Is a rocket engine most efficient when its velocity is equal to the exhaust velocity?

I was reading Dale's answer on this question and from the answer I concluded the following:

1- When the rocket is traveling at a velocity much less than the exhaust speed, most of the chemical energy of the fuel is used in increasing the KE of the exhaust and and a smaller part is used in increasing the KE of the rocket itself.

2- When the rocket is traveling at a velocity = $$1/2$$ the exhaust velocity, the change in the KE of the fuel is zero and therefore all the chemical energy of the burnt fuel is used in increasing the KE of the rocket.

3- When the rocket is traveling at a velocity = the exhaust velocity, the exhaust now is brought to a stop and the change in its KE is at maximum value and therefore ALL of the already gained KE is now gained by rocket and the rocket KE is now being increased by both the chemical energy and the KE of the now-accelerated the fuel.

4- Now if we consider the case in which the rocket is traveling faster than the exhaust speed, the exhaust will not be brought to a stop, it will rather exit the nozzle at a velocity but in the direction of the rocket and therefore the change in its KE is less than in case 3. So the rocket KE is now being increased only by part of this fuel KE, and the chemical energy.

From the above analysis, it seems that case 3 in which the rocket velocity = the exhaust velocity is the most efficient since all the KE of the fuel is utilized in accelerating the rocket. So is this analysis practically correct?

Edit 1: Wikipedia here has this paragraph which I guess confirms what I concluded: "However, a variable exhaust speed can be more energy efficient still. For example, if a rocket is accelerated from some positive initial speed using an exhaust speed equal to the speed of the rocket no energy is lost as kinetic energy of reaction mass, since it becomes stationary. (Theoretically, by making this initial speed low and using another method of obtaining this small speed, the energy efficiency approaches 100%, but requires a large initial mass.)"

Edit 2: I guess this too confirms the analysis.

• This strategy won't work very well at take-off, when the rocket velocity is zero. ;) BTW, you can answer your own question. – PM 2Ring Apr 7 at 10:01

$$\Delta KE_{exhaust}=KE_f - KE_i$$ $$=\frac{1}{2}m(v_f^2-v_i^2)$$ $$=\frac{1}{2}m((v_i-v_e)^2-v_i^2)$$ $$=-m v_i v_e+\frac{1}{2}m v_e^2$$ For case 3 $$v_i=v_e$$ so this becomes the expected $$-\frac{1}{2}m v_e^2$$, but note that for $$v_i>v_e$$ this quantity becomes even more negative. In fact, it becomes arbitrarily large (negative) as $$v_i$$ becomes arbitrarily large.