| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2011 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Find stationary points coordinates |
| Difficulty | Standard +0.3 This is a standard C3 differentiation question covering routine techniques: finding x-intercepts by inspection, applying product rule, sign-change method for locating roots, rearranging equations, and performing iterative calculations. All parts are textbook exercises requiring no novel insight, though part (b) requires careful product rule application and parts (d)-(e) involve standard numerical methods that are slightly above pure recall. |
| Spec | 1.06d Natural logarithm: ln(x) function and properties1.07l Derivative of ln(x): and related functions1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((8-x)=0\) or \(\ln x = 0 \Rightarrow x = 8, 1\) | B1 | Either one of \(x=1\) OR \(x=8\) |
| Coordinates are \(A(1,0)\) and \(B(8,0)\) | B1 | Both \(A(1,\{0\})\) and \(B(8,\{0\})\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Apply product rule: \(u=(8-x),\ v=\ln x,\ \frac{du}{dx}=-1,\ \frac{dv}{dx}=\frac{1}{x}\) | M1 | \(vu' + uv'\) |
| \(f'(x) = -\ln x + \frac{8-x}{x}\) | A1 | Any one term correct |
| A1 | Both terms correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f'(3.5) = 0.032951317...\), \(f'(3.6) = -0.058711623...\) Sign change and \(f'(x)\) continuous, therefore \(x\)-coordinate of \(Q\) lies between 3.5 and 3.6 | M1 | Attempts to evaluate both \(f'(3.5)\) and \(f'(3.6)\) |
| A1 | Both values correct to at least 1 s.f., sign change and conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f'(x)=0 \Rightarrow -\ln x + \frac{8-x}{x} = 0\) | M1 | Setting \(f'(x)=0\) |
| \(\Rightarrow -\ln x + \frac{8}{x} - 1 = 0\) | M1 | Splitting up the numerator and proceeding to \(x=\) |
| \(\Rightarrow \frac{8}{x} = \ln x + 1 \Rightarrow 8 = x(\ln x + 1) \Rightarrow x = \frac{8}{\ln x + 1}\) | A1 | Correct proof, no errors seen in working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x_1 = \frac{8}{\ln(3.55)+1}\) | M1 | Attempt to substitute \(x_0 = 3.55\) into iterative formula |
| \(x_1 = 3.528974374...,\ x_2 = 3.538246011...,\ x_3 = 3.534144722...\) | A1 | Both \(x_1\) awrt 3.529 and \(x_2\) awrt 3.538 |
| \(x_1 = 3.529,\ x_2 = 3.538,\ x_3 = 3.534\) (to 3 d.p.) | A1 | \(x_1, x_2, x_3\) all stated correctly to 3 d.p. |
## Question 5:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(8-x)=0$ or $\ln x = 0 \Rightarrow x = 8, 1$ | B1 | Either one of $x=1$ OR $x=8$ |
| Coordinates are $A(1,0)$ and $B(8,0)$ | B1 | Both $A(1,\{0\})$ and $B(8,\{0\})$ |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Apply product rule: $u=(8-x),\ v=\ln x,\ \frac{du}{dx}=-1,\ \frac{dv}{dx}=\frac{1}{x}$ | M1 | $vu' + uv'$ |
| $f'(x) = -\ln x + \frac{8-x}{x}$ | A1 | Any one term correct |
| | A1 | Both terms correct |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(3.5) = 0.032951317...$, $f'(3.6) = -0.058711623...$ Sign change and $f'(x)$ continuous, therefore $x$-coordinate of $Q$ lies between 3.5 and 3.6 | M1 | Attempts to evaluate both $f'(3.5)$ and $f'(3.6)$ |
| | A1 | Both values correct to at least 1 s.f., sign change and conclusion |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(x)=0 \Rightarrow -\ln x + \frac{8-x}{x} = 0$ | M1 | Setting $f'(x)=0$ |
| $\Rightarrow -\ln x + \frac{8}{x} - 1 = 0$ | M1 | Splitting up the numerator and proceeding to $x=$ |
| $\Rightarrow \frac{8}{x} = \ln x + 1 \Rightarrow 8 = x(\ln x + 1) \Rightarrow x = \frac{8}{\ln x + 1}$ | A1 | Correct proof, no errors seen in working |
### Part (e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x_1 = \frac{8}{\ln(3.55)+1}$ | M1 | Attempt to substitute $x_0 = 3.55$ into iterative formula |
| $x_1 = 3.528974374...,\ x_2 = 3.538246011...,\ x_3 = 3.534144722...$ | A1 | Both $x_1$ awrt 3.529 and $x_2$ awrt 3.538 |
| $x_1 = 3.529,\ x_2 = 3.538,\ x_3 = 3.534$ (to 3 d.p.) | A1 | $x_1, x_2, x_3$ all stated correctly to 3 d.p. |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3ff6824f-9fbf-4b5b-8bab-91332c549b36-08_624_1054_274_447}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of part of the curve with equation $y = \mathrm { f } ( x )$, where
$$\mathrm { f } ( x ) = ( 8 - x ) \ln x , \quad x > 0$$
The curve cuts the $x$-axis at the points $A$ and $B$ and has a maximum turning point at $Q$, as shown in Figure 1.
\begin{enumerate}[label=(\alph*)]
\item Write down the coordinates of $A$ and the coordinates of $B$.
\item Find f'(x).
\item Show that the $x$-coordinate of $Q$ lies between 3.5 and 3.6
\item Show that the $x$-coordinate of $Q$ is the solution of
$$x = \frac { 8 } { 1 + \ln x }$$
To find an approximation for the $x$-coordinate of $Q$, the iteration formula
$$x _ { n + 1 } = \frac { 8 } { 1 + \ln x _ { n } }$$
is used.
\item Taking $x _ { 0 } = 3.55$, find the values of $x _ { 1 } , x _ { 2 }$ and $x _ { 3 }$.
Give your answers to 3 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2011 Q5 [13]}}