CAIE P3 2015 November — Question 2 3 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2015
SessionNovember
Marks3
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 bx=9x. Students need only substitute into the standard formula and calculate two coefficients—pure recall with minimal algebraic manipulation. Easier than average as it requires no problem-solving or insight beyond direct formula application.
Spec1.04c Extend binomial expansion: rational n, |x|<1
2 Given that \(\sqrt [ 3 ] { } ( 1 + 9 x ) \approx 1 + 3 x + a x ^ { 2 } + b x ^ { 3 }\) for small values of \(x\), find the values of the coefficients \(a\) and \(b\).
Either
AnswerMarks
State correct unsimplified \(x^2\) or \(x^3\) termM1
Obtain \(a = -9\)A1
Obtain \(b = 45\)A1
Or
AnswerMarks Guidance
Use chain rule to differentiate twice to obtain form \(k(1 + 9x)^{-\frac{5}{3}}\)M1
Obtain \(f''(x) = -18(1 + 9x)^{-\frac{5}{3}}\) and hence \(a = -9\)A1
Obtain \(f''(x) = 270(1 + 9x)^{-\frac{8}{3}}\) and hence \(b = 45\)A1 [3] 
**Either**

State correct unsimplified $x^2$ or $x^3$ term | M1 |
Obtain $a = -9$ | A1 |
Obtain $b = 45$ | A1 |

**Or**

Use chain rule to differentiate twice to obtain form $k(1 + 9x)^{-\frac{5}{3}}$ | M1 |
Obtain $f''(x) = -18(1 + 9x)^{-\frac{5}{3}}$ and hence $a = -9$ | A1 |
Obtain $f''(x) = 270(1 + 9x)^{-\frac{8}{3}}$ and hence $b = 45$ | A1 | [3]
2 Given that $\sqrt [ 3 ] { } ( 1 + 9 x ) \approx 1 + 3 x + a x ^ { 2 } + b x ^ { 3 }$ for small values of $x$, find the values of the coefficients $a$ and $b$.

\hfill \mbox{\textit{CAIE P3 2015 Q2 [3]}}
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