A) \[a=0\]
B) \[a=1\]
C) \[a=-1\]
D) \[a=2\]
Correct Answer: B
Solution :
Given equation is \[{{x}^{2}}-(a-2)x-a-1=0\] \[\therefore \] \[\alpha +\beta =a-2\] and \[\alpha \beta =-(a+1)\] \[\therefore \] \[{{\alpha }^{2}}+{{\beta }^{2}}={{(\alpha +\beta )}^{2}}-2\alpha \beta \] \[={{(a-2)}^{2}}+2(a+1)\] \[={{a}^{2}}+4-4a+2a+2\] \[={{a}^{2}}-2a+6\] \[={{(a-1)}^{2}}+5\] For least value of \[{{\alpha }^{2}}+{{\beta }^{2}},\,\,a=1.\]
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