Question

The range of the function $f\left(x\right)=\frac{(x-2{)}^2}{(x-1)(x-3)}$ is

• A. $(1,\infty )$
• B. $(-\infty ,1)$
• C. $R-(0,1]$
• D. $(0,1]$

Answer 1

The correct option is A $(1,\infty )$

Range of $\frac{{(x-2)}^2}{(x-1)(x-3)}.$

$y=\frac{x^2+4-4x}{x^2-4x+3}$

$y{x}^2-4xy+3y=x^2+4-4x$

$(y-1)x^2-4x(y-1)-(4-3y)=0$

$D\ge 0$

${[-4(y-1\left)\right]}^2+4(4-3y)(y-1)\ge 0$

$16{(y-1)}^2+4(4y-4-{3y}^2+3y)\ge 0$

$16{(y^2+1-2y)}^2+16y-16-{12y}^2+12y)\ge 0$

${16y}^2+16-32y+16y-16-{12y}^2+12y)\ge 0$

${4y}^2-4y\ge 0$

$y(y-1)\ge 0$

$\therefore y\in (-\infty ,0]\cup [1,\infty ).$