| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| 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 multi-part question combining standard C4 techniques: (i) routine trapezium rule calculation with given values, (ii) direct application of binomial expansion formula to three terms, (iii) integration of exponential functions. All parts are textbook exercises requiring recall and method application rather than problem-solving or insight, making it slightly easier than average. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<11.09f Trapezium rule: numerical integration |
| \(x\) | 1 | 1.5 | 2 |
| \(\sqrt { 1 + \mathrm { e } ^ { - x } }\) | 1.1696 | 1.1060 | 1.0655 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(A \approx 0.5\left[\dfrac{1.1696 + 1.0655}{2} + 1.1060\right]\) | M1 | Correct expression for trapezium rule |
| \(= 1.11\) (3 s.f.) | A1 cao [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((1+e^{-x})^{1/2} = 1 + \frac{1}{2}e^{-x} + \dfrac{\frac{1}{2}\cdot\frac{-1}{2}}{2!}(e^{-x})^2 + \ldots\) | M1 | Binomial expansion with \(p = \frac{1}{2}\) |
| \(\approx 1 + \frac{1}{2}e^{-x} - \frac{1}{8}e^{-2x}\) | A1, E1 [3] | Correct coefficients |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(I = \int_1^2\left(1 + \frac{1}{2}e^{-x} - \frac{1}{8}e^{-2x}\right)dx\) | ||
| \(= \left[x - \frac{1}{2}e^{-x} + \frac{1}{16}e^{-2x}\right]_1^2\) | M1 | integration |
| \(= \left(2 - \frac{1}{2}e^{-2} + \frac{1}{16}e^{-4}\right) - \left(1 - \frac{1}{2}e^{-1} + \frac{1}{16}e^{-2}\right)\) | A1 | substituting limits into correct expression |
| \(= 1.9335 - 0.8245 = 1.11\) (3 s.f.) | A1 [3] |
# Question 6:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $A \approx 0.5\left[\dfrac{1.1696 + 1.0655}{2} + 1.1060\right]$ | M1 | Correct expression for trapezium rule |
| $= 1.11$ (3 s.f.) | A1 cao [2] | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(1+e^{-x})^{1/2} = 1 + \frac{1}{2}e^{-x} + \dfrac{\frac{1}{2}\cdot\frac{-1}{2}}{2!}(e^{-x})^2 + \ldots$ | M1 | Binomial expansion with $p = \frac{1}{2}$ |
| $\approx 1 + \frac{1}{2}e^{-x} - \frac{1}{8}e^{-2x}$ | A1, E1 [3] | Correct coefficients |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $I = \int_1^2\left(1 + \frac{1}{2}e^{-x} - \frac{1}{8}e^{-2x}\right)dx$ | | |
| $= \left[x - \frac{1}{2}e^{-x} + \frac{1}{16}e^{-2x}\right]_1^2$ | M1 | integration |
| $= \left(2 - \frac{1}{2}e^{-2} + \frac{1}{16}e^{-4}\right) - \left(1 - \frac{1}{2}e^{-1} + \frac{1}{16}e^{-2}\right)$ | A1 | substituting limits into correct expression |
| $= 1.9335 - 0.8245 = 1.11$ (3 s.f.) | A1 [3] | |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, $\left( 1 + \mathrm { e } ^ { - x } \right) ^ { \frac { 1 } { 2 } } \approx 1 + \frac { 1 } { 2 } \mathrm { e } ^ { - x } \quad \frac { 1 } { 8 } \mathrm { e } ^ { - 2 x }$.\\
(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 Q6 [8]}}