| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Expand and state validity |
| Difficulty | Moderate -0.8 This is a straightforward application of the binomial expansion formula for fractional powers with standard validity checking. Part (a) requires direct substitution into the formula (1+x)^n = 1 + nx + n(n-1)x²/2! + ... with minimal algebraic manipulation. Part (b) tests basic understanding that |9x| < 1 is required for validity, which is routine knowledge. This is easier than average as it involves no problem-solving or novel insight, just mechanical application of a standard technique. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{\frac{1}{2}(\frac{1}{2}-1)}{2!}(\pm 9x)^2\) or \(\frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)}{3!}(-9x)^3\) | M1 | Attempt at binomial expansion with \(n=\frac{1}{2}\); correct structure for term 3 or term 4 with correct coefficient and correct power of \(x\). |
| \((1-9x)^{\frac{1}{2}} = 1 + \frac{1}{2}(-9x) + \frac{\frac{1}{2}(\frac{1}{2}-1)}{2!}(\pm 9x)^2 + \frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)}{3!}(-9x)^3\) | A1 | Correct unsimplified expression; bracketing must be correct unless implied by subsequent work. |
| \((1-9x)^{\frac{1}{2}} = 1 - \frac{9}{2}x - \frac{81}{8}x^2 - \frac{729}{16}x^3\) | A1 | Fully simplified correct answer. Allow decimal equivalents \(1-4.5x-10.125x^2-45.5625x^3\) if exact. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Expansion is valid for \( | x | < \frac{1}{9}\) and \(x = -\frac{2}{9}\) is outside this range |
# Question 2(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{\frac{1}{2}(\frac{1}{2}-1)}{2!}(\pm 9x)^2$ or $\frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)}{3!}(-9x)^3$ | M1 | Attempt at binomial expansion with $n=\frac{1}{2}$; correct structure for term 3 **or** term 4 with correct coefficient and correct power of $x$. |
| $(1-9x)^{\frac{1}{2}} = 1 + \frac{1}{2}(-9x) + \frac{\frac{1}{2}(\frac{1}{2}-1)}{2!}(\pm 9x)^2 + \frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)}{3!}(-9x)^3$ | A1 | Correct unsimplified expression; bracketing must be correct unless implied by subsequent work. |
| $(1-9x)^{\frac{1}{2}} = 1 - \frac{9}{2}x - \frac{81}{8}x^2 - \frac{729}{16}x^3$ | A1 | Fully simplified correct answer. Allow decimal equivalents $1-4.5x-10.125x^2-45.5625x^3$ if exact. |
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# Question 2(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Expansion is valid for $|x| < \frac{1}{9}$ and $x = -\frac{2}{9}$ **is outside this range** | B1 | Requires an acceptable range of validity AND an acceptable comparison of $-\frac{2}{9}$ or $\frac{2}{9}$ with their range leading to "not valid". Do not accept vague statements without mention of the range. |
---\begin{enumerate}
\item (a) Find, in ascending powers of $x$, the first four terms of the binomial expansion of
\end{enumerate}
$$( 1 - 9 x ) ^ { \frac { 1 } { 2 } }$$
giving each term in simplest form.\\
(b) Give a reason why $x = - \frac { 2 } { 9 }$ should not be used in the expansion to find an approximation to $\sqrt { 3 }$
\hfill \mbox{\textit{Edexcel Paper 1 2024 Q2 [4]}}