Question

Let $f\left(x\right)=\frac{x}{e^x-1}+\frac{x}{2}+1$ and $g$ is either odd or even function. If $fog$ is defined, then

• A. $fog$ is odd function
• B. $fog$ is even function
• C. $f\left(x\right)$ is even function
• D. $f\left(x\right)$ is odd function

Answer 1

Answer: (B), (C)

$f\left(x\right)$ is even function

$f\left(x\right)=\frac{x}{e^x-1}+\frac{x}{2}+1$
$\Rightarrow f(-x)=\frac{-x}{e^{-x}-1}-\frac{x}{2}+1$
$\Rightarrow f(-x)=\frac{-x{e}^x}{1-e^x}-\frac{x}{2}+1$
$\Rightarrow f(-x)=\frac{-2x{e}^x-x+x{e}^x}{2(1-e^x)}+1$
$\Rightarrow f(-x)=\frac{x(1-e^x)-2x}{2(1-e^x)}+1$
$\Rightarrow f(-x)=\frac{x}{e^x-1}+\frac{x}{2}+1$
$\Rightarrow f(-x)=f\left(x\right)$
$\therefore f\left(x\right)$ is even function.

As we know composite of two functions (which are either odd or even) will be even if atleast one of them is even, so $fog$ is even function.