| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Multi-part: volume and area |
| Difficulty | Standard +0.8 This question requires finding a stationary point using product rule differentiation, then solving a more challenging problem where two areas are equal. Part (b) requires integrating a product of powers, finding the total area under the curve, setting up an equation where the area from 0 to k equals half the total area, and solving a polynomial equation. The integration and algebraic manipulation are non-trivial, placing this above average difficulty but within reach of competent P2 students. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08b Integrate x^n: where n != -1 and sums1.08d Evaluate definite integrals: between limits1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = 2x^{\frac{3}{2}}(4-x) = 8x^{\frac{3}{2}} - 2x^{\frac{5}{2}}\) | Expansion | |
| \(\frac{dy}{dx} = 12x^{\frac{1}{2}} - 5x^{\frac{3}{2}}\) | M1, A1 | Complete attempt to differentiate; writes as sum of two terms and reduces power of at least one term by 1. Alternatively by product rule look for \(Ax^{\frac{1}{2}}(4-x) - Bx^{\frac{3}{2}}\). A1: \(12x^{\frac{1}{2}} - 5x^{\frac{3}{2}}\) or \(3x^{\frac{1}{2}}(4-x) - 2x^{\frac{3}{2}}\), may be left unsimplified |
| Stationary point: \(12x^{\frac{1}{2}} - 5x^{\frac{3}{2}} = 0 \Rightarrow x = \frac{12}{5}\) | dM1, A1 | dM1: Sets \(\frac{dy}{dx}=0\) and attempts to solve for non-zero \(x\). A1cso: \(x=\frac{12}{5}\) from correct work; ignore \(x=0\). Watch for incorrect methods e.g. \(\left(12x^{\frac{1}{2}}-5x^{\frac{3}{2}}\right)^2=0 \Rightarrow 144x-25x^3=0\) scores A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int 8x^{\frac{3}{2}} - 2x^{\frac{5}{2}}\,dx = \frac{16}{5}x^{\frac{5}{2}} - \frac{4}{7}x^{\frac{7}{2}}\) | M1, A1 | M1: Complete attempt to integrate; writes as sum of two terms and increases power of at least one term by 1. For integration by parts must proceed through second integration. A1: \(\frac{16}{5}x^{\frac{5}{2}} - \frac{4}{7}x^{\frac{7}{2}}\), may be unsimplified |
| \(\frac{16}{5}k^{\frac{5}{2}} - \frac{4}{7}k^{\frac{7}{2}} = 0 \Rightarrow k = \frac{16\times7}{4\times5} = \frac{28}{5}\) | dM1, A1 | dM1: Substitutes \(x=k\) and sets \(=0\), or applies limits 0 and 4 to find area \(R_1\left(=\frac{1024}{35}\right)\) and sets equal to negative of expression from limits 4 to \(k\). A1: \(k=\frac{28}{5}\); ignore \(k=0\); accept decimal 5.6 |
## Question 9:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = 2x^{\frac{3}{2}}(4-x) = 8x^{\frac{3}{2}} - 2x^{\frac{5}{2}}$ | | Expansion |
| $\frac{dy}{dx} = 12x^{\frac{1}{2}} - 5x^{\frac{3}{2}}$ | M1, A1 | Complete attempt to differentiate; writes as sum of two terms and reduces power of at least one term by 1. Alternatively by product rule look for $Ax^{\frac{1}{2}}(4-x) - Bx^{\frac{3}{2}}$. A1: $12x^{\frac{1}{2}} - 5x^{\frac{3}{2}}$ or $3x^{\frac{1}{2}}(4-x) - 2x^{\frac{3}{2}}$, may be left unsimplified |
| Stationary point: $12x^{\frac{1}{2}} - 5x^{\frac{3}{2}} = 0 \Rightarrow x = \frac{12}{5}$ | dM1, A1 | dM1: Sets $\frac{dy}{dx}=0$ and attempts to solve for non-zero $x$. A1cso: $x=\frac{12}{5}$ from correct work; ignore $x=0$. Watch for incorrect methods e.g. $\left(12x^{\frac{1}{2}}-5x^{\frac{3}{2}}\right)^2=0 \Rightarrow 144x-25x^3=0$ scores A0 |
**(4 marks)**
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int 8x^{\frac{3}{2}} - 2x^{\frac{5}{2}}\,dx = \frac{16}{5}x^{\frac{5}{2}} - \frac{4}{7}x^{\frac{7}{2}}$ | M1, A1 | M1: Complete attempt to integrate; writes as sum of two terms and increases power of at least one term by 1. For integration by parts must proceed through second integration. A1: $\frac{16}{5}x^{\frac{5}{2}} - \frac{4}{7}x^{\frac{7}{2}}$, may be unsimplified |
| $\frac{16}{5}k^{\frac{5}{2}} - \frac{4}{7}k^{\frac{7}{2}} = 0 \Rightarrow k = \frac{16\times7}{4\times5} = \frac{28}{5}$ | dM1, A1 | dM1: Substitutes $x=k$ and sets $=0$, or applies limits 0 and 4 to find area $R_1\left(=\frac{1024}{35}\right)$ and sets equal to negative of expression from limits 4 to $k$. A1: $k=\frac{28}{5}$; ignore $k=0$; accept decimal 5.6 |
**(4 marks total: 8 marks)**
---9.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b822842d-ee62-40ce-a8de-967e556a80a8-26_915_912_255_580}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 is a sketch of the curve $C$ with equation
$$y = 2 x ^ { \frac { 3 } { 2 } } ( 4 - x ) \quad x \geqslant 0$$
The point $P$ is the stationary point of $C$.
\begin{enumerate}[label=(\alph*)]
\item Find, using calculus, the $x$ coordinate of $P$.
The region $R _ { 1 }$, shown shaded in Figure 1, is bounded by $C$ and the $x$-axis.\\
The region $R _ { 2 }$, also shown shaded in Figure 1, is bounded by $C$, the $x$-axis and the line with equation $x = k$, where $k$ is a constant.
Given that the area of $R _ { 1 }$ is equal to the area of $R _ { 2 }$
\item find, using calculus, the exact value of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P2 2024 Q9 [8]}}