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Let a function be defined as f(x) = (c^x + c^-x)/2, for c > 0. What is
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Bunuel
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Let a function be defined as f(x) = (c^x + c^-x)/2, for c > 0. What is [#permalink] New post  22 Jan 2021, 07:00
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Let a function be defined as \(f(x) = \frac{c^x + c^{-x}}{2}\), for \(c > 0\). What is the value of \(f(x + y) + f(x - y) \)?


A. \(2f(x) + 2f(y)\)

B. \(f(x) + f(y)\)

C. \(4f(x)f(y)\)

D. \(2f(x)f(y)\)

E. \(f(x)f(y)\)
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Re: Let a function be defined as f(x) = (c^x + c^-x)/2, for c > 0. What is [#permalink] New post  22 Jan 2021, 17:26
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Bunuel wrote:
Let a function be defined as \(f(x) = \frac{c^x + c^{-x}}{2}\), for \(c > 0\). What is the value of \(f(x + y) + f(x - y) \)?


A. \(2f(x) + 2f(y)\)

B. \(f(x) + f(y)\)

C. \(4f(x)f(y)\)

D. \(2f(x)f(y)\)

E. \(f(x)f(y)\)


\(f(x + y) + f(x - y) = \frac{c^{x+y} + c^{-x-y}}{2} + \frac{c^{x-y} + c^{y-x}}{2} = \frac{c^{x+y} + c^{-x-y} + c^{x-y} + c^{y-x}}{2}\).

We can note the answer needs the four terms summarized as \(c^{\pm x \pm y}\) and has to have the constant \(\frac{1}{2}\).

A and B both do not feature the terms \(c^{\pm x \pm y}\).

C has the four terms but the constant is 1. D has the constant \(\frac{1}{2}\) so D is the answer.

Ans: D
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