Question

# Which of the following functions is an even function ?

A
f(x)=sinx+cosx
B
f(x)=log(1x1+x)
C
f(x)=xex1+x2
D
None of these

Solution

## The correct option is D None of theseThe function $$f(x)$$ is even oif $$f(-x)=f(x)$$ for every $$x$$And $$f$$ is odd if $$f(-x)=-f(x)$$ for every $$x$$$$(1)\ f(x)= \sin x+\cos x$$$$f(-x)=-\sin x + \cos x;$$ neither even nor odd.$$(2)\ f(x)=\log \left (\dfrac{1-x}{1+x} \right )$$$$f(-x)=\log \left (\dfrac{1+x}{1-x} \right )$$ $$= \log(1+x)-\log(1-x)$$ $$=-(\log(1-x)-\log(1+x))$$ $$=-\log\dfrac{1-x}{1+x} = -f(x)$$$$\therefore f$$ is odd$$(3)\ f(x)=\dfrac{x}{e^{x}-1}+\frac{x}{2}$$$$f(-x)=\dfrac{-x}{e^{-x}-1}-\dfrac{x}{2}$$$$=\dfrac{-xe^{x}}{1-e^{x}}-\dfrac{x}{2}$$So, it is neither even nor odd.Hence, none of these is an even function.Maths

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