Question : The value of $\frac{1}{\sqrt{17+12 \sqrt{2}}}$ is closest to _____.
Option 1: 1.4
Option 2: 1.2
Option 3: 0.14
Option 4: 0.17
Correct Answer: 0.17
Solution :
Given expression,
$\frac{1}{\sqrt{17+12 \sqrt{2}}}$
$=\frac{1}{\sqrt{{3^2}+(2\sqrt2)^2+2\times3\times2\sqrt2}}$
$=\frac{1}{\sqrt{(3+2\sqrt2)^2}}$
$=\frac{1}{3+2\sqrt2}$
$=\frac{1}{3+2\sqrt2}\times\frac{3-2\sqrt2}{3-2\sqrt2}$
$=\frac{3-2\sqrt2}{3^2-(2\sqrt2)^2}$ [As $a^2-b^2=(a-b)(a+b)$]
$=\frac{3-2\sqrt2}{9-8}$
$=3-2\sqrt2$
$=3-2\times1.414$
= 3 – 2.828
= 0.172
Hence, the correct answer is 0.17.
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