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