A) 0
B) 1
C) 2
D) 3
Correct Answer: B
Solution :
Given equation is \[{{x}^{2}}-(a-2)x-a-1=0\] Let \[\alpha ,\beta \]be the roots of the given equation. Then, \[\alpha +\beta =a-2,\] \[\alpha \beta =-(\alpha +1)\] \[\therefore \] \[{{\alpha }^{2}}+{{\beta }^{2}}={{(\alpha +\beta )}^{2}}-2\alpha \beta \] \[={{(a-2)}^{2}}+2(a+1)\] \[={{a}^{2}}-2a+6\] \[={{(a-1)}^{2}}+5\] Clearly, \[{{\alpha }^{2}}+{{\beta }^{2}}\ge 5.\] Thus, the minimum value of \[{{\alpha }^{2}}+{{\beta }^{2}}\]is 5, which it attains at \[a=1.\]
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