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The no. of Real Roots of the equation $\displaystyle \frac{\pi^e}{x-e}+\frac{e^\pi}{x-\pi}+\frac{\pi^{\pi}+e^{e}}{x-\pi-e} = 0$

My try:: Let $\displaystyle f(x) = \frac{\pi^e}{x-e}+\frac{e^\pi}{x-\pi}+\frac{\pi^{\pi}+e^{e}}{x-\pi-e}$

Now we will take a interval $x\in \left(e\;,\pi\right)$ and $x\in \left(\pi\;,\pi+e\right)$

$\bullet$ If $x\in (e,\pi),$ Then $(x-e) > 0$ and $(x-\pi) < 0$

So $f(x) = $

Similarly

$\bullet$ If $x\in (\pi,\pi+e),$ Then $(x-\pi) > 0$ and $(x-\pi-e) < 0$

So $f(x) = $

y Question is How Can I check Sign of $f(x)$ in Given Interval and Is Iam Thinking Right.

If Not plz explain me,

Thanks

My try:: Let $\displaystyle f(x) = \frac{\pi^e}{x-e}+\frac{e^\pi}{x-\pi}+\frac{\pi^{\pi}+e^{e}}{x-\pi-e}$

Now we will take a interval $x\in \left(e\;,\pi\right)$ and $x\in \left(\pi\;,\pi+e\right)$

$\bullet$ If $x\in (e,\pi),$ Then $(x-e) > 0$ and $(x-\pi) < 0$

So $f(x) = $

Similarly

$\bullet$ If $x\in (\pi,\pi+e),$ Then $(x-\pi) > 0$ and $(x-\pi-e) < 0$

So $f(x) = $

y Question is How Can I check Sign of $f(x)$ in Given Interval and Is Iam Thinking Right.

If Not plz explain me,

Thanks

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