1
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The differential equation obtained by eliminating arbitrary constant from the equation $y^2=(x+c)^3$ is

A
$\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^3=27 y$
B
$\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^3=-27 y$
C
$8\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^3=27 y$
D
$ 8\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^3+27 y=0$
2
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The decay rate of radium is proportional to the amount present at any time $t$. If initially 60 gms was present and half life period of radium is 1600 years, then the amount of radium present after 3200 years is

A
20 grams
B
15 grams
C
12 grams
D
10 grams
3
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The particular solution of differential equation $\left(1+y^2\right)(1+\log x) \mathrm{d} x+x \mathrm{~d} y=0$ at $x=1, y=1$ is

A
$\log x-\frac{1}{2}(\log x)^2-\tan ^{-1} y=-\frac{\pi}{4}$
B
$\log x+\frac{1}{2}(\log x)^2+\tan ^{-1} y=\frac{\pi}{4}$
C
$\log x-\frac{1}{2}(\log x)^2+\tan ^{-1} y=\frac{\pi}{4}$
D
$\log x+\frac{1}{2}(\log x)^2-\tan ^{-1} y=\frac{\pi}{4}$
4
MHT CET 2024 4th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

Let $y=y(x)$ be the solution of the differential equation $\sin x \frac{\mathrm{~d} y}{\mathrm{~d} x}+y \cos x=4 x, x \in(0, \pi)$. If $y\left(\frac{\pi}{2}\right)=0$, then $y\left(\frac{\pi}{6}\right)$ is equal to

A
$-\frac{4}{9} \pi^2$
B
$\frac{4}{9 \sqrt{3}} \pi^2$
C
$\frac{-8}{9 \sqrt{3}} \pi^2$
D
$-\frac{8}{9} \pi^2$
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