Application Of Derivatives(11th And 12th > Mathematics ) Questions and answers for exam Preparation

11th And 12th > Mathematics

APPLICATION OF DERIVATIVES MCQs

Total Questions : 30 | Page 2 of 3 pages

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

$\frac{1}{2}$
$\frac{1}{5}$
$\frac{1}{10}$
$7$

Discuss Question

Answer: Option C. ->

\frac{1}{10}

:
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

f^{'} (c) = \frac{f (5) - f (2)}{5 - 2} \Rightarrow f^{'} (c) = \frac{\frac{1}{2} - \frac{1}{5}}{3} = \frac{1}{10} .

Hence (c) is the correct answer.

Question 12.

The equation x log x = 3 - x has, in the interval (1, 3),

Exactly one root
Atmost one root
Atleast one root
No root

Discuss Question

Answer: Option C. -> Atleast one root
:
C

Let f (x) = (x - 3) log x
Then, f (1) = - 2 log 1 = 0 and f (3) = (3-3) log 3 = 0. As, (x-3) and log x are continuos and differentiable in [1, 3], therefore (x-3) log x = f (x) is also continuos and differentiable in [1, 3]. Hence, by Rolle's theorem, there exists a value of x in (1, 3) such that
f ' (x) = 0 $\Rightarrow$ log x+(x-3) $\frac{1}{x}$ = 0
$\Rightarrow$ x log x = 3 - x.
Hence (c) is the correct answer.

Question 13.

Let f (x) = sinx + ax + b. Then f(x) = 0 has

only one real root which is positive if $a > 1, b < 0$
only one real root which is negative if $a > 1, b < 0$
only one real root which is negative if $a < - 1, b > 0$
CAN'T SAY ANYTHING

Discuss Question

Answer: Option A. -> only one real root which is positive if

a > 1, b < 0

:
A

f'(x) = - cosx + a, if a $>$ 1,then f(x) entirely increasing. So f(x) =0 has only one real root, which is positive if f(0) $<$ 0 and negative if f(0) $>$ 0.

Similarly when a $<$ -1. Then f(x) entirely decreasing. So f(x) has only one real root which is negative if f(0) $<$ 0 and positive if f(0) $>$ 0

Question 14.

Between any two real roots of the equation $e^{x}$ sin x = 1, the equation $e^{x}$ cos x = - 1 has

Atleast one root
Exactly one root
Atmost one root
No root

Discuss Question

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

Let $\propto, β (\propto< β)$ be any two real roots of
f(x) = e - x - sin x
Then, f $(\propto) = 0 = f (β)$
Moreover, f(x) is continuos and differentiable for $x ε [\propto, β] .$
Hence, from Rolle's thereom, thereom, there exists atleast one x in $\propto, β$ such t
$f^{'} (x) = 0 \Rightarrow - e^{- x} - c o s x = 0 \Rightarrow - e^{- x} (1 + e^{x} c o s x) = 0 \Rightarrow e^{x} c o s x = - 1.$
Hence (a) is the correct answer.

Question 15.

Let f(x) = ${\begin{matrix} 1 + s i n x, x < 0 x^{2} - x + 1, x \geq 0 \end{matrix}$ . Then

f has a local maximum at x = 0
f has a local minimum at x = 0
f is increasing every where
f is decreasing everywhere

Discuss Question

Answer: Option A. -> f has a local maximum at x = 0
:
A

f is continuous at ‘0’ and f' (0-) $>$ 0 and f' ( 0 +) $<$ 0 . Thus f has a local maximum at ‘0’.

Question 16.

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.

$∴ \frac{d y}{d t}$ =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 17.

A ladder20 ft long has one end on the ground and the other end in contact with a vertical wall. The lower end slips along the ground. If the lower end of the ladder is 16 ft away from the wall, upper end is moving $λ$ times as fast as the lower end, then $λ$ is

$\frac{1}{3}$
$\frac{2}{3}$
$\frac{4}{3}$
$\frac{5}{3}$

Discuss Question

Answer: Option C. ->

\frac{4}{3}

:
C
Let OC be the wall. Let AB be the position of the ladder at any time t such that OA =x and OB=y. Length of the ladder AB =20 ft.
In

Δ A O B

A Ladder20 Ft Long Has One End On The Ground And The Other E...

x^{2} + y^{2} = (20)^{2}

\Rightarrow 2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} = 0 ∴ \frac{d y}{d t} = - \frac{x}{y} \frac{d x}{d t} = \frac{- x}{\sqrt{400} - x^{2}} . \frac{d x}{d t} = - \frac{16}{\sqrt{400} - (16)^{2}} . \frac{d x}{d t} = - \frac{4}{3} \frac{d x}{d t}

-ve sign indicates, that when X increases with time, y decreases. Hence, the upper end is moving

\frac{4}{3}

times as fast as the lower end.

Question 18.

The angle of intersection of the curves $y = 2 s i n^{2} x a n d y = c o s 2 x a t x = \frac{π}{6}$ is

$\frac{π}{4}$
$\frac{π}{3}$
$\frac{π}{2}$
$\frac{2 π}{3}$

Discuss Question

Answer: Option B. ->

\frac{π}{3}

:
B and D
We have,

y = s i n^{2} x . . . (1) y = c o s 2 x . . . (2)

And
On differentiating equation (1) w.r.t x, we get

\frac{d y}{d x} = 4 s i n x c o s x {[\frac{d y}{d x}]}_{x - \frac{π}{6}} = 4 (\frac{1}{2}) \frac{\sqrt{3}}{2} = \sqrt{3} = m_{1} (s a y)

On differentiating equation (2) w.r.t x, we get

\frac{d y}{d x} = - 2 s i n 2 x {[\frac{d y}{d x}]}_{x - \frac{π}{6}} = - 2 s i n \frac{π}{3} = - \sqrt{3} = m_{2} (s a y)

Hence, angle between the two curves is

θ = \pm t a n^{- 1} (\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}) = \pm t a n^{- 1} \sqrt{3} = \frac{π}{3} o r \frac{2 π}{3}

Hence (b) is the correct answer.

Question 19.

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

$\frac{1}{2}$
1
$\frac{3}{2}$
2

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

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

Question 20.

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

$\frac{7}{6}$
$\frac{5}{6}$
$\frac{6}{7}$
$1$

Discuss Question

Answer: Option C. ->

\frac{6}{7}

:
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