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Maths M.C.Q (1 Marks)

Integrals · Chhattisgarh Board (CGBSE) STD 12 Science (Chhattisgarh - English Medium). 268 questions.

Questions · Page 5 of 6

M.C.Q (1 Marks)

MCQ 2011 Mark
Choose the correct answer in Exercise : $\text{If}\ \text{f}(\text{x})=\int^{\text{x}}_{0}\text{t}\sin\text{t}\ \text{dt}$, then $\text{f}\text{(x)}$ is
  • A
    $\cos\text{x}+\text{x}\sin\text{x}$
  • ✓
    $\text{x}\sin\text{x}$
  • C
    $\text{x}\cos\text{x}$
  • D
    $\sin\text{x}+\text{x}\cos\text{x}$
Answer
Correct option: B.
$\text{x}\sin\text{x}$
$\text{f}\text{(x)}=\int\limits_{0}^{\text{x}}\text{t}\sin\text{t}\ \text{dt}$
$\therefore\text{f}\ '\text{(x)}=\text{x}\sin\text{x}\ $
$\big[\therefore$ of first fundamental theorem$\big]$
$\text{f}\text{(x)}=\int\limits_{0}^{\text{x}}\text{t}\sin\text{t}\ \text{dt}$
$\therefore\text{f}\ '\text{(x)}=\text{x}\sin\text{x}\ \ $
$\big[\therefore$ of first fundamental theorem$\big]$
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MCQ 2021 Mark
Choose the correct answer in Exercise : $\int^{\sqrt{3}}_{1}\frac{\text{dx}}{1+\text{x}^{2}}\text{ equals}$
  • A
    $\frac{\pi}{3}$
  • B
    $\frac{2\pi}{3}$
  • C
    $\frac{\pi}{6}$
  • ✓
    $\frac{\pi}{12}$
Answer
Correct option: D.
$\frac{\pi}{12}$
$\int\frac{\text{dx}}{1+\text{x}^{2}}=\tan^{-1}\text{x}=\text{F}\text{(x)}$
By second fundamental theorem of calculus, we obtain
$\int\limits_{1}^{\sqrt{3}}\frac{\text{dx}}{1+\text{x}^{2}}=\text{F}(\sqrt{3})-\text{F}(1)$
$=\tan^{-1}\sqrt{3}-\tan^{-1}1$
$=\frac{\pi}{3}-\frac{\pi}{4}$
$=\frac{\pi}{12}$
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MCQ 2031 Mark
Choose the correct answer in Exercises:
$\int\frac{\sin^2\text{x}-\cos^2\text{x}}{\sin^2\text{x}\cos^2\text{x}}\text{ dx}$ is equal to
  • ✓
    $\tan\text{x}+\cot\text{x}+\text{C}$
  • B
    $\tan\text{x}+\text{cosec x}+\text{C}$
  • C
    $-\tan\text{x}+\cot\text{x}+\text{C}$
  • D
    $\tan\text{x}+\sec\text{x}+\text{C}$
Answer
Correct option: A.
$\tan\text{x}+\cot\text{x}+\text{C}$
$\int\frac{\sin^2\text{x}-\cos^2\text{x}}{\sin^2\text{x}\cos^2\text{x}}\text{ dx}$
$=\int\bigg(\frac{\sin^2\text{x}}{\sin^2\text{x}\cos^2\text{x}}-\frac{\cos^2\text{x}}{\sin^2\text{x}\cos^2\text{x}}\bigg)\text{dx}$
$=\int(\sec^2\text{x}-\text{cosec}^2\text{x})\text{ dx}$
$=\tan\text{x}+\cot\text{x}+\text{C}$
Hence, the correct answer is A.
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MCQ 2041 Mark
The derivative of $\text{f(x)}=\int\limits^{\text{x}^3}_{\text{x}^2}\frac{1}{\log_{\text{e}}\text{t}}\text{ dt},(\text{x} > 0),$ is :
  • A
    $\frac{1}{3\ln\text{x}}$
  • B
    $\frac{1}{3\ln\text{x}}-\frac{1}{2\ln\text{x}}$
  • ✓
    $\big(\ln\text{x}\big)^{-1}\text{x}(\text{x}-1)$
  • D
    $\frac{3\text{x}^2}{\ln\text{x}}$
Answer
Correct option: C.
$\big(\ln\text{x}\big)^{-1}\text{x}(\text{x}-1)$
$\text{f}\ '(\text{x})=\frac{1}{\log_\text{e}\text{x}^3}(3\text{x}^2)-\frac{1}{\log_\text{e}\text{x}^2}(2\text{x})$
$=\frac{3\text{x}^2}{3\ln\text{ x}}-\frac{2\text{x}}{2\ln\text{ x}}$
$=\frac{\text{x}^2}{\ln\text{ x}^{-1}}-\frac{\text{x}}{\ln\text{ x}}$
$=\frac{1}{\ln\text{ x}}\text{x}(\text{x}-1)$
$=\big(\ln\text{ x}\big)^{-1}\text{x}(\text{x}-1)$
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MCQ 2051 Mark
$\int\frac{\sin^2\text{x}-\cos^2\text{x}}{\sin^2\text{x}\cos^2\text{x}}\text{dx}$ is equal to :
  • A
    $\tan\text{x}+\cos\text{x}+\text{c}$
  • B
    $\tan\text{x}+\text{cosec}\ \text{x}+\text{c}$
  • ✓
    $\tan\text{x}+\text{cot}\ \text{x}+\text{c}$
  • D
    $\tan\text{x}+\sec\text{x}+\text{c}$
Answer
Correct option: C.
$\tan\text{x}+\text{cot}\ \text{x}+\text{c}$
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MCQ 2061 Mark
$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\sin|\text{x}|\text{dx}$ is equal to :
  • A
    $1$
  • ✓
    $2$
  • C
    $-1$
  • D
    $-2$
Answer
Correct option: B.
