| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Prove inverse hyperbolic logarithmic form |
| Difficulty | Standard +0.8 This is a structured multi-part Further Maths question on hyperbolic functions requiring standard derivations (inverse hyperbolic form, differentiation) and integration by parts. While it involves several techniques and careful algebra, each part follows established methods without requiring novel insight. The integration by parts in (iv) is the most challenging element but is a standard application once the setup is recognized. |
| Spec | 1.08i Integration by parts4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07e Inverse hyperbolic: definitions, domains, ranges4.07f Inverse hyperbolic: logarithmic forms |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\tanh t = \dfrac{e^t - e^{-t}}{e^t + e^{-t}}\) | B1 | Or \(\dfrac{e^{2t}-1}{e^{2t}+1}\). Condone other variables used |
| Correct shape | G1 | |
| Asymptotes at \(y = \pm 1\). Dependent on first G1 | G1 | If text and graph conflict, mark what is shown on the graph |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = \text{artanh}\, x \Rightarrow x = \tanh y \Rightarrow x = \dfrac{e^y - e^{-y}}{e^y + e^{-y}}\) | ||
| \(\Rightarrow x(e^y + e^{-y}) = e^y - e^{-y}\) | M1 | First step in rearrangement. Or \(x = \dfrac{e^{2y}-1}{e^{2y}+1}\). Variables the right way round, clearing fractions |
| \(\Rightarrow e^{-y}(1+x) = e^y(1-x)\) | ||
| \(\Rightarrow e^{2y} = \dfrac{1+x}{1-x}\) | M1, A1 | Obtaining \(e^{2y}\) in terms of \(x\). Dependent on first M1 |
| \(\Rightarrow 2y = \ln\!\left(\dfrac{1+x}{1-x}\right)\) | ||
| \(\Rightarrow \text{artanh}\, x = \dfrac{1}{2}\ln\!\left(\dfrac{1+x}{1-x}\right)\) | E1 | Answer given |
| Valid for \(-1 < x < 1\) | B1 | Independent |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\tanh y = x \Rightarrow \text{sech}^2 y\,\dfrac{dy}{dx} = 1\) | ||
| \(\Rightarrow \dfrac{dy}{dx} = \dfrac{1}{\text{sech}^2 y} = \dfrac{1}{1 - \tanh^2 y}\) | M1 | Differentiating and explicitly attempting to express in terms of \(\tanh y\) |
| \(= \dfrac{1}{1-x^2}\) | E1 | Correctly obtained |
| \(y = \dfrac{1}{2}\ln(1+x) - \dfrac{1}{2}\ln(1-x)\) | ||
| \(\Rightarrow \dfrac{dy}{dx} = \dfrac{1}{2}\times\dfrac{1}{1+x} - \dfrac{1}{2}\times\dfrac{-1}{1-x}\) | M1, A1 | Attempting logarithmic diff. Any correct form |
| \(= \dfrac{1}{2}\times\dfrac{1-x+1+x}{(1+x)(1-x)} = \dfrac{1}{1-x^2}\) | E1 | Convincing manipulation |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^{\frac{1}{2}} \text{artanh}\, x\, dx = \bigl[x\,\text{artanh}\, x\bigr]_0^{\frac{1}{2}} - \int_0^{\frac{1}{2}} \dfrac{x}{1-x^2}\,dx\) | M1 | Using integration by parts, with \(u = \text{artanh}\, x\), \(v'=1\), \(v=x\) |
| This line correct | A1 | Condone omitted limits |
| \(= \dfrac{1}{2}\,\text{artanh}\,\dfrac{1}{2} - \left[-\dfrac{1}{2}\ln(1-x^2)\right]_0^{\frac{1}{2}}\) | A1 | \(-\dfrac{1}{2}\ln(1-x^2)\). Or \(-\dfrac{1}{2}\ln(1-x) - \dfrac{1}{2}\ln(1+x)\) |
| \(= \dfrac{1}{4}\ln\!\left(\dfrac{1+\frac{1}{2}}{1-\frac{1}{2}}\right) + \dfrac{1}{2}\ln\dfrac{3}{4}\) | M1 | Applying limits and using (ii) in result of integration. Must be exact |
| \(= \dfrac{1}{4}\ln 3 + \dfrac{1}{4}\ln\dfrac{9}{16}\) | ||
| \(= \dfrac{1}{4}\ln\dfrac{27}{16}\) | E1 | Convincing manipulation. Answer given |
| [5] |
