2019 AMC 10B Problem 9

Below is the professionally curated solution for Problem 9 of the 2019 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2019 AMC 10B solutions, or check the answer key.

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Concepts:floor and ceiling functionsabsolute valuecasework

Difficulty rating: 1370

9.

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

{1,0} \{-1, 0\}

The set of nonpositive integers

{1,0,1}\{-1, 0, 1\}

{0} \{0\}

The set of nonnegative integers

Solution:

If xx was an integer, then xx=xx=0.\begin{align*}\lfloor|x|\rfloor - |\lfloor x \rfloor| &= |x| -|x|\\&=0.\end{align*}

If xx was positive, then xx=xx=0.\begin{align*}\lfloor|x|\rfloor - |\lfloor x \rfloor| &= \lfloor x \rfloor - \lfloor x \rfloor \\&= 0.\end{align*}

Now, we must look at negative non-integers. If we have a negative non-integer, then x\lfloor|x|\rfloor would negate xx and then round down, while x|\lfloor x \rfloor| would round down then negate it, effectively negating it and rounding up.

The first one rounds down and the second one rounds up, the second one is 11 larger than the first, making f=1.f=-1.

Therefore, the range is {1,0}.\{-1,0\}.

Thus, the answer is A .

Problem 9 in Other Years