A) \[4:1\]
B) \[5:3\]
C) \[4:3\]
D) \[25:9\]
Correct Answer: A
Solution :
According to the question, we can write \[\frac{{{I}_{\max }}}{{{I}_{\min }}}={{\left( \frac{{{a}_{1}}+{{a}_{2}}}{{{a}_{1}}-{{a}_{2}}} \right)}^{2}}\] \[\Rightarrow \] \[\frac{25}{9}={{\left( \frac{{{a}_{1}}+{{a}_{2}}}{{{a}_{1}}-{{a}_{2}}} \right)}^{2}}\] \[\Rightarrow \] \[\frac{5}{3}=\frac{{{a}_{1}}+{{a}_{2}}}{{{a}_{1}}-{{a}_{2}}}\] \[\Rightarrow \] \[5{{a}_{1}}-5{{a}_{2}}=3{{a}_{1}}+3{{a}_{2}}\] \[\Rightarrow \] \[2{{a}_{1}}=8{{a}_{2}}\] \[\Rightarrow \] \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{8}{2}=\frac{4}{1}\]
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