FP3 June 2016 Q3
3.
Given that \(y = (\text{arcoth}\,x)^2\),
| Scheme | Marks |
|---|---|
| \(y = \text{arcoth}\,x \Rightarrow \coth y = x\) or e.g. \(u = \text{arcoth}\,x \Rightarrow \coth u = x\) | B1 |
| \(x = \dfrac{\cosh y}{\sinh y} \Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}y} = \dfrac{\sinh^2 y - \cosh^2 y}{\sinh^2 y}\left(= -\dfrac{1}{\sinh^2 y}\right)\) | M1 |
| \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = -\text{cosech}^2 y = 1 - \coth^2 y\) \(\Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{1 - \coth^2 y} = \dfrac{1}{1 - x^2}\) * | A1* |
| (3) |
Notes
B1: Changes from arcoth to coth correctly. This may be implied by e.g. \(\tanh y = \dfrac{1}{x}\)
M1: Uses \(\coth y = \dfrac{\cosh y}{\sinh y}\) and attempts product or quotient rule
A1*: Correct completion with no errors seen and an intermediate step shown.
(a) Alternative 2
| Scheme | Marks |
|---|---|
| \(y = \text{arcoth}\,x \Rightarrow \coth y = x\) | B1 |
| \(-\text{cosech}^2 y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1\) or \(-\text{cosech}^2 y = \dfrac{\mathrm{d}x}{\mathrm{d}y}\) \(\left(\Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{1}{\text{cosech}^2 y}\right)\) | M1 |
| \(\coth^2 y - 1 = \text{cosech}^2 y \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{1 - x^2}\) * | A1* |
B1: Changes from arcoth to coth correctly. This may be implied by e.g. \(\tanh y = \dfrac{1}{x}\)
M1: \(\pm\text{cosech}^2 y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1\) or \(\pm\text{cosech}^2 y = \dfrac{\mathrm{d}x}{\mathrm{d}y}\)
A1*: Correct completion with no errors seen and an intermediate step shown.
(a) Alternative 3
| Scheme | Marks |
|---|---|
| \(y = \text{arcoth}\,x \Rightarrow \coth y = x\) | B1 |
| \(x = \coth y = \dfrac{e^{2y} + 1}{e^{2y} - 1} \Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}y} = \dfrac{\left(e^{2y} - 1\right)2e^{2y} - \left(e^{2y} + 1\right)2e^{2y}}{\left(e^{2y} - 1\right)^2}\) | M1 |
| \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = \dfrac{-4e^{2y}}{\left(e^{2y} - 1\right)^2} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{e^{4y} - 2e^{2y} + 1}{-4e^{2y}} = \dfrac{e^{2y} - 2 + e^{-2y}}{-4} = -\left(\dfrac{e^y - e^{-y}}{2}\right)^2 = -\sinh^2 y = -\dfrac{1}{\text{cosech}^2 y}\) \(= \dfrac{1}{1 - \coth^2 y} = \dfrac{1}{1 - x^2}\) * Completes correctly with no errors | A1* |
B1: Changes from arcoth to coth correctly This may be implied by e.g. \(\tanh y = \dfrac{1}{x}\)
M1: Expresses \(\coth y\) in terms of exponentials and differentiates
(a) Alternative 4
| Scheme | Marks |
|---|---|
| \(y = \text{arcoth}\,x \Rightarrow \coth y = x\) | B1 |
| \(x = \coth y = \dfrac{e^y + e^{-y}}{e^y - e^{-y}} \Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}y} = \dfrac{\left(e^y - e^{-y}\right)^2 - \left(e^y + e^{-y}\right)^2}{\left(e^y - e^{-y}\right)^2}\) | M1 |
| \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = \dfrac{-4}{\left(e^y - e^{-y}\right)^2} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{1}{\text{cosech}^2 y}\) | |
| \(\coth^2 y - 1 = \text{cosech}^2 y \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{1 - x^2}\) * | A1* |
B1: Changes from arcoth to coth correctly This may be implied by e.g. \(\tanh y = \dfrac{1}{x}\)
M1: Expresses \(\coth y\) in terms of exponentials and differentiates
A1*: Correct completion with no errors seen and an intermediate step shown.
(a) Alternative 5
| Scheme | Marks |
|---|---|
| \(y = \text{arcoth}\,x = \dfrac{1}{2}\ln\left(\dfrac{1 + x}{x - 1}\right)\) | B1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2}\left[\dfrac{x - 1}{x + 1} \times \dfrac{(x - 1) - (x + 1)}{(x - 1)^2}\right]\) or \(\dfrac{1}{2}\ln\left(\dfrac{1 + x}{x - 1}\right) = \dfrac{1}{2}\ln(1 + x) - \dfrac{1}{2}\ln(x - 1)\) \(\Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2(x + 1)} - \dfrac{1}{2(x - 1)}\) | M1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{1 - x^2}\) | A1 |
B1: Correct ln form for arcoth
M1: Attempts to differentiate using the chain rule and quotient rule or writes as two logarithms and differentiates.
A1: Correct completion with no errors seen.
