Question

If $x$ is real, then $\frac{x^2-2x+4}{x^2+2x+4}$ takes values in the interval:

• A. $[1/3,3]$
• B. $(1/3,3)$
• C. $(3,3)$
• D. $(-1/3,3)$

Answer 1

Answer: (A)

Let $y=\frac{x^2-2x+4}{x^2+2x+4}$
$\Rightarrow x^2(y-1)+x(2y+2)+4y-4=0$
Since, $x$ is real therefore its discriminant
$b^2-4ac\ge 0$
$\therefore (2y+2{)}^2-4(y-1)4(y-1)\ge 0$
$\Rightarrow 4y^2+8y+4-16y^2-16+32y\ge 0$
$\Rightarrow -12y^2+40y-12\ge 0$
$\Rightarrow 3y^2-10y+3\le 0$
$\Rightarrow (3y-1)(y-3)\le 0$
$\Rightarrow \frac{1}{3}y\le 3$