The values of $a$ for which the integral $\int_0^2 |x - a|dx \geq 1$ is satisfied are:
Step-by-Step Solution
Key Concept: Split the absolute value integral into cases based on the position of $a$ relative to the integration bounds, then solve the resulting algebraic inequality in each region.
For $a \leq 0$: $\int_0^2 |x-a| dx \geq 1$ gives $\frac{a^2}{2} + 2 - 2a \geq 1$, yielding $a \leq -\frac{1}{2}$. For $0 < a < 2$: splitting the integral $\int_0^a (a-x)dx + \int_a^2 (x-a)dx \geq 1$ gives $\frac{a^2}{2} + 2 - 2a + \frac{a^2}{2} - 1 \geq 0$, so $(a-1)^2 \geq 0$ (always true, with equality at $a=1$). For $a \geq 2$: $\int_0^2 (a-x)dx \geq 1$ gives $a \geq \frac{3}{2}$, satisfied for all $a \geq 2$. Thus $a \in (-\infty, -\frac{1}{2}] \cup [1,2] \cup [2,\infty) = (-\infty, -\frac{1}{2}] \cup [1,\infty)$.
Correct Answer: 1,2,3