1. **Stating the problem:** A body weighing 1 newton is on a rough horizontal plane with a coefficient of friction $\mu = \sqrt{3}$. A horizontal force is applied trying to move the body. We need to find the range in which the resultant reaction force $R$ lies. 2. **Understanding the forces:** - Weight $W = 1$ N acts vertically downward. - Normal reaction $N$ acts vertically upward. - Frictional force $F$ acts horizontally opposing motion, with maximum value $F_{max} = \mu N$. - Applied horizontal force $P$ tries to move the body. 3. **Reaction force components:** The resultant reaction force $R$ is the vector sum of the normal reaction $N$ and frictional force $F$: $$ R = \sqrt{N^2 + F^2} $$ 4. **Since the body is on a horizontal plane and weight balances the normal reaction:** $$ N = W = 1 $$ 5. **Maximum frictional force:** $$ F = \mu N = \sqrt{3} \times 1 = \sqrt{3} $$ 6. **Resultant reaction force magnitude:** $$ R = \sqrt{N^2 + F^2} = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 $$ 7. **Interpretation:** The resultant reaction force $R$ is exactly 2 newtons. **Final answer:** The resultant reaction force $R \in \{2\}$, which corresponds to option (d).