Question

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

• A. An even function
• B. An odd function
• C. A periodic function
• D. Neither an even nor odd function

Answer 1

The correct option is A An even function

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