| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Second derivative relations with hyperbolics |
| Difficulty | Standard +0.3 This is a straightforward Further Maths calculus question requiring standard techniques: proving a derivative formula using the definition of arcoth (or implicit differentiation), then applying chain rule and product rule to find the second derivative. While it involves hyperbolic functions (a Further Maths topic), the actual calculus manipulations are routine and methodical with no novel insight required. The algebraic simplification is manageable, making this slightly easier than average overall. |
| Spec | 4.07e Inverse hyperbolic: definitions, domains, ranges |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \text{arcoth}\, x \Rightarrow \coth y = x\) | B1 | Changes from arcoth to coth correctly. May be implied by e.g. \(\tanh y = \frac{1}{x}\) |
| \(x = \frac{\cosh y}{\sinh y} \Rightarrow \frac{dx}{dy} = \frac{\sinh^2 y - \cosh^2 y}{\sinh^2 y}\) | M1 | Uses \(\coth y = \frac{\cosh y}{\sinh y}\) and attempts product or quotient rule |
| \(\frac{dx}{dy} = -\text{cosech}^2\, y = 1 - \coth^2 y \Rightarrow \frac{dy}{dx} = \frac{1}{1-\coth^2 y} = \frac{1}{1-x^2}\)* | A1* | Correct completion with no errors seen and an intermediate step shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \text{arcoth}\, x \Rightarrow \coth y = x\) | B1 | As above |
| \(-\text{cosech}^2\, y \frac{dy}{dx} = 1\) or \(-\text{cosech}^2\, y = \frac{dx}{dy}\) | M1 | \(\pm\text{cosech}^2\, y \frac{dy}{dx} = 1\) or \(\pm\text{cosech}^2\, y = \frac{dx}{dy}\) |
| \(\coth^2 y - 1 = \text{cosech}^2\, y \Rightarrow \frac{dy}{dx} = \frac{1}{1-x^2}\)* | A1* | Correct completion with no errors seen and intermediate step shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \text{arcoth}\, x = \frac{1}{2}\ln\left(\frac{1+x}{x-1}\right)\) | B1 | Correct ln form for arcoth |
| \(\frac{dy}{dx} = \frac{1}{2}\left[\frac{x-1}{x+1}\times\frac{(x-1)-(x+1)}{(x-1)^2}\right]\) or \(\frac{dy}{dx} = \frac{1}{2(x+1)} - \frac{1}{2(x-1)}\) | M1 | Attempts to differentiate using chain rule and quotient rule, or writes as two logarithms and differentiates |
| \(\frac{dy}{dx} = \frac{1}{1-x^2}\) | A1 | Correct completion with no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(y = \text{arcoth}\, x \Rightarrow \coth y = x\) | B1 | Changes from arcoth to coth correctly; may be implied by e.g. \(\tanh y = \frac{1}{x}\) |
| \(\tanh y = \frac{1}{x} \Rightarrow -\frac{1}{x^2} = \text{sech}^2 y \frac{dy}{dx}\) | M1 | \(\pm\frac{1}{x^2} = \pm\text{sech}^2 y \frac{dx}{dy}\) |
| \(-\frac{1}{x^2} = \left(1 - \frac{1}{x^2}\right)\frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{1}{1-x^2}\) | A1* | Correct completion with no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(y = \text{arcoth}\, x = \text{artanh}\left(\frac{1}{x}\right)\) | B1 | Expresses arcoth in terms of artanh correctly |
| \(\frac{dy}{dx} = \frac{1}{1-\left(\frac{1}{x}\right)^2} \times -x^{-2}\) | M1 | Differentiates using the chain rule |
| \(= \frac{-1}{x^2-1} = \frac{1}{1-x^2}\) | A1* | Correct completion with no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(y = (\text{arcoth}\, x)^2 \Rightarrow \frac{dy}{dx} = \frac{2\,\text{arcoth}\, x}{1-x^2}\) | B1 | Correct first derivative |
