$$\begin{aligned} & f(x)=\frac{2^{2 \mathrm{x}}-1}{2^{2 \mathrm{x}}+1} \\ & =1-\frac{2}{2^{2 \mathrm{x}}+1} \\ & \mathrm{f}^{\prime}(\mathrm{x})=\frac{2}{\left(2^{2 \mathrm{x}}+1\right)^2} \cdot 2 \cdot 2^{2 \mathrm{x}} \cdot \ln 2 \text { i.e always }+\mathrm{ve} \end{aligned}$$

so $f(x)$ is $\uparrow$ function

$$\begin{aligned} & \therefore \mathrm{f}(-\infty)=-1 \\ & \mathrm{f}(\infty)=1 \\ & \therefore \mathrm{f}(\mathrm{x}) \in(-1,1) \neq \text { co-domain } \end{aligned}$$

so function is one-one but not onto