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EAMCET Engineering 2003 Solved Paper

Section: Mathematics

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Question : 135 of 160

Marks: +1, -0

If

f(x)=\;\frac{1}{\sqrt{x+2\sqrt{2x-4}}}+\;\frac{1}{\sqrt{x-2\sqrt{2x-4}}}

x > 2

f(11)

$\;\frac{7}{6}$
$\;\frac{5}{6}$
$\;\frac{6}{7}$
$\;\frac{5}{7}$

Solution:

We have,

f(x)=\;\frac{1}{\sqrt{x+2\sqrt{2x-4}}}+\;\frac{1}{\sqrt{x-2\sqrt{2x-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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