Question

The range of the function $y=\frac{x-1}{(x^2-3x+3)}$ is [a, b] where a, b are respectively

• A. $\frac{1}{3},2$
• B. $\frac{-1}{3},1$
• C. $\frac{-1}{3},2$
• D. $\frac{1}{3},1$

Answer 1

The correct option is B

$\frac{-1}{3},1$

$y=\frac{x-1}{x^2-3x+3}\Rightarrow (x^2-3x+3)y=x-1\Rightarrow y{x}^2-(3y+1)x+(3y+1)=0$
The above equation is a quadratic in x. Since x is real,
$D\ge 0\Rightarrow (3y+1{)}^2-4y(3y+1)\ge 0\Rightarrow 9y^2+6y+1-4y-12y^2\ge 0\Rightarrow -3y^2+2y+1\ge 0\Rightarrow 3y^2-2y-1\le 0(multiplyingby(-1))\Rightarrow y^2-\frac{2y}{3}-\frac{1}{3}\le 0\Rightarrow {(y-\frac{1}{3})}^2-\frac{1}{9}-\frac{1}{3}\le 0\Rightarrow {(y-\frac{1}{3})}^2\le \frac{4}{9}\Rightarrow \frac{-2}{3}\le y-\frac{1}{3}\le \frac{2}{3}\Rightarrow -\frac{1}{3}\le y\le 1\Rightarrow yϵ[\frac{-1}{3},1]\Rightarrow a=-\frac{1}{3},b=1$