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21 Apr 2024, 13:52
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If \(x ≠ 1\) and \(\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = -1\), what is the value of \(\frac{-|x| - 1}{x - 1}\)?
A. \(1\)
B. \(\frac{1}{2}\)
C. \(0\)
D. \(-\frac{1}{2}\)
E. \(-1\)
A. \(1\)
B. \(\frac{1}{2}\)
C. \(0\)
D. \(-\frac{1}{2}\)
E. \(-1\)
21 Apr 2024, 13:52
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Official Solution:
If \(x ≠ 1\) and \(\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = -1\), what is the value of \(\frac{-|x| - 1}{x - 1}\)?
A. \(1\)
B. \(\frac{1}{2}\)
C. \(0\)
D. \(-\frac{1}{2}\)
E. \(-1\)
Notice that if \(x\) is positive, then \(\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = \text{positive} + \text{positive} + \text{positive} + \text{positive} = \text{positive}\), not a negative number \(-1\). So, \(x\) cannot be positive.
When \(x \leq 0\), then \(|x| = -x\), and in this case \(\frac{-|x| - 1}{x - 1} = \frac{-(-x) - 1}{x - 1} = \frac{x - 1}{x - 1} = 1\).
Answer: A
If \(x ≠ 1\) and \(\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = -1\), what is the value of \(\frac{-|x| - 1}{x - 1}\)?
A. \(1\)
B. \(\frac{1}{2}\)
C. \(0\)
D. \(-\frac{1}{2}\)
E. \(-1\)
Notice that if \(x\) is positive, then \(\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = \text{positive} + \text{positive} + \text{positive} + \text{positive} = \text{positive}\), not a negative number \(-1\). So, \(x\) cannot be positive.
When \(x \leq 0\), then \(|x| = -x\), and in this case \(\frac{-|x| - 1}{x - 1} = \frac{-(-x) - 1}{x - 1} = \frac{x - 1}{x - 1} = 1\).
Answer: A
26 Aug 2024, 23:07
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I did not understand this. If modulus sign is removed , the digit will be positive.
For example , -|-5| = -5 right ? Please guide
For example , -|-5| = -5 right ? Please guide
