## Solution.

Let's write out the first few values of the given sequence of functions for $x=4$.

$$
\begin{aligned}
& f_{1}(4)=\frac{1}{2-4}=-\frac{1}{2} \\
& f_{2}(4)=\left(f_{1} \circ f_{1}\right)(4)=\frac{1}{2+\frac{1}{2}}=\frac{2}{5} \\
& f_{3}(4)=\left(f_{1} \circ f_{2}\right)(4)=\frac{1}{2-\frac{2}{5}}=\frac{5}{8} \\
& f_{4}(4)=\left(f_{1} \circ f_{3}\right)(4)=\frac{1}{2-\frac{5}{8}}=\frac{8}{11}=\frac{3 \cdot 4-4}{3 \cdot 4-1}
\end{aligned}
$$

Notice that each term of this sequence is of the form $f_{n}(4)=\frac{3 n-4}{3 n-1}$.

We will prove by mathematical induction that the statement $(*)$ is true for all natural numbers $n$.

For $n=1$, the statement is true, as we have already verified.

1 point

Assume that the statement is true for some natural number $n$, i.e., $f_{n}(4)=\frac{3 n-4}{3 n-1}$, for some $n \in \mathbb{N}$.

We need to prove that the statement is also true for the next natural number, $n+1$. We write:

$f_{n+1}(4)=\frac{1}{2-f_{n}(4)}$, by the assumption $f_{n}(4)=\frac{3 n-4}{3 n-1}$, so

$f_{n+1}(4)=\frac{1}{2-\frac{3 n-4}{3 n-1}}=\frac{3 n-1}{3 n+2}=\frac{3(n+1)-4}{3(n+1)-1}$.

Thus, from the fact that the statement $(*)$ is true for some $n \in \mathbb{N}$, we have shown that it is also true for $n+1$, so the statement $(*)$ is true for all natural numbers $n$.

Therefore, it is also true for $n=2021$, so $f_{2021}(4)=\frac{3 \cdot 2021-4}{3 \cdot 2021-1}=\frac{6059}{6062}$.

Note: A student who did not prove the general statement $(*)$ by induction can receive a maximum of 5 points.

A student could have proven the statement $f_{n}(x)=\frac{n-(n-1)(x)}{(n+1)-n x}$ by induction, which should be graded analogously to the proposed scoring.
## Task B-4.6.

Let $f_{1}(x)=\frac{1}{2-x}, f_{n}(x)=\left(f_{1} \circ f_{n-1}\right)(x), n \geqslant 2$, for all real numbers $x$ for which the given functions are defined. What is $f_{2021}(4)$?