2019 AMC 10B Problem 9

P1-10~3 minPrint

Functions Number Properties Split into exhaustive cases and handle each Test small/specific values or special cases to find or verify the answer

The function $f$ is defined by $f(x) = \lfloor|x|\rfloor - |\lfloor x \rfloor|$ for all real numbers $x$ , where $\lfloor r \rfloor$ denotes the greatest integer less than or equal to the real number $r$ . What is the range of $f$ ?

Show solution without answering

Key insight

Try a positive number, a negative whole number, and a negative number with a decimal part. Only the last kind gives something other than 0, and it always gives -1.

Solution

Floors and absolute values behave differently for negative numbers, so test a few kinds of numbers first.

$x$	$\lfloor \lvert x\rvert \rfloor$	$\lvert \lfloor x \rfloor \rvert$	$f(x)$
$3.7$	$\lfloor 3.7 \rfloor = 3$	$\lvert 3 \rvert = 3$	$0$
$-2$	$\lfloor 2 \rfloor = 2$	$\lvert -2 \rvert = 2$	$0$
$-2.5$	$\lfloor 2.5 \rfloor = 2$	$\lvert -3 \rvert = 3$	$-1$

The table suggests the only values are $0$ and $-1$ . Now check that this is true for every $x$ .

$x$ is $0$ or positive. Then $\lvert x \rvert = x$ and $\lfloor x \rfloor$ is $0$ or positive, so both parts equal $\lfloor x \rfloor$ . So $f(x) = 0$ .
$x$ is a negative whole number, like $-5$ . Both parts equal $5$ , the number without its sign. So $f(x) = 0$ .
$x$ is negative with a decimal part, like $-5.3$ . Then $\lvert x \rvert = 5.3$ and its floor is $5$ . But the floor of $-5.3$ is $-6$ (the floor rounds down, and down means further from zero for negatives), and its absolute value is $6$ . So $f(x) = 5 - 6 = -1$ . The same thing happens for any such number: the first part is one less than the second.

Every real number falls in one of these three groups, so the range is exactly $\{-1, 0\}$ .

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

Why this works

Taking the absolute value first and then rounding down moves a negative number toward zero. Rounding down first moves it away from zero. The two orders disagree by exactly $1$ whenever the number is negative and not whole. Whenever a floor appears, test a positive number, a negative whole number and a negative number with a decimal part; the last one is where surprises happen.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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 $\boxed{\textbf{(A)}\ \{-1, 0\}}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 16: Functions
Introduction to Number Theory, Chapter 1: Integers: The Basics
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)
Introduction to Algebra, Chapter 17: Graphing Functions (§17.2)