FP3 June 2017 Q8
8. The curve \(C\) has equation \[y = \ln\left(\frac{e^x + 1}{e^x - 1}\right), \qquad \ln 2 \leqslant x \leqslant \ln 3\]
| Scheme | Marks |
|---|---|
| Way 1: \(y = \ln\left(\dfrac{e^x + 1}{e^x - 1}\right) = \ln\left(e^x + 1\right) - \ln\left(e^x - 1\right) \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{e^x}{e^x + 1} - \dfrac{e^x}{e^x - 1}\) M1: Uses correct log rule and attempts derivative using chain rule A1: Correct Derivative | M1A1 |
| \(= \dfrac{e^{2x} - e^x - e^{2x} - e^x}{e^{2x} - 1} = \dfrac{-2e^x}{e^{2x} - 1}\) * | dM1A1* |
| (4) |
Notes
dM1: Attempt single fraction and uses \((e^x - 1)(e^x + 1) = e^{2x} - 1\). Dependent on the first method mark.
A1: Completes correctly with no errors
(a) Way 2
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{e^x - 1}{e^x + 1}\left(\dfrac{e^x\left(e^x - 1\right) - e^x\left(e^x + 1\right)}{\left(e^x - 1\right)^2}\right)\) Or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{e^x - 1}{e^x + 1}\left(e^x\left(e^x - 1\right)^{-1} - e^x\left(e^x + 1\right)\left(e^x - 1\right)^{-2}\right)\) | M1A1 |
| \(= \dfrac{1}{e^x + 1}\left(-\dfrac{2e^x}{e^x - 1}\right) = -\dfrac{2e^x}{e^{2x} - 1}\) * | dM1A1* |
M1: Uses chain and quotient or product rules
A1: Correct derivative
dM1: Cancels \(e^x - 1\) and uses \((e^x - 1)(e^x + 1) = e^{2x} - 1\). Dependent on the first method mark.
A1: Completes correctly with no errors
(a) Way 3
| Scheme | Marks |
|---|---|
| \(y = \ln\left(\dfrac{e^x + 1}{e^x - 1}\right) \Rightarrow e^y = \dfrac{e^x + 1}{e^x - 1} \Rightarrow e^y\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{e^x\left(e^x - 1\right) - e^x\left(e^x + 1\right)}{\left(e^x - 1\right)^2}\) M1: Removes logs correctly and differentiates implicitly using chain and quotient rules A1: Correct differentiation | M1A1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{2e^x}{\left(e^x - 1\right)^2} \times \dfrac{e^x - 1}{e^x + 1} = -\dfrac{2e^x}{e^{2x} - 1}\) * | dM1A1 |
dM1: Divides by \(e^y\) in terms of \(x\). Dependent on the first method mark.
A1: Completes correctly with no errors
(a) Way 4
| Scheme | Marks |
|---|---|
| \(y = \ln\left(\dfrac{e^x + 1}{e^x - 1}\right) = \ln\left(\coth\dfrac{1}{2}x\right) \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{\coth\frac{1}{2}x} \times -\dfrac{1}{2}\text{cosech}^2\dfrac{1}{2}x\) M1: Writes as \(\ln\left(\coth\dfrac{1}{2}x\right)\) and differentiates using the chain rule A1: Correct differentiation | M1A1 |
| \(= \left(\dfrac{e^x - 1}{e^x + 1}\right) \times \dfrac{-2e^x}{\left(e^x - 1\right)^2} = -\dfrac{2e^x}{e^{2x} - 1}\) | dM1A1 |
dM1: Substitutes the correct exponential forms. Dependent on the first method mark.
A1: Completes correctly with no errors
(a) Way 5
| Scheme | Marks |
|---|---|
| \(\text{artanh}\,x = \dfrac{1}{2}\ln\left(\dfrac{1 + x}{1 - x}\right) \Rightarrow y = 2\,\text{artanh}\left(e^{-x}\right)\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2}{1 - \left(e^{-x}\right)^2} \times -e^{-x}\) M1: Writes \(y\) correctly in terms of artanh and attempts to differentiate using the chain rule A1: Correct differentiation | M1A1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-2e^{-x}}{1 - e^{-2x}} = \dfrac{-2e^x}{e^{2x} - 1}\) * | dM1A1 |
dM1: Multiplies numerator and denominator by \(e^{2x}\). Dependent on the first method mark.
