 Solution

Fix $\lambda >0$ and choose the positive rootUse the Exponential martingale for Brownian motion $M_t=\exp (\theta B_t-{\theta }^{2}t/2)$. Before the linear-boundary stopping time $\tau$, path continuity gives $B_s<a+bs$. At the boundary there is equality. ConsequentlyThe optional stopping theorem at the bounded stopping time $t\wedge \tau$ yields $\mathbb{E}{M}_{t\wedge \tau }=1$. Separate the two events:The last term is at most $e^{\theta a-\lambda t}$ and tends to zero. The first term converges by the monotone convergence theorem. Taking $e^{-\lambda \infty }=0$ therefore provesThis bounded-stopping argument does not assume in advance that $\tau$ is finite. For $b\le 0$, letting $\lambda \downarrow 0$ makes the right side tend to one, so it also proves $\mathbb{P}(\tau <\infty )=1$. Setting $b=0$ gives the Brownian first-passage Laplace transformThe zero-parameter value follows from the just-proved almost-sure finiteness.