CAIE FP1 2009 June — Question 6 7 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2009
SessionJune
Marks7
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Mark schemeDownload PDF ↗
TopicImplicit equations and differentiation
TypeFind second derivative d²y/dx²
DifficultyStandard +0.8 This is a Further Maths implicit differentiation question requiring both first and second derivatives at a specific point. While the first derivative is straightforward using the product rule and implicit differentiation, finding the second derivative requires careful application of the quotient rule to an expression containing dy/dx, then substituting both coordinates and the first derivative value. This multi-step process with algebraic complexity places it moderately above average difficulty.
Spec1.07s Parametric and implicit differentiation
6 A curve has equation $$( x + y ) \left( x ^ { 2 } + y ^ { 2 } \right) = 1$$ Find the values of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) at the point \(( 0,1 )\).
AnswerMarks
\((1 + y_1)(x^2 + y^2) + (x + y)(2x + 2yy_1) = 0\)B1B1
Puts \(x = 0, y = 1\) to obtain \(y_1 = -1/3\)B1
\((x^2 + y^2)y_2 + 2(1 + y_1)(2x + 2yy_1) + (x + y)(2 + 2y_1^2 + 2yy_2) = 0\)B1B1B1
Puts \(x = 0, y = 1, y_1 = -1/3\) to obtain \(y_2 = -4/9\) (cwo)B1
or
AnswerMarks Guidance
\(x^3 + xy^2 + x^2y + y^3 = 1\) (expanding)
Differentiating
\(3x^2 + y^2 + 2xyy' + 2xy + x^2y' + 3y^2y' = 0\)B1 B1
Sub \((0,1) 1 + 3y' = 0 \Rightarrow y' = -\frac{1}{3}\)B1
Differentiating again
\(6x + 2yy' + 2yy' + 2x(yy'' + (y')^2) + 2xy + x^2y'' + 2xy' + 6yy' + 3y^2y'' = 0\)B1 B1
Sub \((0,1)\) \(\frac{4}{3} + 3y'' = 0 \Rightarrow y' = -\frac{4}{9}\)B1
For those who write \(\frac{dy}{dx} = -\frac{3x^2 + 2xy + y^2}{x^2 + 2xy + 3y^2}\)
4th, 5th and 6th marks are awarded as follows:
AnswerMarks
\(\frac{y\frac{du}{dx}(B1) - u\frac{dy}{dx}(B1)}{y^2(B1)}\) 
$(1 + y_1)(x^2 + y^2) + (x + y)(2x + 2yy_1) = 0$ | B1B1 |

Puts $x = 0, y = 1$ to obtain $y_1 = -1/3$ | B1 |

$(x^2 + y^2)y_2 + 2(1 + y_1)(2x + 2yy_1) + (x + y)(2 + 2y_1^2 + 2yy_2) = 0$ | B1B1B1 |

Puts $x = 0, y = 1, y_1 = -1/3$ to obtain $y_2 = -4/9$ (cwo) | B1 |

**or**

$x^3 + xy^2 + x^2y + y^3 = 1$ (expanding) |

Differentiating |

$3x^2 + y^2 + 2xyy' + 2xy + x^2y' + 3y^2y' = 0$ | B1 | B1 |

Sub $(0,1) 1 + 3y' = 0 \Rightarrow y' = -\frac{1}{3}$ | B1 |

Differentiating again |

$6x + 2yy' + 2yy' + 2x(yy'' + (y')^2) + 2xy + x^2y'' + 2xy' + 6yy' + 3y^2y'' = 0$ | B1 | B1 | B1 |

Sub $(0,1)$ $\frac{4}{3} + 3y'' = 0 \Rightarrow y' = -\frac{4}{9}$ | B1 |

For those who write $\frac{dy}{dx} = -\frac{3x^2 + 2xy + y^2}{x^2 + 2xy + 3y^2}$ |

4th, 5th and 6th marks are awarded as follows:

$\frac{y\frac{du}{dx}(B1) - u\frac{dy}{dx}(B1)}{y^2(B1)}$ |
6 A curve has equation

$$( x + y ) \left( x ^ { 2 } + y ^ { 2 } \right) = 1$$

Find the values of $\frac { \mathrm { d } y } { \mathrm {~d} x }$ and $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ at the point $( 0,1 )$.

\hfill \mbox{\textit{CAIE FP1 2009 Q6 [7]}}
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