| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2011 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find second derivative d²y/dx² |
| Difficulty | Standard +0.8 This is a Further Maths implicit differentiation question requiring both first and second derivatives at a specific point. While the first derivative is standard FP1 fare, finding the second derivative implicitly requires careful differentiation of the first derivative expression (using quotient/product rule on terms containing dy/dx) and then substitution. This is more demanding than typical C4 implicit differentiation and requires multiple steps with careful algebraic manipulation, placing it moderately above average difficulty. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(3x^2 - 6y^2y' = 3y + 3xy'\) | B1B1 | One mark for each side |
| \(12 - 6y' = 3 + 6y' \Rightarrow y' = \frac{3}{4}\) | B1√ | Substitutes \((2,1)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(6x - \{6y^2y'' + 12y(y')^2\} = 3y' + 3y' + 3xy''\) | B1B1, B1 | One mark for differentiating both 1st and 3rd terms; one mark for each of 2nd and 4th terms |
| \(12 - (6y'' + \frac{27}{4}) = \frac{9}{4} + \frac{9}{4} + 6y'' \Rightarrow 12y'' = \frac{3}{4} \Rightarrow y'' = \frac{1}{16}\) | B1 | Substitute \((2,1)\) and \(y'(2) = \frac{3}{4}\) |
# Question 5:
## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $3x^2 - 6y^2y' = 3y + 3xy'$ | B1B1 | One mark for each side |
| $12 - 6y' = 3 + 6y' \Rightarrow y' = \frac{3}{4}$ | B1√ | Substitutes $(2,1)$ | **Part total: 3** |
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $6x - \{6y^2y'' + 12y(y')^2\} = 3y' + 3y' + 3xy''$ | B1B1, B1 | One mark for differentiating both 1st and 3rd terms; one mark for each of 2nd and 4th terms |
| $12 - (6y'' + \frac{27}{4}) = \frac{9}{4} + \frac{9}{4} + 6y'' \Rightarrow 12y'' = \frac{3}{4} \Rightarrow y'' = \frac{1}{16}$ | B1 | Substitute $(2,1)$ and $y'(2) = \frac{3}{4}$ | **Part total: 4** |
---5 The point $P ( 2,1 )$ lies on the curve with equation
$$x ^ { 3 } - 2 y ^ { 3 } = 3 x y$$
Find\\
(i) the value of $\frac { \mathrm { d } y } { \mathrm {~d} x }$ at $P$,\\
(ii) the value of $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ at $P$.
\hfill \mbox{\textit{CAIE FP1 2011 Q5 [7]}}