The problem $$-y^{\prime\prime} +(1+x)y=\lambda y, x\in (0,1); y(0)=y(1)=0$$
has a non zero solution
$1$. for all $\lambda$ $<0$
$2$. for all $\lambda$ $\in$ $[0,1]$
$3$. for some $\lambda $ $\in$ $(2, \infty)$
$4$. for a countable number of $\lambda$'s
I know that this i a regular $Sturm-Liouville$ BVP. So there could be a countable no. of eigenfunctions. And the eigen values are simple. So can I reject options $1$ and $2$. I think option $4$ is right with this reason. I have no idea for option $3$
3 and 4 are both true. As you say, this is a regular Sturm-Liouville problem, in fact the one-dimensional Schrödinger equation with potential $1+x$ on the interval $[0,1]$ and Dirichlet boundary conditions. Such an operator has a countable number of eigenvalues tending to infinity, which makes 3 and 4 true. 1 and 2 are obviously false, and since $1+x>1$ in the interval (except for at 0), all eigenvalues are $>1$. Thus there is not a single eigenvalue $\le1$. Almost any medium level ODE text should prove all this.