| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2017 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Second derivative relations with hyperbolics |
| Difficulty | Challenging +1.2 This is a Further Maths question requiring differentiation of inverse hyperbolic functions and algebraic manipulation. Part (a) involves chain rule application with artanh and simplification to find a constant. Part (b) requires computing the second derivative and verifying an identity using the result from (a). While it involves multiple steps and careful algebra, the techniques are standard for FM students and the 'show that' format provides clear targets, making it moderately above average difficulty. |
| Spec | 1.07d Second derivatives: d^2y/dx^2 notation1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates4.07e Inverse hyperbolic: definitions, domains, ranges4.07f Inverse hyperbolic: logarithmic forms |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{d\left\{\text{artanh}\left(\frac{2x}{1+x^2}\right)\right\}}{dx} = \frac{1}{1-\left(\frac{2x}{1+x^2}\right)^2} \times \frac{2(1+x^2)-4x^2}{(1+x^2)^2}\) | M1A1 | M1: Correct attempt to differentiate artanh using chain and quotient (or product) rule; A1: Correct expression in any form |
| \(1 - \frac{dy}{dx} = 1 - \left(1 + \frac{2}{x^2-1}\right) = \frac{2}{1-x^2}\) | dM1, A1cso | dM1: Attempt \(1 - \frac{dy}{dx}\); A1: cao and cso |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(y = x - \frac{1}{2}\ln\left(\frac{1+\frac{2x}{1+x^2}}{1-\frac{2x}{1+x^2}}\right) = x - \frac{1}{2}\ln\left(\frac{x+1}{x-1}\right)^2 = x - \ln(x+1) + \ln(x-1)\) | ||
| \(\frac{d\{-\ln(x+1)+\ln(x-1)\}}{dx} = \frac{-1}{x+1} + \frac{1}{x-1}\) | M1A1 | M1: Attempt to differentiate; A1: Correct derivative |
| \(1 - \frac{dy}{dx} = 1 - \left(1 + \frac{2}{x^2-1}\right) = \frac{2}{1-x^2}\) | dM1, A1cso | dM1: Attempt \(1-\frac{dy}{dx}\); A1: cao and cso |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(y = x - \ln\left(\frac{x+1}{x-1}\right)\) | Obtained as Way 2 | |
| \(\frac{d\left(\ln\left(\frac{x+1}{x-1}\right)\right)}{dx} = \left(\frac{x-1}{x+1}\right) \times \frac{(x-1)-(x+1)}{(x-1)^2}\) | M1A1 | M1: Attempt to differentiate; A1: Correct derivative |
| \(1 - \frac{dy}{dx} = 1 - \left(1 + \frac{2}{x^2-1}\right) = \frac{2}{1-x^2}\) | dM1, A1cso | dM1: Attempt \(1-\frac{dy}{dx}\); A1: cao and cso |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(u = \text{artanh}\left(\frac{2x}{1+x^2}\right) \Rightarrow \tanh u = \frac{2x}{1+x^2}\) | ||
| \((1-\tanh^2 u)\frac{du}{dx} = \frac{2(1-x^2)}{(1+x^2)^2}\) | ||
| \(\frac{du}{dx} = \frac{2(1-x^2)}{(1+x^2)^2} \times \frac{1}{1 - \frac{4x^2}{(1+x^2)^2}}\) | M1A1 | M1: Attempt to differentiate to obtain \(d(\text{artanh}(\ldots))/dx\) as a function of \(x\); A1: Correct (unsimplified) derivative |
| Reduces to \(\frac{2}{1-x^2}\), then as main scheme | dM1A1cso |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{d^2y}{dx^2} = 2(1-x^2)^{-2} \times -2x \left(= \frac{-4x}{(1-x^2)^2}\right)\) | M1A1ft | M1: Attempt second derivative (quotient/product rule as in (a)); A1ft: Follow through their \(k\) or leave as \(k\) |
| \(\frac{d^2y}{dx^2} + x\left(1-\frac{dy}{dx}\right)^2 = \frac{-4x}{(1-x^2)^2} + x\left(\frac{2}{1-x^2}\right)^2 = 0\) | M1A1cso | M1: Attempt \(\frac{d^2y}{dx^2} + x\left(1-\frac{dy}{dx}\right)^2\) with their value for \(k\) from (a); A1: cso |
