Which of the following is an even function?

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(C) function $f(x)$ is even if $f(-x) = f(x)$.Let us check option $C$:$f(x) = \frac{x}{2^x - 1} + \frac{x}{2} + 1$$f(-x) = \frac{-x}{2^{-x} - 1} + \fr

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$f(x) = \frac{x}{2^x - 1} + \frac{x}{2} + 1$

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(C) function $f(x)$ is even if $f(-x) = f(x)$.Let us check option $C$:$f(x) = \frac{x}{2^x - 1} + \frac{x}{2} + 1$$f(-x) = \frac{-x}{2^{-x} - 1} + \frac{-x}{2} + 1$$f(-x) = \frac{-x}{\frac{1}{2^x} - 1} - \frac{x}{2} + 1$$f(-x) = \frac{-x \cdot 2^x}{1 - 2^x} - \frac{x}{2} + 1$$f(-x) = \frac{x \cdot 2^x}{2^x - 1} - \frac{x}{2} + 1$Since $\frac{x \cdot 2^x}{2^x - 1} = \frac{x(2^x - 1 + 1)}{2^x - 1} = x + \frac{x}{2^x - 1}$,$f(-x) = x + \frac{x}{2^x - 1} - \frac{x}{2} + 1 = \frac{x}{2^x - 1} + \frac{x}{2} + 1 = f(x)$.Thus,$f(x)$ is an even function.

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