MHT CET 2025 23rd April Morning Shift | MHT CET Year Wise Previous Years Questions

MHT CET 2025 23rd April Morning Shift

MCQ (Single Correct Answer)

-0

The maximum value of $x^{2 / 3}+(x-2)^{2 / 3}$ is

$2^{\frac{2}{3}}$

MHT CET 2025 23rd April Morning Shift

MCQ (Single Correct Answer)

-0

$$ \int\left(\frac{x-3}{x^2+9}\right)^2 d x= $$

$\frac{1}{3} \tan ^{-1}\left(\frac{x}{3}\right)-\frac{3}{x^2+9}+\mathrm{c}$, where c is the constant of integration.

$\frac{1}{3} \tan ^{-1}\left(\frac{x}{3}\right)-\frac{1}{x^2+9}+c$, where $c$ is the constant of integration.

$\frac{1}{3} \tan ^{-1}\left(\frac{x}{3}\right)+\frac{3}{x^2+9}+c$, where $c$ is the constant of integration.

$\frac{1}{3} \tan ^{-1}\left(\frac{x}{3}\right)-\frac{1}{x^2+9}+c$, where $c$ is the constant of integration.

MHT CET 2025 23rd April Morning Shift

MCQ (Single Correct Answer)

-0

The point on the curve $4 y^2-4 y+2 x-1=0$ at which the tangent becomes parallel to Y -axis is

$\left(1, \frac{1}{2}\right)$

$\left(\frac{1}{2}, 1\right)$

$\left(-1,-\frac{1}{2}\right)$

$\left(\frac{1}{2}, 0\right)$

MHT CET 2025 23rd April Morning Shift

MCQ (Single Correct Answer)

-0

The value of $\tan ^2\left(\sec ^{-1} 4\right)+\cot ^2\left(\operatorname{cosec}^{-1} 3\right)$ is

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