1
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
$\dfrac{-27}{2}$
2
MHT CET 2026 13th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
If $y = x + e^x$, then $\dfrac{d^2 x}{dy^2}$ is equal to
A
$\dfrac{e^x}{(1 + e^x)^2}$
B
$\dfrac{-e^x}{(1 + e^x)^2}$
C
$\dfrac{e^x}{(1 + e^x)^3}$
D
$\dfrac{-e^x}{(1 + e^x)^3}$
3
MHT CET 2026 13th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
Let $f(x) = x - 5$.
If $g(x) = [f(4h(x) + 3)]^2$ and $h(1) = 4, h'(1) = -2$, then $g'(1) =$ .....
A
$-242$
B
$-224$
C
$-112$
D
$-640$
4
MHT CET 2026 13th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
The equation of the normal to the curve $xy + 7 = 0$ is $Ax + By + C = 0$, then
A
$A > 0, B > 0$ or $A < 0, B < 0$
B
$A > 0, B < 0$
C
$A < 0, B > 0$
D
$C = 0$

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