Question

Let $\alpha$ & $\beta$ be the roots of the equation, $a{x}^2+bx+c=0$ where $1<\alpha <\beta$, then $\underset{x\to m}{lim}\frac{|a{x}^2+bx+c|}{a{x}^2+bx+c}=1$, then which of the following is correct?

• A. $a>0$ & $m>1$
• B. $a<0$ & $m<1$
• C. $a<0$ & $\alpha <m<\beta$
• D. $\frac{|a|}{a}=1$ & $m>\alpha$

Answer 1

Answer: (C)

$a<0$ & $\alpha <m<\beta$

Given that ${lim}_{x\to m}\frac{|a{x}^2+bx+c|}{a{x}^2+bx+c}=1$

$\Rightarrow a{m}^2+bm+c>0$

If $f\left(x\right)>0\Rightarrow a<0$

$f\left(x\right)<0\Rightarrow a>0$

$\Rightarrow a<0$

Since $a<0$

The equation now represents a downward parabola and $\alpha$ & $\beta$ are points where the parabola meets x-axis.

We know that y is positive only between $\alpha$ & $\beta$

(in the graph of a downward parabola)

$\Rightarrow \alpha <m<\beta$