A2 June 2025 Paper 2 Q8
8. Given that
\[y = \cos x\sinh x \qquad x \in \mathbb{R}\]| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = A\sin x\sinh x + B\cos x\cosh x\) | M1 | 1.1b |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\sin x\sinh x + \cos x\cosh x\) and \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = -2\sin x\cosh x\) | A1 | 1.1b |
| \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = C\sin x\cosh x \qquad \dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = D\cos x\cosh x + E\sin x\sinh x\) \(\dfrac{\mathrm{d}^4y}{\mathrm{d}x^4} = F\cos x\sinh x\) (unsimplified) | M1 | 2.1 |
| \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = -2\cos x\cosh x - 2\sin x\sinh x\) and \(\dfrac{\mathrm{d}^4y}{\mathrm{d}x^4} = -4\cos x\sinh x\) | A1 | 1.1b |
| \(\dfrac{\mathrm{d}^4y}{\mathrm{d}x^4} = -4y\) | A1 | 2.1 |
| (5) |
Notes
M1: Uses the product rule to find the correct form for at least the first derivative.
A1: Achieves correct first and second derivatives. Need not be simplified.
M1: Uses the product rule to find the correct form for the second, third and fourth derivatives.
A1: Correct (unsimplified) expressions for the third and fourth derivatives.
A1: Correct fourth derivative in terms of \(y\) from correct work. Note do not accept just \(k = -4\), we must see the \(\dfrac{\mathrm{d}^4y}{\mathrm{d}x^4} = -4y\)
8(a) Alt
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{aligned} u &= \cos x &\qquad v &= \sinh x\\ u^{\prime} &= -\sin x & v^{\prime} &= \cosh x\\ u^{\prime\prime} &= -\cos x & v^{\prime\prime} &= \sinh x\\ u^{\prime\prime\prime} &= \sin x & v^{\prime\prime\prime} &= \cosh x\\ u^{iv} &= \cos x & v^{iv} &= \sinh x\end{aligned}\) | M1 A1 | 1.1b 1.1b |
| \(\dfrac{\mathrm{d}^4y}{\mathrm{d}x^4} = uv^{iv} + 4u^{\prime}v^{\prime\prime\prime} + 6u^{\prime\prime}v^{\prime\prime} + 4u^{\prime\prime\prime}v^{\prime} + u^{iv}v = \ldots\) | M1 | 2.1 |
| \(= \cos x\sinh x - 4\sin x\cosh x - 6\cos x\sinh x + 4\sin x\cosh x + \cos x\sinh x\) | A1 | 1.1b |
| \(\dfrac{\mathrm{d}^4y}{\mathrm{d}x^4} = (-4\cos x\sinh x =) -4y\) | A1 | 2.1 |
| (5) |
There may be attempts via Leibnitz theorem.
M1: Attempts first four derivatives for each of \(\cos x\) and \(\sinh x\)
A1: Correct derivatives.
M1: Applies Leibnitz theorem correctly with their derivatives.
A1: Correct unsimplified expression.
A1: Correct fourth derivative in terms of \(y\).
There may be rare cases where exponential forms are used in part (a). For reference here are the derivatives they should get.
\(y = \dfrac{1}{2}\left(\mathrm{e}^x - \mathrm{e}^{-x}\right)\cos x\)
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2}\left(\mathrm{e}^x + \mathrm{e}^{-x}\right)\cos x - \dfrac{1}{2}\left(\mathrm{e}^x - \mathrm{e}^{-x}\right)\sin x = \dfrac{1}{2}\mathrm{e}^x(\cos x - \sin x) + \dfrac{1}{2}\mathrm{e}^{-x}(\cos x + \sin x)\)
\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = -\left(\mathrm{e}^x + \mathrm{e}^{-x}\right)\sin x\ \left[= \dfrac{1}{2}\mathrm{e}^x(\cos x - \sin x) + \dfrac{1}{2}\mathrm{e}^x(-\sin x - \cos x) - \dfrac{1}{2}\mathrm{e}^{-x}(\cos x + \sin x) + \dfrac{1}{2}\mathrm{e}^{-x}(-\sin x + \cos x)\right]\)
\(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = -\left(\mathrm{e}^x - \mathrm{e}^{-x}\right)\sin x - \left(\mathrm{e}^x + \mathrm{e}^{-x}\right)\cos x\)
\(\dfrac{\mathrm{d}^4y}{\mathrm{d}x^4} = -\left(\mathrm{e}^x + \mathrm{e}^{-x}\right)\sin x - \left(\mathrm{e}^x - \mathrm{e}^{-x}\right)\cos x - \left(\mathrm{e}^x - \mathrm{e}^{-x}\right)\cos x + \left(\mathrm{e}^x + \mathrm{e}^{-x}\right)\sin x = -2\left(\mathrm{e}^x - \mathrm{e}^{-x}\right)\cos x = -4\sinh x\cos x\)
NB Some are misreading the function as \(y = \cosh x\sinh x\) or \(y = \cos x\sin x\). These can be score method marks as long as work is equivalent (ie via product rule), but reduction of the problem to triviality via an identity is less demand and so will score M0. Allow SC A1 for the first A if the equivalent form is reached via the misread to that shown in the scheme.
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}^5y}{\mathrm{d}x^5} = \text{``}{-4}\text{''}\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(\dfrac{\mathrm{d}^5y}{\mathrm{d}x^5} = -\text{``}4\text{''}\cos x\cosh x + \text{``}4\text{''}\sin x\sinh x\) or \(y^{(5)} = \text{``}{-4}\text{''} \times \text{their } y^{\prime}\) | B1ft | 2.2a |
| When \(x = 0\) \(y = 0,\ y^{\prime} = 1,\ y^{\prime\prime} = 0,\ y^{(3)} = -2,\ y^{(4)} = 0,\ y^{(5)} = -4\) Uses their values in the expansion \(y = y(0) + xy^{\prime}(0) + \dfrac{x^2}{2!}y^{\prime\prime}(0) + \dfrac{x^3}{3!}y^{(3)}(0) + \dfrac{x^4}{4!}y^{(4)}(0) + \dfrac{x^5}{5!}y^{(5)}(0) + \ldots\) | M1 | 1.1b |
| \((y =)\ x - \dfrac{x^3}{3} - \dfrac{x^5}{30} + \ldots\) | A1 | 2.5 |
| (3) | ||
| (8 marks) |
Notes
B1ft: Correct fifth derivative or value for the fifth derivative deduced following through on their \(k\) from part (a). If they have stated \(y^{\prime}\) incorrectly allow for \(y^{(5)} = \text{``}{-4}\text{''} \times \text{their } y^{\prime}\) provided their \(y^{\prime}\) is non-zero
M1: Attempts the evaluation of all the derivatives at \(x = 0\) and applies the Maclaurin formula correctly with their values to achieve at least three non-zero terms.
A1: Correct simplified expansion. Accept with \(\mathrm{f}(x) = \ldots\) or with nothing before – mark the expression. Accept as a list of terms.