Question 5 - A-Level Maths

CAIE FP1 2008 November — Question 5 7 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2008
Session	November
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Implicit equations and differentiation
Type	Find second derivative d²y/dx²
Difficulty	Standard +0.3 This is a straightforward implicit differentiation question requiring two applications of the technique. Finding dy/dx involves standard implicit differentiation with product rule, then substituting the given point. Finding d²y/dx² requires differentiating again and substituting known values. While it involves multiple steps, the techniques are routine for Further Maths students and the question provides the first derivative value to verify, reducing potential errors.
Spec	1.07s Parametric and implicit differentiation

5 The curve $C$ has equation $$x ^ { 2 } - x y - 2 y ^ { 2 } = 4 .$$ Show that, at the point $A ( 2,0 )$ on $C , \frac { \mathrm {~d} y } { \mathrm {~d} x } = 2$. Find the value of $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ at $A$.

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Answer	Marks	Guidance
$2x - xy_1 - y - 4yy_1 = 0$	B1
$\Rightarrow \ldots y_1(2) = 2$ (AG)	B1
$2 - xy_2 - y_1 - y_1 - 4y_1^2 - 4yy_2 = 0$	M1A2
$2 - 2y_2(2) - 2 - 2 - 16 - 0 \Rightarrow y_2(2) = -9$	M1A1
OR
$2x - y = (4y + x)\frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{2x - y}{4y + x}$
$\Rightarrow \frac{d^2 y}{dx^2} = \frac{(4x + y)(-2y_1) - (2x - y)(1 + 4y_1)}{(x + 4y)^2}$	M1	Use of quotient rule, A1 $\Rightarrow$ term of numerator, A1 All correct
$= \frac{8 \times 0 - 4 \times 9}{4}$	M1	Substitution of values
$= -9$	A1

$2x - xy_1 - y - 4yy_1 = 0$ | B1 |

$\Rightarrow \ldots y_1(2) = 2$ (AG) | B1 |

$2 - xy_2 - y_1 - y_1 - 4y_1^2 - 4yy_2 = 0$ | M1A2 |

$2 - 2y_2(2) - 2 - 2 - 16 - 0 \Rightarrow y_2(2) = -9$ | M1A1 |

OR |

$2x - y = (4y + x)\frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{2x - y}{4y + x}$ | |

$\Rightarrow \frac{d^2 y}{dx^2} = \frac{(4x + y)(-2y_1) - (2x - y)(1 + 4y_1)}{(x + 4y)^2}$ | M1 | Use of quotient rule, A1 $\Rightarrow$ term of numerator, A1 All correct |

$= \frac{8 \times 0 - 4 \times 9}{4}$ | M1 | Substitution of values |

$= -9$ | A1 |

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5 The curve $C$ has equation

$$x ^ { 2 } - x y - 2 y ^ { 2 } = 4 .$$

Show that, at the point $A ( 2,0 )$ on $C , \frac { \mathrm {~d} y } { \mathrm {~d} x } = 2$.

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

\hfill \mbox{\textit{CAIE FP1 2008 Q5 [7]}}

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Q1 5 Q2 6 Q3 6 Q4 6 Q5 7 Q6 7 Q7 8 Q8 9 Q9 10 Q10 10 Q11 12 Q12 EITHER Q12 OR

Answer	Marks	Guidance
\(2x - xy_1 - y - 4yy_1 = 0\)	B1
\(\Rightarrow \ldots y_1(2) = 2\) (AG)	B1
\(2 - xy_2 - y_1 - y_1 - 4y_1^2 - 4yy_2 = 0\)	M1A2
\(2 - 2y_2(2) - 2 - 2 - 16 - 0 \Rightarrow y_2(2) = -9\)	M1A1
OR
\(2x - y = (4y + x)\frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{2x - y}{4y + x}\)
\(\Rightarrow \frac{d^2 y}{dx^2} = \frac{(4x + y)(-2y_1) - (2x - y)(1 + 4y_1)}{(x + 4y)^2}\)	M1	Use of quotient rule, A1 \(\Rightarrow\) term of numerator, A1 All correct
\(= \frac{8 \times 0 - 4 \times 9}{4}\)	M1	Substitution of values
\(= -9\)	A1