| Exam Board | OCR MEI |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2020 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Small angle approximation |
| Type | Find derivative at origin using approximation |
| Difficulty | Standard +0.3 This question involves routine application of small angle approximations (sin x ≈ x) to simplify expressions, standard integration and differentiation, and interpretation of Newton-Raphson output. While multi-part with several marks, each component uses familiar techniques without requiring novel insight—the approximations lead to straightforward algebra, and part (c) is basic numerical interpretation. |
| Spec | 1.05e Small angle approximations: sin x ~ x, cos x ~ 1-x^2/2, tan x ~ x1.07l Derivative of ln(x): and related functions1.08d Evaluate definite integrals: between limits1.09a Sign change methods: locate roots1.09d Newton-Raphson method |
| \(n\) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| \(\mathrm { x } _ { \mathrm { n } }\) | 1 | 0.958509 | 0.950084 | 0.948261 | 0.94786 | 0.947772 | 0.947753 | 0.947748 |
| \(x\) | \(y\) |
| 0.9477475 | \(- 7.79967 \mathrm { E } - 07\) |
| 0.9477485 | \(- 2.90821 \mathrm { E } - 06\) |
| \(x\) | \(y\) |
| 0.947745 | \(4.54066 \mathrm { E } - 06\) |
| 0.947755 | \(- 1.67417 \mathrm { E } - 05\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sin 2x \approx 2x\) or \(\sin x \approx x\) used | M1 (3.1a) | may see \(\cos x \approx 1 - \frac{x^2}{2}\) |
| \(\int \left(\frac{1}{x}\right)dx\) or \(\int \left(\frac{1}{x} - x\right)dx\) obtained | A1 (1.1) | oe nfww |
| \(F[x] = \ln x\) or \(F[x] = \ln x - \frac{1}{2}x^2\) | A1 (1.1) | intermediate step needed from here to earn final mark |
| \(\ln(0.05) - \ln(0.01) = \ln 5\) or \(\ln(0.05) - \ln(0.01) + 0.0012 \approx \ln 5\) | A1 [4] (3.2a) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| differentiation of \(\frac{1}{x}\) | M1 (2.1) | or differentiation of \(y\) using quotient rule and use of small angle approximation |
| substitution of \(0.01\) and \(-10\,000\) correctly obtained | A1 [2] (1.1) | from \(-\frac{1}{x^2}\) or \(-\frac{1}{x^2} - 1\) oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(4.54066 \times 10^{-6}\) or \(0.00000454066\) | B1 (2.5) | cao |
| no sign change for 6 dp, but sign change for 5 dp or last two iterates agree to 5 dp | E1 (3.1a) | allow sign change between \(0.947745\) and \(0.9477475\) |
| \(0.94775\) | B1 [3] (3.2a) |
## Question 10:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sin 2x \approx 2x$ or $\sin x \approx x$ used | M1 (3.1a) | may see $\cos x \approx 1 - \frac{x^2}{2}$ |
| $\int \left(\frac{1}{x}\right)dx$ or $\int \left(\frac{1}{x} - x\right)dx$ obtained | A1 (1.1) | oe nfww |
| $F[x] = \ln x$ or $F[x] = \ln x - \frac{1}{2}x^2$ | A1 (1.1) | intermediate step needed from here to earn final mark |
| $\ln(0.05) - \ln(0.01) = \ln 5$ or $\ln(0.05) - \ln(0.01) + 0.0012 \approx \ln 5$ | A1 [4] (3.2a) | |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| differentiation of $\frac{1}{x}$ | M1 (2.1) | or differentiation of $y$ using quotient rule and use of small angle approximation |
| substitution of $0.01$ and $-10\,000$ correctly obtained | A1 [2] (1.1) | from $-\frac{1}{x^2}$ or $-\frac{1}{x^2} - 1$ oe |
### Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $4.54066 \times 10^{-6}$ or $0.00000454066$ | B1 (2.5) | cao |
| no sign change for 6 dp, but sign change for 5 dp or last two iterates agree to 5 dp | E1 (3.1a) | allow sign change between $0.947745$ and $0.9477475$ |
| $0.94775$ | B1 [3] (3.2a) | |
---10 In this question you must show detailed reasoning.
The equation of a curve is
$$y = \frac { \sin 2 x - x } { x \sin x }$$
\begin{enumerate}[label=(\alph*)]
\item Use the small angle approximation given in the list of formulae on pages 2-3 of this question paper to show that
$$\int _ { 0.01 } ^ { 0.05 } \mathrm { ydx } \approx \ln 5$$
\item Use the same small angle approximation to show that
$$\frac { d y } { d x } \approx - 10000 \text { at the point where } x = 0.01 \text {. }$$
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.
\begin{table}[h]
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | c | }
\hline
$n$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline
$\mathrm { x } _ { \mathrm { n } }$ & 1 & 0.958509 & 0.950084 & 0.948261 & 0.94786 & 0.947772 & 0.947753 & 0.947748 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 10.1}
\end{center}
\end{table}
Joan carries out some analysis of this output. The results are shown in Fig. 10.2.
\begin{table}[h]
\begin{center}
\begin{tabular}{ | c | c | }
\hline
$x$ & $y$ \\
\hline
0.9477475 & $- 7.79967 \mathrm { E } - 07$ \\
\hline
0.9477485 & $- 2.90821 \mathrm { E } - 06$ \\
\hline
$x$ & $y$ \\
\hline
0.947745 & $4.54066 \mathrm { E } - 06$ \\
\hline
0.947755 & $- 1.67417 \mathrm { E } - 05$ \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 10.2}
\end{center}
\end{table}
\item Consider the information in Fig. 10.1 and Fig. 10.2.
\begin{itemize}
\item Write 4.54066E-06 in standard mathematical notation.
\item State the value of the root as accurately as you can, justifying your answer.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 2 2020 Q10 [9]}}