CAIE P1 2024 November — Question 4 6 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2024
SessionNovember
Marks6
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Mark schemeDownload PDF ↗
TopicBinomial Theorem (positive integer n)
TypeProduct of expansions
DifficultyModerate -0.8 Part (a) is a standard binomial expansion exercise requiring identification of the term where powers of x cancel (r=4 gives x^{-2} term). Part (b) extends this by multiplying by a binomial, requiring students to combine two terms from the expansion - a routine extension but still mechanical application of the binomial theorem with no novel insight required.
Spec1.04a Binomial expansion: (a+b)^n for positive integer n
Find the term independent of \(x\) in the expansion of each of the following:
  1. \(\left(x + \frac{3}{x^2}\right)^6\) [2]
  2. \((4x^3 - 5)\left(x + \frac{3}{x^2}\right)^6\) [4]
Question 4:
AnswerMarks
4(a)6  3  2 65 3  2
15 or  x4   or x6   
AnswerMarks Guidance
2 x2  2 x3B1 OE
May be in a list.
6
Allow  .
4
AnswerMarks Guidance
135B1 Correct term must be identified if in a list.
Allow 135x0.
2
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
4(b)6  3  3 6.5.4 3  3
20 or  x3   or x6  
AnswerMarks Guidance
3 x2  3! x3B1 OE
May be in a list.
540
=
AnswerMarks Guidance
x3B1 1
Identifying term.
x3
This can be implied by sight of 2160 as part of the constant
term.
AnswerMarks Guidance
4540−51 35M1 4their 540−5their1 35
1485A1 Allow 1485x0.
4
AnswerMarks Guidance
QuestionAnswer Marks
5(a)(i)1
 f(−1)=
AnswerMarks Guidance
3B1 Condone 0.333.
1
y
4
2
x
−4 −2 2 4 6 8 10
−2
AnswerMarks Guidance
−4B1 For showing the correct mirror line.
−105(a)(ii) y
4
2
x
−4 −2 2 4 6 8 10
−2
−4
4
AnswerMarks
B1For correct shape: the curves should intersect in the first square
in the third quadrant. To the left of the point of intersection, the
reflection is below the original and crosses the x-axis. To the
right of the point of intersection, the reflection is to the right the
original.
2
AnswerMarks Guidance
−8 −6− 4
−2
2
AnswerMarks Guidance
QuestionAnswer Marks 
Question 4:
--- 4(a) ---
4(a) | 6  3  2 65 3  2
15 or  x4   or x6   
2 x2  2 x3 | B1 | OE
May be in a list.
6
Allow  .
4
135 | B1 | Correct term must be identified if in a list.
Allow 135x0.
2
Question | Answer | Marks | Guidance
--- 4(b) ---
4(b) | 6  3  3 6.5.4 3  3
20 or  x3   or x6  
3 x2  3! x3 | B1 | OE
May be in a list.
540
=
x3 | B1 | 1
Identifying term.
x3
This can be implied by sight of 2160 as part of the constant
term.
4540−51 35 | M1 | 4their 540−5their1 35
1485 | A1 | Allow 1485x0.
4
Question | Answer | Marks | Guidance
5(a)(i) | 1
 f(−1)=

3 | B1 | Condone 0.333.
1
y
4
2
x
−4 −2 2 4 6 8 10
−2
−4 | B1 | For showing the correct mirror line.
−10 | 5(a)(ii) | y
4
2
x
−4 −2 2 4 6 8 10
−2
−4
4
B1 | For correct shape: the curves should intersect in the first square
in the third quadrant. To the left of the point of intersection, the
reflection is below the original and crosses the x-axis. To the
right of the point of intersection, the reflection is to the right the
original.
2
−8 −6 | − | 4 | − | 2 | 2 | 4
−2
2
Question | Answer | Marks | Guidance
Find the term independent of $x$ in the expansion of each of the following:

\begin{enumerate}[label=(\alph*)]
\item $\left(x + \frac{3}{x^2}\right)^6$ [2]

\item $(4x^3 - 5)\left(x + \frac{3}{x^2}\right)^6$ [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2024 Q4 [6]}}
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