| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2015 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Derive stationary point equation |
| Difficulty | Standard +0.3 Part (a) is a routine calculus exercise requiring differentiation of x²ln(x) using the product rule, setting the derivative to zero, and algebraic rearrangement. Parts (b) and (c) are mechanical iteration and substitution. This is a standard multi-part question testing basic differentiation and iteration techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07l Derivative of ln(x): and related functions1.07n Stationary points: find maxima, minima using derivatives1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \frac{x^2\ln x}{3} - 2x + 4 \Rightarrow \frac{dy}{dx} = \frac{2x\ln x}{3} + \frac{x^2}{3x} - 2\) | M1A1, B1 | |
| \(\frac{2x\ln x}{3} + \frac{x^2}{3x} - 2 = 0 \Rightarrow x(2\ln x + 1) = 6\) | dM1 | |
| \(\Rightarrow x = \frac{6}{1+\ln x^2}\) | A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x_1 = \frac{6}{1+\ln(2.27^2)} = \text{awrt } 2.273\) | M1A1 | |
| \(x_2 = \text{awrt } 2.271\) and \(x_3 = \text{awrt } 2.273\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(A = (2.3, 0.9)\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Applying product rule to \(x^2 \ln x\), e.g. \(u = \frac{x^2}{3}, v = \ln x\) giving \(vu' + uv'\) | M1 | Rule must be correct if quoted; implied by \(\frac{x^2 \ln x}{3}\) or \(\frac{x^2}{3} \times \frac{\ln x}{3}\) |
| \(\frac{x^2 \ln x}{3} \rightarrow \frac{2x \ln x}{3} + \frac{x^2}{3x}\) | A1 | Correct unsimplified derivative |
| Derivative of \(-2x+4\) term seen or implied as \(-2\) | B1 | |
| Set \(\frac{dy}{dx} = 0\), take common factor of \(x\), proceed to \(x=\) | dM1 | Dependent on previous M; or state \(\frac{dy}{dx}=0\) and write \(\frac{2x\ln x}{3} + \frac{x^2}{3x} = 2\) |
| \(x = \frac{6}{1 + \ln x^2}\) | A1* | Given answer; all aspects must be correct; \(2\ln x + 1 \Rightarrow \ln x^2 + 1\) may be stated |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts \(\frac{6}{1 + \ln(2.27)^2}\) | M1 | awrt 2.273 implies method |
| \(x_1 =\) awrt \(2.273\) | A1 | Subscript not important; mark as first value given |
| \(x_2 =\) awrt \(2.271\) (3dp) and \(x_3 =\) awrt \(2.273\) (3dp) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Deduces \(x\) coordinate of \(A\) is \(2.3\) | M1 | Sight of \(2.3\) sufficient if values in (b) round to this; or substitutes rounded answer from (b) into \(y\) equation |
| \((2.3, 0.9)\) | A1 | Accept \(x = 2.3\), \(y = 0.9\) |
# Question 10:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \frac{x^2\ln x}{3} - 2x + 4 \Rightarrow \frac{dy}{dx} = \frac{2x\ln x}{3} + \frac{x^2}{3x} - 2$ | M1A1, B1 | |
| $\frac{2x\ln x}{3} + \frac{x^2}{3x} - 2 = 0 \Rightarrow x(2\ln x + 1) = 6$ | dM1 | |
| $\Rightarrow x = \frac{6}{1+\ln x^2}$ | A1* | |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x_1 = \frac{6}{1+\ln(2.27^2)} = \text{awrt } 2.273$ | M1A1 | |
| $x_2 = \text{awrt } 2.271$ and $x_3 = \text{awrt } 2.273$ | A1 | |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $A = (2.3, 0.9)$ | M1 A1 | |
# Question (a) - Differentiation:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Applying product rule to $x^2 \ln x$, e.g. $u = \frac{x^2}{3}, v = \ln x$ giving $vu' + uv'$ | M1 | Rule must be correct if quoted; implied by $\frac{x^2 \ln x}{3}$ or $\frac{x^2}{3} \times \frac{\ln x}{3}$ |
| $\frac{x^2 \ln x}{3} \rightarrow \frac{2x \ln x}{3} + \frac{x^2}{3x}$ | A1 | Correct unsimplified derivative |
| Derivative of $-2x+4$ term seen or implied as $-2$ | B1 | |
| Set $\frac{dy}{dx} = 0$, take common factor of $x$, proceed to $x=$ | dM1 | Dependent on previous M; or state $\frac{dy}{dx}=0$ and write $\frac{2x\ln x}{3} + \frac{x^2}{3x} = 2$ |
| $x = \frac{6}{1 + \ln x^2}$ | A1* | Given answer; all aspects must be correct; $2\ln x + 1 \Rightarrow \ln x^2 + 1$ may be stated |
---
# Question (b) - Iteration:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $\frac{6}{1 + \ln(2.27)^2}$ | M1 | awrt 2.273 implies method |
| $x_1 =$ awrt $2.273$ | A1 | Subscript not important; mark as first value given |
| $x_2 =$ awrt $2.271$ (3dp) **and** $x_3 =$ awrt $2.273$ (3dp) | A1 | |
---
# Question (c) - Coordinates of A:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Deduces $x$ coordinate of $A$ is $2.3$ | M1 | Sight of $2.3$ sufficient if values in (b) round to this; or substitutes rounded answer from (b) into $y$ equation |
| $(2.3, 0.9)$ | A1 | Accept $x = 2.3$, $y = 0.9$ |
---10.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{03548211-79cb-4629-b6ca-aa9dfcc77a33-17_598_736_223_603}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a sketch of part of the curve $C$ with equation
$$y = \frac { x ^ { 2 } \ln x } { 3 } - 2 x + 4 , \quad x > 0$$
Point $A$ is the minimum turning point on the curve.
\begin{enumerate}[label=(\alph*)]
\item Show, by using calculus, that the $x$ coordinate of point $A$ is a solution of
$$x = \frac { 6 } { 1 + \ln \left( x ^ { 2 } \right) }$$
\item Starting with $x _ { 0 } = 2.27$, use the iteration
$$x _ { n + 1 } = \frac { 6 } { 1 + \ln \left( x _ { n } ^ { 2 } \right) }$$
to calculate the values of $x _ { 1 } , x _ { 2 }$ and $x _ { 3 }$, giving your answers to 3 decimal places.
\item Use your answer to part (b) to deduce the coordinates of point $A$ to one decimal place.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2015 Q10 [10]}}