CAIE P3 2011 June — Question 1 4 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2011
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeneralised Binomial Theorem
TypeForm (1+bx)^n expansion
DifficultyModerate -0.8 This is a straightforward application of the binomial expansion formula with n=1/3 and b=-6. Students need only substitute into the standard formula and simplify coefficients—pure recall with minimal algebraic manipulation, making it easier than average but not trivial since fractional indices require care.
Spec1.04c Extend binomial expansion: rational n, |x|<1
1 Expand \(\sqrt [ 3 ] { } ( 1 - 6 x )\) in ascending powers of \(x\) up to and including the term in \(x ^ { 3 }\), simplifying the coefficients.
Question 1:
Either method:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Obtain \(1 + \frac{1}{3}kx\), where \(k = \pm6\) or \(\pm1\)M1
Obtain \(1 - 2x\)A1
Obtain \(-4x^2\)A1
Obtain \(-\frac{40}{3}x^3\) or equivalentA1 [4]
Or method:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Differentiate expression to obtain form \(k(1-6x)^{-\frac{2}{3}}\) and evaluate \(f(0)\) and \(f'(0)\)M1
Obtain \(f'(x) = -2(1-6x)^{-\frac{2}{3}}\) and hence correct first two terms \(1 - 2x\)A1
Obtain \(f''(x) = -8(1-6x)^{-\frac{5}{3}}\) and hence \(-4x^2\)A1
Obtain \(f'''(x) = -80(1-6x)^{-\frac{8}{3}}\) and hence \(-\frac{40}{3}x^3\) or equivalentA1 [4] 
## Question 1:

**Either method:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Obtain $1 + \frac{1}{3}kx$, where $k = \pm6$ or $\pm1$ | M1 | |
| Obtain $1 - 2x$ | A1 | |
| Obtain $-4x^2$ | A1 | |
| Obtain $-\frac{40}{3}x^3$ or equivalent | A1 | [4] |

**Or method:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Differentiate expression to obtain form $k(1-6x)^{-\frac{2}{3}}$ and evaluate $f(0)$ and $f'(0)$ | M1 | |
| Obtain $f'(x) = -2(1-6x)^{-\frac{2}{3}}$ and hence correct first two terms $1 - 2x$ | A1 | |
| Obtain $f''(x) = -8(1-6x)^{-\frac{5}{3}}$ and hence $-4x^2$ | A1 | |
| Obtain $f'''(x) = -80(1-6x)^{-\frac{8}{3}}$ and hence $-\frac{40}{3}x^3$ or equivalent | A1 | [4] |

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1 Expand $\sqrt [ 3 ] { } ( 1 - 6 x )$ in ascending powers of $x$ up to and including the term in $x ^ { 3 }$, simplifying the coefficients.

\hfill \mbox{\textit{CAIE P3 2011 Q1 [4]}}
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