Question

Let f(x) = aˣ (a > 0) be written as f(x) = f₁(x) + f₂(x), where f₁(x) is an even function and f₂(x) is an odd function. Then f₁(x + y) + f₁(x - y) equals (A) 2f₁(x)f₂(y) (B) 2f₁(x) f₁(y)
(C) 2f₁(x + y) f₂(x - y) (D) 2f₁(x + y) f₁(x - y)

Answer 1

The value of  f₁ ( x + y ) + f₁( x - y ) equals 2f₁ ( x ) f₁ ( y )

• Given  f ( x )  =  aˣ ( a  >  0 )
• f ( x ) = f₁ ( x ) + f₂ ( x ) , where f₁ ( x ) is an even function and f₂(x) is an odd function.

We can take

• f1 ( x ) = ( aˣ + $a^{-x}$ ) / 2 since f1 is even

and

• f2 ( x ) = ( aˣ  - $a^{-x}$  )  / 2 since f2 is odd

• f₁( x  + y ) + f₁ ( x - y )  = ( $a^{x + y}$ + $a^{- (x+y)}$ ) / 2 + ( $a^{x-y} + a^{-(x -y)}$ ) / 2

• f₁ ( x +  y )  + f₁ ( x - y ) = ( $a^{x} + a^{-x}$ ) ( $a^{y} + a^{-y}$) / 2

Hence,

• f₁( x + y ) + f₁( x - y ) = 2 f1 ( x ) . f1 ( y )