M.C.Q (1 Marks)

MCQ 11 Mark

The value of $\int \frac{\sec x}{\operatorname{cosec}^2 x} d x$

A
$\sec x-x+c$
B
$\sec x \tan x+c$
C
$\tan x+x^2+c$
✓
$\tan x-x+c$

Answer

Correct option: D.

$\tan x-x+c$

(D)
Correct option is (D) $\tan x-x+c$

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MCQ 21 Mark

$\int \cos ^2 d x$ is equal to:

✓
$\frac{x}{2}+\frac{1}{4} \sin 2 x+C$
B
$x^2+\frac{1}{4} \sin 2 x+ C$
C
$\frac{x}{4}+\frac{1}{2} \sin x+C$
D
$\frac{x^2}{2}+\frac{1}{2} \sin ^2 x+ C$

Answer

Correct option: A.

$\frac{x}{2}+\frac{1}{4} \sin 2 x+C$

(A)
Let $I =\int \cos ^2 d x$
$
\begin{array}{l}
=\int \frac{1+\cos 2 x}{2} d x \\
=\frac{1}{2} x+\frac{1}{4} \sin 2 x+C
\end{array}
$
Hence option (A) is correct.

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MCQ 31 Mark

Integration of $\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)$ with respect to $x$ :

A
$\frac{1}{3} x^{\frac{1}{3}}+2 x^{\frac{1}{2}}+ C$
B
$\frac{2}{3} x^{\frac{2}{3}}+\frac{1}{2} x^2+ C$
✓
$\frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}+ C$
D
$\frac{3}{2} x^{\frac{3}{2}}+\frac{1}{2} x^{\frac{1}{2}}+ C$

Answer

Correct option: C.

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

(C)
Let $I =\int\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right) d x$
$
\begin{array}{l}
=\int\left(x^{\frac{1}{2}}+x^{-\frac{1}{2}}\right) d x \\
=\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+C
\end{array}
$
$
=\frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}+C
$
Hence correct option is (C).

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MCQ 41 Mark

Integration of function $\frac{x}{e^{x^2}}$ with respect to $x$ is equal to :

A
$\frac{1}{2 e^{x^2}}+ C$
B
$\frac{2}{e^{x^2}}+ C$
C
$\frac{-2}{e^{x^2}}+ C$
✓
$\frac{-1}{2 e^{x^2}}+ C$

Answer

Correct option: D.

$\frac{-1}{2 e^{x^2}}+ C$

(D)
$\int \frac{x}{e^{x^2}} d x$
Let
$x^2=t \quad \therefore 2 x d x=d t$
$\Rightarrow \quad x d x=\frac{1}{2} d t$
$
\begin{aligned}
\int \frac{\frac{1}{2} d t}{e^t} & =\frac{1}{2} \int e^{-t} d t \\
& =\frac{-1}{2} e^{-t}+C \\
& =\frac{-1}{2 e^t}+C=\frac{-1}{2 e^{x^2}}+C
\end{aligned}
$
Hence correct option is (D).

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MCQ 51 Mark

The value of $\int \log x d x$ :

A
$\log x-x+c$
B
$1+\log x+c$
✓
$x(\log x-1)+c$
D
$x(\log x+1)+c$

Answer

Correct option: C.

$x(\log x-1)+c$

(C)
$\int_{\text {II }}^1 1 \cdot \log x \cdot d x=\log x \cdot x-\int \frac{1}{x} \times x d x$
$=x \log x-x=x(\log x-1)+c$

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MCQ 61 Mark

$\int_0^1 \log \left(\frac{1}{x}-1\right) d x$ is equal to :

✓
$0$
B
1
C
$\log 2$
D
$\log \frac{3}{2}$

Answer

Correct option: A.

$0$

(A)Let
$
\begin{aligned}
I & =\int_0^1 \log \left(\frac{1}{x}-1\right) d x=\int_0^1 \log \left(\frac{1-x}{x}\right) d x \ldots(1) \\
I & =\int_0^1 \log \left(\frac{1-x}{x}\right) d x \\
& =\int_0^1 \log \left(\frac{1-(1-x)}{(1-x)}\right) d x \text { From Property } P_5
\end{aligned}
$
$
\begin{aligned}
I & =\int_0^1 \log \left(\frac{1-1+x}{(1-x)}\right) d x=\int_0^1 \log \left(\frac{x}{(1-x)}\right) d x \\
I & =-\int_0^1 \log \left(\frac{1-x}{x}\right) d x \\
I & =-I \\
2 I & =0 \therefore I=0
\end{aligned}
$
Hence option (A) is correct.

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MCQ 71 Mark

$\int\left(\sin ^{-1} x+\cos ^{-1} x\right) d x$ is equal to :

A
$\frac{\pi}{2}$
B
$0$
✓
$\frac{\pi}{2} x+ C$
D
1

Answer

Correct option: C.

$\frac{\pi}{2} x+ C$

(C)
$
\begin{aligned}
\int\left(\sin ^{-1} x+\cos ^{-1} x\right) d x=\int \frac{\pi}{2} \cdot d x & =\frac{\pi}{2} \int d x \\
& =\frac{\pi}{2} \cdot x+C
\end{aligned}
$
Hence option (A) is correct.

