Question

Let $ \alpha $ and $ \beta $ be the roots of equation $ {{x}^{2}}-(a-2)x-a-1=0, $ then $ {{\alpha }^{2}}+{{\beta }^{2}} $ assumes the least value, if

• A. $ a=0 $
• B. $ a=1 $
• C. $ a=-1 $
• D. $ a=2 $

Answer 1

Answer: (B)

Given equation is $ {{x}^{2}}-(a-2)x-a-1=0 $
$ \therefore $ $ \alpha +\beta =a-2 $ and $ \alpha \beta =-(a+1) $
$ \therefore $ $ {{\alpha }^{2}}+{{\beta }^{2}}={{(\alpha +\beta )}^{2}}-2\alpha \beta $
$ ={{(a-2)}^{2}}+2(a+1) $
$ ={{a}^{2}}+4-4a+2a+2 $
$ ={{a}^{2}}-2a+6 $
$ ={{(a-1)}^{2}}+5 $
For least value of
$ {{\alpha }^{2}}+{{\beta }^{2}},\,\,a=1. $