| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2017 |
| Session | October |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Find stationary points coordinates |
| Difficulty | Moderate -0.8 This is a straightforward stationary point question requiring routine differentiation of power functions (rewriting √x as x^{1/2}), setting derivative to zero, and solving a simple equation. All steps are standard C1/C2 techniques with no problem-solving insight needed, making it easier than average but not trivial due to the algebraic manipulation required. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Uses \(x^n \to x^{n-1}\) at least once | M1 | Sight of index \(x^{-0.5}\) or \(x^{-\frac{1}{2}}\) or \(x^1\) |
| Either term correct (may be unsimplified). E.g. \(2\times2x^1\) acceptable | A1 | Indices must be tidied up. Do not allow \(2\times2x^{2-1}\) |
| \(\frac{dy}{dx} = \frac{27}{2}x^{-0.5} - 4x\) | A1 | Or exact equivalent e.g. \(13.5 \times \frac{1}{\sqrt{x}} - 4x\). Must be fully tidied. Do not allow \(2\times2x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Sets \(\frac{dy}{dx} = 0\) | M1 | May be implied by subsequent working |
| \(x^{1.5} = \frac{27}{8} \Rightarrow x = \frac{9}{4}\) | dM1, A1 | Dependent on previous M. Correct index work leading to \(x^{\pm1.5}=k\). Also allow squaring route. Correct answer implies previous mark if no incorrect work seen |
| \(x = \frac{9}{4} \Rightarrow y = \frac{243}{8}\) | dM1, A1 | Dependent on first M1 in (b). Substitute \(x\) into \(y\). No need to verify with calculator. Accept exact equivalent (30.375) |
## Question 5:
**(a)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Uses $x^n \to x^{n-1}$ at least once | M1 | Sight of index $x^{-0.5}$ or $x^{-\frac{1}{2}}$ or $x^1$ |
| Either term correct (may be unsimplified). E.g. $2\times2x^1$ acceptable | A1 | Indices must be tidied up. Do not allow $2\times2x^{2-1}$ |
| $\frac{dy}{dx} = \frac{27}{2}x^{-0.5} - 4x$ | A1 | Or exact equivalent e.g. $13.5 \times \frac{1}{\sqrt{x}} - 4x$. Must be fully tidied. Do not allow $2\times2x$ |
**(b)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Sets $\frac{dy}{dx} = 0$ | M1 | May be implied by subsequent working |
| $x^{1.5} = \frac{27}{8} \Rightarrow x = \frac{9}{4}$ | dM1, A1 | Dependent on previous M. Correct index work leading to $x^{\pm1.5}=k$. Also allow squaring route. Correct answer implies previous mark if no incorrect work seen |
| $x = \frac{9}{4} \Rightarrow y = \frac{243}{8}$ | dM1, A1 | Dependent on first M1 in (b). Substitute $x$ into $y$. No need to verify with calculator. Accept exact equivalent (30.375) |5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{bb1becd5-96c1-426d-9b85-4bbc4a61af27-10_678_1076_248_434}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of part of the curve with equation
$$y = 27 \sqrt { x } - 2 x ^ { 2 } , \quad x \in \mathbb { R } , x > 0$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$
The curve has a maximum turning point $P$, as shown in Figure 2.
\item Use the answer to part (a) to find the exact coordinates of $P$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2017 Q5 [8]}}