2021/22 AHSMC Part 1 Problem 14

We first remark that $f(1) = 0$, hence $b = -\dfrac{1}{2} - a$. The given inequality is equivalent to $$-(x-1)^2 \le \frac{1}{2}x^2 + ax - a - \frac{1}{2} \le (x-1)^2,$$ for any real $x$, or $$\frac{3}{2}x^2 + (a-2)x - a + \frac{1}{2} \ge 0 \qquad \text{and} \qquad \frac{1}{2}x^2 - (a+2)x + a + \frac{3}{2} \ge 0,$$ for any real $x$. The second inequality holds if and only if $(a+2)^2 - 2a - 3 \le 0$ or equivalently $(a+1)^2 \le 0$, hence $a = -1$. For this value of $a$ the first inequality holds so $b = \dfrac{1}{2}$ and $f(x) = \dfrac{1}{2}x^2 - x + \dfrac{1}{2}$. Thus $f(4) = \dfrac{9}{2}$. The answer is (B).