If $$n = 7 + 4\sqrt 3 {\text{,}}$$   then the value of $$\left( {\sqrt n + \frac{1}{{\sqrt n }}} \right)$$   is:

A. $${\text{2}}\sqrt 3 $$

B. 4

C. -4

D. $$ - 2\sqrt 3 $$

Answer: Option B

Solution(By Examveda Team)

$$\eqalign{ & n = 7 + 4\sqrt 3 \cr & \Rightarrow n = 4 + 3 + 4\sqrt 3 \cr & \Rightarrow n = {\left( 2 \right)^2} + {\left( {\sqrt 3 } \right)^2} + 2 \times 2 \times \sqrt 3 \cr & \Rightarrow n = {\left( {2 + \sqrt 3 } \right)^2} \cr & \Rightarrow \sqrt n = 2 + \sqrt 3 \cr & \Rightarrow \frac{1}{{\sqrt n }} = 2 - \sqrt 3 \cr & \therefore \sqrt n + \frac{1}{{\sqrt n }} \cr & = 2 + \sqrt 3 + 2 - \sqrt 3 \cr & = 4 \cr} $$