MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 14 - MCQExams.com
CBSE
Class 12 Commerce Applied Mathematics
Indefinite Integrals
Quiz 14
$$\displaystyle \int { { e }^{ \tan ^{ -1 }{ x } } } \left[ \frac { 1+x+x^{ 2 } }{ 1+x^{ 2 } } \right] dx=$$
Report Question
0%
$$\displaystyle x^{ 2 }{ e }^{ \tan ^{ -1 }{ x } }+c$$
0%
$$\displaystyle x{ e }^{ \tan ^{ -1 }{ x } }+c$$
0%
$$\displaystyle \dfrac{{ e }^{ \tan ^{ -1 }{ x } }}{x}+c$$
0%
$$\displaystyle \dfrac{{ e }^{ \tan ^{ -1 }{ x } }}{x^2}+c$$
$$\int {x\left( {\frac{{\ln {a^{{a^{\frac{x}{2}}}}}}}{{3{a^{\frac{{5x}}{2}}}{b^{3x}}}} + \frac{{\ln {b^{{b^{\frac{x}{2}}}}}}}{{2{a^{2x}}{b^{4x}}}}} \right)} dx\left( {where\,a,b \in R} \right)$$ is equal to
Report Question
0%
$$\frac{1}{{6\ln {a^2}{b^3}}}{a^{2x}}{b^{3x}}\ln \frac{{{a^{2x}}{b^{3x}}}}{e} + k$$
0%
$$\frac{1}{{6\ln {a^2}{b^3}}}{a^{2x}}{b^{3x}}\ln \frac{1}{{e{a^{2x}}{b^{3x}}}} + k$$
0%
$$\frac{1}{{6\ln {a^2}{b^3}}}\frac{1}{{{a^{2x}}{b^{3x}}}}\ln \left( {e{a^{2x}}{b^{3x}}} \right) + k$$
0%
$$ - \frac{1}{{6\ln {a^2}{b^3}}}\frac{1}{{{a^{2x}}{b^{3x}}}}\ln \left( {e{a^{2x}}{b^{3x}}} \right) + k$$
$$\displaystyle \int { { e }^{ \tan ^{ -1 }{ x } } } \left[ \dfrac { 1+x+x^{ 2 } }{ 1+x^{ 2 } } \right] dx=$$
Report Question
0%
$$x^{ 2 }{ e }^{ \tan ^{ -1 }{ x } }+c$$
0%
$$x{ e }^{ \tan ^{ -1 }{ x } }+c$$
0%
$${ e }^{ \tan ^{ -1 }{ x } }+c$$
0%
$$\frac { 1 }{ 2 } { e }^{ \tan ^{ -1 }{ x } }+c$$
The value of the integral $$\int\int xy(x+y)dx {\,}dy$$ over the area between $$y=x^2$$ and $$y=x$$ is
Report Question
0%
$$\dfrac{3}{56}$$
0%
$$\dfrac{47}{56}$$
0%
$$\dfrac{33}{56}$$
0%
$$\dfrac{23}{56}$$
$$I=\int { \left\{ \log _{ e }( \log _{ e }{ x }) +\cfrac { 1 }{ { \left( \log _{ e }{ x } \right) }^{ 2 } } \right\} } dx$$ is equal to
Report Question
0%
$$x\log _{ e }( \log _{ e }{ x } )+c$$
0%
$$x\log _{ e }( \log _{ e }{ x })-\cfrac { x }{ { \left( \log _{ e }{ x } \right) }^{ } } +c$$
0%
$$x\log _{ e }{ x } \log _{ e }{ x } +\cfrac { x }{ { \left( \log _{ e }{ x } \right) }^{ } } +c$$
0%
none of these
$$\left[ \int \log ( 1 + \cos x ) - x \tan \frac { x } { 2 } \right] d x$$ is equal to ?
