Correct option is (B) Both roots in \((\alpha, \beta)\)
\((x - \alpha) (x - \beta) + 1 = 0\)
⇒ \(x^2 - (\alpha + \beta)x +(\alpha\beta + 1) =0\)
\(D = b^2 - 4ac = (\alpha + \beta)^2 - 4(\alpha \beta + 1)\)
\(=(\alpha + \beta)^2 - 4\alpha\beta - 4\)
\(= (\alpha - \beta)^2 - 4\)
\(= (\beta - \alpha)^2 - 4\)
\(> 0\) (as \(\beta - \alpha > 2⇒ (b - \alpha)^2 > 4\))
\(\therefore \) Both roots are real and distinct.
Let \(f(x) = (x - \alpha) (x - \beta) + 1\)
\(f(\alpha ) = 1 > 0\)
\(f(\beta) = 1 > 0\)
\(\because f(\alpha) \,f(\beta) > 0\)
\(\therefore \) Either both roots will lie between \((\alpha, \beta)\) or one root lie in \((-\infty, \alpha)\) and other in \((\beta, \infty)\).
Because graph of \(f(x) =x^2 - (\alpha + \beta)x + (\alpha\beta + 1)\) is upward.
