CAIE P1 2024 November — Question 9 10 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2024
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSimultaneous equations
TypeLine intersecting quadratic curve
DifficultyStandard +0.3 This is a slightly above-average A-level question. Part (a) involves substituting a known point into two equations to find k and p, then solving a quadratic to find the second intersection—standard simultaneous equations technique with some algebraic manipulation. Part (b) requires using the discriminant condition for no intersection, which is a common textbook exercise. The algebra is moderately involved but follows well-established methods with no novel insight required.
Spec1.02c Simultaneous equations: two variables by elimination and substitution1.02d Quadratic functions: graphs and discriminant conditions1.02g Inequalities: linear and quadratic in single variable
The equation of a curve is \(y = \frac{1}{3}k^2x^2 - 2kx + 2\) and the equation of a line is \(y = kx + p\), where \(k\) and \(p\) are constants with \(0 < k < 1\).
  1. It is given that one of the points of intersection of the curve and the line has coordinates \(\left(\frac{6}{5}, \frac{3}{5}\right)\). Find the values of \(k\) and \(p\), and find the coordinates of the other point of intersection. [7]
  2. It is given instead that the line and the curve do not intersect. Find the set of possible values of \(p\). [3]
Question 9:
AnswerMarks
9(a)1 25 5 1 
 k2 −2k +2= 
2 4 2 2 
OR 
 
1 k2 25 −2k 5 +2= k 5 +   1 − 5 k   
 2 4 2 2 2 2  
AnswerMarks Guidance
25k2 −40k+12 =0 M1* 5 1
Using  ,  in the curve equation or equating the line and
2 2
5 1 5
the curve and then using x= and p= − k.
2 2 2
Simplify to get a three-term quadratic in k. Condone errors in
simplification.
2
k =
AnswerMarks Guidance
5A1 OE
6
Condone inclusion of k = .
5
1  25
=their  + p p=
AnswerMarks Guidance
2  52DM1* 5 1
Using  ,  and their k in an equation in p.
2 2
Either the line (as shown) or 4p2 +12p+5=0 are the most
likely and solving for p.
1
p=−
AnswerMarks Guidance
2A1 OE
5
Condone inclusion of p=− .
2
2 6 5
x2 − x+ =0  4x2 −60x+125 =0 
 
AnswerMarks Guidance
25 5 2DM1 Equating the line and curve using their k and p and simplify to
get a three-term quadratic [= 0].
25 9
 , 
AnswerMarks Guidance
 2 2A1 A1 OE
25 9
Accept x= , y= .
2 2
AnswerMarks Guidance
QuestionAnswer Marks
9(a)Alternative Method for Question 9(a)
1 25 5 5 
k2 −2k +2= k + p
 
2 4 2 2 
AnswerMarks Guidance
4p2+12p+5 =0 M1* OE
5 1
Using  ,  in the curve equation or equating the line and
2 2
5 1 2
the curve and then using x= and k = − p.
2 5 5
Simplify to get a three-term quadratic in p [= 0].
1
p=− OE
AnswerMarks Guidance
2A1 5
Condone inclusion of p=− .
2
1 5   1
= k+their−   k=
AnswerMarks Guidance
2 2   2DM1* 5 1
Using  ,  and their p in the line equation and solving for
2 2
k.
2
k =
AnswerMarks Guidance
5A1 OE
6
Condone inclusion of k = .
5
2 6 5
x2 − x+ =0  4x2 −60x+125 =0 
 
AnswerMarks Guidance
25 5 2DM1 Equating the line and curve using their k and p and simplify to
get a three-term quadratic [= 0].
25 9
 , 
AnswerMarks Guidance
 2 2A1 A1 OE
25 9
Accept x= , y= .
2 2
7
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
9(b)1  1
k2x2 −2kx+2= kx+ p  k2x2 −3kx+2− p
 
AnswerMarks Guidance
2  2M1* Equate the original equations of the curve and the line and
collect like terms; k and p must still be present.
1
9k2 −4 k2(2− p)
AnswerMarks Guidance
2DM1 Use of b2 −4ac for their quadratic in x to give an expression in
k and p. This expression can come from their equation in (a).
5
p−
AnswerMarks
2A1
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 9:
--- 9(a) ---
9(a) | 1 25 5 1 
 k2 −2k +2= 
2 4 2 2 
OR 
 
1 k2 25 −2k 5 +2= k 5 +   1 − 5 k   
 2 4 2 2 2 2  
25k2 −40k+12 =0  | M1* | 5 1
Using  ,  in the curve equation or equating the line and
2 2
5 1 5
the curve and then using x= and p= − k.
2 2 2
Simplify to get a three-term quadratic in k. Condone errors in
simplification.
2
k =
5 | A1 | OE
6
Condone inclusion of k = .
5
1  25
=their  + p p=
2  52 | DM1* | 5 1
Using  ,  and their k in an equation in p.
2 2
Either the line (as shown) or 4p2 +12p+5=0 are the most
likely and solving for p.
1
p=−
2 | A1 | OE
5
Condone inclusion of p=− .
2
2 6 5
x2 − x+ =0  4x2 −60x+125 =0 
 
25 5 2 | DM1 | Equating the line and curve using their k and p and simplify to
get a three-term quadratic [= 0].
25 9
 , 
 2 2 | A1 A1 | OE
25 9
Accept x= , y= .
2 2
Question | Answer | Marks | Guidance
9(a) | Alternative Method for Question 9(a)
1 25 5 5 
k2 −2k +2= k + p
 
2 4 2 2 
4p2+12p+5 =0  | M1* | OE
5 1
Using  ,  in the curve equation or equating the line and
2 2
5 1 2
the curve and then using x= and k = − p.
2 5 5
Simplify to get a three-term quadratic in p [= 0].
1
p=− OE
2 | A1 | 5
Condone inclusion of p=− .
2
1 5   1
= k+their−   k=
2 2   2 | DM1* | 5 1
Using  ,  and their p in the line equation and solving for
2 2
k.
2
k =
5 | A1 | OE
6
Condone inclusion of k = .
5
2 6 5
x2 − x+ =0  4x2 −60x+125 =0 
 
25 5 2 | DM1 | Equating the line and curve using their k and p and simplify to
get a three-term quadratic [= 0].
25 9
 , 
 2 2 | A1 A1 | OE
25 9
Accept x= , y= .
2 2
7
Question | Answer | Marks | Guidance
--- 9(b) ---
9(b) | 1  1
k2x2 −2kx+2= kx+ p  k2x2 −3kx+2− p
 
2  2 | M1* | Equate the original equations of the curve and the line and
collect like terms; k and p must still be present.
1
9k2 −4 k2(2− p)
2 | DM1 | Use of b2 −4ac for their quadratic in x to give an expression in
k and p. This expression can come from their equation in (a).
5
p−
2 | A1
3
Question | Answer | Marks | Guidance
The equation of a curve is $y = \frac{1}{3}k^2x^2 - 2kx + 2$ and the equation of a line is $y = kx + p$, where $k$ and $p$ are constants with $0 < k < 1$.

\begin{enumerate}[label=(\alph*)]
\item It is given that one of the points of intersection of the curve and the line has coordinates $\left(\frac{6}{5}, \frac{3}{5}\right)$.

Find the values of $k$ and $p$, and find the coordinates of the other point of intersection. [7]

\item It is given instead that the line and the curve do \textbf{not} intersect.

Find the set of possible values of $p$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2024 Q9 [10]}}
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