| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2011 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Maclaurin series for hyperbolic inverse functions |
| Difficulty | Standard +0.8 This is a Further Maths question requiring multiple derivatives of an inverse hyperbolic function (non-trivial differentiation using chain rule and quotient rule repeatedly), then applying Maclaurin series. While the procedure is standard for FP2, the algebraic manipulation in part (i) is substantial and error-prone, and inverse hyperbolic functions are inherently more challenging than standard functions. Slightly above average difficulty even for Further Maths students. |
| Spec | 1.07l Derivative of ln(x): and related functions4.07e Inverse hyperbolic: definitions, domains, ranges4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(x)=\tanh^{-1}x,\; f'(x)=\frac{1}{1-x^2},\; f''(x)=\frac{2x}{(1-x^2)^2}\) | M1 | For quoting \(f'(x)=\frac{1}{1\pm x^2}\) and attempting to differentiate \(f'(x)\) |
| A1 | For \(f''(x)\) correct www | |
| \(f'''(x)=\frac{2(1-x^2)^2-2x\cdot2(1-x^2)\cdot(-2x)}{(1-x^2)^4}\) OR \(\frac{2x\cdot4x}{(1-x^2)^3}+\frac{2}{(1-x^2)^2}\) | M1 | For using quotient OR product rule on \(f''(x)\) |
| \(=\frac{2(1-x^2)^2+8x^2(1-x^2)}{(1-x^2)^4}\) OR \(\frac{8x^2}{(1-x^2)^3}+\frac{2(1-x^2)}{(1-x^2)^3}\) | A1 | For correct unsimplified \(f'''(x)\) |
| \(=\frac{2(1+3x^2)}{(1-x^2)^3}\) | A1 5 | For simplified \(f'''(x)\) www AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(0)=0,\; f'(0)=1,\; f''(0)=0\) | B1\(\checkmark\) | For all values correct (may be implied) f.t. from (i) |
| \(f'''(0)=2 \Rightarrow \tanh^{-1}x = x+\frac{1}{3}x^3\) | M1 | For evaluating \(f'''(0)\) and using Maclaurin expansion |
| A1 3 | For correct series |
## Question 2(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(x)=\tanh^{-1}x,\; f'(x)=\frac{1}{1-x^2},\; f''(x)=\frac{2x}{(1-x^2)^2}$ | M1 | For quoting $f'(x)=\frac{1}{1\pm x^2}$ and attempting to differentiate $f'(x)$ |
| | A1 | For $f''(x)$ correct **www** |
| $f'''(x)=\frac{2(1-x^2)^2-2x\cdot2(1-x^2)\cdot(-2x)}{(1-x^2)^4}$ OR $\frac{2x\cdot4x}{(1-x^2)^3}+\frac{2}{(1-x^2)^2}$ | M1 | For using quotient OR product rule on $f''(x)$ |
| $=\frac{2(1-x^2)^2+8x^2(1-x^2)}{(1-x^2)^4}$ OR $\frac{8x^2}{(1-x^2)^3}+\frac{2(1-x^2)}{(1-x^2)^3}$ | A1 | For correct unsimplified $f'''(x)$ |
| $=\frac{2(1+3x^2)}{(1-x^2)^3}$ | A1 **5** | For simplified $f'''(x)$ **www AG** |
## Question 2(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(0)=0,\; f'(0)=1,\; f''(0)=0$ | B1$\checkmark$ | For all values correct (may be implied) f.t. from **(i)** |
| $f'''(0)=2 \Rightarrow \tanh^{-1}x = x+\frac{1}{3}x^3$ | M1 | For evaluating $f'''(0)$ and using Maclaurin expansion |
| | A1 **3** | For correct series |
---2 It is given that $\mathrm { f } ( x ) = \tanh ^ { - 1 } x$.\\
(i) Show that $\mathrm { f } ^ { \prime \prime \prime } ( x ) = \frac { 2 \left( 1 + 3 x ^ { 2 } \right) } { \left( 1 - x ^ { 2 } \right) ^ { 3 } }$.\\
(ii) Hence find the Maclaurin series for $\mathrm { f } ( x )$, up to and including the term in $x ^ { 3 }$.
\hfill \mbox{\textit{OCR FP2 2011 Q2 [8]}}