Question

The interval in which the function $y=f(x)=\frac{x-1}{x^2-3x+3}$ transforms the real line is

• A. $(0,\infty )$
• B. $(-\infty ,\infty )$
• C. [0, 1]
• D. $[-\frac{1}{3},1]$

Answer 1

Answer: (D)

The interval in which $y=f(x)=\frac{x-1}{x^2-3x+3}$ transforms the real line is known as the range of the function.
Since $y=\frac{x-1}{x^2-3x+3}\therefore y{x}^2-3xy+3y=x-1$
or $y{x}^2-x(1+3y)+3y+1=0$
If $y\ne 0$, it is quadratic in x and $\because$ x is real,
$\therefore (1+3y{)}^2-4y(3y+1)\ge 0$
$\Rightarrow 1+9y^2+6y-12y^2-4y\ge 0$
$\Rightarrow -3y^2+2y+1\ge 0\Rightarrow 3y^2-2y-1\le 0$
$\Rightarrow (3y+1)(y-1)\le 0\Rightarrow -\frac{1}{3}\le y\le 1(y\ne 0)$
Also, y =$\frac{x-1}{x^2-3x+3}\Rightarrow y=0$ for $x=1$
$\therefore y=0$ is also valid.
Hence the reqd. interval is $\left[-\frac{1}{3},1\right]$