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Question

Which of the following functions is an even function ?


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

The correct option is D None of these
The 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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