Question

If $y=\frac{x^2-2x+4}{x^2+2x+4}$ where $x$ can take real values, then

• A. $\frac{1}{3}\le y\le 3$
• B. $3<y\le 5$
• C. $5<y\le 7$
• D. $7<y\le 9$

Answer 1

The correct option is A $\frac{1}{3}\le y\le 3$

$y=\frac{x^2-2x+4}{x^2+2x+4}$

$x^{2}y+2xy+4y=x^2-2x+4$

$x^2(y-1)+2x(y+1)+4y-4$ =0

Since x is real, $D\ge 0$

$4(y+1{)}^2-4(4y-4\left)\right(y-1)\ge 0$

$4y^2+8y+4-16y^2+16y-16+16y\ge 0$

$-12y^2+40y-12\ge 0$

$3y^2-10y+3\le 0$

$(3y-1)(y-3)\le 0$

Hence, $\frac{1}{3}\le y\le 3$