A2 June 2025 Paper 1 Q9
9.
| Scheme | Marks | AO |
|---|---|---|
| \(\sinh x = \left(\dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right)\) and \(\tanh x = \dfrac{\mathrm{e}^{2x} - 1}{\mathrm{e}^{2x} + 1}\) or \(\dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{\mathrm{e}^x + \mathrm{e}^{-x}}\) | B1 | 1.2 |
| \(\dfrac{3}{4}\sinh x = \tanh x + \dfrac{1}{5} \Rightarrow \dfrac{3}{4}\left(\dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right) = \dfrac{\mathrm{e}^{2x} - 1}{\mathrm{e}^{2x} + 1} + \dfrac{1}{5}\) or \(\dfrac{3}{4}\left(\dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right) = \dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{\mathrm{e}^x + \mathrm{e}^{-x}} + \dfrac{1}{5}\) Leading to a quartic equation for \(\mathrm{e}^x\) | M1 | 3.1a |
| \(15\mathrm{e}^{4x} - 48\mathrm{e}^{3x} + 32\mathrm{e}^x - 15 = 0\) * | A1* | 1.1b |
| (3) |
Notes
B1: Recalls the exponential definitions for \(\sinh x\) and at least one of \(\cosh x\) or \(\tanh x\), which may be embedded in their workings.
M1: Substitutes the correct exponential definitions to form an equation leading to a quartic equation for \(\mathrm{e}^x\). Any identities used such as \(\sinh 2x = 2\sinh x\cosh x\) must be correct
A1*: Correct equation following at least one intermediate stage, with no errors seen. They cannot go directly from their substitution to the given answer. cso
| Scheme | Marks | AO |
|---|---|---|
| \(15(3)^4 - 48(3)^3 + 32(3) - 15 = 0\) therefore \(\mathrm{e}^x = 3\) is a solution | B1 | 1.1b |
| (1) |
Notes
B1: Substitutes \(\mathrm{e}^x = 3\) into each term of the equation, shows \(= 0\) and states therefore a solution or writes \(\left(\mathrm{e}^x - 3\right)\) is a factor. Allow a tick, box, QED or appropriate conclusion.
OR
Substitutes \(x = \ln 3\) into each term of the equation, shows \(= 0\) and states therefore a solution or writes \(\left(\mathrm{e}^x - 3\right)\) is a factor. Allow a tick, box, QED or appropriate conclusion.
OR
Factorises their equation:
\(15\mathrm{e}^{4x} - 48\mathrm{e}^{3x} + 32\mathrm{e}^x - 15 = 0\)
\(\Rightarrow \left(\mathrm{e}^x - 3\right)\left(15\mathrm{e}^{3x} - 3\mathrm{e}^{2x} - 9\mathrm{e}^x + 5\right) = 0\)
and states hence \(\mathrm{e}^x = 3\) is a solution
Do not award this mark for simply solving a quartic such as \(15y^4 - 48y^3 + 32y - 15 = 0\) on the calculator and stating \(y = 3\), hence \(\mathrm{e}^x = 3\) is a solution.
| Scheme | Marks | AO |
|---|---|---|
| \((\ln 3,\ 1)\) | B1 | 1.1b |
| (1) |
Notes
B1: States the correct exact coordinates. Allow \(x = \ldots,\ y = \ldots\) Do not accept decimals.
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\int \dfrac{\mathrm{e}^{\frac{1}{x}}}{x^2}\,\mathrm{d}x = \lambda\mathrm{e}^{\frac{1}{x}}\) \(\left\{u = x^{-1} \Rightarrow \dfrac{\mathrm{d}u}{\mathrm{d}x} = -x^{-2} \Rightarrow \displaystyle\int x^{-2}\mathrm{e}^u.\dfrac{\mathrm{d}u}{-x^{-2}} = \displaystyle\int -\mathrm{e}^u\,\mathrm{d}u\right\}\) | M1 | 3.1a |
| \(\displaystyle\int \dfrac{\mathrm{e}^{\frac{1}{x}}}{x^2}\,\mathrm{d}x = -\mathrm{e}^{\frac{1}{x}}\) | A1 | 1.1b |
| \(\displaystyle\int_{-4}^{0} \dfrac{\mathrm{e}^{\frac{1}{x}}}{x^2}\,\mathrm{d}x = \lim_{t \to 0}\left[-\mathrm{e}^{\frac{1}{x}}\right]_{-4}^{t} = \lim_{t \to 0}\left[\left(-\mathrm{e}^{\frac{1}{t}}\right) - \left(-\mathrm{e}^{-\frac{1}{4}}\right)\right]\) | M1 | 2.5 |
| \(t \to 0 \Rightarrow \mathrm{e}^{\frac{1}{t}} \to 0\) therefore \(\displaystyle\int_{-4}^{0} \dfrac{\mathrm{e}^{\frac{1}{x}}}{x^2}\,\mathrm{d}x = \mathrm{e}^{-\frac{1}{4}}\) * | A1* | 2.4 |
| (4) | ||
| (9 marks) |
Notes
M1: Use the substitution \(u = \pm\dfrac{1}{x}\) to obtain an integral of the form \(\displaystyle\int \lambda\mathrm{e}^u\,\mathrm{d}u\) oe and integrates to \(\lambda\mathrm{e}^u\)
Award for a sight of \(\pm\lambda\mathrm{e}^{\frac{1}{x}}\) as the answer to their integral.
Alternatively applies the reverse of the chain rule \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(e^{\frac{1}{x}}\right) = \pm\dfrac{1}{x^2}e^{\frac{1}{x}}\) and integrates to reach \(\pm\lambda\mathrm{e}^{\frac{1}{x}}\)
Answers using integration by parts are unlikely to lead to a correct solution.
A1: Correct integration. Award for correct answer with minimal workings as this can be done by inspection using the reverse of the chain rule. However, withhold this mark for incorrect workings.
M1: Uses correct notation to write the integral as the limit as \(t \to 0\), with the limits of \(t\) (or any such variable) and \(-4\). Applies correctly the limits of \(-4\) and \(t\) with correct limit notation, with limits seen substituted the right way round.
Withhold this mark if there is no evidence of using the limiting process. We must see as a minimum \(\displaystyle\lim_{t \to 0}\) oe at some stage in their work.
A1*: Produces an argument that includes an upper limit that approaches 0.
e.g. States that \(t \to 0 \Rightarrow \mathrm{e}^{\frac{1}{t}} \to 0\) and reaches a value of \(\mathrm{e}^{-\frac{1}{4}}\)
Alternative method: changing limits using \(u = \dfrac{1}{x}\)
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\int_{-4}^{0} \dfrac{\mathrm{e}^{\frac{1}{x}}}{x^2}\,\mathrm{d}x = \displaystyle\int -\mathrm{e}^u\,\mathrm{d}u = \lambda\mathrm{e}^u\) | M1 | |
| \(= \left(\displaystyle\lim_{t \to 0}\right)\left[-\mathrm{e}^u\right]_{-\frac{1}{4}}^{\frac{1}{t}}\) correct limits required | A1 | |
| \(= \displaystyle\lim_{t \to 0}\left[\left(-\mathrm{e}^{\frac{1}{t}}\right) - \left(-\mathrm{e}^{-\frac{1}{4}}\right)\right]\) | M1 | |
| \(t \to 0 \Rightarrow \mathrm{e}^{\frac{1}{t}} \to 0\) therefore \(\displaystyle\int_{-4}^{0} \dfrac{\mathrm{e}^{\frac{1}{x}}}{x^2}\,\mathrm{d}x = \mathrm{e}^{-\frac{1}{4}}\) | A1* |