| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Session | Specimen |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Optimise perimeter or area of 2D region |
| Difficulty | Standard +0.3 This is a straightforward C2 optimization question with clear scaffolding. Part (a) is algebraic expansion of the triangle area formula, part (b) is routine differentiation and second derivative test, and parts (c)-(d) are substitution. The multi-step nature adds slight complexity, but each step uses standard techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C)1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(A = \frac{1}{2}(x+1)(4-x)^2 \sin 30°\) | M1 | Use of \(\frac{1}{2}ab\sin C\) |
| \(= \frac{1}{4}(x+1)(16 - 8x + x^2)\) | M1 | Attempt to multiply out |
| \(= \frac{1}{4}(x^3 - 7x^2 + 8x + 16)\) | A1 cso | |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dA}{dx} = \frac{1}{4}(3x^2 - 14x + 8)\) | M1 A1 | Ignore the \(\frac{1}{4}\) |
| \(\frac{dA}{dx} = 0 \Rightarrow (3x-2)(x-4) = 0\) | M1 | |
| So \(x = \frac{2}{3}\) or \(4\) | A1 | At least \(x = \frac{2}{3}\) or... |
| \(\frac{d^2A}{dx^2} = \frac{1}{4}(6x-14)\), when \(x = \frac{2}{3}\) it is \(< 0\), so maximum | M1 | Any full method |
| So \(x = \frac{2}{3}\) gives maximum area | A1 | Full accuracy |
| (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Maximum area \(= \frac{1}{4}(\frac{5}{3})(\frac{10}{3})^2 = 4.6\) or \(4.63\) or \(4.630\) | B1 | |
| (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Cosine rule: \(QR^2 = (\frac{5}{3})^2 + (\frac{10}{3})^4 - 2 \times \frac{5}{3} \times (\frac{10}{3})^2 \cos 30°\) | M1 A1 | M1 for \(QR\) or \(QR^2\) |
| \(= 94.159\ldots\) | ||
| \(QR = 9.7\) or \(9.70\) or \(9.704\) | A1 | |
| (3) |
# Question 9:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $A = \frac{1}{2}(x+1)(4-x)^2 \sin 30°$ | M1 | Use of $\frac{1}{2}ab\sin C$ |
| $= \frac{1}{4}(x+1)(16 - 8x + x^2)$ | M1 | Attempt to multiply out |
| $= \frac{1}{4}(x^3 - 7x^2 + 8x + 16)$ | A1 cso | |
| | **(3)** | |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dA}{dx} = \frac{1}{4}(3x^2 - 14x + 8)$ | M1 A1 | Ignore the $\frac{1}{4}$ |
| $\frac{dA}{dx} = 0 \Rightarrow (3x-2)(x-4) = 0$ | M1 | |
| So $x = \frac{2}{3}$ or $4$ | A1 | At least $x = \frac{2}{3}$ or... |
| $\frac{d^2A}{dx^2} = \frac{1}{4}(6x-14)$, when $x = \frac{2}{3}$ it is $< 0$, so maximum | M1 | Any full method |
| So $x = \frac{2}{3}$ gives maximum area | A1 | Full accuracy |
| | **(6)** | |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Maximum area $= \frac{1}{4}(\frac{5}{3})(\frac{10}{3})^2 = 4.6$ or $4.63$ or $4.630$ | B1 | |
| | **(1)** | |
## Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Cosine rule: $QR^2 = (\frac{5}{3})^2 + (\frac{10}{3})^4 - 2 \times \frac{5}{3} \times (\frac{10}{3})^2 \cos 30°$ | M1 A1 | M1 for $QR$ or $QR^2$ |
| $= 94.159\ldots$ | | |
| $QR = 9.7$ or $9.70$ or $9.704$ | A1 | |
| | **(3)** | |9.
Figure 3
$$( x + 1 ) ^ { 2 }$$
Figure 3 shows a triangle $P Q R$. The size of angle $Q P R$ is $30 ^ { \circ }$, the length of $P Q$ is $( x + 1 )$ and the length of $P R$ is $( 4 - x ) ^ { 2 }$, where $X \in \Re$.
\begin{enumerate}[label=(\alph*)]
\item Show that the area $A$ of the triangle is given by $A = \frac { 1 } { 4 } \left( x ^ { 3 } - 7 x ^ { 2 } + 8 x + 16 \right)$
\item Use calculus to prove that the area of $\triangle P Q R$ is a maximum when $x = \frac { 2 } { 3 }$.
Explain clearly how you know that this value of $x$ gives the maximum area.
\item Find the maximum area of $\triangle P Q R$.
\item Find the length of $Q R$ when the area of $\triangle P Q R$ is a maximum.
END
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q9 [13]}}