What are the functional groups present in the structure of maltose?

First, let us recall that $$\text{maltose}$$ is a disaccharide obtained by the condensation of two $$\alpha\text{-D-glucopyranose}$$ units. These two monosaccharide rings are linked through their anomeric carbon of the first glucose (C-1) and the hydroxyl group attached to C-4 of the second glucose. This linkage is called an $$\alpha(1\rightarrow 4)$$ glycosidic bond.

An anomeric carbon that is part of a free (unbound) ring possesses the $$-OR$$ and $$-OH$$ pattern that characterises a hemiacetal. In contrast, when that anomeric carbon loses its $$-OH$$ during glycoside formation and now bears two $$-OR$$ groups, it is converted into an acetal.

We have one glucose residue acting as the glycosyl donor; its anomeric $$-OH$$ participates in the condensation and is replaced by the $$-O$$ bridge to the second glucose. Therefore the anomeric carbon of this first glucose now has two $$-OR$$ substituents and is an acetal.

The second glucose residue is the glycosyl acceptor; its own anomeric carbon (C-1) does not participate in the bond formation, so it retains the original $$-OH$$ group on the anomeric carbon along with one $$-OR$$ group already present inside the ring. Hence this carbon still displays the $$-OH$$ + $$-OR$$ pattern of a hemiacetal.

Summarising, in maltose we find:

Thus, the molecule contains one acetal group and one hemiacetal group.

Hence, the correct answer is Option C.