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

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Aicharya · Accepted Answer

Given, $\begin{aligned} f(x)= & \sqrt{x+2 \sqrt{2 x-4}}+\sqrt{x-2 \sqrt{2 x-4}} \ = & \sqrt{(x-2)+2+2 \sqrt{2} \sqrt{x-2}} \ & \quad+\sqrt{(x-2)+2-2 \sqrt{2} \sqrt{x-2}} \ & =|\sqrt{x-2}+\sqrt{2}|+|\sqrt{x-2}-\sqrt{2}| \end{aligned}$ for $x \geq 4$ $\begin{aligned} f(x) & =\sqrt{x-2}+\sqrt{2}+\sqrt{x-2}-\sqrt{2}=2 \sqrt{x-2} \ \therefore \quad & f^{\prime}(x)=\frac{1}{\sqrt{x-2}} \end{aligned}$ So, $10 \times f^{\prime}(102)=10 \times \frac{1}{\sqrt{102-2}}=1$

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

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