| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2022 |
| Session | October |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent equation at a known point on circle |
| Difficulty | Standard +0.3 This is a structured multi-part question requiring standard A-level techniques: finding stationary points via differentiation, determining a circle equation from center and radius, finding a tangent to a circle, and computing an area via integration. Each part follows directly from the previous with clear guidance, making it slightly easier than average despite involving multiple topics. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = x^3 - 5x^2 + 3x + 14 \Rightarrow \frac{dy}{dx} = 3x^2 - 10x + 3 = 0\) | M1 | Differentiates and sets equal to 0. Look for at least 2 of: \(x^3 \to ...x^2\), \(-5x^2 \to ...x\), \(3x \to 3\) |
| Roots are \(3, \frac{1}{3}\); when \(x=3\), \(y = 3^3 - 5 \times 3^2 + 3 \times 3 + 14 = ...\) | dM1 | Solves quadratic and substitutes root into original equation. Depends on first mark |
| Centre is \((3, 5)\) | A1 | Correct coordinates \((3,5)\) or e.g. \(x=3\), \(y=5\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| At \(A\), \(y = 8\) | B1 | Allow this to score anywhere |
| \(r^2 = (2-"3")^2 + ("8"-"5")^2\ (=10)\) | M1 | Attempts to find radius (or radius²) using \((2,"8")\) and their minimum point |
| \((x-3)^2 + (y-5)^2 = 10\) | A1 | e.g. \((x-3)^2+(y-5)^2=(\sqrt{10})^2\) also accepted |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{"8"-"5"}{2-"3"} = ...(-3)\) | M1 | Attempts to find gradient between \(A\) and \(T\). Note: \(-3\) just written down not accepted |
| \(y - "8" = "\frac{1}{3}"(x-2)\) | M1 | Attempts equation of line using \(x=2\), their \(y\) at \(A\), and negative reciprocal of gradient of \(AT\) |
| \(y = \frac{1}{3}x + \frac{22}{3}\) | A1* | With no errors and both previous method marks scored |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int_0^2 x^3 - 5x^2 + 3x + 14\, dx = ...\left(\frac{1}{4}x^4 - \frac{5}{3}x^3 + \frac{3}{2}x^2 + 14x\right)\) | M1 | Attempts to integrate. Award for power increasing by 1 on one of the terms |
| \(\text{Area} = \left[\frac{1}{4}x^4 - \frac{5}{3}x^3 + \frac{3}{2}x^2 + 14x\right]_0^2 - \left(\frac{1}{2}\times\left(\frac{22}{3}+"8"\right)\times 2\right) = ...\) | dM1 | Correct method to find shaded area: substitutes 2 and 0 and subtracts trapezium area \(\frac{1}{2}\times\left(\frac{22}{3}+\text{their }y\text{ at }A\right)\times 2\). Dependent on previous M |
| \(\frac{1}{4}\times16 - \frac{5}{3}\times8 + \frac{3}{2}\times4 + 14\times2 - \frac{46}{3}\) | ||
| \(\frac{74}{3} - \frac{46}{3} = \frac{28}{3}\) | A1 | \(\frac{28}{3}\) or exact equivalent cso |
## Question 9:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = x^3 - 5x^2 + 3x + 14 \Rightarrow \frac{dy}{dx} = 3x^2 - 10x + 3 = 0$ | M1 | Differentiates and sets equal to 0. Look for at least 2 of: $x^3 \to ...x^2$, $-5x^2 \to ...x$, $3x \to 3$ |
| Roots are $3, \frac{1}{3}$; when $x=3$, $y = 3^3 - 5 \times 3^2 + 3 \times 3 + 14 = ...$ | dM1 | Solves quadratic and substitutes root into original equation. Depends on first mark |
| Centre is $(3, 5)$ | A1 | Correct coordinates $(3,5)$ or e.g. $x=3$, $y=5$ |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| At $A$, $y = 8$ | B1 | Allow this to score anywhere |
| $r^2 = (2-"3")^2 + ("8"-"5")^2\ (=10)$ | M1 | Attempts to find radius (or radius²) using $(2,"8")$ and their minimum point |
| $(x-3)^2 + (y-5)^2 = 10$ | A1 | e.g. $(x-3)^2+(y-5)^2=(\sqrt{10})^2$ also accepted |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{"8"-"5"}{2-"3"} = ...(-3)$ | M1 | Attempts to find gradient between $A$ and $T$. Note: $-3$ just written down not accepted |
| $y - "8" = "\frac{1}{3}"(x-2)$ | M1 | Attempts equation of line using $x=2$, their $y$ at $A$, and negative reciprocal of gradient of $AT$ |
| $y = \frac{1}{3}x + \frac{22}{3}$ | A1* | With no errors and both previous method marks scored |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_0^2 x^3 - 5x^2 + 3x + 14\, dx = ...\left(\frac{1}{4}x^4 - \frac{5}{3}x^3 + \frac{3}{2}x^2 + 14x\right)$ | M1 | Attempts to integrate. Award for power increasing by 1 on one of the terms |
| $\text{Area} = \left[\frac{1}{4}x^4 - \frac{5}{3}x^3 + \frac{3}{2}x^2 + 14x\right]_0^2 - \left(\frac{1}{2}\times\left(\frac{22}{3}+"8"\right)\times 2\right) = ...$ | dM1 | Correct method to find shaded area: substitutes 2 and 0 and subtracts trapezium area $\frac{1}{2}\times\left(\frac{22}{3}+\text{their }y\text{ at }A\right)\times 2$. Dependent on previous M |
| $\frac{1}{4}\times16 - \frac{5}{3}\times8 + \frac{3}{2}\times4 + 14\times2 - \frac{46}{3}$ | | |
| $\frac{74}{3} - \frac{46}{3} = \frac{28}{3}$ | A1 | $\frac{28}{3}$ or exact equivalent cso |
---\begin{enumerate}
\item In this question you must show detailed reasoning.
\end{enumerate}
Solutions relying entirely on calculator technology are not acceptable.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6f926d53-c6de-4eb7-9d18-596f61ec26e1-26_723_455_413_804}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows
\begin{itemize}
\item the curve $C _ { 1 }$ with equation $y = x ^ { 3 } - 5 x ^ { 2 } + 3 x + 14$
\item the circle $C _ { 2 }$ with centre $T$
\end{itemize}
The point $T$ is the minimum turning point of $C _ { 1 }$\\
Using Figure 3 and calculus,\\
(a) find the coordinates of $T$
The curve $C _ { 1 }$ intersects the circle $C _ { 2 }$ at the point $A$ with $x$ coordinate 2\\
(b) Find an equation of the circle $C _ { 2 }$
The line $l$ shown in Figure 3, is the tangent to circle $C _ { 2 }$ at $A$\\
(c) Show that an equation of $l$ is
$$y = \frac { 1 } { 3 } x + \frac { 22 } { 3 }$$
The region $R$, shown shaded in Figure 3, is bounded by $C _ { 1 } , l$ and the $y$-axis.\\
(d) Find the exact area of $R$.
\hfill \mbox{\textit{Edexcel P2 2022 Q9 [12]}}