Question

If $f(x)=\frac{10^{x}-10^{-x}}{10^{x}+10^{-x}}+2$, then $f^{-1}(y)=$

• A. $\frac{1}{2} \log _{10}\left(\frac{y-1}{y-3}\right)$
• B. $\frac{1}{2} \log _{10}\left(\frac{y-3}{y-1}\right)$
• C. $\frac{1}{2} \log _{10}\left(\frac{y-1}{3-y}\right)$
• D. $\frac{1}{2} \log _{10}\left(\frac{y-1}{y}\right)$

Answer 1

Answer: (C)

Given $f(x)=\frac{10^{x}-10^{-x}}{10^{x}+10^{-x}}+2$
Let $f(x)=y$
$ \begin{array}{l} \Rightarrow y=\frac{10^{x}-10^{-x}}{10^{x}+10^{-x}}+2 \\ \Rightarrow \frac{y-2}{1}=\frac{10^{x}-10^{-x}}{10^{x}+10^{-x}} \end{array} $
Applying componendo and dividendo on both sides, we get
$ \begin{array}{l} \frac{(y-2)+1}{(y-2)-1}=\frac{\left(10^{x}-10^{-x}\right)+\left(10^{x}+10^{-x}\right)}{\left(10^{x}-10^{-x}\right)-\left(10^{x}+10^{-x}\right)} \\ \Rightarrow \frac{y-1}{y-3}=\frac{10^{x}}{-10^{-x}} \\ \Rightarrow \frac{y-1}{3-y}=10^{2 x} \end{array} $
Taking $\log _{10}$ on both sides, we get
$ \begin{array}{l} \log _{10}\left(\frac{y-1}{3-y}\right)=\log _{10}(10)^{2 x} \\ \Rightarrow \log _{10}\left(\frac{y-1}{3-y}\right)=2 x\left[\because \log (a)^{m}=m \log (a) \& \log _{a} a=1, a \neq 1\right] \\ \Rightarrow x=\frac{1}{2} \log _{10}\left(\frac{y-1}{3-y}\right) \end{array} $
For $f^{-1}(x)$, replacing $x \rightarrow f^{-1}(x) \& y \rightarrow x$, we get
$ f^{-1}(x)=\frac{1}{2} \log _{10}\left(\frac{x-1}{3-x}\right) $