Question 2 - A-Level Maths

Edexcel C4 2008 January — Question 2 7 marks

Exam Board	Edexcel
Module	C4 (Core Mathematics 4)
Year	2008
Session	January
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Generalised Binomial Theorem
Type	Form (a+bx)^n requiring factorisation
Difficulty	Standard +0.3 This is a standard C4 binomial expansion question requiring factorisation of (8-3x)^(1/3) as 2(1-3x/8)^(1/3), then applying the formula with fractional index. Part (b) requires substituting x=0.1 to approximate ∛7.7. While it involves multiple steps and careful arithmetic with fractions, it follows a well-practiced template with no novel problem-solving required, making it slightly easier than average.
Spec	1.04c Extend binomial expansion: rational n, \|x\|<1 1.04d Binomial expansion validity: convergence conditions

2. (a) Use the binomial theorem to expand $$( 8 - 3 x ) ^ { \frac { 1 } { 3 } } , \quad | x | < \frac { 8 } { 3 }$$ in ascending powers of $x$, up to and including the term in $x ^ { 3 }$, giving each term as a simplified fraction.
(b) Use your expansion, with a suitable value of $x$, to obtain an approximation to $\sqrt [ 3 ] { } ( 7.7 )$. Give your answer to 7 decimal places.

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Question 2:

Part (a) – Way 1

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$(8-3x)^{\frac{1}{3}} = (8)^{\frac{1}{3}}\left(1-\frac{3x}{8}\right)^{\frac{1}{3}} = 2\left(1-\frac{3x}{8}\right)^{\frac{1}{3}}$	B1	Takes 8 outside bracket to give any of $(8)^{\frac{1}{3}}$ or 2
$= 2\left\{1 + \left(\frac{1}{3}\right)( x) + \frac{\left(\frac{1}{3}\right)\left(-\frac{2}{3}\right)}{2!}(x)^2 + \frac{\left(\frac{1}{3}\right)\left(-\frac{2}{3}\right)\left(-\frac{5}{3}\right)}{3!}(x)^3 + \ldots\right\}$ with $ \neq 1$	M1	Expands $(1+x)^{\frac{1}{3}}$ to give simplified or un-simplified $1 + \left(\frac{1}{3}\right)(x)$
Correct $\{\ldots\}$ expansion with candidate's followed through $(**x)$	A1$\sqrt{}$	A correct simplified or un-simplified $\{\ldots\}$ expansion
$= 2\left\{1 - \frac{1}{8}x; -\frac{1}{64}x^2 - \frac{5}{1536}x^3 - \ldots\right\}$	A1	Either $2\{1 - \frac{1}{8}x\ldots\}$ or anything that cancels to $2 - \frac{1}{4}x$
$= 2 - \frac{1}{4}x; -\frac{1}{32}x^2 - \frac{5}{768}x^3 - \ldots$	A1	Simplified $-\frac{1}{32}x^2 - \frac{5}{768}x^3$ [5 marks total]

Part (a) – Way 2 (Aliter)

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$(8)^{\frac{1}{3}} + \left(\frac{1}{3}\right)(8)^{-\frac{2}{3}}(x) + \frac{\left(\frac{1}{3}\right)\left(-\frac{2}{3}\right)}{2!}(8)^{-\frac{5}{3}}(x)^2 + \frac{\left(\frac{1}{3}\right)\left(-\frac{2}{3}\right)\left(-\frac{5}{3}\right)}{3!}(8)^{-\frac{8}{3}}(x)^3 + \ldots$ with $ \neq 1$	B1, M1, A1$\sqrt{}$	2 or $(8)^{\frac{1}{3}}$; Expands $(8-3x)^{\frac{1}{3}}$ to give un-simplified or simplified $(8)^{\frac{1}{3}} + \left(\frac{1}{3}\right)(8)^{-\frac{2}{3}}(**x)$
$= 2 - \frac{1}{4}x; -\frac{1}{32}x^2 - \frac{5}{768}x^3 - \ldots$	A1, A1	Anything cancelling to $2 - \frac{1}{4}x$; Simplified $-\frac{1}{32}x^2 - \frac{5}{768}x^3$ [5 marks total]

Part (b)

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$(7.7)^{\frac{1}{3}} \approx 2 - \frac{1}{4}(0.1) - \frac{1}{32}(0.1)^2 - \frac{5}{768}(0.1)^3 - \ldots$	M1	Attempt to substitute $x = 0.1$ into candidate's binomial expansion
$= 2 - 0.025 - 0.0003125 - 0.0000065104166\ldots$
$= 1.97468099\ldots$	A1	awrt 1.9746810 [2 marks total]

