CAIE P1 2023 November — Question 1 4 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2023
SessionNovember
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicBinomial Theorem (positive integer n)
TypeCoefficient relationship between terms
DifficultyStandard +0.3 This is a straightforward binomial coefficient problem requiring students to write out two terms using the binomial theorem, equate them according to the given condition, and solve a simple equation. It's slightly easier than average as it's purely mechanical application of a standard formula with no conceptual challenges or multi-step reasoning.
Spec1.04a Binomial expansion: (a+b)^n for positive integer n

The coefficient of \(x^3\) in the expansion of \((3 + 2ax)^5\) is six times the coefficient of \(x^2\) in the expansion of \((2 + ax)^6\). Find the value of the constant \(a\). [4]

Question 1:
AnswerMarks Guidance
1[Coefficient of x3 from (3+2ax)5 =] 1098a3 =720a3
 B1 May be seen in an expansion or with x3.
[Coefficient of x2 from (2+ax)6 =] 1516a2 =240a2
AnswerMarks Guidance
 B1 May be seen in an expansion or with x2.
( 1098a3) ( 1516a2)
their =6their
720a3 =1440a2
AnswerMarks Guidance
 M1 OE
Equating their coefficient of x3 and 6 × their coefficient of x2.
AnswerMarks Guidance
a= 2A1 Condone extra solution a = 0.
4
AnswerMarks Guidance
QuestionAnswer Marks
Question 1:
1 | [Coefficient of x3 from (3+2ax)5 =] 1098a3 =720a3
  | B1 | May be seen in an expansion or with x3.
[Coefficient of x2 from (2+ax)6 =] 1516a2 =240a2
  | B1 | May be seen in an expansion or with x2.
( 1098a3) ( 1516a2)
their =6their
720a3 =1440a2
  | M1 | OE
Equating their coefficient of x3 and 6 × their coefficient of x2.
a= 2 | A1 | Condone extra solution a = 0.
4
Question | Answer | Marks | Guidance
The coefficient of $x^3$ in the expansion of $(3 + 2ax)^5$ is six times the coefficient of $x^2$ in the expansion of $(2 + ax)^6$.

Find the value of the constant $a$. [4]

\hfill \mbox{\textit{CAIE P1 2023 Q1 [4]}}