Question

Find the range of values of $a$, such that $f\left(x\right)=\frac{a{x}^2+2(a+1)x+9a+4}{x^2-8x+32}$ is always negative.

• A. $a\in (-\infty ,-\frac{3}{4})$
• B. $a\in (-\infty ,-\frac{1}{4})$
• C. $a\in (-\infty ,-\frac{1}{2})$
• D. $a\in (-\infty ,-1)$

Answer 1

The correct option is C $a\in (-\infty ,-\frac{1}{2})$

Considering denominator $x^2-8x+32$

$D<0$ and $a>0$

So denominator is always positive

For $f\left(x\right)$ to be negative, numerator $\Rightarrow a{x}^2+2(a+1)x+9a+4<0$ i.e $a<0$ and $b^2-4ac<0$

$\Rightarrow a{x}^2+2(a+1)x+9a+4<0$

$\Rightarrow$$a<0$ & $4(a+1{)}^2-4a(9a+4)<0$

$\Rightarrow 4(a^2+2a+1-9a^2-4a)<0$

$\Rightarrow 4(-8^2-2a+1)<0$

$8a^2+2a-1>0$

$(4a-1)(2a+1)>0$

$a>\frac{1}{4}$ or $a<-\frac{1}{2}$ but $a<0$

$\therefore a\in (-\infty ,-\frac{1}{2})$