Maltose on treatment with dilute HCl gives:

We first recall what maltose is. Maltose is a disaccharide whose molecular formula is $$\mathrm{C_{12}H_{22}O_{11}}$$ and whose structure consists of two units of $$\alpha\text{-D-glucopyranose}$$ (that is, two $$\alpha$$-D-glucose molecules) joined together through an $$\alpha(1 \rightarrow 4)$$ glycosidic linkage.

Whenever a disaccharide is treated with dilute mineral acid such as $$\text{dil.\,HCl}$$ in the presence of water and heat, the process that occurs is called acid-catalysed hydrolysis. The general chemical fact we use is:

“A glycosidic bond is cleaved by water in the presence of dilute acid, giving back its constituent monosaccharide units.”

Writing the reaction in symbolic form, we have

$$\text{Disaccharide} + \mathrm{H_2O} \xrightarrow[\text{heat}]{\text{dil.\,HCl}} \text{Monosaccharide}_1 + \text{Monosaccharide}_2.$$

Substituting maltose specifically, the equation becomes

$$\mathrm{C_{12}H_{22}O_{11}} + \mathrm{H_2O} \;\xrightarrow[\text{heat}]{\text{dil.\,HCl}}\; \mathrm{C_6H_{12}O_6} + \mathrm{C_6H_{12}O_6}.$$

Each $$\mathrm{C_6H_{12}O_6}$$ appearing on the right-hand side represents a molecule of D-glucose, because both halves of maltose are $$\alpha$$-D-glucose units to begin with. No other monosaccharide (such as fructose or galactose) is present in maltose, so none can appear after hydrolysis. Thus the only product is D-glucose.

So, upon treatment of maltose with dilute $$\mathrm{HCl}$$, we obtain exclusively D-glucose.

Hence, the correct answer is Option A.