The function f  satisfying f(b)−f(a)b−a≠f'(x)  for any x∈(a,b)  is

The function in (a) ,(c) ,(d) satisfying hypothesis of  Lagrange’s mean value theorem so there is c∈(a,b)  such that f(b)−f(a)b−a=f'(c). The function in (b) is differentiable on (1,2)  but not continuous at x=1  and x=2  as limx→1+f(x)=limx→1+x2=1≠f(1)  and limx→2−f(x)=limx→2−x2=4≠f(2). Moreover , f(2)−f(1)2−1=−1  and f'(x)=2x>0  on (1,2)  so there is no x∈(1,2)  such that f(2)−f(1)2−1=f'(x).