CAIE P1 2024 March — Question 6 5 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2024
SessionMarch
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicBinomial Theorem (positive integer n)
TypeProduct with unknown constant to determine
DifficultyModerate -0.3 This is a straightforward binomial expansion problem requiring students to expand (2+ax)^4, multiply by (5-ax), collect x^3 terms, and solve a linear equation for a. While it involves multiple steps and careful algebraic manipulation, it's a standard textbook exercise testing routine application of the binomial theorem with no novel insight required—slightly easier than average.
Spec1.04a Binomial expansion: (a+b)^n for positive integer n

6 It is given that the coefficient of \(x ^ { 3 }\) in the expansion of $$( 2 + a x ) ^ { 4 } ( 5 - a x )$$ is 432 .
Find the value of the constant \(a\).

Question 6:
AnswerMarks Guidance
AnswerMarks Guidance
\(\binom{4}{2}2^2(ax)^2\), \(\binom{4}{3}2^{[1]}(ax)^3\)B1 B1 OE. Expect \(24a^2x^2, 8a^3x^3\) (may be seen in an expansion).
Multiply terms involving \(x^2\) and \(x^3\) by \(5-ax\) to obtain \(x^3\) term\*M1 Must find two products only (may be seen in an expansion).
Equate coefficient of \(x^3\) to 432 and solve for \(a\)DM1 Ignore inclusion of \(x^3\) at this stage.
Obtain \(a=3\) onlyA1
Total: 5
## Question 6:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\binom{4}{2}2^2(ax)^2$, $\binom{4}{3}2^{[1]}(ax)^3$ | B1 B1 | OE. Expect $24a^2x^2, 8a^3x^3$ (may be seen in an expansion). |
| Multiply terms involving $x^2$ and $x^3$ by $5-ax$ to obtain $x^3$ term | \*M1 | Must find two products only (may be seen in an expansion). |
| Equate coefficient of $x^3$ to 432 and solve for $a$ | DM1 | Ignore inclusion of $x^3$ at this stage. |
| Obtain $a=3$ only | A1 | |
| **Total: 5** | | |

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6 It is given that the coefficient of $x ^ { 3 }$ in the expansion of

$$( 2 + a x ) ^ { 4 } ( 5 - a x )$$

is 432 .\\
Find the value of the constant $a$.\\

\hfill \mbox{\textit{CAIE P1 2024 Q6 [5]}}