1
MHT CET 2026 17th April Evening Shift
MCQ (Single Correct Answer)
+2
-0
If $A = \begin{bmatrix} 3 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & 7 \end{bmatrix}$; $B = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{bmatrix}$ then, $(2A + 3B)^{-1} =$ _______
A
$\begin{bmatrix} \dfrac{1}{3} & 0 & 0 \\ 0 & -\dfrac{1}{4} & 0 \\ 0 & 0 & \dfrac{1}{26} \end{bmatrix}$
B
$\begin{bmatrix} \dfrac{1}{3} & 0 & 0 \\ 0 & \dfrac{1}{4} & 0 \\ 0 & 0 & \dfrac{1}{26} \end{bmatrix}$
C
$\begin{bmatrix} \dfrac{1}{3} & 0 & 0 \\ 0 & \dfrac{1}{4} & 0 \\ 0 & 0 & -\dfrac{1}{26} \end{bmatrix}$
D
$\begin{bmatrix} -\dfrac{1}{3} & 0 & 0 \\ 0 & \dfrac{1}{4} & 0 \\ 0 & 0 & -\dfrac{1}{26} \end{bmatrix}$
2
MHT CET 2026 17th April Evening Shift
MCQ (Single Correct Answer)
+2
-0
If $\sin^{-1}(x - 2) + \cos^{-1}(x) + \tan^{-1}(x + 2) + \cot^{-1}(x + 4) = \sec^{-1}(\sqrt{k}) - \dfrac{\pi}{2}$, then $\cos(2\,\text{cosec}^{-1}\sqrt{k - 1}) = \ldots$
A
$\dfrac{15}{16}$
B
$\dfrac{31}{32}$
C
$\dfrac{63}{64}$
D
$\dfrac{7}{8}$
3
MHT CET 2026 17th April Evening Shift
MCQ (Single Correct Answer)
+2
-0
$\sin\left(3\sin^{-1}\left(\dfrac{1}{5}\right)\right) =$
A
$\dfrac{74}{125}$
B
$\dfrac{71}{125}$
C
$\dfrac{3}{5}$
D
$\dfrac{1}{2}$
4
MHT CET 2026 17th April Evening Shift
MCQ (Single Correct Answer)
+2
-0
$\cos^{-1}\left(\cos\dfrac{4\pi}{3}\right) + \sin^{-1}\left(\sin\dfrac{4\pi}{3}\right) = \ldots$
A
$\dfrac{4\pi}{3}$
B
$\dfrac{8\pi}{3}$
C
$\dfrac{\pi}{3}$
D
$\dfrac{3\pi}{2}$

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