Question

Solve this:

$44.\mathrm{The}\mathrm{function}\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{{\mathrm{e}}^{\mathrm{x}}-1}+\frac{\mathrm{x}}{2}+1,\mathrm{is}$
(1) An even function (2) An odd function
(3) A periodic function (4) Neither an even nor odd function

Answer 1

$$\begin{gathered} f\left(x\right)=\frac{x}{e^x-1}+\frac{x}{2}+1 \\ f\left(x\right)=x\left(\frac{1}{e^x-1}+\frac{1}{2}\right)+1 \\ f\left(x\right)=x\left(\frac{2+e^x-1}{2\left(e^x-1\right)}\right)+1 \\ f\left(x\right)=x\left(\frac{e^x+1}{2\left(e^x-1\right)}\right)+1 \\ f\left(-x\right)=-x\left(\frac{e^{-x}+1}{2\left(e^{-x}-1\right)}\right)+1 \\ =-x\left(\frac{{\displaystyle \frac{1}{e^x}}+1}{2\left(\frac{1}{e^x}-1\right)}\right)+1 \\ =-x\left(\frac{{\displaystyle \frac{1+e^x}{e^x}}}{2\left(\frac{1-e^x}{e^x}\right)}\right)+1 \\ =-x\left(\frac{{\displaystyle 1+e^x}}{2\left(1-e^x\right)}\right)+1 \\ =x\left(\frac{{\displaystyle 1+e^x}}{2\left(e^x-1\right)}\right)+1 \\ =f\left(x\right) \\ Henceitisanevenfunction \end{gathered}$$