2006 AMC 10B Problem 7

Attempt Problem 7 of the 2006 AMC 10B 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 2006 AMC 10B solutions, or check the answer key.

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

7.

Which of the following is equivalent to
x1−x−1x\sqrt{\dfrac{x}{1-\dfrac{x-1}{x}}}
when x<0?x \lt 0?
−x-x
xx
11
x2\sqrt{\dfrac{x}{2}}
x−1x\sqrt{-1}
Answer: A
Concepts:radicalabsolute valuealgebraic manipulation
Small hint:
Combine the denominator 1−x−1x1-\dfrac{x-1}{x} into a single fraction
Big hint:
x2=∣x∣,\sqrt{x^2}=|x|, and ∣x∣=−x|x|=-x when x<0x \lt 0
Solution:
The denominator simplifies: 1−x−1x=x−(x−1)x=1x.1-\dfrac{x-1}{x}=\dfrac{x-(x-1)}{x}=\dfrac{1}{x}.
So the expression is x1x=x2=∣x∣.\sqrt{\dfrac{x}{\frac{1}{x}}}=\sqrt{x^2}=|x|. Since x<0,x \lt 0, this equals −x.-x.
Thus, the correct answer is A.
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