Report Question
0%
$$x \tan \frac { x } { 2 }$$
0%
$$\log ( 1 + \cos x )$$
0%
$$x \log ( 1 + \cos x )$$
0%
None of these
$$\int (1+x-x^{-1})e^{x+x^{1}}dx=$$
Report Question
0%
$$(x+1)e^{x+x^{-1}}+c$$
0%
$$(x-1)e^{x+x^{-1}}+c$$
0%
$$-xe^{x+x^{-1}}+c$$
0%
$$xe^{x+x^{-1}}+c$$
Solve: $$\int\dfrac{sin^32x}{cos^52x}dx$$
Report Question
0%
$$\dfrac{{{{\tan }^4}2x}}{8} + C$$
0%
$$\dfrac{{{{\cos}^4}2x}}{8} + C$$
0%
$$\dfrac{{{{\sin}^4}2x}}{8} + C$$
0%
$$\dfrac{{{{\sec}^4}2x}}{8} + C$$
If $$f\left (\dfrac {x - 4}{x + 2}\right ) = 2x + 1, (x\epsilon R - \left \{1, -2\right \})$$m then $$\int f(x) dx$$ is equal to
(where $$C$$ is a constant of integration).
Report Question
0%
$$12\log_{e}|1 - x| + 3x + C$$
0%
$$-12\log_{e}|1 - x| - 3x + C$$
0%
$$12\log_{e}|1 - x| - 3x + C$$
0%
$$-12\log_{e}|1 - x| + 3x + C$$
Evaluate : $$\int { \cfrac { dx }{ x\cos { x } } } $$
Report Question
0%
$$\ln { x } +\cfrac { { x }^{ 2 } }{ 2 } +\cfrac { { 3x }^{ 3 } }{ 3 } +...$$
0%
$$\ln { x } -\cfrac { { x }^{ 2 } }{ 4 } +\cfrac { { 4x }^{ 4 } }{ 16 } +...$$
0%
$$\ln { x } +\cfrac { { x }^{ 2 } }{ 4 } +\cfrac { { 5x }^{ 4 } }{ 96 } +...$$
0%
$$\ln { x } -\cfrac { { x }^{ 2 } }{ 3 } +\cfrac { { x }^{ 4 } }{ 9 } +...$$
$$\int_{}^{} {\frac{{{x^2}}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}dx} $$ would be equal to
Report Question
0%
$$\frac{{\sin x + x\cos x}}{{x\sin x + \cos x}} + c$$
0%
$$\frac{{\sin x - x\cos x}}{{x\sin x + \cos x}} + c$$
0%
$$\frac{{\sin x - x\cos x}}{{x\sin x - \cos x}} + c$$
0%
none of these
$$\displaystyle \int_{}^{} {{e^x}\left( {\frac{{1 - \sin x}}{{1 - \cos x}}} \right)dx} $$ is equal to :
Report Question
0%
$$ - {e^x}\tan \frac{x}{2} + c$$
0%
$$ - {e^x}\cot \frac{x}{2} + c$$
0%
$$ - \frac{1}{2}{e^x}\tan \frac{x}{2} + c$$
0%
$$ - \frac{1}{2}{e^x}\cot \frac{x}{2} + c$$
Explanation
Let $$I=\displaystyle\int{{e}^{x}\dfrac{1-\sin{x}}{1-\cos{x}}dx}$$
$$=\displaystyle\int{{e}^{x}\dfrac{1-2\sin{\dfrac{x}{2}}\cos{\dfrac{x}{2}}}{1-1+2{\sin}^{2}{\dfrac{x}{2}}}dx}$$
$$=\displaystyle\int{{e}^{x}\dfrac{1-2\sin{\dfrac{x}{2}}\cos{\dfrac{x}{2}}}{2{\sin}^{2}{\dfrac{x}{2}}}dx}$$
