2010/11 AHSMC Part 1 Problem 8

Since $\frac{1}{x} \le -3$, $x \le 0$. Multiplying by $-\frac{x}{3}$, we have $-\frac{1}{3} \le x$. Hence the set of all values of $x$ is $\bigl\{-\frac{1}{3} \le x < 0\bigr\}$. The answer is (D).