$2$
$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\sin|\text{x}|\text{dx}$
$=-\int\limits^0_{-\frac{\pi}{2}}\sin\text{x}\text{ dx}+\int\limits^\frac{\pi}{2}_0\sin\text{x}\text{ dx}$
$=-\big[-\cos\text{x}\big]^0_{-\frac{\pi}{2}}+\big[-\cos\text{x}\big]^\frac{\pi}{2}_0$
$=1-0-0+1$
$=2$
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MCQ 2071 Mark
$\lim\limits_{\text{n}\rightarrow\infty}\Big\{\frac{1}{2\text{n}+1}+\frac{1}{2\text{n}+2}+\ .....+\frac{1}{2\text{n}+\text{n}}\Big\}$ is equal to :
  • A
    $\ln\Big(\frac{1}{3}\Big)$
  • B
    $\ln\Big(\frac{2}{3}\Big)$
  • ✓
    $\ln\Big(\frac{3}{2}\Big)$
  • D
    $\ln\Big(\frac{4}{3}\Big)$
Answer
Correct option: C.
$\ln\Big(\frac{3}{2}\Big)$
$\lim\limits_{\text{n}\rightarrow\infty}\Big\{\frac{1}{2\text{n}+1}+\frac{1}{2\text{n}+2}+\ .....\ +\frac{1}{2\text{n}+\text{n}}\Big\}$
$=\lim\limits_{\text{n}\rightarrow\infty}\sum\limits^\text{n}_{\text{r}=1}\frac{1}{2\text{n}+\text{r}}$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{1}{\text{n}}\sum\limits^\text{n}_{\text{r}=1}\frac{1}{2+\frac{\text{r}}{\text{n}}}$
let $\frac{\text{r}}{\text{n}}=\text{x}$
$=\int\limits^\infty_0\frac{1}{2+\text{x}}\text{dx}$
$=\Big[\log\big(2+\text{x}\big)\Big]^\infty_0$
$=\log3-\log2$
$=\log\frac{3}{2}$
$=\ln\Big(\frac{3}{2}\Big)$
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MCQ 2081 Mark
What is the value of $\int_{0}^{\frac{ \pi}{2}}\frac{\sqrt{\tan\text{x}}}{\sqrt{\tan}\text{x}+\sqrt{\cot}\text{x}}\text{dx}?$
  • A
    $\frac{\pi}{2}$
  • ✓
    $\frac{\pi}{4}$
  • C
    $\frac{\pi}{8}$
  • D
    None of these
Answer
Correct option: B.
$\frac{\pi}{4}$
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MCQ 2091 Mark
$\int_0^{\pi / 2} \frac{\sin x-\cos x}{1+\sin x \cos x} d x$ is equal to :
  • A
    $\pi$
  • ✓
    Zero
  • C
    $\int_0^{\pi / 2} \frac{2 \sin x}{1+\sin x \cos x} d x$
  • D
    $\frac{\pi^2}{4}$
Answer
Correct option: B.
Zero
Let $I=\int_0^{\pi / 2} \frac{\sin x-\cos x}{1+\sin x \cos x} d x$
$\Rightarrow I=\int_0^{\pi / 2} \frac{\sin (\pi / 2-x)-\cos (\pi / 2-x)}{1+\sin (\pi / 2-x) \cos (\pi / 2-x)} d x$
$\Rightarrow I=\int_0^{\pi / 2} \frac{\cos x-\sin x}{1+\cos x \cdot \sin x} d x$
Adding $(i)$ and $(ii)$, we get
$2 I=\int_0^{\pi / 2} 0 d x=0$
$\Rightarrow I=0$
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MCQ 2101 Mark
The value of $\int_1^e \log x d x$ is
  • A
    $0$
  • ✓
    $1$
  • C
    $e$
  • D
    $e \log e$
Answer
Correct option: B.
$1$
Let $I=\int_1^e \log x d x$
Using integration by parts, we get $I=[x \log x]_1^e-\int_1^e \frac{1}{x} \cdot x d x$
$I=[x \log x]_1^e-[x]_1^e=e \log e-\log 1-e+1$
$=e-0-e+1=1 \quad[\because \log e=1 \text { and } \log 1=0]$
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MCQ 2111 Mark
If $\int_{-2}^3 x^2 d x=k \int_0^2 x^2 d x+\int_2^3 x^2 d x$, then the value of $k$ is
  • ✓
    $2$
  • B
    $1$
  • C
    $0$
  • D
    $\frac{1}{2}$
Answer
Correct option: A.
$2$
Given, $\int_{-2}^3 x^2 d x=k \int_0^2 x^2 d x+\int_2^3 x^2 d x$
$\Rightarrow\left[\frac{x^3}{3}\right]_{-2}^3=k\left[\frac{x^3}{3}\right]_0^2+\left[\frac{x^3}{3}\right]_2^3$
$\Rightarrow \frac{27}{3}+\frac{8}{3}=k\left(\frac{8}{3}\right)+\frac{27}{3}-\frac{8}{3}$
$\Rightarrow \frac{8 k}{3}=\frac{16}{3}$
$\Rightarrow k=2$
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MCQ 2121 Mark
The value of $\int_0^3 \frac{d x}{\sqrt{9-x^2}}$ is:
  • A
    $\frac{\pi}{6}$
  • B
    $\frac{\pi}{4}$
  • C
    $\frac{\pi}{2}$
  • D
    $\frac{\pi}{18}$
Answer
We have, $\int_0^3 \frac{d x}{\sqrt{9-x^2}}=\left[\sin ^{-1} \frac{x}{3}\right]_0^3=\sin ^{-1} 1-\sin ^{-1} 0=\frac{\pi}{2}$
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MCQ 2131 Mark
For any integer $n$, the value of $\int_{-\pi}^\pi e^{\cos ^2 x} \sin ^3(2 n+1) x d x$ is
  • A
    $-1$
  • ✓
    $0$
  • C
    $1$
  • D
    $2$
Answer
Correct option: B.
