Question

Let $a\in R$. For possible real values of $x$, $\frac{x^2+ax+3}{x^2+x+a}$ takes all real values, then

• A. $4a^3+39<0$
• B. $4a^3+39\ge 0$
• C. $a\ge \frac{1}{4}$
• D. $a<\frac{1}{4}$

Answer 1

Answer: (A), (D)

$a<\frac{1}{4}$

Let $\frac{x^2+ax+3}{x^2+x+a}=y$
$\Rightarrow (1-y)x^2-(y-a)x+(3-ay)=0$
Since $x\in R$
$\therefore (y-a{)}^2-4(1-y\left)\right(3-ay)\ge 0$
$\Rightarrow (1-4a)y^2+(2a+12)y+a^2-12\ge 0\cdots \left(1\right)$

Eq.$\left(1\right)$ is true for all $y\in R$
$\Rightarrow 1-4a>0$ and $D\le 0$
$\Rightarrow a<\frac{1}{4}$ and $4(a+6{)}^2-4(a^2-12\left)\right(1-4a)\le 0$
$\Rightarrow a<\frac{1}{4}$ and $4a^3\le 36a-48$
$\Rightarrow 4a^3<36\left(\frac{1}{4}\right)-48[\because a<\frac{1}{4}]$
$\Rightarrow 4a^3+39<0$