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 9th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $\overline{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overline{\mathrm{a}} \cdot \overline{\mathrm{b}}=1$ and $\overline{\mathrm{a}} \times \overline{\mathrm{b}}=\hat{\mathrm{j}}-\hat{\mathrm{k}}$, then $\overline{\mathrm{b}}$ is

A
$\hat{i}-\hat{j}+\hat{k}$
B
$2 \hat{j}-\hat{k}$
C
$\hat{i}$
D
$2 \hat{\mathrm{i}}$
3
MHT CET 2024 9th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

If the mean and the variance of Binomial variate $X$ are 2 and 1 respectively, then the probability that X takes a value greater than or equal to one is

A
$\frac{1}{16}$
B
$\frac{9}{16}$
C
$\frac{3}{4}$
D
$\frac{15}{16}$
4
MHT CET 2024 9th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The value of $I=\int \frac{(x-1) \mathrm{e}^x}{(x+1)^3} \mathrm{dx}$ is

A
$\frac{-\mathrm{e}^x}{(x+1)^2}+C$, (where $C$ is a constant of integration)
B
$\frac{-x \mathrm{e}^x}{(x+1)^2}+\mathrm{C}$, (where C is a constant of integration)
C
$\frac{x \mathrm{e}^x}{(x+1)^2}+C$, (where C is a constant of integration)
D
$\frac{\mathrm{e}^x}{(x+1)^2}+\mathrm{C}$, (where C is a constant of integration)
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