Question

Range of the function $f\left(x\right)=\frac{x-2}{x^2-4x+3}$ is

• A. $(-\infty ,0)$
• B. $R$
• C. $(0,\infty )$
• D. $R-\left\{0\right\}$

Answer 1

Answer: (A)

$(-\infty ,0)$

Range of function $\frac{x-2}{x^2-4x+3}$

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

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

${yx}^2-4xy-x+3y+2=0$

${yx}^2-(4y+1)x+(3y+2)=0$

$\because D\ge 0$

$⟹{(4y+1)}^2-4(3y+2)y\ge 0$

$⟹{16y}^2+1+8y-{12y}^2-8y\ge 0$

$⟹{4y}^2+1\ge 0$

$\therefore y\in R.$ Because ${4y}^2+1$ is always $posetive.$