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. $\frac{-1\pm \sqrt{5}}{2}$
• B. $\frac{-3\pm \sqrt{5}}{2}$
• C. $\frac{-2\pm \sqrt{5}}{2}$
• D. non of these

Answer 1

Answer: (A), (B)

$\frac{-1\pm \sqrt{5}}{2}$
$\frac{-3\pm \sqrt{5}}{2}$

Since $f\left(x\right)=f\left(\frac{x+1}{x+2}\right)\Rightarrow x=\frac{x+1}{x+2}$

$\Rightarrow x^2+x-1=0\Rightarrow x=\frac{-1\pm \sqrt{5}}{2}$ ...(1)

Since $f\left(x\right)$ ia an even function defined on $[-5,5]$

$\therefore f(-x)=f\left(x\right),\forall x\in [-5,5]$

$\Rightarrow x=-\left(\frac{x+1}{x+2}\right)\Rightarrow x^2+3x+1=0$

$\Rightarrow x=\frac{-3\pm \sqrt{5}}{2}$ ...(2)

From (1) and (2), the values of $x$ are

$\frac{-1\pm \sqrt{5}}{2}$ and $\frac{-3\pm \sqrt{5}}{2}$