Question

Find the range of $f\left(x\right)=\mathrm{sgn}\left(x^2-2x+3\right)$ is

• A. $\left\{-1,0,1\right\}$
• B. $\left\{0,1\right\}$
• C. $\left\{1\right\}$
• D. $\left\{-1,-2,2,3\right\}$

Answer 1

The correct option is C $\left\{1\right\}$

We know that $f\left(x\right)=Sgn\left(y\right)=\{\begin{array}{cc}1& f\left(x\right)>0\\ 0& f\left(x\right)=0\\ -1& f\left(x\right)<1\end{array}$

The given function is $f\left(x\right)=\text{sgn}(x^2-2x+3)=\frac{|x^2-2x+3|}{x^2-2x+3}$

Consider, $x^2-2x+3=(x^2-2x+1)+2=(x-1{)}^2+2>0$

$\text{So the quadratic equation is always positive}$

$f\left(x\right)=\text{sgn}(x^2-2x+3)=\frac{|x^2-2x+3|}{x^2-2x+3}=\frac{x^2-2x+3}{x^2-2x+3}=1$

Hence, C will be correct answer.