The correct options are
B f′(x)=0 for some x∈(a,b)
C f′′(x)=0 for some x∈(a,b)
Applying Rolle's theorem to f on the interval [a,b], we get
f′(c)=0 for some c∈(a,b)
Again f′(a)=0=f′(c) for some x∈(a,c)⊂(a,b)
Therefore, Rolle's theorem is applicable to f′ on the interval (a,c)
Hence, f′′(c1)=0 for some c1∈(a,c)