Which of the following functions is an even f | vedclass

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(B) function $f(x)$ is even if $f(-x) = f(x)$ and odd if $f(-x) = -f(x)$.For option $(A)$: $f(-x) = \frac{a^{-x} + 1}{a^{-x} - 1} = \frac{\frac{1}{a^x

Vedclass · Accepted Answer

$f(x) = x \left( \frac{a^x - 1}{a^x + 1} \right)$

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(B) function $f(x)$ is even if $f(-x) = f(x)$ and odd if $f(-x) = -f(x)$.For option $(A)$: $f(-x) = \frac{a^{-x} + 1}{a^{-x} - 1} = \frac{\frac{1}{a^x} + 1}{\frac{1}{a^x} - 1} = \frac{1 + a^x}{1 - a^x} = -\frac{a^x + 1}{a^x - 1} = -f(x)$. Thus,it is an odd function.For option $(B)$: $f(-x) = (-x) \left( \frac{a^{-x} - 1}{a^{-x} + 1} \right) = (-x) \left( \frac{\frac{1}{a^x} - 1}{\frac{1}{a^x} + 1} \right) = (-x) \left( \frac{1 - a^x}{1 + a^x} \right) = x \left( \frac{a^x - 1}{a^x + 1} \right) = f(x)$. Thus,it is an even function.For option $(C)$: $f(-x) = \frac{a^{-x} - a^x}{a^{-x} + a^x} = -\frac{a^x - a^{-x}}{a^x + a^{-x}} = -f(x)$. Thus,it is an odd function.For option $(D)$: $f(-x) = \sin(-x) = -\sin x = -f(x)$. Thus,it is an odd function.Therefore,the correct option is $(B)$.

Which of the following functions is an even function?

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