| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Factoring out constants first |
| Difficulty | Standard +0.3 This is a straightforward application of the binomial expansion requiring factoring out the constant (√4 = 2) to get 2(1 - 9x/4)^(1/2), then expanding using the standard formula. Part (b) tests understanding of alternating series and convergence, which is conceptual but routine for this topic. Slightly above average due to the factoring step and conceptual part (b), but still a standard textbook-style question. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sqrt{4-9x} = 2\!\left(1\pm\ldots\right)^{\frac{1}{2}}\) | B1 | Takes out factor of 4; writes \(\sqrt{4-9x} = 2(1\pm\ldots)^{\frac{1}{2}}\) or equivalent. |
| \(\left(1 - \frac{9x}{4}\right)^{\frac{1}{2}} = \ldots + \dfrac{\frac{1}{2}\times\left(-\frac{1}{2}\right)\left(-\frac{9x}{4}\right)^2}{2!}\) or \(\ldots + \dfrac{\frac{1}{2}\times\left(-\frac{1}{2}\right)\times\left(-\frac{3}{2}\right)\left(-\frac{9x}{4}\right)^3}{3!}\) | M1 | Attempt at binomial expansion of \((1+ax)^{\frac{1}{2}},\ a\neq 1\), forming term 3 or term 4 with correct structure and binomial coefficients. |
| \(1 + \frac{1}{2}\times\!\left(-\frac{9x}{4}\right) + \dfrac{\frac{1}{2}\times\!\left(-\frac{1}{2}\right)\!\left(-\frac{9x}{4}\right)^2}{2!} + \dfrac{\frac{1}{2}\times\!\left(-\frac{1}{2}\right)\times\!\left(-\frac{3}{2}\right)\!\left(-\frac{9x}{4}\right)^3}{3!}\) | A1 | Correct unsimplified expansion of \(\left(1-\frac{9x}{4}\right)^{\frac{1}{2}}\). |
| \(\sqrt{4-9x} = 2 - \dfrac{9x}{4} - \dfrac{81x^2}{64} - \dfrac{729x^3}{512}\) | A1 | Fully simplified final answer. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States the approximation will be an overestimate since all terms after the first are negative (since \(x>0\)) | B1 | 3.2b — Must state overestimate and give reason that all subsequent terms are negative for \(x>0\). |
## Question 7:
### Part 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sqrt{4-9x} = 2\!\left(1\pm\ldots\right)^{\frac{1}{2}}$ | B1 | Takes out factor of 4; writes $\sqrt{4-9x} = 2(1\pm\ldots)^{\frac{1}{2}}$ or equivalent. |
| $\left(1 - \frac{9x}{4}\right)^{\frac{1}{2}} = \ldots + \dfrac{\frac{1}{2}\times\left(-\frac{1}{2}\right)\left(-\frac{9x}{4}\right)^2}{2!}$ or $\ldots + \dfrac{\frac{1}{2}\times\left(-\frac{1}{2}\right)\times\left(-\frac{3}{2}\right)\left(-\frac{9x}{4}\right)^3}{3!}$ | M1 | Attempt at binomial expansion of $(1+ax)^{\frac{1}{2}},\ a\neq 1$, forming term 3 or term 4 with correct structure and binomial coefficients. |
| $1 + \frac{1}{2}\times\!\left(-\frac{9x}{4}\right) + \dfrac{\frac{1}{2}\times\!\left(-\frac{1}{2}\right)\!\left(-\frac{9x}{4}\right)^2}{2!} + \dfrac{\frac{1}{2}\times\!\left(-\frac{1}{2}\right)\times\!\left(-\frac{3}{2}\right)\!\left(-\frac{9x}{4}\right)^3}{3!}$ | A1 | Correct unsimplified expansion of $\left(1-\frac{9x}{4}\right)^{\frac{1}{2}}$. |
| $\sqrt{4-9x} = 2 - \dfrac{9x}{4} - \dfrac{81x^2}{64} - \dfrac{729x^3}{512}$ | A1 | Fully simplified final answer. |
### Part 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| States the approximation will be an **overestimate** since all terms after the first are negative (since $x>0$) | B1 | 3.2b — Must state overestimate and give reason that all subsequent terms are negative for $x>0$. |\begin{enumerate}
\item (a) Find the first four terms, in ascending powers of $x$, of the binomial expansion of
\end{enumerate}
$$\sqrt { 4 - 9 x }$$
writing each term in simplest form.
A student uses this expansion with $x = \frac { 1 } { 9 }$ to find an approximation for $\sqrt { 3 }$\\
Using the answer to part (a) and without doing any calculations,\\
(b) state whether this approximation will be an overestimate or an underestimate of $\sqrt { 3 }$ giving a brief reason for your answer.
\hfill \mbox{\textit{Edexcel Paper 2 2022 Q7 [5]}}