1957 AMC 12 Problem 29

Attempt Problem 29 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.

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29.

The relation x2(x2−1)≥0x^2(x^2-1)\ge0 is true only for:
Here x≥ax\ge a means that xx can take on all values greater than aa and the value equal to a,a, while x≤ax\le a has a corresponding meaning with “less than.”
x≥1x\ge1
−1≤x≤1-1\le x\le1
x=0,x=0, x=1,x=1, x=−1x=-1
x=0,x=0, x≤−1,x\le-1, x≥1x\ge1
x≥0x\ge0
Answer: D
Concepts:inequalityfactoringzero product property
Small hint:
The factor x2x^2 is positive except at x=0x=0
Big hint:
Away from zero, the sign is controlled by x2−1x^2-1
Solution:
At x=0,x=0, the product is zero. For x≠0,x\ne0, the factor x2x^2 is positive, so the product is nonnegative exactly when x2−1≥0, x^2-1\ge0, or x≤−1x\le-1 or x≥1.x\ge1. Including the isolated value x=0x=0 gives the set in choice D.
Thus, the correct answer is D.
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