Step-by-Step Solution:

Step 1: Analyze the first property of disaccharide X.

The problem states that "A disaccharide $\mathbf{X}$ cannot be oxidised by bromine water." Bromine water is a mild oxidizing agent. It can oxidize aldoses (sugars with a free aldehyde group) but not ketoses or sugars where the anomeric carbon is part of a glycosidic bond (non-reducing sugars). Sugars that can be oxidized by bromine water are called reducing sugars, and they must have a free hemiacetal or hemiketal group which can open up to form an aldehyde or ketone.

Since disaccharide $\mathbf{X}$ cannot be oxidized by bromine water, it must be a non-reducing sugar. In a non-reducing disaccharide, the anomeric carbons of both monosaccharide units are involved in the glycosidic linkage, so there is no free hemiacetal group.

Let's examine the options:

• Option A (Sucrose): It is formed from $\alpha$-D-glucose and $\beta$-D-fructose. The glycosidic linkage is between C1 of glucose (anomeric carbon) and C2 of fructose (anomeric carbon). Since both anomeric carbons are involved in the bond, sucrose has no free hemiacetal group and is a non-reducing sugar.
• Option B (Lactose): It is formed from $\beta$-D-galactose and D-glucose, linked by a $\beta$-1,4 glycosidic bond. The C1 (anomeric carbon) of the glucose unit is free and exists as a hemiacetal. Therefore, lactose is a reducing sugar.
• Option C (Maltose): It is formed from two $\alpha$-D-glucose units, linked by an $\alpha$-1,4 glycosidic bond. The C1 (anomeric carbon) of the second glucose unit is a free hemiacetal. Therefore, maltose is a reducing sugar.
• Option D (Cellobiose): It is formed from two $\beta$-D-glucose units, linked by a $\beta$-1,4 glycosidic bond. The C1 (anomeric carbon) of the second glucose unit is a free hemiacetal. Therefore, cellobiose is a reducing sugar.

Based on the first condition alone, only Option A (Sucrose) is a possible candidate for $\mathbf{X}$.

Step 2: Analyze the second property of disaccharide X.

The problem states that "The acid hydrolysis of $\mathbf{X}$ leads to a laevorotatory solution." Acid hydrolysis breaks the glycosidic bond, yielding the constituent monosaccharides. A laevorotatory solution rotates plane-polarized light to the left (negative specific rotation, $[\alpha]_D < 0$).

Let's analyze the hydrolysis product of sucrose (Option A):

$\text{Sucrose} + H_2O \xrightarrow{H^+} \text{D-Glucose} + \text{D-Fructose}$

We need to check the optical rotation of the resulting equimolar mixture of D-Glucose and D-Fructose.

• The specific rotation of D-Glucose at equilibrium is $[\alpha]_D = +52.7^\circ$ (dextrorotatory).
• The specific rotation of D-Fructose is $[\alpha]_D = -92.4^\circ$ (laevorotatory).

The net specific rotation of the mixture is the sum of the individual contributions:

$[\alpha]_{D, \text{mixture}} = (+52.7^\circ) + (-92.4^\circ) = -39.7^\circ$

Since the net rotation is negative, the solution after hydrolysis is laevorotatory. This phenomenon is called the inversion of sucrose.

This confirms that Option A satisfies the second condition as well.

Step 3: Verify other options (for completeness).

• Hydrolysis of Lactose (B) yields D-Galactose ($[\alpha]_D = +80.2^\circ$) and D-Glucose ($[\alpha]_D = +52.7^\circ$). The resulting solution is strongly dextrorotatory.
• Hydrolysis of Maltose (C) yields two molecules of D-Glucose ($[\alpha]_D = +52.7^\circ$). The resulting solution is dextrorotatory.
• Hydrolysis of Cellobiose (D) yields two molecules of D-Glucose ($[\alpha]_D = +52.7^\circ$). The resulting solution is dextrorotatory.

Conclusion:

Disaccharide $\mathbf{X}$ must be sucrose (Option A) because it is the only non-reducing sugar among the options, and its acid hydrolysis results in a laevorotatory solution.