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16 Sep 2014, 00:24
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If for all numbers \(p\), where \(p \neq 1\), the function \(@\) is defined as \(p@ = \frac{p}{p-1}\), what is the value of \(((x@)@)\), where \(x \neq 1\)?
A. \(\frac {x}{x-1}\)
B. \(\frac{1}{x}\)
C. \(x\)
D. \(2 - x\)
E. \(x - 1\)
A. \(\frac {x}{x-1}\)
B. \(\frac{1}{x}\)
C. \(x\)
D. \(2 - x\)
E. \(x - 1\)
16 Sep 2014, 00:24
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Official Solution:
If for all numbers \(p\), where \(p \neq 1\), the function \(@\) is defined as \(p@ = \frac{p}{p-1}\), what is the value of \(((x@)@)\), where \(x \neq 1\)?
A. \(\frac {x}{x-1}\)
B. \(\frac{1}{x}\)
C. \(x\)
D. \(2 - x\)
E. \(x - 1\)
Approach 1: Number Picking:
If \(x= 2\), then \(x@ = \frac{x}{x-1}=\frac{2}{2 - 1} = 2=x\). Since \(x@ = x\), then \(((x@))@)=x@ = x\).
Approach 2: Algebra:
\(x@ = \frac{x}{x-1}\). Take this value and substitute it for \(x\) to find \(((x@)@)\).
\((x@)@ = (\frac{x}{x-1})@ = \frac{\frac{x}{x-1}}{\frac{x}{x-1} - 1} = \frac{\frac{x}{x-1}}{\frac{x-(x-1)}{x1}} = \frac{\frac{x}{x-1}}{\frac{1}{x-1}} = \frac{x}{x-1}*(x-1) = x\).
Answer: C
If for all numbers \(p\), where \(p \neq 1\), the function \(@\) is defined as \(p@ = \frac{p}{p-1}\), what is the value of \(((x@)@)\), where \(x \neq 1\)?
A. \(\frac {x}{x-1}\)
B. \(\frac{1}{x}\)
C. \(x\)
D. \(2 - x\)
E. \(x - 1\)
Approach 1: Number Picking:
If \(x= 2\), then \(x@ = \frac{x}{x-1}=\frac{2}{2 - 1} = 2=x\). Since \(x@ = x\), then \(((x@))@)=x@ = x\).
Approach 2: Algebra:
\(x@ = \frac{x}{x-1}\). Take this value and substitute it for \(x\) to find \(((x@)@)\).
\((x@)@ = (\frac{x}{x-1})@ = \frac{\frac{x}{x-1}}{\frac{x}{x-1} - 1} = \frac{\frac{x}{x-1}}{\frac{x-(x-1)}{x1}} = \frac{\frac{x}{x-1}}{\frac{1}{x-1}} = \frac{x}{x-1}*(x-1) = x\).
Answer: C
General Discussion
20 Apr 2023, 04:32
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
