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

12th Grade > Physics

CALCULUS MCQs

Total Questions : 29 | Page 3 of 3 pages

Question 21. Find the integral of the given function w.r.t x

y = s i n (8 x) + x

−cos 8x8+x22+c
8 cos 8x8+1+c
cos 8x8+x22+c
cos 8x8+1+c

Discuss Question

Answer: Option A. -> −cos 8x8+x22+c
:
A

\int (s i n (8 x) + x) d x = \int s i n (8 x) d x + \int (x) d x

8x = t
8dx = dt
dx =

\frac{d t}{8}

\Rightarrow \int \frac{s i n t d t}{8} + \frac{x^{2}}{2} + c_{1}

(∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1}) + c

\Rightarrow - \frac{c o s t}{8} + c_{o} + \frac{x^{2}}{2} + c

(∵ \int s i n t d t = - c o s t)

- \frac{c o s 8 x}{8} + \frac{x^{2}}{2} + c

(c = c_{0} + c_{1})

Question 22. If

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

3b4at
4at3b
12abt
none of these

Discuss Question

Answer: Option A. -> 3b4at
:
A
This is called implicit differentiation or indirect differentiation. Here both the terms given are differentiatedindependently 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 23. Find the integral of the given function w.r.t - x

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

x22−lnx+1x+c
x22+lnx+1x+c
x22−lnx−1x+c
None of these

Discuss Question

Answer: Option C. -> x22−lnx−1x+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 24. If

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

−12x
−12x32
−1√x2
None of these

Discuss Question

Answer: Option B. -> −12x32
:
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}}}

Question 25. 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.

π100
π200
12
None of these

Discuss Question

Answer: Option B. -> π200
:
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 26. If y=2sinx+4secx-4cotx, then find

\frac{d y}{d x}

2cosx+4tan2x−4cosecx
−2cosx+4sextanx−4cosec2x
2cosx+4secx tanx+4cosec2x
none of these

Discuss Question

Answer: Option C. -> 2cosx+4secx tanx+4cosec2x
:
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 27. Find the derivative of

y = \sqrt{x^{2} + 1}

x√x2+12
x√x2+1
12√x2+1
none of these

Discuss Question

Answer: Option B. -> x√x2+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 28. You are given a rod of length 'L'. The linear mass density varies as

λ^{'}

λ = a + b x

. Here a & b are constants and the mass of the rod increases as x increases. Find the mass of the rod.

You Are Given A Rod Of Length 'L'. The Linear Mass Density V...

aL2
aL+bL22
L2(a+b)2
None of these

Discuss Question

Answer: Option B. -> aL+bL22
:
B
We cannot calculate the mass of the rod by simply multiplying linear mass density '

^{'} λ^{'}

will the length of the rod as

^{'} λ^{'}

is not constant.
Hence we have to divide entire rod length into a number of elements and add the mass of all elements.

M = d m_{1} + d m_{2} + d m_{3} + . . . . . . .

This step can be written in terms of integration.

M = \int d m

If we take an element dx on the rod at a distance x from the left end of the rod.
Mass of this element

d m = λ . d x

d m = (a + b x) d x

Hence total mass

M = \int d m \Rightarrow M = L \int 0 (a + b x) d x

M = L \int 0 a d x + L \int 0 b x d x = a L \int 0 d x + b L \int 0 x d x = a [x]_{0}^{L} + b {[\frac{x^{2}}{2}]}_{0}^{L} \Rightarrow M = a L + \frac{b L^{2}}{2}

Question 29. Find the derivative of the following function with respect to x.

y = \frac{s i n x}{x}

sinx−xcosxx2
cosxx
cosxx2
xcosx−sinxx2

Discuss Question

Answer: Option D. -> xcosx−sinxx2
:
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}}

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