The function, f(x)=x2nx2nsgn⁡x2n+1e1x−e−1xe1x+e−1xx≠01x=0n∈N is:

f(x)=x2nx2nsgn⁡x2n+1e1x−e−1xe1x+e−1xx≠0 and f(0)=1when     f(x)=x2nx2n2n+1e1x−e−1xe1x+e−1x;x>0f(x)=x2n−x2n2n+1e1x−e−1xe1x+e−1x;x<0 Clearly, f(x)=f(−x). Hence, f(x) is even function.