1
MHT CET 2024 2nd May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The solution of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}=(x-y)^2$ when $y(1)=1$ is

A
$\log \left|\frac{2-y}{2-x}\right|=2(y-1)$
B
$-\log \left|\frac{1+x-y}{1-x+y}\right|=x+y-2$
C
$\log \left|\frac{2-x}{2-y}\right|=x-y$
D
$-\log \left|\frac{1-+xy}{1+x-y}\right|=2(x-1)$
2
MHT CET 2024 2nd May Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $x \frac{\mathrm{~d} y}{\mathrm{~d} x}=y(\log y-\log x+1)$, then general solution of this equation is

A
$\log \left(\frac{x}{y}\right)=\mathrm{cy}$, where c is a constant of integration.
B
$\log \left(\frac{x}{y}\right)=\mathrm{c} x$, where c is a constant of integration.
C
$\log \left(\frac{y}{x}\right)=\mathrm{cy}$, where c is a constant of integration.
D
$\log \left(\frac{y}{x}\right)=c x$, where $c$ is a constant of integration.
3
MHT CET 2024 2nd May Morning Shift
MCQ (Single Correct Answer)
+2
-0

A spherical metal ball at 80$^\circ$C cools in 5 minutes to 60$^\circ$C, in surrounding temperature of 20$^\circ$C, then the temperature of the ball after 20 minutes is approximately

A
(8.15)$^\circ$C
B
(11.85)$^\circ$C
C
(28.15)$^\circ$C
D
(31.85)$^\circ$C
4
MHT CET 2023 14th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If the slope of the tangent of the curve at any point is equal to $$-y+\mathrm{e}^{-x}$$, then the equation of the curve passing through origin is

A
$$y+x \mathrm{e}^x=0$$
B
$$y \mathrm{e}^x+x=0$$
C
$$y \mathrm{e}^x-x=0$$
D
$$y-x \mathrm{e}^x=0$$
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