# Question 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $1 + 2\sinh^2 x - 7\sinh x = 5$ | M1 | Replaces $\cosh 2x$ by $1 + 2\sinh^2 x$ or replaces $\cosh 2x$ with $\cosh^2 x + \sinh^2 x$ and then $\cosh^2 x$ with $1 + \sinh^2 x$. No incorrect identities. |
| $2\sinh^2 x - 7\sinh x - 4 = 0$ | A1 | Correct quadratic |
| $(2\sinh x + 1)(\sinh x - 4) = 0 \Rightarrow \sinh x =$ | M1 | Attempt to solve 3TQ in $\sinh x$ (usual rules) |
| $\sinh x = -\frac{1}{2}, 4$ | A1 | Both (allow un-simplified e.g. $\frac{7\pm 9}{4}$) |
| $\text{arsinh}\, x = \ln\left(x + \sqrt{x^2+1}\right)$ | M1 | Use of the correct log form of arsinh |
| $x = \ln\left(-\frac{1}{2}+\sqrt{\frac{5}{4}}\right),\ \ln\left(4+\sqrt{17}\right)$ | A1, A1 | A1: One correct exact value. A1: Both values correct, no incorrect values. Condone missing brackets. |

**Alternative Method:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(\frac{e^{2x}+e^{-2x}}{2}\right)-7\left(\frac{e^x - e^{-x}}{2}\right)=5$ | M1A1 | M1: Substitutes correct exponential definitions. A1: Correct expression |
| $e^{4x}-7e^{3x}-10e^{2x}-7e^x+1=0$ | M1A1 | M1: Multiplies by $e^{2x}$. A1: Correct quartic in $e^x$ |
| $\left(e^{2x}+e^x-1\right)\left(e^{2x}-8e^x-1\right)=0 \Rightarrow e^x = \ldots \Rightarrow x = \ldots$ | M1 | Solves quartic as far as $e^x = \ldots$, converts to $x$ in terms of ln. Recognisable attempt to solve quartic with at least 4 terms as product of two 3TQ's in $e^x$ |
| $x = \ln\left(-\frac{1}{2}+\sqrt{\frac{5}{4}}\right),\ \ln\left(4+\sqrt{17}\right)$ | A1, A1 | A1: One correct exact value. A1: Both correct. Condone missing brackets |

**Total: 7 marks**

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