Question

If $x$ is real, then $\frac{x^{2}-2 x+4}{x^{2}+2 x+4}$ 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 $y=\frac{x^{2}-2 x+4}{x^{2}+2 x+4}$
$\Rightarrow x^{2}(y-1)+x(2 y+2)+4 y-4=0$
Since, $x$ is real therefore its discriminant
$b^{2}-4 a c \geq 0$
$\therefore (2 y+2)^{2}-4(y-1) 4(y-1) \geq 0$
$\Rightarrow 4 y^{2}+8 y+4-16 y^{2}-16+32 y \geq 0$
$\Rightarrow-12 y^{2}+40 y-12 \geq 0$
$\Rightarrow 3 y^{2}-10 y+3 \leq 0$
$\Rightarrow(3 y-1)(y-3) \leq 0$
$\Rightarrow \frac{1}{3} y \leq 3$