 0
⇒ X² – 2X + 1 + Y² + 4Y + 4 + 2X – 2 – 4y – 8 + 1 = 0
⇒ X² + Y² – 4 = 0
⇒ X² + Y² = 4

Question 13.
Find the equation of the straight lines passing through (1, 3) and
(i) parallel to
(ii) perpendicular to the line passing through the points (3, -5) and (-6, 1).
Solution:
Let A = (3, -5)
B = (-6, 1)
Equation of \(\overline{\mathrm{AB}}\) is
y – y₁ = \(\frac{y_2-y_1}{x_2-x_1}\)(x – x₁)
⇒ y + 5 = \(\frac{1+5}{-6-3}\)(x – 3)
⇒ -9y – 45 = 6x – 18
⇒ 6x + 9y + 27 = 0
⇒ 2x + 3y + 9 = 0

i) Equation or the straight line passing through (1,3) and parallel to \(\overline{\mathrm{AB}}\) is a (x – x₁) + b(y – y₁) = 0
⇒ 2(x – 1) + 3(y – 3) = 0
⇒ 2x – 2 + 3y – 9 = 0
⇒ 2x + 3y – 11 = 0

ii) Equation of the straight line passing through (1, 3) and perpendicular to AB is b(x – x₁) – a(y – y₁) = 0
⇒ 3(x – 1) – 2(y – 3) = 0
⇒ 3x – 3 – 2y + 6 = 0
⇒ 3x – 2y + 3 = 0

Question 14.
If f, given by :

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

Since f is a continuous function on R
∴ f is continuous at x = 1
\(\lim _{x \rightarrow 1-} f(x)\) = f(1) = \(\lim _{x \rightarrow 1_{+}} f(x)\) ……. (1)
\(\lim _{x \rightarrow 1^{-}} f(x)\) = 2
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) (k + 1) = 0
∴ k = 2 (or) – 1

Question 15.
Find the derivative of cos ax using first principle.
Solution:

Question 16.
The volume of a cube is increasing at a rate of 9 cubic centimeters per second. How fast is the surface area increasing when the length of the edge is 10 centimeters ?
Solution:
Let x be the length of the edge of the cube, V be its volume and S be its surface area. Then, V = x³ and S = 6x². Given that rate of change of volume is 9 cm³/sec.

Question 17.
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:
Equations of the curve are x = a (cos t + t sin t)
x = a (cos t + t sin t)
\(\frac{\mathrm{dx}}{\mathrm{dt}}\) = a(-sin t + t cos t) = at cos t dt
y = a (sin t – 1 cos t)

Length of the sub-normal = \(\frac{y_1}{f^{\prime}\left(x_1\right)}\) = \(\frac{a(\sin t-t \cos t)}{\tan t}\)
Length of the sub-normal = |y₁.f'(x₁)|
= |a(sint – tcos t) tan t
= |a tan t (sin t – t cos t|

Section – C

III. Long answer type questions :

1. Attempt ANY FIVE questions.
2. Each question carries SEVEN marks.

Question 18.
Find the ortho center of the triangle with the following vertices : (-2, -1), (6, -1) and (2, 5).
Solution:
Let A = (-2, -1); B = (6, -1); C = (2, 5)
Slope of \(\overline{\mathrm{BC}}\) = \(\frac{5+1}{2-6}\) = \(\frac{-3}{2}\)

Question 19.
Show that the equation : 2x² – 13xy – 7y² + x + 23y – 6 = 0 represents a pair of straight lines, also find the angle between them and the co-ordinates of the point of intersection of the lines.
Solution:
Here a = 2 f = \(\frac{23}{2}\)
b = -7 g = \(\frac{1}{2}\)
c = 6 h = –\(\frac{13}{2}\)
abc + 2fgh – af² – bg² – ch²

The given equation represents a pair of lines

Question 20.
Show that the lines joining the origin to the 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 Y
x² – xy + y² + 3x + 3y – 2 = 0 ……. (1)
Equation of AB is x – y – \(\sqrt{2}\) = 0
x – y = \(\sqrt{2}\)
\(\frac{x-y}{\sqrt{2}}\) = 2 ……. (2)

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

Question 21.
Show that the lines whose d.c’s are given by l + m + n = 0, 2mn + 3nl – 5lm =0 are perpendicular to each other.
Solution:
Given l + m + n = 0 ……. (1)
2mn + 3nl – 5lm = 0 ……… (2)
From (1), l = -(m + n)
Substituting in (2)
2mn – 3n(m + n) + 5m(m + n) = 0
2mn – 3mn – 3n² + 5m² + 5mn = 0
5m² + 4mn – 3n² = 0
\(\frac{m_1 m}{n_1 n_2}\) = \(-\frac{3}{5}\) ⇒ \(\frac{m_1 m_2}{-3}\) = \(\frac{n_1 n_2}{5}\) …… (1)
From (1), n = -(l + m)
Substituting is (2)
-2m (l + m) – 3l (l + m) – 5lm = 0
-2lm – 2m² – 3l² – 3lm – 5lm = 0
3l² + 10lm + 2m² = 0
\(\frac{l_1 l_2}{m_1 m_2}\) = \(\frac{2}{3}\) ⇒ \(\frac{l_1 l_2}{2}\) = \(\frac{m_1 m_2}{3}\) ….. (2)
From.(1) and (2) we get
\(\frac{l_1 l_2}{2}\) = \(\frac{m_1 m_2}{3}\) = \(\frac{n_1 n_2}{-5}\) = 1
l₁l₂ + m₁m₂ + n₁n₂ = 2k + 3k – 5k = 0
The two lines are perpendicular.

Question 22.
If y = tan^-1\(\left[\frac{2 x}{1-x^2}\right]\) + tan^-1\(\left[\frac{3 x-x^3}{1-3 x^2}\right]\) – tan^-1\(\left[\frac{4 x-4 x^3}{1-6 x^2+x^4}\right]\) then \(\frac{d y}{d x}\) = \(\frac{1}{1+x^2}\)
Solution:

Question 23.
Show that the curves 6x² – 5x + 2y = 0 and 4x² + 8y² = 3 touch each other at \(\left(\frac{1}{2}, \frac{1}{2}\right)\).
Solution:
Given curve equations are
6x² – 5x + 2y = 0 ……. (1)
4x² + 8y² = 3 ……. (2)
Differentiating (1) w.r.to ‘x’ on both sides dy
12x – 5 + 2 \(\frac{d y}{d x}\) = 0
⇒ 2 \(\frac{d y}{d x}\) = 5 – 12x

∴ Given curves (1) & (2) touch each other at \(\left(\frac{1}{2}, \frac{1}{2}\right)\).

Question 24.
Find two positive numbers whose sum is 15 so that the sum of their squares is minimum.
Solution:
Let x, y be two positive numbers.
∴ x + y = 15
⇒ y = 15 – x
Let f(x) = x² + y²
= x² + (15 – x)²
= x² + 225 – 30x + x²
= 2x² – 30x + 225
f(x) = 4x – 30
For maxima or minima
f'(x) = 0
⇒ 4x – 30 = 0
⇒ 4x = 30
⇒ 2x = 15
⇒ x = \(\frac{15}{2}\)
f”(x) = 4
f”\(\left(\frac{15}{2}\right)\) = 4 > 0
∴ f has maximum at x = \(\frac{15}{2}\)
∴ y = 15 – \(\frac{15}{2}\) = \(\frac{30-15}{2}\) = \(\frac{15}{2}\)
∴ x = \(\left(\frac{15}{2}\right)\), y = \(\left(\frac{15}{2}\right)\)