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'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 \(n^2=\frac{1}{4}\).


Answer: B
_________________

Hi Bunuel,

Can we now simply solve it like this? Since f(x)=−1/x^2, then from f(m)=−1/16 we'll have that −1/n^4=−1/16, so n^2=4.
As f(n^2) = -1/n^4.

Could you please explain if this is correct?

There should be m instead of n.
_________________

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 \(f(\frac{1}{n})=-\frac{1}{(\frac{1}{n})^2}=-n^2\)
_________________

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

• Theory
  How to Interpret Unfamiliar Symbols
  Functions on GMAT
  A Closer Look at GMAT Function Questions
  
  • Questions
    Operations/functions defining algebraic/arithmetic expressions
    Rounding Functions Problems
    Various Functions Problems
    DS Questions
    PS Questions
  
  

  7. Algebra

  
  
  • Theory
    Math : Algebra 101
    Algebra: Tips and hints
    FOIL on the GMAT: Simplifying and Expanding
    Factoring Quadratics
    Solving Quadratic Equations
    Using Algebra vs. Logic on GMAT Quant Questions
    
    • Questions
      Tough Algebra Set
      Algebra Ability Quiz by e-GMAT
      DS Questions
      PS Questions

For more check Ultimate GMAT Quantitative Megathread

Hope it helps.
_________________

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,

I think this is a high-quality question and I agree with explanation.

I think this is a high-quality question and I agree with explanation.