MHT CET 2025 21st April Evening Shift | MHT CET Year Wise Previous Years Questions

MHT CET 2025 21st April Evening Shift

MCQ (Single Correct Answer)

-0

$$ \int \cos \left(\frac{x}{16}\right) \cdot \cos \left(\frac{x}{8}\right) \cdot \cos \left(\frac{x}{4}\right) \cdot \sin \left(\frac{x}{16}\right) \mathrm{d} x= $$

$\frac{\cos 16 x}{256}+\mathrm{c}$, where c is the constant of integration

$\frac{-\cos 16 x}{256}+c$, where $c$ is the constant of integration

$\frac{\sin 16 x}{256}+c$, where $c$ is the constant of integration

$\frac{-\cos \left(\frac{x}{2}\right)}{4}+\mathrm{c}$, where $c$ is the constant of integration

MHT CET 2025 21st April Evening Shift

MCQ (Single Correct Answer)

-0

$$ \int \frac{x^3}{(x+1)^2} d x= $$

$\frac{x^2}{2}-2 x+3 \log (x+1)+\frac{1}{x+1}+c$ where c is the constant of integration

$\frac{x^2}{2}+2 x-3 \log (x+1)+\frac{1}{x+1}+c$ where c is the constant of integration

$\frac{x^2}{2}-2 x+3 \log (x+1)-\frac{1}{x+1}+\mathrm{c}$, where c is the constant of integratio

$\frac{x^2}{2}-2 x-3 \log (x+1)-\frac{1}{x+1}+\mathrm{c}$, where c is the constant of integration

MHT CET 2025 21st April Evening Shift

MCQ (Single Correct Answer)

-0

A wire of length 8 units is cut into two parts which are bent respectively in the form of a square and a circle. The least value of the sum of the areas so formed is

$\frac{8}{\pi+4}$

$\frac{64}{\pi+4}$

$\frac{2}{\pi+4}$

$\frac{16}{\pi+4}$

MHT CET 2025 21st April Evening Shift

MCQ (Single Correct Answer)

-0

If $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are mutually exclusive and exhaustive events of a sample space $S$ such that $P(B)=\frac{3}{2} P(A)$ and $P(C)=\frac{1}{2} P(B)$, then $P(A)=$

$\frac{4}{13}$

$\frac{6}{13}$

$\frac{8}{13}$

$\frac{3}{13}$

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