CAIE P1 2024 November — Question 7 8 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2024
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAreas by integration
TypeRegion bounded by two curves
DifficultyStandard +0.3 This is a straightforward two-part question combining completing the square (routine algebraic manipulation) with finding area between curves via integration. Part (a) is standard bookwork. Part (b) requires finding intersection points and integrating a quadratic, which are core P1 skills with no novel problem-solving required. Slightly above average difficulty only due to the multi-step nature and 8 total marks.
Spec1.02e Complete the square: quadratic polynomials and turning points1.08e Area between curve and x-axis: using definite integrals
  1. By expressing \(-2x^2 + 8x + 11\) in the form \(-a(x - h)^2 + c\), where \(a\), \(b\) and \(c\) are positive integers, find the coordinates of the vertex of the graph with equation \(y = -2x^2 + 8x + 11\). [3]
  2. \includegraphics{figure_7} The diagram shows part of the curve with equation \(y = -2x^2 + 8x + 11\) and the line with equation \(y = 8x + 9\). Find the area of the shaded region. [5]
Question 7:
AnswerMarks Guidance
7(a)−2 ( (x p)2 q ) or −2(x p)2q M1*
−2 ( (x−2)2 q ) or −2(x−2)2 qDM1
−2(x−2)2+19
AnswerMarks Guidance
and (2, 19)A1 Accept x=2, y=19 or 2, 19.
3
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
7(b)Method 1
x=1B1* Both x co-ordinates for the points of intersection.
Subtract and attempt to integrateM1*
 ( −2x2 +2 ) dx− 2 x3+2x
 
AnswerMarks Guidance
3B1* Both terms correct.
 2  2 
− +2− −2
AnswerMarks Guidance
 3  3 M1 Apply their limits, one positive and one negative, obtained
from equating the line and the curve to their integrated
expression.
8 2
= , 2
AnswerMarks Guidance
3 3DB1 AWRT 2.67 WWW.
−8 8
Condone → .
3 3
11
SC B1 for mistaking triangle for trapezium leading to , i.e.
3
a total of 2/5.
Method 2
AnswerMarks Guidance
x=1B1* Both x co-ordinates for the points of intersection.
Attempt to integrate and subtractM1* The second integral can be replaced with what is clearly their
area of a trapezium.
−2x3 8  8 
 + x2 +11x −  x2 +9x
AnswerMarks Guidance
 3 2  2 B1* OE
All terms correct.
1
(1+17)2
The second integral can be replaced by OE.
2
AnswerMarks Guidance
QuestionAnswer Marks
7(b)    −2 +4+11  −   2 +4−11     − (4+9)−(4−9)
 3  3 M1 Apply their limits, one positive and one negative, obtained
from equating the line and the curve, to their integrated
expressions.
If the trapezium has been used, the second integral can be
replaced by their 18.
8 2
= , 2
AnswerMarks Guidance
3 3DB1 AWRT 2.67 WWW.
−8 8
Condone → .
3 3
11
SC B1 for mistaking triangle for trapezium leading to , i.e.
3
a total of 2/5.
Method 3
AnswerMarks Guidance
x=1B1* Both x co-ordinates for the points of intersection.
Subtract and attempt to integrateM1*
2 8
− (x−2)3− x2 +10x
AnswerMarks Guidance
3 2B1* All terms correct.
2 
−4+10−(18−4−10)
AnswerMarks Guidance
3 M1 Apply their limits, one positive and one negative, obtained
from equating the line and the curve, to their integrated
expression.
8 2
= , 2
AnswerMarks Guidance
3 3DB1 AWRT 2.67 WWW.
QuestionAnswer Marks
7(b)Method 4
x=1B1* Both x co-ordinates for the points of intersection.
Attempt to integrate and subtractM1* The second integral can be replaced with what is clearly their
area of a trapezium.
 2  8 
− (x−2)3+19x −  x2 +9x
AnswerMarks Guidance
 3  2 B1* All terms correct.
1
(1+17)2
The second integral can be replaced with OE.
2
    2 +19  −(18−19)   − (4+9)−(4−9)
AnswerMarks Guidance
3  M1 Apply their limits, one positive and one negative, obtained
from equating the line and the curve, to their integrated
expression.
If the trapezium has been used the second integral can be
replaced with their 18 OE.
8 2
= , 2
AnswerMarks Guidance
3 3DB1 AWRT 2.67 WWW.
−8 8
Condone → .
3 3
11
SC B1 for mistaking triangle for trapezium leading to , i.e.
3
a total of 2/5.
5
AnswerMarks Guidance
QuestionAnswer Marks 
Question 7:
--- 7(a) ---
7(a) | −2 ( (x p)2 q ) or −2(x p)2q | M1* | p0.
−2 ( (x−2)2 q ) or −2(x−2)2 q | DM1
−2(x−2)2+19
and (2, 19) | A1 | Accept x=2, y=19 or 2, 19.
3
Question | Answer | Marks | Guidance
--- 7(b) ---
7(b) | Method 1
x=1 | B1* | Both x co-ordinates for the points of intersection.
Subtract and attempt to integrate | M1*
 ( −2x2 +2 ) dx− 2 x3+2x
 
