| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2017 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Find stationary points coordinates |
| Difficulty | Moderate -0.8 Part (a) is a standard stationary point problem requiring differentiation and solving a simple equation. Parts (b)(i-iii) test understanding of function transformations but require only algebraic manipulation of given information, not calculus. This is routine C1/C2 material with no novel problem-solving required. |
| 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 | Mark | Guidance |
| \(f(x) = 8x^{-1} + \frac{1}{2}x - 5 \Rightarrow f'(x) = -8x^{-2} + \frac{1}{2}\) | M1A1 | M1: \(-8x^{-2}\) or \(\frac{1}{2}\); A1: fully correct \(f'(x) = -8x^{-2} + \frac{1}{2}\) (may be unsimplified) |
| Sets \(-8x^{-2} + \frac{1}{2} = 0 \Rightarrow x = 4\) | M1A1 | M1: sets \(f'(x)=0\) (may be implied) and proceeds to find \(x\); A1: \(x=4\) (allow \(x=\pm4\)) |
| \((4, -1)\) | A1 | Correct coordinates (allow \(x=4, y=-1\)); ignore \((-4,\ldots)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = 2, 8\) | B1 | \(x=2\) and \(x=8\) only; do not accept as coordinates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((4, 1)\) | B1ft | Follow through from (a); accept \((x, y+2)\) from their \((x,y)\); with no other points |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = 2, \frac{1}{2}\) | B1 | Both answers needed; accept \((2,0), \left(\frac{1}{2}, 0\right)\); ignore reference to image of turning point |
## Question 9:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) = 8x^{-1} + \frac{1}{2}x - 5 \Rightarrow f'(x) = -8x^{-2} + \frac{1}{2}$ | M1A1 | M1: $-8x^{-2}$ or $\frac{1}{2}$; A1: fully correct $f'(x) = -8x^{-2} + \frac{1}{2}$ (may be unsimplified) |
| Sets $-8x^{-2} + \frac{1}{2} = 0 \Rightarrow x = 4$ | M1A1 | M1: sets $f'(x)=0$ (may be implied) and proceeds to find $x$; A1: $x=4$ (allow $x=\pm4$) |
| $(4, -1)$ | A1 | Correct coordinates (allow $x=4, y=-1$); ignore $(-4,\ldots)$ |
### Part (b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 2, 8$ | B1 | $x=2$ and $x=8$ **only**; do not accept as coordinates |
### Part (b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(4, 1)$ | B1ft | Follow through from (a); accept $(x, y+2)$ from their $(x,y)$; with no other points |
### Part (b)(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 2, \frac{1}{2}$ | B1 | Both answers needed; accept $(2,0), \left(\frac{1}{2}, 0\right)$; ignore reference to image of turning point |
---9.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f39ade34-32e2-4b5c-b80a-9663c6a65c87-14_609_744_223_593}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a sketch of the curve with equation $y = \mathrm { f } ( x )$ where
$$f ( x ) = \frac { 8 } { x } + \frac { 1 } { 2 } x - 5 , \quad 0 < x \leqslant 12$$
The curve crosses the $x$-axis at $( 2,0 )$ and $( 8,0 )$ and has a minimum point at $A$.
\begin{enumerate}[label=(\alph*)]
\item Use calculus to find the coordinates of point $A$.
\item State
\begin{enumerate}[label=(\roman*)]
\item the roots of the equation $2 \mathrm { f } ( x ) = 0$
\item the coordinates of the turning point on the curve $y = \mathrm { f } ( x ) + 2$
\item the roots of the equation $\mathrm { f } ( 4 x ) = 0$
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2017 Q9 [8]}}