A) \[\frac{2-\sqrt{3}}{2+\sqrt{3}}\]
B) \[\frac{2+\sqrt{3}}{2-\sqrt{3}}\]
C) \[\frac{\sqrt{3}-2}{\sqrt{3}+2}\]
D) \[\frac{\sqrt{3}+2}{\sqrt{3}-2}\]
Correct Answer: B
Solution :
Given A.M.\[=\]2(G.M.) or \[\frac{1}{2}(a+b)=2\sqrt{ab}\] or \[\frac{a+b}{2\sqrt{ab}}=\frac{2}{1}\]\[\Rightarrow \]\[\frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}}=\frac{2+1}{2-1}=\frac{3}{1}\] \[\Rightarrow \] \[\frac{{{(\sqrt{a}+\sqrt{b})}^{2}}}{{{(\sqrt{a}-\sqrt{b})}^{2}}}=\frac{3}{1}\]\[\Rightarrow \]\[\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{3}}{1}\] \[\Rightarrow \]\[\frac{a}{b}={{\left( \frac{\sqrt{3}+1}{\sqrt{3}-1} \right)}^{2}}\] \[\Rightarrow \] \[\frac{a}{b}=\frac{2+\sqrt{3}}{2-\sqrt{3}}\]or \[a:b=(2+\sqrt{3}):(2-\sqrt{3})\].
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