A) 1
B) \[3/2\]
C) 2
D) None
Correct Answer: B
Solution :
If \[\alpha ,\beta \] be the roots of the equation then \[\alpha +\beta =a-2,\] \[\alpha \beta =-\frac{a+1}{2}\] Sum of square of roots \[S={{\alpha }^{2}}+{{\beta }^{2}}={{(\alpha +\beta )}^{2}}-2\alpha \beta \] \[={{(a-2)}^{2}}+(a+1)={{a}^{2}}-3a+5\] \[S={{a}^{2}}-3a+\frac{9}{4}+\frac{11}{4}\] \[S={{\left( a-\frac{3}{2} \right)}^{2}}+\frac{11}{4}\] Clearly S is least when \[a-\frac{3}{2}=0\Rightarrow a=\frac{3}{2}\]
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