A) 0
B) 2
C) \[-1\]
D) \[1\]
Correct Answer: D
Solution :
We have the equation \[{{x}^{2}}-(a-2)x-(a+1)=0\] Let the roots are\[\alpha \]and\[\beta \]. Then, \[\alpha +\beta =\alpha -2\] and \[\alpha \beta =-(\alpha +1)\] Now, \[{{\alpha }^{2}}+{{\beta }^{2}}={{(\alpha +\beta )}^{2}}-2\alpha \beta \] \[={{(a-2)}^{2}}-2[-(a+1)]\] \[={{\alpha }^{2}}+4-4a+2a+2\] \[={{a}^{2}}-2a+6\] \[={{(a-1)}^{2}}+5\] which is minimum for \[a=1\]
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