CAIE P1 2010 November — Question 2 4 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2010
SessionNovember
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicBinomial Theorem (positive integer n)
TypeSingle coefficient given directly
DifficultyModerate -0.8 This is a straightforward binomial expansion question requiring only basic substitution into the binomial coefficient formula. Given one coefficient to find the constant 'a', then calculate another coefficient using the same formula—purely mechanical with no problem-solving insight needed.
Spec1.02v Inverse and composite functions: graphs and conditions for existence

2 In the expansion of \(( 1 + a x ) ^ { 6 }\), where \(a\) is a constant, the coefficient of \(x\) is - 30 . Find the coefficient of \(x ^ { 3 }\).

Question 2:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\((1+ax)^6\): Term in \(x = 6ax\)\(B1\) co
Equate with \(-30 \rightarrow a = -5\)\(B1\sqrt{}\) \(\sqrt{}\) from his answer for \(6ax\)
Term in \(x^3 = \frac{6.5.4}{3!}a^3\)\(B1\) co
\(\rightarrow\) coefficient of \(-2500\)\(B1\sqrt{}\) For \(20 \times a^3\)
## Question 2:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $(1+ax)^6$: Term in $x = 6ax$ | $B1$ | co |
| Equate with $-30 \rightarrow a = -5$ | $B1\sqrt{}$ | $\sqrt{}$ from his answer for $6ax$ |
| Term in $x^3 = \frac{6.5.4}{3!}a^3$ | $B1$ | co |
| $\rightarrow$ coefficient of $-2500$ | $B1\sqrt{}$ | For $20 \times a^3$ |

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2 In the expansion of $( 1 + a x ) ^ { 6 }$, where $a$ is a constant, the coefficient of $x$ is - 30 . Find the coefficient of $x ^ { 3 }$.

\hfill \mbox{\textit{CAIE P1 2010 Q2 [4]}}