The 1-dimensional Hamiltonian of a classical particle of mass m is $$H = \frac{p^2}{2m} e^{-x/a} + V(x)$$​​
where a is a constant with appropriate dimensions. The corresponding Lagrangian is,

Correct option is A

Solution:

$$H = \frac{p^2}{2m} e^{-x/a} - V(x)$$ 

$$L = \dot{x} p - H = \dot{x} p - \frac{p^2}{2m} e^{-x/a} + V(x)$$ 

$$\frac{\partial H}{\partial p} = \dot{x} \Rightarrow \frac{p}{m} e^{-x/a} = \dot{x} \Rightarrow p = m \dot{x} e^{x/a}$$ 

$$\text{so Lagrangian is } \frac{1}{2} m \dot{x}^2 e^{x/a} - V(x)$$.