| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2020 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Area between curve and line |
| Difficulty | Standard +0.8 This question requires finding stationary points by differentiation, using the relationship between them to establish b=3a, then finding the area between a curve and line through integration. It involves multiple algebraic steps, parametric relationships, and careful integration with substitution, making it more challenging than standard area-under-curve questions but still within typical A-level scope. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dy}{dx} = 3x^2 - 4bx + b^2\) | B1 | |
| \(3x^2 - 4bx + b^2 = 0 \rightarrow (3x-b)(x-b) = 0\) | M1 | |
| \(x = \frac{b}{3}\) or \(b\) | A1 | |
| \(a = \frac{b}{3} \rightarrow b = 3a\) | A1 | AG |
| Alternative method: | ||
| \(\frac{dy}{dx} = 3x^2 - 4bx + b^2\) | B1 | |
| Sub \(b = 3a\) and obtain \(\frac{dy}{dx} = 0\) when \(x = a\) and when \(x = 3a\) | M1 | |
| \(\frac{d^2y}{dx^2} = 6x - 12a\) | A1 | |
| \(< 0\) Max at \(x = a\) and \(> 0\) Min at \(x = 3a\). Hence \(b = 3a\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Area under curve \(= \int(x^3 - 6ax^2 + 9a^2x)\,dx\) | M1 | |
| \(\frac{x^4}{4} - 2ax^3 + \frac{9a^2x^2}{2}\) | B2,1,0 | |
| \(\frac{a^4}{4} - 2a^4 + \frac{9a^4}{2} \left(= \frac{11a^4}{4}\right)\) | M1 | M1 for applying limits \(0 \to a\) |
| When \(x = a\), \(y = a^3 - 6a^3 + 9a^3 = 4a^3\) | B1 | |
| Area under line \(= \frac{1}{2}a \times \text{their } 4a^3\) | M1 | |
| Shaded area \(= \frac{11a^4}{4} - 2a^4 = \frac{3}{4}a^4\) | A1 |
## Question 11(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dy}{dx} = 3x^2 - 4bx + b^2$ | B1 | |
| $3x^2 - 4bx + b^2 = 0 \rightarrow (3x-b)(x-b) = 0$ | M1 | |
| $x = \frac{b}{3}$ or $b$ | A1 | |
| $a = \frac{b}{3} \rightarrow b = 3a$ | A1 | AG |
| **Alternative method:** | | |
| $\frac{dy}{dx} = 3x^2 - 4bx + b^2$ | B1 | |
| Sub $b = 3a$ and obtain $\frac{dy}{dx} = 0$ when $x = a$ and when $x = 3a$ | M1 | |
| $\frac{d^2y}{dx^2} = 6x - 12a$ | A1 | |
| $< 0$ Max at $x = a$ and $> 0$ Min at $x = 3a$. Hence $b = 3a$ | A1 | AG |
---
## Question 11(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Area under curve $= \int(x^3 - 6ax^2 + 9a^2x)\,dx$ | M1 | |
| $\frac{x^4}{4} - 2ax^3 + \frac{9a^2x^2}{2}$ | B2,1,0 | |
| $\frac{a^4}{4} - 2a^4 + \frac{9a^4}{2} \left(= \frac{11a^4}{4}\right)$ | M1 | M1 for applying limits $0 \to a$ |
| When $x = a$, $y = a^3 - 6a^3 + 9a^3 = 4a^3$ | B1 | |
| Area under line $= \frac{1}{2}a \times \text{their } 4a^3$ | M1 | |
| Shaded area $= \frac{11a^4}{4} - 2a^4 = \frac{3}{4}a^4$ | A1 | |11\\
\includegraphics[max width=\textwidth, alt={}, center]{aa4c496d-ce5f-4f46-ad37-d901644a9e7c-18_387_920_260_609}
The diagram shows part of the curve with equation $y = x ^ { 3 } - 2 b x ^ { 2 } + b ^ { 2 } x$ and the line $O A$, where $A$ is the maximum point on the curve. The $x$-coordinate of $A$ is $a$ and the curve has a minimum point at ( $b , 0$ ), where $a$ and $b$ are positive constants.
\begin{enumerate}[label=(\alph*)]
\item Show that $b = 3 a$.
\item Show that the area of the shaded region between the line and the curve is $k a ^ { 4 }$, where $k$ is a fraction to be found.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2020 Q11 [11]}}