A1: Completes correctly with no errors
(a) Way 6
| Scheme | Marks |
|---|---|
| \(y = \ln\left(1 + \dfrac{2}{e^x - 1}\right) \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{1 + 2\left(e^x - 1\right)^{-1}} \times -2e^x\left(e^x - 1\right)^{-2}\) M1: Writes \(\dfrac{e^x + 1}{e^x - 1}\) as \(1 + \dfrac{2}{e^x - 1}\) and differentiates using the chain rule A1: Correct differentiation | M1A1 |
| \(= \dfrac{-2e^x}{\left(e^x - 1\right)^2 + 2\left(e^x - 1\right)} = \dfrac{-2e^x}{e^{2x} - 1}\) | dM1A1 |
dM1: Multiplies denominator by \(\left(e^x - 1\right)^2\). Dependent on the first method mark.
A1: Completes correctly with no errors
| Scheme | Marks |
|---|---|
| \(\displaystyle L = \int\sqrt{\left(1 + \left(\pm\frac{2e^x}{e^{2x} - 1}\right)^2\right)}\,\mathrm{d}x\) | M1 |
| \(\displaystyle = \int\sqrt{\left(\frac{e^{4x} - 2e^{2x} + 1 + 4e^{2x}}{\left(e^{2x} - 1\right)^2}\right)}\,\mathrm{d}x\) | dM1 |
| \(\displaystyle L = \int\sqrt{\frac{\left(e^{2x} + 1\right)^2}{\left(e^{2x} - 1\right)^2}}\,\mathrm{d}x = \int\frac{\left(e^{2x} + 1\right)}{\left(e^{2x} - 1\right)}\,\mathrm{d}x\) | A1 |
| \(\displaystyle = \int\coth x\,\mathrm{d}x,\ \int\frac{e^x + e^{-x}}{e^x - e^{-x}}\,\mathrm{d}x,\ \int 1 + \frac{2e^{-2x}}{1 - e^{-2x}}\,\mathrm{d}x,\ \frac{1}{2}\int\frac{2}{u} - \frac{1}{u + 1}\,\mathrm{d}u\ \left(u = e^{2x} - 1\right)\) \(\displaystyle\frac{1}{2}\int\frac{2}{u - 1} - \frac{1}{u}\,\mathrm{d}u\ \left(u = e^{2x}\right),\ \int\frac{1}{u + 1} + \frac{1}{u - 1} - \frac{1}{u}\,\mathrm{d}u\ \left(u = e^x\right),\) \(\displaystyle\frac{1}{2}\int\frac{2}{u - 2} - \frac{1}{u - 1}\,\mathrm{d}u\ \left(u = e^{2x} + 1\right)\) | |
| \(= \left[\ln\sinh x\right],\ \left[\ln\left(e^x - e^{-x}\right)\right],\ \left[x + \ln\left(1 - e^{-2x}\right)\right],\ \left[\ln u - \ln\sqrt{(1 + u)}\right],\) \(\left[\ln(u - 1) - \ln\sqrt{u}\right],\ \left[\ln\dfrac{\left(u^2 - 1\right)}{u}\right],\ \left[\ln(u - 2) - \ln\sqrt{u - 1}\right]\) Correct integration | A1 |
| \(= \ln\sinh(\ln 3) - \ln\sinh(\ln 2)\) \(\left(= \ln\dfrac{4}{3} - \ln\dfrac{3}{4}\right)\) | ddM1 |
| \(= \ln\dfrac{16}{9}\) | A1 |
| (6) | |
| (10 marks) |
Notes
M1: Uses the correct arc length formula with \(\pm\) the result from part (a). Note that we condone the omission of the minus sign on the fraction)
dM1: Attempt single fraction. Dependent on the first method mark.
Note that, for the first 2 marks, the candidate may just work on the integrand e.g. \(\sqrt{\left(1 + \left(\pm\dfrac{2e^x}{e^{2x} - 1}\right)^2\right)} = \sqrt{\left(\dfrac{e^{4x} - 2e^{2x} + 1 + 4e^{2x}}{\left(e^{2x} - 1\right)^2}\right)}\) Would score the first 2 marks.
A1: Correct integral with square root removed. No limits required.
ddM1: Correct use of limits e.g. ln3 and ln2 for \(x\) and e.g. 3 and 8 if \(u = e^{2x} - 1\). They must be the correct limits for their method if they use substitution. Dependent on both previous method marks.
A1: cao