Question - 15

As given :

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

Let's check what $f(-x)$ will be , so replacing $x$ by $-x$ .

$\implies f(-x)=\dfrac{-x}{e^{-x}-1}+\dfrac{-x}{2}+1$

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

$\quad \quad =\dfrac{-x \cdot e^x}{1-e^x}-\dfrac{x}{2}+1$

On adding and subtracting x in the first term Numerator

$\quad \quad =\dfrac{x-x \cdot e^x-x}{1-e^x}-\dfrac{x}{2}+1$

$\quad \quad =\dfrac{x \cdot (1-e^x)}{1-e^x}-\dfrac{x}{1-e^x}-\dfrac{x}{2}+1$

$\quad \quad=x + \dfrac{x}{e^x-1}-\dfrac{x}{2}+1$

$\quad \quad=\dfrac{x}{e^x-1}+\dfrac{x}{2}+1=f(x)$ .

Now, what we found after these calculations that $\boxed{f(-x)=f(x)}$ .

Hence $f(x)$ is an even function.

enjoy!

$For\quad the\quad given\quad function\quad if\quad we\quad take\quad 'x'\quad common\\ from\quad first\quad two\quad terms\quad and\quad take\quad LCM\quad then\\ it\quad will\quad be\quad in\quad the\quad exponential\quad form\quad of\quad sin\quad and\quad cos,\\ Then\quad it\quad becomes\quad easy\quad to\quad comment\quad on\quad \\ odd,even\quad and\quad periodicity.\\ f\left( x \right) =\quad \frac { x }{ { e }^{ x }-1 } +\frac { x }{ 2 } +1\quad =\quad x\left( \frac { 1 }{ { e }^{ x }-1 } +\frac { 1 }{ 2 } \right) +1\\ =\quad \frac { x }{ 2 } \left( \frac { { e }^{ x }+1 }{ { e }^{ x }-1 } \right) +1\quad =\quad \frac { x }{ 2 } \left( \frac { { e }^{ \frac { x }{ 2 } }+{ e }^{ \frac { -x }{ 2 } } }{ { e }^{ \frac { x }{ 2 } }-{ e }^{ \frac { -x }{ 2 } } } \right) +1\\ =\quad -i\quad \frac { x }{ 2 } \quad \cot { \frac { x }{ 2 } } +\quad 1\\ \therefore \quad f(-x)\quad =\quad -i\quad \frac { x }{ 2 } \quad \cot { \frac { x }{ 2 } } +\quad 1=\quad f(x)$