| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Transformation effect on key points |
| Difficulty | Moderate -0.8 This is a straightforward Core 1/2 question requiring basic differentiation of a polynomial plus reciprocal term, solving f'(x)=0 for a stationary point, and applying simple transformations. All steps are routine textbook exercises with no problem-solving insight required, making it easier than average but not trivial due to the multi-part structure. |
| Spec | 1.02w 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 |
| \(f(x) = x^2 + \frac{16}{x} \Rightarrow f'(x) = 2x - \frac{16}{x^2}\) | M1 A1 | M1: \(x^n \to x^{n-1}\) for either term. Accept \(x^2\to x\) or \(\frac{1}{x}\to\frac{1}{x^2}\). A1: correct unsimplified \(f'(x)=2x-\frac{16}{x^2}\). Accept \(f'(x)=2\times x + 16\times-1x^{-2}\). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Setting \(2x - \frac{16}{x^2} = 0 \Rightarrow x = ..\) | M1 | Sets \(f'(x)=0\) and proceeds to \(x=..\) |
| \(x^3 = 8 \Rightarrow x = 2\) | dM1 A1 | dM1: dependent on previous M, scored for \(\times x^2\) to reach \(x^3=k\Rightarrow x=\sqrt[3]{k}\). A1: correctly achieving \(x=2\). Ignore additional solutions. |
| \(A = (2, 12)\) | A1 | Correctly achieving \(A=(2,12)\). Accept \(x=2, y=12\). If additional solutions given \((x>0)\) mark withheld. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(A' = (1, 12)\) | B1ft | Accept on sketch or as \(x=1, y=12\). If \(A=(p,q)\) was incorrect, follow through with \(A'=(p-1,q)\). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(A' = (2, 6)\) | B1ft | Accept on sketch or as \(x=2, y=6\). If \(A=(p,q)\) was incorrect, follow through with \(A'=\!\left(p,\frac{1}{2}q\right)\). Do not allow multiple attempts; mark in order given if not clearly labelled. isw rule suspended for this part. |
# Question 4:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(x) = x^2 + \frac{16}{x} \Rightarrow f'(x) = 2x - \frac{16}{x^2}$ | M1 A1 | M1: $x^n \to x^{n-1}$ for either term. Accept $x^2\to x$ or $\frac{1}{x}\to\frac{1}{x^2}$. A1: correct unsimplified $f'(x)=2x-\frac{16}{x^2}$. Accept $f'(x)=2\times x + 16\times-1x^{-2}$. |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Setting $2x - \frac{16}{x^2} = 0 \Rightarrow x = ..$ | M1 | Sets $f'(x)=0$ and proceeds to $x=..$ |
| $x^3 = 8 \Rightarrow x = 2$ | dM1 A1 | dM1: dependent on previous M, scored for $\times x^2$ to reach $x^3=k\Rightarrow x=\sqrt[3]{k}$. A1: correctly achieving $x=2$. Ignore additional solutions. |
| $A = (2, 12)$ | A1 | Correctly achieving $A=(2,12)$. Accept $x=2, y=12$. If additional solutions given $(x>0)$ mark withheld. |
## Part (c)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $A' = (1, 12)$ | B1ft | Accept on sketch or as $x=1, y=12$. If $A=(p,q)$ was incorrect, follow through with $A'=(p-1,q)$. |
## Part (c)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $A' = (2, 6)$ | B1ft | Accept on sketch or as $x=2, y=6$. If $A=(p,q)$ was incorrect, follow through with $A'=\!\left(p,\frac{1}{2}q\right)$. Do not allow multiple attempts; mark in order given if not clearly labelled. isw rule suspended for this part. |4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b85872d4-00b2-499b-9765-f7559d3de66a-05_716_725_219_603}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of part of the curve with equation $y = \mathrm { f } ( x )$ where
$$f ( x ) = x ^ { 2 } + \frac { 16 } { x } , \quad x > 0$$
The curve has a minimum turning point at $A$.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { f } ^ { \prime } ( x )$.
\item Hence find the coordinates of $A$.
\item Use your answer to part (b) to write down the turning point of the curve with equation
\begin{enumerate}[label=(\roman*)]
\item $y = \mathrm { f } ( x + 1 )$,
\item $y = \frac { 1 } { 2 } \mathrm { f } ( x )$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2014 Q4 [8]}}