$$=\displaystyle\int{{e}^{x}\left(\dfrac{{\csc}^{2}{\dfrac{x}{2}}}{2}-\cot{\dfrac{x}{2}}\right)dx}$$
Let $$f\left(x\right)=-\cot{\dfrac{x}{2}}$$
$$\Rightarrow\,{f}^{\prime}{\left(x\right)}=\dfrac{1}{2}{\csc}^{2}{\dfrac{x}{2}}$$
$$\Rightarrow\,I=\displaystyle\int{{e}^{x}\left(\dfrac{{\csc}^{2}{\dfrac{x}{2}}}{2}-\cot{\dfrac{x}{2}}\right)dx}$$
Using $$\displaystyle\int{{e}^{x}\left(f\left(x\right)+{f}^{\prime}{\left(x\right)}\right)dx}={e}^{x}f\left(x\right)+C$$
$$\Rightarrow\,I=-{e}^{x}\cot{\dfrac{x}{2}}+c$$
If the primitive of $$\frac{1}{{f\left( x \right)}}\,$$ is equal to $$\,{\left\{ {f\left( x \right)} \right\}^2} + c$$,then f(x) is
Report Question
0%
$$x + D$$
0%
$$\frac{x}{2} + d$$
0%
$$\frac{{{x^2}}}{2} + d$$
0%
$${x^2} + d$$
If $$\int \sqrt{\dfrac{x-5}{x-7}}dx=a\sqrt{x^{2}-12x+35+log}|x-6+\sqrt{x^{2}-12x+35}|+C$$ then $$A$$=
Report Question
0%
$$-1$$
0%
$$\dfrac{1}{2}$$
0%
$$-\dfrac{1}{2}$$
0%
$$1$$
If $$\dfrac{d}{dx}\ f(x)=g(x)$$ for $$a\leq x\leq b$$ then $$\int_{b}^{a}f(x)g(x)dx$$ equals to:
Report Question
0%
$$f(2)-f(1)$$
0%
$$g(2)-g(1)$$
0%
$$\dfrac{[f(b)]^{2}-[f(a)]^{2}}{2}$$
0%
$$\dfrac{[g(b)]^{2}-[g(a)]^{2}}{2}$$
The value of $$\displaystyle \int \left ( 3x^{2}\tan \dfrac{1}{x}-x\sec^{2}\dfrac{1}{x} \right )dx$$ is
Report Question
0%
$$x^{3}\tan \dfrac{1}{x}+c$$
0%
$$x^{2}\tan \dfrac{1}{x}+c$$
0%
$$x\tan \dfrac{1}{x}+c$$
0%
$$\tan \dfrac{1}{x}+c$$
Explanation
$$I=\int \left(3x^{2} \tan \dfrac{1}{x}- x \sec^{2} \dfrac{1}{x} \right) dx$$
$$I= \int 3x^{2} \tan \dfrac{1}{x} dx- \int \sec^{2} \dfrac{1}{x}. x \ dx$$
$$I=I_1- I_2$$
$$I_1= \int 3x^{2} \tan \dfrac{1}{x} dx$$
$$=\tan \dfrac{1}{x}. \int 3x^{2} dx- \int \left( \dfrac{d}{dx} \tan \dfrac{1}{x}. \int 3x^{2} dx \right) dx+c$$
$$=\tan \dfrac{1}{x}. x^{3}- \int \sec^{2} \dfrac{1}{x}. \left( \dfrac{-1}{x^{2}} \right). x^{3} dx+c$$
$$=\tan \dfrac{1}{x}. x^{3}+ \int x \sec^{2} \dfrac{1}{x} dx+c$$
Now $$I=I_{1}-I_{2}$$
$$=\tan \dfrac{1}{x}. x^{3}+ \int x. \sec^{2} \dfrac{1}{x} dx- \int x \sec^{2} \dfrac{1}{x} dx+ c$$
$$=x^{2} \tan \dfrac{1}{x}+ c$$.
