MHT CET 2026 11th April Morning Shift | MHT CET Year Wise Previous Years Questions

MHT CET 2026 11th April Morning Shift

MCQ (Single Correct Answer)

-0

The differential equation representing the family of curves $x \sin x + y^3 = 4ax$ is

$\dfrac{dy}{dx} = \dfrac{y^3 + x^2 \cos x}{3xy^2}$

$\dfrac{dy}{dx} = \dfrac{y^3 - x^2 \cos x}{3xy^2}$

$\dfrac{dy}{dx} = \dfrac{y^3 - x^2 \cos x}{3xy}$

$\dfrac{dy}{dx} = \dfrac{y^3 - x^2 \cos x}{3x^2 y}$

MHT CET 2026 11th April Morning Shift

MCQ (Single Correct Answer)

-0

The order and degree of the differential equation $\sqrt{2 + \left(\dfrac{d^2 y}{dx^2}\right)^3} = \left(\dfrac{d^3 y}{dx^3}\right)^{5/2}$ respectively are

$2, 3$

$3, 5$

$3, 2$

$5, 3$

MHT CET 2026 11th April Morning Shift

MCQ (Single Correct Answer)

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The vector $\bar{a} + 3\bar{b}$ is perpendicular to $7\bar{a} - 5\bar{b}$ and the vector $\bar{a} - 4\bar{b}$ is perpendicular to $7\bar{a} - 2\bar{b}$. Then the angle between $\bar{a}$ and $\bar{b}$ is

$\dfrac{\pi}{2}$

$\dfrac{\pi}{4}$

$\dfrac{\pi}{6}$

$\dfrac{\pi}{3}$

MHT CET 2026 11th April Morning Shift

MCQ (Single Correct Answer)

-0

If $7\hat{j} + 10\hat{k}, -\hat{i} + 6\hat{j} + 6\hat{k}$ and $-4\hat{i} + 9\hat{j} + 6\hat{k}$ are the position vectors of the vertices A, B and C repectively of $\triangle ABC$. Then the position vector of the point where the bisector of the angle A meets side BC is

$(2 + 3\sqrt{2})\hat{i} + (3 + 3\sqrt{3})\hat{j} + 6\hat{k}$

$(2 - 3\sqrt{2})\hat{i} + (3 + 3\sqrt{2})\hat{j} + 6\hat{k}$

$(2 - 3\sqrt{2})\hat{i} + (3 - 3\sqrt{2})\hat{j} + 6\hat{k}$

$(2 - 3\sqrt{2})\hat{i} + (3 + 3\sqrt{2})\hat{j} - 6\hat{k}$

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