Question

Let f(x) = a^x (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 + y) f₂(x - y)
• B. 2f₂(x)f(y)
• C. 2f₁(x + y) f₁(x - y)
• D. 2f₁(x)f₁(y)

Answer 1

Answer: (D)

2f₁(x)f₁(y)

Answer 2

Given $f(x) = a^x$ for $a > 0$. We can decompose $f(x)$ into even and odd parts as $f(x) = f_1(x) + f_2(x)$, where $f_1(x)$ is even and $f_2(x)$ is odd.

The even part of $f(x)$ is given by $f_1(x) = \frac{f(x) + f(-x)}{2}$. The odd part is $f_2(x) = \frac{f(x) - f(-x)}{2}$.

For $f(x) = a^x$, we have $f(-x) = a^{-x}$. Thus, $f_1(x) = \frac{a^x + a^{-x}}{2}$ and $f_2(x) = \frac{a^x - a^{-x}}{2}$.

We want to find $f_1(x + y) + f_1(x - y)$. We have $f_1(x+y) = \frac{a^{x+y} + a^{-(x+y)}}{2} = \frac{a^{x+y} + a^{-x-y}}{2}$ $f_1(x-y) = \frac{a^{x-y} + a^{-(x-y)}}{2} = \frac{a^{x-y} + a^{-x+y}}{2}$

Adding these two expressions: $f_1(x+y) + f_1(x-y) = \frac{a^{x+y} + a^{-x-y}}{2} + \frac{a^{x-y} + a^{-x+y}}{2} = \frac{a^{x+y} + a^{-x-y} + a^{x-y} + a^{-x+y}}{2}$

Using properties of exponents: $= \frac{a^x a^y + a^{-x} a^{-y} + a^x a^{-y} + a^{-x} a^y}{2} = \frac{a^x (a^y + a^{-y}) + a^{-x} (a^y + a^{-y})}{2}$ $= \frac{(a^x + a^{-x})(a^y + a^{-y})}{2}$

Since $f_1(x) = \frac{a^x + a^{-x}}{2}$ and $f_1(y) = \frac{a^y + a^{-y}}{2}$, we have $a^x + a^{-x} = 2 f_1(x)$ and $a^y + a^{-y} = 2 f_1(y)$.

Substituting these into the expression: $f_1(x+y) + f_1(x-y) = \frac{(2 f_1(x))(2 f_1(y))}{2} = \frac{4 f_1(x) f_1(y)}{2} = 2 f_1(x) f_1(y)$.

Therefore, $f_1(x + y) + f_1(x - y) = 2f_1(x)f_1(y)$.