CAIE P1 2015 June — Question 9 10 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2015
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTangents, normals and gradients
TypeFind stationary points
DifficultyStandard +0.3 This is a straightforward stationary points question requiring standard differentiation, solving quadratic equations, and using the second derivative test. Part (iii) involves a discriminant condition which is slightly beyond pure routine but still a common A-level technique. The multi-part structure and 10 marks suggest slightly above average difficulty, but all methods are standard bookwork with no novel insight required.
Spec1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx

The equation of a curve is \(y = x^3 + px^2\), where \(p\) is a positive constant.
  1. Show that the origin is a stationary point on the curve and find the coordinates of the other stationary point in terms of \(p\). [4]
  2. Find the nature of each of the stationary points. [3]
Another curve has equation \(y = x^3 + px^2 + px\).
  1. Find the set of values of \(p\) for which this curve has no stationary points. [3]

Question 9:

(i) ---
9
AnswerMarks
(i)y = x3 + px2
dy
= 3x² + 2px
dx
2p
Sets to 0 → x = 0 or −
3
 2p 4p3
→ (0, 0) or  − ,  
AnswerMarks
 3 27 B1
M1
A1 A1
AnswerMarks
[4]cao
Sets differential to 0
cao cao, first A1 for any correct
turning point or any correct pair of
x values. 2nd A1 for 2 complete
TPs
(ii)
AnswerMarks
(iii)d2y
= 6x + 2p
dx2
At (0, 0) → 2p +ve Minimum
 2p 4p3 
At  − ,   → −2p –ve Maximum
 3 27 
y = x3 + px2 + px → 3x² + 2px + p (= 0)
Uses b2 −4ac
→ 4p² − 12p< 0
AnswerMarks
→ 0 <p< 3 aefM1
A1
A1
[3]
B1
M1
A1
AnswerMarks
[3]Other methods include; clear
demonstration of sign change of
gradient, clear reference to the
shape of the curve
www
Any correct use of discriminant
cao (condone ø)
AnswerMarks Guidance
Page 9Mark Scheme Syllabus
Cambridge International AS/A Level – May/June 20159709 11
Question 9:
--- 9
(i) ---
9
(i) | y = x3 + px2
dy
= 3x² + 2px
dx
2p
Sets to 0 → x = 0 or −
3
 2p 4p3
→ (0, 0) or  − ,  
 3 27  | B1
M1
A1 A1
[4] | cao
Sets differential to 0
cao cao, first A1 for any correct
turning point or any correct pair of
x values. 2nd A1 for 2 complete
TPs
(ii)
(iii) | d2y
= 6x + 2p
dx2
At (0, 0) → 2p +ve Minimum
 2p 4p3 
At  − ,   → −2p –ve Maximum
 3 27 
y = x3 + px2 + px → 3x² + 2px + p (= 0)
Uses b2 −4ac
→ 4p² − 12p< 0
→ 0 <p< 3 aef | M1
A1
A1
[3]
B1
M1
A1
[3] | Other methods include; clear
demonstration of sign change of
gradient, clear reference to the
shape of the curve
www
Any correct use of discriminant
cao (condone ø)
Page 9 | Mark Scheme | Syllabus | Paper
Cambridge International AS/A Level – May/June 2015 | 9709 | 11
The equation of a curve is $y = x^3 + px^2$, where $p$ is a positive constant.

\begin{enumerate}[label=(\roman*)]
\item Show that the origin is a stationary point on the curve and find the coordinates of the other stationary point in terms of $p$. [4]
\item Find the nature of each of the stationary points. [3]
\end{enumerate}

Another curve has equation $y = x^3 + px^2 + px$.

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Find the set of values of $p$ for which this curve has no stationary points. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2015 Q9 [10]}}