Question

If f is an even function defined on the interval (-5,5), then the real values of x, satisfying the equation $f\left(x\right)=f\left(\frac{x+1}{x+2}\right)$ are ___.

• A. $\frac{3\pm \sqrt{5}}{2}$
• B. $\frac{5\pm \sqrt{3}}{2}$
• C. $\frac{-3\pm \sqrt{5}}{2}$
• D. $\frac{-5\pm \sqrt{3}}{2}$

Answer 1

Answer: (A), (C)

The correct options are
A

$\frac{3\pm \sqrt{5}}{2}$

C

$\frac{-3\pm \sqrt{5}}{2}$

Given that $f\left(x\right)=f\left(\frac{x+1}{x+2}\right)$ and f is an even function
$\therefore f\left(x\right)=f(-x)=f\left(\frac{-x+1}{-x+2}\right)\Rightarrow x=\frac{-x+1}{-x+2}\Rightarrow x^2-3x+1=0\Rightarrow x=\frac{3\pm \sqrt{5}}{2}Alsof\left(x\right)=f\left(\frac{x+1}{x+2}\right)=f(-x)\Rightarrow \frac{x+1}{x+2}=-x\Rightarrow x^2+3x+1=0\Rightarrow x=\frac{-3\pm \sqrt{5}}{2}\therefore$ Four values of x are
$\frac{3+\sqrt{5}}{2},\frac{3-\sqrt{5}}{2},\frac{-3+\sqrt{5}}{2},\frac{-3-\sqrt{5}}{2}$