JEE Main 2022 — Mathematics Calculus

Given, $\frac{dy}{dx}\propto \frac{-y}{x}$dxdy​∝x−y​

$\Rightarrow \frac{dy}{dx}=\frac{-Ky}{x}$⇒dxdy​=x−Ky​ (where $K$K is proportionality constant)

Now integrating both side,

$$\Rightarrow \int \frac{dy}{y}=-K\int \frac{dx}{x}$$⇒∫ydy​=−K∫xdx​

$$\Rightarrow \mathrm{ln}|y|=-K\mathrm{ln}|x|+C$$⇒ln∣y∣=−Kln∣x∣+C

If the above equation satisfy $(1,2)$(1,2)

$$\Rightarrow \mathrm{ln}2=-K\times 0+C\Rightarrow C=\mathrm{ln}2$$⇒ln2=−K×0+C⇒C=ln2

So, $\mathrm{ln}|y|=-K\mathrm{ln}|x|+\mathrm{ln}2$ln∣y∣=−Kln∣x∣+ln2

Now it also passes through $(8,1)$(8,1)

$$\Rightarrow \mathrm{ln}1=-K\mathrm{ln}8+\mathrm{ln}2\Rightarrow K=\frac{1}{3}$$⇒ln1=−Kln8+ln2⇒K=31​

So, equation becomes $\mathrm{ln}|y|=-\frac{1}{3}\mathrm{ln}|x|+\mathrm{ln}2$ln∣y∣=−31​ln∣x∣+ln2

So, at $x=\frac{1}{8}$x=81​

$$\Rightarrow \mathrm{ln}|y|=-\frac{1}{3}\mathrm{ln}(\frac{1}{8})+\mathrm{ln}2=2\mathrm{ln}2$$⇒ln∣y∣=−31​ln(81​)+ln2=2ln2

$$\Rightarrow |y|=4$$⇒∣y∣=4