1974 AMC 12 Problem 27

Attempt Problem 27 of the 1974 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1974 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

27.

If f(x)=3x+2f(x)=3x+2 for all real x,x, then the statement:
“∣f(x)+4∣<a|f(x)+4|\lt a whenever ∣x+2∣<b|x+2|\lt b and a>0a\gt0 and b>0b\gt0”
is true when
b≤a3b\le \frac{a}{3}
b>a3b\gt \frac{a}{3}
a≤b3a\le \frac{b}{3}
a>b3a\gt \frac{b}{3}
The statement is never true.
Answer: A
Concepts:absolute valuefunctioninequality
Small hint:
Factor f(x)+4f(x)+4 in terms of x+2x+2
Big hint:
The condition becomes 3∣x+2∣<a3|x+2|\lt a whenever ∣x+2∣<b|x+2|\lt b
Solution:
We have ∣f(x)+4∣=∣3x+6∣=3∣x+2∣. |f(x)+4|=|3x+6|=3|x+2|. If ∣x+2∣<b|x+2|\lt b and b≤a3,b\le \frac{a}{3}, then 3∣x+2∣<3b≤a,3|x+2|\lt3b\le a, as required. If b>a3,b\gt \frac{a}{3}, one can choose ∣x+2∣|x+2| strictly between a3\frac{a}{3} and b,b, so the statement fails. Thus the exact condition is b≤a3.b\le \frac{a}{3}.
Therefore, the correct answer is A.
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