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