If $$g(x)$$ is a differentiable function satisfying $$\dfrac{d}{dx}{g(x)}=g(x)$$ and $$g(0)=1,$$ then $$\int { g\left( x \right) } \left( \dfrac { 2-sin2x }{ 1-cos2x } \right) dx$$ is equal to
Report Question
0%
$$g(x)cot$$ $$x+C$$
0%
$$-g(x)cot$$ $$x+C$$
0%
$$\dfrac{g(x)}{1-cos2x}+C$$
0%
$$None$$ $$of$$ $$these$$
Let $$f:R\longrightarrow R,g : R\longrightarrow R$$ be continuous functions. then the value of integeral.
$$\int _{ \ell n\lambda }^{ \ell n/\lambda }{ \frac { f\left( \dfrac { { x }^{ 2 } }{ 4 } \right) \left[ f\left( x \right) -f\left( -x \right) \right] }{ g\left( \dfrac { { x }^{ 2 } }{ 4 } \right) \left[ g\left( x \right) +g\left( -x \right) \right] } } dx$$ is:
Report Question
0%
depend on $$\lambda$$
0%
a non-zero constant
0%
zero
0%
1
If $$I=\int { \dfrac { 1 }{ { x }^{ 4 }\sqrt { { a }^{ 2 }+{ x }^{ 2 } } } dx, } $$ then $$I$$ equals
Report Question
0%
$$\dfrac { 1 }{ { a }^{ 4 } } =\left\{ \dfrac { 1 }{ x } \sqrt { { a }^{ 2 }+{ x }^{ 2 } } -\dfrac { 1 }{ { 3x }^{ 3 } } \sqrt { { a }^{ 2 }{ +x }^{ 2 } } \right\} +C$$
0%
$$\dfrac { 1 }{ { a }^{ 4 } } =\left\{ \dfrac { 1 }{ x } \sqrt { { a }^{ 2 }+{ x }^{ 2 } } -\dfrac { 1 }{ { 3x }^{ 3 } } ({ a }^{ 2 }{ x }^{ 2 })^{3/2} \right\} +C$$
0%
$$\dfrac { 1 }{ { a }^{ 4 } } =\left\{ \dfrac { 1 }{ x } \sqrt { { a }^{ 2 }+{ x }^{ 2 } } -\dfrac { 1 }{ { 2 }\sqrt { x } } ({ a }^{ 2 }{ x }^{ 2 })^{3/2} \right\} +C$$
0%
None of these
Evaluate: $$\displaystyle \int \dfrac { 1 } { x ^ { 2 } \left( x ^ { 4 } + 1 \right) ^ { \frac { 3 } { 4 } } } d x ; x = 0$$
Report Question
0%
$$\dfrac { \left( x ^ { 4 } - 1 \right) ^ { \frac { 1 } { 4 } } } { x } + c$$
0%
$$-\dfrac { \left( x ^ { 4 } + 1 \right) ^ { \frac { 1 } { 4 } } } { x } + c$$
0%
$$\dfrac { \sqrt { x ^ { 4 } + 1 } } { x } + c$$
0%
None of these
Explanation
$$I=\displaystyle\int \frac{dx}{x^{2}(1+x^{4})^{3/4}}$$
$$=\displaystyle\int \frac{dx}{x^{2}\dfrac{(1+x^{4})^{3/4}.x^3}{x^{3}}}$$
$$I=\displaystyle\int \frac{dx}{x^{5}(\dfrac{(1+x^{4})^{1/4}}{x})^{3}}$$
$$u=\dfrac{(1+x^{4})^{1/4}}{x}$$
$$(ux)^{4}= x^{4}+1$$
$$u^{4}= 1+\dfrac{1}{x^{4}}$$
$$4u^{3}du = -\dfrac{4\, dx}{x^{5}}$$
$$u^{3}du=-\dfrac{dx}{x^{5}}$$
