$\begin{array}{l} I f f i s a n e v e n f u n c t i o n d e f i n e d o n t h e int e r v a l [- 5, 5], t h e n t h e r e a l v a l u e s o f x \\ s a t i s f y i n g t h e e q u a t i o n f (x) = f (\frac{x + 1}{x + 2}) a r e \end{array}$

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

$\frac{- 2 \pm \sqrt{5}}{2}, \frac{- 4 \pm \sqrt{5}}{2}$

$\frac{- 4 \pm \sqrt{5}}{2}, \frac{- 1 \pm \sqrt{5}}{2}$

$\frac{- 3 \pm \sqrt{5}}{2}, \frac{- 4 \pm \sqrt{5}}{2}$

answer is A.

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Detailed Solution

$\begin{array}{l} G i v e n f (x) = f (\frac{x + 1}{x + 2}) \\ \Rightarrow x = \frac{x + 1}{x + 2} \\ \Rightarrow x^{2} + 2 x = x + 1 \\ \Rightarrow x^{2} + x - 1 = 0 \\ x = \frac{- 1 \pm \sqrt{5}}{2} \\ a n d f (x) i s e v e n \\ ∴ f (- x) = f (x) \\ \Rightarrow f (\frac{1 - x}{2 - x}) = f (x) \\ \Rightarrow \frac{1 - x}{2 - x} = x \\ \Rightarrow 1 - x = 2 x - x^{2} \\ \Rightarrow x^{2} - 3 x + 1 = 0 \\ \Rightarrow x = \frac{3 \pm \sqrt{5}}{2} \end{array}$