A) $ 2\sqrt{3} $
B) $ 4 $
C) $ -4 $
D) $ -2\sqrt{3} $
Correct Answer: B
$ \therefore $ $ \frac{1}{\sqrt{n}}=\frac{1}{2+\sqrt{3}} $ [reciprocal] $ =\frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}} $ [rationatising] $ =2-\sqrt{3} $ … (ii) On adding Eqs. (i) and (ii), we get $ \sqrt{n}+\frac{1}{\sqrt{n}}=2+\sqrt{3}+2-\sqrt{3}=4 $ Alternate Method Given, $ n=7+4\sqrt{3} $
$ \therefore $ $ \frac{1}{n}=\frac{1}{7+4\sqrt{3}}=\frac{1}{7+4\sqrt{3}}\times \frac{7-4\sqrt{3}}{7-4\sqrt{3}} $ [rationalising] $ =\frac{7-4\sqrt{3}}{{{(7)}^{2}}-{{(4\sqrt{3})}^{2}}} $ $ [\because (a+b)(a-b)=a^{2}-b^{2}] $ $ =\frac{7-4\sqrt{3}}{49-48}=7-4\sqrt{3} $ We know that, $ {{( \sqrt{n}+\frac{1}{\sqrt{n}} )}^{2}}=( n+\frac{1}{n}+2\times \sqrt{n}\times \frac{1}{\sqrt{n}} ) $
$ \Rightarrow $ $ ( \sqrt{n}+\frac{1}{\sqrt{n}} )=\sqrt{n+\frac{1}{n}+2} $ $ =\sqrt{7+4\sqrt{3}+7-4\sqrt{3}+2} $ $ =\sqrt{14+2}=\sqrt{16}=4 $
BETA
