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 1 of 3 pages

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 2.

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 ?

$- 3, - \frac{1}{3}$
$- 3, \frac{1}{3}$
$3, - \frac{1}{3}$
$3, \frac{1}{3}$

Discuss Question

Answer: Option D. ->

3, \frac{1}{3}

:
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 3.

The two curves $x^{3} - 3 x y^{2} + 2 = 0 a n d 3 x^{2} y - y^{3} - 2 = 0$

Cut at right angles
Touch each other
Cut at an angle π/3
Cut at an angle π/4

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Answer: Option A. -> Cut at right angles
:
A

x^{3} - 3 x y^{2} + 2 = 0... (1) 3 x^{2} y - y^{3} - 2 = 0... (2)

On differentiating equations (1) and (2) w.r.t x, we obtain

{(\frac{d y}{d x})}_{c 1} = \frac{x^{2} - y^{2}}{2 x y} a n d {(\frac{d y}{d x})}_{c 2} = \frac{- 2 x y}{x^{2} - y^{2}}

Since

m_{1} . m_{2} = - 1

.Therefore the two curves cut at right angles.
Hence (a) is the correct answer.

Question 4.

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

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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 5.

If the tangent to the curve $\sqrt{x} + \sqrt{y} = \sqrt{a}$ at any point on it cuts the axes OX and OY at P and Q respectively, then OP +OQ is

$\frac{a}{2}$
a
2a
4a

Discuss Question

Answer: Option B. -> a
:
B

\sqrt{x} + \sqrt{y} = \sqrt{a} . . . . . (i)

\Rightarrow \frac{1}{2 \sqrt{x}} + \frac{1}{2 \sqrt{y}} \frac{d y}{d x} = 0

If The Tangent To The Curve √x+√y=√a At Any Point On I...

∴ \frac{d y}{d x} = - \frac{\sqrt{y}}{\sqrt{x}}

Equation of tangent at

(x_{1} y_{1}) i s y - y_{1} = - \frac{\sqrt{y_{1}}}{\sqrt{x_{1}}} (x - x_{1})

\Rightarrow \frac{x}{\sqrt{x_{1}}} + \frac{y}{\sqrt{y_{1}}} = \sqrt{a}; \Rightarrow o p = \sqrt{a} \sqrt{x_{1}}, O Q = \sqrt{a} \sqrt{y_{1}} ∴ O P + O Q = \sqrt{a}

Question 6.

If the function $f (x) = 2 x^{3} - 9 a x^{2} + 12 a^{2} x + 1,$ where a > 0, attains its maximum and minimum at p and q respectively such that $p^{2} = q$ , then a equals

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

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Answer: Option C. -> 2
:
C
We have,

f (x) = 2 x^{3} - 9 a x^{2} + 12 a^{2} x + 1 ∴ f (x) = 6 x^{2} - 18 a x + 12 a^{2} = 0 \Rightarrow 6 [x^{2} - 3 a x + 2 a^{2}] = 0 \Rightarrow x^{2} - 3 a x + 2 a^{2} = 0 \Rightarrow x^{2} - 2 a x - a x + 2 a^{2} = 0 \Rightarrow x (x - 2 a) - a (x - 2 a) = 0 \Rightarrow (x - a) (x - 2 a) = 0 \Rightarrow x = a, x = 2 a

Now,

f^{'} (x) = 12 x - 18 a

∴ f^{'} (a) = 12 a - 18 a = - 6 a < 0 ∴ f (x)

will be maximum at x = a
i.e. p = a
Also,

f^{'} (2 a) = 24 a - 18 a = 6 a ∴ f (x)

will be minimum at x = 2a
i.e.q = 2a
Given,

p^{2} = q

\Rightarrow a^{2} = 2 a \Rightarrow a = 2.

Hence (c) is the correct answer.

Question 7.

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

$(x - a)^{2 n + 2}$
$b - 1 - (x + 1 - a)^{2 n + 1}$
$b - (x - a)^{2 n + 2}$
$(x - a)^{2 n + 2} - b .$

Discuss Question

Answer: Option C. ->

b - (x - a)^{2 n + 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 8.

The point in the interval $[0, 2 π]$ where $f (x) = e^{x} s i n x$ has maximum slope, is

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

Discuss Question

Answer: Option B. ->

\frac{π}{2}

:
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 9.

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

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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 10.

The approximate value of square root of 25.2 is

5.01
5.02
5.03
5.04

Discuss Question

Answer: Option B. -> 5.02
:
B

Let f (x) = $\sqrt{x}$
Now, $f (x + δ x) - f (x) = f^{'} (x) . δ x = \frac{δ x}{2 \sqrt{x}}$
We may write, 25.2 = 25 + 0.2
Taking x = 25 and $δ x = 0.2$ We have
$f (25.2) - f (25) = \frac{0.2}{2 \sqrt{25}} = 0.02 ∴ f (25.2) = f (25) + 0.02 = \sqrt{25} + 0.02 = 5.02 \Rightarrow \sqrt{(25.2)} = 5.02$