Let $f(x+y)+f(x-y)=2\,f(x)f(y), ∀\,y∈R$ and f(0) = k then:
I. f(x) even if k = 1
II. f(x) is odd, if k = 0
III. f(x) is always odd
IV. f(x) is neither even nor odd for any value of k
The correct choice is:
Answer & explanation
Correct answer: option 3
$f(x+y)+f(x-y)=2\,f(x)f(y),\,f(0)=k$
at $x=0,y=0$
$⇒f(0)+f(0)=2f(0)f(0)⇒2f(0)=2f^2(0)$
$f(0)=k=0,1$
let $x=0$
$⇒f(y)+f(-y)=2f(0)f(y)$ when $k=0$
$f(y)+f(-y)=0$ odd function
when $k = 1$
so $f(y)+f(-y)=2f(y)$
$⇒f(-y)=f(y)$ even function