MHT CET 2025 22nd April Evening Shift | Differential Equations Question 18 | Mathematics

A wet substance in the open air loses its moisture at a rate proportional to the moisture content. If a sheet, hung in the open air, loses half its moisture during the first hour, then $90 \%$ of the moisture will be lost in ________ hours.

$2 \log _2 10$

$\frac{4 \log 10}{\log 2}$

$\quad \log _2 10$

$\frac{3 \log 10}{\log 2}$

MHT CET 2025 22nd April Evening Shift

MCQ (Single Correct Answer)

-0

The general solution of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}+\sin \left(\frac{x+y}{2}\right)=\sin \left(\frac{x-y}{2}\right)$ is

$\log \tan \left(\frac{y}{2}\right)=\mathrm{c}-2 \sin \frac{x}{2}$, where c is the constant of integration

$\log \tan \left(\frac{y}{4}\right)=\mathrm{c}-2 \sin \left(\frac{x}{2}\right)$, where c is the constant of integration

$\log \left[\tan \left(\frac{y}{2}+\frac{\pi}{4}\right)\right]=\mathrm{c}-2 \sin x$, where c is the constant of integration

$\log \left[\tan \left(\frac{y}{4}+\frac{\pi}{4}\right)\right]=\mathrm{c}-2 \sin \frac{x}{2}$, where c is the constant of integration

MHT CET 2025 22nd April Evening Shift

MCQ (Single Correct Answer)

-0

The equation of the curve passing through $\left(2, \frac{9}{2}\right)$ and having the slope $\left(1-\frac{1}{x^2}\right)$ at $(x, y)$ is

$x y=x^2+2 x+1$

$x y=x^2+x+2$

$x y=x^2+x+5$

$x y=x^2+2 x+5$

MHT CET 2025 22nd April Morning Shift

MCQ (Single Correct Answer)

-0

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

$\frac{1}{3}$

$\frac{2}{3}$

$-\frac{1}{3}$

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