A) \[m+\sqrt{{{m}^{2}}-{{n}^{2}}}:n+\sqrt{{{m}^{2}}-{{n}^{2}}}\]
B) \[m+\sqrt{{{m}^{2}}-{{n}^{2}}}:m-\sqrt{{{m}^{2}}-{{n}^{2}}}\]
C) \[m+\sqrt{{{m}^{2}}+{{n}^{2}}}:n+\sqrt{{{m}^{2}}+{{n}^{2}}}\]
D) \[m+\sqrt{{{m}^{2}}+{{n}^{2}}}:m-\sqrt{{{m}^{2}}-{{n}^{2}}}\]
Correct Answer: B
Solution :
If a and b are two numbers, then \[AM=\frac{a+b}{2}\]and \[GM=\sqrt{ab},\] then according to question \[\frac{AM}{GM}=\frac{m}{n}\] \[\Rightarrow \] \[\frac{\frac{a+b}{2}}{\sqrt{ab}}=\frac{m}{n}\] \[\Rightarrow \] \[\frac{a+b}{2\sqrt{ab}}=\frac{m}{n}\] On squaring both sides, we get \[\frac{{{(a+b)}^{2}}}{4ab}=\frac{{{m}^{2}}}{{{n}^{2}}}\] Applying dividendo rule, we get \[\frac{{{a}^{2}}+{{b}^{2}}+2ab}{{{a}^{2}}+{{b}^{2}}+2ab-4ab}=\frac{{{m}^{2}}}{{{m}^{2}}-{{n}^{2}}}\] \[\Rightarrow \]\[\frac{{{(a+b)}^{2}}}{{{(a-b)}^{2}}}=\frac{{{m}^{2}}}{({{m}^{2}}-{{n}^{2}})}\] \[\Rightarrow \]\[\frac{a+b}{a-b}=\frac{m}{\sqrt{{{m}^{2}}-{{n}^{2}}}}\] Applying componendo and dividendo rule, we get \[\frac{a+b+a-b}{a+b-a+b}=\frac{m+\sqrt{{{m}^{2}}-{{n}^{2}}}}{m-\sqrt{{{m}^{2}}-{{n}^{2}}}}\] \[\Rightarrow \]\[\frac{2a}{2b}=\frac{m+\sqrt{{{m}^{2}}-{{n}^{2}}}}{m-\sqrt{{{m}^{2}}-{{n}^{2}}}}\] \[\Rightarrow \]\[\frac{a}{b}=\frac{m+\sqrt{{{m}^{2}}-{{n}^{2}}}}{m-\sqrt{{{m}^{2}}-{{n}^{2}}}}\]
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