$0$
$\text {Let } f(x)=e^{\cos ^2 x} \sin ^3(2 n+1) x$
$f(-x)=e^{\cos ^2(-x)} \sin ^3(2 n+1)(-x)$
$=-e^{\cos ^2 x} \sin ^3(2 n+1) x=-f(x)$
$\therefore \int_{-\pi}^\pi e^{\cos ^2 x} \sin ^3(2 n+1) x d x=0$
$(\because f(-x)=-f(x))$
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MCQ 2141 Mark
Which of these is equal to $\int e^{(x \log 5)} e^x d x$, where $C$ is the constant of integration?
  • ✓
    $\frac{(5 e)^x}{\log 5 e}+C$
  • B
    $\log 5^x+x+C$
  • C
    $5^x \varepsilon^x+C$
  • D
    $(5 e)^x \log x+C$
Answer
Correct option: A.
$\frac{(5 e)^x}{\log 5 e}+C$
$\frac{(5 e)^x}{\log 5 e}+C$
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MCQ 2151 Mark
$\int_0^4\left(e^{2 x}+x\right) d x$ is equal to
  • A
    $\frac{15+e^8}{2}$
  • B
    $\frac{16-e^8}{2}$
  • ✓
    $\frac{e^8-15}{2}$
  • D
    $\frac{-e^8-15}{2}$
Answer
Correct option: C.
$\frac{e^8-15}{2}$
$\text {Let } I=\int_0^4\left(e^{2 x}+x\right)$
$d x=\left[\frac{e^{2 x}}{2}+\frac{x^2}{2}\right]_0^4$
$=\frac{e^8}{2}+\frac{16}{2}-\frac{e^0}{2}-0$
$=\frac{e^8}{2}+\frac{16}{2}-\frac{1}{2}$
$=\frac{e^8+15}{2}$
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MCQ 2161 Mark
$\int_{-1}^1 \frac{|x-2|}{x-2} d x, x \neq 2$ is equal to
  • A
    1
  • B
    -1
  • C
    2
  • D
    -2
Answer
Let $I=\int_{-1}^1 \frac{|x-2|}{x-2} d x$
$
=\int_{-1}^1 \frac{-(x-2)}{x-2} d x=\int_{-1}^1-1 \cdot d x=[-x]_{-1}^1=-[1-(-1)]=-2
$
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MCQ 2171 Mark
$\int_0^{\frac{\pi}{6}} \sec ^2\left(x-\frac{\pi}{6}\right) d x$ is equal to :
  • ✓
    $\frac{1}{\sqrt{3}}$
  • B
    $-\frac{1}{\sqrt{3}}$
  • C
    $\sqrt{3}$
  • D
    $-\sqrt{3}$
Answer
Correct option: A.
$\frac{1}{\sqrt{3}}$
$\text {We have, } \int_0^{\pi / 6} \sec ^2\left(x-\frac{\pi}{6}\right) d x$
$=\left[\tan \left(x-\frac{\pi}{6}\right)\right]_0^{\pi / 6}$
$=\tan \left(\frac{\pi}{6}-\frac{\pi}{6}\right)-\tan \left(0-\frac{\pi}{6}\right)$
$=\tan 0^{\circ}-\tan \left(-\frac{\pi}{6}\right) $
$\left(\because \tan 0^{\circ}=0 \text { and } \tan (-\theta)=-\tan \theta\right)$
$=0+\tan (\pi / 6)=\tan \frac{\pi}{6}$
$=\frac{1}{\sqrt{3}}$
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MCQ 2181 Mark
The value of $\int_0^{\pi / 6} \sin 3 x d x$ is:
  • A
    $-\frac{\sqrt{3}}{2}$
  • B
    $-\frac{1}{3}$
  • C
    $\frac{\sqrt{3}}{2}$
  • ✓
    $\frac{1}{3}$
Answer
Correct option: D.
$\frac{1}{3}$
$\text {Let } I=\int_0^{\pi / 6} \sin 3 x d x$
$=\frac{-1}{3}[\cos 3 x]_0^{\pi / 6}$
$=\frac{-1}{3}\left[\cos \frac{\pi}{2}-\cos 0\right]=\frac{-1}{3}(0-1)=\frac{1}{3}$
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MCQ 2191 Mark
$\int e^{5 \log x} d x$ is equal to
  • A
    $\frac{x^5}{5}+C$
  • ✓
    $\frac{x^6}{6}+C$
  • C
    $5 x^4+C$
  • D
    $6 x^5+C$
Answer
Correct option: B.
$\frac{x^6}{6}+C$
$\text {Let } I=\int e^{5 \log x} d x$
$=\int e^{\log x^5} d x=\int x^5 d x $
$\left[\because e^{\log x}=x\right]$
$=\frac{x^6}{6}+C$
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MCQ 2201 Mark
$\int \frac{\sec x}{\sec x-\tan x} d x \text { equals }$
  • A
    $\sec x-\tan x+c$
  • B
    $\sec x+\tan x+c$
  • ✓
    $\tan x-\sec x+c$
  • D
    $-(\sec x+\tan x)+c$
Answer
Correct option: C.
$\tan x-\sec x+c$
$\text {Let } I=\int \frac{\sec x}{\sec x-\tan x} d x$
$=\int \frac{\sec x(\sec x+\tan x)}{(\sec x-\tan x)(\sec x+\tan x)} d x=\int\left(\frac{\sec ^2 x+\sec x \tan x}{\sec ^2 x-\tan ^2 x}\right) d x$
$=\int \sec ^2 x d x+\int \sec x \tan x d x \quad\left[\because \sec ^2 x-\tan ^2 x=1\right]$
$=\tan x+\sec x+c$
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MCQ 2211 Mark
If $\frac{d}{d x}(f(x))=\log x$, then $f(x)$ equals:
  • A
    $-\frac{1}{x}+C$
  • ✓
    $x(\log x-1)+C$
  • C
    $x(\log x+x)+C$
  • D
    $\frac{1}{x}+C$
Answer
Correct option: B.
