1
MHT CET 2025 19th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

The last column in the truth table of the statement pattern $[\mathrm{p} \rightarrow(\mathrm{q} \wedge \sim \mathrm{p})] \vee[(\mathrm{p} \vee \sim \mathrm{q}) \wedge \mathrm{p}]$ is

A
TTTF
B
TFFF
C
TTTT
D
FFTT
2
MHT CET 2025 19th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

A straight line through the origin $O$ meets the line $3 y=10-4 x$ and $8 x+6 y+5=0$ at the points $A$ and B respectively. Then O divides the segment $A B$ in the ratio

A
$4: 1$
B
$2: 3$
C
$1: 5$
D
$1: 3$
3
MHT CET 2025 19th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

The position of a point in time $t$ is given by $x=\mathrm{a}+\mathrm{bt}-\mathrm{ct}^2, y=\mathrm{at}+\mathrm{bt}^2$. It's resultant acceleration at time $t$ in seconds is given by

A
$\mathrm{b}-\mathrm{c}$ unit $/$ seconds $^2$
B
$\mathrm{b}+\mathrm{c}$ unit/seconds ${ }^2$
C
$2 \mathrm{~b}-2 \mathrm{c}$ unit/seconds ${ }^2$
D
$2 \sqrt{b^2+c^2}$ unit/seconds ${ }^2$
4
MHT CET 2025 19th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

Let $\overline{\mathrm{OA}}=\overline{\mathrm{a}}, \overline{\mathrm{OB}}=\overline{\mathrm{b}}$ and if the vector along the angle bisector of $\angle \mathrm{AOB}$ is given by $x \frac{\overline{\mathrm{a}}}{|\overline{\mathrm{a}}|}+y \frac{\overline{\mathrm{~b}}}{|\overline{\mathrm{~b}}|}$ then

A
$x-y=0$
B
$x+y=0$
C
$x=2 y$
D
$y=2 x$

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