4. Determine all real numbers $a$ for which no number $x$ satisfying

$$
a x^{2}+\left(1-a^{2}\right) x-a>0
$$

has an absolute value greater than 2.
Solution. For any $a \geq 0$, there always exist real values of $x$ with arbitrarily large absolute values that satisfy the given inequality. Therefore, let us assume that $a<0$. Then the condition stated in the problem is equivalent to the condition that both roots of the equation

$$
a x^{2}+\left(1-a^{2}\right) x-a=0
$$

by absolute value are less than or equal to 2 (it is easy to check that its roots are always real and distinct). Since these roots are $x_{1}=a$ and $x_{2}=-\frac{1}{a}$, we get that the stated condition is satisfied if and only if $-2 \leq a \leq-\frac{1}{2}$.

## III year