A) 6
B) 68
C) \[\frac{3}{68}\]
D) \[-\frac{68}{3}\]
Correct Answer: D
Solution :
The given equation is \[{{\operatorname{m}}^{2}}{{x}^{2}}+\left( 2m-{{m}^{2}} \right)x+3=0\] \[\therefore \,\,\,\alpha +\beta =-\frac{2m-{{m}^{2}}}{{{m}^{2}}}=\frac{m-2}{m}\,\,and\,\,\alpha \beta =\frac{3}{{{m}^{2}}}\] Now \[\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{4}{3}\Rightarrow \frac{{{\alpha }^{2}}+{{\beta }^{2}}}{\alpha \beta }=\,\,\frac{4}{3}\Rightarrow \frac{{{(\alpha +\beta )}^{2}}-2\alpha \beta }{\alpha \beta }=\frac{4}{3}\] Substituting the values, we get \[\frac{{{\left( \frac{m-2}{m} \right)}^{2}}-2.\frac{3}{{{m}^{2}}}}{\frac{3}{{{m}^{2}}}}=\frac{4}{3}\] \[\Rightarrow \,\,\,\frac{{{m}^{2}}-4m+4-6}{3}=\frac{4}{3}\Rightarrow {{m}^{2}}-4m-6=0\] \[{{\operatorname{m}}_{1}}\,\,and\,\,{{m}_{2}}\] are roots of this equation, therefore \[{{\operatorname{m}}_{1}}+{{m}_{2}}=4\,\,and\,\,{{m}_{1}}{{m}_{2}}=-\,6\] The given expression is, \[\frac{\operatorname{m}_{1}^{2}}{{{m}_{2}}}+\frac{m_{2}^{2}}{{{m}_{1}}}=\,\,\frac{m_{1}^{3}+m_{2}^{3}}{{{m}_{1}}{{m}_{2}}}\] \[=\,\,\frac{{{({{m}_{1}}+{{m}_{2}})}^{3}}-\,\,3{{m}_{1}}{{m}_{2}}\,({{m}_{1}}+{{m}_{2}})}{{{m}_{1}}{{m}_{2}}}\] \[=\,\,\frac{{{(4)}^{3}}-3.(-6).(4)}{-\,6}=-\frac{68}{3}\]
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