Recall that \(\boxed{m=\tan\theta} \).

The cause of the problem: What we're after is the

**range** of the particle. Recall that the range of the particle is

**maximised** when \(\theta = 45^\circ\). In other words, as you increase \(\theta\) from \(0^\circ\) up to \(45^\circ\), the range increases. But as you increase \(\theta\) from \(45^\circ\) to \(90^\circ\), the range decreases.

What I will do now is sketch \( R = \frac{40\tan\theta}{1+\tan^2\theta} \) on GeoGebra, with the four angles above included. (You can actually prove that in fact, \(R = 20\sin 2\theta\)).

The problem: If we just plug \(m=0.8\) and \(m=1.2\) blindly into \( \frac{40m}{1+m^2} \), we will miss out on the fact that there's a stationary point at \(45^\circ\). Plugging \( m = 0.8\) and \(m = 1.2\) will give our ranges \(R = 19.512\) and \(R = 19.672\). This would make us misbelieve that the width of the interval is just 19.672 - 19.512 = 0.160.

But in reality, when \(\theta=45^\circ\) and hence \(m=1\), we see \(R = 20\). This is actually a longer range! Hence the width of the interval is actually 20 - 19.512 = 0.488

**Moral of the story: If you're given an interval, you need to consider any STATIONARY POINTS as well as the endpoints.**Remark: This problem does not occur for the other interval, since we don't bump into any more stationary points.