| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2018 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Deduce inequality solutions from sketch |
| Difficulty | Standard +0.3 This is a multi-part question testing standard curve sketching and transformations. Part (a) requires routine differentiation using product rule and solving f'(x)=0. Parts (b)-(e) test understanding of horizontal translations, which is a core C1/C2 topic. All steps are algorithmic with no novel problem-solving required, making this slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02w Graph transformations: simple transformations of f(x)1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(x)=(x-2)^2(2x+1)=2x^3-7x^2+4x+4\) | M1 | Expand brackets; must have four-term cubic |
| \(f'(x) = 6x^2 - 14x + 4\) | M1 A1 | M1: differentiate to quadratic (power reduced by one). A1: completely correct |
| Set \(f'(x)=0\), solve 3TQ e.g. \(2(3x-1)(x-2)=0\) | M1 | Derivative must be 3TQ expression |
| \(x=\frac{1}{3}\) (with \(x=2\)) | A1 | Allow exact equivalences including recurring decimals; may include \(x=2\) |
| \(\Rightarrow \left(\frac{1}{3},\ \frac{125}{27}\right)\) | dM1 A1 | dM1: substitute \(x=\frac{1}{3}\) into \(f(x)\). A1: \(x=\frac{1}{3}\), \(y=\frac{125}{27}\) exact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y=(x-1)^2(2x+3)\) | B1 | cao; must be in form \(y=\ldots\) or \(f(x)=\ldots\); do not allow left in form \((x+1-2)^2(2(x+1)+1)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| When \(x=0\), \(y=3\) | M1 A1 | M1: put \(x=0\) into new function. A1: \(y=3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((1,0)\) and \(\left(-\dfrac{2}{3},\ \dfrac{125}{27}\right)\) | M1 A1ft | M1: either coordinate pair correct (follow through from point \(P\)). A1ft: both pairs correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Sketch: same shape as before but translated left | M1 | Shape same as before (+ve cubic) but moved; tolerant on position of maximum |
| Maximum in second quadrant, minimum on \(+ve\) \(x\)-axis, graph in three quadrants | A1 | Translated to the left; maximum in 2nd quadrant, minimum on \(+ve\) \(x\)-axis |
| \((1,0)\) and \((-1.5, 0)\) marked on \(x\)-axis | A1 | Mark as 1 and \(-1.5\) on \(x\)-axis |
# Question 14:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(x)=(x-2)^2(2x+1)=2x^3-7x^2+4x+4$ | M1 | Expand brackets; must have four-term cubic |
| $f'(x) = 6x^2 - 14x + 4$ | M1 A1 | M1: differentiate to quadratic (power reduced by one). A1: completely correct |
| Set $f'(x)=0$, solve 3TQ e.g. $2(3x-1)(x-2)=0$ | M1 | Derivative must be 3TQ expression |
| $x=\frac{1}{3}$ (with $x=2$) | A1 | Allow exact equivalences including recurring decimals; may include $x=2$ |
| $\Rightarrow \left(\frac{1}{3},\ \frac{125}{27}\right)$ | dM1 A1 | dM1: substitute $x=\frac{1}{3}$ into $f(x)$. A1: $x=\frac{1}{3}$, $y=\frac{125}{27}$ exact |
**[7 marks]**
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y=(x-1)^2(2x+3)$ | B1 | cao; must be in form $y=\ldots$ or $f(x)=\ldots$; do not allow left in form $(x+1-2)^2(2(x+1)+1)$ |
**[1 mark]**
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| When $x=0$, $y=3$ | M1 A1 | M1: put $x=0$ into new function. A1: $y=3$ |
**[2 marks]**
## Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(1,0)$ and $\left(-\dfrac{2}{3},\ \dfrac{125}{27}\right)$ | M1 A1ft | M1: either coordinate pair correct (follow through from point $P$). A1ft: both pairs correct |
**[2 marks]**
## Part (e):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Sketch: same shape as before but translated left | M1 | Shape same as before (+ve cubic) but moved; tolerant on position of maximum |
| Maximum in second quadrant, minimum on $+ve$ $x$-axis, graph in three quadrants | A1 | Translated to the **left**; maximum in 2nd quadrant, minimum on $+ve$ $x$-axis |
| $(1,0)$ and $(-1.5, 0)$ marked on $x$-axis | A1 | Mark as 1 and $-1.5$ on $x$-axis |
**[3 marks]**14.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ce06b37a-aa57-4256-bec8-7277c8a9fc65-40_611_1214_219_548}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of the curve $C _ { 1 }$ with equation $y = \mathrm { f } ( x )$ where
$$f ( x ) = ( x - 2 ) ^ { 2 } ( 2 x + 1 ) , \quad x \in \mathbb { R }$$
The curve crosses the $x$-axis at $\left( - \frac { 1 } { 2 } , 0 \right)$, touches it at $( 2,0 )$ and crosses the $y$-axis at ( 0,4 ). There is a maximum turning point at the point marked $P$.
\begin{enumerate}[label=(\alph*)]
\item Use $\mathrm { f } ^ { \prime } ( x )$ to find the exact coordinates of the turning point $P$.
A second curve $C _ { 2 }$ has equation $y = \mathrm { f } ( x + 1 )$.
\item Write down an equation of the curve $C _ { 2 }$ You may leave your equation in a factorised form.
\item Use your answer to part (b) to find the coordinates of the point where the curve $C _ { 2 }$ meets the $y$-axis.
\item Write down the coordinates of the two turning points on the curve $C _ { 2 }$
\item Sketch the curve $C _ { 2 }$, with equation $y = \mathrm { f } ( x + 1 )$, giving the coordinates of the points where the curve crosses or touches the $x$-axis.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2018 Q14 [15]}}