Let f(x)=ex−12nsinnxalog1+xan for x≠016n for x=0and f is a continuous at x=0 then the value of a is
16
2
8
4
Ltx=0 fx=Ltx=0 ex−1x2nxansinnxa×xanlog1+xan.a2n=a2n=f0
∴a2n=16n⇒a=4