A) \[\frac{1+\beta }{\sqrt{\beta }}\]
B) \[\sqrt{\frac{1+\beta }{\beta }}\]
C) \[\frac{1+\beta }{2\sqrt{\beta }}\]
D) \[\frac{2\sqrt{\beta }}{1+\beta }\]
Correct Answer: D
Solution :
Given, \[\frac{{{l}_{1}}}{{{l}_{2}}}=\beta =\frac{{{a}^{2}}}{{{b}^{2}}}\] \[\Rightarrow \] \[\frac{a}{b}=\sqrt{\beta }\] \[{{l}_{\max }}={{(a+b)}^{2}}\] and \[{{l}_{\min }}={{(a-b)}^{2}}\] \[\therefore \] \[\frac{{{l}_{\max }}-{{l}_{\min }}}{{{l}_{\max }}+{{l}_{\min }}}=\frac{{{(a+b)}^{2}}-{{(a-b)}^{2}}}{{{(a+b)}^{2}}+{{(a-b)}^{2}}}\] \[=\frac{4ab}{2({{a}^{2}}+{{b}^{2}})}=\frac{2ab}{({{a}^{2}}+{{b}^{2}})}\] \[=\frac{2b.b\sqrt{\beta }}{{{b}^{2}}+\beta {{b}^{2}}}=\frac{2\sqrt{\beta }}{1+\beta }\]
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