JEE MAIN Indefinite Integration Previous Year Questions (PYQs) – Page 1 of 6
A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations
Study Stuff for MCA Examinations
Showing 52 questions
Let $\mathrm{I}(x)=\int \frac{d x}{(x-11)^{\frac{11}{13}}(x+15)^{\frac{15}{13}}}$. If $\mathrm{I}(37)-\mathrm{I}(24)=\frac{1}{4}\left(\frac{1}{\mathrm{~b}^{\frac{1}{13}}}-\frac{1}{\mathrm{c}^{\frac{1}{13}}}\right), \mathrm{b}, \mathrm{c} \in \mathcal{N}$, then $3(\mathrm{~b}+\mathrm{c})$ is equal to
1
2
3
4
Solution
Let $t=\frac{x-11}{x+15}$
Then, $dx=\frac{26}{(1-t)^2}dt$
Also, $(x-11)^{\frac{11}{13}}(x+15)^{\frac{15}{13}}=t^{\frac{11}{13}}(x+15)^2$
and $x+15=\frac{26}{1-t}$
So, $I=\int \frac{\frac{26}{(1-t)^2}dt}{t^{\frac{11}{13}}\cdot \frac{26^2}{(1-t)^2}}$
$I=\frac{1}{26}\int t^{-\frac{11}{13}}dt$
$I=\frac{1}{26}\cdot\frac{13}{2}t^{\frac{2}{13}}$
$I=\frac14\left(\frac{x-11}{x+15}\right)^{\frac{2}{13}}$
Now, $I(37)-I(24)=\frac14\left[\left(\frac{26}{52}\right)^{\frac{2}{13}}-\left(\frac{13}{39}\right)^{\frac{2}{13}}\right]$
$=\frac14\left[\left(\frac12\right)^{\frac{2}{13}}-\left(\frac13\right)^{\frac{2}{13}}\right]$
$=\frac14\left[\frac{1}{4^{\frac{1}{13}}}-\frac{1}{9^{\frac{1}{13}}}\right]$
So, $b=4,\ c=9$
$3(b+c)=3(4+9)=39$
$\boxed{39}$
Then, $dx=\frac{26}{(1-t)^2}dt$
Also, $(x-11)^{\frac{11}{13}}(x+15)^{\frac{15}{13}}=t^{\frac{11}{13}}(x+15)^2$
and $x+15=\frac{26}{1-t}$
So, $I=\int \frac{\frac{26}{(1-t)^2}dt}{t^{\frac{11}{13}}\cdot \frac{26^2}{(1-t)^2}}$
$I=\frac{1}{26}\int t^{-\frac{11}{13}}dt$
$I=\frac{1}{26}\cdot\frac{13}{2}t^{\frac{2}{13}}$
$I=\frac14\left(\frac{x-11}{x+15}\right)^{\frac{2}{13}}$
Now, $I(37)-I(24)=\frac14\left[\left(\frac{26}{52}\right)^{\frac{2}{13}}-\left(\frac{13}{39}\right)^{\frac{2}{13}}\right]$
$=\frac14\left[\left(\frac12\right)^{\frac{2}{13}}-\left(\frac13\right)^{\frac{2}{13}}\right]$
$=\frac14\left[\frac{1}{4^{\frac{1}{13}}}-\frac{1}{9^{\frac{1}{13}}}\right]$
So, $b=4,\ c=9$
$3(b+c)=3(4+9)=39$
$\boxed{39}$
The integral $\displaystyle \int (1+x-\frac{1}{x})e^{x+\frac{1}{x}}\,dx$ is equal to
1
2
3
4
Solution
If $\displaystyle \int x^{5}\,e^{-4x^{3}}\,dx=\dfrac{1}{48}\,e^{-4x^{3}}\,f(x)+C$, where $C$ is a constant of integration, then $f(x)$ is equal to –
1
2
3
4
Solution
