A) 2
B) 1
C) 3
D) 0
Correct Answer: B
Solution :
Let \[\alpha ,\ \beta \] be the roots of the equation \[{{x}^{2}}-(a-2)x-a+1=0,\] then \[\alpha +\beta =a-2,\,\,\alpha \beta =-a+1\] \[\therefore z={{\alpha }^{2}}+{{\beta }^{2}}={{(\alpha +\beta )}^{2}}-2\alpha \beta \] \[={{(a-2)}^{2}}+2(a-1)={{a}^{2}}-2a+2\] \[\frac{dz}{da}=2a-2=0\Rightarrow a=1\] \[\frac{{{d}^{2}}z}{d{{a}^{2}}}=2>0,\] so z has minima at \[a=1\] So \[{{\alpha }^{2}}+{{\beta }^{2}}\] has least value for \[a=1\]. This is because we have only one stationary value at which we have minima. Hence \[a=1\].
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