Question

If $f\left(x\right)=\frac{x^2-2x+4}{x^2+2x+4},x\in R$, then prove that the range of $f\left(x\right)$ is

• A. $\therefore y\in [\frac{1}{2},2].$
• B. $\therefore y\in [\frac{1}{3},3].$
• C. $\therefore y\in [\frac{1}{3},2].$
• D. $\therefore y\in [\frac{1}{2},3].$

Answer 1

Answer: (B)

$\therefore y\in [\frac{1}{3},3].$

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

$⟹{yx}^2+2xy+4y=x^2-2x+4$

$⟹x^2(y-1)+2xy+2x+4y-4=0$

$⟹x^2(y-1)+2x(y+1)+4(y-1)=0$

$D\ge 0$

$⟹4{(y+1)}^2-4\times 4{(y-1)}^2\ge 0$

$⟹4{(y+1)}^2-4\times 4{(y-1)}^2\ge 0$

$⟹y^2+1+2y-{4y}^2-4+8y\ge 0$

$⟹{-3y}^2+10y-3\ge 0$

$⟹{3y}^2-10y+3\le 0$

$⟹{3y}^2-9y-y+3\le 0$

$⟹(3y-1)(y-3)\le 0$

$\therefore y\in [\frac{1}{3},3].$