JEE Main 2026 — Functions Question with Solution

Given $f(x)=\frac{x-1}{x+1}$. Let us find the first few compositions of $f(x)$:

$f^2(x)=f(f(x))=\frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1}=\frac{x-1-x-1}{x-1+x+1}=\frac{-2}{2x}=-\frac{1}{x}$

$f^3(x)=f(f^2(x))=f\left(-\frac{1}{x}\right)=\frac{-\frac{1}{x}-1}{-\frac{1}{x}+1}=\frac{-1-x}{-1+x}=\frac{x+1}{1-x}$

$f^4(x)=f(f^3(x))=f\left(\frac{x+1}{1-x}\right)=\frac{\frac{x+1}{1-x}-1}{\frac{x+1}{1-x}+1}=\frac{x+1-1+x}{x+1+1-x}=\frac{2x}{2}=x$

Since $f^4(x)=x$, the sequence of functions is periodic with a period of $4$.

Therefore, $f^{26}(x)=f^{4\times 6+2}(x)=f^2(x)=-\frac{1}{x}$.

We are given $g(x)+f^{26}(x)=0$, which implies:
$g(x)-\frac{1}{x}=0\Rightarrow g(x)=\frac{1}{x}$

We need to find the area of the region bounded by the curves $y=\frac{1}{x}$, $y=x-\frac{3}{2}$ (from $2y=2x-3$), $y=0$, and $x=4$.

First, find the point of intersection of $y=\frac{1}{x}$ and $y=x-\frac{3}{2}$:
$\frac{1}{x}=x-\frac{3}{2}\Rightarrow 2x^2-3x-2=0$
$(2x+1)(x-2)=0$
Since $x\in (1,\infty )$, we get $x=2$.

The line $y=x-\frac{3}{2}$ intersects the x-axis ($y=0$) at $x=\frac{3}{2}$.

The required area $A$ is bounded by $y=x-\frac{3}{2}$ from $x=\frac{3}{2}$ to $x=2$, and by $y=\frac{1}{x}$ from $x=2$ to $x=4$.

$A={\int }_{3/2}^2\left(x-\frac{3}{2}\right)dx+{\int }_2^4\frac{1}{x}dx$

Evaluating the first integral:
${\int }_{3/2}^2\left(x-\frac{3}{2}\right)dx={\left[\frac{1}{2}{\left(x-\frac{3}{2}\right)}^2\right]}_{3/2}^2=\frac{1}{2}{\left(2-\frac{3}{2}\right)}^2-0=\frac{1}{2}{\left(\frac{1}{2}\right)}^2=\frac{1}{8}$

Evaluating the second integral:
${\int }_2^4\frac{1}{x}dx=[{\log }_{e}x{]}_2^4={\log }_{e}4-{\log }_{e}2={\log }_{e}2$

Total Area = $\frac{1}{8}+{\log }_{e}2$