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

2010 AMC 12A Problem 4

Attempt Problem 4 of the 2010 AMC 12A 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 2010 AMC 12A solutions, or check the answer key.

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

x\lt0,

then which of the following must be positive?

\dfrac{x}{|x|}

-x^2

-2^x

-x^{-1}

\sqrt[3]{x}

Test each choice with a negative value such as

x=-1

Rewrite

-x^{-1}

-\dfrac1x;

dividing into a negative number flips the sign

Choice (D) is

-x^{-1}=-\dfrac1x.

When

x\lt0,

\dfrac1x\lt0,

-\dfrac1x\gt0.

Testing

x=-1

shows the other choices need not be positive:

\dfrac{x}{|x|}=-1,

-x^2=-1,

-2^x=-\tfrac12,

and

\sqrt[3]{x}=-1.

Thus, D is the correct answer.