| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2023 |
| Session | October |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Determine if function is increasing/decreasing |
| Difficulty | Standard +0.3 This is a straightforward P2 question requiring chain rule differentiation of x^(1/2), solving dy/dx > 0 for an inequality, then finding area using integration. All techniques are standard with no novel insight required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{4}{3}x - \frac{9}{2\sqrt{x}}\) | M1, A1 | M1: Attempts \(\frac{dy}{dx}\) with one index correct; allow \(x^2 \to x^1\) or \(\sqrt{x} \to x^{-\frac{1}{2}}\); A1: correct derivative, may be left unsimplified |
| Attempts to solve \(\frac{dy}{dx} > 0 \Rightarrow x^{\frac{3}{2}} > \frac{27}{8} \Rightarrow x > \frac{9}{4}\) | dM1, A1 | dM1: Attempts to find where \(\frac{dy}{dx}>0\), proceeds to \(x^{\frac{3}{2}}>k\) or \(x^3>k\) following correct squaring; also allow \(\frac{dy}{dx} \geq 0\); A1: States \(x>\frac{9}{4}\) or \(x \geq \frac{9}{4}\) following both M marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\int \frac{2}{3}x^2 - 9\sqrt{x} + 13\, dx = \frac{2}{9}x^3 - 6x^{\frac{3}{2}} + 13x \quad (+c)\) | M1 | Attempts to integrate; scored for increasing a correct index by 1; \(13 \to 13x\) counts |
| Area \(= \left[\frac{2}{9}x^3 - 6x^{\frac{3}{2}} + 13x\right]_0^9 = 117\) | dM1, A1 | dM1: Uses limits of 9 and 0; A1: Achieves 117; must see M1 and dM1 with correct integration; calculator integration not allowed |
| Substitutes \(x=9\) into \(\frac{dy}{dx} = \frac{4}{3}x - \frac{9}{2\sqrt{x}} \Rightarrow \frac{dy}{dx} = \frac{21}{2}\) | M1 | Substitutes \(x=9\) into their \(\frac{dy}{dx}\) which must have one index correct |
| Finds where tangent meets \(x\)-axis: \(0 - 40 = \frac{21}{2}(x-9) \Rightarrow x = \frac{109}{21}\) | dM1, A1 | dM1: Correct method finding where tangent meets \(x\)-axis using point \((9,40)\), their gradient, and \(y=0\); A1: \(x=\frac{109}{21}\) |
| Area of \(R = \left[\frac{2}{9}x^3 - 6x^{\frac{3}{2}} + 13x\right]_0^9 - \frac{1}{2} \times 40 \times \left(9 - \frac{109}{21}\right)\) | ddM1 | Complete method for area of \(R\); dependent on all previous M marks; typically area under curve minus area of triangle |
| \(= 117 - \frac{1600}{21} = \frac{857}{21}\) | A1 | Area \(R = \frac{857}{21}\) |
# Question 9:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{dy}{dx} = \frac{4}{3}x - \frac{9}{2\sqrt{x}}$ | M1, A1 | M1: Attempts $\frac{dy}{dx}$ with one index correct; allow $x^2 \to x^1$ or $\sqrt{x} \to x^{-\frac{1}{2}}$; A1: correct derivative, may be left unsimplified |
| Attempts to solve $\frac{dy}{dx} > 0 \Rightarrow x^{\frac{3}{2}} > \frac{27}{8} \Rightarrow x > \frac{9}{4}$ | dM1, A1 | dM1: Attempts to find where $\frac{dy}{dx}>0$, proceeds to $x^{\frac{3}{2}}>k$ or $x^3>k$ following correct squaring; also allow $\frac{dy}{dx} \geq 0$; A1: States $x>\frac{9}{4}$ or $x \geq \frac{9}{4}$ following both M marks |
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $\int \frac{2}{3}x^2 - 9\sqrt{x} + 13\, dx = \frac{2}{9}x^3 - 6x^{\frac{3}{2}} + 13x \quad (+c)$ | M1 | Attempts to integrate; scored for increasing a correct index by 1; $13 \to 13x$ counts |
| Area $= \left[\frac{2}{9}x^3 - 6x^{\frac{3}{2}} + 13x\right]_0^9 = 117$ | dM1, A1 | dM1: Uses limits of 9 and 0; A1: Achieves 117; must see M1 and dM1 with correct integration; calculator integration not allowed |
| Substitutes $x=9$ into $\frac{dy}{dx} = \frac{4}{3}x - \frac{9}{2\sqrt{x}} \Rightarrow \frac{dy}{dx} = \frac{21}{2}$ | M1 | Substitutes $x=9$ into their $\frac{dy}{dx}$ which must have one index correct |
| Finds where tangent meets $x$-axis: $0 - 40 = \frac{21}{2}(x-9) \Rightarrow x = \frac{109}{21}$ | dM1, A1 | dM1: Correct method finding where tangent meets $x$-axis using point $(9,40)$, their gradient, and $y=0$; A1: $x=\frac{109}{21}$ |
| Area of $R = \left[\frac{2}{9}x^3 - 6x^{\frac{3}{2}} + 13x\right]_0^9 - \frac{1}{2} \times 40 \times \left(9 - \frac{109}{21}\right)$ | ddM1 | Complete method for area of $R$; dependent on all previous M marks; typically area under curve minus area of triangle |
| $= 117 - \frac{1600}{21} = \frac{857}{21}$ | A1 | Area $R = \frac{857}{21}$ |
---9.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{66abdef1-072e-41eb-a933-dd51a96330ff-24_803_1050_251_511}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
Figure 3 shows a sketch of part of the curve $C$ with equation
$$y = \frac { 2 } { 3 } x ^ { 2 } - 9 \sqrt { x } + 13 \quad x \geqslant 0$$
\begin{enumerate}[label=(\alph*)]
\item Find, using calculus, the range of values of $x$ for which $y$ is increasing.
The point $P$ lies on $C$ and has coordinates (9, 40).\\
The line $l$ is the tangent to $C$ at the point $P$.\\
The finite region $R$, shown shaded in Figure 3, is bounded by the curve $C$, the line $l$, the $x$-axis and the $y$-axis.
\item Find, using calculus, the exact area of $R$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P2 2023 Q9 [12]}}