Question

Equation $\frac{{\pi }^e}{x-e}+\frac{e^{\pi }}{x-\pi }+\frac{{\pi }^{\pi }{e}^e}{x-\pi -e}=0$ has

• A. One real roots in $(e,\pi )$ and other in $(\pi -e,e)$
• B. One real roots in $(e,\pi )$ and other in $(\pi ,\pi +e)$
• C. Two real roots in $(\pi -e,\pi +e)$
• D. No real roots

Answer 1

The correct option is B One real roots in $(e,\pi )$ and other in $(\pi ,\pi +e)$

$f\left(x\right)=\frac{{\pi }^e}{x-e}+\frac{e^{\pi }}{x-\pi }+\frac{{\pi }^{\pi }+e^e}{x-\pi -e}=0$
$=\frac{{\pi }^e(x-\pi )(x-\pi -e)+e^{\pi }(x-e)(x-\pi -e)+({\pi }^{\pi }+e^e)(x-e)(x-\pi )}{(x-e)(x-\pi )(x-\pi -e)}=0$
$={\pi }^e(x-\pi )(x-\pi -e)+e^{\pi }(x-e)(x-\pi -e)+({\pi }^{\pi }+e^e)(x-e)(x-\pi )=0$
$f\left(e\right)={\pi }^e(e-\pi )(-\pi )={\pi }^{e+1}(\pi -e)$
$f\left(\pi \right)=e^{\pi }(\pi -e)-e=-e^{\pi +1}(\pi -e)$
$f(\pi +e)=({\pi }^{\pi }+e^e)\left(\pi \right)\left(e\right)$
$clearly$
$e<\pi <\pi +e$
$Hence$
$f\left(e\right)>0,f\left(\pi \right)<0,f(\pi +e)>0$
$\therefore f\left(x\right)hasonerealroot(e,\pi )andother(\pi ,\pi +e)$