1
MHT CET 2025 21st April Morning Shift
MCQ (Single Correct Answer)
+2
-0

The logical statement

$$ [\sim(\sim p \vee q) \vee(p \wedge r) \wedge(\sim q \wedge r)] $$

is equivalent to

A
$(\mathrm{p} \wedge \mathrm{r}) \wedge \sim \mathrm{q}$
B
$(\sim p \wedge \sim q) \wedge r$
C
$\sim p \vee r$
D
$\quad(p \wedge \sim q) \vee r$
2
MHT CET 2025 21st April Morning Shift
MCQ (Single Correct Answer)
+2
-0

The L.P.P. , minimize $z=30 x+20 y, x+y \leq 8$, $x+2 y \geq 4,6 x+4 y \geq 12, x \geqslant 0, y \geqslant 0$ has

A
a unique solution
B
infinitely many solutions
C
minimum value at $(4,0)$
D
minimum value at $(8,0)$
3
MHT CET 2025 21st April Morning Shift
MCQ (Single Correct Answer)
+2
-0

If the distance of the point $\mathrm{P}(1,-2,1)$ from the plane $x+2 y-2 z=\alpha$, where $\alpha>0$ is 5 units, then the foot of the perpendicular from P to the plane is

A
$\left(2, \frac{2}{3}, \frac{-10}{3}\right)$
B
$\left(\frac{8}{3}, \frac{7}{3}, \frac{-4}{3}\right)$
C
$\left(\frac{4}{3}, \frac{2}{3}, \frac{-8}{3}\right)$
D
$\left(\frac{8}{3}, \frac{4}{3}, \frac{-7}{3}\right)$
4
MHT CET 2025 21st April Morning Shift
MCQ (Single Correct Answer)
+2
-0

The equation of the plane containing the line $\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-4}{-2}$ and the point $(0,5,0)$ is

A
$2 x-4 y-3 z+20=0$
B
$2 x+8 y+11 z-40=0$
C
$8 x-5 y+z+25=0$
D
$\quad x-4 y+3 z+20=0$
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