A2 June 2019 Paper 2 Q1
1.
| Scheme | Marks | AO |
|---|---|---|
| \(y = \tanh^{-1}(x) \Rightarrow \tanh y = x \Rightarrow x = \dfrac{\sinh y}{\cosh y} = \dfrac{\mathrm{e}^y - \mathrm{e}^{-y}}{\mathrm{e}^y + \mathrm{e}^{-y}}\) | M1 A1 | 2.1 1.1b |
| Note that some candidates only have one variable and reach e.g. \(x = \dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{\mathrm{e}^x + \mathrm{e}^{-x}}\) or \(\tanh x = \dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{\mathrm{e}^x + \mathrm{e}^{-x}}\) Allow this to score M1A1 | ||
| \(x\left(\mathrm{e}^{2y} + 1\right) = \mathrm{e}^{2y} - 1 \Rightarrow \mathrm{e}^{2y}(1 - x) = 1 + x \Rightarrow \mathrm{e}^{2y} = \dfrac{1 + x}{1 - x}\) | M1 | 1.1b |
| \(\mathrm{e}^{2y} = \dfrac{1 + x}{1 - x} \Rightarrow 2y = \ln\left(\dfrac{1 + x}{1 - x}\right) \Rightarrow y = \dfrac{1}{2}\ln\left(\dfrac{1 + x}{1 - x}\right)\) * | A1* | 2.1 |
| Note that \(\mathrm{e}^{2y}(x - 1) + x + 1 = 0\) can be solved as a quadratic in \(\mathrm{e}^y\): \(\mathrm{e}^y = \dfrac{-\sqrt{0 - 4(x - 1)(x + 1)}}{2(x - 1)} = \dfrac{-\sqrt{4(1 - x)(x + 1)}}{2(x - 1)} = \dfrac{2\sqrt{(1 - x)(x + 1)}}{2(1 - x)}\) \(= \dfrac{\sqrt{(x + 1)}}{\sqrt{(1 - x)}} \Rightarrow y = \dfrac{1}{2}\ln\dfrac{(x + 1)}{(1 - x)}\) * Score M1 for an attempt at the quadratic formula to make \(\mathrm{e}^y\) the subject (condone \(\pm\sqrt{\ldots}\)) and A1* for a correct solution that rejects the positive root at some point and deals with the \((x - 1)\) bracket correctly | ||
| \(k = 1\) or \(-1 \lt x \lt 1\) | B1 | 1.1b |
| (5) |
Notes
If you come across any attempts to use calculus to prove the result – send to review
M1: Begins the proof by expressing tanh in terms of exponentials and forms an equation in exponentials.
The exponential form can be any of \(\dfrac{\left(\mathrm{e}^y - \mathrm{e}^{-y}\right)/2}{\left(\mathrm{e}^y + \mathrm{e}^{-y}\right)/2},\ \dfrac{\mathrm{e}^y - \mathrm{e}^{-y}}{\mathrm{e}^y + \mathrm{e}^{-y}},\ \dfrac{\mathrm{e}^{2y} - 1}{\mathrm{e}^{2y} + 1}\)
Allow any variables to be used but the final answer must be in terms of \(x\). Allow alternative notation for \(\tanh^{-1}x\) e.g. artanh, arctanh.
A1: Correct expression for “\(x\)” in terms of exponentials
M1: Full method to make \(\mathrm{e}^{2\text{“}y\text{”}}\) the subject of the formula. This must be correct algebra so allow sign errors only.
A1*: Completes the proof by using logs correctly and reaches the printed answer with no errors.
Allow e.g. \(\dfrac{1}{2}\ln\left(\dfrac{x + 1}{1 - x}\right),\ \dfrac{1}{2}\ln\dfrac{x + 1}{1 - x},\ \dfrac{1}{2}\ln\left|\dfrac{x + 1}{1 - x}\right|\). Need to see \(\tanh^{-1}x = \dfrac{1}{2}\ln\left(\dfrac{1 + x}{1 - x}\right)\) as a conclusion but allow if the proof concludes that \(y = \dfrac{1}{2}\ln\left(\dfrac{1 + x}{1 - x}\right)\) with \(y\) defined as \(\tanh^{-1}x\) earlier.
