The wave function corresponding to energy level $n$ is
$\begin{align*}\psi_n = \left(\text{const} \right) H_n \left(\xi \right) e^{-\xi^2/2}\end{align*}$
where $\xi = \left(\text{const} \right) x$ and $H_n$ is the $n$ th Hermite polynomial. These wavefunctions represent all possible solutions to the harmonic oscillator potential that go to $0$ when $\left|\xi\right|$ or equivalent $\left|x\right|$ goes to infinity. Therefore, these wavefunctions represent all possuble solutions to the harmonic oscillator potential on a domain that is infinite in at least one direction (like in this problem). Now, Hermite polynomials are even functions if $n$ is even and odd functions if $n$ is odd. Therefore, only those $\psi_n$ where $n$ is odd satisfy the boundary condition $\psi_n\left(0 \right) = 0$ . Therefore, only these $\psi_n$ are solutions for the potential in this problem. Therefore, answer (E) is correct.

Solution to 1992 Problem 89