FP3 June 2018 Q1
1.
| Scheme | Marks |
|---|---|
| \(\left(\tanh x = \dfrac{\sinh x}{\cosh x}\right) = \dfrac{\frac{e^x - e^{-x}}{2}}{\frac{e^x + e^{-x}}{2}}\) or \(\dfrac{\frac{e^{2x} - 1}{2e^x}}{\frac{e^{2x} + 1}{2e^x}}\) | M1 |
| \(= \dfrac{e^x - e^{-x}}{e^x + e^{-x}} = \dfrac{e^{2x} - 1}{e^{2x} + 1}\) * | A1* |
| (2) |
Notes
M1: Substitutes the correct exponential forms. Note that the \(\tanh x = \dfrac{\sinh x}{\cosh x}\) may be implied.
A1*: Correct proof with no errors or omissions or notational errors such as using sin for sinh
Note that the question says “starting from the definitions of sinh\(x\) and cosh\(x\) in terms of exponentials” so:
\(\sinh x = \dfrac{e^x - e^{-x}}{2},\ \cosh x = \dfrac{e^x + e^{-x}}{2},\ \tanh x = \dfrac{\sinh x}{\cosh x} = \dfrac{e^x - e^{-x}}{e^x + e^{-x}} = \dfrac{e^{2x} - 1}{e^{2x} + 1}\) scores M1A1
BUT
\(\tanh x = \dfrac{\sinh x}{\cosh x} = \dfrac{e^x - e^{-x}}{e^x + e^{-x}} = \dfrac{e^{2x} - 1}{e^{2x} + 1}\) scores M0A0 as sinh\(x\) and cosh\(x\) have not been defined
| Scheme | Marks |
|---|---|
| \(y = \text{artanh}\,\theta \Rightarrow \tanh y = \theta \Rightarrow \theta = \dfrac{e^{2y} - 1}{e^{2y} + 1}\) \(\theta\left(e^{2y} + 1\right) = e^{2y} - 1 \Rightarrow e^{2y}(\theta - 1) = -1 - \theta \Rightarrow e^{2y} = \dfrac{1 + \theta}{1 - \theta}\) M1 for setting \(\theta = \dfrac{e^{2y} - 1}{e^{2y} + 1}\) or any other variables for \(\theta\) and \(y\) e.g. \(y = \dfrac{e^{2x} - 1}{e^{2x} + 1}\) and uses correct processing (allow sign errors only) to make e2”y” or e”y” the subject | M1 |
| \(e^{2y} = \dfrac{1 + \theta}{1 - \theta} \Rightarrow 2y = \ln\left(\dfrac{1 + \theta}{1 - \theta}\right)\) | dM1 |
| \(y = \dfrac{1}{2}\ln\left(\dfrac{1 + \theta}{1 - \theta}\right)\) * or \(\dfrac{1}{2}\ln\dfrac{1 + \theta}{1 - \theta}\) * Correct completion with no errors. Must be in terms of \(\theta\) for this mark but allow “mixed” variables for the M’s. This mark should be withheld if there are any errors such as the appearance of a “tan” instead of “tanh” and/or missing variables. The proof does need to convey that \(\text{artanh}\,\theta = \dfrac{1}{2}\ln\left(\dfrac{1 + \theta}{1 - \theta}\right)\) So if \(y\) has been defined as \(\text{artanh}\,\theta\) and the proof ends \(y = \dfrac{1}{2}\ln\left(\dfrac{1 + \theta}{1 - \theta}\right)\), this is acceptable. So must be in terms of \(\theta\) for the A mark but allow other variables to be used for the M’s. Allow arctanh, artanh, tanh-1 etc. for the inverse | A1* |
| (3) | |
| (5 marks) |
Notes
dM1: Removes e correctly by taking ln’s. Dependent on the first method mark.
(b) Alternative
| Scheme | Marks |
|---|---|
| \(\text{artanh}\,\theta = \dfrac{1}{2}\ln\left(\dfrac{1 + \theta}{1 - \theta}\right) \Rightarrow \theta = \tanh\left(\dfrac{1}{2}\ln\left(\dfrac{1 + \theta}{1 - \theta}\right)\right)\) | |
| \(\theta = \dfrac{e^{\ln\left(\frac{1 + \theta}{1 - \theta}\right)} - 1}{e^{\ln\left(\frac{1 + \theta}{1 - \theta}\right)} + 1}\) | M1 |
| \(= \dfrac{\frac{1 + \theta}{1 - \theta} - 1}{\frac{1 + \theta}{1 - \theta} + 1} = \dfrac{1 + \theta - 1 + \theta}{1 + \theta + 1 - \theta} = \theta\) | dM1 |
| \(\Rightarrow \text{artanh}\,\theta = \dfrac{1}{2}\ln\left(\dfrac{1 + \theta}{1 - \theta}\right)\) * Allow \(\dfrac{1}{2}\ln\dfrac{1 + \theta}{1 - \theta}\) * | A1* |
| (3) |
M1: Uses part (a) to express \(\theta\) in terms of e
dM1: Removes e’s and ln’s correctly. Dependent on the first method mark.
A1*: Obtains \(\theta = \theta\) with no errors and makes a conclusion. Must be in terms of \(\theta\) for this mark but allow a different variable for the M’s. This mark should be withheld if there are any errors such as the appearance of a “tan” instead of “tanh” and/or missing variables.
Attempts that assume \(\text{artanh}\,\theta = \dfrac{\text{arsinh}\,\theta}{\text{arcosh}\,\theta}\) score no marks in (b)