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16 Sep 2014, 00:42
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B
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Difficulty:
55%
(hard)
Question Stats:
59% (02:05) correct
41%
(01:51)
wrong
based on 165
sessions
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The function \(f\) is defined by \(f(x) = - \frac{1}{x^2}\) for all nonzero numbers \(x\). If \(f(m) = - \frac{1}{16}\) and \(f(mn) = f(\frac{1}{n})\), what is the value of \(n^2\)?
A. \(\frac{1}{16}\)
B. \(\frac{1}{4}\)
C. \(\frac{1}{2}\)
D. \(2\)
E. \(4\)
A. \(\frac{1}{16}\)
B. \(\frac{1}{4}\)
C. \(\frac{1}{2}\)
D. \(2\)
E. \(4\)
16 Sep 2014, 00:42
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Official Solution:
The function \(f\) is defined by \(f(x) = - \frac{1}{x^2}\) for all nonzero numbers \(x\). If \(f(m) = - \frac{1}{16}\) and \(f(mn) = f(\frac{1}{n})\), what is the value of \(n^2\)?
A. \(\frac{1}{16}\)
B. \(\frac{1}{4}\)
C. \(\frac{1}{2}\)
D. \(2\)
E. \(4\)
Since \(f(x) = - \frac{1}{x^2}\), then from \(f(m) = - \frac{1}{16}\) we have that \(-\frac{1}{m^2}=-\frac{1}{16}\), so \(m^2=16\).
In the same way, from \(f(mn) = f(\frac{1}{n})\) we have that \(-\frac{1}{(mn)^2}=-n^2\), which simplifies to \(n^4=\frac{1}{m^2}\).
Since \(m^2=16\), then \(n^4=\frac{1}{m^2}=\frac{1}{16}\), which gives \(n^2=\frac{1}{4}\).
Answer: B
The function \(f\) is defined by \(f(x) = - \frac{1}{x^2}\) for all nonzero numbers \(x\). If \(f(m) = - \frac{1}{16}\) and \(f(mn) = f(\frac{1}{n})\), what is the value of \(n^2\)?
A. \(\frac{1}{16}\)
B. \(\frac{1}{4}\)
C. \(\frac{1}{2}\)
D. \(2\)
E. \(4\)
Since \(f(x) = - \frac{1}{x^2}\), then from \(f(m) = - \frac{1}{16}\) we have that \(-\frac{1}{m^2}=-\frac{1}{16}\), so \(m^2=16\).
In the same way, from \(f(mn) = f(\frac{1}{n})\) we have that \(-\frac{1}{(mn)^2}=-n^2\), which simplifies to \(n^4=\frac{1}{m^2}\).
Since \(m^2=16\), then \(n^4=\frac{1}{m^2}=\frac{1}{16}\), which gives \(n^2=\frac{1}{4}\).
Answer: B
