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Difference between revisions of "1974 AHSME Problems/Problem 27"

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Latest revision as of 11:44, 5 July 2013

Problem

If $f(x)=3x+2$ for all real $x$, then the statement: "$|f(x)+4|<a$ whenever $|x+2|<b$ and $a>0$ and $b>0$" is true when

$\mathrm{(A)}\ b\le a/3\qquad\mathrm{(B)}\ b > a/3\qquad\mathrm{(C)}\ a\le b/3\qquad\mathrm{(D)}\ a > b/3\\ \qquad\mathrm{(E)}\ \text{The statement is never true.}$

Solution

Plugging in $f(x)=3x+2$ into $|f(x)+4|<a$, we get $|3x+6|=|3(x+2)|=3|x+2|<a$. Therefore, $|x+2|<\frac{a}{3}$. Therefore, we want $|x+2|<b\implies |x+2|<\frac{a}{3}$. This is clearly only true when $b\le\frac{a}{3}, \boxed{\text{A}}$.

See Also

1974 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 26 		Followed by
Problem 28
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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