| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find dy/dx at a point |
| Difficulty | Standard +0.8 This is a Further Maths implicit differentiation question requiring multiple derivatives at a point to build a Maclaurin series. Part (a) involves straightforward implicit differentiation and substitution, but part (b) requires finding the second derivative implicitly (which involves product rule and chain rule on already complex expressions) then evaluating at x=0. The algebraic manipulation is substantial and error-prone, placing it moderately above average difficulty. |
| Spec | 1.07s Parametric and implicit differentiation4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{d}{dx}\left((x+1)y\right) = (x+1)y' + y\) | B1 | Differentiates \((x+1)y\) correctly |
| \(\frac{d}{dx}\left((x+y+1)^3\right) = 3(x+y+1)^2(y'+1)\) | B1 | Differentiates \((x+y+1)^3\) correctly |
| \((1)y' + 3(y'+1) = 0 \Rightarrow y' = -\frac{3}{4}\) | B1 | Substitutes \((0,0)\), AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((x+1)y'' + 2y'\) | B1 | Differentiates \((x+1)y' + y\) |
| \(+3(x+y+1)^2\,y'' + 6(1+y')^2(x+y+1) = 0\) | B1 B1 | Differentiates \(3(x+y+1)^2(1+y')\). Expect \(6(x+y+1)(1+y') + 6(x+y+1)(1+y')y'\) instead of \(6(1+y')^2(x+y+1)\) |
| \(y'' + 2\left(-\frac{3}{4}\right) + 3y'' + 6\left(\frac{1}{4}\right)^2 = 0\) | M1 | Substitutes \((0,0)\) |
| \(y'' = \frac{9}{32}\) | A1 | |
| \(y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 = -\frac{3}{4}x + \frac{9}{64}x^2\) | M1 A1 | Finds Maclaurin's series |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{d}{dx}\left((x+1)y\right) = (x+1)y' + y$ | B1 | Differentiates $(x+1)y$ correctly |
| $\frac{d}{dx}\left((x+y+1)^3\right) = 3(x+y+1)^2(y'+1)$ | B1 | Differentiates $(x+y+1)^3$ correctly |
| $(1)y' + 3(y'+1) = 0 \Rightarrow y' = -\frac{3}{4}$ | B1 | Substitutes $(0,0)$, AG |
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(x+1)y'' + 2y'$ | B1 | Differentiates $(x+1)y' + y$ |
| $+3(x+y+1)^2\,y'' + 6(1+y')^2(x+y+1) = 0$ | B1 B1 | Differentiates $3(x+y+1)^2(1+y')$. Expect $6(x+y+1)(1+y') + 6(x+y+1)(1+y')y'$ instead of $6(1+y')^2(x+y+1)$ |
| $y'' + 2\left(-\frac{3}{4}\right) + 3y'' + 6\left(\frac{1}{4}\right)^2 = 0$ | M1 | Substitutes $(0,0)$ |
| $y'' = \frac{9}{32}$ | A1 | |
| $y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 = -\frac{3}{4}x + \frac{9}{64}x^2$ | M1 A1 | Finds Maclaurin's series |5 The variables $x$ and $y$ are such that $y = 0$ when $x = 0$ and
$$( x + 1 ) y + ( x + y + 1 ) ^ { 3 } = 1$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \mathrm { dy } } { \mathrm { dx } } = - \frac { 3 } { 4 }$ when $x = 0$.
\item Find the Maclaurin's series for $y$ up to and including the term in $x ^ { 2 }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2022 Q5 [10]}}