| \(\frac{d^2y}{dx^2} = \frac{2}{1-x^2}(1-x^2)^{-1} + 4x\,\text{arcoth}\, x \cdot (1-x^2)^{-2}\) | M1A1 | M1: attempts product/quotient rule on expression of form \(\frac{k\,\text{arcoth}\, x}{1-x^2}\); Product rule requires \(\pm P(1-x^2)^{-2} \pm Qx\,\text{arcoth}\, x(1-x^2)^{-2}\); Quotient rule requires \(\frac{\pm P \pm Qx\,\text{arcoth}\, x}{(1-x^2)^2}\) |
| \((1-x^2)\frac{d^2y}{dx^2} - 2x\frac{dy}{dx} = (1-x^2)\left(\frac{4x\,\text{arcoth}\, x+2}{(1-x^2)^2}\right) - 2x\left(\frac{2\,\text{arcoth}\, x}{1-x^2}\right)\) | M1 | Substitutes first and second derivatives into lhs of DE, or multiplies through by \((1-x^2)\) and replaces \(2(\text{arcoth}\, x)\times\frac{1}{1-x^2}\) by \(\frac{dy}{dx}\) |
| \((1-x^2)\frac{d^2y}{dx^2} - 2x\frac{dy}{dx} = \frac{2}{1-x^2}\) | A1cso | Correct conclusion with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(y = (\text{arcoth}\, x)^2 \Rightarrow \frac{dy}{dx} = 2(\text{arcoth}\, x)\times\frac{1}{1-x^2}\) | B1 | Correct first derivative |
| \((1-x^2)\frac{dy}{dx} = 2\,\text{arcoth}\, x \Rightarrow (1-x^2)\frac{d^2y}{dx^2} - 2x\frac{dy}{dx} = \ldots\) | M1A1 | M1: multiplies through by \(1-x^2\) and attempts product rule on \((1-x^2)\frac{dy}{dx}\); requires \((1-x^2)\frac{d^2y}{dx^2} \pm Px\frac{dy}{dx}\); A1: correct differentiation |
| \(\frac{d(2\,\text{arcoth}\, x)}{dx} = \frac{2}{1-x^2}\) | M1 | Differentiates rhs using result from part (a) |
| \((1-x^2)\frac{d^2y}{dx^2} - 2x\frac{dy}{dx} = \frac{2}{1-x^2}\) | A1cso | Correct conclusion with no errors |
## Question 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \text{arcoth}\, x \Rightarrow \coth y = x$ | B1 | Changes from arcoth to coth correctly. May be implied by e.g. $\tanh y = \frac{1}{x}$ |
| $x = \frac{\cosh y}{\sinh y} \Rightarrow \frac{dx}{dy} = \frac{\sinh^2 y - \cosh^2 y}{\sinh^2 y}$ | M1 | Uses $\coth y = \frac{\cosh y}{\sinh y}$ and attempts product or quotient rule |
| $\frac{dx}{dy} = -\text{cosech}^2\, y = 1 - \coth^2 y \Rightarrow \frac{dy}{dx} = \frac{1}{1-\coth^2 y} = \frac{1}{1-x^2}$* | A1* | Correct completion with no errors seen and an intermediate step shown |
**Alternative 2:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \text{arcoth}\, x \Rightarrow \coth y = x$ | B1 | As above |
| $-\text{cosech}^2\, y \frac{dy}{dx} = 1$ or $-\text{cosech}^2\, y = \frac{dx}{dy}$ | M1 | $\pm\text{cosech}^2\, y \frac{dy}{dx} = 1$ or $\pm\text{cosech}^2\, y = \frac{dx}{dy}$ |
| $\coth^2 y - 1 = \text{cosech}^2\, y \Rightarrow \frac{dy}{dx} = \frac{1}{1-x^2}$* | A1* | Correct completion with no errors seen and intermediate step shown |
**Alternative 5:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \text{arcoth}\, x = \frac{1}{2}\ln\left(\frac{1+x}{x-1}\right)$ | B1 | Correct ln form for arcoth |
| $\frac{dy}{dx} = \frac{1}{2}\left[\frac{x-1}{x+1}\times\frac{(x-1)-(x+1)}{(x-1)^2}\right]$ or $\frac{dy}{dx} = \frac{1}{2(x+1)} - \frac{1}{2(x-1)}$ | M1 | Attempts to differentiate using chain rule and quotient rule, or writes as two logarithms and differentiates |
| $\frac{dy}{dx} = \frac{1}{1-x^2}$ | A1 | Correct completion with no errors seen |
