$\begin{array}{l} I f [x] r e p r e s e n t g r e a t e s t int e g e r \leq x a n d [α, β] i s t h e s e t o f a l l r e a l v a l u e s o f x f o r w h i c h t h e r e a l f u n c t i o n \\ f (x) = \frac{\sqrt{3 + x} + \sqrt{3 - x}}{\sqrt{[x] + 2}} i s d e f i n e d, t h e n f^{2} (α + 1) + 5 f^{2} (β) = \end{array}$

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Detailed Solution

$\begin{array}{l} f (x) = \frac{\sqrt{3 + x} + \sqrt{3 - x}}{\sqrt{[x] + 2}} i s d e f i n e d i f f \\ 3 + x \geq 0 a n d 3 - x \geq 0 a n d [x] + 2 > 0 \\ \Rightarrow x \geq - 3, \Rightarrow 3 \geq x \Rightarrow x \leq 3, \Rightarrow [x] > - 2 \Rightarrow x \geq - 1 \\ ∴ x \in [- 1, 3] = [α, β] \\ α = - 1, β = 3 \\ ∴ f^{2} (α + 1) + 5 f^{2} (β) = f^{2} (0) + 5 f^{2} (3) \\ = {[\frac{\sqrt{3} + \sqrt{3}}{\sqrt{2}}]}^{2} + 5 {[\frac{\sqrt{6} + 0}{\sqrt{5}}]}^{2} \\ = \frac{4 (3)}{2} + 5 (\frac{6}{5}) \\ = 6 + 6 \\ = 12 \end{array}$