| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2008 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Factoring out constants before expansion |
| Difficulty | Standard +0.3 This is a standard C4 binomial expansion question with routine steps: expand (1-x)^(1/4), factor out constants from (81-16x)^(1/4) to get 3(1-16x/81)^(1/4), substitute, and apply to a numerical approximation. All steps are algorithmic with no novel insight required, making it slightly easier than average. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| \((81-16x)^{\frac{1}{4}} = 81^{\frac{1}{4}} + \frac{1}{4}81^{-\frac{3}{4}}(-16x) + \frac{1}{4}\left(-\frac{3}{4}\right)\frac{1}{2}81^{-\frac{7}{4}}(-16x)^2\) | M1, A1 | using \((a+bx)^n\) from FB; condone one error |
| \(= \left(3 - \frac{4}{27}x - \frac{8}{729}x^2\right)\) | A1 | 3 |
## Question 4(a)(ii):
$(81-16x)^{\frac{1}{4}} = 81^{\frac{1}{4}} + \frac{1}{4}81^{-\frac{3}{4}}(-16x) + \frac{1}{4}\left(-\frac{3}{4}\right)\frac{1}{2}81^{-\frac{7}{4}}(-16x)^2$ | M1, A1 | using $(a+bx)^n$ from FB; condone one error
$= \left(3 - \frac{4}{27}x - \frac{8}{729}x^2\right)$ | A1 | 3 | CSO completely correct
---4
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Obtain the binomial expansion of $( 1 - x ) ^ { \frac { 1 } { 4 } }$ up to and including the term in $x ^ { 2 }$.
\item Hence show that $( 81 - 16 x ) ^ { \frac { 1 } { 4 } } \approx 3 - \frac { 4 } { 27 } x - \frac { 8 } { 729 } x ^ { 2 }$ for small values of $x$.
\end{enumerate}\item Use the result from part (a)(ii) to find an approximation for $\sqrt [ 4 ] { 80 }$, giving your answer to seven decimal places.
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2008 Q4 [7]}}