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

The functions 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)