MHT CET 2026 15th April Evening Shift | MHT CET Year Wise Previous Years Questions

MHT CET 2026 15th April Evening Shift

MCQ (Single Correct Answer)

-0

The point on the curve $9y^2 = x^3$ where the normal to the curve makes equal intercepts with the co-ordinate axes is

$\left(-4, \dfrac{8}{3}\right)$

$\left(4, \dfrac{8}{3}\right)$

$\left(-4, \dfrac{-8}{3}\right)$

$\left(-4, \dfrac{3}{8}\right)$

MHT CET 2026 15th April Evening Shift

MCQ (Single Correct Answer)

-0

A spherical mothball has initial radius 3 cm. Due to evaporation, the radius of the ball reduces to 1 cm in 4 months. In how many months would the mothball evaporate completely if the volume is lost at a rate proportional to the surface area ?

$6$ months

$8$ months

$10$ months

$12$ months

MHT CET 2026 15th April Evening Shift

MCQ (Single Correct Answer)

-0

$\int \dfrac{x^2\, \text{d}x}{(x^2 + 2)(x^2 + 5)} =$

$-\left(\dfrac{\sqrt{2}}{3}\tan^{-1}\dfrac{x}{\sqrt{2}} + \dfrac{\sqrt{5}}{3}\tan^{-1}\dfrac{x}{\sqrt{5}}\right) + c$, where c is the constant of integration

$\left(\dfrac{\sqrt{2}}{3}\tan^{-1}\dfrac{x}{\sqrt{2}} + \dfrac{\sqrt{5}}{3}\tan^{-1}\dfrac{x}{\sqrt{5}}\right) + c$, where c is the constant of integration

$\dfrac{\sqrt{2}}{3}\tan^{-1}\left(\dfrac{x}{\sqrt{2}}\right) - \dfrac{\sqrt{5}}{3}\tan^{-1}\left(\dfrac{x}{\sqrt{5}}\right) + c$, where c is the constant of integration

$-\dfrac{\sqrt{2}}{3}\tan^{-1}\left(\dfrac{x}{\sqrt{2}}\right) + \dfrac{\sqrt{5}}{3}\tan^{-1}\left(\dfrac{x}{\sqrt{5}}\right) + c$, where c is the constant of integration

MHT CET 2026 15th April Evening Shift

MCQ (Single Correct Answer)

-0

The value of $\int \dfrac{\sin^6 x + \cos^6 x}{\sin^2 x \cdot \cos^2 x}\, \text{d}x$ is

$\cot x + \cot^2 x + c$, where c is the constant of integration

$\tan x - \cot x - 3x + c$, where c is the constant of integration

$2\sin^5 x + 2\cos^5 x + c$, where c is the constant of integration

$\sin^3 x + 2\cos^3 x + c$, where c is the constant of integration

PREVIOUS NEXT