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If the function f is defined by f(p) = p^2 + 1/p^2 for all non-zero nu

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If the function f is defined by f(p) = p^2 + 1/p^2 for all non-zero nu  [#permalink]

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01 Oct 2018, 21:54
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Difficulty:

55% (hard)

Question Stats:

59% (01:36) correct 41% (01:57) wrong based on 54 sessions

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If the function f is defined by $$f(p) = p^2 + \frac{1}{p^2}$$ for all non-zero numbers p, then $$(f(−\frac{1}{√p}))^2 =$$

(A) $$f(p) + 2$$

(B) $$\frac{2}{f(p^2)}$$

(C) $$(\frac{1}{f(√p)})^2$$

(D) $$1 − (f(√p))^2$$

(E) $$f(p) - 2$$

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Re: If the function f is defined by f(p) = p^2 + 1/p^2 for all non-zero nu  [#permalink]

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01 Oct 2018, 23:21
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Substituting p = $$\frac{-1}{\sqrt{p}}$$ in f(p)=$$p^2$$+$$\frac{1}{p^2}$$

We get, $$(p+\frac{1}{p})^2$$ = f(p)+2.

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Re: If the function f is defined by f(p) = p^2 + 1/p^2 for all non-zero nu  [#permalink]

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02 Oct 2018, 05:03
Well.. here the answer is not given.

When substituted, we get (1+p^2)^2/ p^2
resulting in f(p)+2.
Re: If the function f is defined by f(p) = p^2 + 1/p^2 for all non-zero nu   [#permalink] 02 Oct 2018, 05:03
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