The function f:(-,) (-,1), defined by f(x)=2^x-2^-x2^x+2^-x is :

Question

The function $f:(-\infty,\infty)	o (-\infty,1)$, defined by $f(x)=\frac{{2}^{x}-{2}^{-x}}{{2}^{x}+{2}^{-x}}$ is :

Accepted Answer

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

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

Solutions · 1