Question

The inverse of the function $f\left( x \right)=\dfrac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}}+2$ is given by
\[\begin{align}
  & A.\log {{\left( \dfrac{\left( x-2 \right)}{\left( x-1 \right)} \right)}^{\dfrac{1}{2}}} \\
 & B.\dfrac{1}{2}{{\log }_{e}}\left( \dfrac{\left( x-1 \right)}{\left( 3-x \right)} \right) \\
 & C.\log {{\left( \dfrac{x}{2-x} \right)}^{\dfrac{1}{2}}} \\
 & D.\log {{\left( \dfrac{\left( x-1 \right)}{\left( 3-x \right)} \right)}^{\dfrac{1}{2}}} \\
\end{align}\]

Answer 1

Hint:
 To solve this question, we will first substitute $f\left( x \right)=y$ after simplifying f(x) as $f\left( x \right)=y\Rightarrow {{f}^{-1}}\left( y \right)=x$. So, we will then try to compute value of x from $y=f\left( x \right)$. After computing value of x we will use ${{f}^{-1}}\left( y \right)=x$ to get answer in terms of y and finally we will replace y by x to get answer in term of x.

Complete step by step answer:
We are given f(x) as \[f\left( x \right)=\dfrac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}}+2\]
First of all we will simplify f(x).
If a function $g:A\to B$ has $g\left( a \right)=b$ then inverse of g is given by ${{g}^{-1}}$ where ${{g}^{-1}}\left( b \right)=a$
Here, \[f\left( x \right)=\dfrac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}}+2\]
Let us use ${{e}^{-x}}=\dfrac{1}{{{e}^{x}}}$ in above, we get:
\[\begin{align}
  & \Rightarrow f\left( x \right)=\dfrac{{{e}^{x}}-\dfrac{1}{{{e}^{x}}}}{{{e}^{x}}+\dfrac{1}{{{e}^{x}}}}+2 \\
 & \Rightarrow f\left( x \right)=\dfrac{\dfrac{{{e}^{2x}}-1}{{{e}^{x}}}}{\dfrac{{{e}^{2x}}+1}{{{e}^{x}}}}+2 \\
\end{align}\]
Cancelling ${{e}^{x}}$ we have:
\[\Rightarrow f\left( x \right)=\dfrac{{{e}^{2x}}-1}{{{e}^{2x}}+1}+2\]
To find inverse of f we substitute $f\left( x \right)=y$
Put \[f\left( x \right)=y\Rightarrow x={{f}^{-1}}\left( y \right)\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}\]
Put $f\left( x \right)=y$
\[\begin{align}
  & \Rightarrow y=\dfrac{{{e}^{2x}}-1}{{{e}^{2x}}+1}+2 \\
 & \Rightarrow y-2=\dfrac{{{e}^{2x}}-1}{{{e}^{2x}}+1} \\
\end{align}\]
Cross multiplying the above equation, we get:
\[\begin{align}
  & \Rightarrow \left( y-2 \right)\left( {{e}^{2x}}+1 \right)=\left( {{e}^{2x}}-1 \right) \\
 & \Rightarrow {{e}^{2x}}y-2{{e}^{2x}}+y-2={{e}^{2x}}-1 \\
 & \Rightarrow {{e}^{2x}}\left( y-2 \right)-{{e}^{2x}}=-1+2-y \\
 & \Rightarrow {{e}^{2x}}\left( y-2-1 \right)=-y+1 \\
 & \Rightarrow {{e}^{2x}}\left( y-3 \right)=1-y \\
\end{align}\]
Dividing by (y-3) we get:
\[\Rightarrow {{e}^{2x}}=\dfrac{1-y}{y-3}\]
Now as \[\begin{align}
  & \ln {{e}^{x}}=x \\
 & \Rightarrow \ln {{e}^{2x}}=2x \\
\end{align}\]
Taking $\ln \Rightarrow {{\log }_{e}}$ on both sides we get:
\[\begin{align}
  & \Rightarrow \ln \left( {{e}^{2x}} \right)=\ln \left( \dfrac{1-y}{y-3} \right)={{\log }_{e}}\left( \dfrac{1-y}{y-3} \right) \\
 & \Rightarrow 2x={{\log }_{e}}\left( \dfrac{1-y}{y-3} \right) \\
 & \Rightarrow x=\dfrac{1}{2}{{\log }_{e}}\left( \dfrac{1-y}{y-3} \right) \\
\end{align}\]
Now because $m\log n=\log {{n}^{m}}$ using this in above we get:
\[\Rightarrow x={{\log }_{e}}{{\left( \dfrac{1-y}{y-3} \right)}^{\dfrac{1}{2}}}\]
Now from equation (i) we had ${{f}^{-1}}\left( y \right)=x$
\[\Rightarrow {{f}^{-1}}\left( y \right)={{\log }_{e}}{{\left( \dfrac{1-y}{y-3} \right)}^{\dfrac{1}{2}}}\]
Now replacing y from x in above and using $\log {{n}^{m}}=m\log n$ we have:
\[\Rightarrow {{f}^{-1}}\left( x \right)=\dfrac{1}{2}{{\log }_{e}}\left( \dfrac{1-x}{x-3} \right)=\dfrac{1}{2}{{\log }_{e}}\left( \dfrac{x-1}{3-x} \right)\]
The value of inverse of f is \[\dfrac{1}{2}{{\log }_{e}}\left( \dfrac{1-x}{x-3} \right)=\dfrac{1}{2}{{\log }_{e}}\left( \dfrac{x-1}{3-x} \right)\]
So option B is correct.

Note:
Students should not get confused between step of using $m\log n=\log {{n}^{m}}$
The answer to this question can also be \[{{f}^{-1}}\left( x \right)={{\log }_{e}}{{\left( \dfrac{1-x}{x-3} \right)}^{\dfrac{1}{2}}}\] but it was not in any option given, so we have again used $m\log n=\log {{n}^{m}}$ to get option matched. Anyways \[{{f}^{-1}}\left( x \right)=\dfrac{1}{2}{{\log }_{e}}\left( \dfrac{1-x}{x-3} \right)\text{ and }{{f}^{-1}}\left( x \right)={{\log }_{e}}{{\left( \dfrac{1-x}{x-3} \right)}^{\dfrac{1}{2}}}\] both are correct.