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

MHT CET 2025 23rd April Evening Shift

MCQ (Single Correct Answer)

-0

If $x=\mathrm{e}^{\tan ^{-1}\left(\frac{y-x^2}{x^2}\right)}$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ at $x=1$ is

MHT CET 2025 23rd April Evening Shift

MCQ (Single Correct Answer)

-0

The difference between the maximum value and minimum value of objective function $\mathrm{z}=3 x+5 y$ subject to constraints $x+3 y \leq 60$, $x+y \geq 10, x-y \geq 0, x, y \geq 0$ is

MHT CET 2025 23rd April Evening Shift

MCQ (Single Correct Answer)

-0

If $\bar{a}=\frac{1}{\sqrt{10}}(3 \hat{i}+\hat{k})$ and $\bar{b}=\frac{1}{7}(2 \hat{i}+3 \hat{j}-6 \hat{k})$, then the value of $(2 \overline{\mathrm{a}}-\overline{\mathrm{b}}) \cdot((\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times(\overline{\mathrm{a}}+2 \overline{\mathrm{~b}}))=$

-3

-5

MHT CET 2025 23rd April Evening Shift

MCQ (Single Correct Answer)

-0

In a triangle with one of the angles $120^{\circ}$, the lengths of the sides form an A.P. If length of the greatest side is 7 m , then the area of the triangle is

$\frac{15 \sqrt{3}}{4} \mathrm{~m}^2$

$\frac{15 \sqrt{3}}{2} \mathrm{~m}^2$

$\frac{15}{2} \mathrm{~m}^2$

$\frac{15}{4} \mathrm{~m}^2$

PREVIOUS NEXT