5. Find the greatest value of the parameter a for which the equation:

$$
(|x-2|+2 a)^{2}-3(|x-2|+2 a)+4 a(3-4 a)=0 \text { has three solutions. }
$$

In the answer, specify the greatest of them.
Selection (online) stage of the "Step into the Future" School Students' Olympiad in the subject of Mathematics

Let $|\mathrm{x}-2|+2 \mathrm{a}=\mathrm{t}$, then

$$
t^{2}-3 t+4 a(3-4 a)=0
$$

$t^{2}-2 \cdot \frac{3}{2} \mathrm{t}+\frac{4}{9}-\frac{4}{9}+12 \mathrm{a}-16 a^{2}=0$

$\left(\mathrm{t}-\frac{3}{2}\right)^{2}-\left(\frac{9}{4}-12 \mathrm{a}+16 a^{2}\right)=0$;

$\left(\mathrm{t}-\frac{3}{2}\right)^{2}-\left(4 \mathrm{a}-\frac{3}{2}\right)^{2}=0 ;\left(\mathrm{t}-\frac{3}{2}+4 \mathrm{a}-\frac{3}{2}\right) \cdot\left(\mathrm{t}-\frac{3}{2}-4 \mathrm{a}+\frac{3}{2}\right)=0$;

$(t+4 a-3) \cdot(t-4 a)=0$

We get $(|x-2|+2 a+4 a-3) \cdot(|x-2|+2 a-4 a)=0$

$$
\begin{aligned}
& |\mathrm{x}-2|+6 \mathrm{a}-3=0 \\
& |\mathrm{x}-2|-2 \mathrm{a}=0 ;
\end{aligned} \quad\left[\begin{array}{l}
a=-\frac{1}{6} \cdot|x-2|+\frac{1}{2} \\
a=\frac{1}{2} \cdot|x-2|
\end{array}\right.
$$

Let's plot the graphs of the equations

$a=-\frac{1}{6}|x-2|+\frac{1}{2}$ and

$\mathrm{a}=\frac{1}{2}|\mathrm{x}-2|$ in the XOA system

The equation has three solutions when $\mathrm{a}=0, \mathrm{a}=0.5$

Answer: 0.5

![](https://cdn.mathpix.com/cropped/2024_05_06_d112a5cdf1fb6afebffeg-3.jpg?height=911&width=1308&top_left_y=1486&top_left_x=705)

Selection (online) stage of the "Step into the Future" School Students' Olympiad in the subject of Mathematics

6. The test consists of questions with 4 answer options, only one of which is correct. With a probability of $2 / 3$, Ilya knows the correct answer to a question, otherwise he randomly selects an answer option. If Ilya gave the correct answer to a question, what is the probability that he guessed the answer to this question? (Round the answer to the nearest hundredth)

## Solution:

Ilya could have given the correct answer in two ways:

1) If he knew the answer, the probability was $\frac{2}{3} \cdot 1=\frac{2}{3}$
2) If he guessed the answer to this question, the probability was $\frac{1}{3} \cdot \frac{1}{4}=\frac{1}{12}$

We get the answer to the problem: $\frac{\frac{1}{12}}{\frac{2}{3}+\frac{1}{12}}=\frac{1}{9}$

Answer: 0.11