## Task $9 / 72$

Let $a$ and $b$ be real numbers with $a \neq 0,|a| \neq 1$. We are looking for all functions $f(x)$ such that for every real $x \neq 0$ the following relationship holds:

$$
a x \cdot f(x)+\frac{1}{a x} \cdot f\left(\frac{1}{x}\right)=b
$$
Since the relationship should hold for all real $x \neq 0$, it must also hold for $\frac{1}{x}$:

$$
\frac{a}{x} \cdot f\left(\frac{1}{x}\right)+\frac{x}{a} \cdot f(x)=b
$$

Therefore,

$$
\frac{a}{x} \cdot f\left(\frac{1}{x}\right)+\frac{x}{a} \cdot f(x)=a x \cdot f(x)+\frac{1}{a x} \cdot f\left(\frac{1}{x}\right) \rightarrow\left[x f(x)-\frac{1}{x} f\left(\frac{1}{x}\right)\right] \cdot\left(a-\frac{1}{a}\right)=0
$$

Since $|a| \neq 1$, it follows that $a-\frac{1}{a} \neq 0$. Thus, $x f(x)=\frac{1}{x} f\left(\frac{1}{x}\right)$. Substituting this into the required condition, we get

$$
a x \cdot f(x)+\frac{x}{a} \cdot f(x)=b \rightarrow f(x)=\frac{1}{x} \cdot \frac{a b}{a^{2}+1}
$$

A check confirms the correctness of the result.