A) \[\frac{145}{12}\]
B) \[-\frac{145}{12}\]
C) \[\frac{145}{7}\]
D) \[\frac{-145}{15}\]
Correct Answer: B
Solution :
| m and n zeroes of \[3{{x}^{2}}+11x-4\], |
| \[m+n=-\frac{11}{3},\,m\,.\,n=-\frac{4}{3}\] |
| \[\Rightarrow \,\frac{m}{n}+\frac{n}{m}=\frac{{{m}^{2}}+{{n}^{2}}}{m\,.\,n}=\frac{{{\left( m+n \right)}^{2}}-2m\,.\,n}{mn}\] |
| \[\Rightarrow \,\,\frac{{{\left( \frac{-11}{3} \right)}^{2}}-2\times \left( -\frac{4}{3} \right)}{-\frac{4}{3}}=-\frac{145}{12}\] |
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