MHT CET 2026 16th April Evening Shift | MHT CET Year Wise Previous Years Questions

MHT CET 2026 16th April Evening Shift

MCQ (Single Correct Answer)

-0

The equation of the curve passing through the point (0, 1) and having a slope of the tangent at point $P(x, y)$ is equal to $\dfrac{y}{x + y}$, is

$x = e^{-\frac{y}{x}}$

$y = e^{-\frac{y}{x}} + 1$

$x = e^{\frac{x}{y}} - 1$

$y = e^{\frac{x}{y}}$

MHT CET 2026 16th April Evening Shift

MCQ (Single Correct Answer)

-0

If $f(x)$ is a polynomial such that $f(x) = [f'(x)]^2$ and $f(2) = 0$, then $f(-2) = \ldots$

$1$

$-1$

$4$

$-4$

MHT CET 2026 16th April Evening Shift

MCQ (Single Correct Answer)

-0

The equation of a line in cartesian form passing through (0, 0, 0) and (4, 3, c) and parallel to $\vec{a} \times \vec{b}$ where $\vec{a} = 2\hat{i} + \hat{j} + 2\hat{k}$, $\vec{b} = 3\hat{i} - 4\hat{j}$ is

$\dfrac{x-4}{8} = \dfrac{y-3}{6} = \dfrac{z + \frac{11}{2}}{11}$

$\dfrac{x-4}{8} = \dfrac{y+3}{6} = \dfrac{z + \frac{11}{2}}{-11}$

$\dfrac{x-4}{8} = \dfrac{y-3}{6} = \dfrac{z + \frac{11}{2}}{-11}$

$\dfrac{x+4}{8} = \dfrac{y-3}{6} = \dfrac{z + \frac{11}{2}}{-11}$

MHT CET 2026 16th April Evening Shift

MCQ (Single Correct Answer)

-0

If $\bar{a} = 4\hat{i} + \hat{j} + \hat{k}$, $\bar{b} = 2\hat{i} + \hat{j} + 2\hat{k}$ and $\bar{c} = 3\hat{i} + 4\hat{j} + 5\hat{k}$, then $(\bar{a} + \bar{b}) \cdot (\bar{b} + \bar{c}) = $

$30$

$21$

$61$

$10$

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