Let f(x) be a real function not identically zero in Z, such that f(x+y2n+1)=f(x)+{f(y)}2n+1,n∈N  and x,y∈R . If  f1(0) ≥ 0  then f1(6) is equal to

f(0)=0 f(1)=f(0)+{f(1)}2n+1f(1)[f(1)2n−1]=0f(1)=0 or 1 or =−1Now, f1(0)=ltx→0f(x)−f(0)n=ltx→0f(x)x≥0⇒f(x)≥0 for x>0 ∴f(1)≠−1Also f(1)=0 is not true (∵f(2)=f(3)=.......0)⇒f1=1 ∴f(x) is a liner function passing  through (0,0) ⇒f(x)=x        ⇒f1(6)=1