If f(x) = sqrt{x + 2 sqrt{2x - 4}} + sqrt{x - | vedclass

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(A) Given,$f(x) = \sqrt{x + 2 \sqrt{2x - 4}} + \sqrt{x - 2 \sqrt{2x - 4}}$.We can rewrite the expression inside the square roots by setting $u = \sqrt

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(A) Given,$f(x) = \sqrt{x + 2 \sqrt{2x - 4}} + \sqrt{x - 2 \sqrt{2x - 4}}$.We can rewrite the expression inside the square roots by setting $u = \sqrt{x-2}$,so $x-2 = u^2$ and $x = u^2 + 2$.Then $2x-4 = 2(x-2) = 2u^2$,so $\sqrt{2x-4} = \sqrt{2}u$.Substituting these into $f(x)$:$f(x) = \sqrt{u^2 + 2 + 2\sqrt{2}u} + \sqrt{u^2 + 2 - 2\sqrt{2}u}$$f(x) = \sqrt{(u + \sqrt{2})^2} + \sqrt{(u - \sqrt{2})^2}$$f(x) = |u + \sqrt{2}| + |u - \sqrt{2}| = |\sqrt{x-2} + \sqrt{2}| + |\sqrt{x-2} - \sqrt{2}|$.For $x \geq 4$,$\sqrt{x-2} \geq \sqrt{2}$,so $|\sqrt{x-2} - \sqrt{2}| = \sqrt{x-2} - \sqrt{2}$.Thus,$f(x) = \sqrt{x-2} + \sqrt{2} + \sqrt{x-2} - \sqrt{2} = 2\sqrt{x-2}$.Differentiating with respect to $x$:$f^{\prime}(x) = 2 	imes \frac{1}{2\sqrt{x-2}} = \frac{1}{\sqrt{x-2}}$.Therefore,$10 	imes f^{\prime}(102) = 10 	imes \frac{1}{\sqrt{102-2}} = 10 	imes \frac{1}{\sqrt{100}} = 10 	imes \frac{1}{10} = 1$.

If $f(x) = \sqrt{x + 2 \sqrt{2x - 4}} + \sqrt{x - 2 \sqrt{2x - 4}}$,then the value of $10 \times f^{\prime}(102) =$

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