MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 12 Commerce Maths Integrals Quiz 11 - MCQExams.com
CBSE
Class 12 Commerce Maths
Integrals
Quiz 11
What is $$\int _ { 1 } ^ { 3 } \left| 1 - x ^ { 4 } \right| d x$$ equal to $$?$$
Report Question
0%
$$- 232 / 5$$
0%
$$- 116 / 5$$
0%
$$ 116 / 5$$
0%
$$ 232 / 5$$
The value of $$\int _{ 3/2 }^{ 2 }{ { \left( \cfrac { x-1 }{ 3-x } \right) }^{ 1/2 } } dx$$ is
Report Question
0%
$$\cfrac { \sqrt { 3 } }{ 2 } -1+\cfrac { \pi }{ 6 } $$
0%
$$\cfrac { \sqrt { 3 } }{ 2 } -1+\cfrac { \pi }{ 8 } $$
0%
$$\cfrac { \sqrt { 3 } }{ 2 } +1-\cfrac { \pi }{ 6 } $$
0%
None of these
The value of integral $$\int _{ a }^{ b }{ \frac { \left| x \right| }{ x } dx,\quad a<b }$$ is
Report Question
0%
$$\left| a \right| -\left| b \right|$$
0%
$$\left| b \right| -\left| a \right|$$
0%
$$\left| a \right| -b$$
0%
$$\left| b \right| -a$$
The value of the definite integral $$\int _0^1 (1+e^{-x^2})dx$$ is
Report Question
0%
$$-1$$
0%
$$2$$
0%
$$1+e^{-1}$$
0%
None of the above
Let $$J=\int _{ 0 }^{ \infty }{ \dfrac { Inx }{ 1+{ x }^{ 3 } } } dx$$ and $$K=\int _{ 0 }^{ \infty }{ \dfrac { xInx }{ 1+{ x }^{ 3 } } } dx$$ then
Report Question
0%
$$J+K=0$$
0%
$$J-K=0$$
0%
$$J+K<0$$
0%
$$J+K>0$$
The integral $$\int^{\pi/3}_{\pi/6} \dfrac{f(x)}{f(x) +f(\dfrac{\pi}{2} -x)} dx$$ is equal to
Report Question
0%
$$\dfrac{\pi}{12}$$
0%
$$\dfrac{12}{\pi}$$
0%
$$\dfrac{\pi}{13}$$
0%
$$\dfrac{13}{\pi}$$
If $$a > 0$$ and $$A=\displaystyle \int^{a}_{0}\cos^{-1}xdx$$, then $$\int ^{a}_{-a}(\cos^{-1}x-\sin^{1}\sqrt {1-x^{2}})dx=\pi a-\lambda A$$. Then $$\lambda$$
Report Question
0%
$$0$$
0%
$$3$$
0%
$$2$$
0%
$$none\ of\ these$$
$$\displaystyle \int_{\pi/4}^{\pi/2}{\sqrt{2+\sqrt{2+2\cos 4x}}dx}$$ is equal to
Report Question
0%
$$\sqrt{2}$$
0%
$$\sqrt{2(\sqrt{2-1})}$$
0%
$$2$$
0%
$$none\ of\ these$$
$$\int {{{\left( {1 + {x^4}} \right)}^{\dfrac{1}{4}}}dx = A\;In\;\left( {\dfrac{{{{\left( {1 + {x^4}} \right)}^{\dfrac{1}{4}}} + x}}{{{{\left( {1 + {x^4}} \right)}^{\dfrac{1}{4}}} - x}}} \right) + B\;\tan {\;^{ - 1}}\left( {\dfrac{{{{\left( {1 + {x^4}} \right)}^{\dfrac{1}{4}}}}}{x}} \right) + C,\;then\;A + B = } $$
Report Question
0%
1/4
0%
1/2
0%
$$ - \dfrac{1}{4}$$
0%
$$ - \dfrac{1}{2}$$
$$\displaystyle \int_{4}^{5}e^{(x + 5)^2} dx + 3\int_{1/3}^{2/3}e^{9\left ( x \dfrac{2}{3} \right )^2}dx$$ is equal to