# Question (a): Derivative of arcoth x
| Working | Mark | Notes |
|---------|------|-------|
| $y = \text{arcoth}\, x \Rightarrow \coth y = x$ | B1 | Changes from arcoth to coth correctly; may be implied by e.g. $\tanh y = \frac{1}{x}$ |
| $\tanh y = \frac{1}{x} \Rightarrow -\frac{1}{x^2} = \text{sech}^2 y \frac{dy}{dx}$ | M1 | $\pm\frac{1}{x^2} = \pm\text{sech}^2 y \frac{dx}{dy}$ |
| $-\frac{1}{x^2} = \left(1 - \frac{1}{x^2}\right)\frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{1}{1-x^2}$ | A1* | Correct completion with no errors seen |
**Alternative 7:**
| Working | Mark | Notes |
|---------|------|-------|
| $y = \text{arcoth}\, x = \text{artanh}\left(\frac{1}{x}\right)$ | B1 | Expresses arcoth in terms of artanh correctly |
| $\frac{dy}{dx} = \frac{1}{1-\left(\frac{1}{x}\right)^2} \times -x^{-2}$ | M1 | Differentiates using the chain rule |
| $= \frac{-1}{x^2-1} = \frac{1}{1-x^2}$ | A1* | Correct completion with no errors seen |
---
# Question (b): Second derivative proof
| Working | Mark | Notes |
|---------|------|-------|
| $y = (\text{arcoth}\, x)^2 \Rightarrow \frac{dy}{dx} = \frac{2\,\text{arcoth}\, x}{1-x^2}$ | B1 | Correct first derivative |
| $\frac{d^2y}{dx^2} = \frac{2}{1-x^2}(1-x^2)^{-1} + 4x\,\text{arcoth}\, x \cdot (1-x^2)^{-2}$ | M1A1 | M1: attempts product/quotient rule on expression of form $\frac{k\,\text{arcoth}\, x}{1-x^2}$; Product rule requires $\pm P(1-x^2)^{-2} \pm Qx\,\text{arcoth}\, x(1-x^2)^{-2}$; Quotient rule requires $\frac{\pm P \pm Qx\,\text{arcoth}\, x}{(1-x^2)^2}$ |
| $(1-x^2)\frac{d^2y}{dx^2} - 2x\frac{dy}{dx} = (1-x^2)\left(\frac{4x\,\text{arcoth}\, x+2}{(1-x^2)^2}\right) - 2x\left(\frac{2\,\text{arcoth}\, x}{1-x^2}\right)$ | M1 | Substitutes first and second derivatives into lhs of DE, **or** multiplies through by $(1-x^2)$ and replaces $2(\text{arcoth}\, x)\times\frac{1}{1-x^2}$ by $\frac{dy}{dx}$ |
| $(1-x^2)\frac{d^2y}{dx^2} - 2x\frac{dy}{dx} = \frac{2}{1-x^2}$ | A1cso | Correct conclusion with no errors |
**Alternative 1:**
| Working | Mark | Notes |
|---------|------|-------|
| $y = (\text{arcoth}\, x)^2 \Rightarrow \frac{dy}{dx} = 2(\text{arcoth}\, x)\times\frac{1}{1-x^2}$ | B1 | Correct first derivative |
| $(1-x^2)\frac{dy}{dx} = 2\,\text{arcoth}\, x \Rightarrow (1-x^2)\frac{d^2y}{dx^2} - 2x\frac{dy}{dx} = \ldots$ | M1A1 | M1: multiplies through by $1-x^2$ and attempts product rule on $(1-x^2)\frac{dy}{dx}$; requires $(1-x^2)\frac{d^2y}{dx^2} \pm Px\frac{dy}{dx}$; A1: correct differentiation |
| $\frac{d(2\,\text{arcoth}\, x)}{dx} = \frac{2}{1-x^2}$ | M1 | Differentiates rhs using result from part (a) |
| $(1-x^2)\frac{d^2y}{dx^2} - 2x\frac{dy}{dx} = \frac{2}{1-x^2}$ | A1cso | Correct conclusion with no errors |
---3. (a) Prove that
$$\frac { \mathrm { d } ( \operatorname { arcoth } x ) } { \mathrm { d } x } = \frac { 1 } { 1 - x ^ { 2 } }$$
Given that $y = ( \operatorname { arcoth } x ) ^ { 2 }$,\\
(b) show that
$$\left( 1 - x ^ { 2 } \right) \frac { d ^ { 2 } y } { d x ^ { 2 } } - 2 x \frac { d y } { d x } = \frac { k } { 1 - x ^ { 2 } }$$
where $k$ is a constant to be determined.
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\hfill \mbox{\textit{Edexcel FP3 2016 Q3 [8]}}