| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2018 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Optimise perimeter or area of 2D region |
| Difficulty | Standard +0.3 This is a straightforward optimization problem requiring students to set up an expression for PQ by subtracting y-coordinates, then differentiate and solve dP/dt = 0. The algebra is simple (cubic minus linear), differentiation is routine, and finding the maximum requires only basic calculus techniques. Slightly easier than average due to clear setup and standard method. |
| Spec | 1.02q Use intersection points: of graphs to solve equations1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| 3(i) | P is (t, 5t) Q is (t, t(9 – t²)) → 4t – t³ | B1 B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks |
|---|---|
| 3(ii) | ( ) |
| Answer | Marks | Guidance |
|---|---|---|
| dt | B1FT | B1FT for differentiation of their PQ, which MUST be a cubic |
| Answer | Marks | Guidance |
|---|---|---|
| √3 | M1 | Setting their differential of PQ to 0 and attempt to solve for t |
| Answer | Marks | Guidance |
|---|---|---|
| 3√3 9 | A1 | Allow 3.08 awrt. If answer comes from wrong method in (i) |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 3:
--- 3(i) ---
3(i) | P is (t, 5t) Q is (t, t(9 – t²)) → 4t – t³ | B1 B1 | B1 for both y coordinates which can be implied by
subsequent working. B1 for PQ allow 4t – t³ or t³– 4t .
Note: 4x – x3 from equating line and curve 0/2 even if x then
replaced by t.
[2]
Question | Answer | Marks | Guidance
--- 3(ii) ---
3(ii) | ( )
d PQ
= 4 – 3t²
dt | B1FT | B1FT for differentiation of their PQ, which MUST be a cubic
d ( )
expression, but can be f x from (i) but not the equation
dx
of the curve.
2
= 0 → t = +
√3 | M1 | Setting their differential of PQ to 0 and attempt to solve for t
or x.
16 16 3
→ Maximum PQ = or
3√3 9 | A1 | Allow 3.08 awrt. If answer comes from wrong method in (i)
award A0.
Correct answer from correct expression by T&I scores 3/3.
3
Question | Answer | Marks | Guidance\includegraphics{figure_3}
The diagram shows part of the curve $y = x(9 - x^2)$ and the line $y = 5x$, intersecting at the origin $O$ and the point $R$. Point $P$ lies on the line $y = 5x$ between $O$ and $R$ and the $x$-coordinate of $P$ is $t$. Point $Q$ lies on the curve and $PQ$ is parallel to the $y$-axis.
\begin{enumerate}[label=(\roman*)]
\item Express the length of $PQ$ in terms of $t$, simplifying your answer. [2]
\item Given that $t$ can vary, find the maximum value of the length of $PQ$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2018 Q3 [5]}}