2021 AMC 10 A Fall

Complete problem set with solutions and individual problem pages

Problem 16 Hard

The graph of f(x)=|\lfloor x\rfloor|-|\lfloor 1-x\rfloor \mid is symmetric about which of the following? (Here \lfloor x\rfloor is the greatest integer not exceeding x(2021 AMC Fall 10A, Question #16)

  • A.

    the y-axis

  • B.

    the line x=1

  • C.

    the origin

  • D.

    the point \left(\frac{1}{2}, 0\right)

  • E.

    the point (1,0)

Answer:D

Solution1:

1. \lfloor x+n\rfloor=\lfloor x\rfloor+n and \lceil x+n\rceil=\lceil x\rceil+n 2. \lfloor-x\rfloor=-\lceil x\rceil 3. \lceil x\rceil-\lfloor x\rfloor= \begin{cases}0 & \text { if } x \in \mathbb{Z} \\ 1 & \text { if } x \notin \mathbb{Z}\end{cases} We rewrite f(x) as \begin{aligned} f(x) &=|\lfloor x\rfloor|-|\lfloor 1-x\rfloor| \\ &=|\lfloor x\rfloor|-|-\lceil x-1\rceil| \\ &=|\lfloor x\rfloor|-|-\lceil x\rceil+1| \end{aligned} We apply casework to the value of x : 1. x \in \mathbb{Z}^{-} It follows that f(x)=-x-(-x+1)=-1. 2. x=0 It follows that f(x)=0-1=-1. 3. x \in \mathbb{Z}^{+} It follows that f(x)=x-(x-1)=1. 4. x \notin \mathbb{Z} and x<0 It follows that f(x)=-\lfloor x\rfloor-(-\lceil x\rceil+1)=(\lceil x\rceil-\lfloor x\rfloor)-1=0. 5. x \notin \mathbb{Z} and 0<x<1 It follows that f(x)=0-0=0. 6. x \notin \mathbb{Z} and x>1 It follows that f(x)=\lfloor x\rfloor-(\lceil x\rceil-1)=(\lfloor x\rfloor-\lceil x\rceil)+1=0. Together, we have f(x)= \begin{cases}-1 & \text { if } x \in \mathbb{Z}^{-} \cup\{0\} \\ 1 & \text { if } x \in \mathbb{Z}^{+} \\ 0 & \text { if } x \notin \mathbb{Z}\end{cases} so the graph of f(x) is symmetric about (D) the point \left(\frac{1}{2}, 0\right). Alternatively, we can eliminate (A), (B), (C), and (E) once we finish with Case 3 . This leaves us with (D).

Solution 2:

Denote x=a+b, where a \in \mathbb{Z} and b \in[0,1). Hence, a is the integer part of x and b is the decimal part of x.

Case 1: b=0. We have f(x)=|\lfloor x\rfloor|-|\lfloor 1-x\rfloor| =|a|-|1-a| = \begin{cases}a-(a-1) & \text { if } a \in \mathbb{Z} \text { and } a \geq 1 \\ -1 & \text { if } a=0 \\ -a-(1-a) & \text { if } a \in \mathbb{Z} \text { and } a \leq-1\end{cases} = \begin{cases}1 & \text { if } a \in \mathbb{Z} \text { and } a \geq 1 \\ -1 & \text { if } a=0 \\ -1 & \text { if } a \in \mathbb{Z} \text { and } a \leq-1\end{cases} = \begin{cases}1 & \text { if } a \in \mathbb{Z} \text { and } a \geq 1 \\ -1 & \text { if } a \in \mathbb{Z} \text { and } a \leq 0\end{cases}

Case 2: b \neq 0. We have \begin{aligned} f(x) &=|\lfloor x\rfloor|-|\lfloor 1-x\rfloor| \\ &=|a|-|\lfloor 1-a-b\rfloor| \\ &=|a|-|\lfloor-a+(1-b)\rfloor| \\ &=|a|-|-a| \\ &=0 . \end{aligned} Therefore, the graph of f(x) is symmetric through the point \left(\frac{1}{2}, 0\right). Therefore, the answer is (D) the point \left(\frac{1}{2}, 0\right)

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