If [x] represents the greatest integer leq x | vedclass

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(C) The given function is $f(x)=\frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{[x]+2}}$.For $f(x)$ to be defined,the following conditions must be satisfied:$1. \

Vedclass · Accepted Answer

$12$

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(C) The given function is $f(x)=\frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{[x]+2}}$.For $f(x)$ to be defined,the following conditions must be satisfied:$1. \ 3+x \geq 0 \Rightarrow x \geq -3$$2. \ 3-x \geq 0 \Rightarrow x \leq 3$$3. \ [x]+2 > 0 \Rightarrow [x] > -2$Since $[x] > -2$,the smallest integer value $[x]$ can take is $-1$. Thus,$x \geq -1$.Combining these conditions,we get $x \in [-1, 3]$.Given that the interval is $[\alpha, \beta]$,we have $\alpha = -1$ and $\beta = 3$.Now,we calculate $f^2(\alpha+1) + 5f^2(\beta) = f^2(0) + 5f^2(3)$.$f(0) = \frac{\sqrt{3+0} + \sqrt{3-0}}{\sqrt{[0]+2}} = \frac{2\sqrt{3}}{\sqrt{2}} = \sqrt{6}$. So,$f^2(0) = 6$.$f(3) = \frac{\sqrt{3+3} + \sqrt{3-3}}{\sqrt{[3]+2}} = \frac{\sqrt{6} + 0}{\sqrt{3+2}} = \frac{\sqrt{6}}{\sqrt{5}}$. So,$f^2(3) = \frac{6}{5}$.Therefore,$f^2(0) + 5f^2(3) = 6 + 5 	imes \frac{6}{5} = 6 + 6 = 12$.Hence,option $C$ is correct.

If $[x]$ represents the 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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