A) \[2n\]
B) \[{{n}^{3}}\]
C) \[{{n}^{2}}\]
D) \[2{{n}^{2}}\]
Correct Answer: C
Solution :
\[\alpha +\beta =1+{{n}^{2}}\]; \[\alpha \beta =\frac{1}{2}(1+{{n}^{2}}+{{n}^{4}})\] \ \[{{\alpha }^{2}}+{{\beta }^{2}}={{(\alpha +\beta )}^{2}}-2\alpha \beta \] \[={{(1+{{n}^{2}})}^{2}}-2.\frac{1}{2}(1+{{n}^{2}}+{{n}^{4}})\] \[=1+{{n}^{4}}+2{{n}^{2}}-1-{{n}^{2}}-{{n}^{4}}\]Þ \[{{\alpha }^{2}}+{{\beta }^{2}}={{n}^{2}}\].
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