1
MHT CET 2026 19th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
The inverse of matrix $\begin{bmatrix} 1+pq & p & 0 \\ q & 1+pq & p \\ 0 & q & 1 \end{bmatrix}$ is ...
A
$\begin{bmatrix} 1+pq & p & 0 \\ q & 1+pq & p \\ 0 & q & 1 \end{bmatrix}$
B
$\begin{bmatrix} 1 & p & p^2 \\ q & 1+pq & p+p^2q \\ q^2 & q+pq^2 & 1+pq+p^2q^2 \end{bmatrix}$
C
$\begin{bmatrix} 1 & -p & p^2 \\ -q & 1+pq & -(p+p^2q) \\ q^2 & -(q+pq^2) & 1+pq+p^2q^2 \end{bmatrix}$
D
$\begin{bmatrix} 1 & -p & p^2 \\ -q & 1+pq & p+p^2q \\ q^2 & q+pq^2 & 1+pq+p^2q^2 \end{bmatrix}$
2
MHT CET 2026 19th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
$\sum\limits_{k=1}^{2026} \sin^{-1}\left(\cos\dfrac{k\pi}{4}\right) =$
A
$-\dfrac{\pi}{4}$
B
$0$
C
$\dfrac{\pi}{4}$
D
$\dfrac{\pi}{2}$
3
MHT CET 2026 19th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
If $\sin^{-1}\left(\tan\dfrac{\pi}{4}\right) - \sin^{-1}\left(\sqrt{\dfrac{3}{x}}\right) = \dfrac{\pi}{6}$ then $x$ is a root of the equation
A
$x^2 - x - 6 = 0$
B
$x^2 - x - 12 = 0$
C
$x^2 + x - 12 = 0$
D
$x^2 + x - 6 = 0$
4
MHT CET 2026 19th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
If $f(x) = x^2$ and $g(x) = [x^2]$ where $[\cdot]$ represents the greatest integer function then, $(f \circ g)\left(\dfrac{3}{2}\right) + (g \circ f)\left(\dfrac{3}{2}\right)$ is equal to ...
A
$2$
B
$5$
C
$9$
D
$10$

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