JEE Main 2024 — Mathematics Calculus

Given: $\frac{\mathrm{dx}}{\mathrm{dt}}+\mathrm{ax}=0$dtdx​+ax=0

$$\Rightarrow \frac{\mathrm{dx}}{x}=-\mathrm{adt}$$⇒xdx​=−adt

Integrating both sides,

$$\Rightarrow \int \frac{\mathrm{dx}}{x}=-a\int \mathrm{dt}$$⇒∫xdx​=−a∫dt

$$\Rightarrow \mathrm{log}|x|=-\mathrm{at}+c$$⇒log∣x∣=−at+c

At $t=0,x=2$t=0,x=2

$$\Rightarrow \mathrm{log}2=0+c$$⇒log2=0+c

$$\Rightarrow \mathrm{logx}=-\mathrm{at}+\mathrm{log}2$$⇒logx=−at+log2

$$\Rightarrow \mathrm{logx}-\mathrm{log}2=-\mathrm{at}$$⇒logx−log2=−at

$$\Rightarrow \mathrm{log}(\frac{x}{2})=-\mathrm{at}$$⇒log(2x​)=−at

$$\Rightarrow \frac{x}{2}={e}^{-\mathrm{at}}$$⇒2x​=e−at

$$\Rightarrow x=2{e}^{-\mathrm{at}}...(i)$$⇒x=2e−at...(i)

It is also given that, $\frac{\mathrm{dy}}{\mathrm{dt}}+\mathrm{by}=0$dtdy​+by=0

$$\Rightarrow \frac{\mathrm{dy}}{y}=-\mathrm{bdt}$$⇒ydy​=−bdt

$$\Rightarrow \mathrm{log}|y|=-\mathrm{bt}+\lambda$$⇒log∣y∣=−bt+λ

At, $t=0,y=1$t=0,y=1

$$\Rightarrow 0=0+\lambda$$⇒0=0+λ

$$\Rightarrow y={e}^{-\mathrm{bt}}...(\mathrm{ii})$$⇒y=e−bt...(ii)

According to question,

$$3y(1)=2x(1)$$3y(1)=2x(1)

$$\Rightarrow 3{e}^{-b}=2(2{e}^{-a})$$⇒3e−b=2(2e−a)

$$\Rightarrow {e}^{a-b}=\frac{4}{3}$$⇒ea−b=34​

$$\text{For}x(t)=y(t)$$Forx(t)=y(t)

$$\Rightarrow 2{e}^{-\mathrm{at}}={e}^{-\mathrm{bt}}$$⇒2e−at=e−bt

$$\Rightarrow 2={e}^{(a-b)t}$$⇒2=e(a−b)t

$$\Rightarrow 2={(\frac{4}{3})}^{t}$$⇒2=(34​)t

$$\Rightarrow \mathrm{log}(2)=t\times \mathrm{log}(\frac{4}{3})$$⇒log(2)=t×log(34​)

$$\Rightarrow {\mathrm{log}}_{\frac{4}{3}}2=t$$⇒log34​​2=t