A) \[25: 16\]
B) \[5:3\]
C) \[16: 1\]
D) \[25:9\]
Correct Answer: C
Solution :
\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{3}{5}\] \[\therefore \,\,\,\,\,\,\frac{{{\operatorname{I}}_{max}}}{{{\operatorname{I}}_{min}}}=\frac{{{\left( {{a}_{1}}+{{a}_{2}} \right)}^{2}}}{{{\left( {{a}_{1}}-{{a}_{2}} \right)}^{2}}}=\frac{{{\left( 3+5 \right)}^{2}}}{{{\left( 3-5 \right)}^{2}}}=\frac{16}{1}\]
You need to login to perform this action.
You will be redirected in
3 sec
