| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2018 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Find constant using stationary point |
| Difficulty | Moderate -0.3 This is a straightforward multi-part calculus question requiring standard techniques: setting derivative to zero at a stationary point to find a constant, integrating to find the curve equation, and using the second derivative test. While it involves multiple steps, each step uses routine A-level methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0 = 9a + 3a^2\) | M1 | Sub \(\frac{dy}{dx} = 0\) and \(x = 3\) |
| \(a = -3\) only | A1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = -3x^2 + 9x \rightarrow y = -x^3 + \frac{9x^2}{2}\ (+c)\) | M1A1FT | Attempt integration. \(\frac{1}{3}ax^3 + \frac{1}{2}a^2x^2\) scores M1. Ft on *their* \(a\) |
| \(9\frac{1}{2} = -27 + 40\frac{1}{2} + c\) | DM1 | Sub \(x=3, y=9\frac{1}{2}\). Dependent on \(c\) present |
| \(c = -4\) | A1 | Expect \(y = -x^3 + \frac{9x^2}{2} - 4\) |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{d^2y}{dx^2} = -6x + 9\) | M1 | \(2ax + a^2\) scores M1 |
| At \(x=3\), \(\frac{d^2y}{dx^2} = -9 < 0\) MAX AG | A1 | Requires at least one of \(-9\) or \(< 0\). Other methods possible |
| Total: 2 |
## Question 6:
**Part (i):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0 = 9a + 3a^2$ | M1 | Sub $\frac{dy}{dx} = 0$ and $x = 3$ |
| $a = -3$ only | A1 | |
| | **Total: 2** | |
**Part (ii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = -3x^2 + 9x \rightarrow y = -x^3 + \frac{9x^2}{2}\ (+c)$ | M1A1FT | Attempt integration. $\frac{1}{3}ax^3 + \frac{1}{2}a^2x^2$ scores M1. Ft on *their* $a$ |
| $9\frac{1}{2} = -27 + 40\frac{1}{2} + c$ | DM1 | Sub $x=3, y=9\frac{1}{2}$. Dependent on $c$ present |
| $c = -4$ | A1 | Expect $y = -x^3 + \frac{9x^2}{2} - 4$ |
| | **Total: 4** | |
**Part (iii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{d^2y}{dx^2} = -6x + 9$ | M1 | $2ax + a^2$ scores M1 |
| At $x=3$, $\frac{d^2y}{dx^2} = -9 < 0$ MAX **AG** | A1 | Requires at least one of $-9$ or $< 0$. Other methods possible |
| | **Total: 2** | |
---6 A curve has a stationary point at $\left( 3,9 \frac { 1 } { 2 } \right)$ and has an equation for which $\frac { \mathrm { d } y } { \mathrm {~d} x } = a x ^ { 2 } + a ^ { 2 } x$, where $a$ is a non-zero constant.\\
(i) Find the value of $a$.\\
\includegraphics[max width=\textwidth, alt={}, center]{d178603a-f59a-4986-b5ab-b47eceedb2fc-08_67_1569_461_328}\\
(ii) Find the equation of the curve.\\
(iii) Determine, showing all necessary working, the nature of the stationary point.\\
\hfill \mbox{\textit{CAIE P1 2018 Q6 [8]}}