1
MHT CET 2026 13th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
Let $A = \begin{bmatrix} -5 & -3 \\ 2 & 1 \end{bmatrix}$. The Row transformation $R_1 \rightarrow R_1 + 3R_2$ will transform matrix A into
A
an upper triangular matrix
B
a lower triangular matrix
C
an identity matrix
D
a singular matrix
2
MHT CET 2026 13th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
If $A = \begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}_{3 \times 3}$, and $A^{-1} = \begin{bmatrix} \gamma & -1 & 1 \\ \alpha & 6 & -5 \\ \beta & -2 & 2 \end{bmatrix}_{3 \times 3}$, then $|\alpha \cdot \beta \cdot \gamma| = $ _____ (where $|\cdot|$ denotes the absolute value)
A
$125$
B
$220$
C
$225$
D
$-225$
3
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
$\dfrac{2x}{\sqrt{1 - x^4}}$
4
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
$\dfrac{4\pi}{3}$

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