M.C.Q (1 Marks) - Page 6 - Maths STD 12 Science Questions - Vidyadip

Questions · Page 6 of 6

M.C.Q (1 Marks)

MCQ 2511 Mark

Ramkali is trying to find the solution of the following definite integrals :
(i) $\int_0^{2 \pi} \frac{d x}{e^{\sin x}+1}$
(ii) $\int_0^1 x^2 d x$
(iii) $\int_0^1 e^x d x$
Which of the above integrals solved by using of definite integral properties?

✓
Only (i)
B
Only (i) and (ii)
C
Only (ii) and (iii)
D
None of the above

Answer

Correct option: A.

Only (i)

(a) : (i) Let $I=\int_0^{2 \pi} \frac{d x}{e^{\sin x}+1}$ ....(1)
$\Rightarrow \quad I=\int_0^{2 \pi} \frac{d x}{e^{\sin (2 \pi-x)}+1} \quad\left(\because \int_0^a f(x) d x=\int_0^a f(a-x) d x\right)$
$\Rightarrow \quad I=\int_0^{2 \pi} \frac{d x}{e^{-\sin x}+1} \Rightarrow I=\int_0^{2 \pi} \frac{e^{\sin x}}{e^{\sin x}+1} d x$ ....(2)
Adding (1) and (2), we get
$
2 I=\int_0^{2 \pi} 1 \cdot d x=2 \pi \quad \therefore \quad I=\pi
$
(ii) $\int_0^1 x^2 d x=\left[\frac{x^3}{3}\right]_0^1=\frac{1}{3}$
(iii) $\int_0^1 e^x d x=\left[e^x\right]_0^1=e^1-1$

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MCQ 2521 Mark

Evaluate : $\int_0^2 e^{3-4 x} d x$

A
$\frac{-1}{4}\left(e^5-e^3\right)$
B
$\frac{1}{4}\left(e^5-e^3\right)$
C
$\frac{1}{4}\left(e^{-5}-e^3\right)$
✓
$\frac{-1}{4}\left(e^{-5}-e^3\right)$

Answer

Correct option: D.

$\frac{-1}{4}\left(e^{-5}-e^3\right)$

We have$, \int_0^2 e^{3-4 x} d x=\left[\frac{e^{3-4 x}}{-4}\right]_0^2$
$=-\frac{1}{4}\left[e^{3-8}-e^{3-0}\right]=\frac{-1}{4}\left[e^{-5}-e^3\right]$

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MCQ 2531 Mark

Evaluate: $\int \frac{\sqrt{x}}{\sqrt{a^3-x^3}} d x$

A
$\frac{3}{2} \sin ^{-1}\left(\frac{x}{a}\right)^{3 / 2}+C$
✓
$\frac{2}{3} \sin ^{-1}\left(\frac{x}{a}\right)^{3 / 2}+C$
C
$\frac{2}{3} \cos ^{-1}\left(\frac{x}{a}\right)^{3 / 2}+C$
D
$\frac{3}{2} \cos ^{-1}\left(\frac{x}{a}\right)^{3 / 2}+C$

Answer

Correct option: B.

$\frac{2}{3} \sin ^{-1}\left(\frac{x}{a}\right)^{3 / 2}+C$

Let $I=\int \frac{\sqrt{x}}{\sqrt{a^3-x^3}} d x$
Put $x^{3 / 2}=t \Rightarrow \frac{3}{2} x^{1 / 2} d x=d t$
$\therefore I=\frac{2}{3} \int \frac{d t}{\sqrt{a^3-t^2}}$
$=\frac{2}{3} \int \frac{d t}{\sqrt{\left(a^{3 / 2}\right)^2-t^2}}$
$=\frac{2}{3}\left[\sin ^{-1}\left(\frac{t}{a^{3 / 2}}\right)\right]+C$
$=\frac{2}{3}\left[\sin ^{-1}\left(\frac{x^{3 / 2}}{a^{3 / 2}}\right)\right]+C$
$=\frac{2}{3} \sin ^{-1}\left(\frac{x}{a}\right)^{3 / 2}+C$

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MCQ 2541 Mark

Which of these is equal to $\int_0^1\left\{e^x+\sin \frac{\pi x}{4}\right\} d x$ ?

