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:

I, II

$f(x+y)+f(x-y)=2\,f(x)f(y),\,f(0)=k$

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