Calculus(11th And 12th > Physics ) Questions and answers for exam Preparation

11th And 12th > Physics

CALCULUS MCQs

Total Questions : 30 | Page 1 of 3 pages

Question 1.

Find the derivative of $y = \sqrt{x^{2} + 1}$

$\frac{x \sqrt{x^{2} + 1}}{2}$
$\frac{x}{\sqrt{x^{2} + 1}}$
$\frac{1}{2 \sqrt{x^{2} + 1}}$
none of these

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Answer: Option B. ->

\frac{x}{\sqrt{x^{2} + 1}}

:
B

$y = \sqrt{x^{2} + 1}$
$x^{2} + 1 = u . . . . (1)$
$\Rightarrow \frac{d u}{d x} = 2 x . . . . (2)$
$\Rightarrow y = \sqrt{u}$
Now, $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$
= $\frac{1}{2 \sqrt{u}} \times \frac{d u}{d x}$
= $\frac{1}{2 \sqrt{x^{2} + 1}} \times 2 x$ From (1) & (2)
$\frac{d y}{d x} = \frac{x}{\sqrt{x^{2} + 1}}$

Question 2.

If $x = a t^{4}, y = b t^{3}, t h e n \frac{d y}{d x} = ?$

$\frac{3 b}{4 a t}$
$\frac{4 a t}{3 b}$
12abt
none of these

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Answer: Option A. ->

\frac{3 b}{4 a t}

:
A

This is called implicit differentiation or indirect differentiation. Here both the terms given are differentiated independently w.r.t a third variable and then combined together. Differentiating x and y w.r.t 't'
$\frac{d x}{d t} = 4 a t^{3}, \frac{d y}{d t} = 3 b t^{2}$
Diving $\frac{d y}{d t} b y \frac{d x}{d t}, \frac{d y}{d x} = \frac{d y}{d t} \times \frac{d t}{d x} = \frac{3 b t^{2}}{4 a t^{3}} = \frac{3 b}{4 a t}$

Question 3.

Find the derivative of the following function with respect to x. $y = \frac{s i n x}{x}$

$\frac{s i n x - x c o s x}{x^{2}}$
$\frac{c o s x}{x}$
$\frac{c o s x}{x^{2}}$
$\frac{x c o s x - s i n x}{x^{2}}$

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Answer: Option D. ->

\frac{x c o s x - s i n x}{x^{2}}

:
D

$y = \frac{s i n x}{x}$
$\frac{d y}{d x} = \frac{x \frac{d}{d x} (s i n x) - s i n x (\frac{d x}{d x})}{x^{2}}$
= $\frac{x c o s x - s i n x}{x^{2}}$

Question 4.

If y=2sinx+4secx-4cotx, then find $\frac{d y}{d x}$

$2 c o s x + 4 t a n^{2} x - 4 c o s e c x$
$- 2 c o s x + 4 s e x t a n x - 4 c o s e c^{2} x$
$2 c o s x + 4 s e c x t a n x + 4 c o s e c^{2} x$
none of these

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Answer: Option C. ->

2 c o s x + 4 s e c x t a n x + 4 c o s e c^{2} x

:
C
y= 2sinx + 4secx - 4cotx

\frac{d y}{d x} = \frac{d (2 s i n x + 4 s e c x - 4 c o t x)}{d x}

\frac{d (2 s i n x)}{d x} + \frac{d (4 s e c x)}{d x} + \frac{d (- 4 c o t x)}{d x}

\frac{2 d s i n x}{d x} + \frac{4 d (s e c x)}{d x} - \frac{4 d (c o t x)}{d x}

= 2 c o s x + 4 s e c x t a n x - 4 (- c o s e c^{2} x)

(s i n c e \frac{d s i n x}{d x} = c o s x; \frac{d s e c x}{d x} = s e c x t a n x; \frac{d c o t x}{d x} = - c o s e c^{2} x)

\frac{d y}{d x} = 2 c o s x + 4 s e c x t a n x + 4 c o s e c^{2} x

Question 5.

If $y = x^{2} + 5 x^{\frac{3}{2}} + \frac{2}{x}, t h e n \frac{d y}{d x} =$ ?

