Question

The range of $\frac{x^2-2x+3}{x^2-2x-8},x\in R-\{-2,4\}$ is ?

Answer 1

$Letf\left(x\right)=\frac{x^2-2x+3}{x^2-2x-8},x\in R-\{-2,4\}$

Here, the domain of f(x) will be real value of $x$

and for range let,

$y=\frac{x^2-2x+3}{x^2-2x-8}\Rightarrow x^2-2x+3=y(x^2-2x-8)\Rightarrow x^2(y-1)-2x(y-1)-(8y+3)=0$

For $x$ to be real the determinant of equation must be greater than zero

$\Rightarrow b^2-4ac$ $\Rightarrow {[-2(y-1\left)\right]}^2-4$. $(y-1)(-(8y+3\left)\right)$ $\Rightarrow 9y^2-7y-2$ $\Rightarrow (9y+2)(y-1)$ $\Rightarrow y$ $\frac{2}{9}ory$ $\Rightarrow y\in [\frac{-2}{9},\infty )]$ $ory\in [1,\infty ]$