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

12th Grade > Mathematics

APPLICATION OF DERIVATIVES MCQs

Total Questions : 58 | Page 5 of 6 pages

Question 41. A function f such that

f (a) = f^{''} (a) = . . . . . . f^{2 n} (a) = 0

and f has a local maximum value b at x = a, if f (x) is

(x−a)2n+2
b−1−(x+1−a)2n+1
b−(x−a)2n+2
(x−a)2n+2−b.

Discuss Question

Answer: Option C. -> b−(x−a)2n+2
:
C
For local maximum or local minimum odd derivative must be equal to zero.
For local maxima, even derivative must be negative.
Since maximum value at x = a is b.

∴ f (x) = b - (x - a)^{2 n + 2} (∵ f^{2 n + 2} (a) = - v e)

Hence (c) is the correct answer.

Question 42. The point in the interval

[0, 2 π]

where

f (x) = e^{x} s i n x

has maximum slope, is

π4
π2
π
3π2

Discuss Question

Answer: Option B. -> π2
:
B
We have,

f^{'} (x) = e^{x} + c o s x + s i n x e^{x} A n d f^{'} (x) = - s i n x e^{x} + c o s x e^{x} + c o s x e^{x} + s i n x c o s x e^{x} . N o w, f^{'} (x) = 2 c o s x c o s x e^{x} = 0 \Rightarrow c o s x = 0 \Rightarrow x = \frac{π}{2} . A l s o, f^{'} (x) = - 2 s i n x e^{x} + 2 c o s x e^{x} = - v e

∴

Slope is maximum at

x = \frac{π}{2} .

Hence (b) is the correct answer.

Question 43. 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 44. Let f (x) = sinx + ax + b. Then f(x) = 0 has

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

Discuss Question

Answer: Option A. -> only one real root which is positive if a > 1, b
:
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)

>

Question 45. 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.

Question 46. 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 47. The number of values of x where the function f(x) = 2 (cos 3x + cos

\sqrt{3 x}

attains its maximum, is

1
2
0
Infinite

Discuss Question

Answer: Option A. -> 1
:
A
We have,

f (x) = 2 (c o s 3 x + c o s \sqrt{3} x) = 4 c o s (\frac{3 + \sqrt{3}}{2}) x c o s (\frac{3 - \sqrt{3}}{2}) x ⩽ 4

and it is equal to 4 when both cos

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

Are equal to 1 for a value of x. This is possible only when x = 0.
Hence (a) is the correct answer.

Question 48. 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 49. 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} ∣ ∣∣ ∣,

where

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 50. The slope of the tangent to the curve