# Question 2:

## Part (a) – Way 1

| Answer/Working | Marks | Guidance |
|---|---|---|
| $(8-3x)^{\frac{1}{3}} = (8)^{\frac{1}{3}}\left(1-\frac{3x}{8}\right)^{\frac{1}{3}} = 2\left(1-\frac{3x}{8}\right)^{\frac{1}{3}}$ | B1 | Takes 8 outside bracket to give any of $(8)^{\frac{1}{3}}$ or 2 |
| $= 2\left\{1 + \left(\frac{1}{3}\right)(** x) + \frac{\left(\frac{1}{3}\right)\left(-\frac{2}{3}\right)}{2!}(**x)^2 + \frac{\left(\frac{1}{3}\right)\left(-\frac{2}{3}\right)\left(-\frac{5}{3}\right)}{3!}(**x)^3 + \ldots\right\}$ with $** \neq 1$ | M1 | Expands $(1+**x)^{\frac{1}{3}}$ to give simplified or un-simplified $1 + \left(\frac{1}{3}\right)(**x)$ |
| Correct $\{\ldots\}$ expansion with candidate's followed through $(**x)$ | A1$\sqrt{}$ | A correct simplified or un-simplified $\{\ldots\}$ expansion |
| $= 2\left\{1 - \frac{1}{8}x; -\frac{1}{64}x^2 - \frac{5}{1536}x^3 - \ldots\right\}$ | A1 | Either $2\{1 - \frac{1}{8}x\ldots\}$ or anything that cancels to $2 - \frac{1}{4}x$ |
| $= 2 - \frac{1}{4}x; -\frac{1}{32}x^2 - \frac{5}{768}x^3 - \ldots$ | A1 | Simplified $-\frac{1}{32}x^2 - \frac{5}{768}x^3$ [5 marks total] |

## Part (a) – Way 2 (Aliter)

| Answer/Working | Marks | Guidance |
|---|---|---|
| $(8)^{\frac{1}{3}} + \left(\frac{1}{3}\right)(8)^{-\frac{2}{3}}(**x) + \frac{\left(\frac{1}{3}\right)\left(-\frac{2}{3}\right)}{2!}(8)^{-\frac{5}{3}}(**x)^2 + \frac{\left(\frac{1}{3}\right)\left(-\frac{2}{3}\right)\left(-\frac{5}{3}\right)}{3!}(8)^{-\frac{8}{3}}(**x)^3 + \ldots$ with $** \neq 1$ | B1, M1, A1$\sqrt{}$ | 2 or $(8)^{\frac{1}{3}}$; Expands $(8-3x)^{\frac{1}{3}}$ to give un-simplified or simplified $(8)^{\frac{1}{3}} + \left(\frac{1}{3}\right)(8)^{-\frac{2}{3}}(**x)$ |
| $= 2 - \frac{1}{4}x; -\frac{1}{32}x^2 - \frac{5}{768}x^3 - \ldots$ | A1, A1 | Anything cancelling to $2 - \frac{1}{4}x$; Simplified $-\frac{1}{32}x^2 - \frac{5}{768}x^3$ [5 marks total] |

## Part (b)

| Answer/Working | Marks | Guidance |
|---|---|---|
| $(7.7)^{\frac{1}{3}} \approx 2 - \frac{1}{4}(0.1) - \frac{1}{32}(0.1)^2 - \frac{5}{768}(0.1)^3 - \ldots$ | M1 | Attempt to substitute $x = 0.1$ into candidate's binomial expansion |
| $= 2 - 0.025 - 0.0003125 - 0.0000065104166\ldots$ | | |
| $= 1.97468099\ldots$ | A1 | awrt 1.9746810 [2 marks total] |

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2. (a) Use the binomial theorem to expand

$$( 8 - 3 x ) ^ { \frac { 1 } { 3 } } , \quad | x | < \frac { 8 } { 3 }$$

in ascending powers of $x$, up to and including the term in $x ^ { 3 }$, giving each term as a simplified fraction.\\
(b) Use your expansion, with a suitable value of $x$, to obtain an approximation to $\sqrt [ 3 ] { } ( 7.7 )$. Give your answer to 7 decimal places.\\

\hfill \mbox{\textit{Edexcel C4 2008 Q2 [7]}}

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Q1 6 Q2 7 Q3 5 Q4 9 Q5 9 Q6 11 Q7 15 Q8 13

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
\((8-3x)^{\frac{1}{3}} = (8)^{\frac{1}{3}}\left(1-\frac{3x}{8}\right)^{\frac{1}{3}} = 2\left(1-\frac{3x}{8}\right)^{\frac{1}{3}}\)	B1	Takes 8 outside bracket to give any of \((8)^{\frac{1}{3}}\) or 2
\(= 2\left\{1 + \left(\frac{1}{3}\right)( x) + \frac{\left(\frac{1}{3}\right)\left(-\frac{2}{3}\right)}{2!}(x)^2 + \frac{\left(\frac{1}{3}\right)\left(-\frac{2}{3}\right)\left(-\frac{5}{3}\right)}{3!}(x)^3 + \ldots\right\}\) with \( \neq 1\)	M1	Expands \((1+x)^{\frac{1}{3}}\) to give simplified or un-simplified \(1 + \left(\frac{1}{3}\right)(x)\)
Correct \(\{\ldots\}\) expansion with candidate's followed through \((**x)\)	A1\(\sqrt{}\)	A correct simplified or un-simplified \(\{\ldots\}\) expansion
\(= 2\left\{1 - \frac{1}{8}x; -\frac{1}{64}x^2 - \frac{5}{1536}x^3 - \ldots\right\}\)	A1	Either \(2\{1 - \frac{1}{8}x\ldots\}\) or anything that cancels to \(2 - \frac{1}{4}x\)
\(= 2 - \frac{1}{4}x; -\frac{1}{32}x^2 - \frac{5}{768}x^3 - \ldots\)	A1	Simplified \(-\frac{1}{32}x^2 - \frac{5}{768}x^3\) [5 marks total]