| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Solve exponential equation via iteration |
| Difficulty | Standard +0.3 This is a standard C3 iteration question with routine differentiation, normal line calculation, and fixed-point iteration. Parts (a)-(b) are straightforward chain rule and normal line work. Part (c) involves standard sign-change verification and applying a given iteration formula with calculator work, plus drawing a cobweb diagram—all textbook exercises requiring no novel insight. Slightly easier than average due to the structured guidance. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.09a 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 |
| \(\frac{dy}{dx} = \frac{-2}{2e-x}\) | M1 | Attempt at chain rule/differentiation of \(\ln\) |
| \(\frac{dy}{dx} = \frac{-2}{2e-x}\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| At \(x = e\): \(\frac{dy}{dx} = \frac{-2}{2e-e} = \frac{-2}{e}\) | M1 | Substituting \(x=e\) into their derivative |
| Gradient of normal \(= \frac{e}{2}\) | M1 | Using \(m_1 m_2 = -1\) |
| At \(x = e\): \(y = 2\ln(2e-e) = 2\ln e = 2\) | A1 | Correct point \((e, 2)\) |
| Normal: \(y - 2 = \frac{e}{2}(x - e)\) | A1 | Correct equation of normal |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Let \(f(x) = 2\ln(2e-x) - x\) | ||
| \(f(1) = 2\ln(2e-1) - 1 = 2(1.4786...) - 1 = 1.957... > 0\) | M1 | Evaluating at both endpoints |
| \(f(3) = 2\ln(2e-3) - 3 = 2(0.3120...) - 3 = -2.376... < 0\) | A1 | Both correct with sign change, conclusion stated |
| Sign change implies root between 1 and 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x_2 = 2\ln(2e - 1) = 2.957\) | M1 | Correct iteration applied |
| \(x_3 = 2\ln(2e - 2.957) = 1.751\) | A1 | Both values correct to 3 d.p. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Staircase diagram: from \(x_1\) on \(x\)-axis, vertical line to curve, horizontal to \(y=x\), vertical to curve, horizontal to \(y=x\) | M1 | Correct cobweb/staircase structure shown |
| \(x_2\) and \(x_3\) correctly marked on \(x\)-axis | A1 | Positions consistent with calculated values |
# Question 2:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{-2}{2e-x}$ | M1 | Attempt at chain rule/differentiation of $\ln$ |
| $\frac{dy}{dx} = \frac{-2}{2e-x}$ | A1 | Correct answer |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| At $x = e$: $\frac{dy}{dx} = \frac{-2}{2e-e} = \frac{-2}{e}$ | M1 | Substituting $x=e$ into their derivative |
| Gradient of normal $= \frac{e}{2}$ | M1 | Using $m_1 m_2 = -1$ |
| At $x = e$: $y = 2\ln(2e-e) = 2\ln e = 2$ | A1 | Correct point $(e, 2)$ |
| Normal: $y - 2 = \frac{e}{2}(x - e)$ | A1 | Correct equation of normal |
## Part (c)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $f(x) = 2\ln(2e-x) - x$ | | |
| $f(1) = 2\ln(2e-1) - 1 = 2(1.4786...) - 1 = 1.957... > 0$ | M1 | Evaluating at both endpoints |
| $f(3) = 2\ln(2e-3) - 3 = 2(0.3120...) - 3 = -2.376... < 0$ | A1 | Both correct with sign change, conclusion stated |
| Sign change implies root between 1 and 3 | | |
## Part (c)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x_2 = 2\ln(2e - 1) = 2.957$ | M1 | Correct iteration applied |
| $x_3 = 2\ln(2e - 2.957) = 1.751$ | A1 | Both values correct to 3 d.p. |
## Part (c)(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Staircase diagram: from $x_1$ on $x$-axis, vertical line to curve, horizontal to $y=x$, vertical to curve, horizontal to $y=x$ | M1 | Correct cobweb/staircase structure shown |
| $x_2$ and $x_3$ correctly marked on $x$-axis | A1 | Positions consistent with calculated values |
---2 A curve has equation $y = 2 \ln ( 2 \mathrm { e } - x )$.
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item Find an equation of the normal to the curve $y = 2 \ln ( 2 \mathrm { e } - x )$ at the point on the curve where $x = \mathrm { e }$.\\[0pt]
[4 marks]
\item The curve $y = 2 \ln ( 2 \mathrm { e } - x )$ intersects the line $y = x$ at a single point, where $x = \alpha$.
\begin{enumerate}[label=(\roman*)]
\item Show that $\alpha$ lies between 1 and 3 .
\item Use the recurrence relation
$$x _ { n + 1 } = 2 \ln \left( 2 \mathrm { e } - x _ { n } \right)$$
with $x _ { 1 } = 1$ to find the values of $x _ { 2 }$ and $x _ { 3 }$, giving your answers to three decimal places.
\item Figure 1, on the opposite page, shows a sketch of parts of the graphs of $y = 2 \ln ( 2 \mathrm { e } - x )$ and $y = x$, and the position of $x _ { 1 }$.
On Figure 1, draw a cobweb or staircase diagram to show how convergence takes place, indicating the positions of $x _ { 2 }$ and $x _ { 3 }$ on the $x$-axis.\\[0pt]
[2 marks]
\section*{(c)(iii)}
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\includegraphics[alt={},max width=\textwidth]{57412ec0-ad97-4418-8ba8-93f1f7d8aac1-05_864_1284_1802_386}
\end{center}
\end{figure}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C3 2014 Q2 [12]}}