Answer
a. See explanations.
b. Yes.
Work Step by Step
a. The given conditions indicate that the Mean Value Theorem can be used. For $x$ in $(-\infty,1)$, $f'(c)=\frac{f(1)-f(x)}{1-x}\lt0$ which gives $f(1)-f(x)\lt0$ or $f(x)\gt f(1)=1$. For $x$ in $(1,\infty)$, $f'(c)=\frac{f(x)-f(1)}{x-1}\gt0$ which gives $f(x)\gt f(1)=1$. Thus $f(x)\geq1$ for all $x$.
b. We have $\lim_{x\to1^-}f'(x)\leq0$ and $\lim_{x\to1^+}f'(x)\geq0$. Since it is given that $f'(1)$ exists, the two limits need to be equal and we have $f'(0)=0$
