JEE Main 2026 — 8 April, Evening Shift — Question 35

$f^{2}(x)=f(f(x))=\frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1}=\frac{-1}{x}$ $f^{3}(x)=f\left(f^{2}(x)\right)=f\left(\frac{-1}{x}\right)=\frac{1+x}{1-x}$ $f^{4}(x)=f\left(f^{3}(x)\right)=f\left(f\left(\frac{-1}{x}\right)\right)=x$ $\mathrm{f}^{5}(\mathrm{x})=\mathrm{f}\left(\mathrm{f}^{4}(\mathrm{x})\right)=\mathrm{f}(\mathrm{x})$ $f^{6}(x)=f\left(f^{5}(x)\right)=f(f(x))=\frac{-1}{x}$ . • $\mathrm{f}^{26}(\mathrm{x})=\frac{-1}{\mathrm{x}}$ $\mathrm{g}(\mathrm{x})=\frac{1}{\mathrm{x}}$ $\mathrm{x}-3 / 2=1 / \mathrm{x} \Rightarrow 2 \mathrm{x}^{2}-3 \mathrm{x}-2=0$ $\Rightarrow \mathrm{x}=-1 / 2,2$ Req. area $=\int_{2}^{4} \frac{1}{\mathrm{x}} \mathrm{dx}+\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}$ $=\frac{1}{8}+(\ell \mathrm{nx})_{2}^{4}=\frac{1}{8}+\ell \mathrm{n} 2$