Question

Which of the following functions is an even function ?

• A. $f\left(x\right)=\sin x+\cos x$
• B. $f\left(x\right)=\log \left(\frac{1-x}{1+x}\right)$
• C. $f\left(x\right)=\frac{x}{e^x-1}+\frac{x}{2}$
• D. None of these

Answer 1

The correct option is D None of these

The function $f\left(x\right)$ is even oif $f(-x)=f\left(x\right)$ for every $x$

And $f$ is odd if $f(-x)=-f\left(x\right)$ for every $x$

$\left(1\right)f\left(x\right)=\sin x+\cos x$

$f(-x)=-\sin x+\cos x;$ neither even nor odd.

$\left(2\right)f\left(x\right)=\log \left(\frac{1-x}{1+x}\right)$

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

$=\log (1+x)-\log (1-x)$

$=-\left(\log \right(1-x)-\log (1+x\left)\right)$

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

$\therefore f$ is odd

$\left(3\right)f\left(x\right)=\frac{x}{e^x-1}+\frac{x}{2}$

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

$=\frac{-x{e}^x}{1-e^x}-\frac{x}{2}$

So, it is neither even nor odd.

Hence, none of these is an even function.