Question

If $f:R\to R$ satisfies condition $e^{x}f\left(y\right)+e^{y}f\left(x\right)=2e^{x+y}-e^{x-y}-e^{y-x}\forall x,yϵR,$ then

• A. $f\left(x\right)$ is onto function
• B. $f\left(x\right)$ is one-one function
• C. $f\left(x\right)$ is an odd function
• D. $f\left(\left|x\right|\right)$ & $\left|f\left(x\right)\right|$ are identical

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $f\left(x\right)$ is onto function
B $f\left(x\right)$ is one-one function
C $f\left(x\right)$ is an odd function
D $f\left(\left|x\right|\right)$ & $\left|f\left(x\right)\right|$ are identical

We have,
$e^{x}f\left(y\right)+e^{y}f\left(x\right)=2e^{x+y}-e^{x-y}-e^{y-x}\forall x,yϵR,$
Put $x=y\Rightarrow 2e^{x}f\left(x\right)=2e^{2x}-2$
$f\left(x\right)=e^x-e^{-x}$
$f(-x)=-f\left(x\right)\Rightarrow$ odd function
$f^{\prime }\left(x\right)=e^x+e^{-x}>0$
$\Rightarrow$ one-one function
Also $f\left(\infty \right)\to \infty$, $f(-\infty )\to -\infty$
$\Rightarrow$ Range is $(-\infty ,\infty )=R=$ co-domain of f
Hence f is also an onto.
Also $f\left(|x|\right)=\{\begin{array}{c}f\left(x\right),x\ge 0\\ f(-x)=-f\left(x\right),x<0\end{array}$
and $|f\left(x\right)|=\{\begin{array}{c}f\left(x\right),f\left(x\right)\ge 0\Rightarrow x\ge 0\\ -f\left(x\right),f\left(x\right)<0\Rightarrow x<0\end{array}$
Thus $f|\left(x|\right)$ and $|f\left(x\right)|$ are identical