IB Mathematics SL 5.6 Local maximum and minimum points AI HL Paper 1- Exam Style Questions- New Syllabus

(a)
Set \(w=0\): \(p^3 – 4p^2 + 3p = 0\)
Factor out \(p\): \(p(p^2 – 4p + 3) = 0\)
Factor the quadratic: \(p(p-3)(p-1) = 0\)
Possible values: \(p = 0, 1, 3\).

(b)
We are looking for values of \(w > 0\) where the horizontal line \(y=w\) intersects the curve \(w(p)\) at only one positive \(p\) location.
Consider the local maximum and minimum. Using calculus or a GDC:
Find stationary points: \(\frac{dw}{dp} = 3p^2 – 8p + 3 = 0\).
Using the quadratic formula: \(p = \frac{8 \pm \sqrt{64 – 36}}{6} = \frac{8 \pm \sqrt{28}}{6} \approx 0.451, 2.215\).
Calculate \(w\) at the local maximum (the smaller \(p\)):
\(p \approx 0.451 \Rightarrow w \approx 0.631\).
The curve goes up to \(0.631\), down to a local minimum (negative value), and then up again.
For \(w\) values above the local maximum (\(w > 0.631\)), the line intersects the curve only once for \(p > 0\) (on the right branch).
Therefore, for there to be only one positive value of \(p\) (ignoring the complex roots or negative/zero roots that don’t apply), \(w\) must be greater than the local maximum.
Answer: \(w > 0.631\) (approx).
(Note: There is also a small region between \(w=0\) and the local max where there are 3 positive roots. Above the max, there is only 1).
Conclusion: \(w > 0.631\).