A) 0
B) 1
C) 2
D) 3
E) \[-2\]
Correct Answer: B
Solution :
Let\[\alpha \]and\[\beta \]be the roots of given equation so that \[\alpha +\beta =a-2\]and \[\alpha \beta =(a-1)\] Let\[S={{\alpha }^{2}}+{{\beta }^{2}},\]thus \[S={{(\alpha +\beta )}^{2}}-2\alpha \beta ={{(a-2)}^{2}}+2(a-1)\] \[\Rightarrow \]\[S={{a}^{2}}-2a+2\] On differentiating w.r.t. a \[\therefore \] \[\frac{dS}{da}=2a-2\] Put\[\frac{dS}{da}=0\]for maxima or minima \[\therefore \] \[a=1\] Also \[\frac{{{d}^{2}}S}{d{{a}^{2}}}=2>0\] \[\therefore \] \[S\]is minimum when\[a=1\]
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