What is the set formed by all positive real values of $x$ such that $(x-1)(x-2)(x-3)<0$?
(a) $(1,2)$
(b) $(1,3)$
(c) $(0,1) \cup(2,3)$
(d) $(0,3)$
(e) $(0,2)$
The correct option is (c).

For the product of three factors to be negative, we must have two positive factors and one negative factor, or all three negative. The possibilities are as follows.

$(+)(+)(-)=-$ $(+)(-)(+)=-$ $(-)(+)(+)=-$ $(-)(-)(-)=-$

1) $\underbrace{(x-1)}_{+} \underbrace{(x-2)}_{+} \underbrace{(x-3)}_{-}$.

This is equivalent to $x>1, x>2$ and $x<3$, which means $2<x<3$.
2) $\underbrace{(x-1)}_{+} \underbrace{(x-2)}_{-} \underbrace{(x-3)}_{+}$.

This is equivalent to $x>1, x<2$ and $x>3$, which is not possible. Therefore, this case cannot occur.
3) $\underbrace{(x-1)}_{-} \underbrace{(x-2)}_{+} \underbrace{(x-3)}_{+}$.

This is equivalent to $x<1, x>2$ and $x>3$, which is not possible. Therefore, this case cannot occur.
![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-057.jpg?height=438&width=642&top_left_y=2214&top_left_x=1272)

4) $\underbrace{(x-1)}_{-} \underbrace{(x-2)}_{-} \underbrace{(x-3)}_{-}$.

This is equivalent to $x<1, x<2$ and $x<3$, or $x<1$.

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-058.jpg?height=186&width=583&top_left_y=318&top_left_x=1205)

Therefore, the only real numbers $x$ satisfying $(x-1)(x-2)(x-3)<0$ are the real numbers $x$ such that $x<1$ or $2<x<3$. Thus, the union of intervals $(0,1) \cup(2,3)$ is the set formed by all positive real numbers $x$ such that $(x-1)(x-2)(x-3)<0$.