Difference between revisions of "2019 AMC 10B Problems/Problem 9"
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− | ==Problem== | + | == Problem == |
+ | The function <math>f</math> is defined by <cmath>f(x) = \lfloor|x|\rfloor - |\lfloor x \rfloor|</cmath>for all real numbers <math>x</math>, where <math>\lfloor r \rfloor</math> denotes the greatest integer less than or equal to the real number <math>r</math>. What is the range of <math>f</math>? | ||
+ | |||
+ | <math>\textbf{(A) } \{-1, 0\} \qquad\textbf{(B) } \text{The set of nonpositive integers} \qquad\textbf{(C) } \{-1, 0, 1\} \qquad\textbf{(D) } \{0\} \qquad \textbf{(E) } \text{The set of nonnegative integers} </math> | ||
+ | |||
+ | == Solution 1 == | ||
+ | There are four cases we need to consider here. | ||
+ | |||
+ | '''Case 1''': <math>x</math> is a positive integer. Without loss of generality, assume <math>x=1</math>. Then <math>f(1) = 1 - 1 = 0</math>. | ||
+ | |||
+ | '''Case 2''': <math>x</math> is a positive fraction. Without loss of generality, assume <math>x=\frac{1}{2}</math>. Then <math>f\left(\frac{1}{2}\right) = 0 - 0 = 0</math>. | ||
+ | |||
+ | '''Case 3''': <math>x</math> is a negative integer. Without loss of generality, assume <math>x=-1</math>. Then <math>f(-1) = 1 - 1 = 0</math>. | ||
− | + | '''Case 4''': <math>x</math> is a negative fraction. Without loss of generality, assume <math>x=-\frac{1}{2}</math>. Then <math>f\left(-\frac{1}{2}\right) = 0 - 1 = -1</math>. | |
+ | |||
+ | Thus the range of the function <math>f</math> is <math>\boxed{\textbf{(A) } \{-1, 0\}}</math>. | ||
+ | |||
+ | ~IronicNinja | ||
+ | |||
+ | == Solution 2 == | ||
+ | It is easily verified that when <math>x</math> is an integer, <math>f(x)</math> is zero. We therefore need only to consider the case when <math>x</math> is not an integer. | ||
+ | |||
+ | When <math>x</math> is positive, <math>\lfloor x\rfloor \geq 0</math>, so | ||
+ | <cmath>\begin{split}f(x)&=\lfloor|x|\rfloor-|\lfloor x\rfloor| \\ | ||
+ | &=\lfloor x\rfloor-\lfloor x\rfloor \\ | ||
+ | &=0\end{split}</cmath> | ||
+ | |||
+ | When <math>x</math> is negative, let <math>x=-a-b</math> be composed of integer part <math>a</math> and fractional part <math>b</math> (both <math>\geq 0</math>): | ||
+ | <cmath>\begin{split}f(x)&=\lfloor|-a-b|\rfloor-|\lfloor -a-b\rfloor| \\ | ||
+ | &=\lfloor a+b\rfloor-|-a-1| \\ | ||
+ | &=a-(a+1)=-1\end{split}</cmath> | ||
+ | |||
+ | Thus, the range of x is <math>\boxed{\textbf{(A) } \{-1, 0\}}</math>. | ||
+ | |||
+ | ''Note'': One could solve the case of <math>x</math> as a negative non-integer in this way: | ||
+ | <cmath>\begin{split}f(x)&=\lfloor|x|\rfloor-|\lfloor x\rfloor| \\ | ||
+ | &=\lfloor -x\rfloor-|-\lfloor -x\rfloor-1| \\ | ||
+ | &=\lfloor -x\rfloor-(\lfloor -x\rfloor+1) = -1\end{split}</cmath> | ||
+ | |||
+ | == Solution 3 (Formal) == | ||
+ | Let {<math>x</math>} denote the fractional part of <math>x</math>; for example, {<math>2.7</math>}<math>= 0.7</math>, and {<math>-1.3</math>}<math>= 0.3</math>. | ||
+ | Then for <math> x \geq 0</math>, <math> x = \lfloor x \rfloor +</math> {<math>x</math>} and for <math> x < 0</math>, <math> x = \lfloor x \rfloor + 1 - </math>{<math>x</math>}. | ||
+ | |||
+ | Now we can rewrite <math>\lfloor |x| \rfloor - |\lfloor x \rfloor|</math>, breaking the expression up based on whether <math> x \geq 0 </math> or <math> x < 0</math>. | ||
+ | |||
+ | For <math>x \geq 0</math>, the above expression is equal to <math> \lfloor |\lfloor x \rfloor + </math>{<math>x</math>}<math>| \rfloor - | \lfloor \lfloor x \rfloor + </math> {<math>x</math>}<math> \rfloor | \implies \lfloor \lfloor x \rfloor + </math>{<math>x</math>}<math> \rfloor - | \lfloor x \rfloor | </math> | ||
+ | |||
+ | <math> \implies \lfloor x \rfloor - \lfloor x \rfloor = \mathbf{0} </math>. | ||
+ | |||
+ | For <math> x < 0</math>, the expression is equal to <math> \lfloor |\lfloor x \rfloor + 1 - </math>{<math>x</math>}<math>| \rfloor - | \lfloor \lfloor x \rfloor + 1 - </math> {<math>x</math>}<math> \rfloor |</math> | ||
