Question

If $x$ is real then the function $f\left(x\right)=\left(\frac{x^2-2x+4}{x^2+2x+4}\right)$ lies in the interval

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

Answer 1

The correct option is A $[1/3,3]$

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

Discriminant of $x^2-2x+4$ is $\Delta ={(-2)}^2-4\times 1\times 4=4-16=-12<0\therefore x\in R$

Discriminant of $x^2+2x+4$ is $\Delta =2^2-4\times 1\times 4=4-16=-12<0\therefore x\in R$

$\Rightarrow y(x^2+2x+4)=x^2-2x+4$

$\Rightarrow (y-1)x^2+2(y+1)x+4(y-1)=0$ is quadratic in $x$ for all $x\in R$

$\Rightarrow \Delta \ge 0$

$\Rightarrow 4{(y+1)}^2-4\times (y-1)4(y-1)\ge 0$

$\Rightarrow {(y+1)}^2-{(2y-2)}^2\ge 0$

$\Rightarrow (y+1+2y-2)(y+1-2y+2)\ge 0$

$\Rightarrow (3y-1)(-y+3)\ge 0$

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

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