Question

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

• A. $\frac{-3\pm \sqrt{5}}{2},\frac{3\pm \sqrt{5}}{2}$
• B. $\frac{-2\pm \sqrt{3}}{2},\frac{3\pm \sqrt{5}}{2}$
• C. $\frac{2\pm \sqrt{5}}{2},\frac{-3\pm \sqrt{5}}{2}$
• D. None of these

Answer 1

The correct option is A $\frac{-3\pm \sqrt{5}}{2},\frac{3\pm \sqrt{5}}{2}$

Given $f\left(x\right)=f\left(\frac{x+1}{x+2}\right)$
$\therefore f(-x)=f\left(\frac{-x+1}{-x+2}\right)$
$\Rightarrow x=\frac{-x+1}{-x+2}$ and $-x=\frac{x+1}{x+2}$
$\Rightarrow x^2-3x+1=0$ and $x^2+3x+1=0$
$\Rightarrow {(x-\frac{3}{2})}^2=-1+\frac{9}{4}$ and ${(x+\frac{3}{2})}^2=-1+\frac{9}{4}$
$\Rightarrow {(x-\frac{3}{2})}^2=\left(\frac{\sqrt{5}}{2}\right)$ and ${(x+\frac{3}{2})}^2={\left(\frac{\sqrt{5}}{2}\right)}^2$
$\Rightarrow x=\frac{3\pm \sqrt{5}}{2}$ and $x=\frac{-3\pm \sqrt{5}}{2}$