Difference between revisions of "2022 AMC 10A Problems/Problem 6"
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<cmath>\begin{align*} | <cmath>\begin{align*} | ||
\left|a-2-\sqrt{(a-1)^2}\right| &= \left|a-2-|a-1|\right| \\ | \left|a-2-\sqrt{(a-1)^2}\right| &= \left|a-2-|a-1|\right| \\ | ||
− | &=\left|a-2-(-a | + | &=\left|a-2-(1-a)\right| \\ |
&=\left|2a-3\right| \\ | &=\left|2a-3\right| \\ | ||
&=\boxed{\textbf{(A) } 3-2a}. | &=\boxed{\textbf{(A) } 3-2a}. | ||
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-MathWizard09 | -MathWizard09 | ||
+ | |||
+ | == Solution 3 == | ||
+ | The given function is continuous, so assume that <math>a=0.</math> Then, the given expression simplifies to <math>3.</math> | ||
+ | |||
+ | We test each of the answer choices and get <math>\textbf{(A) } 3-2a</math> or <math>\textbf{(E) } 3.</math> | ||
+ | |||
+ | We test <math>x = - 1000</math> and get <math>\left|-1000-2- \text{positive} \right| \ne 3 \implies \boxed{\textbf{(A) } 3-2a}.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
==Video Solution 1 (Quick and Easy)== | ==Video Solution 1 (Quick and Easy)== | ||
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~Education, the Study of Everything | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution 2== | ||
+ | https://youtu.be/XWTTtL9kW98 | ||
== See Also == | == See Also == |
Latest revision as of 11:13, 25 December 2023
Contents
Problem
Which expression is equal to for
Solution 1
We have ~MRENTHUSIASM
Solution 2
Assume that Then, the given expression simplifies to : Then, we test each of the answer choices to see which one is equal to :
The only answer choice equal to for is
-MathWizard09
Solution 3
The given function is continuous, so assume that Then, the given expression simplifies to
We test each of the answer choices and get or
We test and get
vladimir.shelomovskii@gmail.com, vvsss
Video Solution 1 (Quick and Easy)
~Education, the Study of Everything
Video Solution 2
See Also
2022 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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.