
Problem 9.1. Let $f(x)=x^{2}-2 a x-a^{2}-\frac{3}{4}$ where $a$ is real parameter. Find all values of $a$ such that the inequality $|f(x)| \leq 1$ holds for any $x$ in the interval $[0,1]$.


Solution: Let $M$ and $m$ be the maximum and minimum values of $f(x)$ in the interval $[0,1]$. Then the condition of the problem is equivalent to $M \leq 1$ and $m \geq-1$. There are three cases to consider.

Case 1: $a \in[0,1]$. Then $m=f(a)=-2 a^{2}-\frac{3}{4}$ and $M=f(0)=$ $-a^{2}-\frac{3}{4}$ or $M=f(1)=-a^{2}-2 a+\frac{1}{4}$. It follows from $m \geq-1$ and $M \leq 1$ that $a \in\left[-\frac{\sqrt{2}}{4}, \frac{\sqrt{2}}{4}\right]$ and $a \in\left(-\infty,-\frac{3}{2}\right] \cup\left[-\frac{1}{2}, \infty\right)$. In this case the solution is $a \in\left[0, \frac{\sqrt{2}}{4}\right]$.

Case 2: $a1$. Now $m=f(1)$ and $M=f(0)$. It follows from $m \geq-1$ and $M \leq 1$ that $a \in\left[-\frac{5}{2}, \frac{1}{2}\right]$ which is a contradiction with $a>1$.

Thus the solution is $a \in\left[-\frac{1}{2}, \frac{\sqrt{2}}{4}\right]$.