Let $\int x^3 \sin x \mathrm{~d} x=g(x)+C$, where $C$ is the constant of integration. If $8\left(g\left(\frac{\pi}{2}\right)+g^{\prime}\left(\frac{\pi}{2}\right)\right)=\alpha \pi^3+\beta \pi^2+\gamma, \alpha, \beta, \gamma \in Z$, then $\alpha+\beta-\gamma$ equals :
1
2
3
4
Solution
If $\int f(x)\,dx=\psi(x)$, then $\int x^{5}f(x^{3})\,dx$ is equal to
1
2
3
4
Solution
Let $I(x)=\displaystyle \int \frac{x^{2}\big(x\sec^{2}x+\tan x\big)}{(x\tan x+1)^{2}}\,dx.$
If $I(0)=0$, then $I\!\left(\frac{\pi}{4}\right)$ is equal to:
1
2
3
4
Solution
If $\displaystyle \int \frac{\sqrt{\,1-x^{2}\,}}{x^{4}}\,dx = A(x)\left(\sqrt{\,1-x^{2}\,}\right)^{m} + C$, for a suitable chosen integer $m$ and a function $A(x)$, where $C$ is a constant of integration, then $(A(x))^{m}$ equals :
1
2
3
4
Solution
If $\displaystyle \int (\sin x)^{-\frac{11}{2}} (\cos x)^{\frac{5}{2}} dx = -\frac{p_1}{q_1}(\cot x)^{\frac{9}{2}} - \frac{p_2}{q_2}(\cot x)^{\frac{5}{2}} - \frac{p_3}{q_3}(\cot x)^{\frac{1}{2}} + \frac{p_4}{q_4}(\cot x)^{-\frac{3}{2}} + C$,
where $p_i, q_i$ are positive integers with $\gcd(p_i,q_i)=1$ for $i = 1,2,3,4$ and $C$ is the constant of integration, then
$\displaystyle \frac{15p_1p_2p_3p_4}{q_1q_2q_3q_4}$ is equal to ______.
1
2
3
4
Solution
$\displaystyle \int (\tan x)^{-\frac{11}{2}} \sec^8 x , dx$
$= \int (\tan x)^{-\frac{11}{2}} (1+\tan^2 x)^3 \sec^2 x , dx$
Put $\tan x = t$
$\Rightarrow \int t^{-\frac{11}{2}} (1+t^2)^3 , dt$
$= \int \left(t^{-\frac{11}{2}} + 3t^{-\frac{7}{2}} + 3t^{-\frac{3}{2}} + t^{\frac{1}{2}}\right) dt$
$= -\frac{2}{9}t^{-\frac{9}{2}} - \frac{6}{5}t^{-\frac{5}{2}} - 2t^{-\frac{1}{2}} + \frac{2}{3}t^{\frac{3}{2}} + C$
Back substitute $t = \cot x$
$\Rightarrow p_1 = 2,; p_2 = 6,; p_3 = 2,; p_4 = 2$
$q_1 = 9,; q_2 = 5,; q_3 = 1,; q_4 = 3$
$\displaystyle \frac{15p_1p_2p_3p_4}{q_1q_2q_3q_4} = \frac{15 \cdot 2 \cdot 6 \cdot 2 \cdot 2}{9 \cdot 5 \cdot 1 \cdot 3} = 16$
If $f\left( {{{x - 4} \over {x + 2}}} \right) = 2x + 1,$ (x $ \in $ R $-${1, $-$ 2}), then $\int f \left( x \right)dx$ is equal to :
(where C is a constant of integration)
(where C is a constant of integration)
1
2
3
4
Solution
If $\displaystyle \int \frac{x+1}{\sqrt{2x-1}}\,dx = f(x)\,\sqrt{2x-1}+C$, where $C$ is a constant of integration, then $f(x)$ is equal to :
1
2
3
4
Solution
JEE MAIN
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
JEE MAIN
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
Ask Your Question or Put Your Review.
