1
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.
2
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.
3
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
$\frac{\pi}{6}$
4
MHT CET 2024 2nd May Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $\mathrm{a}>0$ and $\mathrm{z}=\frac{(1+\mathrm{i})^2}{\mathrm{a}-\mathrm{i}}, \mathrm{i}=\sqrt{-1}$, has magnitude $\sqrt{\frac{2}{5}}$ then $\bar{z}$ is equal to

A
$\frac{1}{5}-\frac{3}{5} \mathrm{i}$
B
$-\frac{1}{5}-\frac{3}{5} \mathrm{i}$
C
$-\frac{1}{5}+\frac{3}{5} \mathrm{i}$
D
$-\frac{3}{5}-\frac{1}{5} \mathrm{i}$
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