1974 AHSME Problems/Problem 27

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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 (ProblemsAnswer KeyResources)
Preceded by
Problem 26
Followed by
Problem 28
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