1
MHT CET 2026 13th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
If $y = \tan^{-1}\left\{\dfrac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{\sqrt{1 + x^2} + \sqrt{1 - x^2}}\right\}$, where $|x| < 1$, then $\dfrac{dy}{dx}$ is equal to
A
$\dfrac{-x}{\sqrt{1 - x^4}}$
B
$\dfrac{x}{\sqrt{1 - x^4}}$
C
$\dfrac{-2x}{\sqrt{1 - x^4}}$
D
$\dfrac{2x}{\sqrt{1 - x^4}}$
2
MHT CET 2026 13th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
The value of $\tan^{-1}(\sqrt{3}) + \sec^{-1}(-2) - \sin^{-1}\left(-\dfrac{1}{2}\right)$ is
A
$\dfrac{5\pi}{6}$
B
$\dfrac{2\pi}{3}$
C
$\dfrac{7\pi}{6}$
D
$\dfrac{4\pi}{3}$
3
MHT CET 2026 13th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
The domain of the function $f(x) = e^{\sqrt{5x - 3 - 2x^2}}$ is
A
$(1, 2)$
B
$\left(-1, \dfrac{3}{2}\right)$
C
$\left[1, \dfrac{3}{2}\right]$
D
$(-1, 0)$
4
MHT CET 2026 13th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
If $f(x) = \begin{cases} \dfrac{(8 - 2x)^{\frac{1}{3}} - 2}{3 - (243 + 5x)^{\frac{1}{5}}}, & \text{if } x \neq 0 \\ k, & \text{if } x = 0 \end{cases}$ is continuous at $x = 0$, then $k =$
A
$\dfrac{5}{2}$
B
$-\dfrac{5}{2}$
C
$\dfrac{27}{2}$
D
$\dfrac{-27}{2}$

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