A) \[\frac{2}{5}\]
B) \[\frac{3}{5}\]
C) \[\frac{4}{5}\]
D) None of the above
Correct Answer: B
Solution :
\[\alpha +\beta =1,\,\,\,\alpha \beta =-1\] \[\begin{align} & {{\left( \alpha -\beta \right)}^{2}}={{\left( \alpha +\beta \right)}^{2}}-4\alpha \beta \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=1-4\times (-1)=5 \\ \end{align}\] Now, \[\frac{{{\alpha }^{2}}+{{\beta }^{2}}}{\left( {{\alpha }^{2}}-{{\beta }^{2}} \right)\left( \alpha -\beta \right)}=\frac{{{\left( \alpha -\beta \right)}^{2}}+2\alpha \beta }{\left( \alpha +\beta \right){{\left( \alpha -\beta \right)}^{2}}}\] = \[\frac{5+2\times (-1)}{1\times 5}=\frac{5-2}{5}=\frac{\mathbf{3}}{\mathbf{5}}\]
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