MHT CET 2026 19th April Evening Shift | MHT CET Year Wise Previous Years Questions

MHT CET 2026 19th April Evening Shift

MCQ (Single Correct Answer)

-0

The area (in square units) of the region bounded by the circle $x^2 + y^2 = 9$ and the parabola $y^2 \leq 8x$ is...

$\dfrac{8\sqrt{2}}{3} + \dfrac{9\pi}{2} - 2\sqrt{2} - 9\sin^{-1}\dfrac{1}{3}$

$\dfrac{8\sqrt{2}}{3} + \dfrac{9\pi}{2} + 2\sqrt{2} + 9\sin^{-1}\dfrac{1}{3}$

$\dfrac{4\sqrt{2}}{3} + \dfrac{9\pi}{4} - \sqrt{2} - \dfrac{9}{2}\sin^{-1}\dfrac{1}{3}$

$\dfrac{4\sqrt{2}}{3} + \dfrac{9\pi}{4} + \sqrt{2} + \dfrac{9}{2}\sin^{-1}\dfrac{1}{3}$

MHT CET 2026 19th April Evening Shift

MCQ (Single Correct Answer)

-0

The differential equation of all lines where the length of the normal from the origin is p and the inclination of the normal is $\alpha$ is... (where p and $\alpha$ are arbitrary constants)

$\dfrac{d^2y}{dx^2} = 0$

$\dfrac{dy}{dx} = 0$

$\dfrac{dy}{dx} = -\cot\alpha$

$\dfrac{d^2y}{dx^2} = \csc^2\alpha$

MHT CET 2026 19th April Evening Shift

MCQ (Single Correct Answer)

-0

If the solution of the differential equation $(1 + x^3)\dfrac{dy}{dx} + 6x^2y = 1 + x^2$ is $y = \dfrac{1}{(1 + x^3)^s}\left[x + \dfrac{x^p}{p} + \dfrac{x^q}{q} + \dfrac{x^r}{r} + c\right]$, then the LCM of $p, q, r$ and $s$ is...

$1$

$6$

$4$

$12$

MHT CET 2026 19th April Evening Shift

MCQ (Single Correct Answer)

-0

The solution of the differential equation $\dfrac{dy}{dx} = \cos(x + y)$ is...

$\cot\left(\dfrac{x + y}{2}\right) = x + c$

$\tan\left(\dfrac{x + y}{2}\right) = x + c$

$-\sin(x + y) = x + c$

$\sec(x - y) + x = c$