| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Maclaurin series for inverse hyperbolics |
| Difficulty | Challenging +1.2 This is a Further Maths question requiring differentiation of inverse hyperbolic functions and Maclaurin series construction. Part (i) involves chain rule application with the standard derivative of tanh^(-1), which is systematic but requires careful algebra. Part (ii) is straightforward once the derivatives are found. While this requires knowledge beyond standard A-level, it's a routine application of techniques for FP2 students with no novel problem-solving required. |
| Spec | 4.07d Differentiate/integrate: hyperbolic functions4.07e Inverse hyperbolic: definitions, domains, ranges4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{1}{1-\left(\frac{1-x}{3+x}\right)^2}\times\frac{-(3+x)-(1-x)}{(3+x)^2}\) | B1 | Sight of standard diffn for \(\tanh^{-1}x\) |
| M1 | Fn of fn and diffn of quotient | |
| A1 | Soi correct quotient (i.e. correct expression for 2nd part) | |
| \(\Rightarrow\frac{dy}{dx} = \left(\frac{-4}{(3+x)^2-(1-x)^2}\right) = \frac{k}{1+x}\) | A1 | |
| \(\Rightarrow\frac{dy}{dx} = \frac{-1}{2(1+x)}\) | A1 | Correct for \(y'\) |
| \(\Rightarrow\frac{d^2y}{dx^2} = \frac{1}{2(1+x)^2}\) | A1 | \(2^{nd}\) diffn (NB AG) |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| When \(x=0\), \(y = \tanh^{-1}\frac{1}{3}\) or \(\frac{1}{2}\ln 2\) or \(\ln\sqrt{2}\) | B1 | For \(1^{st}\) value (needs to be exact) |
| \(\frac{dy}{dx} = -\frac{1}{2}\), \(\frac{d^2y}{dx^2} = \frac{1}{2}\) | B1 | For both |
| \(\Rightarrow y = \tanh^{-1}\frac{1}{3}+\left(-\frac{1}{2}\right)x+\left(\frac{1}{2}\right)\frac{x^2}{2}\) | M1 | Use of correct Maclaurin's series |
| \(= \tanh^{-1}\frac{1}{3} - \frac{1}{2}x + \frac{x^2}{4}\) | A1 | Accept 0.347 |
| [4] |
# Question 3(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \frac{1}{1-\left(\frac{1-x}{3+x}\right)^2}\times\frac{-(3+x)-(1-x)}{(3+x)^2}$ | B1 | Sight of standard diffn for $\tanh^{-1}x$ |
| | M1 | Fn of fn **and** diffn of quotient |
| | A1 | Soi correct quotient (i.e. correct expression for 2nd part) |
| $\Rightarrow\frac{dy}{dx} = \left(\frac{-4}{(3+x)^2-(1-x)^2}\right) = \frac{k}{1+x}$ | A1 | |
| $\Rightarrow\frac{dy}{dx} = \frac{-1}{2(1+x)}$ | A1 | Correct for $y'$ |
| $\Rightarrow\frac{d^2y}{dx^2} = \frac{1}{2(1+x)^2}$ | A1 | $2^{nd}$ diffn (NB AG) |
| **[6]** | | |
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# Question 3(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $x=0$, $y = \tanh^{-1}\frac{1}{3}$ or $\frac{1}{2}\ln 2$ or $\ln\sqrt{2}$ | B1 | For $1^{st}$ value (needs to be exact) |
| $\frac{dy}{dx} = -\frac{1}{2}$, $\frac{d^2y}{dx^2} = \frac{1}{2}$ | B1 | For both |
| $\Rightarrow y = \tanh^{-1}\frac{1}{3}+\left(-\frac{1}{2}\right)x+\left(\frac{1}{2}\right)\frac{x^2}{2}$ | M1 | Use of correct Maclaurin's series |
| $= \tanh^{-1}\frac{1}{3} - \frac{1}{2}x + \frac{x^2}{4}$ | A1 | Accept 0.347 |
| **[4]** | | |
---3 It is given that $\mathrm { f } ( x ) = \tanh ^ { - 1 } \left( \frac { 1 - x } { 3 + x } \right)$ for $x > - 1$.\\
(i) Show that $\mathrm { f } ^ { \prime \prime } ( x ) = \frac { 1 } { 2 ( x + 1 ) ^ { 2 } }$.\\
(ii) Hence find the Maclaurin series for $\mathrm { f } ( x )$ up to and including the term in $x ^ { 2 }$.
\hfill \mbox{\textit{OCR FP2 2013 Q3 [10]}}