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MCQ 81 Mark

Find the value of $\int \sin ^2 x d x$ :

A
$\frac{x}{2}+\frac{\sin 2 x}{4}+C$
✓
$\frac{x}{2}-\frac{\sin 2 x}{4}+C$
C
$\frac{x}{2}+\frac{\cos 2 x}{4}+ C$
D
$\frac{x}{2}-\frac{\cos 2 x}{4}+C$

Answer

Correct option: B.

$\frac{x}{2}-\frac{\sin 2 x}{4}+C$

(B)
$
\begin{array}{l}
\int \sin ^2 x d x=\int\left(\frac{1-\cos 2 x}{2}\right) d x \\
\Rightarrow \quad \frac{1}{2} \int d x-\frac{1}{2} \int \cos 2 x d x \\
\Rightarrow \quad \frac{1}{2} x-\frac{1}{2} \frac{\sin 2 x}{2}+C \\
=\frac{1}{2} x-\frac{1}{4} \sin 2 x+C
\end{array}
$
Hence option (B) is correct.

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MCQ 91 Mark

$\int \frac{d x}{x^2-9}$ equal to :

✓
$\frac{1}{6} \log \left(\frac{x-3}{x+3}\right)+C$
B
$\frac{1}{6} \log \left(\frac{x+3}{x-3}\right)+C$
C
$\frac{1}{3} \log \left(\frac{x-3}{x+3}\right)+C$
D
$\frac{1}{3} \log \left(\frac{x+3}{x-3}\right)+ C$

Answer

Correct option: A.

$\frac{1}{6} \log \left(\frac{x-3}{x+3}\right)+C$

(A)
$
\begin{aligned}
I= & \int \frac{d x}{x^2-9}=\int \frac{d x}{(x)^2-(3)^2} \\
& \int \frac{d x}{x^2-a^2}=\frac{1}{2 a} \log \left(\frac{x-a}{x+a}\right)+C \\
\therefore \quad I & =\frac{1}{2 \times 3} \log \left(\frac{x-3}{x+3}\right)+C \\
I & =\frac{1}{6} \log \left(\frac{x-3}{x+3}\right)+C
\end{aligned}
$
Hence option (A) is correct.

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MCQ 101 Mark

$\int \frac{d x}{x \log _e x}$ is equal to :

A
$|\log x|+C$
B
$\frac{1}{x}+ C$
✓
$\log |\log x|+C$
D
$-\frac{1}{x^2}+C$

Answer

Correct option: C.

$\log |\log x|+C$

(C)
$\int \frac{d x}{x \log _e x}$
Let $\log _{ e } x=t$
$
\begin{aligned}
\frac{1}{x} d x & =d t \\
\int \frac{d t}{t} & =\log |t|+C \\
& =\log |\log x|+C
\end{aligned}
$
Hence option (C) is correct.

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MCQ 111 Mark

$\int_0^{\pi / 4} \tan ^2 x d x$ is equal to :

✓
$1-\frac{\pi}{4}$
B
$1+\frac{\pi}{4}$
C
$-1+\frac{\pi}{4}$
D
$-1-\frac{\pi}{4}$

Answer

Correct option: A.

$1-\frac{\pi}{4}$

(A)
$
\begin{aligned}
\int_0^{\pi / 4} \tan ^2 x d x & =\int_0^{\pi / 4}\left(\sec ^2 x-1\right) d x \\
& =\int_0^{\pi / 4} \sec ^2 x d x-\int_0^{\pi / 4} d x \\
& =(\tan x)_0^{\frac{\pi}{4}}-(x)_0^{\frac{\pi}{4}} \\
& =\left(\tan \frac{\pi}{4}-\tan 0\right)-\left(\frac{\pi}{4}-0\right) \\
& =\left(\tan \frac{\pi}{4}-0-\frac{\pi}{4}\right)=1-\frac{\pi}{4}
\end{aligned}
$
Hence option (A) is correct.

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MCQ 121 Mark

$\int \frac{d x}{1-e^x}$ is equal to :

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

Answer

Correct option: A.

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

(A)
$\int \frac{d x}{1-e^x}$
Let $1-e^x=t$
$
\begin{aligned}
\therefore-e^x d x & =d t \\
d x & =\frac{d t}{-e^x}=\frac{d t}{t-1}
\end{aligned}
$
$
\begin{array}{l}
\int \frac{d t}{t(t-1)} \\
\begin{aligned}
\Rightarrow \int\left(\frac{1}{t-1}-\frac{1}{t}\right) d t =\log (t-1)-\log t+C \\
=\log \frac{(t-1)}{t}+C \\
=\log \left(\frac{1-e^x-1}{1-e^x}\right)+C \\
=\log \left(\frac{e^x}{e^x-1}\right)=-\log \left(\frac{e^x-1}{e^x}\right)+C \\
=-\log \left(1-e^{-x}\right)+C
\end{aligned}
\end{array}
$
Hence option (A) is correct.

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Maths M.C.Q (1 Marks)

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