A disaccharide X cannot be oxidised by bromine water. The acid hydrolysis of X leads to a laevorotatory solution. The disaccharide X is

Bromine water is a very mild oxidising agent. It oxidises only those carbohydrates that possess a free aldehydic $$-CHO$$ group (or its equivalent $$\alpha$$-hydroxy-ketone) present in the open-chain (hemiacetal/hemiketal) form. Such carbohydrates are called reducing sugars.

If a disaccharide does not react with bromine water, both of its anomeric carbon centres must be tied up in the glycosidic linkage, so that no monomer can open out to the free $$-CHO$$ form. Hence X is a non-reducing disaccharide.

Among the common disaccharides asked in JEE, the only non-reducing member is sucrose (glucose-α-1→2-β-fructose). Lactose and maltose are reducing; cellobiose is also reducing.

To cross-check, let us look at the optical activity on hydrolysis.
Sucrose $$\xrightarrow[\text{H}_2O]{\text{acid}}$$ glucose + fructose

The specific rotations are:
  • D-glucose : $$+52.5^\circ$$
  • D-fructose : $$-92.0^\circ$$

Stoichiometrically 1 mol sucrose gives 1 mol each of glucose and fructose. The net rotation after complete hydrolysis is therefore

$$\alpha_{\text{mix}}=\frac{(+52.5)+(-92.0)}{2}= -19.75^\circ$$

The mixture is laevorotatory. The change of sign from dextrorotatory sucrose $$(+66.5^\circ)$$ to a laevorotatory hydrolysate is called inversion; the product mixture is known as invert sugar.

Thus all given observations are satisfied only by sucrose.

Option A which is: sucrose