CAIE FP1 2010 June — Question 1 5 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2010
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImplicit equations and differentiation
TypeFind second derivative d²y/dx²
DifficultyStandard +0.8 This Further Maths question requires implicit differentiation of a non-standard equation involving (dy/dx)³, then differentiating again to find d²y/dx². Students must first solve a cubic equation, then carefully apply the chain rule and product rule in the second differentiation. The unusual form and multi-step nature make it moderately challenging but still within standard FP1 scope.
Spec1.07s Parametric and implicit differentiation

1 The variables \(x\) and \(y\) are such that \(y = - 1\) when \(x = 1\) and $$x ^ { 2 } + y ^ { 2 } + \left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) ^ { 3 } = 29$$ Find the values of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) when \(x = 1\).

AnswerMarks
\(1 + 1 + (y')^3 = 29 \Rightarrow (y')^3 = 27 \Rightarrow y' = 3\)B1
\(2x + 2yy' + 3(y')^2 y'' = 0\)M1A1, A1
\(2 - 6 + 27y'' = 0 \Rightarrow y'' = \frac{4}{27}\)A1
[5]
$1 + 1 + (y')^3 = 29 \Rightarrow (y')^3 = 27 \Rightarrow y' = 3$ | B1 |

$2x + 2yy' + 3(y')^2 y'' = 0$ | M1A1, A1 |

$2 - 6 + 27y'' = 0 \Rightarrow y'' = \frac{4}{27}$ | A1 |

| [5] |
1 The variables $x$ and $y$ are such that $y = - 1$ when $x = 1$ and

$$x ^ { 2 } + y ^ { 2 } + \left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) ^ { 3 } = 29$$

Find the values of $\frac { \mathrm { d } y } { \mathrm {~d} x }$ and $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ when $x = 1$.

\hfill \mbox{\textit{CAIE FP1 2010 Q1 [5]}}