3 | B1* | Both terms correct.
 2  2 
− +2− −2
 3  3  | M1 | Apply their limits, one positive and one negative, obtained
from equating the line and the curve to their integrated
expression.
8 2
= , 2
3 3 | DB1 | AWRT 2.67 WWW.
−8 8
Condone → .
3 3
11
SC B1 for mistaking triangle for trapezium leading to , i.e.
3
a total of 2/5.
Method 2
x=1 | B1* | Both x co-ordinates for the points of intersection.
Attempt to integrate and subtract | M1* | The second integral can be replaced with what is clearly their
area of a trapezium.
−2x3 8  8 
 + x2 +11x −  x2 +9x
 3 2  2  | B1* | OE
All terms correct.
1
(1+17)2
The second integral can be replaced by OE.
2
Question | Answer | Marks | Guidance
7(b) |     −2 +4+11  −   2 +4−11     − (4+9)−(4−9)
 3  3  | M1 | Apply their limits, one positive and one negative, obtained
from equating the line and the curve, to their integrated
expressions.
If the trapezium has been used, the second integral can be
replaced by their 18.
8 2
= , 2
3 3 | DB1 | AWRT 2.67 WWW.
−8 8
Condone → .
3 3
11
SC B1 for mistaking triangle for trapezium leading to , i.e.
3
a total of 2/5.
Method 3
x=1 | B1* | Both x co-ordinates for the points of intersection.
Subtract and attempt to integrate | M1*
2 8
− (x−2)3− x2 +10x
3 2 | B1* | All terms correct.
2 
−4+10−(18−4−10)

3  | M1 | Apply their limits, one positive and one negative, obtained
from equating the line and the curve, to their integrated
expression.
8 2
= , 2
3 3 | DB1 | AWRT 2.67 WWW.
Question | Answer | Marks | Guidance
7(b) | Method 4
x=1 | B1* | Both x co-ordinates for the points of intersection.
Attempt to integrate and subtract | M1* | The second integral can be replaced with what is clearly their
area of a trapezium.
 2  8 
− (x−2)3+19x −  x2 +9x
 3  2  | B1* | All terms correct.
1
(1+17)2
The second integral can be replaced with OE.
2
    2 +19  −(18−19)   − (4+9)−(4−9)
3   | M1 | Apply their limits, one positive and one negative, obtained
from equating the line and the curve, to their integrated
expression.
If the trapezium has been used the second integral can be
replaced with their 18 OE.
8 2
= , 2
3 3 | DB1 | AWRT 2.67 WWW.
−8 8
Condone → .
3 3
11
SC B1 for mistaking triangle for trapezium leading to , i.e.
3
a total of 2/5.
5
Question | Answer | Marks | Guidance
\begin{enumerate}[label=(\alph*)]
\item By expressing $-2x^2 + 8x + 11$ in the form $-a(x - h)^2 + c$, where $a$, $b$ and $c$ are positive integers, find the coordinates of the vertex of the graph with equation $y = -2x^2 + 8x + 11$. [3]

\item \includegraphics{figure_7}

The diagram shows part of the curve with equation $y = -2x^2 + 8x + 11$ and the line with equation $y = 8x + 9$.

Find the area of the shaded region. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2024 Q7 [8]}}
This paper (10 questions)
View full paper
Q1 5 Q2 5 Q3 5 Q4 6 Q5 10 Q6 6 Q7 8 Q8 10 Q9 10 Q10 10