## Question 1:

### Part (a)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $4\sinh^3 x + 3\sinh x \equiv 4\left(\frac{e^x - e^{-x}}{2}\right)^3 + 3\left(\frac{e^x - e^{-x}}{2}\right)$ | M1 | 2.1 — Begins proof by expressing $\sinh x$ correctly in terms of exponentials and makes progress cubing the bracket. Award for obtaining form $Ae^{3x} + Be^x + Ce^{-x} + De^{-3x}$ (terms need not be collected) |
| $\equiv 4\left(\frac{e^{3x} - 3e^x + 3e^{-x} - e^{-3x}}{8}\right) + 3\left(\frac{e^x - e^{-x}}{2}\right)$ | | |
| $\equiv \frac{e^{3x} - e^{-3x}}{2} \equiv \sinh 3x$ * | A1* | 1.1b — Fully correct proof with no errors. Must see $= \sinh 3x$ or e.g. LHS = RHS |
| | **(2)** | |

### Part (b)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sinh 3x = 19\sinh x \Rightarrow 4\sinh^3 x + 3\sinh x = 19\sinh x$ | M1 | 3.1a — Uses part (a), collects terms and attempts to factorise or cancel $\sinh x$. Can be implied if they go straight from cubic to writing **all** correct answers for $\sinh x$ including zero |
| $4\sinh x(\sinh^2 x - 4) = 0$ | | |
| $\sinh x = 0 \Rightarrow x = 0$ | B1 | 2.2a — Deduces root $x=0$; allow $\ln 1$ but not $\ln\left[0 + \ln\sqrt{0+1}\right]$ |
| $\sinh^2 x = 4 \Rightarrow \sinh x = \pm 2$ | M1 | 1.1b — Proceeds to $\sinh x = a$ and uses correct logarithmic form of arsinh to obtain at least one exact value for $x$. Alternatively substitutes exponential form and solves cubic in $e^x$ |
| $\Rightarrow x = \ln\left(\pm 2 + \sqrt{(\pm 2)^2 + 1}\right)$ | | |
| $x = \ln(2+\sqrt{5})$ **or** $x = \ln(-2+\sqrt{5})$ oe e.g. $x = -\ln(2+\sqrt{5})$ | A1 | 1.1b — One correct non-zero solution |
| $x = \ln(2+\sqrt{5})$ **and** $x = \ln(-2+\sqrt{5})$ | A1 | 1.1b — Both correct non-zero solutions and no incorrect other solutions (isw if they evaluate incorrectly). Allow $x = \ln(\sqrt{5}\pm 2)$ oe e.g. $x = \pm\ln(2+\sqrt{5})$ or $\frac{1}{2}\ln(9\pm 4\sqrt{5})$ |
| | **(5)** | |
| | **Total: 7 marks** | |

# Question 1 (Alternative - Exponential Form):

| Working | Marks | Guidance |
|---------|-------|----------|
| $\frac{e^{3x}-e^{-3x}}{2}=19\left(\frac{e^x-e^{-x}}{2}\right)$ leading to $e^{3x}-e^{-3x}-19e^x+19e^{-x}=0$ | M1 | Substitutes exponential form and proceeds to four-term cubic equation in $e^{2x}=0$ |
| $e^{6x}-1-19e^{4x}+19e^{2x}=0$ | M1 | For cubic in $e^{2x}=0$ |
| $x=0$, deduced from $e^{2x}=1$ | B1 | Allow $\ln 1$ but not $\ln\left[0+\ln\sqrt{0+1}\right]$ |
| $(e^{2x}-1)(e^{4x}-18e^{2x}+1)=0$ | M1 | Factorise cubic, obtain exact values for $e^{2x}$, then take logs |
| $e^{2x}=1,\ 9\pm4\sqrt{5}$ | A1 | One correct non-zero solution |
| $x=0,\ x=\frac{1}{2}\ln\left(9\pm4\sqrt{5}\right)$ | A1 | Both correct non-zero solutions, no incorrect solutions |

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