| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Substitute expression for variable |
| Difficulty | Standard +0.8 This question requires understanding the validity conditions of binomial expansions (|x| < 4 and |x³| < 4), then recognizing that at x = -2, we have |-2³| = 8 > 4, violating the convergence condition. While the setup is standard, identifying why the expansion fails at the boundary requires conceptual understanding beyond routine application. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{\sqrt{4+x}} = \frac{1}{2}\left(1+\frac{x}{4}\right)^{-\frac{1}{2}}\) | M1 | Writes in form to which binomial expansion can be applied; accept \(A\left(1+\frac{x}{4}\right)^{-\frac{1}{2}}\) |
| \(\approx \frac{1}{2}\left[1+\left(-\frac{1}{2}\right)\frac{x}{4}+\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}\left(\frac{x}{4}\right)^2\right]\) | M1 | Uses binomial expansion for \((1+kx)^{\pm\frac{1}{2}}\) with at least two terms correct (can be unsimplified) |
| \(\approx \frac{1}{2} - \frac{1}{16}x + \frac{3}{256}x^2\) | A1 | Obtains correct simplified answer; no need to expand brackets; CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{\sqrt{4-x^3}} \approx \frac{1}{2} - \frac{1}{16}(-x^3) + \frac{3}{256}(-x^3)^2\) | M1 | Substitutes \(-x^3\) in their three term expansion from part (a) |
| \(\approx \frac{1}{2} + \frac{x^3}{16} + \frac{3x^6}{256}\) | A1F | Obtains correct expansion; FT their (a) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int_0^1 \frac{1}{\sqrt{4-x^3}}\,dx \approx \int_0^1 \frac{1}{2} + \frac{x^3}{16} + \frac{3x^6}{256}\,dx\) | M1 | Uses their three term expansion as integrand; ignore limits PI by next mark |
| \(\approx \left[\frac{x}{2} + \frac{x^4}{64} + \frac{3x^7}{1792}\right]_0^1\) | M1 | Integrates (at least two terms correct) |
| \(\approx \frac{1}{2} + \frac{1}{64} + \frac{3}{1792} \approx 0.5172991\) | A1 | Obtains correct value; CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Each term in the expansion is positive. | E1 | Explains that each term in the expansion is positive |
| So increasing the terms will increase the estimated value, hence the value must be an underestimate. | R1 | Deduces that increasing number of terms will increase estimated value and that value must be an underestimate; condone inference if evidence given i.e. value calculated and compared |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| The binomial expansion is valid for \( | x | < \sqrt[3]{4}\) |
| \(2 > \sqrt[3]{4}\) | E1 | Compares integral lower limit with validity of correct expansion; CAO |
# Question 6:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{\sqrt{4+x}} = \frac{1}{2}\left(1+\frac{x}{4}\right)^{-\frac{1}{2}}$ | M1 | Writes in form to which binomial expansion can be applied; accept $A\left(1+\frac{x}{4}\right)^{-\frac{1}{2}}$ |
| $\approx \frac{1}{2}\left[1+\left(-\frac{1}{2}\right)\frac{x}{4}+\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}\left(\frac{x}{4}\right)^2\right]$ | M1 | Uses binomial expansion for $(1+kx)^{\pm\frac{1}{2}}$ with at least two terms correct (can be unsimplified) |
| $\approx \frac{1}{2} - \frac{1}{16}x + \frac{3}{256}x^2$ | A1 | Obtains correct simplified answer; no need to expand brackets; CAO |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{\sqrt{4-x^3}} \approx \frac{1}{2} - \frac{1}{16}(-x^3) + \frac{3}{256}(-x^3)^2$ | M1 | Substitutes $-x^3$ in their three term expansion from part (a) |
| $\approx \frac{1}{2} + \frac{x^3}{16} + \frac{3x^6}{256}$ | A1F | Obtains correct expansion; FT their (a) |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int_0^1 \frac{1}{\sqrt{4-x^3}}\,dx \approx \int_0^1 \frac{1}{2} + \frac{x^3}{16} + \frac{3x^6}{256}\,dx$ | M1 | Uses their three term expansion as integrand; ignore limits PI by next mark |
| $\approx \left[\frac{x}{2} + \frac{x^4}{64} + \frac{3x^7}{1792}\right]_0^1$ | M1 | Integrates (at least two terms correct) |
| $\approx \frac{1}{2} + \frac{1}{64} + \frac{3}{1792} \approx 0.5172991$ | A1 | Obtains correct value; CAO |
## Part (d)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Each term in the expansion is positive. | E1 | Explains that each term in the expansion is positive |
| So increasing the terms will increase the estimated value, hence the value must be an underestimate. | R1 | Deduces that increasing number of terms will increase estimated value and that value must be an underestimate; condone inference if evidence given i.e. value calculated and compared |
## Part (d)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| The binomial expansion is valid for $|x| < \sqrt[3]{4}$ | B1F | States validity of their binomial expansion for part (b); provided their $k \neq \pm 1$ |
| $2 > \sqrt[3]{4}$ | E1 | Compares integral lower limit with validity of correct expansion; CAO |
---6
\begin{enumerate}[label=(\alph*)]
\item Find the first three terms, in ascending powers of $x$, of the binomial expansion of $\frac { 1 } { \sqrt { 4 + x } }$\\
6
\item Hence, find the first three terms of the binomial expansion of $\frac { 1 } { \sqrt { 4 - x ^ { 3 } } }$\\
6 (d) (i) Edward, a student, decides to use this method to find a more accurate value for the integral by increasing the number of terms of the binomial expansion used.
Explain clearly whether Edward's approximation will be an overestimate, an underestimate, or if it is impossible to tell.\\[0pt]
[2 marks]\\
6 (d) (ii) Edward goes on to use the expansion from part (b) to find an approximation for $\int _ { - 2 } ^ { 0 } \frac { 1 } { \sqrt { 4 - x ^ { 3 } } } \mathrm {~d} x$
Explain why Edward's approximation is invalid.
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2018 Q6 [12]}}