JEE Main 2022 — Application Of Derivatives Question with Solution

Total students = 100

At t = 0 (zero day), infected student = 2

Let at t = t day infected student = x

$$\therefore$$ At t = t day non infected student = (100 $$-$$ x)

Rate of infection $$=\frac{dx}{dt}$$

Given, $$\frac{dx}{dt}\propto x(100-x)$$

$$\Rightarrow \int \frac{dx}{x(100-x)}=\int kdt$$

$$\Rightarrow \frac{1}{100}\int \frac{100-x+x}{x(100-x)}dx=kt+c$$

$$\Rightarrow \frac{1}{100}\int \left(\frac{1}{x}+\frac{1}{100-x}\right)dx=kt+c$$

$$\Rightarrow \frac{1}{100}\left[\ln x-\ln (100-x)\right]=kt+c$$

$$\Rightarrow \frac{1}{100}\ln \frac{x}{100-x}=kt+c$$ ...... (1)

Given, At, t = 0, x = 2

$$\therefore$$ $$\frac{1}{100}\ln \frac{2}{98}=c$$

Putting value of c in equation (1), we get

$$\frac{1}{100}\ln \frac{x}{100-x}=kt+\frac{1}{100}\ln \frac{2}{98}$$

$$\Rightarrow \frac{1}{100}\ln \frac{x}{100-x}-\frac{1}{100}\ln \frac{2}{98}=kt$$

$$\Rightarrow \frac{1}{100}\ln \frac{x\times 98}{2(100-x)}=kt$$

Given, At t = 4, x = 30

$$\therefore$$ $$\frac{1}{100}\ln \frac{30\times 98}{2(70)}=k\times 4$$

$$\Rightarrow k=\frac{1}{400}\ln 21$$

$$\therefore$$ $$\frac{1}{100}\ln \frac{x\times 98}{2(100-x)}=t\times \frac{1}{400}\times \ln 21$$

Now, when t = 8, then r = ?

$$\frac{1}{100}\ln \frac{49x}{(100-x)}=8\times \frac{1}{400}\times \ln 21$$

$$\Rightarrow \ln \frac{49x}{(100-x)}=2\ln 21$$

$$\Rightarrow \frac{49x}{100-x}=21^2$$

$$\Rightarrow \frac{x}{100-x}=\frac{21\times 21}{49}$$

$$\Rightarrow \frac{x}{100-x}=9$$

$$\Rightarrow x=900-9x$$

$$\Rightarrow 10x=900$$

$$\Rightarrow x=90$$