Question

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

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

Answer 1

Answer: (C)

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

first of all find out the value of$f(-x)$

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

since, $f(-x)=f\left(x\right)$ which imply that the given function is even.