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58% (01:17) correct 42% (01:11) wrong based on 114 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. \(4\)
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'll have that \(-\frac{1}{m^2}=-\frac{1}{16}\), so \(m^2=16\).
The same way, from \(f(mn) = f(\frac{1}{n})\) we'll 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 \(n^2=\frac{1}{4}\).
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. \(4\)
What sub-topic/tag does this question fall under? I am looking to solve similar question types both from the GMAT Club and the OG.
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. \(4\)
What sub-topic/tag does this question fall under? I am looking to solve similar question types both from the GMAT Club and the OG.
I think this is a high-quality question and I agree with explanation. I am quite sure that I marked Option B , but it shows that I have marked Option C. I just hope it is a technical error. If it is kindly look into it,
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58% (01:17) correct 
