JEE Main 2025 Jan 24 Shift 2, Mathematics Q2: Invertible And Composite Functions

Rewrite the function in a more transparent form by multiplying numerator and denominator by $2^x$:

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

Monotonicity: As $x$ increases, $2^{2x}+1$ strictly increases, so $\frac{2}{2^{2x}+1}$ strictly decreases, so $f(x)$ is strictly increasing on $\mathbb{R}$. Hence $f$ is one-one.

Range:

$\underset{x\to -\infty }{\lim }f(x)=1-\frac{2}{0+1}=-1,\underset{x\to \infty }{\lim }f(x)=1-0=1$

Since $f$ is continuous and strictly increasing, its range is the open interval $(-1,1)$. The given codomain is $(-\infty ,1)$, which strictly contains $(-1,1)$, so $f$ is not onto.

Correct option: (1)