Question

Let $f\left(x\right)=e^{\left\{e^x\text{sgn}x\right\}}$ and $g\left(x\right)=e^{\left[e^x\text{sgn}x\right]},x\in R$ where $\{.\}$ and $[.]$ denote greatest integer function and fractional part function, respectively. Also $h\left(x\right)=\log \left(f\right(x\left)\right)+\log \left(g\right(x\left)\right),$ then for real $x,h\left(x\right)$ is

• A. an odd function.
• B. an even function.
• C. neither odd nor an even function.
• D. both odd as well as even function.

Answer 1

The correct option is C neither odd nor an even function.

$h\left(x\right)=\log \left(f\right(x\left)\right)+\log \left(g\right(x\left)\right)$
$\Rightarrow h\left(x\right)=\left\{e^x\text{sgn}x\right\}+\left[e^x\text{sgn}x\right]$
$\Rightarrow h\left(x\right)=e^x\text{sgn}x=\{\begin{array}{c}{e}^x,\text{if}x>0\\ 0,\text{if}x=0\\ -e^x,\text{if}x<0\end{array}$
$\Rightarrow h(-x)=e^{-x}\text{sgn}-x=\{\begin{array}{c}-e^{-x},\text{if}x>0\\ 0,\text{if}x=0\\ e^{-x},\text{if}x<0\end{array}$
Clearly, $h\left(x\right)+h(-x)\ne 0\forall x$ and neither $h(-x)=h\left(x\right)$
$\therefore h\left(x\right)$ is neither even nor odd function.