1
MHT CET 2024 9th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The value of $\mathrm{I}=\mathrm{I}=\int_{\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1+\mathrm{e}^{-x}} \mathrm{~d} x$ is equal to

A
$\frac{\pi^2}{4}-2$
B
$\frac{\pi^2}{4}+2$
C
$\pi^2-\mathrm{e}^{\frac{\pi}{2}}$
D
$\pi^2+e^{\frac{\pi}{2}}$
2
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The value of $\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{1}{\sin 2 x\left(\tan ^5 x+\cot ^5 x\right)} d x$ is

A
$\frac{1}{5}\left(\frac{\pi}{4}-\tan ^{-1}\left(\frac{1}{3 \sqrt{3}}\right)\right)$
B
$\frac{1}{2}\left(\frac{\pi}{4}-\tan ^{-1}\left(\frac{1}{9 \sqrt{3}}\right)\right)$
C
$\frac{1}{10}\left(\frac{\pi}{4}-\tan ^{-1}\left(\frac{1}{9 \sqrt{3}}\right)\right)$
D
$\frac{1}{10}\left(\frac{\pi}{4}-\tan ^{-1}\left(\frac{1}{3 \sqrt{3}}\right)\right)$
3
MHT CET 2024 4th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The integral $\int_{\frac{-1}{2}}^{\frac{1}{2}}\left([x]+\log _{\mathrm{e}}\left(\frac{1+x}{1-x}\right)\right) \mathrm{d} x$, where $[x]$ represent greatest integer function, equals

A
$-\frac{1}{2}$
B
$\log _{\mathrm{c}}\left(\frac{1}{2}\right)$
C
$\frac{1}{2}$
D
$ 2 \log _{\mathrm{e}}\left(\frac{1}{2}\right)$
4
MHT CET 2024 3rd May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The value of the integral $\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\left(x^2+\log \frac{\pi-x}{\pi+x}\right) \cos x d x$ is equal to

A
0
B
$\frac{\pi^2}{2}-4$
C
$\frac{\pi^2}{2}$
D
$\frac{\pi^2}{2}+4$
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