$2 x + \frac{15}{2 \sqrt{x}} - \frac{2}{x^{2}}$
$2 x + \frac{10}{3} x + 2 l n x$
$2 x + \frac{15}{\sqrt{x}} + 2 l n x$
none of these

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Answer: Option D. -> none of these
:
D

Differentiating both sides w.r.t. 'x'
$\frac{d y}{d x} = \frac{d}{d x} [x^{2} + 5 x^{\frac{3}{2}} + \frac{2}{x}]$
Using the linearity property of the differentiation, we get = $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [5 x^{\frac{3}{2}}] + \frac{d}{d x} [\frac{2}{x}]$
Taking constants out, = $\frac{d}{d x} [x^{2}] + 5 \frac{d [x^{\frac{3}{2}}]}{d x} + 2 \frac{d}{d x} [\frac{1}{x}]$
= $2 x + 5. \frac{3}{2} x^{\frac{1}{2}} + 2. (- 1) x^{- 2} = 2 x + \frac{15}{2} x^{\frac{1}{2}} - \frac{2}{x^{2}}$

Question 6.

If $y = x^{5}, t h e n \frac{d y}{d x}$ =?

$5 x^{4}$
$x^{5}$
$\frac{x^{6}}{6}$
none of these

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Answer: Option A. ->

5 x^{4}

:
A

Given $y = x^{5}$
Differentiating both sides w.r.t. 'x', Using $\frac{d x^{n}}{d x} = n . x^{n - 1}$
$\frac{d y}{d x} = \frac{d}{d x} [x^{5}] = 5 x^{5 - 1} = 5 x^{4}$

Question 7.

A curve is represented by y=sin x. If x is changed from $\frac{π}{3} t o \frac{π}{3} + \frac{π}{100}$ , find approximately the change in y.

$\frac{π}{100}$
$\frac{π}{200}$
$\frac{1}{2}$
None of these

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Answer: Option B. ->

\frac{π}{200}

:
B

y= sinx
$\frac{d y}{d x}$ is instantaneous rate of change = cosx
Question is asking change in y i.e. $△$ y
given $△$ x i.e., $\frac{π}{3} t o \frac{π}{3} + \frac{π}{100}$
$\Rightarrow △ x = \frac{π}{100}$
Since rate is asked to be assumed approximately the same as instantaneous rate so
$\frac{d y}{d x} = \frac{△ y}{△ x}$
$\frac{d y}{d x} a t x = \frac{π}{3} = c o s \frac{π}{3} = \frac{1}{2}$
$\frac{1}{2} = \frac{△ y}{\frac{π}{100}}, △ y = \frac{π}{200}$

Question 8.

Find the integral of the given function w.r.t - x

$y = x - \frac{1}{x} + \frac{1}{x^{2}}$

$\frac{x^{2}}{2} - l n x + \frac{1}{x} + c$
$\frac{x^{2}}{2} + l n x + \frac{1}{x} + c$
$\frac{x^{2}}{2} - l n x - \frac{1}{x} + c$
None of these

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Answer: Option C. ->

\frac{x^{2}}{2} - l n x - \frac{1}{x} + c

:
C

$\int y d x = \int x d x - \int \frac{1}{x} d x + \int \frac{1}{x^{2}} d x$

= $\frac{x^{2}}{2} - l n x - \frac{x^{- 2 + 1}}{- 2 + 1} + c$

= $\frac{x^{2}}{2} - l n x - \frac{1}{x} + c$

Question 9.

If $y = \frac{1}{\sqrt{x}}, t h e n \frac{d y}{d x}$ =?

$\frac{- 1}{2 x}$
$\frac{- 1}{2 x^{\frac{3}{2}}}$
$\frac{- 1 \sqrt{x}}{2}$
None of these

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Answer: Option B. ->

\frac{- 1}{2 x^{\frac{3}{2}}}

:
B

Differentiating both sides w.r.t. x,
$\frac{d y}{d x} = \frac{d [(x)^{\frac{- 1}{2}}]}{d x} = - \frac{1}{2} x^{\frac{- 3}{2}} = - \frac{1}{2 x^{\frac{3}{2}}}$