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AP EAPCET 19-Aug-2021 Shift 2 Paper

Section: Mathematics

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Question : 1 of 160

Marks: +1, -0

The real valued function

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

R \setminus \{0\}

is _______

An odd function
An even function
Both even & odd function
Neither even nor odd function

Solution:

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

Now,

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

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

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

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

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

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

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

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

∴

f(x)

is an even function.

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