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(D) The given function is $y = \frac{|x-x^2|}{x^2-x}$.First,we simplify the expression inside the absolute value: $x-x^2 = -x(x-1)$.Thus,$|x-x^2| = |-

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(D) The given function is $y = \frac{|x-x^2|}{x^2-x}$.First,we simplify the expression inside the absolute value: $x-x^2 = -x(x-1)$.Thus,$|x-x^2| = |-x(x-1)| = |x| \cdot |x-1|$.The denominator is $x^2-x = x(x-1)$.So,$y = \frac{|x| \cdot |x-1|}{x(x-1)}$.The function is defined for $x eq 0$ and $x eq 1$.Case $1$: If $x > 1$,then $x > 0$ and $x-1 > 0$. Thus,$|x| = x$ and $|x-1| = x-1$. $y = \frac{x(x-1)}{x(x-1)} = 1$.Case $2$: If $0 0$ and $x-1 $y = \frac{x \cdot -(x-1)}{x(x-1)} = -1$.Case $3$: If $x $y = \frac{-x \cdot -(x-1)}{x(x-1)} = \frac{x(x-1)}{x(x-1)} = 1$.Combining these,the function is $y = 1$ for $x 1$,and $y = -1$ for $0 < x < 1$. The points $x=0$ and $x=1$ are excluded from the domain.

Which one of the following best represents the graph of $y = \frac{|x-x^2|}{x^2-x}$?

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