Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.8 AIME Prove it China

Let $a$ , $b$ be real numbers, and $f(x) = a x + b$ satisfies $|f(x)| \le 1$ for any $x \in [0, 1]$ . Then the maximum of $a b$

is ______.

Solution

It is easy to find that $a = f(1) - f(0)$ , $b = f(0)$ . Then
$ab = f(0) \cdot (f(1) - f(0)) = - \left( f(0) - \frac{1}{2} f(1) \right)^2 + \frac{1}{4} (f(1))^2 \le \frac{1}{4} (f(1))^2 \le \frac{1}{4}.$
When $2 f(0) = f(1) = \pm 1$ , i.e., $a = b = \pm \frac{1}{2}$ , we get $a b = \frac{1}{4}$ .
Therefore, the maximum of $a b$ is $\frac{1}{4}$ .
The answer is $\frac{1}{4}$ .

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