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AP EAMCET Engineering 20 Apr 2019 Shift 1 Paper

Section: Mathematics

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Question : 1 of 160

Marks: +1, -0

f (x) =

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

\cos^3 \frac{x}{2}

on R - {0}

One one function
Bisection
Algebraic function
Even function

Solution:

Consider the function,

f(x)=\;\;\frac{x}{e^{x}-1}+\;\;\frac{x}{2}+2 \cos^{3}\;\;\frac{x}{2}

Then

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

=\;\;\frac{-x e^{x}}{1-e^{x}}-\;\;\frac{x}{2}+2 \cos^{3}\;\;\frac{x}{2}

=\;\;\frac{x e^{x}}{e^{x}-1}-\;\;\frac{x}{2}+2 \cos^{3}\;\;\frac{x}{2}

=\;\;\frac{x(e^{x}-1+1)}{e^{x}-1}-\;\;\frac{x}{2}+2 \cos^{3}\;\;\frac{x}{2}

Further simplify the above,

f(-x)=x+\;\;\frac{x}{e^{x}-1}-\;\;\frac{x}{2}+2 \cos^{3}\;\;\frac{x}{2}

=\;\;\frac{x}{e^{x}-1}+\;\;\frac{x}{2}+2 \cos^{3}\;\;\frac{x}{2}

=f(x), \forall x \in R-\{0\}

This implies

f(x)

is an even function.

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