CAIE P3 2013 June — Question 5 6 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2013
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImplicit equations and differentiation
TypeFind stationary points
DifficultyChallenging +1.2 This is a standard implicit differentiation problem requiring finding dy/dx, setting it to zero for stationary points, and solving for x. While it involves algebraic manipulation of a cubic implicit equation with a parameter, the technique is routine for P3 level and follows a predictable method. The 6 marks reflect moderate algebraic complexity rather than conceptual difficulty.
Spec1.07s Parametric and implicit differentiation

\includegraphics{figure_5} The diagram shows the curve with equation $$x^3 + xy^2 + ay^2 - 3ax^2 = 0,$$ where \(a\) is a positive constant. The maximum point on the curve is \(M\). Find the \(x\)-coordinate of \(M\) in terms of \(a\). [6]

AnswerMarks Guidance
EITHER: State \(2ay\frac{dy}{dx}\) as derivative of \(ay^2\)B1
State \(y^2 + 2xy\frac{dy}{dx}\) as derivative of \(xy^2\)B1
Equate derivative of LHS to zero and set \(\frac{dy}{dx}\) equal to zeroM1
Obtain \(3x^2 + y^2 - 6ax = 0\), or horizontal equivalentA1
Eliminate \(y\) and obtain an equation in \(x\)M1
Solve for \(x\) and obtain answer \(x = \sqrt{3a}\)A1
OR1: Rearrange equation in the form \(y^2 = \frac{3ax^2 - x^3}{x + a}\) and attempt differentiation of one sideB1
Use correct quotient or product rule to differentiate RHSM1
Obtain correct derivative of RHS in any formA1
Set \(\frac{dy}{dx}\) equal to zero and obtain an equation in \(x\)M1
Obtain a correct horizontal equation free of surdsA1
Solve for \(x\) and obtain answer \(x = \sqrt{3a}\)A1
OR2: Rearrange equation in the form \(y = \left(\frac{3ax^2 - x^3}{x + a}\right)^{\frac{1}{2}}\) and differentiation of RHSB1
Use correct quotient or product rule and chain ruleM1
Obtain correct derivative in any formA1
Equate derivative to zero and obtain an equation in \(x\)M1
Obtain a correct horizontal equation free of surdsA1
Solve for \(x\) and obtain answer \(x = \sqrt{3a}\)A1 [6]
**EITHER:** State $2ay\frac{dy}{dx}$ as derivative of $ay^2$ | B1 |
State $y^2 + 2xy\frac{dy}{dx}$ as derivative of $xy^2$ | B1 |
Equate derivative of LHS to zero and set $\frac{dy}{dx}$ equal to zero | M1 |
Obtain $3x^2 + y^2 - 6ax = 0$, or horizontal equivalent | A1 |
Eliminate $y$ and obtain an equation in $x$ | M1 |
Solve for $x$ and obtain answer $x = \sqrt{3a}$ | A1 |

**OR1:** Rearrange equation in the form $y^2 = \frac{3ax^2 - x^3}{x + a}$ and attempt differentiation of one side | B1 |
Use correct quotient or product rule to differentiate RHS | M1 |
Obtain correct derivative of RHS in any form | A1 |
Set $\frac{dy}{dx}$ equal to zero and obtain an equation in $x$ | M1 |
Obtain a correct horizontal equation free of surds | A1 |
Solve for $x$ and obtain answer $x = \sqrt{3a}$ | A1 |

**OR2:** Rearrange equation in the form $y = \left(\frac{3ax^2 - x^3}{x + a}\right)^{\frac{1}{2}}$ and differentiation of RHS | B1 |
Use correct quotient or product rule and chain rule | M1 |
Obtain correct derivative in any form | A1 |
Equate derivative to zero and obtain an equation in $x$ | M1 |
Obtain a correct horizontal equation free of surds | A1 |
Solve for $x$ and obtain answer $x = \sqrt{3a}$ | A1 | [6]
\includegraphics{figure_5}

The diagram shows the curve with equation
$$x^3 + xy^2 + ay^2 - 3ax^2 = 0,$$
where $a$ is a positive constant. The maximum point on the curve is $M$. Find the $x$-coordinate of $M$ in terms of $a$. [6]

\hfill \mbox{\textit{CAIE P3 2013 Q5 [6]}}