5. Depending on the real parameter $a$, determine how many different real solutions the equation

$$
x^{3}-3 x=a \text {. }
$$

## NATIONAL MATHEMATICS COMPETITION FOR HIGH SCHOOL STUDENTS, 10. 3. 2017.

## First Grade - A Category
5. The solutions of the observed equation are the zeros of the polynomial $f(x)=x^{3}-3 x-a$. Since this is a third-degree polynomial, it has at most 3 real zeros. Considering $\lim _{x \rightarrow-\infty} f(x)=$

| 1 | 1 | 1 | 1 |
| :---: | :---: | :---: | :---: |
| 1 | 1 | 1 | -1 |
| 1 | 0 | -1 | -1 |
| 0 | -1 | -1 | -1 |

Ok 2018 4B 4 $-\infty$ and $\lim _{x \rightarrow \infty} f(x)=\infty$, and since the polynomial is a continuous function, it follows that $f(x)$ must have at least one zero. The derivative of this function is $f^{\prime}(x)=3 x^{2}-3=3(x-1)(x+1)$, so since the derivative has zeros at points $x=-1$ and $x=1$, it is negative on the interval $(-1,1)$, and positive otherwise, we conclude that $f(x)$ increases on the intervals $(-\infty,-1)$ and $(1, \infty)$, decreases on $(-1,1)$, has a local maximum at $x=-1$, and a local minimum at $x=1$. On each of the intervals $(-\infty,-1),(-1,1)$, and $(1, \infty)$, the polynomial $f(x)$ can have at most one zero, due to monotonicity on those intervals. The polynomial will have a total of three distinct real zeros if and only if $f(-1)>0$ and $f(1)<0$, i.e., $2-a>0$ and $-2-a<0$,
it will have two distinct real zeros (one of which is a double zero) if and only if $f(-1)=0$ or $f(1)=0$, and otherwise (i.e., if and only if the values of $f(-1)$ and $f(1)$ are different from 0 and of the same sign) only one real zero.

Thus, the observed equation has three distinct real solutions for $a \in (-2,2)$, two distinct real solutions for $a \in\{-2,2\}$, and only one real solution for $a \in(-\infty,-2) \cup(2, \infty)$.

## SOLUTIONS TO PROBLEMS FROM THE NATIONAL MATHEMATICS COMPETITION FOR HIGH SCHOOL STUDENTS, 10. 3. 2018.

## First Grade - A category