MHT CET 2025 22nd April Evening Shift | MHT CET Year Wise Previous Years Questions

MHT CET 2025 22nd April Evening Shift

MCQ (Single Correct Answer)

-0

In a box containing 100 apples, 10 are defective. The probability that in a sample of 6 apples, 3 are defective is

0.1548

0.1458

0.01854

0.01458

MHT CET 2025 22nd April Evening Shift

MCQ (Single Correct Answer)

-0

The value of the integral $\int_1^2 \frac{x \mathrm{~d} x}{(x+2)(x+3)}$ is

$\quad \log \left(\frac{125}{16}\right)$

$\quad \log \left(\frac{1024}{1125}\right)$

$\quad \log \left(\frac{16}{125}\right)$

$\quad \log \left(\frac{1125}{1024}\right)$

MHT CET 2025 22nd April Evening Shift

MCQ (Single Correct Answer)

-0

The general solution of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}+\sin \left(\frac{x+y}{2}\right)=\sin \left(\frac{x-y}{2}\right)$ is

$\log \tan \left(\frac{y}{2}\right)=\mathrm{c}-2 \sin \frac{x}{2}$, where c is the constant of integration

$\log \tan \left(\frac{y}{4}\right)=\mathrm{c}-2 \sin \left(\frac{x}{2}\right)$, where c is the constant of integration

$\log \left[\tan \left(\frac{y}{2}+\frac{\pi}{4}\right)\right]=\mathrm{c}-2 \sin x$, where c is the constant of integration

$\log \left[\tan \left(\frac{y}{4}+\frac{\pi}{4}\right)\right]=\mathrm{c}-2 \sin \frac{x}{2}$, where c is the constant of integration

MHT CET 2025 22nd April Evening Shift

MCQ (Single Correct Answer)

-0

Four defective oranges are accidentally mixed with sixteen good ones. Three oranges are drawn from the mixed lot. The probability distribution of defective oranges is

$$ \begin{array}{|c|c|c|c|c|} \hline \mathrm{X} & 0 & 1 & 2 & 3 \\ \hline \mathrm{P}(\mathrm{X}=x) & \frac{28}{57} & \frac{8}{95} & \frac{8}{19} & \frac{1}{285} \\ \hline \end{array} $$

$$ \begin{array}{|c|c|c|c|c|} \hline \mathrm{X} & 0 & 1 & 2 & 3 \\ \hline \mathrm{P}(\mathrm{X}=x) & \frac{28}{57} & \frac{8}{19} & \frac{8}{95} & \frac{1}{285} \\ \hline \end{array} $$

$$ \begin{array}{|c|c|c|c|c|} \hline \mathrm{X} & 0 & 1 & 2 & 3 \\ \hline \mathrm{P}(\mathrm{X}=x) & \frac{28}{57} & \frac{8}{95} & \frac{1}{285} & \frac{8}{19} \\ \hline \end{array} $$

$$ \begin{array}{|c|c|c|c|c|} \hline \mathrm{X} & 0 & 1 & 2 & 3 \\ \hline \mathrm{P}(\mathrm{X}=x) & \frac{1}{285} & \frac{8}{95} & \frac{8}{19} & \frac{28}{57} \\ \hline \end{array} $$

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