JEE MainMathematicsToughFunctions

Let f(x)=a^(a>0) be written as $f(x) = g(x) + h(x),$ where
g(x) is an even function and h(x) is an odd function then the value of
the $g(x + y) + g(x - y)$ is

Show answer and solution

g(x) = \frac{1}{2}(a^{x} + a^{- x})

h(x) = \frac{1}{2}(a^{x} - a^{- x})

g(x + y) + g(x - y) = \frac{1}{2}(a^{x + y} + a^{- (x + y)}) + \frac{1}{2}(a^{x - y} + a^{- x + y})

= 2.\frac{1}{4}\lbrack a^{x}(a^{y} + a^{- y}) + a^{- x}(a^{y} + a^{- y})\rbrack

= 2\lbrack\frac{1}{2}(a^{x} + a^{- x}) \cdot \frac{1}{2}(a^{y} + a^{- y})\rbrack

= 2\lbrack\frac{1}{2}(a^{x} + a^{- x}) \cdot \frac{1}{2}(a^{y} + a^{- y})\rbrack

= 2g(x)g(y)

Source: Grass-G / M-17 Mathematics Paper 26 Nov Second.docx

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