A
$e+1+\frac{2 \sqrt{2}}{\pi}+\frac{4}{\pi}$
✓
$e-1-\frac{2 \sqrt{2}}{\pi}+\frac{4}{\pi}$
C
$e+1-\frac{2 \sqrt{2}}{\pi}+\frac{4}{\pi}$
D
$e-1+\frac{2 \sqrt{2}}{\pi}-\frac{4}{\pi}$

Answer

Correct option: B.

$e-1-\frac{2 \sqrt{2}}{\pi}+\frac{4}{\pi}$

We have$, \int_0^1\left\{e^x+\sin \frac{\pi x}{4}\right\} d x$
$=\left[e^x\right]_0^1+\frac{4}{\pi}\left[-\cos \frac{\pi}{4} x\right]_0^1$
$=e-1-\frac{4}{\sqrt{2} \pi}+\frac{4}{\pi}$
$=e-1-\frac{2 \sqrt{2}}{\pi}+\frac{4}{\pi}$

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MCQ 2551 Mark

Which of these is equal to $\int x^2(a x+b)^{-2} d x$, where $C$ is the constant of integration?

A
$\frac{1}{a^3}\left(a x+b-\frac{b^2}{a x+b}-2 b \log (a x+b)\right)+C$
B
$\frac{1}{a^3}\left(a x+b+\frac{b^2}{a x+b}-2 b \log (a x+b)\right)+C$
C
$\frac{1}{a^3}\left(a x+b+\frac{b^2}{a x+b}+2 b \log (a x+b)\right)+C$
D
$\frac{1}{a^3}\left(a x+b-\frac{b^2}{a x+b}+2 b \log (a x+b)\right)+C$

Answer

14. (a) : Let $I=\int \frac{x^2}{(a x+b)^2} d x$
Put $a x+b=t \Rightarrow d x=\frac{1}{a} d t$
$
\begin{aligned}
\therefore \quad I & =\frac{1}{a^3} \int \frac{(t-b)^2}{t^2} d t=\frac{1}{a^3} \int\left(1+\frac{b^2}{t^2}-\frac{2 b}{t}\right) d t \\
& =\frac{1}{a^3}\left(t-\frac{b^2}{t}-2 b \log t\right)+C \\
& =\frac{1}{a^3}\left(a x+b-\frac{b^2}{a x+b}-2 b \log (a x+b)\right)+C
\end{aligned}
$

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MCQ 2561 Mark

Evaluate: $\int \frac{\cot x}{\sqrt[3]{\sin x}} d x$

✓
$\frac{-3}{\sqrt[3]{\sin x}}+C$
B
$\frac{-2}{\sin ^3 x}+C$
C
$\frac{3}{\sin ^{1 / 3} x}+C$
D
None of these

Answer

Correct option: A.

$\frac{-3}{\sqrt[3]{\sin x}}+C$

Let $I=\int \frac{\cot x}{\sqrt[3]{\sin x}} d x=\int \frac{\cos x}{\sin ^{1 / 3} x \cdot \sin x} d x$
$=\int \frac{\cos x}{\sin ^{4 / 3} x} d x=\int \sin ^{-4 / 3} x \cdot \cos x d x$
Put $\sin x=t \Rightarrow \cos x d x=d t$
$\Rightarrow I=\int t^{-4 / 3} d t=\frac{t^{-1 / 3}}{-1 / 3}+C$
$=\frac{-3}{\sqrt[3]{\sin x}}+C$

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MCQ 2571 Mark

Evaluate: $\int_0^{\pi / 4} \tan ^3 x d x$

A
$(1-\log 2)$
B
$(1+\log 2)$
✓
$\frac{1}{2}(1-\log 2)$
D
$\frac{1}{2}(1+\log 2)$

Answer

Correct option: C.

