1
MHT CET 2025 5th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

ABCD is a quadrilateral with $\overline{\mathrm{AB}}=\overline{\mathrm{a}}, \overline{\mathrm{AD}}=\overline{\mathrm{b}}$ and $\overline{\mathrm{AC}}=2 \overline{\mathrm{a}}+3 \overline{\mathrm{~b}}$. If its area is $\alpha$ times the area of the parallelogram with $\mathrm{AB}, \mathrm{AD}$ as adjacent sides, then the value of $\alpha$ is

A

$\frac{1}{2}$

B

$\frac{5}{2}$

C

$\frac{3}{2}$

D

2

2
MHT CET 2025 5th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The equation of a line passing through the point $(-1,2,3)$ and perpendicular to the lines $\frac{x}{2}=\frac{y-1}{-3}=\frac{z+2}{-2}$ and $\frac{x+3}{-1}=\frac{y+3}{2}=\frac{z-1}{3}$ is

A

$\frac{x+1}{5}=\frac{y-2}{-4}=\frac{z+3}{1}$

B

$\frac{x+1}{5}=\frac{y+2}{4}=\frac{z+3}{1}$

C

$\frac{x+1}{5}=\frac{y-2}{4}=\frac{z-3}{-1}$

D

$\frac{x+1}{1}=\frac{y-2}{4}=\frac{z-3}{3}$

3
MHT CET 2025 5th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $A=\left[\begin{array}{rrr}1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]$ then $A(I+\operatorname{adj} A)=$

A

$\left[\begin{array}{ccc}9 & -2 & 2 \\ 0 & 10 & -3 \\ 3 & -2 & 11\end{array}\right]$

B

$\left[\begin{array}{ccc}8 & -2 & 2 \\ 0 & 9 & -3 \\ 3 & -2 & 10\end{array}\right]$

C

$\left[\begin{array}{rrr}9 & -2 & 2 \\ 0 & 10 & -3 \\ 3 & -2 & 12\end{array}\right]$

D

$\left[\begin{array}{ccc}3 & 2 & -2 \\ 0 & 10 & 3 \\ -3 & 2 & 12\end{array}\right]$

4
MHT CET 2025 5th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

A line L is passing through points $\mathrm{A}(1,3,2)$ and $\mathrm{B}(2,2,1)$. If mirror image of point $\mathrm{P}(1,1,-1)$ in the line L is $(x, y, z)$ then $x+y+\mathrm{z}=$

A

$\frac{10}{3}$

B

$\frac{13}{3}$

C

$\frac{14}{3}$

D

$\frac{23}{3}$

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