## Question 4:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\tanh t = \dfrac{e^t - e^{-t}}{e^t + e^{-t}}$ | B1 | Or $\dfrac{e^{2t}-1}{e^{2t}+1}$. Condone other variables used |
| Correct shape | G1 | |
| Asymptotes at $y = \pm 1$. Dependent on first G1 | G1 | If text and graph conflict, mark what is shown on the graph |
| **[3]** | | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = \text{artanh}\, x \Rightarrow x = \tanh y \Rightarrow x = \dfrac{e^y - e^{-y}}{e^y + e^{-y}}$ | | |
| $\Rightarrow x(e^y + e^{-y}) = e^y - e^{-y}$ | M1 | First step in rearrangement. Or $x = \dfrac{e^{2y}-1}{e^{2y}+1}$. Variables the right way round, clearing fractions |
| $\Rightarrow e^{-y}(1+x) = e^y(1-x)$ | | |
| $\Rightarrow e^{2y} = \dfrac{1+x}{1-x}$ | M1, A1 | Obtaining $e^{2y}$ in terms of $x$. Dependent on first M1 |
| $\Rightarrow 2y = \ln\!\left(\dfrac{1+x}{1-x}\right)$ | | |
| $\Rightarrow \text{artanh}\, x = \dfrac{1}{2}\ln\!\left(\dfrac{1+x}{1-x}\right)$ | E1 | Answer given |
| Valid for $-1 < x < 1$ | B1 | Independent |
| **[5]** | | |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\tanh y = x \Rightarrow \text{sech}^2 y\,\dfrac{dy}{dx} = 1$ | | |
| $\Rightarrow \dfrac{dy}{dx} = \dfrac{1}{\text{sech}^2 y} = \dfrac{1}{1 - \tanh^2 y}$ | M1 | Differentiating and explicitly attempting to express in terms of $\tanh y$ |
| $= \dfrac{1}{1-x^2}$ | E1 | Correctly obtained |
| $y = \dfrac{1}{2}\ln(1+x) - \dfrac{1}{2}\ln(1-x)$ | | |
| $\Rightarrow \dfrac{dy}{dx} = \dfrac{1}{2}\times\dfrac{1}{1+x} - \dfrac{1}{2}\times\dfrac{-1}{1-x}$ | M1, A1 | Attempting logarithmic diff. Any correct form |
| $= \dfrac{1}{2}\times\dfrac{1-x+1+x}{(1+x)(1-x)} = \dfrac{1}{1-x^2}$ | E1 | Convincing manipulation |
| **[5]** | | |
### Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^{\frac{1}{2}} \text{artanh}\, x\, dx = \bigl[x\,\text{artanh}\, x\bigr]_0^{\frac{1}{2}} - \int_0^{\frac{1}{2}} \dfrac{x}{1-x^2}\,dx$ | M1 | Using integration by parts, with $u = \text{artanh}\, x$, $v'=1$, $v=x$ |
| This line correct | A1 | Condone omitted limits |
| $= \dfrac{1}{2}\,\text{artanh}\,\dfrac{1}{2} - \left[-\dfrac{1}{2}\ln(1-x^2)\right]_0^{\frac{1}{2}}$ | A1 | $-\dfrac{1}{2}\ln(1-x^2)$. Or $-\dfrac{1}{2}\ln(1-x) - \dfrac{1}{2}\ln(1+x)$ |
| $= \dfrac{1}{4}\ln\!\left(\dfrac{1+\frac{1}{2}}{1-\frac{1}{2}}\right) + \dfrac{1}{2}\ln\dfrac{3}{4}$ | M1 | Applying limits and using (ii) in result of integration. Must be exact |
| $= \dfrac{1}{4}\ln 3 + \dfrac{1}{4}\ln\dfrac{9}{16}$ | | |
| $= \dfrac{1}{4}\ln\dfrac{27}{16}$ | E1 | Convincing manipulation. Answer given |
| **[5]** | | |
---4 (i) Define tanh $t$ in terms of exponential functions. Sketch the graph of $\tanh t$.\\
(ii) Show that $\operatorname { artanh } x = \frac { 1 } { 2 } \ln \left( \frac { 1 + x } { 1 - x } \right)$. State the set of values of $x$ for which this equation is valid.\\
(iii) Differentiate the equation $\tanh y = x$ with respect to $x$ and hence show that the derivative of $\operatorname { artanh } x$ is $\frac { 1 } { 1 - x ^ { 2 } }$.
Show that this result may also be obtained by differentiating the equation in part (ii).\\
(iv) By considering $\operatorname { artanh } x$ as $1 \times \operatorname { artanh } x$ and using integration by parts, show that
$$\int _ { 0 } ^ { \frac { 1 } { 2 } } \operatorname { artanh } x \mathrm {~d} x = \frac { 1 } { 4 } \ln \frac { 27 } { 16 }$$
\hfill \mbox{\textit{OCR MEI FP2 2012 Q4 [18]}}