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Question
If \[f\left( x \right) = \frac{2^x + 2^{- x}}{2}\] , then f(x + y) f(x − y) is equal to
Options
(a) \[\frac{1}{2}\left[ f\left( 2x \right) + f\left( 2y \right) \right]\]
(b) \[\frac{1}{2}\left[ f\left( 2x \right) - f\left( 2y \right) \right]\]
(c) \[\frac{1}{4}\left[ f\left( 2x \right) + f\left( 2y \right) \right]\]
(d) \[\frac{1}{4}\left[ f\left( 2x \right) - f\left( 2y \right) \right]\]
Solution
(a) \[\frac{1}{2}\left[ f\left( 2x \right) + f\left( 2y \right) \right]\]
Given: \[f\left( x \right) = \frac{2^x + 2^{- x}}{2}\] Now,
f(x + y) f(x − y) = \[\left( \frac{2^{x + y} + 2^{- x - y}}{2} \right)\left( \frac{2^{x - y} + 2^{- x + y}}{2} \right)\]
⇒ f(x + y) f(x − y) = \[\frac{1}{4}\left( 2^{2x} + 2^{- 2y} + 2^{2y} + 2^{- 2x} \right)\] ⇒ f(x + y) f(x − y) = \[\frac{1}{2}\left( \frac{2^{2x} + 2^{- 2x}}{2} + \frac{2^{2y} + 2^{- 2y}}{2} \right)\]
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