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M20-26

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M20-26 [#permalink]

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New post 16 Sep 2014, 01:09
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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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Re M20-26 [#permalink]

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New post 16 Sep 2014, 01:09
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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. \(2f(x)\)

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

Answer: B
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M20-26 [#permalink]

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New post 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!
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Re: M20-26 [#permalink]

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Re: M20-26 [#permalink]

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New post 11 Sep 2017, 04:23
Can you explain how you get 1/x^2 - x^2?
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Re: M20-26 [#permalink]

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New post 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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M20-26 [#permalink]

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New post 18 Oct 2017, 23:06
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\(\)
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   [#permalink] 18 Oct 2017, 23:06
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