2. For what values of $a$ does the equation $x \cdot|x-a|=1$ have three distinct solutions?
Answer: $a>2$.

## Solution:

We will construct the graph of the function $y=x|x-a|$ and find the number of intersection points with the horizontal line $y=1$ for different values of the parameter (more precisely, we will determine for which values of the parameter there will be three intersection points).

When $a=0$, the function $y=x|x-a|(y=x|x|)$ is strictly increasing, and the graph intersects the line $y=1$ at a single point.

When $a>0$, we will construct the graph of the function $y=x|x-a|-$ schematically, as shown in the figure on the right. The line $y=1$ intersects this graph at three points only if the vertex of the parabola $\left(a / 2, a^{2} / 4\right)$ lies above it.

This means that $a^{2} / 4>1$. From this, for $a>0$, we get the answer: $a>2$.

![](https://cdn.mathpix.com/cropped/2024_05_06_507ced5c03720ddcb08bg-2.jpg?height=345&width=482&top_left_y=1027&top_left_x=1455)

Similarly, we consider the values $a<0$ (the graph is symmetric to the one considered above with respect to the origin), for which the line $y=1$ has only one intersection point with the graph of the function $y=x|x-a|$.

## Instructions for Checking:

Answer without justification - 3 points.

If the case $a=0$ is omitted, then 3 points.

If the case $a=0$ is considered as part of some other case (for example, $a \leq 0$), it is naturally not considered omitted.