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Difference between revisions of "2022 AMC 10A Problems/Problem 6"

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We know that ,
 
We know that ,
  
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\(\begin{aligned} & |x|=x, \text { if } x>0 \\ & =-x, \text { if } x<0 \\ & \text { Now, }\left|a-2-\sqrt{(a-1)^2}\right|=\left|a-2-\sqrt{(1-a)^2}\right|, \text { as }  a<0 \\ & =|a-2-(1-a)| \\ & =|2 a-3| \\ & =-(2 a-3) \text { as } a<0 \\ & =3-2 a\end{aligned}\) \\
+
\(\begin{aligned} & |x|=x, \text { if } x>0 \\ & =-x, \text { if } x<0 \\ & \text { Now, }\left|a-2-\sqrt{(a-1)^2}\right|=\left|a-2-\sqrt{(1-a)^2}\right|, \text { as }  a<0 \\ & =|a-2-(1-a)| \\ & =|2 a-3| \\ & =-(2 a-3) \text { as } a<0 \\ & =3-2 a\end{aligned}\)  
  
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So, the correct choice is option (A).
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So, the correct choice is option <math>\boxed{\textbf{(A) } 3-2a}.</math>
  
 
~KENJAKURA
 
~KENJAKURA
 +
 +
== Solution 5 ==
 +
 +
Because the degree of the expression is <math>1</math>, and all the solutions are of degree <math>1</math> or <math>0</math>, we can assume the function is linear for <math>a<0</math> and try values to find what the function is.
 +
 +
We can first say from Solution 1 that:
 +
<cmath>\left|a-2-\sqrt{(a-1)^2}\right| = \left|a-2-|a-1|\right|</cmath>
 +
Then, we can try values:
 +
\begin{array}{cccc}
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a & \text{Plug in values} & = & \text{Result} \\
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\hline
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-1 & \left|-1-2-\left|-1-1\right|\right| & = & \left|-3-2\right| = 5 \\
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-2 & \left|-2-2-\left|-2-1\right|\right| & = & \left|-4-3\right| = 7 \\
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-3 & \left|-3-2-\left|-3-1\right|\right| & = & \left|-5-4\right| = 9 \\
 +
\end{array}
 +
We can see that this is a linear function. We can find that the slope is <math>m = \frac{7-5}{-2-(-1)} = \frac{2}{-1} = -2.</math>
 +
Using the first pair:
 +
\begin{aligned}
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y & = mx+b \\
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5 & = -2(-1)+b \\
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5 & = 2+b \\
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3 & = b \\
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y & = -2x+3 \\
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\end{aligned}
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 +
We take that <math>y</math> equals the expression, and <math>x</math> is <math>a</math>. This gives <math>-2a+3</math>, or <math>3-2a</math>.
 +
Testing to make sure the answer is correct (like that in Solution 2) is a good practice to have and shows that the correct choice must be <math>\boxed{\textbf{(A) } 3-2a}.</math>
 +
 +
~Eugenius
  
 
==Video Solution 1 (Quick and Easy)==
 
==Video Solution 1 (Quick and Easy)==

Latest revision as of 20:53, 31 July 2025

Problem

Which expression is equal to \[\left|a-2-\sqrt{(a-1)^2}\right|\] for $a<0?$

$\textbf{(A) } 3-2a \qquad \textbf{(B) } 1-a \qquad \textbf{(C) } 1 \qquad \textbf{(D) } a+1 \qquad \textbf{(E) } 3$

Solution 1

We have \begin{align*} \left|a-2-\sqrt{(a-1)^2}\right| &= \left|a-2-|a-1|\right| \\ &=\left|a-2-(1-a)\right| \\ &=\left|2a-3\right| \\ &=\boxed{\textbf{(A) } 3-2a}. \end{align*} ~MRENTHUSIASM

Solution 2

Assume that $a=-1.$ Then, the given expression simplifies to $5$: \begin{align*} \left|a-2-\sqrt{(a-1)^2}\right| &= \left|-1-2-\sqrt{(-1-1)^2}\right| \\ &= \left|-1-2-\sqrt{4}\right| \\ &= \left|-1-2-2\right| \\ &= 5. \end{align*} Then, we test each of the answer choices to see which one is equal to $5$:

$\textbf{(A) } 3-2a = 3-2\cdot(-1) = 3+2 = 5.$

$\textbf{(B) } 1-a = 1-(-1) = 2 \neq 5.$

$\textbf{(C) } 1 \neq 5.$

$\textbf{(D) } a+1 = -1+1 = 0 \neq 5.$

$\textbf{(E) } 3 \neq 5.$

The only answer choice equal to $5$ for $a=-1$ is $\boxed{\textbf{(A) } 3-2a}.$

-MathWizard09

Solution 3

The given function is continuous, so assume that $a=0.$ Then, the given expression simplifies to $3.$

We test each of the answer choices and get $\textbf{(A) } 3-2a$ or $\textbf{(E) } 3.$

We test $x = - 1000$ and get $\left|-1000-2- \text{positive} \right| \ne 3 \implies \boxed{\textbf{(A) } 3-2a}.$

vladimir.shelomovskii@gmail.com, vvsss

Solution 4

We know that ,

\(\begin{aligned} & |x|=x, \text { if } x>0 \\ & =-x, \text { if } x<0 \\ & \text { Now, }\left|a-2-\sqrt{(a-1)^2}\right|=\left|a-2-\sqrt{(1-a)^2}\right|, \text { as } a<0 \\ & =|a-2-(1-a)| \\ & =|2 a-3| \\ & =-(2 a-3) \text { as } a<0 \\ & =3-2 a\end{aligned}\)

So, the correct choice is option $\boxed{\textbf{(A) } 3-2a}.$

~KENJAKURA

Solution 5

Because the degree of the expression is $1$, and all the solutions are of degree $1$ or $0$, we can assume the function is linear for $a<0$ and try values to find what the function is.

We can first say from Solution 1 that: \[\left|a-2-\sqrt{(a-1)^2}\right| = \left|a-2-|a-1|\right|\] Then, we can try values: \begin{array}{cccc} a & \text{Plug in values} & = & \text{Result} \\ \hline -1 & \left|-1-2-\left|-1-1\right|\right| & = & \left|-3-2\right| = 5 \\ -2 & \left|-2-2-\left|-2-1\right|\right| & = & \left|-4-3\right| = 7 \\ -3 & \left|-3-2-\left|-3-1\right|\right| & = & \left|-5-4\right| = 9 \\ \end{array} We can see that this is a linear function. We can find that the slope is $m = \frac{7-5}{-2-(-1)} = \frac{2}{-1} = -2.$ Using the first pair: \begin{aligned} y & = mx+b \\ 5 & = -2(-1)+b \\ 5 & = 2+b \\ 3 & = b \\ y & = -2x+3 \\ \end{aligned}

We take that $y$ equals the expression, and $x$ is $a$. This gives $-2a+3$, or $3-2a$. Testing to make sure the answer is correct (like that in Solution 2) is a good practice to have and shows that the correct choice must be $\boxed{\textbf{(A) } 3-2a}.$

~Eugenius

Video Solution 1 (Quick and Easy)

https://youtu.be/ZWHzdrW4rvw

~Education, the Study of Everything

Video Solution 2

https://youtu.be/XWTTtL9kW98

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

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