Confusing supremum of a function question Let $f(x)$ be continuous on $D = [a,b]$ where $0&lt;b-a &lt; \infty$

T/F, $f(x) &lt; F^* = supf$(on domain D) $\forall x \in [a,b]$, but there is no $x_0$ in $[a,b]$ where $f(x_0) = F^*$

I do not know where to start with this problem, because it doesn't seem to make any logical sense to me.
It's simply asking if it is true or false that $f(x) &lt; \sup f ([a,b])$ and never equal. 

The way I see it there are three possible awswers:

TRUE. It is always true for all continuous f on all closed intervals.

FALSE. It is never true for any continuous f on any closed interval.

AMBIGUOUS.  It is true for some not for others.

Try a few.  Let $f (x) = 2-x^2$  on [-1,2].  What do you think sup f is.  Does f (x) = sup f ever?  Can you prove the ambiguous case never happens? Or find samples were the statement occurs and another where it doesn't?
Consider the function $f(x) = -1/x$ on the interval $[1, \infty)$.  The supremum of this function is $0$ even though there is no $x$ on the interval such that $f(x) = 0$.  But, this is an open interval, not a closed interval.  The question may have to do with whether or not it is possible for this to be the case for a function defined over a closed interval.

Hint: take a look at the Extreme Value Theorem.