$x(\log x-1)+C$
We have, $\frac{d}{d x}(f(x))=\log x$
On integrating both sides, we get
$\int \frac{d}{d x}(f(x))=\int 1 \cdot \log x d x$
$\Rightarrow f(x)=\log x \int 1 \cdot d x-\int\left(\frac{d}{d x}(\log x) \int 1 \cdot d x\right) d x$
${[\text { Integrating by parts }]}$
$\Rightarrow f(x)=x \cdot \log x-\int \frac{1}{x} x x d x$
$\Rightarrow f(x)=x \log x-x+C$
$\Rightarrow f(x)=x(\log x-1)+C$
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MCQ 2221 Mark
$\int \frac{\cos 2 x}{\sin ^2 x \cdot \cos ^2 x} d x$ is equal to
  • A
    $\tan x-\cot x+C$
  • ✓
    $-\cot x-\tan x+C$
  • C
    $\cot x+\tan x+C$
  • D
    $\tan x-\cot x-C$
Answer
Correct option: B.
$-\cot x-\tan x+C$
$\text {Let, } I=\int \frac{\cos 2 x}{\sin ^2 x \cdot \cos ^2 x} d x$
$=\int\left(\frac{\cos ^2 x-\sin ^2 x}{\sin ^2 x \cdot \cos ^2 x}\right) d x \left[\because \cos 2 \theta=\cos ^2 \theta-\sin ^2 \theta\right]$
$=\int \frac{1}{\sin ^2 x} d x-\int \frac{1}{\cos ^2 x} d x$
$=\int \operatorname{cosec}^2 x d x-\int \sec ^2 x d x$
$=-\cot x-\tan x+C$
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MCQ 2231 Mark
$\int_{-\pi / 4}^{\pi / 4} \sec ^2 x d x$ is equal to
  • A
    $-1$
  • B
    $0$
  • C
    $1$
  • ✓
    $2$
Answer
Correct option: D.
$2$
$\text {Let } I=\int_{-\pi / 4}^{\pi / 4} \sec ^2 x d x$
$=[\tan x]_{-\pi / 4}^{\pi / 4}$
$=\tan \frac{\pi}{4}-\tan \left(-\frac{\pi}{4}\right)$
$=1+1$
$=2$
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MCQ 2241 Mark
$\int_0^{\pi / 8} \tan ^2(2 x) d x$ is equal to
  • ✓
    $\frac{4-\pi}{8}$
  • B
    $\frac{4+\pi}{8}$
  • C
    $\frac{4-\pi}{4}$
  • D
    $\frac{4-\pi}{2}$
Answer
Correct option: A.
$\frac{4-\pi}{8}$
$\text {Let } I=\int_0^{\pi / 8} \tan ^2(2 x) d x=\int_0^{\pi / 8}\left(\sec ^2(2 x)-1\right) d x$
$=\left(\frac{1}{2} \tan 2 x-x\right)_0^{\pi / 8}$
$=\frac{1}{2} \tan 2\left(\frac{\pi}{8}\right)-\frac{\pi}{8}$
$=\frac{1}{2} \tan \frac{\pi}{4}-\frac{\pi}{8}$
$=\frac{1}{2}-\frac{\pi}{8}=\frac{4-\pi}{8}$
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MCQ 2251 Mark
$\int \frac{e^x}{x+1}[1+(x+1) \log (x+1)] d x$ equals
  • A
    $\frac{e^x}{x+1}+c$
  • B
    $e^x \frac{x}{x+1}+c$
  • C
    $e^x \log (x+1)+e^x+c$
  • ✓
    $e^x \log (x+1)+c$
Answer
Correct option: D.
$e^x \log (x+1)+c$
 Let $I=\int \frac{e^x}{x+1}[1+(x+1) \log (x+1)] d x$
$=\int e^x\left[\frac{1}{x+1}+\log (x+1)\right] d x$
It is of the form $\int e^x\left[f(x)+f^{\prime}(x) d x\right]$,
where $f(x)=\log (x+1)$ and $f^{\prime}(x)=\frac{1}{x+1}$
So, $I=e^x \log (x+1)+C$
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MCQ 2261 Mark
$\int e^x\left(\frac{x \log x+1}{x}\right) d x$ is equal to
  • A
    $\log \left(e^x \log x\right)+c$
  • B
    $\frac{e^x}{x}+c$
  • C
    $x \log x+e^x+c$
  • ✓
    $e^x \log x+c$
Answer
Correct option: D.
$e^x \log x+c$
$\text {Let } I=\int e^x\left(\log x+\frac{1}{x}\right) d x$
$\Rightarrow I=e^x \log x+c $
$\left(\because \int e^x\left[f(x)+f^{\prime}(x)\right] d x=e^x f(x)+c\right)$
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MCQ 2271 Mark
$\int \frac{e^x(1+x)}{\cos ^2\left(x e^x\right)} d x$ is equal to
  • ✓
    $\tan \left(x e^x\right)+c$
  • B
    $\cot \left(x e^x\right)+c$
  • C
    $\cot \left(e^x\right)+c$
  • D
    $\tan \left[e^x(1+x)\right]+c$
Answer
Correct option: A.
$\tan \left(x e^x\right)+c$
$\text {Let } I=\int \frac{e^x(1+x)}{\cos ^2\left(x e^x\right)} d x$
$\text { Put } x e^x=t$
$\Rightarrow\left(x e^x+e^x\right) d x=d t$
$\Rightarrow e^x(x+1) d x=d t$
$\therefore I=\int \frac{d t}{\cos ^2 t}=\int \sec ^2 t d t=\tan t+c$
$=\tan \left(x e^x\right)+c$
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MCQ 2281 Mark
$\int x^2 e^{x^3} d x$ equals
  • ✓
    $\frac{1}{3} e^{x^3}+C$
  • B
    $\frac{1}{3} e^{x^4}+C$
  • C
    $\frac{1}{2} e^{x^3}+C$
  • D
    $\frac{1}{2} e^{x^2}+C$
Answer
Correct option: A.
