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If $f(x)=\frac{1}{\sqrt{x+2 \sqrt{2 x-4}}}+\frac{1}{\sqrt{x-2 \sqrt{2 x-4}}}$ for $x>2$, then $f(11)$ is equal to
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Verified Answer
The correct answer is:
$\frac{6}{7}$
We have,
$f(x)=\frac{1}{\sqrt{x+2 \sqrt{2 x-4}}}+\frac{1}{\sqrt{x-2 \sqrt{2 x-4}}}$
$f(11)=\frac{1}{\sqrt{11+2 \sqrt{18}}}+\frac{1}{\sqrt{11-2 \sqrt{18}}}$
$=\frac{1}{\sqrt{11+6 \sqrt{2}}}+\frac{1}{\sqrt{11-6 \sqrt{2}}}$
$=\frac{1}{\sqrt{9+2+6 \sqrt{2}}}+\frac{1}{\sqrt{9+2-6 \sqrt{2}}}$
$=\frac{1}{\sqrt{(3+\sqrt{2})^2}}+\frac{1}{\sqrt{(3-\sqrt{2})^2}}$
$=\frac{1}{3+\sqrt{2}}+\frac{1}{3-\sqrt{2}}$
$=\frac{3-\sqrt{2}+3+\sqrt{2}}{9-2}=\frac{6}{7}$
$f(x)=\frac{1}{\sqrt{x+2 \sqrt{2 x-4}}}+\frac{1}{\sqrt{x-2 \sqrt{2 x-4}}}$
$f(11)=\frac{1}{\sqrt{11+2 \sqrt{18}}}+\frac{1}{\sqrt{11-2 \sqrt{18}}}$
$=\frac{1}{\sqrt{11+6 \sqrt{2}}}+\frac{1}{\sqrt{11-6 \sqrt{2}}}$
$=\frac{1}{\sqrt{9+2+6 \sqrt{2}}}+\frac{1}{\sqrt{9+2-6 \sqrt{2}}}$
$=\frac{1}{\sqrt{(3+\sqrt{2})^2}}+\frac{1}{\sqrt{(3-\sqrt{2})^2}}$
$=\frac{1}{3+\sqrt{2}}+\frac{1}{3-\sqrt{2}}$
$=\frac{3-\sqrt{2}+3+\sqrt{2}}{9-2}=\frac{6}{7}$
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