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

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

$\frac{x^2}{2}-2 x+3 \log (x+1)+\frac{1}{x+1}+c$ 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 $P(A)=$

$\frac{4}{13}$

$\frac{6}{13}$

$\frac{8}{13}$

$\frac{3}{13}$

MHT CET 2025 21st April Evening Shift

MCQ (Single Correct Answer)

-0

Angle between the parabola $y^2=4(x-1)$ and $x^2+4(y-3)=0$ at the common end of their latus rectum is

$\frac{\pi}{2}$

$\frac{\pi}{4}$

$\frac{\pi}{6}$

$\frac{\pi}{3}$