| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area of sector/segment problems |
| Difficulty | Standard +0.3 This is a straightforward optimization problem with guided steps. Part (a) requires setting up volume and surface area formulas for a sector-based prism (standard A-level geometry), part (b) is routine differentiation and solving, and part (c) is standard second derivative test. The question provides the target formula and breaks the problem into clear sub-parts, making it slightly easier than average. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sets up equation using volume \(= 240\), e.g. \(\frac{1}{2}r^2 \times 0.8h = 240 \Rightarrow h = \frac{600}{r^2}\) | M1, A1 | 3.4, 1.1b - Condone \(kr^2h = 240 \Rightarrow h = \ldots\) or \(rh = \ldots\) |
| Attempts to substitute \(h = \frac{600}{r^2}\) into \((S=)\frac{1}{2}r^2\times 0.8 + \frac{1}{2}r^2\times 0.8 + 2rh + 0.8rh\) | dM1 | 3.4 |
| \(S = 0.8r^2 + 2.8rh = 0.8r^2 + 2.8\times\frac{600}{r} = 0.8r^2 + \frac{1680}{r}\) | A1* | 2.1 - Correct work leading to given result |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left(\frac{dS}{dr}\right) = 1.6r - \frac{1680}{r^2}\) | M1, A1 | 3.1a, 1.1b - Derivative of form \(pr \pm \frac{q}{r^2}\) where \(p,q\) non-zero constants. No requirement to see \(\frac{dS}{dr}\) notation |
| Sets \(\frac{dS}{dr} = 0 \Rightarrow r^3 = 1050\), \(r =\) awrt \(10.2\) | dM1, A1 | 2.1, 1.1b - Sets/implies \(\frac{dS}{dr}=0\), proceeds to \(mr^3=n\), \(m\times n>0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts to substitute positive \(r\) from (b) into \(\left(\frac{d^2S}{dr^2}\right) = 1.6 + \frac{3360}{r^3}\) and considers value or sign | M1 | 1.1b - Condone \(\frac{d^2S}{dr^2}\) written as \(\frac{d^2y}{dx^2}\) for this mark only |
| \(\frac{d^2S}{dr^2} = 1.6 + \frac{3360}{r^3}\) with \(\frac{d^2S}{dr^2}\bigg | _{r=10.2} = 5 > 0\) proving minimum value of \(S\) | A1 |
# Question 15:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sets up equation using volume $= 240$, e.g. $\frac{1}{2}r^2 \times 0.8h = 240 \Rightarrow h = \frac{600}{r^2}$ | M1, A1 | 3.4, 1.1b - Condone $kr^2h = 240 \Rightarrow h = \ldots$ or $rh = \ldots$ |
| Attempts to substitute $h = \frac{600}{r^2}$ into $(S=)\frac{1}{2}r^2\times 0.8 + \frac{1}{2}r^2\times 0.8 + 2rh + 0.8rh$ | dM1 | 3.4 |
| $S = 0.8r^2 + 2.8rh = 0.8r^2 + 2.8\times\frac{600}{r} = 0.8r^2 + \frac{1680}{r}$ | A1* | 2.1 - Correct work leading to given result |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(\frac{dS}{dr}\right) = 1.6r - \frac{1680}{r^2}$ | M1, A1 | 3.1a, 1.1b - Derivative of form $pr \pm \frac{q}{r^2}$ where $p,q$ non-zero constants. No requirement to see $\frac{dS}{dr}$ notation |
| Sets $\frac{dS}{dr} = 0 \Rightarrow r^3 = 1050$, $r =$ awrt $10.2$ | dM1, A1 | 2.1, 1.1b - Sets/implies $\frac{dS}{dr}=0$, proceeds to $mr^3=n$, $m\times n>0$ |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts to substitute positive $r$ from (b) into $\left(\frac{d^2S}{dr^2}\right) = 1.6 + \frac{3360}{r^3}$ and considers value or sign | M1 | 1.1b - Condone $\frac{d^2S}{dr^2}$ written as $\frac{d^2y}{dx^2}$ for this mark only |
| $\frac{d^2S}{dr^2} = 1.6 + \frac{3360}{r^3}$ with $\frac{d^2S}{dr^2}\bigg|_{r=10.2} = 5 > 0$ proving minimum value of $S$ | A1 | 1.1b - Can argue without finding value: $\frac{d^2S}{dr^2} = 1.6+\frac{3360}{r^3}>0$ as $r>0$. Do NOT allow $\frac{d^2y}{dx^2}$ for this mark |15.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{129adfbb-98fa-4e88-b636-7b4d111f3349-42_444_739_244_662}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
A company makes toys for children.\\
Figure 5 shows the design for a solid toy that looks like a piece of cheese.\\
The toy is modelled so that
\begin{itemize}
\item face $A B C$ is a sector of a circle with radius $r \mathrm {~cm}$ and centre $A$
\item angle $B A C = 0.8$ radians
\item faces $A B C$ and $D E F$ are congruent
\item edges $A D , C F$ and $B E$ are perpendicular to faces $A B C$ and $D E F$
\item edges $A D , C F$ and $B E$ have length $h \mathrm {~cm}$
\end{itemize}
Given that the volume of the toy is $240 \mathrm {~cm} ^ { 3 }$
\begin{enumerate}[label=(\alph*)]
\item show that the surface area of the toy, $S \mathrm {~cm} ^ { 2 }$, is given by
$$S = 0.8 r ^ { 2 } + \frac { 1680 } { r }$$
making your method clear.
Using algebraic differentiation,
\item find the value of $r$ for which $S$ has a stationary point.
\item Prove, by further differentiation, that this value of $r$ gives the minimum surface area of the toy.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 1 2022 Q15 [10]}}