$$I=\displaystyle\int \dfrac{-u^{3}du}{u^{3}}=-\int du$$
$$= -u+c$$
$$=-\dfrac{(1+x^{4})^{1/4}}{x}+c$$
If $$ \int\dfrac{x^3-6x^2+11x-6}{\sqrt{x^2+4x+3}}dx=(Ax^2+bx+c)\sqrt{x^2+4x+3}+\lambda \int\dfrac{dx}{\sqrt{x^2+4x+3}}$$ , then value of 'A' is
Report Question
0%
$$\dfrac{1}{3}$$
0%
$$1$$
0%
$$3$$
0%
$$-1/3$$
If $$x\in \left(0, \dfrac{\pi}{2}\right)$$ then $$\displaystyle \int{e^{\dfrac{-x}{2}}\dfrac{\sqrt{1-\sin x}}{1+cos x}dx}$$=
Report Question
0%
$$e^{-\pi/2}\sec\dfrac{x}{2}+c$$
0%
$$-e^{\dfrac{-x}{2}}\sec\dfrac{x}{2}+c$$
0%
$$e^{\dfrac{x}{2}}\sec\dfrac{x}{2}+c$$
0%
$$-e^{\dfrac{x}{2}}\sec\dfrac{x}{2}+c$$
$$\int _ { 0 } ^ { \pi / 4 } \tan ^ { 2 } x d x$$ equals -
Report Question
0%
$$\pi / 4$$
0%
$$1 + ( \pi / 4 )$$
0%
$$1 - ( \pi / 4 )$$
0%
$$1 - ( \pi / 2 )$$
Evaluate
$$\int \dfrac{dx}{\sqrt{1-x}}$$
Report Question
0%
$$\sin^{-1} \sqrt{x}$$
0%
$$-\sin^{-1} \sqrt{x}+c$$
0%
$$2\sqrt{1-x}+c$$
0%
$$-2\sqrt{1-x}+c$$
Let $$[\cdot]$$ denote the greatest integer function then the value of $$\displaystyle\int^{1.5}_0x[x^2]dx$$ is?
Report Question
0%
$$0$$
0%
$$\dfrac{3}{2}$$
0%
$$\dfrac{7}{4}$$
0%
$$\dfrac{5}{4}$$
The integral $$\int \dfrac { \sin ^{ { 2 } } x\cos ^{ { 2 } } x }{ \left( \sin ^{ { 5 } } x+\cos ^{ { 3 } } x\sin ^{ { 2 } } x+\sin ^{ { 3 } } x\cos ^{ { 2 } } x+\cos ^{ { 5 } } x \right) ^{ { 2 } } } dx$$ is equal to :
Report Question
0%
$$\dfrac { - 1 } { 1 + \cot ^ { 3 } x } + C$$
0%
$$\dfrac { 1 } { 3 \left( 1 + \tan ^ { 3 } x \right) } + C$$
0%
$$\dfrac { - 1 } { 3 \left( 1 + \tan ^ { 3 } x \right) } + C$$
0%
$$\dfrac { 1 } { 1 + \cot ^ { 3 } x } + C$$
The integral $$\int \dfrac{3x^{13} + 2x^{11}}{(2x^4 + 3x^2 + 1)^4}dx$$ is equal to: (where C is a constant of integration)
Report Question
0%
$$\dfrac{x^4}{(2x^4 + 3x^2 + 1)^3} + C$$
0%
$$\dfrac{x^{12}}{6(2x^4 + 3x^2 + 1)^3} + C$$
0%
$$\dfrac{x^4}{6(2x^4 + 3x^2 + 1)^3} + C$$
0%
$$\dfrac{x^{12}}{(2x^4 + 3x^2 + 1)^3} + C$$
Explanation
$$\int \dfrac{3x^{13} + 2x^{11}}{(2x^4 + 3x^2 + 1)^4}dx$$
$$\int \dfrac{\lgroup \dfrac{3}{x^3} + \dfrac{2}{x^5} \rgroup dx}{\lgroup 2 + \dfrac{3}{x^2} + \dfrac{1}{x^4} \rgroup^4}$$
Let $$\lgroup 2 + \dfrac{3}{x^2} + \dfrac{1}{x^4} \rgroup = t$$