Report Question
0%
1
0%
-1
0%
0
0%
2
Let $$ 1 _ { n } = \int _ { 0 } ^ { \frac { 1 } { 2 } } \frac { 1 } { \sqrt { 1 - x ^ { n } } } d x $$ where $$ n > 2 , $$ then
Report Question
0%
$$
I _ { n } < \frac { \pi } { 6 }
$$
0%
$$
I _ { n } > \frac { \pi } { 6 }
$$
0%
$$
\mathrm { I } _ { \mathrm { n } } < \frac { 1 } { 2 }
$$
0%
$$
I _ { n } > \frac { 1 } { 2 }
$$
If $$\int _{ 0 }^{ b-c }{ f\left( x+c \right) dx=k\int _{ b }^{ c }{ f\left( x \right) dx } } $$ then k=
Report Question
0%
0
0%
1
0%
2
0%
-1
If $$l_1-\int _{ 0 }^{ 1 }{ { 2 }^{ { x }^{ 2 } } } ,{ l }_{ 2 }-\int _{ 0 }^{ 1 }{ { 2 }^{ { x }^{ 3 } } } dx,{ l }_{ 3 }-\int _{ 1 }^{ 2 }{ { 2 }^{ { x }^{ 2 } } } dx$$ and $$l_4-\int _{ 1 }^{ 2 }{ { 2 }^{ { x }^{ 3 } } } dx$$ then
Report Question
0%
$$l_2 > l_1$$
0%
$$l_1 > l_2$$
0%
$$l_3-l_2$$
0%
None of these
Find the value of the equation : $$\int _ { \ln \lambda } ^ { \ln \left( \frac { 1 } { \lambda } \right) } \dfrac { f \left( \dfrac { x ^ { 2 } } { 3 } \right) ( f ( x ) + f ( - x ) ) } { g \left( 3 x ^ { 2 } \right) ( g ( x ) - g ( - x ) ) } d x =?$$ where $$\lambda > 1$$
Report Question
0%
$$0$$
0%
$$1$$
0%
$$\lambda$$
0%
$$1 / \lambda$$
Find the value of the equation $$\int _ { 0 } ^ { \infty } \dfrac { d x } { \left( x + \sqrt { x ^ { 2 } + 1 } \right) ^ { 3 } } =?$$
Report Question
0%
$$3 / 8$$
0%
$$1/8$$
0%
$$-3/8$$
0%
$$-1/8$$
$$7 \left( \int_{\pi}^{0}\dfrac{x^{4}(1-x)^{4} dx}{1+x^{2}}+\pi \right)$$ is equal to
Report Question
0%
$$21$$
0%
$$22$$
0%
$$23$$
0%
$$none\ of\ these$$
$$\int _{ 0 }^{ 1 }{ \sqrt { \cfrac { 1-x }{ 1+x } } dx } $$, is equal to:
Report Question
0%
$$\cfrac { \pi } { 4 } - 1$$
0%
$$\cfrac { \pi } { 2 } - 1$$
0%
$$\frac { \pi } { 4 } + 1$$
0%
$$\cfrac { \pi } { 2 } + 1$$
$$\int _ { 0 } ^ { 1 } \sqrt { \cfrac { 1 - x } { 1 + x } } d x ,$$ is equal to
Report Question
0%
$$\cfrac { \pi } { 4 } - 1$$
0%
$$\cfrac { \pi } { 2 } - 1$$
0%
$$\cfrac { \pi } { 4 } + 1$$
0%
$$\cfrac { \pi } { 2 } + 1$$
$$\frac { 1 }{ \pi } \int _{ -2 }^{ 2 }{ \frac { 1 }{ 4+{ x }^{ 2 } } dx= } $$
Report Question
0%
$$0$$
0%
$$\frac { 1 }{ 4 } $$
0%
$$\frac { \pi }{ 4 } $$
0%
$$\frac { 1 }{ 2 } $$
$$\int _{ 0 }^{ 8 }{ \left[ \sqrt { t } \right] } dt$$ at equals to (where [.] greatest integer function.)
