If f(x+k)+f(x)=0 ∀ x∈R and k∈R+, then the period of f(x) is
0
π2
2K
Does not exist
f(x+k)+f(x)=0
x is replaced by x+k
then we get f(x+2k)+f(x+k)=0
⇒f(x+2k)=−f(x+k)
⇒f(x+2k)=f(x)