CAIE P1 2024 June — Question 1 5 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2024
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCompleting the square and sketching
TypeSolve quadratic by substitution
DifficultyModerate -0.8 Part (a) is a routine completing the square exercise with a leading coefficient, requiring factoring out 3 then standard manipulation. Part (b) is a straightforward substitution (y = x²) followed by solving a quadratic and taking square roots. Both parts are standard textbook procedures with no problem-solving insight required, making this easier than average but not trivial due to the multi-step nature and need for exact form answers.
Spec1.02e Complete the square: quadratic polynomials and turning points1.02f Solve quadratic equations: including in a function of unknown
  1. Express \(3y^2 - 12y - 15\) in the form \(3(y + a)^2 + b\), where \(a\) and \(b\) are constants. [2]
  2. Hence find the exact solutions of the equation \(3x^4 - 12x^2 - 15 = 0\). [3]
Question 1:
AnswerMarks
1(a)3y22
27 or a2, b27B1 B1
2
AnswerMarks Guidance
1(b) x2 2 2 9 leading to x2 23 M1
x2 1 or x2 5M1 Allow omission of -1 if ±3 seen.
x   5A1 B1 SC if M1M1 not awarded. Ignore ± i, i, –i, √–1.
Use of calculator with no working scores 0/3.
Alternative method for Question 1(b)
AnswerMarks Guidance
3 x 4 – 12 x 2 – 15 = 0 leading to 3  x2 5  x2 1  [= 0](M1)
x2 1 or x2 5(M1) Allow omission of –1 if factors seen. Factorising or other
valid method.
AnswerMarks Guidance
x   5(A1) B1 SC if M1M1 not scored. Ignore ± i, i, –i, √–1.
Use of calculator with no working scores 0/3.
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 1:
--- 1(a) ---
1(a) | 3y22
27 or a2, b27 | B1 B1
2
--- 1(b) ---
1(b) |  x2 2 2 9 leading to x2 23 | M1 | Must be x2 unless substitution is clear.
x2 1 or x2 5 | M1 | Allow omission of -1 if ±3 seen.
x   5 | A1 | B1 SC if M1M1 not awarded. Ignore ± i, i, –i, √–1.
Use of calculator with no working scores 0/3.
Alternative method for Question 1(b)
3 x 4 – 12 x 2 – 15 = 0 leading to 3  x2 5  x2 1  [= 0] | (M1)
x2 1 or x2 5 | (M1) | Allow omission of –1 if factors seen. Factorising or other
valid method.
x   5 | (A1) | B1 SC if M1M1 not scored. Ignore ± i, i, –i, √–1.
Use of calculator with no working scores 0/3.
3
Question | Answer | Marks | Guidance
\begin{enumerate}[label=(\alph*)]
\item Express $3y^2 - 12y - 15$ in the form $3(y + a)^2 + b$, where $a$ and $b$ are constants. [2]

\item Hence find the exact solutions of the equation $3x^4 - 12x^2 - 15 = 0$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2024 Q1 [5]}}
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Q1 5 Q2 6 Q3 5 Q4 3 Q5 6 Q6 7 Q7 8 Q8 8 Q9 6 Q10 8 Q11 13