1957 AMC 12 Problem 3

Attempt Problem 3 of the 1957 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 1957 AMC 12 solutions, or check the answer key.

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

3.

The simplest form of 1−11+a1−a1-\dfrac{1}{1+\dfrac{a}{1-a}} is:
aa if a≠0a\ne0
11
aa if a≠−1a\ne-1
1−a1-a with no restriction on aa
aa if a≠1a\ne1
Answer: E
Concepts:rational equationalgebraic manipulation
Small hint:
Combine 1+a1−a1+\frac{a}{1-a} before taking its reciprocal
Big hint:
Retain every restriction imposed by the original denominator
Solution:
The original expression requires a≠1.a\ne1. For such a,a, 1+a1−a=1−a+a1−a=11−a. \begin{aligned} 1+\frac{a}{1-a} &=\frac{1-a+a}{1-a}\\ &=\frac1{1-a}. \end{aligned} Therefore the expression is 1−(1−a)=a.1-(1-a)=a. The restriction a≠1a\ne1 remains.
Thus, the correct answer is E.
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