A) \[\frac{1}{5}\]
B) \[\frac{1}{20}\]
C) \[\frac{1}{10}\]
D) \[\frac{3}{20}\]
Correct Answer: C
Solution :
It is given that for two positive numbers a and b \[\frac{a+b}{2}=2\sqrt{ab}\] \[\Rightarrow \] \[a+b=4\sqrt{ab}\] \[\Rightarrow \] \[{{(a+b)}^{2}}=16ab\] \[\Rightarrow \] \[\frac{{{a}^{2}}+{{b}^{2}}+2ab}{2ab}=\frac{8}{1}\] (By componendo only) \[\left[ If\frac{a}{b}=\frac{c}{d},then\frac{a-b}{b}=\frac{c-d}{d} \right]\] \[\Rightarrow \] \[\frac{{{a}^{2}}+{{b}^{2}}}{2ab}=\frac{7}{1}\] Further applying componendo and dividend rule, we get \[\frac{{{a}^{2}}+{{b}^{2}}+2ab}{{{a}^{2}}+{{b}^{2}}-2ab}=\frac{7+1}{7-1}\] \[\Rightarrow \] \[\frac{{{(a+b)}^{2}}}{{{(a-b)}^{2}}}=\frac{8}{6}=\frac{4}{3}\] \[\Rightarrow \] \[\frac{a+b}{a-b}=\frac{2}{\sqrt{3}}\] Again applying componendo and dividendo, we get \[\frac{2a}{2b}=\frac{2+\sqrt{3}}{2-\sqrt{3}}\] \[\Rightarrow \] \[\frac{a}{b}=\frac{2+\sqrt{3}}{2-\sqrt{3}}\]
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