Report Question
0%
28
0%
11
0%
2
0%
8
If $$f\left( x \right) =\int _{ 0 }^{ 1 }{ \left( xf\left( t \right) +1 \right) dt,then\int _{ 0 }^{ 3 }{ f\left( x \right) dx=12 } } $$
because
Statement-2: f(x) = 3x + 1
Report Question
0%
Statements-1 is true, statements-2 is true and statements-2 is correct explanation for statement-1.
0%
Statements-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statements-1.
0%
Statements-1 is true, statements-2 is false.
0%
Statements-1 is false, statements-2 is true.
If $${ I }_{ m }=\overset { e }{ \underset { 1 }{ \int } } (lnx)^{ m }dx,$$ where $$m\epsilon N,$$then $${ I }_{ 10 }+10{ I }_{ 9 }$$ is equal to-
Report Question
0%
$${ e }^{ 10 }$$
0%
$$\frac { { e }^{ 10 } }{ 10 } $$
0%
e
0%
e-1
If for every integer n, $$\int _{ n }^{ n+1 }{ f(x)dx={ n }^{ 2 } } $$, then the value of $$\int _{ -2 }^{ 4 }{ f(x)dx } $$ is -
Report Question
0%
16
0%
14
0%
19
0%
None of these.
$$\displaystyle E = \int_{R}^{\infty}\dfrac{GMm}{x^2}$$ dx , (where $$G$$ , $$M$$ , $$m$$ are constants ) equal to
Report Question
0%
$$-\dfrac{GMm}{R^2}$$
0%
$$+\dfrac{GMm}{R^2}$$
0%
$$-\dfrac{GMm}{R}$$
0%
$$+\dfrac{GMm}{R}$$
$$\int _{ 0 }^{ 4036 }{ \dfrac { { 2 }^{ x } }{ { 2 }^{ x }+{ 1 }^{ 4036-x } } } dx=............$$
Report Question
0%
2018
0%
4035
0%
2017
0%
-2015
If $${ I }_{ 1 }=\int _{ x }^{ 1 }{ \cfrac { 1 }{ 1+{ t }^{ 2 } } } dt$$ and $${ I }_{ 2 }=\int _{ 1 }^{ 1/x }{ \cfrac { 1 }{ 1+{ t }^{ 2 } } } dt$$ for x > 0, then
Report Question
0%
$${ I }_{ 1 }={ I }_{ 2 }$$
0%
$${ I }_{ 1 }>{ I }_{ 2 }$$
0%
$${ I }_{ 2 }>{ I }_{ 1 }$$
0%
None of these.
The value of the integral $$\int _{ \dfrac { 1 }{ 3 } }^{ 1 }{ \dfrac { \left( x-{ x }^{ 3 } \right) ^{ \dfrac { 1 }{ 3 } } }{ { x }^{ 4 } } }dx $$ is
Report Question
0%
6
0%
0
0%
3
0%
4
$$\overset { -5 }{ \underset { -4 }{ \int } } e^(x + 5)^2 dx + 3 \overset { 2/3}{ \underset { 1/3 }{ \int } } e^9(9(x-2/3)^2$$ dx is equal toi
Report Question
0%
$$e^5$$
0%
$$e^4$$
0%
$$3e^2$$
0%
0
If $$z=x+3i$$ then value of $$\displaystyle\int^4_2\left[arg\left|\dfrac{z-i}{z+i}\right|\right]dx$$, where $$[.]$$ denotes the greatest integer function, is?
