1
MHT CET 2025 5th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The solution of $\log \left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)=2 x-5 y, y(0)=0$ is

A

$\quad 2 \mathrm{e}^{2 x}+5 \mathrm{e}^{5 y}=6$

B

$\quad 5 \mathrm{e}^{2 x}-2 \mathrm{e}^{5 y}=3$

C

$\quad 2 \mathrm{e}^{2 x}-5 \mathrm{e}^{5 y}=6$

D

$5 \mathrm{e}^{2 x}+2 \mathrm{e}^{5 y}=3$

2
MHT CET 2025 5th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The integrating factor of the differential equation $x \frac{\mathrm{~d} y}{\mathrm{~d} x}+y \log x=x \cdot \mathrm{e}^x x^{-\frac{1}{2}} \log x(x>0)$ is

A

$\quad(\log x)^x$

B

$x^{\log x}$

C

$(\sqrt{x})^{\log x}$

D

$e^{\sqrt{x} \log x}$

3
MHT CET 2025 5th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

ABCD is a quadrilateral with $\overline{\mathrm{AB}}=\overline{\mathrm{a}}, \overline{\mathrm{AD}}=\overline{\mathrm{b}}$ and $\overline{\mathrm{AC}}=2 \overline{\mathrm{a}}+3 \overline{\mathrm{~b}}$. If its area is $\alpha$ times the area of the parallelogram with $\mathrm{AB}, \mathrm{AD}$ as adjacent sides, then the value of $\alpha$ is

A

$\frac{1}{2}$

B

$\frac{5}{2}$

C

$\frac{3}{2}$

D

2

4
MHT CET 2025 5th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The equation of a line passing through the point $(-1,2,3)$ and perpendicular to the lines $\frac{x}{2}=\frac{y-1}{-3}=\frac{z+2}{-2}$ and $\frac{x+3}{-1}=\frac{y+3}{2}=\frac{z-1}{3}$ is

A

$\frac{x+1}{5}=\frac{y-2}{-4}=\frac{z+3}{1}$

B

$\frac{x+1}{5}=\frac{y+2}{4}=\frac{z+3}{1}$

C

$\frac{x+1}{5}=\frac{y-2}{4}=\frac{z-3}{-1}$

D

$\frac{x+1}{1}=\frac{y-2}{4}=\frac{z-3}{3}$

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