| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Factoring out constants before expansion |
| Difficulty | Standard +0.3 This is a straightforward binomial expansion question requiring factoring out constants to get the form (1+x)^n, followed by conceptual understanding of convergence conditions. Part (a) is routine manipulation, while part (b) tests understanding of validity (|x|<4) and convergence rate (smaller |x| gives better approximation) without requiring calculation. Slightly easier than average due to the conceptual nature of part (b) requiring reasoning rather than computation. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{\sqrt{4-x}} = (4-x)^{-\frac{1}{2}} = 4^{-\frac{1}{2}} \times (1 \pm \ldots)\) | M1 | Strategy of expanding \(\frac{1}{\sqrt{4-x}}\) using binomial. Must see \(4^{-\frac{1}{2}}\) and an expansion. |
| Uses correct binomial expansion \((1+ax)^n = 1 + nax + \frac{n(n-1)}{2}a^2x^2 + \ldots\) | M1 | Condone sign slips and \(a\) not being squared in term 3. Condone \(a = \pm 1\) |
| \(\left(1-\frac{x}{4}\right)^{-\frac{1}{2}} = 1 + \left(-\frac{1}{2}\right)\left(-\frac{x}{4}\right) + \frac{\left(-\frac{1}{2}\right)\times\left(-\frac{3}{2}\right)}{2}\left(-\frac{x}{4}\right)^2\) | A1 | Correct unsimplified. FYI simplified form: \(1 + \frac{x}{8} + \frac{3x^2}{128}\) |
| \(\frac{1}{\sqrt{4-x}} = \frac{1}{2} + \frac{1}{16}x + \frac{3}{256}x^2\) | A1 | Ignore subsequent terms. Allow commas between terms. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States \(x = -14\) and gives valid reason, e.g. expansion only valid for \( | x | < 4\) and \( |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States \(x = -\frac{1}{2}\) and gives valid reason, e.g. it is closest to zero (gives more accurate approximation) | B1 | Any evaluations of the expansions are irrelevant. Look for suitable value and suitable reason. |
## Question 4:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{\sqrt{4-x}} = (4-x)^{-\frac{1}{2}} = 4^{-\frac{1}{2}} \times (1 \pm \ldots)$ | M1 | Strategy of expanding $\frac{1}{\sqrt{4-x}}$ using binomial. Must see $4^{-\frac{1}{2}}$ and an expansion. |
| Uses correct binomial expansion $(1+ax)^n = 1 + nax + \frac{n(n-1)}{2}a^2x^2 + \ldots$ | M1 | Condone sign slips and $a$ not being squared in term 3. Condone $a = \pm 1$ |
| $\left(1-\frac{x}{4}\right)^{-\frac{1}{2}} = 1 + \left(-\frac{1}{2}\right)\left(-\frac{x}{4}\right) + \frac{\left(-\frac{1}{2}\right)\times\left(-\frac{3}{2}\right)}{2}\left(-\frac{x}{4}\right)^2$ | A1 | Correct unsimplified. FYI simplified form: $1 + \frac{x}{8} + \frac{3x^2}{128}$ |
| $\frac{1}{\sqrt{4-x}} = \frac{1}{2} + \frac{1}{16}x + \frac{3}{256}x^2$ | A1 | Ignore subsequent terms. Allow commas between terms. |
### Part (b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| States $x = -14$ and gives valid reason, e.g. expansion only valid for $|x| < 4$ and $|-14| > 4$ | B1 | Do not allow "$-14$ is too big" or "$x=-14, |x|<4$" without reference to validity of expansion |
### Part (b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| States $x = -\frac{1}{2}$ and gives valid reason, e.g. it is closest to zero (gives more accurate approximation) | B1 | Any evaluations of the expansions are irrelevant. Look for suitable value and suitable reason. |\begin{enumerate}
\item (a) Find the first three terms, in ascending powers of $x$, of the binomial expansion of
\end{enumerate}
$$\frac { 1 } { \sqrt { 4 - x } }$$
giving each coefficient in its simplest form.
The expansion can be used to find an approximation to $\sqrt { 2 }$\\
Possible values of $x$ that could be substituted into this expansion are:
\begin{itemize}
\item $x = - 14$ because $\frac { 1 } { \sqrt { 4 - x } } = \frac { 1 } { \sqrt { 18 } } = \frac { \sqrt { 2 } } { 6 }$
\item $x = 2$ because $\frac { 1 } { \sqrt { 4 - x } } = \frac { 1 } { \sqrt { 2 } } = \frac { \sqrt { 2 } } { 2 }$
\item $x = - \frac { 1 } { 2 }$ because $\frac { 1 } { \sqrt { 4 - x } } = \frac { 1 } { \sqrt { \frac { 9 } { 2 } } } = \frac { \sqrt { 2 } } { 3 }$\\
(b) Without evaluating your expansion,\\
(i) state, giving a reason, which of the three values of $x$ should not be used\\
(ii) state, giving a reason, which of the three values of $x$ would lead to the most accurate approximation to $\sqrt { 2 }$
\end{itemize}
\hfill \mbox{\textit{Edexcel Paper 1 2019 Q4 [6]}}