Question

Let $f\left(x\right)=a{x}^2+bx+c$. lf $f(-1)<1,f\left(1\right)>-1,f\left(3\right)<-4$ and $a\ne 0$, then

• A. $a>0$
• B. $a>1$
• C. $a<-\frac{1}{8}$
• D. $-\frac{1}{8}<a<0$

Answer 1

Answer: (C)

$a<-\frac{1}{8}$

$f(-1)<1\Rightarrow a-b+c<1$-------1)
$f\left(1\right)>-1\Rightarrow a+b+c>-1$-------2)
$f\left(3\right)<-4\Rightarrow 9a+3b+c<-4$-------3)
Solving inequalities 1) and 3)
We get $12a+4c<-1$
From 1) and 2) $a+c=0.$
$\Rightarrow 8a<-1$
$a<\frac{-1}{8}$