If \(x ∈ (a, b)\) satisfies the inequality \(\dfrac{x - 3}{x^2 + 3x + 2} \geq 1\), then the largest possible value of \(b - a\) is

(x-3)/(x^2+3x+2) - 1 = -(x^2+2x+5)/((x+1)(x+2)) >= 0. Numerator is always negative, so denominator must be negative: -2 < x < -1. Largest b-a = 1.