11f(0)=f(1)
and f'(1/4)=0
f'(x)=-f'(1-x)
f"(1-x)=f"(x)
putting x=0 and x=1 we get the option A
=>2f'(1/2)=0
so f'(1/2)=0
option B is correct
S3 will be correct only if f(x+1/2) is even f(x)
it is an even f(x) from -1/2 to 1/2
so C satisfies the condition
as the graph is symmetrical about the line x=1/2 and f(x)=f(1-x)
we get D