| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Approximation for small x |
| Difficulty | Standard +0.3 This is a standard C4 binomial expansion question with routine application of $(1-x)^{-3}$ and straightforward algebraic manipulation. Part (i) is direct expansion, part (ii) requires choosing $x=1/10$, part (iii) involves algebraic rearrangement and another expansion, and part (iv) tests understanding of convergence conditions. While multi-part, each step follows textbook procedures with no novel 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 | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((1-x)^{-3}=1+-3.-x+\frac{-3.-4}{2}(-x)^2+...\) oe; accept \(3x\) for \(-3.-x\) &/or \(-x^2\) or \((x)^2\) for \((-x)^2\) | M1 | As result is given, this expansion must be shown and then simplified. For alternative methods such as expanding \((1-x)^3\) and multiplying by \(x+3x^2+6x^3\) or using long division, consult TL |
| multiplication by \(x\) to produce AG | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Clear indication that \(x=0.1\) is to be substituted | M1 | e.g. \(0.1+3(0.1)^2+6(0.1)^3\) stated. Calculator value \(\rightarrow\) M0 |
| (estimated value is) \(0.1+3(0.1)^2+6(0.1)^3=\underline{0.136}\) | A1 | (0.13717... is calculator value of \(\frac{100}{729}\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Sight of \(1-x=x\left(\frac{1}{x}-1\right)\) or \(1-x=-x\left(1-\frac{1}{x}\right)\) or \(\left(\frac{1}{x}-1\right)^3=-\left(1-\frac{1}{x}\right)^3\) or \(\left(\frac{1}{x}-1\right)^{-3}=-\left(1-\frac{1}{x}\right)^{-3}\) or equivalent | B1 | |
| Complete satisfactory explanation (no reference to style) www | B1 | (Answer Given) |
| \(\left[1+(-3)(-\frac{1}{x})+\frac{(-3)(-4)}{2}\left(-\frac{1}{x}\right)^2+...\right]\) | M1 | Simplified expansion may be quoted – it may have come from result in part (i). Answer for this expansion is not AG. |
| \(\rightarrow-\frac{1}{x^2}-\frac{3}{x^3}-\frac{6}{x^4}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Must say "Not suitable" and one of following: Either: requires \(\left\ | \frac{1}{x}\right\ | <1\), which is not true if \(x=0.1\). Or: substitution of positive/small value of \(x\) in the expansion gives a negative/large value (which cannot be an approximation to 100/729). |
## Question 10(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(1-x)^{-3}=1+-3.-x+\frac{-3.-4}{2}(-x)^2+...$ oe; accept $3x$ for $-3.-x$ &/or $-x^2$ or $(x)^2$ for $(-x)^2$ | M1 | As result is given, this expansion must be shown and then simplified. For alternative methods such as expanding $(1-x)^3$ and multiplying by $x+3x^2+6x^3$ or using long division, consult TL |
| multiplication by $x$ to produce **AG** | A1 | |
---
## Question 10(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Clear indication that $x=0.1$ is to be substituted | M1 | e.g. $0.1+3(0.1)^2+6(0.1)^3$ stated. Calculator value $\rightarrow$ M0 |
| (estimated value is) $0.1+3(0.1)^2+6(0.1)^3=\underline{0.136}$ | A1 | (0.13717... is calculator value of $\frac{100}{729}$) |
---
## Question 10(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Sight of $1-x=x\left(\frac{1}{x}-1\right)$ or $1-x=-x\left(1-\frac{1}{x}\right)$ or $\left(\frac{1}{x}-1\right)^3=-\left(1-\frac{1}{x}\right)^3$ or $\left(\frac{1}{x}-1\right)^{-3}=-\left(1-\frac{1}{x}\right)^{-3}$ or equivalent | B1 | |
| Complete satisfactory explanation (no reference to style) www | B1 | (Answer Given) |
| $\left[1+(-3)(-\frac{1}{x})+\frac{(-3)(-4)}{2}\left(-\frac{1}{x}\right)^2+...\right]$ | M1 | Simplified expansion may be quoted – it may have come from result in part (i). Answer for this expansion is not **AG**. |
| $\rightarrow-\frac{1}{x^2}-\frac{3}{x^3}-\frac{6}{x^4}$ | A1 | |
---
## Question 10(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Must say "Not suitable" and one of following: Either: requires $\left\|\frac{1}{x}\right\|<1$, which is not true if $x=0.1$. Or: substitution of positive/small value of $x$ in the expansion gives a negative/large value (which cannot be an approximation to 100/729). | B1 | This B1 is dep on $x=0.1$ used in (ii). Or "because $\frac{1}{x}>1$". Or "it gives $-63100$". If choice given, do not ignore incorrect comments, but ignore irrelevant/unhelpful ones |
The image provided appears to be only the **back cover/contact page** of an OCR mark scheme document. It contains only OCR's contact information, address, and organizational details — there is **no mark scheme content** (no questions, answers, mark allocations, or guidance notes) on this page.
To extract mark scheme content, please share the **interior pages** of the document that contain the actual marking guidance tables.10 (i) Show that $\frac { x } { ( 1 - x ) ^ { 3 } } \approx x + 3 x ^ { 2 } + 6 x ^ { 3 }$ for small values of $x$.\\
(ii) Use this result, together with a suitable value of $x$, to obtain a decimal estimate of the value of $\frac { 100 } { 729 }$.\\
(iii) Show that $\frac { x } { ( 1 - x ) ^ { 3 } } = - \frac { 1 } { x ^ { 2 } } \left( 1 - \frac { 1 } { x } \right) ^ { - 3 }$. Hence find the first three terms of the binomial expansion of $\frac { x } { ( 1 - x ) ^ { 3 } }$ in powers of $\frac { 1 } { x }$.\\
(iv) Comment on the suitability of substituting the same value of $x$ as used in part (ii) in the expansion in part (iii) to estimate the value of $\frac { 100 } { 729 }$.
\hfill \mbox{\textit{OCR C4 2013 Q10 [9]}}