1
MHT CET 2023 12th May Evening Shift
+2
-0

The solution of $$\mathrm{e}^{y-x} \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{y(\sin x+\cos x)}{(1+y \log y)}$$ is

A
$$\frac{\mathrm{e}^y}{y}=\mathrm{e}^x \sin x+\mathrm{c}$$, where $$\mathrm{c}$$ is a constant of integration.
B
$$\mathrm{e}^y \log y=\mathrm{e}^x \cos x+\mathrm{c}$$, where $$\mathrm{c}$$ is a constant of integration.
C
$$\mathrm{e}^y \log y=\mathrm{e}^x \sin x+\mathrm{c}$$, where $$\mathrm{c}$$ is a constant of integration.
D
$$\mathrm{e}^y y=\mathrm{e}^x \sin x+\mathrm{c}$$, where $$\mathrm{c}$$ is a constant of integration.
2
MHT CET 2023 12th May Evening Shift
+2
-0

Water flows from the base of rectangular tank, of depth 16 meters. The rate of flow of the water is proportional to the square root of depth at any time $$\mathrm{t}$$. If depth is $$4 \mathrm{~m}$$ when $$\mathrm{t}=2$$ hours, then after 3.5 hours the depth (in meters) is

A
0
B
0.25
C
0.5
D
3
3
MHT CET 2023 12th May Evening Shift
+2
-0

If $$(2+\sin x) \frac{\mathrm{d} y}{\mathrm{~d} x}+(y+1) \cos x=0$$ and $$y(0)=1$$, then $$y\left(\frac{\pi}{2}\right)$$ is

A
$$\frac{-2}{3}$$
B
$$\frac{-1}{3}$$
C
$$\frac{4}{3}$$
D
$$\frac{1}{3}$$
4
MHT CET 2023 12th May Morning Shift
+2
-0

The decay rate of radio active material at any time $$t$$ is proportional to its mass at that time. The mass is 27 grams when $$t=0$$. After three hours it was found that 8 grams are left. Then the substance left after one more hour is

A
$$\frac{27}{8}$$ grams
B
$$\frac{81}{4}$$ grams
C
$$\frac{16}{3}$$ grams
D
$$\frac{16}{9}$$ grams
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