Let $A = [-2, 4]$, $B = \{x: x^2 - ax - 4 \leq 0\}$. If $B \subseteq A$, then the range of real $a$ is:
Step-by-Step Solution
Key Concept: The solution set $B$ is an interval between the roots of the quadratic, so $B \subseteq A$ requires both roots to lie within $A = [-2, 4]$.
For $B \subseteq A = [-2, 4]$, we need all solutions of $x^2 - ax - 4 \leq 0$ to lie within $[-2, 4]$. The quadratic $f(x) = x^2 - ax - 4$ has roots $x = \frac{a \pm \sqrt{a^2 + 16}}{2}$. Let $r_1 = \frac{a - \sqrt{a^2 + 16}}{2}$ (smaller root) and $r_2 = \frac{a + \sqrt{a^2 + 16}}{2}$ (larger root). Then $B = [r_1, r_2]$ and we need $-2 \leq r_1$ and $r_2 \leq 4$. From $r_1 \geq -2$: $\frac{a - \sqrt{a^2 + 16}}{2} \geq -2$ gives $a \geq 0$. From $r_2 \leq 4$: $\frac{a + \sqrt{a^2 + 16}}{2} \leq 4$ gives $a < 3$. Therefore, $a \in [0, 3)$.
Correct Answer: 4