$\frac{1}{3} e^{x^3}+C$
$\text {Let } I=\int x^2 e^{x^3} d x$
$\text { Put } x^3=t$
$\Rightarrow 3 x^2 d x=d t$
$\therefore I=\int e^t \frac{d t}{3}=\frac{1}{3} e^t+C$
$=\frac{1}{3} e^{x^3}+C$
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MCQ 2291 Mark
Evaluate: $\int \frac{\left(x^4-x\right)^{\frac{1}{4}}}{x^5} d x$
  • ✓
    $\frac{4}{15}\left(1-\frac{1}{x^3}\right)^{\frac{5}{4}}+C$
  • B
    $\frac{-4}{15}\left(1-\frac{1}{x^3}\right)^{\frac{5}{4}}+C$
  • C
    $\frac{2}{15}\left(1-\frac{1}{x^3}\right)^{\frac{5}{4}}+C$
  • D
    $\frac{-2}{15}\left(1-\frac{1}{x^3}\right)^{\frac{5}{4}}+C$
Answer
Correct option: A.
$\frac{4}{15}\left(1-\frac{1}{x^3}\right)^{\frac{5}{4}}+C$
(a) : Let $I=\int \frac{\left(x^4-x\right)^{\frac{1}{4}}}{x^5} d x$
$
\Rightarrow \quad I=\int \frac{x\left(1-\frac{1}{x^3}\right)^{\frac{1}{4}}}{x^5} d x=\int \frac{\left(1-\frac{1}{x^3}\right)^{\frac{1}{4}}}{x^4} d x
$
Put $1-\frac{1}{x^3}=t \Rightarrow \frac{3}{x^4} d x=d t$
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MCQ 2301 Mark
Evaluate: $\int_0^2(x-[x]) d x$
  • A
    $0$
  • B
    $-1$
  • ✓
    $1$
  • D
    $2$
Answer
Correct option: C.
$1$
Let $I=\int_0^2(x-[x]) d x=\int_0^2 x d x-\int_0^2[x] d x$
$=\left[\frac{x^2}{2}\right]_0^2-\int_0^1[x] d x-\int_1^2[x] d x=\frac{4}{2}-\int_0^1 0 d x-\int_1^2 1 d x$
$=2-0-[x]_1^2=2-[2-1]=2-1=1$
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MCQ 2311 Mark
Using fundamental theorem of calculus, which of following integrals can be solved.
(i) $\int_0^1 x^3 d x$
(ii) $\int x e^x d x$
(iii) $\int_2^3 \frac{1}{x} d x$
  • A
    Only (i)
  • B
    Only (ii)
  • ✓
    Only (i) and (iii)
  • D
    All (i), (ii) and (iii)
Answer
Correct option: C.
Only (i) and (iii)
(c) : (i) $\int_0^1 x^3 d x=\left[\frac{x^4}{4}\right]_0^1=\frac{1}{4}$
(ii) $\int x e^x d x=x e^x-e^x+C$
(iii) $\int_2^3 \frac{1}{x} d x=[\log x]_2^3=\log \frac{3}{2}$
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MCQ 2321 Mark
Evaluate: $\int\left(e^{x \log a}+e^{a \log x}+e^{a \log a}\right) d x$
  • A
    $a^x \log a+\frac{x^{a+1}}{a+1}+\frac{a^a}{x}+C$
  • B
    $a^x \log a+(a+1) x^{a+1}+a^a x+C$
  • ✓
    $\frac{a^x}{\log a}+\frac{x^{a+1}}{a+1}+a^a x+C$
  • D
    None of these
Answer
Correct option: C.
$\frac{a^x}{\log a}+\frac{x^{a+1}}{a+1}+a^a x+C$
Let $I=\int\left(e^{x \log a}+e^{a \log x}+e^{a \log a}\right) d x$
$=\int\left(e^{\log a^x}+e^{\log x^a}+e^{\log a^a}\right) d x=\int\left(a^x+x^a+a^a\right) d x$
$=\frac{a^x}{\log a}+\frac{x^{a+1}}{a+1}+a^a x+c \quad\left[\because e^{\log y=y]}\right]$
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MCQ 2331 Mark
Evaluate: $\int_{-\pi}^\pi x^{10} \sin ^7 x d x$
  • A
    1
  • B
    2
  • C
    -1
  • ✓
    0
Answer
Correct option: D.
0
(d) : Let $I=\int_{-\pi}^\pi x^{10} \sin ^7 x d x$
Let $f(x)=x^{10} \sin ^7 x$
and $f(-x)=(-x)^{10}[\sin (-x)]^7=-x^{10} \sin ^7 x=-f(x)$
$\therefore f(x)$ is an odd function.
$\therefore \quad I=\int_{-\pi}^\pi x^{10} \sin ^7 x d x=0$
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MCQ 2341 Mark
Evaluate: $\int \frac{\sin x}{1+\sin x} d x$
  • A
    $\sec x-\tan x+C$
  • B
    $\sec x+\tan x+x+C$
  • C
    $\sec x+\tan x+C$
  • ✓
    $\sec x-\tan x+x+C$
Answer
Correct option: D.
$\sec x-\tan x+x+C$
(d) : Let $I=\int \frac{\sin x}{1+\sin x} d x=\int \frac{\sin x(1-\sin x)}{(1+\sin x)(1-\sin x)} d x$
$=\int \frac{\sin x-\sin ^2 x}{\cos ^2 x} d x=\int \sec x \tan x d x-\int \tan ^2 x d x$
$=\int \sec x \tan x d x-\int\left(\sec ^2 x-1\right) d x=\sec x-\tan x+x+C$
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MCQ 2351 Mark
Evaluate : $\int \frac{x^3}{x+2} d x$
  • A
    $\frac{x^3}{3}-x^2-4 x-8 \log |x+2|+C$
  • ✓
    $\frac{x^3}{3}-x^2+4 x-8 \log |x+2|+C$
  • C
    $\frac{x^3}{3}+x^2+4 x+8 \log |x+2|+C$
  • D
    $\frac{x^3}{3}+x^2+4 x-8 \log |x+2|+C$
Answer
Correct option: B.