Note that use of \(\text{arcoth}\,x = \dfrac{1}{\text{artanh}\,x}\left(= \dfrac{1}{\frac{1}{2}\ln\left(\frac{1 + x}{1 - x}\right)}\right)\) scores no marks
(a) Alternative 6
| Scheme | Marks |
|---|---|
| \(y = \text{arcoth}\,x \Rightarrow \coth y = x\) | B1 |
| \(\tanh y = \dfrac{1}{x} \Rightarrow -\dfrac{1}{x^2} = \text{sech}^2 y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) | M1 |
| \(-\dfrac{1}{x^2} = \left(1 - \dfrac{1}{x^2}\right)\dfrac{\mathrm{d}y}{\mathrm{d}x} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{1 - x^2}\) | A1* |
B1: Changes from arcoth to coth correctly This may be implied by e.g. \(\tanh y = \dfrac{1}{x}\)
M1: \(\pm\dfrac{1}{x^2} = \pm\text{sech}^2 y\dfrac{\mathrm{d}x}{\mathrm{d}y}\)
A1*: Correct completion with no errors seen.
(a) Alternative 7
| Scheme | Marks |
|---|---|
| \(y = \text{arcoth}\,x = \text{artanh}\left(\dfrac{1}{x}\right)\) | B1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{1 - \left(\frac{1}{x}\right)^2} \times -x^{-2}\) | M1 |
| \(= \dfrac{-1}{x^2 - 1} = \dfrac{1}{1 - x^2}\) | A1* |
B1: Expresses arcoth in terms of artanh correctly
M1: Differentiates using the chain rule
A1*: Correct completion with no errors seen.
| Scheme | Marks |
|---|---|
| \(y = (\text{arcoth}\,x)^2 \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = 2(\text{arcoth}\,x) \times \dfrac{1}{1 - x^2}\) | B1 |
| \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = \dfrac{2}{1 - x^2}\left(1 - x^2\right)^{-1} + 4x\,\text{arcoth}\,x \times \left(1 - x^2\right)^{-2}\) \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = \dfrac{2\left(1 - x^2\right) \times \frac{1}{1 - x^2} + 2\,\text{arcoth}\,x \times 2x}{\left(1 - x^2\right)^2}\left(= \dfrac{4x\,\text{arcoth}\,x + 2}{\left(1 - x^2\right)^2}\right)\) | M1A1 |
| \(\left(1 - x^2\right)\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} - 2x\dfrac{\mathrm{d}y}{\mathrm{d}x} = \left(1 - x^2\right)\left(\dfrac{4x\,\text{arcoth}\,x + 2}{\left(1 - x^2\right)^2}\right) - 2x \times \left(\dfrac{2\,\text{arcoth}\,x}{1 - x^2}\right)\) or \(\left(1 - x^2\right)\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = \dfrac{2}{1 - x^2} + 2x \times \left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)\) | M1 |
| \(\left(1 - x^2\right)\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} - 2x\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2}{1 - x^2}\) | A1cso |
| (5) | |
| (8 marks) |
Notes
B1: Correct first derivative
M1: Attempts product or quotient rule on an expression of the form \(\dfrac{k\,\text{arcoth}\,x}{1 - x^2}\)
Product rule requires \(\pm P\left(1 - x^2\right)^{-2} \pm Qx\,\text{arcoth}\,x\left(1 - x^2\right)^{-2}\) oe
Quotient rule requires \(\dfrac{\pm P \pm Qx\,\text{arcoth}\,x}{\left(1 - x^2\right)^2}\) oe
A1: Correct second derivative
M1: Substitutes their first and second derivatives into the lhs of the differential equation or multiplies through by \((1 - x^2)\) and replaces \(2(\text{arcoth}\,x) \times \dfrac{1}{1 - x^2}\) by \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
A1cso: Correct conclusion with no errors
(b) Alternative 1
| Scheme | Marks |
|---|---|
| \(y = (\text{arcoth}\,x)^2 \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = 2(\text{arcoth}\,x) \times \dfrac{1}{1 - x^2}\) | B1 |
| \(\left(1 - x^2\right)\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2\,\text{arcoth}\,x \Rightarrow \left(1 - x^2\right)\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} - 2x\dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots\) | M1A1 |
| \(\dfrac{\mathrm{d}(2\,\text{arcoth}\,x)}{\mathrm{d}x} = \dfrac{2}{1 - x^2}\) | M1 |
| \(\left(1 - x^2\right)\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} - 2x\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2}{1 - x^2}\) | A1cso |
B1: Correct first derivative
M1: Multiplies through by \(1 - x^2\) and attempts product rule on \(\left(1 - x^2\right)\dfrac{\mathrm{d}y}{\mathrm{d}x}\). Requires \(\left(1 - x^2\right)\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} \pm Px\dfrac{\mathrm{d}y}{\mathrm{d}x}\) oe
A1: Correct differentiation
M1: Differentiates rhs using the result from part (a)
A1cso: Correct conclusion with no errors
(b) Alternative 2
| Scheme | Marks |
|---|---|
| \(y = (\text{arcoth}\,x)^2 \Rightarrow y^{\frac{1}{2}} = \text{arcoth}\,x \Rightarrow \dfrac{1}{2}y^{-\frac{1}{2}}\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{1 - x^2}\) | B1 |
| \(\dfrac{1}{2}y^{-\frac{1}{2}}\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} - \dfrac{1}{4}y^{-\frac{3}{2}}\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = \dfrac{2x}{\left(1 - x^2\right)^2}\) | M1A1 |
| Then substitute as before to obtain \(\dfrac{2}{1 - x^2}\) | M1A1cso |
B1: Correct differentiation
M1: Correct use of product rule to give \(py^{-\frac{1}{2}}\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} - qy^{-\frac{3}{2}}\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2\)
A1: \(\dfrac{1}{2}y^{-\frac{1}{2}}\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} - \dfrac{1}{4}y^{-\frac{3}{2}}\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = \dfrac{2x}{\left(1 - x^2\right)^2}\)