Mathematics - Functions Question with Solution | TestHub

$\begin{array}{rl}& \mathrm{f}(\mathrm{x})=\frac{2^{2\mathrm{x}}-1}{2^{2\mathrm{x}}+1}\\ & =1-\frac{2}{2^{2\mathrm{x}}+1}\\ & {\mathrm{f}}^{\prime }(\mathrm{x})=\frac{2}{{\left(2^{2\mathrm{x}}+1\right)}^2}\cdot 2.2^{2\mathrm{x}}\cdot \ln 2\text{ i.e always }+\mathrm{ve}\end{array}$ so $f(x)$ is $\uparrow$ function $\begin{array}{rl}& \therefore \mathrm{f}(-\infty )=-1\\ & \mathrm{f}(\infty )=1\\ & \therefore \mathrm{f}(\mathrm{x})\in (-1,1)\ne \text{ co-domain }\end{array}$ so function is one-one but not onto