| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2016 |
| Session | October |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Find stationary points of standard polynomial |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question testing standard C1/C2 techniques: factorising a cubic to find roots, differentiating a polynomial to find stationary points, and understanding horizontal translations. All parts are routine textbook exercises requiring no problem-solving insight, though the multiple parts and coordinate calculations place it slightly below average difficulty rather than being trivial. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02w Graph transformations: simple transformations of f(x)1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(x) = \frac{x^3 - 9x^2 - 81x}{27} = 0 \Rightarrow x(x^2 - 9x - 81) = 0\) | M1 | Attempts to solve \(f(x)=0\) by taking out factor of \(x\) and obtaining quadratic factor |
| \(x = \frac{9 \pm \sqrt{81+324}}{2}\) | dM1 | Uses formula or completing the square for their three term quadratic; factorisation is M0 |
| \(x = \frac{9 \pm \sqrt{405}}{2}\) or \(x = \frac{9 \pm 9\sqrt{5}}{2}\) | A1, A1 | First A1: one correct solution (need not be fully simplified); second A1: two correct solutions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Differentiates correctly and sets \(= 0\): \(f'(x) = 3x^2 - 18x - 81 = 0\) | M1, A1 | M1: differentiates to 3 term quadratic; A1: differentiates correctly and sets derivative \(= 0\) |
| Solves \(f'(x) = 0 \Rightarrow x = 9\) and \(x = -3\) | dM1 A1 | dM1: solves quadratic to give two solutions; A1: gives both 9 and \(-3\) |
| Substitutes one of their values into \(f(x)\): \(x=9 \Rightarrow y=-27\) and \(x=-3 \Rightarrow y=5\) | ddM1 | Substitutes at least one \(x\) value from \(f'(x)=0\) into \(f(x)\) |
| \(x=9,\ y=-27\) and \(x=-3,\ y=5\) | A1 | Both \(-27\) and \(5\); do not require coordinates or correct attribution to \(C\) or \(D\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a = 9\) | B1 | For \(a=9\) only; no follow through |
## Question 12:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) = \frac{x^3 - 9x^2 - 81x}{27} = 0 \Rightarrow x(x^2 - 9x - 81) = 0$ | M1 | Attempts to solve $f(x)=0$ by taking out factor of $x$ and obtaining quadratic factor |
| $x = \frac{9 \pm \sqrt{81+324}}{2}$ | dM1 | Uses formula or completing the square for their three term quadratic; factorisation is M0 |
| $x = \frac{9 \pm \sqrt{405}}{2}$ or $x = \frac{9 \pm 9\sqrt{5}}{2}$ | A1, A1 | First A1: one correct solution (need not be fully simplified); second A1: two correct solutions |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Differentiates correctly and sets $= 0$: $f'(x) = 3x^2 - 18x - 81 = 0$ | M1, A1 | M1: differentiates to 3 term quadratic; A1: differentiates correctly and sets derivative $= 0$ |
| Solves $f'(x) = 0 \Rightarrow x = 9$ and $x = -3$ | dM1 A1 | dM1: solves quadratic to give two solutions; A1: gives both 9 and $-3$ |
| Substitutes one of their values into $f(x)$: $x=9 \Rightarrow y=-27$ and $x=-3 \Rightarrow y=5$ | ddM1 | Substitutes at least one $x$ value from $f'(x)=0$ into $f(x)$ |
| $x=9,\ y=-27$ and $x=-3,\ y=5$ | A1 | Both $-27$ and $5$; do not require coordinates or correct attribution to $C$ or $D$ |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = 9$ | B1 | For $a=9$ only; no follow through |
---12.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{53865e15-3838-4551-b507-fe49549b87db-32_748_883_274_477}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Diagram not drawn to scale
Figure 2 shows a sketch of the curve with equation $y = \mathrm { f } ( x )$, where
$$f ( x ) = \frac { x ^ { 3 } - 9 x ^ { 2 } - 81 x } { 27 }$$
The curve crosses the $x$-axis at the point $A$, the point $B$ and the origin $O$. The curve has a maximum turning point at $C$ and a minimum turning point at $D$.
\begin{enumerate}[label=(\alph*)]
\item Use algebra to find exact values for the $x$ coordinates of the points $A$ and $B$.
\item Use calculus to find the coordinates of the points $C$ and $D$.
The graph of $y = \mathrm { f } ( x + a )$, where $a$ is a constant, has its minimum turning point on the $y$-axis.
\item Write down the value of $a$.\\
\includegraphics[max width=\textwidth, alt={}, center]{53865e15-3838-4551-b507-fe49549b87db-35_29_37_182_1914}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2016 Q12 [11]}}