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Bunuel
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Bunuel
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d975490
Could you please elaborate on the following:
f(mn)=f(1/n) then we get -1/(mn)^2= - n^2. Don't understand who we got that result

\(f(x) = - \frac{1}{x^2}\), hence \(f(mn) = -\frac{1}{(mn)^2}\) and \(f(\frac{1}{n})=-\frac{1}{(\frac{1}{n})^2}=-n^2\)
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Bunuel
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.
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Bunuel
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Bunuel
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.

Functions and algebra.

13. Functions




For more check Ultimate GMAT Quantitative Megathread

Hope it helps.
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I think this is a high-quality question and I agree with explanation.
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I think this is a high-quality question and I agree with explanation.
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I think this is a high-quality question and I agree with explanation.
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Why don't we have negative root of n^2?
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Bunuel
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gmatbd
Why don't we have negative root of n^2?

Recall that all numbers on the GMAT are real numbers by default, so n^2, the square of some number, n, cannot be negative: \((real \ number)^2 \geq 0\).
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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.
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I like the solution - it’s helpful.
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