$$-\dfrac{1}{2}\int\dfrac{dt}{t^4} = \dfrac{1}{6t^3} + C \Rightarrow \dfrac{x^{12}}{6(2x^4 + 3x^2 + 1)^4}+ C$$
The value of $$\displaystyle \int^{\pi/4}_{-\pi /4} \dfrac{dx}{sec^2x(1-sinx)}$$ is
Report Question
0%
$$\pi / 4$$
0%
$$\pi$$
0%
$$\pi / 2$$
0%
$$2 \pi$$
Evaluate: $$\int { \sqrt { \dfrac { x }{ 4-{ x }^{ 3 } } } } dx$$
Report Question
0%
$$\dfrac { 2 }{ 3 } \sin ^{ -1 }{ \left( \dfrac { { x }^{ \dfrac { 3 }{ 2 } } }{ 2 } \right) +c } $$
0%
$$\dfrac { 2 }{ 3 } \sin ^{ -1 }{ \left( { x }^{ \dfrac { 3 }{ 2 } } \right) +c } $$
0%
$$2\sin ^{ -1 }{ \left( \dfrac { { x }^{ \dfrac { 3 }{ 2 } } }{ 2 } \right) +c } $$
0%
$$\dfrac { 1 }{ 3 } \sin ^{ -1 }{ \left( \dfrac { { x }^{ \dfrac { 3 }{ 2 } } }{ 2 } \right) +c } $$
If $$\displaystyle\int \dfrac{\cos x \, dx}{\sin^3x(1+\sin^6x)^{2/3}} = f(x)(1 + \sin^6x)^{1/\alpha} + c$$
Where $$c$$ is a constant of integration, then $$\lambda f\left(\dfrac{\pi}{3}\right)$$ is equal to:
Report Question
0%
$$\dfrac{9}{8}$$
0%
$$-2$$
0%
$$2$$
0%
$$-\dfrac{9}{8}$$
Explanation
$$\displaystyle\int \dfrac{\cos x \, dx}{\sin^3x(1+\sin^6x)^{2/3}} = f(x)(1 + \sin^6x)^{1/\alpha} + c.....given$$
Let $$\sin x = t$$
$$\cot x dx= dt$$
$$\displaystyle I= \int \dfrac{dt}{t^3(1+t^6)^{2/3}}$$
$$\displaystyle I=\int \dfrac{dt}{t^3.t^4\left(1 + \dfrac{1}{t^6}\right)^{2/3}}$$
$$\displaystyle I=\int \dfrac{dt}{t^7\left(1 + \dfrac{1}{t^6}\right)^{2/3}}$$
Put $$ 1+\dfrac{1}{t^6}=r^3$$
$$\Rightarrow \dfrac{dt}{t^7} = \dfrac{-1}{2}r^2dr$$
$$= \dfrac{-1}{2}\displaystyle\int \dfrac{r^2dr}{r^2}$$
$$=\dfrac{-1}{2}r +c$$
$$= \dfrac{-1}{2}\left(\dfrac{\sin^6x + 1}{\sin^6x}\right)^{1/3}+c .......$$As $$r=\left (1+\dfrac{1}{t^6}\right )^{1/3}$$ and $$t=\sin x$$
$$=\dfrac{-1}{2\sin^2x} (1 + \sin^6x)^{1/3}+c$$
$$f(x) = -\dfrac{1}{2} cosec^{2}x$$ and $$\lambda = 3$$
Now for $$x=\dfrac{\pi}3$$
$$cosec \dfrac{\pi}3=\dfrac{2}{\sqrt3}$$
$$\lambda f \left(\dfrac{\pi}{3}\right) = \dfrac{-1}2\times \left (\dfrac{2}{\sqrt3}\right )^2\times 3$$
$$\boxed{\lambda f \left(\dfrac{\pi}{3}\right) = -2}.....Answer$$
Hence option $$'B'$$ is the answer.
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page
