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# M20-26

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Math Expert
Joined: 02 Sep 2009
Posts: 49303

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16 Sep 2014, 01:09
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Difficulty:

15% (low)

Question Stats:

74% (00:47) correct 26% (01:25) wrong based on 134 sessions

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If $$f(x) = \frac{x^4 - 1}{x^2}$$, what is $$f(\frac{1}{x})$$ in terms of $$f(x)$$?

A. $$f(x)$$
B. $$-f(x)$$
C. $$\frac{1}{f(x)}$$
D. $$-\frac{1}{f(x)}$$
E. $$2f(x)$$

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Math Expert
Joined: 02 Sep 2009
Posts: 49303

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16 Sep 2014, 01:09
Official Solution:

If $$f(x) = \frac{x^4 - 1}{x^2}$$, what is $$f(\frac{1}{x})$$ in terms of $$f(x)$$?

A. $$f(x)$$
B. $$-f(x)$$
C. $$\frac{1}{f(x)}$$
D. $$-\frac{1}{f(x)}$$
E. $$2f(x)$$

$$f(\frac{1}{x}) = \frac{\frac{1}{x^4} - 1}{\frac{1}{x^2}} = \frac{1}{x^2} - x^2 = \frac{1 - x^4}{x^2} = -f(x)$$.

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Joined: 05 Jul 2016
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14 Jul 2016, 11:27
Hello,

Can you please explain how the last step works. ie how does (1-x^4)/x^2= -f(x)

Thanks!
Math Expert
Joined: 02 Sep 2009
Posts: 49303

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15 Jul 2016, 07:35
2
abhilash53 wrote:
Hello,

Can you please explain how the last step works. ie how does (1-x^4)/x^2= -f(x)

Thanks!

$$\frac{1 - x^4}{x^2} =- \frac{x^4 - 1}{x^2}=-f(x)$$.
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Joined: 29 Jul 2017
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11 Sep 2017, 04:23
Can you explain how you get 1/x^2 - x^2?
Math Expert
Joined: 02 Sep 2009
Posts: 49303

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11 Sep 2017, 04:46
1
xlgoh1992 wrote:
Can you explain how you get 1/x^2 - x^2?

$$f(\frac{1}{x}) = \frac{\frac{1}{x^4} - 1}{\frac{1}{x^2}} =(\frac{1}{x^4} - 1)*x^2 =\frac{1}{x^2} - x^2 = \frac{1 - x^4}{x^2} = -f(x)$$.
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Joined: 08 Jun 2015
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Location: India
GMAT 1: 640 Q48 V29
GMAT 2: 700 Q48 V38
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18 Oct 2017, 05:54
Substitute (1/x) in f(x). You will get -f(x). The answer is option B.
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18 Oct 2017, 23:06
2

Bunuel wrote:
If $$f(x) = \frac{x^4 - 1}{x^2}$$, what is $$f(\frac{1}{x})$$ in terms of $$f(x)$$?

A. $$f(x)$$
B. $$-f(x)$$
C. $$\frac{1}{f(x)}$$
D. $$-\frac{1}{f(x)}$$
E. $$2f(x)$$

Just plug in $$\frac{1}{2}$$ and $$2$$, then compare the answers.

For $$f(\frac{1}{2})$$, you had get $$(\frac{1}{16} -1) / \frac{1}{4} = \frac{-15}{4}$$

and for $$f(2)$$, you had get $$\frac{15}{4}$$ i.e $$\frac{(16-1)}{4}$$

Thus $$f(\frac{1}{x})$$ = $$-f(x)$$, which is B.

alternatively, algebraically, $$f(x) = \frac{x^4 - 1}{x^2}$$ = $$x^2 - \frac{1}{x^2}$$

One can quickly see the pattern and know that the function is negatively symmetrical for the reciprocal i.e let $$x^2 =\frac{1}{2}$$, then $$x^2 - \frac{1}{x^2}$$= negative and positive same value if $$x^2 =2$$. Again, B.
M20-26 &nbs [#permalink] 18 Oct 2017, 23:06
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# M20-26

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