## Question 1:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Applies $\cosh x = \frac{e^x + e^{-x}}{2}$ and expands bracket to at least 4 different and no more than 5 different terms of correct form | M1 | Allow unsimplified terms e.g. $(e^x)^3 e^{-x}$. May see $8\left(\frac{e^x+e^{-x}}{2}\right)^2\left(\frac{e^x+e^{-x}}{2}\right)^2$ but must attempt to expand |
| $= \frac{1}{2}(e^{4x}+e^{-4x})+4\left(\frac{e^{2x}+e^{-2x}}{2}\right)+3$ | M1 | Collects appropriate terms and reaches form $\cosh 4x + p\cosh 2x + q$ or obtains values of $p$ and $q$ |
| $= \cosh 4x + 4\cosh 2x + 3$ | A1 | Correct expression or values e.g. $p=4$ and $q=3$ |

**Note:** No marks available in (a) if exponentials are not used but note that they may appear in combination with the use of hyperbolic identities.

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### Part (b) — Way 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses result from part (a) and $\cosh 2x = \pm 2\cosh^2 x \pm 1$ to obtain a quadratic equation in $\cosh^2 x$ | M1 | Or uses $\cosh 4x = \pm 2\cosh^2 2x \pm 1$ and $\cosh 2x = \pm 2\cosh^2 x \pm 1$ to obtain quadratic in $\cosh^2 x$ |
| $8\cosh^4 x - 42\cosh^2 x + 27 = 0$ | A1 | Correct 3TQ in $\cosh^2 x$ |
| $\cosh^2 x = \frac{9}{2}, \frac{3}{4}$ | M1 | Solves 3TQ in $\cosh^2 x$ to obtain $\cosh^2 x = k$ $(k \in \mathbb{R}$ and $>1)$. May be implied by their values |
| Takes square root to obtain $\cosh x = k$ $(k>1)$ and applies correct logarithmic form for arcosh or uses correct exponential form for $\cosh x$ to obtain at least one value for $x$ | M1 | The root(s) must be real to score this mark |
| $x = \pm\ln\!\left(\frac{3\sqrt{2}}{2}+\frac{\sqrt{14}}{2}\right)$ | A1 | Both correct and exact including brackets. Accept simplified equivalents e.g. $x=\ln\!\left(\frac{3}{\sqrt{2}}+\frac{\sqrt{7}}{\sqrt{2}}\right)$ but withhold if additional answers given unless they are the same |

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### Part (b) — Way 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Applies $\cosh 4x = \pm 2\cosh^2 2x \pm 1$ to obtain quadratic in $\cosh 2x$ | M1 | |
| $2\cosh^2 2x - 17\cosh 2x + 8 = 0$ | A1 | Correct 3TQ in $\cosh 2x$ |
| $\cosh 2x = 8, \frac{1}{2}$ | M1 | Solves 3TQ to obtain $\cosh 2x = k$ $(k \in \mathbb{R}$ and $>1)$ |
| Applies correct logarithmic form for arcosh from $\cosh 2x = k$ $(k>1)$ or uses correct exponential form to obtain at least one value for $2x$ | M1 | The root(s) must be real |
| $x = \pm\frac{1}{2}\ln(8+3\sqrt{7})$ or $x = \pm\ln(8+3\sqrt{7})^{\frac{1}{2}}$ | A1 | Both correct and exact with brackets. Accept simplified equivalents e.g. $x=\frac{1}{2}\ln(8+\sqrt{63})$ but withhold if additional answers given |

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### Part (b) — Way 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{e^{4x}+e^{-4x}}{2} - \frac{17}{2}(e^{2x}+e^{-2x})+9=0 \Rightarrow e^{8x}-17e^{6x}+18e^{4x}-17e^{2x}+1=0$ | M1A1 | M1: Applies correct exponential forms and attempts quartic in $e^{2x}$. A1: Correct equation |
| $e^{2x} = 8 \pm 3\sqrt{7}, \ldots$ | M1 | Solves and proceeds to value for $e^{2x}$ where $e^{2x}>1$ and real |
| $e^{2x} = 8\pm 3\sqrt{7} \Rightarrow 2x = \ln(8\pm 3\sqrt{7})$ | M1 | Takes ln to obtain at least one value for $2x$. Roots must be real |
| $x = \frac{1}{2}\ln(8\pm 3\sqrt{7})$ or $x = \ln(8\pm 3\sqrt{7})^{\frac{1}{2}}$ | A1 | Both correct and exact with brackets |

**Total: 8 marks**