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

12th Grade > Mathematics

APPLICATION OF DERIVATIVES MCQs

Total Questions : 58 | Page 2 of 6 pages

Question 11. 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 12.

f (x) = x^{2} - 4 | x | a n d g (x) = {\begin{matrix} m i n {f (t) : - 6 \leq t \leq x}, x ϵ [- 6, 0] m a x {f (t) : 0 \leq t \leq x}, x ϵ [0, 6] \end{matrix}, t h a n g (x)

Exactly one point of local minima
Exactly one point of local maxima
No point of local maxima but exactly one point of local minima
Neither a point of local maxima nor minima

Discuss Question

Answer: Option D. -> Neither a point of local maxima nor minima
:
D
Bold line represents the graph of y = g(x) clearly g(x) has neither a point of local maxima nor a point of local minima.

F(x) = x2 − 4|x| and g(x) = minf(t) : −6 �...

Question 13.

f (x) = x^{2} - 4 | x | a n d g (x) = {\begin{matrix} m i n 1 f (t) : - 6 \leq t \leq x}, x ϵ [- 6, 0] m a x 1 f (t) : 0 \leq t \leq x}, x ϵ [0, 6] \end{matrix}, t h a n g (x)

Exactly one point of local minima
Exactly one point of local maxima
No point of local maxima but exactly one point of local minima
Neither a point of local maxima nor minima

Discuss Question

Answer: Option D. -> Neither a point of local maxima nor minima
:
D
Bold line represents the graph of y = g(x) clearly g(x) has neither a point of local maxima nor a point of local minima.

F(x) = x2 − 4|x| and g(x) = min1f(t) : −6 �...

Question 14.

f (x) = \frac{x}{l o g_{e} x}, x \neq 1

, is decreasing in interval

(0, e)
(1, e)
(e, ∞)
R

Discuss Question

Answer: Option A. -> (0, e)
:
A
f'(x) =

\frac{l o g_{e} x . 1 - x . \frac{1}{x}}{(l o g_{e} x)^{2}}

\frac{(l o g_{e} x - 1)}{(l o g_{e} x)^{2}} < 0

It is decreasing at (0, e) – {1}

F(x) = xloge x, x ≠ 1, Is Decreasing In Interval

Question 15. Number of possible tangents to the curve

y = c o s (x + y), - 3 π \leq x \leq 3 π

that are parallel to the line x+2y = 0, is

Discuss Question

Answer: Option C. -> 3
:
C
We have, y = cos (x + y)

\frac{d y}{d x} = s i n (x + y) (1 + \frac{d y}{d x})

Since, the tangents are parallel to the line x + 2y = 0

- \frac{1}{2} = - s i n (x + y) (1 - \frac{1}{2}) \Rightarrow s i n (x + y) = 1 \Rightarrow x + y = \frac{π}{2}, \frac{5 π}{2}, \frac{3 π}{2} - 1 \leq y \leq 1.

Hence (c) is the correct answer.

Question 16. The tangent to the curve x = a

\sqrt{c o s 2 θ} c o s θ, y = a \sqrt{c o s 2 θ} s i n θ

at the point corresponding to

θ = \frac{π}{6}

parallel to the x-axis
parallel to the y-axis
parallel to line y = x
3X-4Y+2=0

Discuss Question

Answer: Option A. -> parallel to the x-axis
:
A

\frac{d x}{d θ} = - a \sqrt{c o s 2 θ} s i n θ + \frac{- a c o s θ s i n θ}{\sqrt{c o s 2 θ}} = - a \frac{(c o s 2 θ s i n θ + c o s θ s i n 2 θ)}{\sqrt{c o s 2 θ}} = \frac{- a s i n 3 θ}{\sqrt{c o s 2 θ}} = \frac{d y}{d θ} = a \sqrt{c o s 2 θ} c o s θ - a \frac{s i n θ s i n 2 θ}{\sqrt{c o s 2 θ}} = \frac{a c o s 3 θ}{\sqrt{c o s 2 θ}}

Hence

\frac{d y}{d x} = - c o t 3 θ \Rightarrow \frac{d y}{d x} |_{θ = \frac{π}{6}}

= 0
So the tangent to the curve at

θ = \frac{π}{6}

is parallel to the x-axis.

Question 17. If

f " (x) > 0 \forall x ϵ R

then for any two real numbers

x_{1} a n d x_{2}, (x_{1} \neq x_{2})

f(x1 + x22) > f(x1) + f(x2)2
f(x1 + x22)
f′(x1 + x22) > f′(x1) + f′(x2)2
f′(x1 + x22)

Discuss Question

Answer: Option B. -> f(x1 + x22)
:
B
Let A =

(x_{1}, f (x_{1}))

and B =

(x_{2}, f (x_{2}))

be any two points on the graph of y = f(x).
Since f"(x)

>

0, in the graph of the function tangent will always lie below the curve. Hence chord AB will lie completely above the graph of y = f(x).
Hence

\frac{f (x_{1}) + f (x_{2})}{2} > f (\frac{x_{1} + x_{2}}{2})

Question 18. The distance moved by the particle in time t is given by

x = t^{3} - 12 t^{2} + 6 t + 8

. At the instant when its acceleration is zero, then the velocity is

Discuss Question

Answer: Option B. -> -42
:
B
We have,

x = t^{3} - 12 t^{2} + 6 t + 8

\Rightarrow \frac{d x}{d t} = 3 t^{2} - 24 t + 6 a n d \frac{d^{2} x}{d t^{2}} = 6 t - 24

Now, Acceleration =0

\Rightarrow \frac{d^{2} x}{d t^{2}} = 0 \Rightarrow 6 t - 24 = 0 \Rightarrow t = 4

Att=4, we have
Velocity =

{(\frac{d x}{d t})}_{r - 4} = 3 \times 4^{2} - 24 \times 4 + 6 = - 42

.
Hence (b) is the correct answer.

Question 19. Let S be the set of real values of parameter

λ

for which the equation

f (x) = 2 x^{3} - 3 (2 + λ) x^{2} + 12 λ x

has exactly one local maximum and exactly one local minimum. Then S is a subset of

(−4, ∞)
(−3, 3)
(3, ∞)
R

Discuss Question

Answer: Option C. -> (3, ∞)
:
C

f (x) = 2 x^{3} - 3 (2 + λ) x^{2} + 12 λ x \Rightarrow f^{'} (x) = 6 x^{2} - 6 (2 + λ) x + 12 λ f^{'} (x) = 0 \Rightarrow x = 2, λ

If f(x) has exactly one local maximum and exactly one local minimum, then

λ \neq 2.

Question 20. For what values of x is the rate of increase of

x^{3} - 5 x^{2} + 5 x + 8

is twice rate of increase of x ?

−3,−13
−3,13
3,−13
3,13

Discuss Question

Answer: Option D. -> 3,13
:
D
Let

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

. Then,

\frac{d y}{d x} = (3 x^{2} - 10 x + 5) \frac{d x}{d t} W h e n \frac{d y}{d t} = 2 \frac{d x}{d t}, w e h a v e (3 x^{2} - 10 x + 5) \frac{d x}{d t} = 2 \frac{d x}{d t} \Rightarrow 3 x^{2} - 10 x + 3 = 0 \Rightarrow (3 x - 1) (x - 3) = 0 \Rightarrow x = 3, \frac{1}{3}

.
Hence (d) is the correct answer.

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