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 ______.

Options

• 2f₁(x)f₁(y)

• 2f₁(x + y)f₁(x – y)

• 2f₁(x)f₂(y)

• 2f₁(x + y)f₂(x – y)

Answer 1

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 `underlinebb(2f_1(x)f_1(y))`.

Explanation:

Every function f(x) can be represented as f(x) = `(f(x) + f(-x))/2 + (f(x) - f(x))/2`, where `(f(x) + f(-x))/2` is even function and `(f(x) - f(-x))/2` is odd function.

f(x) = a^x

So, f₁(x) = `(a^x + a^-x)/2`

∴ f₁(x + y) + f₁(x − y)

= `(a^(x + y) + a^(-(x + y)))/2 + (a^(x - y) + a^(-(x - y)))/2`

= `(a^x(a^y + a^-y) + a^-x(a^y + a^-y))/2`

= `((a^x + a^-x)(a^y + a^-y))/2`

= 2f₁(x)f₁(y)