= Solution
Because $v\in L^1(0,1)$, its <indefinite integral> $V$ is bounded, so $e^{-V}$ is bounded above and away from zero. Hence $\psi$ extends continuously and strictly increasingly to $[0,1]$. The quadratic-variation clock of $\psi(X)$ is
$$
\int_0^te^{-2V(X_s)}ds,
$$
whose rate is bounded above and away from zero before exit. The <Dambis-Dubins-Schwarz theorem> therefore identifies $\psi(X)$, up to an equivalent time change, with Brownian motion in the bounded interval $(0,\psi(1))$; in particular, $\mathcal T<\infty$ almost surely.
The bounded stopped local martingale $\psi(X_{t\wedge\mathcal T})$ is a martingale. If $q=\mathbb P_x(X_{\mathcal T}=0)$, the <optional sampling theorem> gives
$$
\psi(x)=q\psi(0)+(1-q)\psi(1)=(1-q)\psi(1).
$$
Therefore
$$
\mathbb P_x(\mathcal T<\infty,X_{\mathcal T}=0)
=\frac{\psi(1)-\psi(x)}{\psi(1)}>0.
$$
This is the <boundary hitting probability from a diffusion scale function>.
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