| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2018 |
| Session | September |
| Marks | 13 |
| Topic | Fixed Point Iteration |
| Type | Apply iteration to find root (pure fixed point) |
| Difficulty | Standard +0.3 This is a standard Newton-Raphson question requiring routine differentiation, algebraic manipulation to derive the formula, iterative calculation (calculator work), and a straightforward change of sign verification. While multi-part, each component uses well-practiced techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.06d Natural logarithm: ln(x) function and properties1.07l Derivative of ln(x): and related functions1.09a Sign change methods: locate roots1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| (i)(a) | \(f'(x) = \frac{3}{3x+1} - 1\) | B1 |
| \(\frac{3}{3x+1} - 1 = 0\) | M1 | Equate to 0 and solve for \(x\) (Must be seen) |
| \(x = \frac{2}{3}\) | A1 | Obtain \(x = \frac{2}{3}\) |
Total: [3]
| Answer | Marks | Guidance |
|---|---|---|
| (i)(b) | E.g. The tangent at \(a\) will be parallel to the x-axis so will never intersect the x-axis to give \(x_2\). | E1 |
Total: [1]
| Answer | Marks | Guidance |
|---|---|---|
| (ii) | \(x_{n+1} = x_n - \frac{\ln(3x_n + 1) - x_n}{3}\) or \(x_{n+1} = x_n - \frac{1}{3x_n + 1} - 1\) | B1 |
| M1 | Attempt rearrangement | |
| \(x_{n+1} = \frac{x_n(2 - 3x_n) - (3x_n + 1)\ln(3x_n + 1) + x_n(3x_n + 1)}{2 - 3x_n}\) | ||
| \(x_{n+1} = \frac{2x_n - 3x_n^2 - (3x_n + 1)\ln(3x_n + 1) + 3x_n^2 + x_n}{2 - 3x_n}\) | ||
| \(x_{n+1} = \frac{3x_n - (3x_n + 1)\ln(3x_n + 1)}{2 - 3x_n}\) | A1 | Obtain given N-R formula |
Total: [3]
| Answer | Marks | Guidance |
|---|---|---|
| (iii) | \(x_2 = 2.54518\) | B1 |
| \(x_3 = 1.94865\), \(x_4 = 1.90416\), \(x_5 = 1.90381\), \(x_6 = 1.90381\) | M1 | Use correct iterative process to find at least two further values (At least 4sf) |
| \(\beta = 1.9038\) | A1 | Obtain \(\beta = 1.9038\) (From at least 5 iterates, to at least 5sf) |
Total: [3]
| Answer | Marks | Guidance |
|---|---|---|
| (iv) | \(\ln(3 \times 1.90375 + 1) - 1.90375 = 0.000035223\) | M1 |
| \(\ln(3 \times 1.90385 + 1) - 1.90385 = -0.000020077\) | A1* | Obtain both correct values (dep*A1) |
| \(f(1.90375) > 0\) and \(f(1.90385) < 0\) so by sign change \(1.90375 < \beta < 1.90385\) so must 1.9038 correct to 5sf | Must refer to sign change |
Total: [3]
(i)(a) | $f'(x) = \frac{3}{3x+1} - 1$ | B1 | Correct $f'(x)$
| $\frac{3}{3x+1} - 1 = 0$ | M1 | Equate to 0 and solve for $x$ (Must be seen)
| $x = \frac{2}{3}$ | A1 | Obtain $x = \frac{2}{3}$
Total: [3]
(i)(b) | E.g. The tangent at $a$ will be parallel to the x-axis so will never intersect the x-axis to give $x_2$. | E1 | Correct reason referring to the tangent at $a$ (A sketch is allowable if annotated.)
Total: [1]
(ii) | $x_{n+1} = x_n - \frac{\ln(3x_n + 1) - x_n}{3}$ or $x_{n+1} = x_n - \frac{1}{3x_n + 1} - 1$ | B1 | State correct N-R formula (Allow no subscripts)
| | M1 | Attempt rearrangement
| $x_{n+1} = \frac{x_n(2 - 3x_n) - (3x_n + 1)\ln(3x_n + 1) + x_n(3x_n + 1)}{2 - 3x_n}$ | | |
| $x_{n+1} = \frac{2x_n - 3x_n^2 - (3x_n + 1)\ln(3x_n + 1) + 3x_n^2 + x_n}{2 - 3x_n}$ | | |
| $x_{n+1} = \frac{3x_n - (3x_n + 1)\ln(3x_n + 1)}{2 - 3x_n}$ | A1 | Obtain given N-R formula
Total: [3]
(iii) | $x_2 = 2.54518$ | B1 | Correct first iterate (At least 4sf)
| $x_3 = 1.94865$, $x_4 = 1.90416$, $x_5 = 1.90381$, $x_6 = 1.90381$ | M1 | Use correct iterative process to find at least two further values (At least 4sf)
| $\beta = 1.9038$ | A1 | Obtain $\beta = 1.9038$ (From at least 5 iterates, to at least 5sf)
Total: [3]
(iv) | $\ln(3 \times 1.90375 + 1) - 1.90375 = 0.000035223$ | M1 | Attempt values either side of root
| $\ln(3 \times 1.90385 + 1) - 1.90385 = -0.000020077$ | A1* | Obtain both correct values (dep*A1)
| $f(1.90375) > 0$ and $f(1.90385) < 0$ so by sign change $1.90375 < \beta < 1.90385$ so must 1.9038 correct to 5sf | | Must refer to sign change
Total: [3]10\\
\includegraphics[max width=\textwidth, alt={}, center]{e3942549-bfc0-432a-bf49-7d01d44af01a-7_579_764_255_651}
The diagram shows the graph of $\mathrm { f } ( x ) = \ln ( 3 x + 1 ) - x$, which has a stationary point at $x = \alpha$. A student wishes to find the non-zero root $\beta$ of the equation $\ln ( 3 x + 1 ) - x = 0$ using the Newton-Raphson method.
\begin{enumerate}[label=(\roman*)]
\item (a) Determine the value of $\alpha$.\\
(b) Explain why the Newton-Raphson method will fail if $\alpha$ is used as the initial value.
\item Show that the Newton-Raphson iterative formula for finding $\beta$ can be written as
$$x _ { n + 1 } = \frac { 3 x _ { n } - \left( 3 x _ { n } + 1 \right) \ln \left( 3 x _ { n } + 1 \right) } { 2 - 3 x _ { n } } .$$
\item Apply the iterative formula in part (ii) with initial value $x _ { 1 } = 1$ to find the value of $\beta$ correct to 5 significant figures. You should show the result of each iteration.
\item Use a change of sign method to verify that the value of $\beta$ found in part (iii) is correct to 5 significant figures.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/01 2018 Q10 [13]}}