Question

If $f\left(x\right)$ is continuous such that $\left|f\right(x\left)\right|\le 1,\forall x\in R$ and $g\left(x\right)=\frac{e^{f\left(x\right)}-e^{\left|f\right(x\left)\right|}}{e^{f\left(x\right)}+e^{\left|f\right(x\left)\right|}}$ then range of $g\left(x\right)$ is

• A. $[0,1]$
• B. $[0,\frac{e^2+1}{e^2-1}]$
• C. $[0,\frac{e^2-1}{e^2+1}]$
• D. $[\frac{1-e^2}{1+e^2},0]$

Answer 1

Answer: (D)

$[\frac{1-e^2}{1+e^2},0]$

$g\left(x\right)=\frac{e^{f\left(x\right)}-e^{\left|f\right(x\left)\right|}}{e^{f\left(x\right)},+e^{\left|f\right(x\left)\right|}},-1\le x\le 1$

For, $0\le x\le 1$,

$g\left(x\right)=0$

For $-1\le f\left(x\right)<0$

$g\left(x\right)=\frac{e^{f\left(x\right)}-e^{-f\left(x\right)}}{e^{f\left(x\right)}+e^{f\left(x\right)}}=\frac{e^{2f\left(x\right)}-1}{e^{2f\left(x\right)}+1}$

$=1-\frac{2}{e^{2f\left(x\right)}+1}$

For, $-1\le f\left(x\right)<0$,

$g\left(x\right)\in [\frac{1-e^2}{1+e^2},0]$

For, $-1\le f\left(x\right)<1$,

$g\left(x\right)\in [\frac{1-e^2}{1+e^2},0]$