| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2021 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Product of separate expansions |
| Difficulty | Standard +0.3 This is a standard binomial expansion question requiring routine application of the generalised binomial theorem. Part (a) involves straightforward expansion with fractional power, part (b) tests knowledge of validity conditions (|x| < 8/3), and part (c) requires multiplying two expansions to find a specific coefficient—a common A-level technique but with multiple steps. Slightly above average due to the fractional index and the product manipulation in part (c), but still a textbook-style question with no novel insight required. |
| Spec | 1.02y Partial fractions: decompose rational functions1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left(1 - \frac{3}{8}x\right)^{\frac{1}{3}} = 1 + \left(\frac{1}{3}\right)\left(-\frac{3}{8}x\right) + \dfrac{\left(\frac{1}{3}\right)\left(-\frac{2}{3}\right)\left(-\frac{3}{8}x\right)^2}{2!}\) | B1 | Obtain correct first two terms; allow unsimplified second term including product of two fractions |
| M1 | Attempt third term in expansion of \(\left(1 - \frac{3}{8}x\right)^{\frac{1}{3}}\); allow BOD if no brackets or no negative sign | |
| \(= 1 - \frac{1}{8}x - \frac{1}{64}x^2\) | A1 | Correct third term; allow unsimplified fraction as coefficient but must be single term |
| \((8-3x)^{\frac{1}{3}} = 8^{\frac{1}{3}}\left(1 - \frac{3}{8}x\right)^{\frac{1}{3}} = 2\left(1 - \frac{3}{8}x\right)^{\frac{1}{3}}\) | B1FT | Correct expansion of \((8-3x)^{\frac{1}{3}}\); FT as \(2\times\) their expansion (at least two terms); bracket expanded and fractions simplified |
| \((8-3x)^{\frac{1}{3}} = 2 - \frac{1}{4}x - \frac{1}{32}x^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \( | x | < \frac{8}{3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((1+2x)^{-2} = 1 + (-2)(2x) + \dfrac{(-2)(-3)}{2!}(2x)^2\) | M1 | Attempt first three terms of expansion; must be expanding \((1+2x)^{-2}\); allow BOD if no brackets on \(2x\) |
| \(= 1 - 4x + 12x^2\) | A1 | Obtain correct first three terms; allow unsimplified fraction for coeff of third term |
| \((2\times 12) + \left(-\frac{1}{4}\times -4\right) + \left(-\frac{1}{32}\times 1\right)\) | M1 | Attempt all 3 relevant products; finding 3 appropriate terms from product of two 3-term expansions; if part of full expansion then M1 when reqd 3 products and no others are combined |
| \(\frac{799}{32}\) or \(24\frac{31}{32}\) | A1 | Any exact equivalent, including 24.96875; condone \(x^2\) still present |
# Question 6:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left(1 - \frac{3}{8}x\right)^{\frac{1}{3}} = 1 + \left(\frac{1}{3}\right)\left(-\frac{3}{8}x\right) + \dfrac{\left(\frac{1}{3}\right)\left(-\frac{2}{3}\right)\left(-\frac{3}{8}x\right)^2}{2!}$ | B1 | Obtain correct first two terms; allow unsimplified second term including product of two fractions |
| | M1 | Attempt third term in expansion of $\left(1 - \frac{3}{8}x\right)^{\frac{1}{3}}$; allow BOD if no brackets or no negative sign |
| $= 1 - \frac{1}{8}x - \frac{1}{64}x^2$ | A1 | Correct third term; allow unsimplified fraction as coefficient but must be single term |
| $(8-3x)^{\frac{1}{3}} = 8^{\frac{1}{3}}\left(1 - \frac{3}{8}x\right)^{\frac{1}{3}} = 2\left(1 - \frac{3}{8}x\right)^{\frac{1}{3}}$ | B1FT | Correct expansion of $(8-3x)^{\frac{1}{3}}$; FT as $2\times$ their expansion (at least two terms); bracket expanded and fractions simplified |
| $(8-3x)^{\frac{1}{3}} = 2 - \frac{1}{4}x - \frac{1}{32}x^2$ | | |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $|x| < \frac{8}{3}$ | B1 | Allow any equivalent eg $-\frac{8}{3} < x < \frac{8}{3}$; must be strict inequality; must be condition for $x$, so B0 for $|3x| < 8$ |
## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(1+2x)^{-2} = 1 + (-2)(2x) + \dfrac{(-2)(-3)}{2!}(2x)^2$ | M1 | Attempt first three terms of expansion; must be expanding $(1+2x)^{-2}$; allow BOD if no brackets on $2x$ |
| $= 1 - 4x + 12x^2$ | A1 | Obtain correct first three terms; allow unsimplified fraction for coeff of third term |
| $(2\times 12) + \left(-\frac{1}{4}\times -4\right) + \left(-\frac{1}{32}\times 1\right)$ | M1 | Attempt all 3 relevant products; finding 3 appropriate terms from product of two 3-term expansions; if part of full expansion then M1 when reqd 3 products and no others are combined |
| $\frac{799}{32}$ or $24\frac{31}{32}$ | A1 | Any exact equivalent, including 24.96875; condone $x^2$ still present |
---6
\begin{enumerate}[label=(\alph*)]
\item Find the first three terms in the expansion of $( 8 - 3 x ) ^ { \frac { 1 } { 3 } }$ in ascending powers of $x$.
\item State the range of values of $x$ for which the expansion in part (a) is valid.
\item Find the coefficient of $x ^ { 2 }$ in the expansion of $\frac { ( 8 - 3 x ) ^ { \frac { 1 } { 3 } } } { ( 1 + 2 x ) ^ { 2 } }$.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/01 2021 Q6 [9]}}