Difference between revisions of "2019 AMC 10B Problems/Problem 9"
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− | ==Problem== | + | == Problem == |
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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>? | 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\} | + | <math>\textbf{(A) } \{-1, 0\} \qquad\textbf{(B) } \text{The set of nonpositive integers} \qquad\textbf{(C) } \{-1, 0, 1\} \qquad\textbf{(D) } \{0\}</math> <br> <math>\textbf{(E) } \text{The set of nonnegative integers} </math> |
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+ | == Solution 1 == | ||
There are four cases we need to consider here. | There are four cases we need to consider here. | ||
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Thus the range of the function <math>f</math> is <math>\boxed{\textbf{(A) } \{-1, 0\}}</math>. | Thus the range of the function <math>f</math> is <math>\boxed{\textbf{(A) } \{-1, 0\}}</math>. | ||
− | ~IronicNinja | + | ~IronicNinja |
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− | |||
+ | == 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. | 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. | ||
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&=a-(a+1)=-1\end{split}</cmath> | &=a-(a+1)=-1\end{split}</cmath> | ||
− | Thus, the range of | + | 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: | ''Note'': One could solve the case of <math>x</math> as a negative non-integer in this way: | ||
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&=\lfloor -x\rfloor-(\lfloor -x\rfloor+1) = -1\end{split}</cmath> | &=\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>}. | ||
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+ | 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> | ||
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+ | <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> \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>. | ||
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+ | 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 | ||
+ | |||
+ | == 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 23:36, 31 October 2021
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
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 |
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