JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Sets Relations and Functions

Rewrite $f(x)=\frac{2^x-2^{-x}}{2^x+2^{-x}}$ by multiplying numerator and denominator by $2^x$:


Differentiating:


so $f$ is strictly increasing on all of $\mathbb{R}$, hence one-one.

For the range, as $x\to -\infty$, $2^{2x}\to 0$ so $f(x)\to -1$; as $x\to \infty$, $2^{2x}\to \infty$ so $f(x)\to 1$. Since $f$ is strictly increasing and continuous, its range is exactly $(-1,1)$.

The given codomain is $(-\infty ,1)$, but the actual range $(-1,1)$ is a proper subset of it, so $f$ does not hit every value in the codomain — it is not onto.

So $f$ is one-one but not onto.