A) \[(2+\sqrt{3}):(2-\sqrt{3})\]
B) \[(2-\sqrt{3}):(2+\sqrt{3})\]
C) \[(2+\sqrt{3}):(2-\sqrt{3})\]
D) \[(2-\sqrt{3}):(2+\sqrt{3})\]
Correct Answer: A
Solution :
\[a+b:\sqrt{ab}=4:1\] \[\frac{a+b}{2\sqrt{ab}}=\frac{4}{2}=\frac{2}{1}\] On applying C and D Rule, we have \[\frac{{{(\sqrt{a}+\sqrt{b})}^{2}}}{{{(\sqrt{a}-\sqrt{b})}^{2}}}=\frac{3}{1}or\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{3}}{1}\] Again applying C and D Rule, we have \[\frac{2\sqrt{a}}{2\sqrt{b}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}\] Squaring both sides. \[\therefore \,\,\,\,\frac{a}{b}=\frac{{{(\sqrt{3}+1)}^{2}}}{{{(\sqrt{3}-1)}^{2}}}=\frac{3+1+2\sqrt{3}}{3+1-2\sqrt{3}}\] \[=\frac{4+2\sqrt{3}}{4-2\sqrt{3}}\] \[\therefore \,\,\,\]
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