$\frac{x^3}{3}-x^2+4 x-8 \log |x+2|+C$
(b) : Let $I=\int \frac{x^3}{x+2} d x$
Dividing $x^3$ by $x+2$, we get
$
=\int\left(x^2-2 x+4-\frac{8}{x+2}\right) d x=\frac{x^3}{3}-x^2+4 x-8 \log |x+2|+C
$
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MCQ 2361 Mark
Evaluate: $\int \sec ^2(7-4 x) d x$
  • A
    $\frac{1}{4} \tan (7-4 x)+C$
  • B
    $\frac{1}{4} \tan (7+4 x)+C$
  • C
    $\frac{-1}{4} \tan (7+4 x)+C$
  • ✓
    $\frac{-1}{4} \tan (7-4 x)+C$
Answer
Correct option: D.
$\frac{-1}{4} \tan (7-4 x)+C$
(d) : Let $I=\int \sec ^2(7-4 x) d x$
Put $7-4 x=t \Rightarrow d x=\frac{-1}{4} d t$
$
\therefore \quad I=\int \frac{\sec ^2 t}{-4} d t \Rightarrow I=\frac{\tan t}{-4}+C=\frac{\tan (7-4 x)}{-4}+C
$
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MCQ 2371 Mark
Evaluate: $\int[\sin (\log x)+\cos (\log x)] d x$
  • ✓
    $x \sin (\log x)+C$
  • B
    $\sin (\log x)+C$
  • C
    $x \cos (\log x)+C$
  • D
    $\cos (\log x)+C$
Answer
Correct option: A.
$x \sin (\log x)+C$
(a) : Let $I=\int[\sin (\log x)+\cos (\log x)] d x$
Put $\log x=t \Rightarrow x=e^t \quad \Rightarrow d x=e^t d t$
$
\begin{aligned}
\therefore \quad I & =\int(\sin t+\cos t) e^t d t=e^t \sin t+C \\
& =x \sin (\log x)+C\left[\because\left[f(x)+f^{\prime}(x)\right] e^x d x=e^x f(x)+C\right)
\end{aligned}
$
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MCQ 2381 Mark
Evaluate: $\int \frac{d x}{5-8 x-x^2}$
  • A
    $\frac{1}{\sqrt{21}} \log \left|\frac{\sqrt{21}+x+4}{\sqrt{21}-x-4}\right|+C$
  • ✓
    $\frac{1}{2 \sqrt{21}} \log \left|\frac{\sqrt{21}+x+4}{\sqrt{21}-x-4}\right|+C$
  • C
    $\frac{1}{\sqrt{21}} \log \left|\frac{\sqrt{21}-x-4}{\sqrt{21}+x+4}\right|+C$
  • D
    $\frac{1}{2 \sqrt{21}} \log \left|\frac{\sqrt{21}-x-4}{\sqrt{21}+x+4}\right|+C$
Answer
Correct option: B.
$\frac{1}{2 \sqrt{21}} \log \left|\frac{\sqrt{21}+x+4}{\sqrt{21}-x-4}\right|+C$
Let $I=\int \frac{d x}{5-8 x-x^2}=\int \frac{d x}{21-(x+4)^2}$
$=\int \frac{d x}{(\sqrt{21})^2-(x+4)^2}=\frac{1}{2 \sqrt{21}} \log \left|\frac{\sqrt{21}+x+4}{\sqrt{21}-x-4}\right|+C$
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MCQ 2391 Mark
Evaluate: $\int \tan x \tan 2 x \tan 3 x d x$
  • A
    $\frac{1}{3} \log |\sec 3 x|-\log |\sec x|+C$
  • B
    $\log |\sec 3 x|-\frac{1}{2} \log |\sec 2 x|+C$
  • C
    $\log |\sec x|-\frac{1}{2} \log |\sec 3 x|+\frac{1}{2} \log |\sec 2 x|+C$
  • ✓
    $\frac{1}{3} \log |\sec 3 x|-\frac{1}{2} \log |\sec 2 x|-\log |\sec x|+C$
Answer
Correct option: D.
$\frac{1}{3} \log |\sec 3 x|-\frac{1}{2} \log |\sec 2 x|-\log |\sec x|+C$
Let $I=\int \tan x \tan 2 x \tan 3 x d x$
$\text { Since, } \tan 3 x=\tan (2 x+x)=\frac{\tan 2 x+\tan x}{1-\tan x \tan 2 x}$
$\Rightarrow \tan x \tan 2 x \tan 3 x=\tan 3 x-\tan 2 x-\tan x ...(i)$
$\therefore I=\int(\tan 3 x-\tan 2 x-\tan x) d x \ ($From $(i))$
$=\frac{1}{3} \log |\sec 3 x|-\frac{1}{2} \log |\sec 2 x|-\log |\sec x|+C$
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MCQ 2401 Mark
Evaluate: $\int\left(5 x^3+2 x^{-5}-7 x+\frac{1}{\sqrt{x}}+\frac{5}{x}\right) d x$
  • A
    $\frac{5 x^4}{4}-\frac{1}{2 x^4}-\frac{7 x^2}{2}+2 \sqrt{x}-5 \log |x|+C$
  • ✓
    $\frac{5 x^4}{4}-\frac{1}{2 x^4}-\frac{7 x^2}{2}+2 \sqrt{x}+5 \log |x|+C$
  • C
    $\frac{5 x^4}{4}+\frac{1}{2 x^4}+\frac{7 x^2}{2}+2 \sqrt{x}+5 \log |x|+C$
  • D
    $\frac{5 x^4}{4}+\frac{1}{2 x^4}+\frac{7 x^2}{2}+2 \sqrt{x}-5 \log |x|+C$
Answer
Correct option: B.
