JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Sets Relations and Functions

For $f\circ g$ to be defined, we need $g(x)>0$, since $f(x)={\log }_{e}x$ requires a positive argument.

The denominator $2x^2-2x+1=2{\left(x-\frac{1}{2}\right)}^2+\frac{1}{2}$ is strictly positive for every real $x$, so the sign of $g(x)$ is decided entirely by the numerator.

Write the numerator as

Since $(x^2-x+1{)}^2\ge 0$, the numerator is at least $1$ for every real $x$, hence always positive.

With both numerator and denominator positive for every real $x$, $g(x)>0$ holds for all $x\in \mathbb{R}$. The domain of $f\circ g$ is $\mathbb{R}$.