## Task B-2.5.

The function $f$ is defined by the rule of assignment $f(x)=x^{2}-\sqrt{b} \cdot x+c$, where $b$ and $c$ are real parameters. What is the probability that, upon random selection of parameters $b$ and $c$ from the interval $[0,10]$, the minimum value of the given function will be greater than or equal to 2 and less than or equal to 3?
## Solution.

The space of all elementary events is the set

$$
\Omega=\left\{(b, c) \in \mathbb{R}^{2} \mid 0 \leqslant b \leqslant 10,0 \leqslant c \leqslant 10\right\} .
$$

Obviously, $m(\Omega)=100$, where $m$ measures the area of the set $\Omega$. In the coordinate plane where the parameter $b$ is in the interval $[0,10]$ on the $x$-axis, and the parameter $c$ is in the interval $[0,10]$ on the $y$-axis, this is a square with a side length of 10, or an area of $100 \text{ sq. units}$.

Let the event $A \subseteq \Omega$ be the set of all ordered pairs $(b, c)$ that satisfy the conditions of the problem. Determine the set $A$.

The minimum value of the function $f(x)$ is $\frac{4 c-b}{4}$.

According to the conditions of the problem, it must hold that $2 \leqslant \frac{4 c-b}{4} \leqslant 3$, or $4 c-b \leqslant 12, 4 c-b \geqslant 8$, so

$$
A=\{(b, c) \in \Omega \mid 4 c-b \leqslant 12, 4 c-b \geqslant 8\}
$$

In the coordinate plane, the set $A$ is represented by a parallelogram as shown in the figure:

![](https://cdn.mathpix.com/cropped/2024_05_30_aa38f3c63b8013e55583g-13.jpg?height=857&width=870&top_left_y=291&top_left_x=593)

Now, let's determine the area of the set $A$. Note that this is a parallelogram with a side length of 1 and a height of 10, so $m(A)=10$.

Finally, we get

$$
p(A)=\frac{m(A)}{m(\Omega)}=\frac{10}{100}=0.1
$$

## NATIONAL MATHEMATICS COMPETITION

## 3rd grade - high school - B variant

## May 11, 2022.