Question

For every possible $x\in R$, If $\frac{x^2+2x+a}{x^2+4x+3a}$ can take all real values, then

• A. $a\in (0,1)$
• B. $a\in [-1,1]$
• C. $a\in (-1,1)$
• D. $a\in [0,1]$

Answer 1

The correct option is A $a\in (0,1)$

Let $y=\frac{x^2+2x+a}{x^2+4x+3a}$
$\Rightarrow x^2(y-1)+2(2y-1)x+a(3y-1)=0$

$\because x\in R$, so $D\ge 0$
$\Rightarrow 4(2y-1{)}^2-4(y-1\left)a\right(3y-1)\ge 0$ for all $y\in R$
$\Rightarrow (4-3a)y^2-4(1-a)y+(1-a)\ge 0$ for all $y\in R$
$\Rightarrow 4-3a>0$ and $D<0$
$(\because f(x)\ge 0$ for all $x\Rightarrow a>0$ and $D<0)$

$a<\frac{4}{3}$ and $16(1-a{)}^2-4(1-a\left)\right(4-3a)<0$
$\Rightarrow a<\frac{4}{3}$ and $a(a-1)<0$
$\Rightarrow 0<a<1$