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Q.
The interval in which the function $y = f(x) = \frac {x-1} {x^2-3x+3} $ transforms the real line is
Relations and Functions
Solution:
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 $\, y=\frac{x-1}{x^2 -3x +3} \therefore yx^2 -3xy+3y =x-1$
or $ yx^2-x (1+3y)+3y+1=0$
If $y \neq 0$, it is quadratic in x and $\because$ x is real,
$\therefore (1 +3y)^2 -4y(3y+1) \geq \,0$
$\Rightarrow 1 + 9y^2 + 6y - 12y^2 - 4y \geq \,0$
$ \Rightarrow -3y^2 + 2y + 1 \geq\,0 \Rightarrow 3y^2-2y-1\,\leq\,0$
$\Rightarrow (3y + 1) (y - 1) \leq \,0 \Rightarrow -\frac{1}{3}\,\leq y \leq 1 (y \neq 0)$
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 $\left[-\frac{1}{3},1\right]$