Question

lf $x$ is real then the value of $\frac{x^2-2x+9}{x^2+2x+9}$ is

• A. $2$
• B. $<\frac{1}{2}$
• C. between $\frac{1}{2}$ and $2$
• D. $(2,3)$

Answer 1

The correct option is C between $\frac{1}{2}$ and $2$

$Let\frac{x^2-2x+9}{x^2+2x+9}=y\Rightarrow x^2(1-y)-x(2+2y)+9(1-y)=0\Delta =(2+2y{)}^2-(6-6y{)}^2\ge 0\because x\in R\therefore (8-4y)(8y-4)\ge 0\Rightarrow 8-4y\ge 0\Rightarrow y<2.or8y-4\ge 0\Rightarrow y\ge \frac{1}{2}$
$\therefore$ Value of $x$ lies between $\frac{1}{2}$ and $2$.