| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2007 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Coefficient of x^n in product |
| Difficulty | Standard +0.3 This is a straightforward application of the binomial expansion for (1+x)^(1/2) with x = e^(-x), requiring students to identify the first three terms and then integrate term-by-term. The trapezium rule part is routine. While it involves multiple steps, each step follows standard procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<11.09f Trapezium rule: numerical integration4.04c Scalar product: calculate and use for angles |
| \(x\) | 1 | 1.5 | 2 |
| \(\sqrt { 1 + \mathrm { e } ^ { - x } }\) | 1.1696 | 1.1060 | 1.0655 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Trapezium rule: \(h = 0.5\) | M1 | |
| \(\approx \frac{0.5}{2}(1.1696 + 2(1.1060) + 1.0655)\) | ||
| \(= \frac{1}{4}(4.4471) = 1.11\) | A1 | 3 s.f. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((1+e^{-x})^{1/2} = 1 + \frac{1}{2}(e^{-x}) + \frac{\frac{1}{2}\cdot(-\frac{1}{2})}{2!}(e^{-x})^2 + ...\) | M1 | Binomial expansion |
| \(= 1 + \frac{1}{2}e^{-x} - \frac{1}{8}e^{-2x} + ...\) | M1 A1 | Correct second term and third term |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int_1^2 \left(1 + \frac{1}{2}e^{-x} - \frac{1}{8}e^{-2x}\right)dx\) | M1 | |
| \(= \left[x - \frac{1}{2}e^{-x} + \frac{1}{16}e^{-2x}\right]_1^2\) | M1 | |
| \(= \left(2 - \frac{1}{2}e^{-2} + \frac{1}{16}e^{-4}\right) - \left(1 - \frac{1}{2}e^{-1} + \frac{1}{16}e^{-2}\right)\) | ||
| \(\approx 1.108 \approx 1.11\) | A1 | 3 s.f. |
## Question 6(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Trapezium rule: $h = 0.5$ | M1 | |
| $\approx \frac{0.5}{2}(1.1696 + 2(1.1060) + 1.0655)$ | | |
| $= \frac{1}{4}(4.4471) = 1.11$ | A1 | 3 s.f. |
## Question 6(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(1+e^{-x})^{1/2} = 1 + \frac{1}{2}(e^{-x}) + \frac{\frac{1}{2}\cdot(-\frac{1}{2})}{2!}(e^{-x})^2 + ...$ | M1 | Binomial expansion |
| $= 1 + \frac{1}{2}e^{-x} - \frac{1}{8}e^{-2x} + ...$ | M1 A1 | Correct second term and third term |
## Question 6(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_1^2 \left(1 + \frac{1}{2}e^{-x} - \frac{1}{8}e^{-2x}\right)dx$ | M1 | |
| $= \left[x - \frac{1}{2}e^{-x} + \frac{1}{16}e^{-2x}\right]_1^2$ | M1 | |
| $= \left(2 - \frac{1}{2}e^{-2} + \frac{1}{16}e^{-4}\right) - \left(1 - \frac{1}{2}e^{-1} + \frac{1}{16}e^{-2}\right)$ | | |
| $\approx 1.108 \approx 1.11$ | A1 | 3 s.f. |
---$$+ \mu \left( \begin{array} { r }
1 \\
0 \\
- 2
\end{array} \right)$$
Find the acute angle between the lines.
6 Two students are trying to evaluate the integral $\int _ { 1 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { - x } } \mathrm {~d} x$.\\
Sarah uses the trapezium rule with 2 strips, and starts by constructing the following table.
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\hline
$x$ & 1 & 1.5 & 2 \\
\hline
$\sqrt { 1 + \mathrm { e } ^ { - x } }$ & 1.1696 & 1.1060 & 1.0655 \\
\hline
\end{tabular}
\end{center}
(i) Complete the calculation, giving your answer to 3 significant figures.
Anish uses a binomial approximation for $\sqrt { 1 + \mathrm { e } ^ { - x } }$ and then integrates this.\\
(ii) Show that, provided $\mathrm { e } ^ { - x }$ is suitably small, $\left( 1 + \mathrm { e } ^ { - x } \right) ^ { \frac { 1 } { 2 } } \approx 1 + \frac { 1 } { 2 } \mathrm { e } ^ { - x } - \frac { 1 } { 8 } \mathrm { e } ^ { - 2 x }$.\\
(iii) Use this result to evaluate $\int _ { 1 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { - x } } \mathrm {~d} x$ approximately, giving your answer to 3 significant figures.
\hfill \mbox{\textit{OCR MEI C4 2007 Q6 [8]}}