A2 June 2025 Paper 2 Q8
8 A function \(\mathrm{f}(x)\) is defined by \(\mathrm{f}(x) = x\sinh 2x\).
| Scheme | Marks | AO |
|---|---|---|
| Basis case: \(\left(\text{LHS} = \frac{\mathrm{d}^0\mathrm{f}}{\mathrm{d}x^0} = \mathrm{f} = x\sinh 2x\right)\) \(\text{RHS} = 4^0(x\sinh 2x + 0 \times \cosh 2x)\) \(= x\sinh 2x\) \(\therefore\) LHS = RHS oe so true for \(n = 0\) | B1 | 2.1 |
| Assume true for \(n = k\): \(\dfrac{\mathrm{d}^{2k}\mathrm{f}}{\mathrm{d}x^{2k}} = 4^k(x\sinh 2x + k\cosh 2x)\) Then for \(n = k + 1\): \(\therefore \dfrac{\mathrm{d}^{2(k+1)}\mathrm{f}}{\mathrm{d}x^{2(k+1)}} = \dfrac{\mathrm{d}^2}{\mathrm{d}x^2}\left(\dfrac{\mathrm{d}^{2k}\mathrm{f}}{\mathrm{d}x^{2k}}\right)\) \(= \dfrac{\mathrm{d}^2}{\mathrm{d}x^2}\left(4^k(x\sinh 2x + k\cosh 2x)\right)\) | M1 | 3.1a |
| \(= 4^k\frac{\mathrm{d}}{\mathrm{d}x}(\sinh 2x + 2x\cosh 2x + 2k\sinh 2x)\) \(\left(= 4^k\frac{\mathrm{d}}{\mathrm{d}x}((2k + 1)\sinh 2x + 2x\cosh 2x)\right)\) | M1 | 1.1 |
| \(= 4^k(2(2k + 1)\cosh 2x + 2\cosh 2x + 4x\sinh 2x)\) | M1 | 1.1 |
| \(= 4^k(4(k + 1)\cosh 2x + 4x\sinh 2x)\) \(= (4)^k(4)((k + 1)\cosh 2x + x\sinh 2x)\) \(= 4^{k+1}(x\sinh 2x + (k + 1)\cosh 2x)\) | A1 | 2.1 |
| So true for \(n = k \Rightarrow\) true for \(n = k + 1\) True for \(\boldsymbol{n = 0} \Rightarrow\) true for \(n \geq 0\) | A1 | 2.4 |
| [6] |
Notes
Throughout this question condone single line slips to sin or cos provided recovered.
B1: May see \(n = 1\) in addition but this mark is for \(n = 0\) only. Must see LHS = RHS \(\Rightarrow\) true for \(n = 0\) oe (but may appear later e.g. in conclusion)
M1: Sets up and uses inductive hypothesis by showing an intent to differentiate twice and using the inductive assumption at least once. Complete statement might not be seen until after first or even second differentiation.
Must have statement in terms of some other variable than \(n\) in all three places consistently.
Condone minor slips in notation e.g. \(\frac{\mathrm{d}^2\boldsymbol{f}}{\mathrm{d}x^2}\left(\frac{\mathrm{d}^{2k}\mathrm{f}}{\mathrm{d}x^{2k}}\right)\) for all but the final A mark.
M1: Differentiates using product rule. Condone sign errors, missing 2s, \(\frac{1}{2}\) instead of 2 but not \(\frac{1}{2}x^2\) as the derivative of \(x\).
May be seen separately. Could be done with exponential form but M1 awarded on correct return to hyperbolic form (which could be after later differentiation).
M1: Differentiates again. Same guidance. Or \(= 4^k(4k\cosh 2x + 4\cosh 2x + 4x\sinh 2x)\)
Allow both differentiations in one step if fully correct.
A1: Rewrites to the correct form.
A1: Clear conclusion for induction process with implication stated. Dependent on all previous M and A marks www and fully correct notation throughout.
Allow this mark if the basis case was given as \(n = 1\) only but their conclusion must match their basis case (i.e. \(n = 1 \Rightarrow n \geq 1\)).
| Scheme | Marks | AO |
|---|---|---|
| \(n = 4 \Rightarrow \left.\dfrac{\mathrm{d}^8\mathrm{f}}{\mathrm{d}x^8}\right|_{x=0} = 4^4(0\sinh 0 + 4\cosh 0)\) | M1 | 3.1a |
| Coefficient of \(x^8\) is \(\frac{\mathrm{f}^{(8)}(0)}{8!}\) | M1 | 2.2a |
| \(= \dfrac{1024}{40320} = \dfrac{8}{315}\) | A1 | 1.1 |
| [3] |
Notes
M1: \(= 1024\) Chooses correct value of \(n\) and uses formula to find value for appropriate derivative when \(x = 0\).
M1: Using the correct rule for finding the coefficient of \(x^8\). This is independent of the previous mark and \(\mathrm{f}^{(8)}(0)\) need not be correct or even evaluated but must be \(x = 0\).
A1: Must be simplified. Do not accept eg 0.0254 etc but ISW once 8/315 seen unless it is clear that a different value is intended (eg 8 from multiplying by 315).
Allow with \(x^8\) e.g. \(\frac{8}{315}x^8\). Answer only with no working scores 0/3.
| Scheme | Marks | AO |
|---|---|---|
| \(\sinh 2x = \dfrac{\mathrm{e}^{2x} - \mathrm{e}^{-2x}}{2} \ldots\) | B1 | 1.1 |
| … so using general term for given series for \(\mathrm{e}^x\), \(x^7\) term is \(= \dfrac{1}{2}\left(\dfrac{(2x)^7}{7!} - \dfrac{(-2x)^7}{7!}\right)\) | M1 | 3.1a |
| So coefficient of \(x^8\) term in \(x\sinh 2x\) is \(\frac{1}{2}\left(\frac{2^7}{7!} - \frac{(-2)^7}{7!}\right) = \frac{1}{2}\left(\frac{128 + 128}{5040}\right) = \frac{128}{5040} = \frac{8}{315}\) | A1 | 2.2a |
| [3] |
Notes
B1: Uses exponential definition of \(\sinh x\) with \(2x\) as argument…
M1: … and given general term (or complete series) for \(\mathrm{e}^x\) to try to find (coefficient of) \(x^7\) term of \(\sinh 2x\) (or \(x^8\) term of \(x\sinh 2x\)).
A1: No conclusion necessary. Must be in same form as answer to (b) (ie explicit verification) (but allow this mark for a correct answer if no solution in (b)).
Allow SC A1 if (b) and (c) both given as eg 0.0254. Allow with \(x^8\) e.g. \(\frac{8}{315}x^8\)