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If $[x]$ represent greatest integer $\leq x$ and $[\alpha, \beta]$ is the set of all real values of $x$ for which the real function $f(x)=\frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{[x]+2}}$ is defined, then $f^2(\alpha+1)+5 f^2(\beta)=$
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Verified Answer
The correct answer is:
$12$
The given function $f(x)=\frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{[x]+2}}$,
then function $f$ is defined if
$3+x \geq 0,3-x \geq 0$ and $[x]+2>0$
$\begin{aligned} & \Rightarrow \quad x \geq-3, x \leq 3 \text { and }[x]>-2 \\ & \Rightarrow \quad x \geq-3, x \leq 3 \text { and } x \in[-1, \infty) \\ & \Rightarrow \quad x \in[-1,3]=[\alpha, \beta] \text { (given) } \\ & \text { So, } \alpha=-1 \text { and } \beta=3\end{aligned}$
$\therefore f^2(\alpha+1)+5 f^2(\beta)=f^2(0)+5 f^2(3)$
$\begin{aligned} & =\left(\frac{\sqrt{3}+\sqrt{3}}{\sqrt{2}}\right)^2+5\left(\frac{\sqrt{6}+\sqrt{0}}{\sqrt{5}}\right)^2=\left(\frac{4 \times 3}{2}\right)+5\left(\frac{6}{5}\right) \\ & =6+6=12\end{aligned}$
Hence, option (c) is correct.
then function $f$ is defined if
$3+x \geq 0,3-x \geq 0$ and $[x]+2>0$
$\begin{aligned} & \Rightarrow \quad x \geq-3, x \leq 3 \text { and }[x]>-2 \\ & \Rightarrow \quad x \geq-3, x \leq 3 \text { and } x \in[-1, \infty) \\ & \Rightarrow \quad x \in[-1,3]=[\alpha, \beta] \text { (given) } \\ & \text { So, } \alpha=-1 \text { and } \beta=3\end{aligned}$
$\therefore f^2(\alpha+1)+5 f^2(\beta)=f^2(0)+5 f^2(3)$
$\begin{aligned} & =\left(\frac{\sqrt{3}+\sqrt{3}}{\sqrt{2}}\right)^2+5\left(\frac{\sqrt{6}+\sqrt{0}}{\sqrt{5}}\right)^2=\left(\frac{4 \times 3}{2}\right)+5\left(\frac{6}{5}\right) \\ & =6+6=12\end{aligned}$
Hence, option (c) is correct.
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