CAIE P3 2016 November — Question 4 7 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2016
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImplicit equations and differentiation
TypeFind stationary points
DifficultyStandard +0.8 This requires implicit differentiation of a product of three terms, setting dy/dx=0 to find stationary points, then solving the resulting system of equations. The algebraic manipulation is non-trivial, particularly handling the cubic constant and proving uniqueness, making it moderately challenging but within reach of a well-prepared P3 student.
Spec1.07n Stationary points: find maxima, minima using derivatives1.07s Parametric and implicit differentiation
4 The equation of a curve is \(x y ( x - 6 y ) = 9 a ^ { 3 }\), where \(a\) is a non-zero constant. Show that there is only one point on the curve at which the tangent is parallel to the \(x\)-axis, and find the coordinates of this point.
Question 4:
EITHER EITHER:
AnswerMarks Guidance
Answer/WorkingMark Guidance
State \(2xy + x^2\frac{dy}{dx}\), or equivalent, as derivative of \(x^2y\)B1
State \(6y^2 + 12xy\frac{dy}{dx}\), or equivalent, as derivative of \(6xy^2\)B1
OR (product rule on LHS directly):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Differentiating LHS using correct product rule, state term \(xy(1 - 6\frac{dy}{dx})\), or equivalentB1
State term \((y + x\frac{dy}{dx})(x - 6y)\), or equivalentB1
Answer/WorkingMark Guidance
Equate attempted derivative of LHS to zero and set \(\frac{dy}{dx}\) equal to zeroM1*
Obtain horizontal equation, e.g. \(6y^2 - 2xy = 0\) (from correct work only)A1
Explicitly reject \(y = 0\) as a possibility \(py^2 - qxy = 0\)A1
Obtain an equation in \(x\) or \(y\)DM1
Obtain answer \((-3a, -a)\)A1
OR (quotient rule):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Rearrange to \(y = \frac{9a^3}{x(x-6y)}\) and use correct quotient rule to obtain \(-\frac{9a^3}{x^2(x-6y)^2} \times \ldots\)B1
State term \((x-6y) + x(1-6y')\), or equivalentB1
Justify division by \(x(x-6y)\)B1
Set \(\frac{dy}{dx}\) equal to zeroM1*
Obtain horizontal equation, e.g. \(6y^2 - 2xy = 0\) (from correct work only)A1
Obtain an equation in \(x\) or \(y\)DM1
Obtain answer \((-3a, -a)\)A1 
## Question 4:

### EITHER EITHER:

| Answer/Working | Mark | Guidance |
|---|---|---|
| State $2xy + x^2\frac{dy}{dx}$, or equivalent, as derivative of $x^2y$ | B1 | |
| State $6y^2 + 12xy\frac{dy}{dx}$, or equivalent, as derivative of $6xy^2$ | B1 | |

### OR (product rule on LHS directly):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Differentiating LHS using correct product rule, state term $xy(1 - 6\frac{dy}{dx})$, or equivalent | B1 | |
| State term $(y + x\frac{dy}{dx})(x - 6y)$, or equivalent | B1 | |

| Answer/Working | Mark | Guidance |
|---|---|---|
| Equate attempted derivative of LHS to zero and set $\frac{dy}{dx}$ equal to zero | M1* | |
| Obtain horizontal equation, e.g. $6y^2 - 2xy = 0$ (from correct work only) | A1 | |
| Explicitly reject $y = 0$ as a possibility $py^2 - qxy = 0$ | A1 | |
| Obtain an equation in $x$ or $y$ | DM1 | |
| Obtain answer $(-3a, -a)$ | A1 | |

### OR (quotient rule):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Rearrange to $y = \frac{9a^3}{x(x-6y)}$ and use correct quotient rule to obtain $-\frac{9a^3}{x^2(x-6y)^2} \times \ldots$ | B1 | |
| State term $(x-6y) + x(1-6y')$, or equivalent | B1 | |
| Justify division by $x(x-6y)$ | B1 | |
| Set $\frac{dy}{dx}$ equal to zero | M1* | |
| Obtain horizontal equation, e.g. $6y^2 - 2xy = 0$ (from correct work only) | A1 | |
| Obtain an equation in $x$ or $y$ | DM1 | |
| Obtain answer $(-3a, -a)$ | A1 | |

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4 The equation of a curve is $x y ( x - 6 y ) = 9 a ^ { 3 }$, where $a$ is a non-zero constant. Show that there is only one point on the curve at which the tangent is parallel to the $x$-axis, and find the coordinates of this point.

\hfill \mbox{\textit{CAIE P3 2016 Q4 [7]}}
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