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2010 AMC 12A Problem 4

Below is the professionally curated solution for Problem 4 of the 2010 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AMC 12A solutions, or check the answer key.

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Concepts:inequalityexponentabsolute value

Difficulty rating: 1070

4.

If x<0,x\lt0, then which of the following must be positive?

x∣x∣\dfrac{x}{|x|}

−x2-x^2

−2x-2^x

−x−1-x^{-1}

x3\sqrt[3]{x}

Solution:

Choice (D) is −x−1=−1x.-x^{-1}=-\dfrac1x. When x<0,x\lt0, 1x<0,\dfrac1x\lt0, so −1x>0.-\dfrac1x\gt0.

Testing x=−1x=-1 shows the other choices need not be positive: x∣x∣=−1,\dfrac{x}{|x|}=-1, −x2=−1,-x^2=-1, −2x=−12,-2^x=-\tfrac12, and x3=−1.\sqrt[3]{x}=-1.

Thus, D is the correct answer.

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