$\frac{5 x^4}{4}-\frac{1}{2 x^4}-\frac{7 x^2}{2}+2 \sqrt{x}+5 \log |x|+C$
We have $\int\left(5 x^3+2 x^{-5}-7 x+\frac{1}{\sqrt{x}}+\frac{5}{x}\right) d x$
$=5 \int x^3 d x+2 \int x^{-5} d x-7 \int x d x+\int x^{-1 / 2} d x+5 \int \frac{1}{x} d x$
$=5 \cdot \frac{x^4}{4}+2 \cdot \frac{x^{-4}}{(-4)}-7 \cdot \frac{x^2}{2}+\frac{x^{1 / 2}}{(1 / 2)}+5 \log |x|+C$
$=\frac{5 x^4}{4}-\frac{1}{2 x^4}-\frac{7 x^2}{2}+2 \sqrt{x}+5 \log |x|+C$
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MCQ 2411 Mark
Evaluate: $\int \frac{x^3-x^2+x-1}{x-1} d x$
  • ✓
    $\frac{x^3}{3}+x+C$
  • B
    $x^3+x+C$
  • C
    $\frac{x^3}{3}+x^2+C$
  • D
    $\frac{x^3}{3}+C$
Answer
Correct option: A.
$\frac{x^3}{3}+x+C$
Let $ I=\int \frac{x^3-x^2+x-1}{x-1} d x$
$=\int \frac{x^2(x-1)+1(x-1)}{x-1}=\int \frac{\left(x^2+1\right)(x-1)}{x-1} d x$
$=\int\left(x^2+1\right) d x=\frac{1}{3} x^3+x+C$
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MCQ 2421 Mark
Evaluate: $\int \frac{x-4}{(x-2)^3} \cdot e^x d x$
  • A
    $\frac{e^x}{(x-2)^3}+C$
  • B
    $\frac{-e^x}{(x-2)^3}+C$
  • ✓
    $\frac{e^x}{(x-2)^2}+C$
  • D
    $\frac{-e^x}{(x-2)^2}+C$
Answer
Correct option: C.
$\frac{e^x}{(x-2)^2}+C$
$ \int \frac{x-4}{(x-2)^3} \cdot e^x d x=\int\left[\frac{x-2}{(x-2)^3}-\frac{2}{(x-2)^3}\right] e^x d x$
$=\int\left[\frac{1}{(x-2)^2}-\frac{2}{(x-2)^3}\right] e^x d x=\frac{e^x}{(x-2)^2}+C$
${\left[\because \int\left[f(x)+f^{\prime}(x)\right] e^x d x=e^x f(x)+C\right]}$
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MCQ 2431 Mark
Evaluate : $\int \frac{d x}{\sqrt{x^2-3 x+2}}$
  • A
    $\log \left|\left(x+\frac{3}{2}\right)+\sqrt{x^2-3 x+2}\right|+C$
  • ✓
    $\log \left|\left(x-\frac{3}{2}\right)+\sqrt{x^2-3 x+2}\right|+C$
  • C
    $\log \left|\left(x-\frac{3}{2}\right)-\sqrt{x^2-3 x+2}\right|+C$
  • D
    $\log \left|\left(x+\frac{3}{2}\right)-\sqrt{x^2-3 x+2}\right|+C$
Answer
Correct option: B.
$\log \left|\left(x-\frac{3}{2}\right)+\sqrt{x^2-3 x+2}\right|+C$
We have, $\int \frac{d x}{\sqrt{x^2-3 x+2}}$
$=\int \frac{d x}{\sqrt{\left(x^2-3 x+\frac{9}{4}\right)-\frac{1}{4}}}$
$=\int \frac{d x}{\sqrt{\left(x-\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2}}$
$=\log \left|\left(x-\frac{3}{2}\right)+\sqrt{x^2-3 x+2}\right|+C$
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MCQ 2441 Mark
Evaluate: $\int_0^1 \frac{x \tan ^{-1} x}{\left(1+x^2\right)^{3 / 2}} d x$
  • A
    $\frac{4-\pi}{2 \sqrt{2}}$
  • B
    $\frac{4+\pi}{2 \sqrt{2}}$
  • ✓
    $\frac{4-\pi}{4 \sqrt{2}}$
  • D
    None of these
Answer
Correct option: C.
$\frac{4-\pi}{4 \sqrt{2}}$
(c): Let $\int_0^1 \frac{x \tan ^{-1} x}{\left(1+x^2\right)^{3 / 2}} d x$ Put $\tan ^{-1} x=\theta \Rightarrow x=\tan \theta \Rightarrow d x=\sec ^2 \theta d \theta$ When, $x=0 \Rightarrow \theta=0$ and $x=1 \Rightarrow \theta=\frac{\pi}{4}$ $I=\int_0^1 \frac{x \tan ^{-1} x}{\left(1+x^2\right)^{3 / 2}} d x=\int_0^{\pi / 4} \frac{\theta \tan \theta}{\sec ^3 \theta} \sec ^2 \theta d \theta$$=\int_0^{\pi / 4} \theta \sin \theta d \theta=[-\theta \cos \theta]_0^{\pi / 4}-\int_0^{\pi / 4}(-\cos \theta) d \theta$[Integrating by parts]
$=[-\theta \cos \theta]_0^{\pi / 4}+[\sin \theta]_0^{\pi / 4}=\frac{4-\pi}{4 \sqrt{2}}$
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MCQ 2451 Mark
Find the value of $\int_{-\pi / 2}^{\pi / 2}|\sin x| d x$.
  • A
    0
  • B
    1
  • C
    2
  • D
    3
Answer
(c) : $\because|\sin x|$ is an even function.$\therefore \quad \int_{-\pi / 2}^{\pi / 2}|\sin x| d x=2 \int_0^{\pi / 2}|\sin x| d x=2 \int_0^{\pi / 2} \sin x d x$ $=-2[\cos x]_0^{\pi / 2}=-2(0-1)=2$

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MCQ 2461 Mark
Evaluate: $\int \frac{\left(a^x+b^x\right)^2}{a^x b^x} d x$
  • ✓
    $\frac{\left(\frac{a}{b}\right)^x}{\log \frac{a}{b}}+\frac{\left(\frac{b}{a}\right)^x}{\log \frac{b}{a}}+2 x+C, a \neq b$
  • B
    $\frac{\left(\frac{a}{b}\right)^x}{\log \frac{a}{b}}+\frac{\left(\frac{b}{a}\right)^x}{\log \frac{a}{b}}+2 x+C, a \neq b$
  • C
    $\left(\frac{a}{b}\right)^x+\left(\frac{b}{a}\right)^x+2 x+C, a \neq b$
  • D
    None of these
Answer
Correct option: A.
