A) 0
B) 1
C) m
D) \[{{m}^{2}}\]
Correct Answer: D
Solution :
\[\alpha +\beta =\frac{2({{m}^{2}}+1)}{2}={{m}^{2}}+1\] .....(i) and \[\alpha \beta =\frac{{{m}^{4}}+{{m}^{2}}+1}{2}\] .....(ii) Therefore \[{{\alpha }^{2}}+{{\beta }^{2}}={{(\alpha +\beta )}^{2}}-2\alpha \beta \] \[={{({{m}^{2}}+1)}^{2}}-2\frac{({{m}^{4}}+{{m}^{2}}+1)}{2}\] \[={{m}^{4}}+2{{m}^{2}}+1-{{m}^{4}}-{{m}^{2}}-1={{m}^{2}}\]You need to login to perform this action.
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