## Task B-4.4.

If $f: \mathbb{R} \backslash\{0\} \rightarrow \mathbb{R}, f(x)=(x+a)\left(\frac{1}{x}+b\right)$ is an odd function and $f\left(a^{2}\right)+f(a)=\frac{35}{6}$, determine the real numbers $a$ and $b$.
## Solution.

Since the function $f$ is an odd function, for all real numbers $x$ in its domain, $f(-x) = -f(x)$ holds.

Let $x = 1$. (Any real number different from 0 or generally $x$ can be taken.)

Then $f(-1) = -f(1)$, which means $(-1 + a)(-1 + b) = -(1 + a)(1 + b)$.

After simplifying this expression, we get $ab = -1$.

From the conditions $f(a^2) + f(a) = \frac{35}{6}$ and $b = -\frac{1}{a}$, we have:

\[
\left(a^2 + a\right)\left(\frac{1}{a^2} - \frac{1}{a}\right) + (a + a)\left(\frac{1}{a} - \frac{1}{a}\right) = \frac{35}{6}
\]

which simplifies to the equation:

\[
\frac{1}{a} - a = \frac{35}{6}, \text{ or } 6a^2 + 35a - 6 = 0.
\]

The solutions to this equation are $a_1 = -6$ and $a_2 = \frac{1}{6}$,

1 point

and from $ab = -1$ we get $b_1 = \frac{1}{6}$ and $b_2 = -6$.