AQA Paper 3 Specimen — Question 5 11 marks

Exam BoardAQA
ModulePaper 3 (Paper 3)
SessionSpecimen
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeneralised Binomial Theorem
TypeDirect single expansion substitution
DifficultyModerate -0.3 This is a straightforward binomial expansion question requiring standard application of the formula for fractional powers, followed by routine substitution and a basic validity check. Part (a) is direct recall, part (b) is simple substitution (recognizing 1.18 = 1 + 0.03×6), and part (c) tests understanding that |x| < 1 is required for convergence. The question is slightly easier than average as it's entirely procedural with no problem-solving or novel insight required.
Spec1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions

  1. Find the first three terms, in ascending powers of \(x\), in the binomial expansion of \((1 + 6x)^{\frac{1}{3}}\) [2 marks]
  2. Use the result from part (a) to obtain an approximation to \(\sqrt[3]{1.18}\) giving your answer to 4 decimal places. [2 marks]
  3. Explain why substituting \(x = \frac{1}{2}\) into your answer to part (a) does not lead to a valid approximation for \(\sqrt[3]{4}\). [1 mark]

Question 5:

AnswerMarks
5(a)Uses binomial expansion,
with at least two terms
AnswerMarks Guidance
correct, may be un-simplifiedAO1.1a M1
(1+6x)3 ≈1+ ⋅6x+ ⋅ ⋅
3 3 3 2
1
(1+6x)3 ≈1+2x−4x2
Obtains correct simplified
AnswerMarks Guidance
answerAO1.1b A1
(b)Determines the correct value
for x and substitutes this into
AnswerMarks Guidance
‘their’ answer to part (a)AO3.1a M1
31.18 1 + 2(0.03) – 4(0.03)2
1.0564
Obtains correct
approximation for ‘their’
answer to part (a)
FT allowed only if M1 from
part (a) and M1 from part (b)
AnswerMarks Guidance
have been awardedAO1.1b A1F
(c)Explains the limitation of the
expansion found in part (a)
1
with reference to x =
AnswerMarks Guidance
2AO2.4 E1
 13
Although 1+6×  = 3 4
 2
1
x = cannot be used since the
2
1
expansion is only valid for x <
6
AnswerMarks Guidance
Total5
QMarking Instructions AO
Question 5:
--- 5(a) ---
5(a) | Uses binomial expansion,
with at least two terms
correct, may be un-simplified | AO1.1a | M1 | 1 1 1 −2 (6x)2
(1+6x)3 ≈1+ ⋅6x+ ⋅ ⋅
3 3 3 2
1
(1+6x)3 ≈1+2x−4x2
Obtains correct simplified
answer | AO1.1b | A1
(b) | Determines the correct value
for x and substitutes this into
‘their’ answer to part (a) | AO3.1a | M1 | x=0.03
31.18 1 + 2(0.03) – 4(0.03)2
1.0564
≈
≈
Obtains correct
approximation for ‘their’
answer to part (a)
FT allowed only if M1 from
part (a) and M1 from part (b)
have been awarded | AO1.1b | A1F
(c) | Explains the limitation of the
expansion found in part (a)
1
with reference to x =
2 | AO2.4 | E1 | 1
 13
Although 1+6×  = 3 4
 2
1
x = cannot be used since the
2
1
expansion is only valid for x <
6
Total | 5
Q | Marking Instructions | AO | Marks | Typical Solution
\begin{enumerate}[label=(\alph*)]
\item Find the first three terms, in ascending powers of $x$, in the binomial expansion of $(1 + 6x)^{\frac{1}{3}}$ [2 marks]

\item Use the result from part (a) to obtain an approximation to $\sqrt[3]{1.18}$ giving your answer to 4 decimal places. [2 marks]

\item Explain why substituting $x = \frac{1}{2}$ into your answer to part (a) does not lead to a valid approximation for $\sqrt[3]{4}$. [1 mark]
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 3  Q5 [11]}}