The function $f:(-\infty, \infty) \rightarrow(-\infty, 1)$, defined by $f(x)=\frac{2^x-2^{-x}}{2^x+2^{-x}}$ is :
Solution
<p>$$\begin{aligned}
& f(x)=\frac{2^{2 \mathrm{x}}-1}{2^{2 \mathrm{x}}+1} \\
& =1-\frac{2}{2^{2 \mathrm{x}}+1} \\
& \mathrm{f}^{\prime}(\mathrm{x})=\frac{2}{\left(2^{2 \mathrm{x}}+1\right)^2} \cdot 2 \cdot 2^{2 \mathrm{x}} \cdot \ln 2 \text { i.e always }+\mathrm{ve}
\end{aligned}$$</p>
<p>so $f(x)$ is $\uparrow$ function</p>
<p>$$\begin{aligned}
& \therefore \mathrm{f}(-\infty)=-1 \\
& \mathrm{f}(\infty)=1 \\
& \therefore \mathrm{f}(\mathrm{x}) \in(-1,1) \neq \text { co-domain }
\end{aligned}$$</p>
<p>so function is one-one but not onto</p>
About this question
Subject: Mathematics · Chapter: Sets, Relations and Functions · Topic: Sets and Operations
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