| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Implicit differentiation for d²y/dx² |
| Difficulty | Standard +0.8 This FP3 question combines numerical methods (improved Euler) with implicit differentiation to find d²y/dx² and Taylor series expansion. Part (a) is routine application of a given formula. Part (b)(i) requires careful implicit differentiation of a composite function. Part (b)(ii) demands systematic calculation of derivatives at x=0 to build a series expansion. The multi-step nature, need for precision in algebraic manipulation, and integration of multiple techniques places this above average difficulty, though it follows standard FP3 patterns without requiring exceptional insight. |
| Spec | 1.07s Parametric and implicit differentiation1.09f Trapezium rule: numerical integration4.08a Maclaurin series: find series for function4.10a General/particular solutions: of differential equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(k_1 = 0.1[(1)\ln(2)] = 0.1\ln 2 \ (= 0.069\ldots)\) | M1 | \(k_1 = 0.1[(2(0)+1)\ln(0+2)]\) OE seen or used |
| \(k_2 = 0.1 \times f(0.1, \ 2 + k_1) = 0.1(0.2+1)\ln(0.1 + 2 + 0.2693\ldots)\) | M1 | \(k_2 = 0.1(0.2+1)\ln(0.1 + 2 + \text{c's } k_1)\) OE |
| \(= 0.12\ln(2.1693\ldots) = 0.092(9293\ldots)\) | A1 | AWFW 0.092 to 0.093 inclusive. PI by final answer 2.081 or 2.0811… |
| \(y(0.1) = 2 + \frac{1}{2}(0.069(3\ldots) + 0.092(9\ldots))\) | m1 | \(2 + \frac{1}{2}(\text{c's } k_1 + \text{c's } k_2)\) but dep on previous two M marks scored. PI by 2.081 or 2.0811… |
| \(= 2.081\) (to 3dp) | A1 | CAO Must be 2.081 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{\text{d}}{\text{d}x}(\ln(x+y)) = \frac{1 + \frac{dy}{dx}}{x+y}\) seen or used | B1 | \(\frac{\text{d}}{\text{d}x}(\ln(x+y)) = \frac{1+\frac{dy}{dx}}{x+y}\) seen or used |
| Product rule used | M1 | Product rule used |
| \(2\ln(x+y) + (2x+1) \times \frac{1+(2x+1)\ln(x+y)}{x+y}\) | A1 | ACF for \(\frac{d^2y}{dx^2}\) in terms of \(x\) and \(y\) only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y(0) = 2, \quad y'(0) = \ln 2\) | B1 | \(y(0) = 2\) and \(y'(0) = \ln 2\) seen or used |
| \(y''(0) = 2\ln 2 + \frac{1 + \ln 2}{2}\) | B1F | Seen or used, ft on c's \(\frac{d^2y}{dx^2}\), an exact value at some stage which may be left unsimplified |
| \((y(x) =) \ 2 + x\ln 2 + x^2\left(\frac{1 + 5\ln 2}{4}\right)\cdots\) | B1F | \(2 + x(\text{c's } y'(0)) + \frac{x^2}{2}(\text{c's } y''(0))\) values must be exact but may be unsimplified |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y(0.1) = 2 + 0.0693\ldots + 0.01116\ldots = 2.080(479\ldots) = 2.080\) to 3dp | B1 | Must be 2.080 and dep on values in (b)(ii) being correctly found |
# Question 3:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $k_1 = 0.1[(1)\ln(2)] = 0.1\ln 2 \ (= 0.069\ldots)$ | M1 | $k_1 = 0.1[(2(0)+1)\ln(0+2)]$ OE seen or used |
| $k_2 = 0.1 \times f(0.1, \ 2 + k_1) = 0.1(0.2+1)\ln(0.1 + 2 + 0.2693\ldots)$ | M1 | $k_2 = 0.1(0.2+1)\ln(0.1 + 2 + \text{c's } k_1)$ OE |
| $= 0.12\ln(2.1693\ldots) = 0.092(9293\ldots)$ | A1 | AWFW 0.092 to 0.093 inclusive. PI by final answer 2.081 or 2.0811… |
| $y(0.1) = 2 + \frac{1}{2}(0.069(3\ldots) + 0.092(9\ldots))$ | m1 | $2 + \frac{1}{2}(\text{c's } k_1 + \text{c's } k_2)$ but dep on previous two M marks scored. PI by 2.081 or 2.0811… |
| $= 2.081$ (to 3dp) | A1 | CAO Must be 2.081 |
**Total: 5 marks**
## Part (b)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{\text{d}}{\text{d}x}(\ln(x+y)) = \frac{1 + \frac{dy}{dx}}{x+y}$ seen or used | B1 | $\frac{\text{d}}{\text{d}x}(\ln(x+y)) = \frac{1+\frac{dy}{dx}}{x+y}$ seen or used |
| Product rule used | M1 | Product rule used |
| $2\ln(x+y) + (2x+1) \times \frac{1+(2x+1)\ln(x+y)}{x+y}$ | A1 | ACF for $\frac{d^2y}{dx^2}$ in terms of $x$ and $y$ only |
**Total: 3 marks**
## Part (b)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y(0) = 2, \quad y'(0) = \ln 2$ | B1 | $y(0) = 2$ and $y'(0) = \ln 2$ seen or used |
| $y''(0) = 2\ln 2 + \frac{1 + \ln 2}{2}$ | B1F | Seen or used, ft on c's $\frac{d^2y}{dx^2}$, an exact value at some stage which may be left unsimplified |
| $(y(x) =) \ 2 + x\ln 2 + x^2\left(\frac{1 + 5\ln 2}{4}\right)\cdots$ | B1F | $2 + x(\text{c's } y'(0)) + \frac{x^2}{2}(\text{c's } y''(0))$ values must be exact but may be unsimplified |
**Total: 3 marks**
## Part (b)(iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y(0.1) = 2 + 0.0693\ldots + 0.01116\ldots = 2.080(479\ldots) = 2.080$ to 3dp | B1 | Must be 2.080 and dep on values in (b)(ii) being correctly found |
**Total: 1 mark**
---3
\begin{enumerate}[label=(\alph*)]
\item It is given that $y ( x )$ satisfies the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$
where
$$\mathrm { f } ( x , y ) = ( 2 x + 1 ) \ln ( x + y )$$
and
$$y ( 0 ) = 2$$
Use the improved Euler formula
$$y _ { r + 1 } = y _ { r } + \frac { 1 } { 2 } \left( k _ { 1 } + k _ { 2 } \right)$$
where $k _ { 1 } = h \mathrm { f } \left( x _ { r } , y _ { r } \right)$ and $k _ { 2 } = h \mathrm { f } \left( x _ { r } + h , y _ { r } + k _ { 1 } \right)$ and $h = 0.1$, to obtain an approximation to $y ( 0.1 )$, giving your answer to three decimal places.
\item It is given that $y ( x )$ satisfies the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = ( 2 x + 1 ) \ln ( x + y )$$
and $y = 2$ when $x = 0$.
\begin{enumerate}[label=(\roman*)]
\item Use implicit differentiation to find $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$, giving your answer in terms of $x$ and $y$.
\item Hence find the first three non-zero terms in the expansion, in ascending powers of $x$, of $y ( x )$. Give your answer in an exact form.
\item Use your answer to part (b)(ii) to obtain an approximation to $y ( 0.1 )$, giving your answer to three decimal places.\\[0pt]
[1 mark]
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2016 Q3 [12]}}