Options A and B are definitely wrong, condition on $f(n)$ can’t be independent of $a$ and $b$ in any case, it should take both $a$ and $b$ into account.
Option C is correct. standard case of master theorem, if $f(n)$ is polynomial time smaller than $O(n^{\log_ba})$, then $T(n) = \Theta(n^{\log_ba})$. (see case $1$ below).
$\textbf{Theorem 4.1 (Master theorem)}$
Let $a \geq 1$ and $b>1$ be constants, let $f(n)$ be a function, and let $T(n)$ be defined on the nonnegative integers by the recurrence
$$T(n) = aT(n/b) + f(n),$$
where we interpret $n/b$ to mean either $\left \lfloor n/b \right \rfloor$ or $\left \lceil n/b \right \rceil.$ Then $T(n)$ has the following asymptotic bounds:
- If $f(n) = O(n^{\log_{b}a-\varepsilon})$ for some constant $\varepsilon >0,$ then $T(n) = \Theta (n^{\log_{b}a}).$
- If $f(n) = \Theta (n^{\log_{b}a}),$ then $T(n) = \Theta(n^{\log_{b}a}\lg n).$
- If $f(n) = \Omega(n^{\log_{b}a+\varepsilon})$ for some constant $\varepsilon > 0,$ and if $af(n/b)\leq cf(n)$ for some constant $c < 1$ and all sufficiently large $n,$ then $T(n) = \Theta (f(n)).$
Reference
Option D is wrong, (see case $2$ above).
A good slide to understand master theorem and the idea behind it.