CAIE FP1 2009 June — Question 3 6 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2009
SessionJune
Marks6
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TopicCentre of Mass 2
TypeCentre of mass with given condition
DifficultyChallenging +1.2 This is a standard centroid calculation requiring integration to find area and first moment, then solving the given condition ȳ = a. While it involves multiple steps (finding area, finding moment, setting up equation), the techniques are routine for Further Maths students and the algebra is straightforward once the standard formulas are applied.
Spec4.08d Volumes of revolution: about x and y axes6.04d Integration: for centre of mass of laminas/solids
3 The equation of a curve is \(y = \lambda x ^ { 2 }\), where \(\lambda > 0\). The region bounded by the curve, the \(x\)-axis and the line \(x = a\), where \(a > 0\), is denoted by \(R\). The \(y\)-coordinate of the centroid of \(R\) is \(a\). Show that \(\lambda = \frac { 10 } { 3 a }\).
AnswerMarks
\(\bar{y} = 2\frac{\int_0^a y^2 dx}{\int_0^a y dx}\)M1
\(\int_0^a y dx = \lambda a^3/3\)B1
\(\int_0^a y^2 dx = \lambda^2 a^5/5\)M1A1
\(\bar{y} = ... = 3\lambda a^2/10\)A1
\(\bar{y} = a \Rightarrow \lambda = \frac{10}{3a}\) (AG)A1
Finding \(\bar{x}\) is not a misread and scores M0 
$\bar{y} = 2\frac{\int_0^a y^2 dx}{\int_0^a y dx}$ | M1 |

$\int_0^a y dx = \lambda a^3/3$ | B1 |

$\int_0^a y^2 dx = \lambda^2 a^5/5$ | M1A1 |

$\bar{y} = ... = 3\lambda a^2/10$ | A1 |

$\bar{y} = a \Rightarrow \lambda = \frac{10}{3a}$ (AG) | A1 |

Finding $\bar{x}$ is not a misread and scores M0 |
3 The equation of a curve is $y = \lambda x ^ { 2 }$, where $\lambda > 0$. The region bounded by the curve, the $x$-axis and the line $x = a$, where $a > 0$, is denoted by $R$. The $y$-coordinate of the centroid of $R$ is $a$. Show that $\lambda = \frac { 10 } { 3 a }$.

\hfill \mbox{\textit{CAIE FP1 2009 Q3 [6]}}
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