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If $f$ is an even function defined on the interval $(-5, 5)$, then four real values of $x$ satisfying the equation $f(x) = f\left( {\frac{{x + 1}}{{x + 2}}} \right)$ are
$\frac{{ - 3 - \sqrt 5 }}{2},\frac{{ - 3 + \sqrt 5 }}{2},\frac{{3 - \sqrt 5 }}{2},\frac{{3 + \sqrt 5 }}{2}$
$\frac{{ - 5 + \sqrt 3 }}{2},\frac{{ - 3 + \sqrt 5 }}{2},\frac{{3 + \sqrt 5 }}{2},\frac{{3 - \sqrt 5 }}{2}$
$\frac{{3 - \sqrt 5 }}{2},\frac{{3 + \sqrt 5 }}{2},\frac{{ - 3 - \sqrt 5 }}{2},\frac{{5 + \sqrt 3 }}{2}$
$ - 3 - \sqrt 5 , - 3 + \sqrt 5 ,3 - \sqrt 5 ,3 + \sqrt 5$
Solution
Since $f$ is even function
$f(-x)=f(x), \forall x \in(-5,5).$
We are given that $f(x)=f\left(\frac{x+1}{x+2}\right)$
$\Rightarrow f(-x)=f\left(\frac{-x+1}{-x+2}\right) \Rightarrow f(x)=f\left(\frac{-x+1}{-x+2}\right)$
$[\because \quad f(-x)=f(x)]$
To find the values of $x,$ we set
$x=\frac{-x+1}{-x+2} \Rightarrow x=\frac{3 \pm \sqrt{9-4}}{2}=\frac{3 \pm \sqrt{5}}{2}$
Also $f(x)=f\left(\frac{x+1}{x+2}\right)=f(-x)$
To find the values of $x,$ we set
$-x=\frac{x+1}{x+2} \Rightarrow x=\frac{-3 \pm \sqrt{9-4}}{2}=\frac{-3 \pm \sqrt{5}}{2}$
Thus the four required values of $x$ are
$\frac{-3-\sqrt{5}}{2}, \frac{-3+\sqrt{5}}{2}, \frac{3-\sqrt{5}}{2}, \frac{3+\sqrt{5}}{2}.$
