| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2009 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Factoring out constants before expansion |
| Difficulty | Moderate -0.3 This is a straightforward application of the binomial theorem with fractional powers. Part (a)(i) is routine recall of the standard expansion formula, part (a)(ii) requires the simple technique of factoring out a constant (√4 = 2), and part (b) is a direct substitution (x=1). While it involves multiple steps, each step follows a standard procedure with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions |
| Answer | Marks |
|---|---|
| \((1-x)^{\frac{1}{2}} = 1 + \frac{1}{2}(-x) + px^2 = 1 - \frac{1}{2}x - \frac{1}{8}x^2\) | M1, A1 (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sqrt{4-x} = 2\left(1 - \frac{x}{4}\right)^{\frac{1}{2}}\) | B1 | or \((4)^{\frac{1}{2}}\left(1-\frac{x}{4}\right)^{\frac{1}{2}}\) |
| \(= (2)\left(1 - \frac{1}{2}\left(\frac{x}{4}\right) - \frac{1}{8}\left(\frac{x}{4}\right)^2\right)\) | M1 | \(x\) replaced by \(\frac{x}{4}\); condone missing \(()\) |
| \(= 2 - \frac{x}{4} - \frac{x^2}{64}\) | A1 (3 marks) | CAO or decimal equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| \((4-x)^{\frac{1}{2}} = 4^{\frac{1}{2}} + \frac{1}{2}\times 4^{-\frac{1}{2}}(-x) + \frac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2}4^{-\frac{3}{2}}(-x)^2 = 2 - \frac{x}{4} - \frac{x^2}{64}\) | (M1), (A1), (A1) (3 marks) | Use of \((a+x)^n\) from formula book; condone missing brackets and 1 error |
| Answer | Marks | Guidance |
|---|---|---|
| \(x=1\): \(\sqrt{3} \approx 2 - \frac{1}{4} - \frac{1}{64} = 1.734\) (3 dp) | M1, A1 (2 marks) | \(x=1\) used in their expansion; CSO |
## Question 4:
### Part (a)(i)
| $(1-x)^{\frac{1}{2}} = 1 + \frac{1}{2}(-x) + px^2 = 1 - \frac{1}{2}x - \frac{1}{8}x^2$ | M1, A1 (2 marks) | |
### Part (a)(ii)
| $\sqrt{4-x} = 2\left(1 - \frac{x}{4}\right)^{\frac{1}{2}}$ | B1 | or $(4)^{\frac{1}{2}}\left(1-\frac{x}{4}\right)^{\frac{1}{2}}$ |
| $= (2)\left(1 - \frac{1}{2}\left(\frac{x}{4}\right) - \frac{1}{8}\left(\frac{x}{4}\right)^2\right)$ | M1 | $x$ replaced by $\frac{x}{4}$; condone missing $()$ |
| $= 2 - \frac{x}{4} - \frac{x^2}{64}$ | A1 (3 marks) | CAO or decimal equivalent |
**Alternative:**
| $(4-x)^{\frac{1}{2}} = 4^{\frac{1}{2}} + \frac{1}{2}\times 4^{-\frac{1}{2}}(-x) + \frac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2}4^{-\frac{3}{2}}(-x)^2 = 2 - \frac{x}{4} - \frac{x^2}{64}$ | (M1), (A1), (A1) (3 marks) | Use of $(a+x)^n$ from formula book; condone missing brackets and 1 error |
### Part (b)
| $x=1$: $\sqrt{3} \approx 2 - \frac{1}{4} - \frac{1}{64} = 1.734$ (3 dp) | M1, A1 (2 marks) | $x=1$ used in their expansion; CSO |
---4
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the binomial expansion of $( 1 - x ) ^ { \frac { 1 } { 2 } }$ up to and including the term in $x ^ { 2 }$.\\
(2 marks)
\item Hence obtain the binomial expansion of $\sqrt { 4 - x }$ up to and including the term in $x ^ { 2 }$.\\
(3 marks)
\end{enumerate}\item Use your answer to part (a)(ii) to find an approximate value for $\sqrt { 3 }$. Give your answer to three decimal places.
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2009 Q4 [7]}}