The function \(f:(-\infty, \infty) \rightarrow(-\infty, 1)\), defined by \(f(x)=\frac{2^x-2^{-x}}{2^x+2^{-x}}\) is : [JE…

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

On differentiating, we get

\(\mathrm{f}^{\prime}(\mathrm{x})=\frac{2}{\left(2^{2 \mathrm{x}}+1\right)^2} \cdot 2 \cdot 2^{2 \mathrm{x}} \cdot \ln 2 \)

Since \(\mathrm{f}^{\prime}(\mathrm{x})>0\)

so \(\mathrm{f}(\mathrm{x})\) is increasing function

\( \therefore \mathrm{f}(-\infty)=-1 \)
\( f(\infty)=1 \)
\( \therefore \mathrm{f}(\mathrm{x}) \in(-1,1) \neq \) co-domain

so function is one-one but not onto.