− | <math>\ | + | <math> \implies \lfloor - \lfloor x \rfloor - 1 + </math>{<math>x</math>}<math> \rfloor - | \lfloor x \rfloor | \implies - \lfloor x \rfloor - 1 - (- \lfloor x \rfloor) = \mathbf{-1}</math>. |
− | + | Therefore the only two possible values for <math>f(x)</math>, and thus the range of the function, is <math>\boxed{\textbf{(A) } \{-1, 0\}}</math>. | |
− | + | ~KingRavi | |
− | + | == Solution 4 == | |
+ | We have 2 cases: either <math>x</math> is positive or <math>x</math> is negative. | ||
− | |||
− | Case | + | '''Case 1 - x is positive:''' |
− | + | Let <math>x = n + f</math>, where <math>n</math> is a positive integer and <math>f</math> is a positive real number between 0 and 1. We have | |
+ | <cmath>\lfloor |x| \rfloor = \lfloor |n+f| \rfloor = \lfloor n+f \rfloor = n</cmath> and | ||
+ | <cmath>|\lfloor x \rfloor| = |\lfloor n+f \rfloor| = |n| = n.</cmath> | ||
+ | <math>n-n=0</math>, so the possible value of <math>f(x)</math> if <math>x</math> is positive is <math>0</math>. | ||
− | |||
− | + | '''Case 2 - x is negative: ''' | |
− | == | + | Let <math>x = -n - f</math>, where <math>n</math> is a positive integer and <math>f</math> is a positive real number between 0 and 1. We have |
+ | <cmath>\lfloor |x| \rfloor = \lfloor |-n-f| \rfloor = \lfloor n+f \rfloor = n</cmath> and | ||
+ | <cmath>|\lfloor x \rfloor| = |\lfloor -n-f \rfloor| =|-n|\:or\: |-n-1|= n \:or\: n+1.</cmath> | ||
− | + | <math>n-n=0</math> and <math>n-(n+1) = -1</math>, so the possible values of <math>f(x)</math> if <math>x</math> is negative are <math>0</math> and <math>-1.</math> | |
− | |||
− | |||
− | |||
− | |||
− | + | Hence, the possible values of <math>f(x)</math> are <math>0</math> and <math>-1</math>, so the answer is <math>\boxed{\textbf{(A) } \{-1, 0\}}</math>. ~azc1027 | |
− | |||
− | |||
− | |||
− | + | ==Video Solution== | |
+ | https://youtu.be/LffjyNNqf14 | ||
+ | ~Education, the Study of Everything | ||
− | + | == Video Solution == | |
− | + | https://youtu.be/PgqjsTkNYdc | |
− | |||
− | |||
− | + | ~savannahsolver | |
+ | == See Also == | ||
{{AMC10 box|year=2019|ab=B|num-b=8|num-a=10}} | {{AMC10 box|year=2019|ab=B|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 16:07, 4 July 2023
Contents
Problem
The function is defined by
for all real numbers
, where
denotes the greatest integer less than or equal to the real number
. What is the range of
?
Solution 1
There are four cases we need to consider here.
Case 1: is a positive integer. Without loss of generality, assume
. Then
.
Case 2: is a positive fraction. Without loss of generality, assume
. Then
.
Case 3: is a negative integer. Without loss of generality, assume
. Then
.
Case 4: is a negative fraction. Without loss of generality, assume
. Then
.
Thus the range of the function is
.
~IronicNinja
Solution 2
It is easily verified that when is an integer,
is zero. We therefore need only to consider the case when
is not an integer.
When is positive,
, so
When is negative, let
be composed of integer part
and fractional part
(both
):
Thus, the range of x is .
Note: One could solve the case of as a negative non-integer in this way:
Solution 3 (Formal)
Let {} denote the fractional part of
; for example, {
}
, and {
}
.
Then for
,
{
} and for
,
{
}.
Now we can rewrite , breaking the expression up based on whether
or
.
For , the above expression is equal to
{
}
{
}
{
}
.
For , the expression is equal to
{
}
{
}
{
}
.
Therefore the only two possible values for , and thus the range of the function, is
.
~KingRavi
Solution 4
We have 2 cases: either is positive or
is negative.
Case 1 - x is positive:
Let , where
is a positive integer and
is a positive real number between 0 and 1. We have
and
, so the possible value of
if
is positive is
.
Case 2 - x is negative:
Let , where
is a positive integer and
is a positive real number between 0 and 1. We have
and
and
, so the possible values of
if
is negative are
and
Hence, the possible values of are
and
, so the answer is
. ~azc1027
Video Solution
~Education, the Study of Everything
Video Solution
~savannahsolver
See Also
2019 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.