Question

Let $f$ be a function defined implicitly by the equation $\frac{1-e^{f\left(x\right)}}{1+e^{f\left(x\right)}}=x.$ If $g$ be the inverse of $f,$ then which of the following options is INCORRECT ?

• A. $g\left(x\right)$ is an odd function.
• B. $g\left(x\right)$ is strictly decreasing function.
• C. $g\left(x\right)$ is differentiable at all points.
• D. Range of $g\left(x\right)$ is $[-1,1]$.

Answer 1

The correct option is D Range of $g\left(x\right)$ is $[-1,1]$.

Given, $\frac{1-e^{f\left(x\right)}}{1+e^{f\left(x\right)}}=x$
$f$ and $g$ are inverse of each other.
So, $g\left(f\right(x\left)\right)=x$
$\Rightarrow \frac{1-e^{f\left(x\right)}}{1+e^{f\left(x\right)}}=g\left(f\right(x\left)\right)$
$\therefore g\left(x\right)=\frac{1-e^x}{1+e^x}$

Now, $g(-x)=\frac{1-e^{-x}}{1+e^{-x}}=-\frac{1-e^x}{1+e^x}$
$\Rightarrow g\left(x\right)=-g(-x)$
Hence, $g\left(x\right)$ is an odd function.

$g^{\prime }\left(x\right)=\frac{-2e^x}{(1+e^x{)}^2}=-2e^x(1+e^x{)}^{-2}$
$\Rightarrow g^{\prime }\left(x\right)<0\forall x\in R$
$\therefore g\left(x\right)$ is strictly decreasing function and differentiable at all points.

$g\left(x\right)=\frac{1-e^x}{1+e^x}=\frac{2}{1+e^x}-1$
$\Rightarrow 1<1+e^x<\infty$
$\Rightarrow 0<\frac{1}{1+e^x}<1$
$\Rightarrow 0<\frac{2}{1+e^x}<2$
$\Rightarrow -1<\frac{2}{1+e^x}-1<1$
$\therefore$ Range of $g\left(x\right)$ is $(-1,1)$