| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Factoring out constants first |
| Difficulty | Moderate -0.3 This is a straightforward application of the binomial expansion requiring factoring out constants first. Part (i) is algebraic manipulation, part (ii) is routine substitution into the binomial formula with n=-1/2, and part (iii) requires knowing the standard validity condition |x/25|<1. Slightly easier than average due to being a standard textbook exercise with clear steps. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{\sqrt{25-x}} = (25-x)^{-\frac{1}{2}}\) | B1 | For index |
| \(= 25^{-\frac{1}{2}}\left(1 - \frac{x}{25}\right)^{-\frac{1}{2}}\) | B1 | For correct factorisation and \(25^{-\frac{1}{2}}\) |
| \(= \frac{1}{5}\left(1 - \frac{x}{25}\right)^{-\frac{1}{2}}\) | ||
| Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(= \frac{1}{5}\left(1 + \frac{-\frac{1}{2}}{1}\cdot\left[-\frac{x}{25}\right] + \frac{\frac{-1}{2}\cdot\frac{-3}{2}}{1.2}\cdot\left[-\frac{x}{25}\right]^2 + \frac{\frac{-1}{2}\cdot\frac{-3}{2}\cdot\frac{-5}{2}}{1.2.3}\cdot\left[-\frac{x}{25}\right]^3 + \ldots\right)\) | M1, A2 | A1 for first three terms, A1 for the fourth |
| \(= \frac{1}{5} + \frac{x}{250} + \frac{3x^2}{25000} + \frac{x^3}{250000} + \ldots\) | ||
| Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(-25 < x < 25\) | B1 | |
| Total: 1 mark |
## Question 7(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{\sqrt{25-x}} = (25-x)^{-\frac{1}{2}}$ | B1 | For index |
| $= 25^{-\frac{1}{2}}\left(1 - \frac{x}{25}\right)^{-\frac{1}{2}}$ | B1 | For correct factorisation and $25^{-\frac{1}{2}}$ |
| $= \frac{1}{5}\left(1 - \frac{x}{25}\right)^{-\frac{1}{2}}$ | | |
| **Total: 2 marks** | | |
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## Question 7(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $= \frac{1}{5}\left(1 + \frac{-\frac{1}{2}}{1}\cdot\left[-\frac{x}{25}\right] + \frac{\frac{-1}{2}\cdot\frac{-3}{2}}{1.2}\cdot\left[-\frac{x}{25}\right]^2 + \frac{\frac{-1}{2}\cdot\frac{-3}{2}\cdot\frac{-5}{2}}{1.2.3}\cdot\left[-\frac{x}{25}\right]^3 + \ldots\right)$ | M1, A2 | A1 for first three terms, A1 for the fourth |
| $= \frac{1}{5} + \frac{x}{250} + \frac{3x^2}{25000} + \frac{x^3}{250000} + \ldots$ | | |
| **Total: 3 marks** | | |
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## Question 7(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $-25 < x < 25$ | B1 | |
| **Total: 1 mark** | | |
---7 (i) Show that $\frac { 1 } { \sqrt { 25 - x } } = \frac { 1 } { 5 } \left( 1 - \frac { x } { 25 } \right) ^ { - \frac { 1 } { 2 } }$.\\
(ii) Hence expand $\frac { 1 } { \sqrt { 25 - x } }$ in ascending powers of $x$ up to and including the term in $x ^ { 3 }$.\\
(iii) Write down the range of values of $x$ for which the expansion is valid.
\hfill \mbox{\textit{OCR MEI C4 Q7 [6]}}