A previous answer uses an intuitive "let's use a linear function with a minimal derivative" solution.
For a rigorous argument, one can use the mean value theorem (in undergraduate mathematics, this is considered a "basic" theorem). This theorem says that, if $f$ is continuously differentiable*, then there must be an $x\in[1,5]$ (that is, an $x$ between 1 and 5) such that:
First note that I do not know what $x$ is. It is definitely in the range $[1,5]$, but the mean value theorem doesn't tell us what it is.
Also notice that I'm using a strict equality here - mathematicians like that, because we can now solve for $f(5)$ (which is good, to say the least):
But now it is easy! If $f'(x)\geq7$ (for /any/ $x$), then $4f'(x)\geq28$, so $4f'(x)+f(1)\geq28+f(1)=25$, thus answer a) would be correct.
*There are many functions which are not continuously differentiable, but we usually do not consider them. Also, the question implicitly stated that the second derivative exists for all $x$ (namely, for all $x$ it has some value greater than or equal to 7), which is sufficient for the mean value theorem.