Report Question
0%
$$3\sqrt{2}$$
0%
$$6\sqrt{3}$$
0%
$$\sqrt{6}$$
0%
None
The area of the region bounded by the lines $$x = 1, x = 2$$, and the curves $$x(y - e^x) = \sin x$$ and $$2xy = 2 \sin x + x^3$$ is
Report Question
0%
$$e^2 - e - \dfrac{1}{6}$$
0%
$$e^2 - e - \dfrac{7}{6}$$
0%
$$e^2 - e + \dfrac{1}{6}$$
0%
$$e^2 - e + \dfrac{7}{6}$$
Explanation
Given:$$x\left(y-{e}^{x}\right)=\sin{x}$$
$$\Rightarrow\,xy-x{e}^{x}=\sin{x}$$
$$\Rightarrow\,xy=\sin{x}+x{e}^{x}$$
$$\Rightarrow\,y=\dfrac{\sin{x}}{x}+{e}^{x}$$
Given:$$2xy=2\sin{x}+{x}^{3}$$
$$\Rightarrow\,y=\dfrac{\sin{x}}{x}+\dfrac{{x}^{2}}{2}$$
$$A=\displaystyle\int_{1}^{2}{\left\{\dfrac{\sin{x}}{x}+{e}^{x}-\left(\dfrac{\sin{x}}{x}+\dfrac{{x}^{2}}{2}\right)\right\}dx}$$
$$=\displaystyle\int_{1}^{2}{\left({e}^{x}-\dfrac{{x}^{2}}{2}\right)dx}$$
$$=\left[{e}^{x}-\dfrac{{x}^{3}}{6}\right]_{1}^{2}$$
$$={e}^{2}-\dfrac{8}{6}-\left(e-\dfrac{1}{6}\right)$$
$$={e}^{2}-e-\dfrac{7}{6}$$
Suppose $$I_1=\displaystyle \int_{0}^{\pi/2} \cos(\pi \sin^2 x)dx;I_2=\displaystyle \int_{0}^{\pi/2} \cos(2\pi \sin^2x)dx$$ and $$I_3=\displaystyle \int_{0}^{\pi/2} \cos(\pi \sin x)dx $$ then
Report Question
0%
$$I_1=0$$
0%
$$I_2+I_3=0$$
0%
$$I_1+I_2+I_3=0$$
0%
$$I_2=I_3$$
Explanation
We have $$I_1=\displaystyle \int_{0}^{\pi/2} cos(\pi sin^2x)dx$$
using $$\int_a^b f(x) \ dx=\int_a^b f(a+b-x) \ dx $$
$$I_1=\displaystyle \int_{0}^{\pi/2} cos(\pi \sin^2(\dfrac {\pi}2- x))dx$$
$$I_1=\displaystyle \int_{0}^{\pi/2} cos(\pi cos^2x)dx$$
On adding,$$ 2I_1=\displaystyle \int_{0}^{\pi/2} cos(\pi sin^2x)+cos(\pi cos^2x )dx$$
$$= \displaystyle \int_{0}^{\pi/2} 2cos\left( \dfrac{\pi}{2}\right).cos\left( \dfrac{\pi}{2}cos2x\right)dx $$
$$\Rightarrow I_1=0$$
Now,
$$I_2=\displaystyle \int_{0}^{\pi/2} \cos(2\pi \sin^2x)dx$$
$$I_2=\displaystyle \int_{0}^{\pi/2} cos\left\{ \pi(1-\cos2x)\right\}dx$$
$$ =-\displaystyle \int_{0}^{\pi/2} \cos(\pi \cos2x)dx $$
$$=-\dfrac{1}{2}\displaystyle \int_{0}^{\pi/4} cos(\pi cos t)dt \qquad (put 2x=t) $$
$$=-\dfrac{2}{2} \displaystyle \int_{0}^{\pi/2} cos(\pi cost)dt =-I_3$$
$$\Rightarrow I_2+I_3=0$$
$$I_3=-\displaystyle \int_{0}^{\pi/2} cos(\pi sint)dt$$
$$\therefore I_2+I_3=0$$
Hence, $$I_1+I_2+I_3=0$$
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
31
32
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
31
32
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 Maths Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page