2019 AMC 10B Problem 9

Split by the sign and integrality of $x$.

Case 1: $x \ge 0$. Then $|x| = x$ and $\lfloor x \rfloor \ge 0$, so $f(x) = \lfloor x \rfloor - \lfloor x \rfloor = 0$.

Case 2: $x$ is a negative integer. Then $|x| = -x$ is an integer, so $\lfloor |x| \rfloor = -x$, and $|\lfloor x \rfloor| = |x| = -x$. Again $f(x) = 0$.

Case 3: $x$ is negative and not an integer. Write $x = -k - \theta$ with $k$ a nonnegative integer and $0 < \theta < 1$. Then $|x| = k + \theta$, so $\lfloor |x| \rfloor = k$. But $\lfloor x \rfloor = -k - 1$, whose absolute value is $k + 1$. Hence $f(x) = k - (k+1) = -1$.

For example, $f(-2.5) = \lfloor 2.5 \rfloor - |\lfloor -2.5 \rfloor| = 2 - 3 = -1$, while $f(3.7) = 3 - 3 = 0$.

The range is $\{-1, 0\}$. The answer is $\boxed{\textbf{(A)}\ \{-1, 0\}}$.

Absolute value and floor do not commute for negative inputs: $\lfloor |x| \rfloor$ truncates toward zero, while $|\lfloor x \rfloor|$ rounds away from zero, and the two differ by exactly $1$ whenever $x$ has a fractional part. Whenever a floor function appears, test a positive number, a negative integer, and a negative non-integer; the third case is where surprises live.

• Testing only positive values or negative integers, concluding the range is {0}, and missing the negative non-integer case.
• Believing the difference can grow with $|x|$ (choice B); the fractional part contributes at most one unit regardless of size.