1
MHT CET 2026 13th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
With usual notations, in $\triangle ABC$, if $2a^2 = b^2 + c^2$, then $\dfrac{\cos 3A}{\cos A} + 2 =$
A
$\dfrac{b^2 - c^2}{2bc}$
B
$\left(\dfrac{b^2 - c^2}{2bc}\right)^2$
C
$\left(\dfrac{c^2 - b^2}{bc}\right)^2$
D
$\dfrac{c^2 - b^2}{bc}$
2
MHT CET 2026 13th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
With usual notations, in $\triangle ABC$, $(b - c)^2 \cos^2 \dfrac{A}{2} + (b + c)^2 \sin^2 \dfrac{A}{2} =$
A
$a^2$
B
$b^2 - c^2$
C
$a^2 + b^2 + c^2$
D
$0$
3
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
4
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
$-225$

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