B1: Correct value for \(k\) or writes \(-1 \lt x \lt 1\)
Alternative: Way 2
| Scheme | Marks | AO |
|---|---|---|
| \(\tanh^{-1}x = \dfrac{1}{2}\ln\left(\dfrac{1 + x}{1 - x}\right) \Rightarrow x = \tanh\left(\dfrac{1}{2}\ln\left(\dfrac{1 + x}{1 - x}\right)\right) = \dfrac{\mathrm{e}^{\ln\frac{1+x}{1-x}} - 1}{\mathrm{e}^{\ln\frac{1+x}{1-x}} + 1}\) | M1 A1 | 2.1 1.1b |
| \(x = \dfrac{\mathrm{e}^{\ln\frac{1+x}{1-x}} - 1}{\mathrm{e}^{\ln\frac{1+x}{1-x}} + 1} = \dfrac{\dfrac{1 + x}{1 - x} - 1}{\dfrac{1 + x}{1 - x} + 1} = x\) Hence true, QED, tick etc. | M1 A1 | 1.1b 2.1 |
M1: Starts with result, takes tanh of both sides and expresses in terms of exponentials
A1: Correct expression
M1: Eliminates exponentials and logs and simplifies
A1: Correct result (i.e. \(x = x\)) with conclusion
B1: Correct value for \(k\) or writes \(-1 \lt x \lt 1\)
| Scheme | Marks | AO |
|---|---|---|
| \(2x = \tanh\left(\ln\sqrt{2 - 3x}\right) \Rightarrow \tanh^{-1}(2x) = \ln\sqrt{2 - 3x}\) | M1 | 3.1a |
| \(\dfrac{1}{2}\ln\left(\dfrac{1 + 2x}{1 - 2x}\right) = \dfrac{1}{2}\ln(2 - 3x) \Rightarrow \dfrac{1 + 2x}{1 - 2x} = 2 - 3x\) | M1 | 2.1 |
| \(6x^2 - 9x + 1 = 0\) | A1 | 1.1b |
| \(6x^2 - 9x + 1 = 0 \Rightarrow x = \ldots\) | M1 | 1.1b |
| \(x = \dfrac{9 - \sqrt{57}}{12}\) | A1 | 3.2a |
| (5) | ||
| (10 marks) |
Notes
M1: Adopts a correct strategy by taking \(\tanh^{-1}\) of both sides
M1: Makes the link with part (a) by replacing artanh\((2x)\) with \(\dfrac{1}{2}\ln\left(\dfrac{1 + 2x}{1 - 2x}\right)\) and demonstrates the use of the power law of logs to obtain an equation with logs removed correctly.
A1: Obtains the correct 3TQ
M1: Solves their 3TQ using a correct method (see General Guidance – if no working is shown (calculator) and the roots are correct for their quadratic, allow M1)
A1: Correct value with the other solution rejected (accept rejection by omission) so \(x = \dfrac{9 \pm \sqrt{57}}{12}\) scores A0 unless the positive root is rejected
Alternative for first 2 marks of (b)
| Scheme | Marks | AO |
|---|---|---|
| \(2x = \tanh\left(\ln\sqrt{2 - 3x}\right) \Rightarrow 2x = \dfrac{\mathrm{e}^{2\ln\sqrt{2 - 3x}} - 1}{\mathrm{e}^{2\ln\sqrt{2 - 3x}} + 1}\) | M1 | 3.1a |
| \(\Rightarrow \dfrac{2 - 3x - 1}{2 - 3x + 1} = 2x\) | M1 | 2.1 |
M1: Adopts a correct strategy by expressing tanh in terms of exponentials
M1: Demonstrates the use of the power law of logs to obtain an equation with logs removed correctly