10.5. Determine all functions $f: R \rightarrow R$, which simultaneously satisfy the conditions:

1) $|f(x)| \geq 1$, for any real number $x$.
2) $f(x+y)=\frac{f(x)+f(y)}{1+f(x) \cdot f(y)}$, for any real numbers $x$ and $y$.
Solution: Since the second relation holds for any real values of the numbers $x$ and $y$, $f(x) \cdot f(y) \neq -1, (\forall) x \in R, (\forall) y \in R$.

The second relation is equivalent to $f(x+y) \cdot (1+f(x) \cdot f(y)) = f(x) + f(y), (\forall) x \in R, (\forall) y \in R$.

In particular, for $y=0$, we obtain $f(x) \cdot (1+f(x) \cdot f(0)) = f(x) + f(0) \Leftrightarrow$

$\Leftrightarrow f(0) \cdot (1-f^2(x)) = 0, (\forall) x \in R$.

Since $|f(x)| \geq 1$, for any real number $x$, then $f(0) \neq 0$.

We obtain that $1-f^2(x) = 0, (\forall) x \in R$.

We will show that the function $f$ can be identically equal to 1 or identically equal to -1 on $R$.

If we assume that for some real $x$ we have $f(x) = 1$, and for some real $y$, $y \neq x$, we will have $f(y) = -1$, we will obtain $1 + f(x) \cdot f(y) = 0$, so the second relation will not be satisfied.

Answer: $f: R \rightarrow R, f(x) = 1$ or $f: R \rightarrow R, f(x) = -1$