JEE Main 2022 — Mathematics Calculus

Given, $\frac{dy}{dx}+\frac{{2}^{x-y}({2}^{y}-1)}{{2}^{x}-1}=0$dxdy​+2x−12x−y(2y−1)​=0 $x,y>0,y(1)=1$x,y>0,y(1)=1

Now rearranging and integrating both side of $\frac{dy}{dx}=-\frac{{2}^{x}({2}^{y}-1)}{{2}^{y}({2}^{x}-1)}$dxdy​=−2y(2x−1)2x(2y−1)​, we get

$$\Rightarrow \int \frac{{2}^{y}}{{2}^{y}-1}dy=-\int \frac{{2}^{x}}{{2}^{x}-1}dx$$⇒∫2y−12y​dy=−∫2x−12x​dx

$$\Rightarrow \frac{1}{\mathrm{ln}2}\int \frac{{2}^{y}\mathrm{ln}2}{{2}^{y}-1}dy=-\frac{1}{{\mathrm{ln}}^{2}}\int \frac{{2}^{x}\mathrm{ln}2}{{2}^{x}-1}dx$$⇒ln21​∫2y−12yln2​dy=−ln21​∫2x−12xln2​dx

$$\Rightarrow \frac{1}{\mathrm{ln}2}\mathrm{ln}|{2}^{y}-1|=\frac{-1}{\mathrm{ln}2}\mathrm{ln}|{2}^{x}-1|+C$$⇒ln21​ln∣2y−1∣=ln2−1​ln∣2x−1∣+C

At $x=1,y=1$x=1,y=1

Putting this values in above equation we get $C=0$C=0

So, $\mathrm{ln}|{2}^{y}-1|+\mathrm{ln}|{2}^{x}-1|=0$ln∣2y−1∣+ln∣2x−1∣=0

$$\Rightarrow ({2}^{x}-1)({2}^{y}-1)=1$$⇒(2x−1)(2y−1)=1

$$\Rightarrow {2}^{y}-1=\frac{1}{{2}^{x}+1}$$⇒2y−1=2x+11​

At $x=2$x=2

$$\Rightarrow {2}^{y}=\frac{1}{3}+1=\frac{4}{3}$$⇒2y=31​+1=34​

$$y={\mathrm{log}}_{2}\frac{4}{3}={\mathrm{log}}_{2}4-{\mathrm{log}}_{2}3=2-{\mathrm{log}}_{2}3$$y=log2​34​=log2​4−log2​3=2−log2​3