$\frac{1}{2}(1-\log 2)$

$\text { : Let } I= \int_0^{\pi / 4} \tan ^3 x d x=\int_0^{\pi / 4}\left(\sec ^2 x-1\right) \tan x d x$
$ =\int_0^{\pi / 4} \sec ^2 x \tan x d x-\int_0^{\pi / 4} \tan x d x$
Put $\tan x=t$ in first integral
$\Rightarrow \sec ^2 x d x=d t$
When
$x=0$
$\Rightarrow t=0$
$\therefore x=\pi / 4$
$\Rightarrow t=1$
$I=\int_0^1 t d t-\int_0^{\pi / 4} \tan x d x=\left[\frac{t^2}{2}\right]_0^1-[\log |\sec x|]_0^{\pi / 4}$
$=\left(\frac{1}{2}-0\right)-\log \left|\sec \frac{\pi}{4}\right|+\log |\sec 0|=\frac{1}{2}(1-\log 2)$

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MCQ 2581 Mark

Evaluate: $\int \sqrt{(x-3)(5-x)} d x$

A
$\frac{1}{2}(x-4) \sqrt{(x-3)(5-x)}+\frac{1}{2} \cos ^{-1}(x-4)+C$
✓
$\frac{1}{2}(x-4) \sqrt{(x-3)(5-x)}+\frac{1}{2} \sin ^{-1}(x-4)+C$
C
$\frac{1}{2} \sqrt{(x-3)(5-x)}+\frac{1}{2} \sin ^{-1}(x-4)+C$
D
None of these

Answer

Correct option: B.

$\frac{1}{2}(x-4) \sqrt{(x-3)(5-x)}+\frac{1}{2} \sin ^{-1}(x-4)+C$

Let $ I=\int \sqrt{(x-3)(5-x)} d x=\int \sqrt{-x^2+8 x-15} d x$
$\Rightarrow I=\int \sqrt{-\left\{x^2-8 x+16-16+15\right\}} d x$
$\Rightarrow I=\int \sqrt{-\left\{(x-4)^2-1^2\right\}} d x=\int \sqrt{1^2-(x-4)^2} d x$
$\Rightarrow I=\frac{1}{2}(x-4) \sqrt{(x-3)(5-x)}+\frac{1}{2} \sin ^{-1}\left(\frac{x-4}{1}\right)+C$

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MCQ 2591 Mark

Evaluate: $\int \sin ^3 x \cos ^3 x d x$

A
$\frac{-1}{32}\left\{\frac{-3}{2} \cos 2 x+\frac{1}{6} \cos 6 x\right\}+C$
B
$\frac{1}{32}\left\{\frac{-3}{2} \cos 6 x+\frac{1}{6} \cos 2 x\right\}+C$
✓
$\frac{1}{32}\left\{\frac{-3}{2} \cos 2 x+\frac{1}{6} \cos 6 x\right\}+C$
D
None of these

Answer

Correct option: C.

$\frac{1}{32}\left\{\frac{-3}{2} \cos 2 x+\frac{1}{6} \cos 6 x\right\}+C$

Let $I=\int \sin ^3 x \cos ^3 x d x$
$\Rightarrow I=\frac{1}{8} \int(2 \sin x \cos x)^3 d x$
$\Rightarrow I=\frac{1}{8} \int \sin ^3 2 x d x $
$\Rightarrow I=\frac{1}{8} \int \frac{3 \sin 2 x-\sin 6 x}{4} d x$
$\Rightarrow I=\frac{1}{32}\left\{-\frac{3}{2} \cos 2 x+\frac{1}{6} \cos 6 x\right\}+C$

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MCQ 2601 Mark

Which of these is equal to $\int 2^{(x+3)} d x$, where $C$ is the constant of integration?

A
$\frac{2^x}{\log 2}+C$
B
$\frac{2^3}{\log 2}+C$
✓
$\frac{2^{(x+3)}}{\log 2}+C$
D
$\frac{2^{(x-3)}}{\log 2}+C$

Answer

Correct option: C.

$\frac{2^{(x+3)}}{\log 2}+C$

(c): $\int 2^{(x+3)} d x=\int 2^x \cdot 2^3 d x=8 \int 2^x d x$
$=8 \cdot \frac{2^x}{\log 2}+C=\frac{2^{(x+3)}}{\log 2}+C$

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MCQ 2611 Mark

Evaluate: $\int 2^{2^{2^x}} 2^{2^x} 2^x d x$

✓
$\frac{1}{(\log 2)^3} 2^{2^{2^x}}+C$
B
$\frac{1}{(\log 2)^3} 2^{2^x}+C$
C
$\frac{1}{(\log 2)^2} 2^{2^x}+C$
D
$\frac{1}{(\log 2)^4} 2^{2^{2^x}}+C$

Answer

Correct option: A.

