1
MHT CET 2025 25th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

Consider the probability distribution

$$ \begin{array}{|l|l|l|l|l|l|} \hline \mathrm{X}=x & 1 & 2 & 3 & 4 & 5 \\ \hline \mathrm{P}(\mathrm{X}=x) & \mathrm{K} & 2 \mathrm{~K} & \mathrm{~K}^2 & 2 \mathrm{~K} & 5 \mathrm{~K}^2 \\ \hline \end{array} $$

Then the value of $\mathrm{P}(\mathrm{X}>2)$ is

A
$\frac{7}{12}$
B
$\frac{1}{36}$
C
$\frac{1}{2}$
D
$\frac{23}{36}$
2
MHT CET 2025 25th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

The equation of the curve passing through origin and satisfying $\left(1+x^2\right) \frac{\mathrm{d} y}{\mathrm{~d} x}+2 x y=4 x^2$ is

A
$y\left(1+x^2\right)=4 x^3$
B
$4\left(1+x^2\right)=4+y^2$
C
$3 y\left(1+x^2\right)=4 x^3$
D
$1+y^2=4 x^3+1$
3
MHT CET 2025 25th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $y=\tan ^{-1}\left(\frac{12 x-64 x^3}{1-48 x^2}\right)$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}=$

A
$\frac{3}{1+16 x^2}$
B
$\frac{4}{1+16 x^2}$
C
$\frac{12}{1+16 x^2}$
D
$\frac{1}{1+16 x^2}$
4
MHT CET 2025 25th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

The order of the differential equation whose general solution is given by $y=\left(\mathrm{C}_1+\mathrm{C}_2\right) \sin \left(x+\mathrm{C}_3\right)-\mathrm{C}_4 \mathrm{e}^{x+\mathrm{C}_5}$ is (where $\mathrm{C}_1, \mathrm{C}_2, \mathrm{C}_3, \mathrm{C}_4, \mathrm{C}_5$ are arbitrary constants)

A
5
B
4
C
2
D
3
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