AQA AS Paper 1 2018 June — Question 4 5 marks

Exam BoardAQA
ModuleAS Paper 1 (AS Paper 1)
Year2018
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicBinomial Theorem (positive integer n)
TypeNumerical approximation using expansion
DifficultyModerate -0.8 Part (a) is a straightforward binomial expansion with a small positive integer power requiring direct application of the formula or Pascal's triangle. Part (b) is a routine substitution exercise (recognizing 0.994 = 1 - 0.002 and setting x = 0.002). Both parts are standard textbook exercises with no problem-solving required, making this easier than average but not trivial since it requires correct technique across 5 marks.
Spec1.04a Binomial expansion: (a+b)^n for positive integer n

  1. Find the first three terms in the expansion of \((1 - 3x)^4\) in ascending powers of \(x\). [3 marks]
  2. Using your expansion, approximate \((0.994)^4\) to six decimal places. [2 marks]

Question 4:

AnswerMarks Guidance
4(a)Expands (1 – 3x)4 showing
a 1, a (±3x) term and a (±3x)2 termAO1.1a M1
1 – 12x + 54x2
Uses correct coefficients,
4 & 6 (PI by correct answer, ignore
AnswerMarks Guidance
sign errors)AO1.1b A1
Obtains totally correct expansion
AnswerMarks Guidance
Ignore extra termsAO1.1b A1
(b)Selects value to use for x and
substitutes it in their expansion.
Correct value, or solves
AnswerMarks Guidance
1 – 3x = 0.994AO2.2a M1
1 – 12×0.002 + 54×0.0022
=0.976216
Evaluates expression
0.976216 CAO for three terms or
0.976215 CAO for four or five
AnswerMarks Guidance
terms used and shownAO1.1b A1
Total5
QMarking Instructions AO
Question 4:
--- 4(a) ---
4(a) | Expands (1 – 3x)4 showing
a 1, a (±3x) term and a (±3x)2 term | AO1.1a | M1 | 1 + 4(–3x) + 6(–3x)2
1 – 12x + 54x2
Uses correct coefficients,
4 & 6 (PI by correct answer, ignore
sign errors) | AO1.1b | A1
Obtains totally correct expansion
Ignore extra terms | AO1.1b | A1
(b) | Selects value to use for x and
substitutes it in their expansion.
Correct value, or solves
1 – 3x = 0.994 | AO2.2a | M1 | Need x = 0.002
1 – 12×0.002 + 54×0.0022
=0.976216
Evaluates expression
0.976216 CAO for three terms or
0.976215 CAO for four or five
terms used and shown | AO1.1b | A1
Total | 5
Q | Marking Instructions | AO | Marks | Typical Solution
\begin{enumerate}[label=(\alph*)]
\item Find the first three terms in the expansion of $(1 - 3x)^4$ in ascending powers of $x$.
[3 marks]

\item Using your expansion, approximate $(0.994)^4$ to six decimal places.
[2 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA AS Paper 1 2018 Q4 [5]}}