CAIE FP1 2016 November — Question 8 11 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2016
SessionNovember
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImplicit equations and differentiation
TypeFind second derivative d²y/dx²
DifficultyChallenging +1.2 This is a Further Maths question requiring implicit differentiation to find dy/dx, setting it to zero for stationary points, then finding d²y/dx² implicitly to determine nature. While it involves multiple steps and implicit second derivatives (which many students find challenging), the algebraic manipulation is straightforward once the method is known, and the question provides significant guidance by giving the relationship x = -2y upfront. Slightly above average difficulty due to the Further Maths context and second derivative requirement, but remains a standard textbook-style question.
Spec1.07s Parametric and implicit differentiation

8 A curve \(C\) has equation \(x ^ { 2 } + 4 x y - y ^ { 2 } + 20 = 0\). Show that, at stationary points on \(C , x = - 2 y\). Find the coordinates of the stationary points on \(C\), and determine their nature by considering the value of \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) at the stationary points.

Question 8:
AnswerMarks Guidance
AnswerMarks Guidance
\(2x + 4(xy' + y) - 2yy' = 0\ (*)\)M1A1
\(y' = 0 \Rightarrow 2x + 4y = 0 \Rightarrow x = -2y\) (AG)A1
At stationary points \(4y^2 - 8y^2 - y^2 + 20 = 0 \Rightarrow 5y^2 = 20 \Rightarrow y = \pm 2\)M1A1
Coordinates of stationary points are \((4, -2)\) and \((-4, 2)\)A1
From \((*)\): \(x + 2(xy' + y) - yy' = 0\)
Differentiating: \(1 + 2(xy'' + y' + y') - (yy'' + [y']^2) = 0\)M1A1 or by quotient rule
At \((4, -2)\) with \(y' = 0\): \(1 + 8y'' + 2y'' = 0 \Rightarrow y'' = -\dfrac{1}{10} \Rightarrow\) maximumM1A1
At \((-4, 2)\) with \(y' = 0\): \(1 - 8y'' - 2y'' = 0 \Rightarrow y'' = \dfrac{1}{10} \Rightarrow\) minimumA1
## Question 8:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $2x + 4(xy' + y) - 2yy' = 0\ (*)$ | M1A1 | |
| $y' = 0 \Rightarrow 2x + 4y = 0 \Rightarrow x = -2y$ (AG) | A1 | |
| At stationary points $4y^2 - 8y^2 - y^2 + 20 = 0 \Rightarrow 5y^2 = 20 \Rightarrow y = \pm 2$ | M1A1 | |
| Coordinates of stationary points are $(4, -2)$ and $(-4, 2)$ | A1 | |
| From $(*)$: $x + 2(xy' + y) - yy' = 0$ | | |
| Differentiating: $1 + 2(xy'' + y' + y') - (yy'' + [y']^2) = 0$ | M1A1 | or by quotient rule |
| At $(4, -2)$ with $y' = 0$: $1 + 8y'' + 2y'' = 0 \Rightarrow y'' = -\dfrac{1}{10} \Rightarrow$ maximum | M1A1 | |
| At $(-4, 2)$ with $y' = 0$: $1 - 8y'' - 2y'' = 0 \Rightarrow y'' = \dfrac{1}{10} \Rightarrow$ minimum | A1 | |

---
8 A curve $C$ has equation $x ^ { 2 } + 4 x y - y ^ { 2 } + 20 = 0$. Show that, at stationary points on $C , x = - 2 y$.

Find the coordinates of the stationary points on $C$, and determine their nature by considering the value of $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ at the stationary points.

\hfill \mbox{\textit{CAIE FP1 2016 Q8 [11]}}