| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2016 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Inverse functions (inverse trig/hyperbolic) |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring differentiation of inverse trig functions, verification of a differential equation, derivation of a Maclaurin series using successive differentiation (not standard series), and error analysis. While systematic, it demands careful algebraic manipulation and understanding of series approximations beyond typical A-level, placing it moderately above average difficulty. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs4.08a Maclaurin series: find series for function4.08g Derivatives: inverse trig and hyperbolic functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| (i) \(\frac{dy}{dx}=\frac{2}{1+4x^2}\) | B1 | For first differentiation |
| \(\frac{d^2y}{dx^2}=-2(1+4x^2)^{-2}\times8x=\frac{-16x}{(1+4x^2)^2}=-4x\left(\frac{dy}{dx}\right)^2\) | M1 | Differentiate again and making comparison |
| A1 | ||
| [3] | ||
| (ii) When \(x=0\): \(y=0,\ \frac{dy}{dx}=2,\ \frac{d^2y}{dx^2}=0\) | B1 | Soi by final answer |
| \(\Rightarrow \frac{d^3y}{dx^3}+4\left(\frac{dy}{dx}\right)^2+8x\left(\frac{dy}{dx}\right)\frac{d^2y}{dx^2}=0\) | M1 | Differentiate the equation given |
| When \(x=0\): \(\frac{d^3y}{dx^3}+16=0\) | A1 | For \(-16\) www |
| \(\Rightarrow y=2x-\frac{8x^3}{3}\) | A1 | Final answer; SC4 final formula from formula book including sight of \((2x)^3\); SC2 if final form only seen (no working); SC0 if final form only and wrong |
| [4] | ||
| (iii) \(x=\frac{1}{2}\Rightarrow \tan^{-1}1=\frac{\pi}{4}\) | B1 | Soi |
| In series \(x=1-\frac{1}{3}=\frac{2}{3}\); \(\Rightarrow\) Estimate for \(\pi=\frac{8}{3}=2.666\ldots\); correct to 1sf \(=3\) | B1 | For showing to 1sf \(\pi=3\) which is correct but to 2sf \(\pi=2.7\) which is not, www |
| [2] |
# Question 5:
| Answer | Mark | Guidance |
|--------|------|----------|
| **(i)** $\frac{dy}{dx}=\frac{2}{1+4x^2}$ | B1 | For first differentiation |
| $\frac{d^2y}{dx^2}=-2(1+4x^2)^{-2}\times8x=\frac{-16x}{(1+4x^2)^2}=-4x\left(\frac{dy}{dx}\right)^2$ | M1 | Differentiate again and making comparison |
| | A1 | |
| | **[3]** | |
| **(ii)** When $x=0$: $y=0,\ \frac{dy}{dx}=2,\ \frac{d^2y}{dx^2}=0$ | B1 | Soi by final answer |
| $\Rightarrow \frac{d^3y}{dx^3}+4\left(\frac{dy}{dx}\right)^2+8x\left(\frac{dy}{dx}\right)\frac{d^2y}{dx^2}=0$ | M1 | Differentiate the equation given |
| When $x=0$: $\frac{d^3y}{dx^3}+16=0$ | A1 | For $-16$ www |
| $\Rightarrow y=2x-\frac{8x^3}{3}$ | A1 | Final answer; SC4 final formula from formula book including sight of $(2x)^3$; SC2 if final form only seen (no working); SC0 if final form only and wrong |
| | **[4]** | |
| **(iii)** $x=\frac{1}{2}\Rightarrow \tan^{-1}1=\frac{\pi}{4}$ | B1 | Soi |
| In series $x=1-\frac{1}{3}=\frac{2}{3}$; $\Rightarrow$ Estimate for $\pi=\frac{8}{3}=2.666\ldots$; correct to 1sf $=3$ | B1 | For showing to 1sf $\pi=3$ which is correct but to 2sf $\pi=2.7$ which is not, www |
| | **[2]** | |5 It is given that $y = \tan ^ { - 1 } 2 x$.\\
(i) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ and show that $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 x \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } = 0$.\\
(ii) Find the Maclaurin series for $y$ up to and including the term in $x ^ { 3 }$. Show all your working.\\
(iii) The result in part (ii), together with the value $x = \frac { 1 } { 2 }$, is used to find an estimate for $\pi$. Show that this estimate is only correct to 1 significant figure.
\hfill \mbox{\textit{OCR FP2 2016 Q5 [9]}}