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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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Bunuel
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If the function f is defined by f(p) = p^2 + 1/p^2 for all non-zero nu [#permalink] New post   01 Oct 2018, 22:54
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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] New post   02 Oct 2018, 00:21
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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.

A is the answer.
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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] New post   08 Apr 2020, 06:22
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Bunuel wrote:
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) \(f(p) - 2\)


Solution


    • \(f(p) =p^2 + \frac{1}{p^2}\)
    o \(f(\frac{-1}{\sqrt{p}})= (\frac{-1}{\sqrt{p}})^2 + \frac{1}{(\frac{-1}{\sqrt{p}})^2} = \frac{1}{p} + p\)
    • \((f(\frac{-1}{\sqrt{p}}))^2 = (\frac{1}{p} + p)^2 = \frac{1}{p^2} + (p)^2 +2*p*\frac{1}{p} = p^2 + \frac{1}{p^2} + 2 = f(p) + 2\)
Hence, the correct answer is Option A.
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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] New post   11 Apr 2020, 06:45
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Bunuel wrote:
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) \(f(p) - 2\)


First, f(-1/√p) = (-1/√p)^2 + 1/(-1/√p)^2 = 1/p + 1/(1/p) = 1/p + p.

Next, (1/p + p)^2 = 1/p^2 + 2(1/p)(p) + p^2 = p^2 + 1/p^2 + 2.

So, (f(-1/√p))^2 = p^2 + 1/p^2 + 2. However, since f(p) = p^2 + 1/p^2, then we see that (f(-1/√p))^2 = f(p) + 2.

Answer: A
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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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