this ques is extremely easy.
let h(x)= ln(f(x)+f’(x)+...+f(n)(x))
then h(b)-h(a)= ln[(f(b)+f’(b)+...+f(n)(b))/(ln(f(a)+f’(a)+...+f(n)(a))]=b – a (given)
so that [h(b)-h(a)]/[b-a] = 1
however, by rolles theorem [h(b)-h(a)]/[b-a] = h’(c) for some c in (a,b)
or 1= h’(c)
but h(x)= ln(f(x)+f’(x)+...+f(n)(x)) which implies h’(x)= [1/(f(x)+f’(x)+...+f(n)(x))]*[f’(x)+f”(x)+...+f(n+1)(x)]
so h’(c)= [1/(f(c)+f’(c)+...+f(n)(c))]*[f’(c)+f”(c)+...+f(n+1)(c)] = 1
or f’(c)+f”(c)+...+f(n+1)(c)= f(c)+f’(c)+...+f(n)
or f(n+1)(c)=f(c)
hence statement 2 correctly explains stat 1.