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

If the vectors $\overline{\mathrm{a}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}, \overline{\mathrm{b}}=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overline{\mathrm{c}}=\mathrm{mi}+\mathrm{j}+\mathrm{nk}$ are mutually perpendicular, then $(\mathrm{m}, \mathrm{n})$ is

A
$(3,-2)$
B
$(-2,3)$
C
$(2,-3)$
D
$(-3,2)$
2
MHT CET 2024 2nd May Morning Shift
MCQ (Single Correct Answer)
+2
-0

$$\int \log (1+x)^{1+x} \mathrm{~d} x=$$

A
$(1+x)^2 \log (1+x)-\frac{1}{2}+\mathrm{c}$, where c is a constant of integration.
B
$\frac{(1+x)^2}{2} \log (1+x)+\mathrm{c}$, where c is a constant of integration.
C
$\frac{(1+x)^2}{2}\left[\log (1+x)-\frac{1}{2}\right]+\mathrm{c}$, where c is a constant of integration.
D
$\frac{1+x}{2} \log (1+x)+\mathrm{c}$, where c is a constant of integration.
3
MHT CET 2024 2nd May Morning Shift
MCQ (Single Correct Answer)
+2
-0

$$\int\left(\frac{x+2}{x+4}\right)^2 \cdot e^x \mathrm{~d} x=$$

A
$\mathrm{e}^x\left(\frac{x}{x+4}\right)+\mathrm{c}$, where c is a constant of integration.
B
$\mathrm{e}^x\left(\frac{x+2}{x+4}\right)+\mathrm{c}$, where c is a constant of integration.
C
$\mathrm{e}^x\left(\frac{x-2}{x+4}\right)+\mathrm{c}$, where c is a constant of integration.
D
$\mathrm{e}^x\left(\frac{2 x}{x+4}\right)+\mathrm{c}$, where c is a constant of integration.
4
MHT CET 2024 2nd May Morning Shift
MCQ (Single Correct Answer)
+2
-0

In $\triangle A B C$, with usual notations, if $\frac{1}{b+c}+\frac{1}{c+a}=\frac{3}{a+b+c}$, then $m \angle C$ is equal to

A
$\frac{\pi}{3}$
B
$\frac{\pi}{2}$
C
$\frac{\pi}{4}$
D
$\frac{\pi}{6}$
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