| Part (b) Total: 4 marks | Question Total: 8 marks |
## Question 3: $y = x - \text{artanh}\left(\frac{2x}{1+x^2}\right)$
### Part (a) — Way 1
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{d\left\{\text{artanh}\left(\frac{2x}{1+x^2}\right)\right\}}{dx} = \frac{1}{1-\left(\frac{2x}{1+x^2}\right)^2} \times \frac{2(1+x^2)-4x^2}{(1+x^2)^2}$ | M1A1 | M1: Correct attempt to differentiate artanh using chain and quotient (or product) rule; A1: Correct expression in any form |
| $1 - \frac{dy}{dx} = 1 - \left(1 + \frac{2}{x^2-1}\right) = \frac{2}{1-x^2}$ | dM1, A1cso | dM1: Attempt $1 - \frac{dy}{dx}$; A1: cao and cso |
### Part (a) — Way 2
| Working/Answer | Mark | Guidance |
|---|---|---|
| $y = x - \frac{1}{2}\ln\left(\frac{1+\frac{2x}{1+x^2}}{1-\frac{2x}{1+x^2}}\right) = x - \frac{1}{2}\ln\left(\frac{x+1}{x-1}\right)^2 = x - \ln(x+1) + \ln(x-1)$ | | |
| $\frac{d\{-\ln(x+1)+\ln(x-1)\}}{dx} = \frac{-1}{x+1} + \frac{1}{x-1}$ | M1A1 | M1: Attempt to differentiate; A1: Correct derivative |
| $1 - \frac{dy}{dx} = 1 - \left(1 + \frac{2}{x^2-1}\right) = \frac{2}{1-x^2}$ | dM1, A1cso | dM1: Attempt $1-\frac{dy}{dx}$; A1: cao and cso |
### Part (a) — Way 3
| Working/Answer | Mark | Guidance |
|---|---|---|
| $y = x - \ln\left(\frac{x+1}{x-1}\right)$ | | Obtained as Way 2 |
| $\frac{d\left(\ln\left(\frac{x+1}{x-1}\right)\right)}{dx} = \left(\frac{x-1}{x+1}\right) \times \frac{(x-1)-(x+1)}{(x-1)^2}$ | M1A1 | M1: Attempt to differentiate; A1: Correct derivative |
| $1 - \frac{dy}{dx} = 1 - \left(1 + \frac{2}{x^2-1}\right) = \frac{2}{1-x^2}$ | dM1, A1cso | dM1: Attempt $1-\frac{dy}{dx}$; A1: cao and cso |
### Part (a) — Way 4
| Working/Answer | Mark | Guidance |
|---|---|---|
| $u = \text{artanh}\left(\frac{2x}{1+x^2}\right) \Rightarrow \tanh u = \frac{2x}{1+x^2}$ | | |
| $(1-\tanh^2 u)\frac{du}{dx} = \frac{2(1-x^2)}{(1+x^2)^2}$ | | |
| $\frac{du}{dx} = \frac{2(1-x^2)}{(1+x^2)^2} \times \frac{1}{1 - \frac{4x^2}{(1+x^2)^2}}$ | M1A1 | M1: Attempt to differentiate to obtain $d(\text{artanh}(\ldots))/dx$ as a function of $x$; A1: Correct (unsimplified) derivative |
| Reduces to $\frac{2}{1-x^2}$, then as main scheme | dM1A1cso | |
**Part (a) Total: 4 marks**
### Part (b)
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{d^2y}{dx^2} = 2(1-x^2)^{-2} \times -2x \left(= \frac{-4x}{(1-x^2)^2}\right)$ | M1A1ft | M1: Attempt second derivative (quotient/product rule as in (a)); A1ft: Follow through their $k$ or leave as $k$ |
| $\frac{d^2y}{dx^2} + x\left(1-\frac{dy}{dx}\right)^2 = \frac{-4x}{(1-x^2)^2} + x\left(\frac{2}{1-x^2}\right)^2 = 0$ | M1A1cso | M1: Attempt $\frac{d^2y}{dx^2} + x\left(1-\frac{dy}{dx}\right)^2$ with their value for $k$ from (a); A1: cso |
**Part (b) Total: 4 marks | Question Total: 8 marks**3. Given that
$$y = x - \operatorname { artanh } \left( \frac { 2 x } { 1 + x ^ { 2 } } \right)$$
\begin{enumerate}[label=(\alph*)]
\item show that
$$1 - \frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { k } { 1 - x ^ { 2 } }$$
where $k$ is a constant to be found.
\item Hence, or otherwise, show that
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + x \left( 1 - \frac { \mathrm { d } y } { \mathrm {~d} x } \right) ^ { 2 } = 0$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2017 Q3 [8]}}