Geometrical isomerism (cis-trans isomerism) is exhibited by alkenes only when each doubly bonded carbon atom is attached to two different atoms or groups (e.g., $\text{XYC=CXY}$, $\text{XYC=CXZ}$, or $\text{XYC=CZW}$) .

Let's evaluate the given options:

1. Pent-1-ene ($\text{CH}_2\text{=CH-CH}_2\text{CH}_2\text{CH}_3$): The terminal carbon atom (C1) is attached to two identical hydrogen atoms, so it will not show geometrical isomerism.
2. 2,3-Dimethylbut-2-ene ($\text{(CH}_3)_2\text{C=C(CH}_3)_2$): Both doubly bonded carbon atoms are attached to two identical methyl groups.
3. 2-Methylprop-1-ene ($\text{CH}_2\text{=C(CH}_3)_2$): The C1 carbon has two identical hydrogen atoms, and C2 has two identical methyl groups.
4. 3,4-Dimethylhex-3-ene ($\text{CH}_3\text{CH}_2\text{-C(CH}_3)\text{=C(CH}_3)\text{-CH}_2\text{CH}_3$): Each doubly bonded carbon atom is attached to two different groups (one methyl group and one ethyl group). Therefore, it can exist as cis and trans isomers and exhibits geometrical isomerism.