2006/07 AHSMC Part 1 Problem 16

Note that $p(-2) < p(-1) > p(0) < p(1) > p(2)$. Hence the graph of this fourth degree polynomial opens down. Let $p(x) = -x^4 + ax^3 + bx^2 + cx + d$. Then $d = p(0) = 2$. From $p(\pm 2) = 2$, we have $-16 \pm 8a + 4b \pm 2c + 2 = 2$. Hence $4b - 16 = \pm(8a + 2c)$. This is only possible if both sides are equal to $0$, so that $b = 4$. Similarly, from $p(\pm 1) = 5$, we have $25a + 5c = 0$. Combining with $8a + 2c = 0$, we have $a = c = 0$, so that $p(x) = -x^4 + 4x^2 + 2 = 6 - (x^2 - 2)^2$. It follows that the maximum value of $p(x)$ is $6$, occurring when $x^2 - 2 = 0$ or $x = \pm\sqrt{2}$. The answer is (b).