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21 Sep 2021, 07:53
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If \(x ≠ 1\), what is the value of \(\frac{-|x| - 1}{x - 1}\)?
(1) \(\sqrt{x^6} > x^3\)
(2) \(\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = -1\)
(1) \(\sqrt{x^6} > x^3\)
(2) \(\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = -1\)
21 Sep 2021, 07:53
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Official Solution:
If \(x ≠ 1\), what is the value of \(\frac{-|x| - 1}{x - 1}\)?
Work on the stem before moving to the statements.
If \(x \leq 0\), then \(|x|=-x\) and in this case we'd have: \(\frac{-|x| - 1}{x - 1}=\frac{-(-x) - 1}{x - 1}=\frac{x - 1}{x - 1}=1\). So, if \(x\) is 0 or negative then the value of the fraction is 1.
If \(x > 0\), then \(|x|=x\) and in this case we'd have: \(\frac{-|x| - 1}{x - 1}=\frac{-x - 1}{x - 1}\). So, in this case the value of the fraction would depend on the value of \(x\).
(1) \(\sqrt{x^6} > x^3\). Take the square root from the left hand side: \(|x^3| > x^3\). For this to hold true, \(x\) must be negative, because if \(x\) is 0 or positive, then \(|x^3|\) would be equal to \(x^3\). From above, we know that if \(x < 0\), then \(\frac{-|x| - 1}{x - 1}=\frac{-(-x) - 1}{x - 1}=\frac{x - 1}{x - 1}=1\). Sufficient.
(2) \(\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = -1\). Notice that if \(x\) is positive, then \(\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = positive +positive+positive+positive=positive\), not negative number \(-1\). So, \(x\) cannot be positive. From above, we know that if \(x\) is not positive, then \(\frac{-|x| - 1}{x - 1}=\frac{-(-x) - 1}{x - 1}=\frac{x - 1}{x - 1}=1\). Sufficient.
Answer: D
If \(x ≠ 1\), what is the value of \(\frac{-|x| - 1}{x - 1}\)?
Work on the stem before moving to the statements.
If \(x \leq 0\), then \(|x|=-x\) and in this case we'd have: \(\frac{-|x| - 1}{x - 1}=\frac{-(-x) - 1}{x - 1}=\frac{x - 1}{x - 1}=1\). So, if \(x\) is 0 or negative then the value of the fraction is 1.
If \(x > 0\), then \(|x|=x\) and in this case we'd have: \(\frac{-|x| - 1}{x - 1}=\frac{-x - 1}{x - 1}\). So, in this case the value of the fraction would depend on the value of \(x\).
(1) \(\sqrt{x^6} > x^3\). Take the square root from the left hand side: \(|x^3| > x^3\). For this to hold true, \(x\) must be negative, because if \(x\) is 0 or positive, then \(|x^3|\) would be equal to \(x^3\). From above, we know that if \(x < 0\), then \(\frac{-|x| - 1}{x - 1}=\frac{-(-x) - 1}{x - 1}=\frac{x - 1}{x - 1}=1\). Sufficient.
(2) \(\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = -1\). Notice that if \(x\) is positive, then \(\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = positive +positive+positive+positive=positive\), not negative number \(-1\). So, \(x\) cannot be positive. From above, we know that if \(x\) is not positive, then \(\frac{-|x| - 1}{x - 1}=\frac{-(-x) - 1}{x - 1}=\frac{x - 1}{x - 1}=1\). Sufficient.
Answer: D
21 Apr 2024, 13:59
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