$\frac{\left(\frac{a}{b}\right)^x}{\log \frac{a}{b}}+\frac{\left(\frac{b}{a}\right)^x}{\log \frac{b}{a}}+2 x+C, a \neq b$
(a) : We have, $\int \frac{\left(a^x+b^x\right)^2}{a^x b^x} d x=\int \frac{a^{2 x}+b^{2 x}+2 a^x b^x}{a^x b^x} d x$
$
=\int\left(\left(\frac{a}{b}\right)^x+\left(\frac{b}{a}\right)^x+2\right) d x=\frac{\left(\frac{a}{b}\right)^x}{\log \frac{a}{b}}+\frac{\left(\frac{b}{a}\right)^x}{\log \frac{b}{a}}+2 x+C, a \neq b
$
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MCQ 2471 Mark
Evaluate: $\int \frac{d x}{\sqrt{1-2 x-x^2}}$
  • A
    $\frac{1}{\sqrt{2}} \sin ^{-1}\left(\frac{1+x}{\sqrt{2}}\right)+C$
  • B
    $\frac{1}{\sqrt{2}} \log (1+x)+C$
  • ✓
    $\sin ^{-1}\left(\frac{1+x}{\sqrt{2}}\right)+C$
  • D
    $\frac{1}{\sqrt{2}} \log \left(\frac{1+x}{\sqrt{2}}\right)+C$
Answer
Correct option: C.
$\sin ^{-1}\left(\frac{1+x}{\sqrt{2}}\right)+C$
Let $ I=\int \frac{d x}{\sqrt{1-\left(x^2+2 x\right)}}=\int \frac{d x}{\sqrt{2-\left(x^2+2 x+1\right)}}$
$=\int \frac{d x}{\sqrt{2-(1+x)^2}}=\int \frac{d x}{\sqrt{(\sqrt{2})^2-(1+x)^2}}$
Put $1+x=z \Rightarrow d x=d z$
$\therefore I=\int \frac{d z}{\sqrt{(\sqrt{2})^2-z^2}}=\sin ^{-1} \frac{z}{\sqrt{2}}+C$
$=\sin ^{-1}\left(\frac{1+x}{\sqrt{2}}\right)+C$
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MCQ 2481 Mark
Evaluate: $\int \frac{\sec ^2 x}{2+\tan x} d x$
  • A
    $\log |\tan x|+C$
  • B
    $\log |2-\tan x|+C$
  • ✓
    $\log |2+\tan x|+C$
  • D
    none of these
Answer
Correct option: C.
$\log |2+\tan x|+C$
(c) : Let $I=\int \frac{\sec ^2 x}{2+\tan x} d x$
Put $2+\tan x=t \Rightarrow \sec ^2 x d x=d t$
$
\therefore \quad I=\int \frac{d t}{t}=\log |t|+C=\log |2+\tan x|+C
$
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MCQ 2491 Mark
Evaluate: $\int \frac{1}{\sin x+\sqrt{3} \cos x} d x$
  • ✓
    $\frac{1}{2} \log \left|\tan \left(\frac{x}{2}+\frac{\pi}{6}\right)\right|+C$
  • B
    $\frac{1}{2} \log \left|\tan \frac{x}{2}\right|+C$
  • C
    $\frac{1}{2} \log \left|\tan \left(\frac{x}{2}-\frac{\pi}{6}\right)\right|+C$
  • D
    $\frac{1}{2} \log \left|\tan \left(x-\frac{\pi}{6}\right)\right|+C$
Answer
Correct option: A.
$\frac{1}{2} \log \left|\tan \left(\frac{x}{2}+\frac{\pi}{6}\right)\right|+C$
Let $I=\int \frac{1}{\sin x+\sqrt{3} \cos x} d x$
$=\frac{1}{2} \int \frac{d x}{\frac{1}{2} \sin x+\frac{\sqrt{3}}{2} \cos x}$
$\Rightarrow I=\frac{1}{2} \int \frac{1}{\sin \left(x+\frac{\pi}{3}\right)} d x$
$=\frac{1}{2} \int \operatorname{cosec}\left(x+\frac{\pi}{3}\right) d x$
$\Rightarrow I=\frac{1}{2} \log \left|\tan \left(\frac{x}{2}+\frac{\pi}{6}\right)\right|+C$
${\left[\because \int \operatorname{cosec} x d x=\log \left|\tan \frac{x}{2}\right|+C\right]}$
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MCQ 2501 Mark
Evaluate: $\int \frac{10 x^9+10^x \log _e 10}{10^x+x^{10}} d x$
  • A
    $\log _e\left(10^x-x^{10}\right)+C$
  • ✓
    $\log _e\left(10^x+x^{10}\right)+C$
  • C
    $\log _e\left(10^x+x^9\right)+C$
  • D
    $\log _e\left(10^x-x^9\right)+C$
Answer
Correct option: B.
$\log _e\left(10^x+x^{10}\right)+C$
(b) : Let $I=\int \frac{10 x^9+10^x \log _e 10}{10^x+x^{10}} d x$ Put $10^x+x^{10}=t$
$
\begin{aligned}
\Rightarrow \quad & \left(10^x \log _e 10+10 x^9\right) d x=d t \\
\therefore \quad & I=\int \frac{10 x^9+10^x \log _e 10}{10^x+x^{10}} d x=\int \frac{d t}{t} \\
& \quad=\log _e t+C=\log _e\left(10^x+x^{10}\right)+C
\end{aligned}
$
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M.C.Q (1 Marks) - Page 5 - Maths STD 12 Science Questions - Vidyadip