1
MHT CET 2025 21st April Morning Shift
MCQ (Single Correct Answer)
+2
-0

$$ \int \frac{\mathrm{d} x}{x\left(x^2+1\right)}= $$

A
$\log (x)-\frac{1}{2} \log \left(x^2+1\right)+\mathrm{c}$, where c is the constant of integration.
B
$\frac{1}{2} \log (x)-\log \left(x^2+1\right)+\mathrm{c}$, where c is the constant of integration.
C
$\log (x)+\frac{1}{2} \log \left(x^2+1\right)+\mathrm{c}$, where c is the constant of integration.
D
$-\log (x)-\frac{1}{2} \log \left(x^2+1\right)+\mathrm{c}$, where c is the constant of integration.
2
MHT CET 2025 21st April Morning Shift
MCQ (Single Correct Answer)
+2
-0

The joint equation of the bisectors of the angles between the lines $x=5$ and $y=3$ is

A
$x^2-y^2-10 x+6 y+16=0$
B
$x^2+y^2-10 x-6 y-16=0$
C
$x^2+y^2+10 x+6 y-16=0$
D
$x^2+y^2-5 x-2 y-7=0$
3
MHT CET 2025 21st April Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $\sin ^{-1} \frac{1}{3}+\sin ^{-1} \frac{3}{5}+\sin ^{-1} x=\frac{\pi}{2}$ then $x=$

A
$\frac{8 \sqrt{2}+3}{5}$
B
$\frac{8 \sqrt{2}-3}{5}$
C
$\frac{8 \sqrt{2}+3}{15}$
D
$\frac{8 \sqrt{2}-3}{15}$
4
MHT CET 2025 21st April Morning Shift
MCQ (Single Correct Answer)
+2
-0

$\int \sqrt{x^2-6 x-16} \mathrm{~d} x$ equals

A

$\left(\frac{x-3}{2}\right) \sqrt{x^2-6 x-16} +\frac{5}{2} \log \left(x-3+\sqrt{x^2-6 x-16}\right)+c$

where c is the constant of integration

B

$$ \begin{aligned} & \left(\frac{x-3}{2}\right) \sqrt{x^2-6 x-16} -\frac{25}{2} \log \left(x-3+\sqrt{x^2-6 x-16}\right)+c \end{aligned} $$

where c is the constant of integration

C

$\left(\frac{x-3}{2}\right) \sqrt{x^2-6 x-16}+\frac{25}{2} \log \left(x-3+\sqrt{x^2-6 x-16}\right)+c $

where c is the constant of integration

D

$$ \begin{aligned} \left(\frac{x-3}{2}\right) \sqrt{x^2-6 x-16} & -\frac{5}{2} \log \left(x-3+\sqrt{x^2-6 x-16}\right)+c \end{aligned} $$

where $c$ is the constant of integration

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