CAIE FP1 2012 November — Question 10 12 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2012
SessionNovember
Marks12
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TopicImplicit equations and differentiation
TypeFind stationary points
DifficultyChallenging +1.2 This is a Further Maths question requiring implicit differentiation of a cubic curve to find stationary points. While it involves multiple steps (implicit differentiation, setting dy/dx=0, solving simultaneous equations, and determining nature), the techniques are standard for FP1. The algebraic manipulation is moderately challenging but follows a predictable pattern for this topic.
Spec1.07n Stationary points: find maxima, minima using derivatives1.07s Parametric and implicit differentiation
10 The curve \(C\) has equation \(x ^ { 3 } + y ^ { 3 } = 3 x y\), for \(x > 0\) and \(y > 0\). Find a relationship between \(x\) and \(y\) when \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\). Find the exact coordinates of the turning point of \(C\), and determine the nature of this turning point.
Question 10:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(3x^2 + 3y^2y' = 3y + 3xy'\)B1B1 Two separate B marks for implicit differentiation
\(\Rightarrow \frac{dy}{dx} = \frac{y-x^2}{y^2-1} = 0\)M1 Equates \(\frac{dy}{dx}\) to zero
\(\Rightarrow y = x^2\)A1 Obtains relationship
\(\Rightarrow xy + y^3 = 3xy \Rightarrow y^3 = 2xy \Rightarrow y^2 = 2x \quad (y \neq 0)\)M1 Substitutes for \(y\)
\(\Rightarrow x^4 = 2x \Rightarrow x^3 = 2 \Rightarrow x = 2^{\frac{1}{3}}\) and \(\Rightarrow y = 2^{\frac{2}{3}} \quad (x \neq 0)\)A1A1 Obtains \(x\) and \(y\)
\(6x + 3y^2y'' + 6y(y')^2 = 3y' + 3xy'' + 3y'\)B1B1 Differentiates again
Uses \(y' = 0\): \(\Rightarrow 6x = y''(3x - 3y^2) \Rightarrow y'' = \frac{2x}{x(1-x^3)}\)M1
\(\Rightarrow y'' = \frac{2}{1-2} = -2 \Rightarrow\) maxA1 Part marks: 4, 8 
# Question 10:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $3x^2 + 3y^2y' = 3y + 3xy'$ | B1B1 | Two separate B marks for implicit differentiation |
| $\Rightarrow \frac{dy}{dx} = \frac{y-x^2}{y^2-1} = 0$ | M1 | Equates $\frac{dy}{dx}$ to zero |
| $\Rightarrow y = x^2$ | A1 | Obtains relationship |
| $\Rightarrow xy + y^3 = 3xy \Rightarrow y^3 = 2xy \Rightarrow y^2 = 2x \quad (y \neq 0)$ | M1 | Substitutes for $y$ |
| $\Rightarrow x^4 = 2x \Rightarrow x^3 = 2 \Rightarrow x = 2^{\frac{1}{3}}$ and $\Rightarrow y = 2^{\frac{2}{3}} \quad (x \neq 0)$ | A1A1 | Obtains $x$ and $y$ |
| $6x + 3y^2y'' + 6y(y')^2 = 3y' + 3xy'' + 3y'$ | B1B1 | Differentiates again |
| Uses $y' = 0$: $\Rightarrow 6x = y''(3x - 3y^2) \Rightarrow y'' = \frac{2x}{x(1-x^3)}$ | M1 | |
| $\Rightarrow y'' = \frac{2}{1-2} = -2 \Rightarrow$ max | A1 | **Part marks: 4, 8** |

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10 The curve $C$ has equation $x ^ { 3 } + y ^ { 3 } = 3 x y$, for $x > 0$ and $y > 0$. Find a relationship between $x$ and $y$ when $\frac { \mathrm { d } y } { \mathrm {~d} x } = 0$.

Find the exact coordinates of the turning point of $C$, and determine the nature of this turning point.

\hfill \mbox{\textit{CAIE FP1 2012 Q10 [12]}}
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Q1 4 Q2 4 Q3 5 Q4 6 Q5 6 Q6 7 Q7 8 Q8 9 Q9 12 Q10 12 Q11 13 Q12 EITHER Q12 OR