$\frac{1}{(\log 2)^3} 2^{2^{2^x}}+C$

(a) : Let $I=\int 2^{2^{2^x}} 2^{2^x} 2^x d x$
Let $2^{2^{2^x}}=t \Rightarrow 2^{2^{2^x}} 2^{2^x} 2^x(\log 2)^3 d x=d t$
$
\Rightarrow I=\int \frac{1}{(\log 2)^3} d t=\frac{1}{(\log 2)^3} t+C=\frac{1}{(\log 2)^3} 2^{2^{2^x}}+C
$

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MCQ 2621 Mark

Evaluate: $\int_2^4 \frac{\left(x^2+x\right)}{\sqrt{2 x+1}} d x$

A
$57-5 \sqrt{5}$
B
$\frac{57-\sqrt{5}}{5}$
C
$\frac{57+5 \sqrt{5}}{5}$
✓
$\frac{57-5 \sqrt{5}}{5}$

Answer

Correct option: D.

$\frac{57-5 \sqrt{5}}{5}$

We have, $\int_2^4 \frac{\left(x^2+x\right)}{\sqrt{2 x+1}} d x$
Integrating by parts, we get
$\int_2^4 \frac{\left(x^2+x\right)}{\sqrt{2 x+1}} d x=\left[\left(x^2+x\right) \cdot \sqrt{2 x+1}\right]_2^4-\int_2^4(2 x+1) \cdot \sqrt{2 x+1} d x$
$=(60-6 \sqrt{5})-\int_2^4(2 x+1)^{3 / 2} d x$
$=(60-6 \sqrt{5})-\frac{1}{5} \cdot\left[(2 x+1)^{5 / 2}\right]_2^4$
$=(60-6 \sqrt{5})-\left(\frac{243}{5}-5 \sqrt{5}\right)$
$=\left(\frac{57}{5}-\sqrt{5}\right)=\left(\frac{57-5 \sqrt{5}}{5}\right)$

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MCQ 2631 Mark

Evaluate: $\int \frac{\cos x}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^3} d x$

A
$\frac{2}{\cos \frac{x}{2}+\sin \frac{x}{2}}+C$
B
$\frac{-2}{\cos \frac{x}{2}-\sin \frac{x}{2}}+C$
✓
$\frac{-2}{\cos \frac{x}{2}+\sin \frac{x}{2}}+C$
D
$\frac{2}{\cos \frac{x}{2}-\sin \frac{x}{2}}+C$

Answer

Correct option: C.

$\frac{-2}{\cos \frac{x}{2}+\sin \frac{x}{2}}+C$

We have,
$ \int \frac{\cos x}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^3} d x=\int \frac{\cos ^2(x / 2)-\sin ^2(x / 2)}{\{\cos (x / 2)+\sin (x / 2)\}^3} d x$
$\text { Put } t=\cos \frac{x}{2}+\sin \frac{x}{2} $
$\Rightarrow 2 d t=\left[\cos \frac{x}{2}-\sin \frac{x}{2}\right] d x$
$\Rightarrow \int \frac{\cos (x / 2)-\sin (x / 2)}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^2} d x=2 \int \frac{1}{t^2} d t$
$\quad=\frac{-2}{t}+C=\frac{-2}{\cos (x / 2)+\sin (x / 2)}+C$

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MCQ 2641 Mark

Which of these is equal to $\int x e^{x^2} d x$, where $C$ is the constant of integration?

A
$-\frac{e^{x^2}}{2}+C$
✓
$\frac{e^{x^2}}{2}+C$
C
$\frac{e^x}{2}+C$
D
$-\frac{e^x}{2}+C$

Answer

Correct option: B.

