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1. If $\sqrt{2}$ : $(1 + \sqrt{3})$ :: $\sqrt{6}$ : $x$ , then $x$ is equal to -

A. $\sqrt{3} + 3$

B. $1 - \sqrt{3}$

C. $1 + \sqrt{3}$

D. $\sqrt{3} - 3$

Solution:

\begin{aligned} = \sqrt{2} : (1 + \sqrt{3}) :: \sqrt{6} : x \\ \Rightarrow \sqrt{2} x = \sqrt{6} (1 + \sqrt{3}) \\ \Rightarrow x = \frac{\sqrt{6} (1 + \sqrt{3})}{\sqrt{2}} \\ = \sqrt{3} (1 + \sqrt{3}) \\ = \sqrt{3} + 3 \end{aligned}

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