List of top Mathematics Questions asked in MHT CET

Evaluate the integral: \[ \int \frac{x}{x+2}\, dx \]

For the curve \( y = 3x^3 - 3x^2 + 1 \) at \( x = 1 \), find the equation of the tangent.

If \( \vec{a} \) and \( \vec{b} \) are unit vectors and the angle between them is \(60^\circ\), find \( |\vec{a}-\vec{b}| \).

Find the general solution of the differential equation \( \dfrac{dy}{dx} + \dfrac{y}{x} = x^2 \).

Find the value of \( k \) if the function \( f(x) = \dfrac{k\sin x + 2\cos x}{\sin x + \cos x} \) is increasing for all \( x \).

The solution of the differential equation $\frac{dy}{dx} = \frac{x+y}{x-y}$ is

The coordinates of the foot of the perpendicular from the origin to the plane $2x - 3y - 6z = 4$ are

In $\triangle ABC$, if $2a^2 = b^2 + c^2$, then the value of $\frac{\cos 3A}{\cos A} + 2$ is

The value of $\tan^{-1}(\sqrt{3}) + \sec^{-1}(-2) - \sin^{-1}\left(-\frac{1}{2}\right)$ is

Evaluate the integral: \(\int \frac{2x}{x^2 - 5x + 4} \, dx\)

Evaluate the integral: \(\int \frac{\sin x}{\sin 4x} \, dx\)

For the curve \(y = 3x^3 - 3x^2 + 1\) at \(x = 1\), find the equation of the tangent.

A plane is formed by the axes whose centroid is \(\left(2, -\frac{2}{3}, \frac{1}{2}\right)\). Find the distance of the plane from the origin.

Evaluate: \(\int_0^3 \sqrt{9 - x^2} \, dx\).

If \(\tan^{-1}(-1) + \tan^{-1}(5) + \tan^{-1}(3) + \tan^{-1}\left(\frac{1}{4}\right) = \pi + \tan^{-1}\left(\frac{\alpha}{2}\right)\), find \(\alpha\).

Determine the shortest distance between the lines \( \dfrac{x-1}{2} = \dfrac{y-2}{3} = \dfrac{z-3}{4} \) and \( \dfrac{x-2}{3} = \dfrac{y-4}{4} = \dfrac{z-5}{5} \).

Evaluate the integral: \[ \int_{3}^{5} |x-4|\,dx \]

Evaluate the integral: \[ \int \frac{\sqrt{\tan x}}{\sin x \cos x} \, dx \]

From the following options, the nearest line to the origin is ....

The slopes of the lines represented by $6x^2 + 2hxy + y^2 = 0$ are in the ratio 2 : 3, then $h = \dots$

If $\theta$ is an obtuse angle between vectors $\vec{a}$ and $\vec{b}$ such that $|\vec{a}| = 5, |\vec{b}| = 3$ and $|\vec{a} \times \vec{b}| = 5\sqrt{5}$ then $\vec{a} \cdot \vec{b} = \dots$