Let f(x)=a^{x}(a>0) be written as f(x)=f_{1}(x)+f_{2}(x), where f_{1}(x) is an even function and f_{2}(x) is an odd function. Then f_{1}(x+y)+f_{1}(x-y) equals:

  • Option 1)

    2f_{1}(x)f_{1}(y)

  • Option 2)

    2f_{1}(x+y)f_{1}(x-y)

  • Option 3)

    2f_{1}(x)f_{2}(y)

  • Option 4)

    2f_{1}(x+y)f_{2}(x-y)

 

Answers (1)

f(x)=a^{x}, a>0

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

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

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

Now,

f_{1}(x+y)+f_{2}(x-y)

=\frac{a^{x+y}+a^{-x-y}}{2}+\frac{a^{x-y}+a^{-x+y}}{2}

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

=f_{1}(x)\times 2f_{1}(y)

=2f_{1}(x)\times f_{1}(y)

 

 


Option 1)

2f_{1}(x)f_{1}(y)

Option 2)

2f_{1}(x+y)f_{1}(x-y)

Option 3)

2f_{1}(x)f_{2}(y)

Option 4)

2f_{1}(x+y)f_{2}(x-y)

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