1
MHT CET 2024 11th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

A random variable x takes the values $0,1,2$, $3, \ldots$ with probability $\mathrm{P}(\mathrm{X}=x)=\mathrm{k}(x+1)\left(\frac{1}{5}\right)^x$, where k is a constant, then $\mathrm{P}(\mathrm{X}=0)$ is

A
$\frac{16}{25}$
B
$\frac{7}{25}$
C
$\frac{19}{25}$
D
$\frac{18}{25}$
2
MHT CET 2024 11th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $A\left[\begin{array}{ll}2 & 1 \\ 7 & 4\end{array}\right]$ then $\left(A^2-5 A\right)^{-1}$ is

A
$\left(-\frac{1}{4}\right)\left[\begin{array}{cc}-3 & 1 \\ 7 & -1\end{array}\right]$
B
$\left(\frac{1}{4}\right)\left[\begin{array}{cc}-3 & 1 \\ 7 & -1\end{array}\right]$
C
$\left(\frac{1}{4}\right)\left[\begin{array}{ll}3 & 1 \\ 7 & 1\end{array}\right]$
D
$\left(\frac{1}{-4}\right)\left[\begin{array}{ll}3 & -1 \\ 7 & -1\end{array}\right]$
3
MHT CET 2024 11th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The following statement $(\mathrm{p} \rightarrow \mathrm{q}) \rightarrow((\sim \mathrm{p} \rightarrow \mathrm{q}) \rightarrow \mathrm{q})$ is

A
a fallacy.
B
equivalent to $(\sim \mathrm{p}) \rightarrow \mathrm{q}$.
C
equivalent to $\mathrm{p} \rightarrow(\sim \mathrm{q})$.
D
a tautology.
4
MHT CET 2024 11th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

Let $\omega=-\frac{1}{2}+\mathrm{i} \frac{\sqrt{3}}{2}, \mathrm{i}=\sqrt{-1}$, then the value of $\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & -1-\omega^2 & \omega^2 \\ 1 & \omega^2 & \omega^4\end{array}\right|$ is

A
$3 \omega$
B
$3 \omega(\omega-1)$
C
$3 \omega^2$
D
$3 \omega(1-\omega)$
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