$\frac{e^{x^2}}{2}+C$

(b) : Let $I=\int x e^{x^2} d x$
Put $x^2=t \Rightarrow 2 x d x=d t \Rightarrow x d x=\frac{d t}{2}$
$\therefore \quad I=\frac{1}{2} \int e^t d t=\frac{e^t}{2}+C=\frac{e^{t^2}}{2}+C$

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MCQ 2651 Mark

Evaluate : $\int_2^4 \frac{x}{x^2+1} d x$

✓
$\frac{1}{2} \log \left(\frac{17}{5}\right)$
B
$\frac{1}{2} \log \left(\frac{5}{17}\right)$
C
$\log \left(\frac{17}{5}\right)$
D
$\log \left(\frac{5}{17}\right)$

Answer

Correct option: A.

$\frac{1}{2} \log \left(\frac{17}{5}\right)$

Let $I=\int_2^4 \frac{x}{x^2+1} d x$
Put $x^2+1=t $
$\Rightarrow 2 x d x=d t $
$\Rightarrow x d x=\frac{1}{2} d t$
Also, $x=2 $
$\Rightarrow t=5$ and $x=4 $
$\Rightarrow t=17$
$\therefore I=\frac{1}{2} \int_5^{17} \frac{d t}{t}=\frac{1}{2}[\log t]_5^{17}$
$=\frac{1}{2}[\log 17-\log 5]=\frac{1}{2} \log \left(\frac{17}{5}\right)$

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MCQ 2661 Mark

Find the value of $\int \frac{\sin ^2 x-\cos ^2 x}{\sin ^2 x \cos ^2 x} d x$.

A
$\tan x-\cot x+C$
B
$-\tan x+\cot x+C$
✓
$\tan x+\cot x+C$
D
$-\tan x-\cot x+C$

Answer

Correct option: C.

$\tan x+\cot x+C$

We have$, \int \frac{\sin ^2 x-\cos ^2 x}{\sin ^2 x \cos ^2 x} d x$
$=\int\left(\sec ^2 x-\operatorname{cosec}^2 x\right) d x$
$=\tan x+\cot x+C$

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MCQ 2671 Mark

Evaluate: $\int\left(2^x+2^{-x}\right)^2 d x$

A
$\frac{1}{2 \log 2}\left(2^{2 x}-2^{-2 x}\right)+C$
✓
$\frac{1}{2 \log 2}\left(2^{2 x}-2^{-2 x}\right)+2 x+C$
C
$\frac{1}{2 \log 2}\left(2^{2 x}+2^{-2 x}\right)+2 x+C$
D
$\frac{1}{2 \log 2}\left(2^{2 x}+2^{-2 x}\right)+C$

Answer

Correct option: B.

$\frac{1}{2 \log 2}\left(2^{2 x}-2^{-2 x}\right)+2 x+C$

We have$, \int\left(2^x+2^{-x}\right)^2 d x=\int\left(2^{2 x}+2^{-2 x}+2\right) d x$
$=\frac{2^{2 x}}{(\log 2) \times 2}+\frac{2^{-2 x}}{(\log 2)(-2)}+2 \cdot x+C$
$=\frac{1}{2 \log 2}\left(2^{2 x}-2^{-2 x}\right)+2 x+C$

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MCQ 2681 Mark

Evaluate:$\int\left(3 \sin x-2 \cos x+4 \sec ^2 x-5 \operatorname{cosec}^2 x\right) d x$

✓
$-3 \cos x-2 \sin x+4 \tan x+5 \cot x+C$
B
$3 \cos x+2 \sin x+4 \tan x+5 \cot x+C$
C
$-3 \cos x+2 \sin x-4 \tan x-5 \cot x+C$
D
$-3 \cos x-2 \sin x-4 \tan x-5 \cot x+C$

Answer

Correct option: A.

$-3 \cos x-2 \sin x+4 \tan x+5 \cot x+C$

Let $I=\int\left(3 \sin x-2 \cos x+4 \sec ^2 x-5 \operatorname{cosec}^2 x\right) d x$
$\Rightarrow I=3 \int \sin x d x-2 \int \cos x d x+4 \int \sec ^2 x d x-5 \int \operatorname{cosec}^2 x d x$
$\Rightarrow I=-3 \cos x-2 \sin x+4 \tan x+5 \cot x+C$

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M.C.Q (1 Marks) - Page 6 - Maths STD 12 Science Questions - Vidyadip