Application Of Derivatives(12th Grade > Mathematics ) Questions and answers for exam Preparation

12th Grade > Mathematics

APPLICATION OF DERIVATIVES MCQs

Total Questions : 58 | Page 3 of 6 pages

Question 21. The value of m for which the area of the triangle included between the axes and any tangent to the curve

x^{m} y = b^{m}

is constant is

Discuss Question

Answer: Option B. -> 1
:
B

∴ x^{m} y = b^{m}

Taking logarithm

m l o g_{e} x + l o g_{e} y = m l o g_{e} b ∴ \frac{m}{x} + \frac{1}{y} \frac{d y}{d x} = 0 ∴ \frac{d y}{d x} = - \frac{m y}{x}

Equation of tangent at (x,y) is

Y - y = - \frac{m y}{x} (X - x) x Y - x y = - m y X + m x y \Rightarrow m y X = x Y = x y (1 + m) \Rightarrow \frac{\frac{X}{x (1 + m)}}{m} + \frac{x}{x (1 + m)} = 1

∴

Area of triangle OAB

= \frac{1}{2} . O A . O B

The Value Of M For Which The Area Of The Triangle Included ...

= \frac{1}{2} ∣ ∣ \frac{x (1 + m)}{m} ∣ ∣ | y (1 + m) | = \frac{| x y | (1 + m)^{2}}{2 | m |} F o r m = 1, = \frac{| x y | (4)}{2} = 2 | x y | (∵ x y = b) = 2 | b | =

constant

Question 22. If a + b +c = 0, then the equation

3 a x^{2} + 2 b x + c = 0

has, in the interval (0, 1)

Atleast one root
Atmost one root
No root
exactly one root

Discuss Question

Answer: Option A. -> Atleast one root
:
A
Let

f (x) = a_{n} X^{n} + a_{n - 1} X^{n - 1} + . . . + a_{2} x^{2} + a_{1} x + a_{0}

Which is a polynomial function in x of degree n. Hence f(x) is continuos and differentiable for all x.
Let

\infty < β

. We given, f

(\infty) = 0 = f (β)

.
By Rolle's theorem, f' (c) = 0 for some value c,

\infty < c < β

.
Hence the equation

F^{'} (x) = n a_{n} x^{n - 1} + (n - 1) a_{n - 1} x^{n - 2} + . . . + a_{1} = 0

has atleast one root between

\infty a n d β

.
Hence (c) is the correct answer.

Question 23. A man of height 2m walks directly away from a lamp of height 5m, on a level road at 3 m/s. The rate at which the length of his shadow is increasing is

1m/s
2m/s
3m/s
4m/s

Discuss Question

Answer: Option B. -> 2m/s
:
B
Let be the lamp and PQ be the man and OQ=x metre be his shadow and let MQ =y metre.

A Man Of Height 2m Walks Directly Away From A Lamp Of Height...

∴ \frac{d y}{d t}

=speed of the man
=3 m/s (given)

∵ Δ O P Q

and

Δ O L M

are similar

∴ \frac{O M}{O Q} = \frac{L M}{P Q} \Rightarrow \frac{x + y}{x} = \frac{5}{2} \Rightarrow y = \frac{3}{2} x ∴ \frac{d y}{d t} = \frac{3}{2} \frac{d x}{d t} \Rightarrow 3 = \frac{3}{2} \frac{d x}{d t} \Rightarrow \frac{d x}{d t} = 2 m / s

Question 24. If

f (x) = ∣ ∣∣ ∣ \begin{matrix} s i n x & s i n a & s i n b c o s x & c o s a & c o s b t a n x & t a n a & t a n b \end{matrix} ∣ ∣∣ ∣, w h e r e 0 < a < b < \frac{π}{2}

then the equation

f^{'} (x) = 0

has in the interval

(a, b)

Atleast one root
Atmost one root
No root
exactly one root

Discuss Question

Answer: Option A. -> Atleast one root
:
A
Here

f (a) = ∣ ∣∣ ∣ \begin{matrix} s i n a & s i n a & s i n b c o s a & c o s a & c o s b t a n a & t a n a & t a n b \end{matrix} ∣ ∣∣ ∣ = 0. Also f (b) = 0.

Moreover, as sin x, cos x and tan x are continuos and differentiable in (a, b) for 0 < a < b <

\frac{π}{2}

, therefore f(x) is also continuos and differentiable in [a, b]. Hence, by Rolle's theorem, there exists atleast one real number c in (a, b) such that f ' (c) = 0.
Hence (a) is the correct answer.

Question 25. The minimum value of the function defined by f(x) = minimum {x, x+1,2 - x } is

Discuss Question

Answer: Option D. -> 32
:
D

∵

f(x) = maximum{x, x + 1, 2-x}

The Minimum Value Of The Function Defined By F(x) = Minimum ...

∴

Minimum value of function

= \frac{3}{2}

Question 26. The slope of the tangent to the curve

x = t^{2} + 3 t - 8, y = 2 t^{2} - 2 t - 5

at the point t = 2 is

Discuss Question

Answer: Option C. -> 67
:
C
We have,

\frac{d x}{d t} = 2 t + 3 a n d \frac{d y}{d t} = 4 t - 2 \frac{d y}{d x} = \frac{d y / d t}{d x / d t} = \frac{4 t - 2}{2 t + 3}

Thus, slope of the tangent to the curve at the point t = 2 is

{[\frac{d y}{d x}]}_{t - 2} = \frac{4 (2) - 2}{2 (2) + 3} = \frac{6}{7}

Thus, slope of the tangent to the curve at the point t = 2 is
Hence (c) is the correct answer

Question 27.

f (x) = c o s (\frac{π}{x}), x \neq 0

increases in the interval

(12k+1, 12k)
(12k+2, 12k+1)
(12k+1, 12k+3)
(12k+1, 12k+3)

Discuss Question

Answer: Option A. -> (12k+1, 12k)
:
A
f(x) = cos

(\frac{π}{x})

Thus
f(x) = cos

(\frac{π}{x})

increases in

\frac{1}{3} \to \frac{1}{2}

generalizing it

(\frac{1}{2 k + 1}, \frac{1}{2 k})

F(x) = cos (πx), x ≠ 0 Increases In The Interval

Question 28. If f (x) is differentiable in the interval [2, 5], where

f (2) = \frac{1}{5}

and

f (5) = \frac{1}{2}

, then there exists a number c, 2 < c < 5 for which f ' (c) is equal to

Discuss Question

Answer: Option C. -> 110
:
C
As f (x) is differentiable in [2 , 5], therefore, it is also continuos in [2, 5]. Hence, by mean value theorem, there exists a real number c in (2, 5) such that