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1962 AHSME Problems/Problem 33: Difference between revisions

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==Solution==
==Solution==
{{solution}}
This inequality can be split into two inequalities:
<math>2\le x-1\le5</math> or <math>2\le1-x\le5</math>. Solving for x gives
<math>3\le x\le6</math> or <math>-4\le x\le-1</math>. <math>\boxed{\textbf{(A)}}</math>

Latest revision as of 16:45, 17 April 2014

Problem

The set of $x$-values satisfying the inequality $2 \leq |x-1| \leq 5$ is:

$\textbf{(A)}\ -4\leq x\leq-1\text{ or }3\leq x\leq 6\qquad$

$\textbf{(B)}\ 3\leq x\leq 6\text{ or }-6\leq x\leq-3\qquad\textbf{(C)}\ x\leq-1\text{ or }x\geq 3\qquad$

$\textbf{(D)}\ -1\leq x\leq 3\qquad\textbf{(E)}\ -4\leq x\leq 6$


Solution

This inequality can be split into two inequalities: $2\le x-1\le5$ or $2\le1-x\le5$. Solving for x gives $3\le x\le6$ or $-4\le x\le-1$. $\boxed{\textbf{(A)}}$

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