A) a = 0
B) a = 1
C) a=-1
D) a = 2
Correct Answer: B
Solution :
[b]: Given, \[\alpha ,\beta \]are the roots of equation \[{{x}^{2}}-\left( a-2 \right)x-a-1=0\] \[\therefore \]\[{{\alpha }^{2}}+{{\beta }^{2}}=a-2\]and\[\alpha \beta =-(a+1)\] \[{{\alpha }^{2}}+{{\beta }^{2}}={{(\alpha +\beta )}^{2}}-2\alpha \beta \] \[={{(a-2)}^{2}}+2(a+1)\] \[={{a}^{2}}-2a+6={{a}^{2}}-2a+1+5\] \[={{(a-1)}^{2}}+5\ge 5\] \[\therefore \]\[{{\alpha }^{2}}+{{\beta }^{2}}\]is least if \[{{(a-1)}^{2}}=0\Rightarrow a=1\].
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