question_answer

Let a, b, c be real. If \[a{{x}^{2}}+bx+c=0\]has two real roots \[\alpha \]and\[\beta ,\]where \[a<-1\]and \[\beta >1,\]then \[1+\frac{c}{a}+\left| \frac{b}{a} \right|\]is

A)  \[<0\]                  

B)         \[>0\]                  

C) \[\le 0\]               

D)         none of these

Correct Answer: A

Solution :

From figure it is clear that if \[a>0,\]and \[f(-1)<0\] and \[f(1)<0\]and if \[a<0,f(-1)>0\]and \[f(1)>0.\] In both cases, \[af(-1)<0\]and \[af(1)<0.\] \[\Rightarrow \]\[a(a-b+c)<0\]and \[a(a+b+c)<0\] On dividing by \[{{a}^{2}},\] \[\Rightarrow \]               \[1-\frac{b}{a}+\frac{c}{a}<0\] and        \[1+\frac{b}{a}+\frac{c}{a}<0\] Combining both equations, we get \[1\pm \frac{b}{a}+\frac{c}{a}<0\]                 \[\Rightarrow \]               \[1+\left| \frac{b}{a} \right|+\frac{c}{a}<0\]