Question

Let a, b, c be real and $a{x}^2+bx+c=0$ has two real roots α and β where $\alpha <–1and\beta >1$, then $1+\frac{c}{a}+\left|\frac{b}{a}\right|$

• A. < 2
• B. < 1
• C. < 0
• D. None of these

Answer 1

The correct option is C

< 0

Since, $\alpha <-1and\beta >1$

$\therefore \alpha +\lambda =-1and\beta =1+\mu where\lambda ,\mu >0Now,1+\frac{c}{a}+\left|\frac{b}{a}\right|=1+\alpha \beta +|\alpha -\beta |=1+(-1,-\lambda )(1+\mu )+|-1-\lambda +1+\mu |=1-1-\mu -\lambda -\lambda \mu +|\mu -\lambda |\{\begin{array}{c}-\mu -\lambda -\lambda \mu +\mu -\lambda ,if\mu >\lambda \\ -\mu -\lambda -\lambda \mu +\lambda -\mu ,if\lambda >\mu \end{array}\therefore 1+\frac{c}{a}+\left|\frac{b}{a}\right|=-2\lambda -\lambda \mu or-2\mu -\lambda \mu Inbothcases1+\frac{c}{a}+\left|\frac{b}{a}\right|<0(\because \lambda ,\mu >0)$