If x ∈ (a, b) satisfies the inequality, $$\dfrac{x-3}{x^2+3x+2}\ge1$$ then the largest possible value of b - a is

$$\dfrac{x-3}{x^2+3x+2}\ge1$$

or, $$\dfrac{x-3}{x^2+3x+2}-1\ge0$$

or, $$\dfrac{x-3-x^2-3x-2}{x^2+3x+2}\ge0$$

or, $$\dfrac{-x^2-2x-5}{x^2+3x+2}\ge0$$

or, $$\left(x^2+2x+5\right)\left(x^2+3x+2\right)\le0$$

Now, $$\left(x^2+2x+5\right)=\left(x+1\right)^2+4>0$$ always

So, $$x^2+3x+2\le0$$

or, $$\left(x+2\right)\left(x+1\right)\le0$$

or, $$-2\le x\le-1$$

So, $$x$$ lies in the interval $$(-2,-1)$$

So, $$a=-2$$ and $$b=-1$$

So, $$b-a=-1-(-2)=1$$