CAIE FP1 2013 June — Question 4 8 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2013
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImplicit equations and differentiation
TypeFind second derivative d²y/dx²
DifficultyStandard +0.3 This is a standard implicit differentiation question requiring two applications of the technique. Finding dy/dx involves routine application of the chain and product rules, and the second derivative requires careful but methodical differentiation of the first result. While it requires multiple steps and careful algebra, it follows a well-practiced procedure with no novel insight needed, making it slightly easier than average for Further Maths students.
Spec1.07s Parametric and implicit differentiation

4 Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { 4 } { 3 }\) at the point \(A ( 1 , - 2 )\) on the curve with equation $$y ^ { 3 } - 3 x ^ { 2 } y + 2 = 0$$ and find the value of \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) at \(A\).

Question 4:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(3y^2y' - (3x^2y' + 6xy) = 0\)B1B1 B1 for 1st term, \(=0\) allow recovery
At \((1,-2)\): \(12y' - (3y'-12) = 0 \Rightarrow 9y' = -12 \Rightarrow y' = -\frac{4}{3}\) (AG)B1
\(3y^2y'' + 6y(y')^2 - (6xy' + 3x^2y'' + 6xy' + 6y) = 0\)B1B1B1 B1 for each pair of terms; 3rd mark includes \(=0\) but allow recovery
\(12y'' - 12\times\frac{16}{9} - \left(-8 + 3y'' + 6\times\frac{-4}{3} - 12\right) = 0\)M1 Substitutes values
\(\Rightarrow 9y'' = -\frac{20}{3} \Rightarrow y'' = -\frac{20}{27}\) (Allow \(-0.741\))A1 8 marks total
## Question 4:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $3y^2y' - (3x^2y' + 6xy) = 0$ | B1B1 | B1 for 1st term, $=0$ allow recovery |
| At $(1,-2)$: $12y' - (3y'-12) = 0 \Rightarrow 9y' = -12 \Rightarrow y' = -\frac{4}{3}$ (AG) | B1 | |
| $3y^2y'' + 6y(y')^2 - (6xy' + 3x^2y'' + 6xy' + 6y) = 0$ | B1B1B1 | B1 for each pair of terms; 3rd mark includes $=0$ but allow recovery |
| $12y'' - 12\times\frac{16}{9} - \left(-8 + 3y'' + 6\times\frac{-4}{3} - 12\right) = 0$ | M1 | Substitutes values |
| $\Rightarrow 9y'' = -\frac{20}{3} \Rightarrow y'' = -\frac{20}{27}$ (Allow $-0.741$) | A1 | 8 marks total |

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4 Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { 4 } { 3 }$ at the point $A ( 1 , - 2 )$ on the curve with equation

$$y ^ { 3 } - 3 x ^ { 2 } y + 2 = 0$$

and find the value of $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ at $A$.

\hfill \mbox{\textit{CAIE FP1 2013 Q4 [8]}}