Task B-3.3. Determine the real numbers $a, b$ and $c$ for which the function

$$
f(x)=a x^{2}+b x-c \sin x \cos x
$$

is odd.
## Solution.

$$
\begin{aligned}
& f(x)=a x^{2}+b x-c \sin x \cos x=a x^{2}+b x-\frac{c}{2} \sin 2 x \\
& f(-x)=a(-x)^{2}+b(-x)-\frac{c}{2} \sin 2(-x)=a x^{2}-b x+\frac{c}{2} \sin 2 x
\end{aligned}
$$

From $f(-x)=-f(x)$ for every $x \in \mathbb{R}$, it follows that

$$
\begin{aligned}
a x^{2}-b x+\frac{c}{2} \sin 2 x & =-a x^{2}-b x+\frac{c}{2} \sin 2 x \\
a x^{2} & =-a x^{2}
\end{aligned}
$$

(2 points)

This is satisfied for all real numbers $x$ if

$$
a=0, \quad b \in \mathbb{R}, \quad c \in \mathbb{R}
$$