October 2020 Paper 2 Q10
10 In this question you must show detailed reasoning.
The equation of a curve is
\(y = \dfrac{\sin 2x - x}{x\sin x}\).
\(\displaystyle\int_{0.01}^{0.05} y\,\mathrm{d}x \approx \ln 5\). [4]
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} \approx -10000\) at the point where \(x = 0.01\). [2]
The equation \(y = 0\) has a root near \(x = 1\). Joan uses the Newton-Raphson method to find this root. The output from the spreadsheet she uses is shown in Fig. 10.1.
| \(n\) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|---|
| \(x_n\) | 1 | 0.958509 | 0.950084 | 0.948261 | 0.94786 | 0.947772 | 0.947753 | 0.947748 |
Fig. 10.1
Joan carries out some analysis of this output. The results are shown in Fig. 10.2.
| \(x\) | \(y\) |
|---|---|
| 0.9477475 | –7.79967E–07 |
| 0.9477485 | –2.90821E–06 |
| \(x\) | \(y\) |
| 0.947745 | 4.54066E–06 |
| 0.947755 | –1.67417E–05 |
Fig. 10.2
- Write 4.54066E–06 in standard mathematical notation.
- State the value of the root as accurately as you can, justifying your answer.
| Scheme | Marks | AO |
|---|---|---|
| \(\sin 2x \approx 2x\) or \(\sin x \approx x\) used | M1 | 3.1a |
| \(\displaystyle\int \left(\frac{1}{x}\right)\mathrm{d}x\) or \(\displaystyle\int \left(\frac{1}{x} - x\right)\mathrm{d}x\) obtained oe nfww | A1 | 1.1 |
| \(\mathrm{F}[x] = \ln x\) oe or \(\mathrm{F}[x] = \ln x - \frac{1}{2}x^2\) oe | A1 | 1.1 |
| \(\ln(0.05) - \ln(0.01) = \ln 5\) oe or \(\ln(0.05) - \ln(0.01) - 0.0012 \approx \ln 5\) oe | A1 | 3.2a |
| [4] |
Notes
M1: may see \(\cos x \approx 1 - \frac{x^2}{2}\)
A1: intermediate step needed from here to earn final mark
(corrected from the printed mark scheme: the last line is printed as \(\ln(0.05) - \ln(0.01) + 0.0012 \approx \ln 5\); with \(\mathrm{F}[x] = \ln x - \frac{1}{2}x^2\) the 0.0012 is subtracted)
| Scheme | Marks | AO |
|---|---|---|
| differentiation of their \(\frac{1}{x}\) | M1 | 2.1 |
| substitution of 0.01 and \(-10\,000\) correctly obtained | A1 | 1.1 |
| [2] |
Notes
M1: or differentiation of \(y\) using quotient rule and use of small angle approximation
A1: from \(-\frac{1}{x^2}\) or \(-\frac{1}{x^2} - 1\) oe
| Scheme | Marks | AO |
|---|---|---|
| \(4.54066 \times 10^{-6}\) or 0.00000454066 cao | B1 | 2.5 |
| (no sign change for 6 dp), but sign change for 5 dp or last two iterates agree to 5dp | E1 | 3.1a |
| 0.94775 | B1 | 3.2a |
| [3] |
Notes
E1: allow sign change between 0.947745 and 0.9477475