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# M10-21

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Math Expert
Joined: 02 Sep 2009
Posts: 43335

Kudos [?]: 139529 [0], given: 12794

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15 Sep 2014, 23:42
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Difficulty:

65% (hard)

Question Stats:

51% (01:18) correct 49% (01:18) wrong based on 105 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. $$4$$
[Reveal] Spoiler: OA

_________________

Kudos [?]: 139529 [0], given: 12794

Math Expert
Joined: 02 Sep 2009
Posts: 43335

Kudos [?]: 139529 [1], given: 12794

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15 Sep 2014, 23:42
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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'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 $$n^2=\frac{1}{4}$$.

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Kudos [?]: 139529 [1], given: 12794

Intern
Joined: 02 Jan 2014
Posts: 41

Kudos [?]: 23 [0], given: 75

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28 Oct 2015, 20:00
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?

Kudos [?]: 23 [0], given: 75

Math Expert
Joined: 02 Sep 2009
Posts: 43335

Kudos [?]: 139529 [0], given: 12794

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28 Oct 2015, 23:59
Celerma wrote:
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.
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Kudos [?]: 139529 [0], given: 12794

Intern
Joined: 30 Apr 2017
Posts: 4

Kudos [?]: 0 [0], given: 29

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25 Sep 2017, 23:52
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

Kudos [?]: 0 [0], given: 29

Math Expert
Joined: 02 Sep 2009
Posts: 43335

Kudos [?]: 139529 [0], given: 12794

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25 Sep 2017, 23:56
d975490 wrote:
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 $$f(\frac{1}{n})=-\frac{1}{(\frac{1}{n})^2}=-n^2$$
_________________

Kudos [?]: 139529 [0], given: 12794

Intern
Joined: 03 May 2014
Posts: 15

Kudos [?]: [0], given: 3

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31 Oct 2017, 08:41
Bunuel wrote:
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.

Kudos [?]: [0], given: 3

Math Expert
Joined: 02 Sep 2009
Posts: 43335

Kudos [?]: 139529 [0], given: 12794

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31 Oct 2017, 08:46
Edofarmer wrote:
Bunuel wrote:
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.
_________________

Kudos [?]: 139529 [0], given: 12794

Re: M10-21   [#permalink] 31 Oct 2017, 08:46
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# M10-21

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