# Question 3:

## Ways 1 & 2
| Answer/Working | Marks | Guidance |
|---|---|---|
| $4\tanh x - \operatorname{sech} x = 1 \Rightarrow 4\frac{\sinh x}{\cosh x} - \frac{1}{\cosh x} = 1 \Rightarrow 4\sinh x - 1 - \cosh x = 0 \Rightarrow 4\frac{e^x-e^{-x}}{2} - 1 - \frac{e^x+e^{-x}}{2} = 0$ | M1 | Replaces **one** hyperbolic function with correct exponential equivalent. Allow correct replacement of just e.g. $\sinh x$ using $\tanh x = \frac{\sinh x}{\cosh x}$. May follow errors but do not allow further marks if original equation was reduced to one in a single hyperbolic function |
| $3e^{2x} - 2e^x - 5 = 0$ | M1 A1 | M1: Obtains equation which if terms collected is a 3TQ (or 2TQ with no constant) in $e^x$. A1: Correct 3TQ |
| $e^x = \frac{2 \pm \sqrt{4+60}}{6} \left(\Rightarrow \frac{2+8}{6} = \frac{5}{3}\right)$ | M1 A1 | M1: Solves 3TQ (or 2TQ with no constant) in $e^x$. A1: Any correct unsimplified expression for $e^x$ including the positive root. Must be exact |
| $x = \ln\frac{5}{3}$ | A1 | $\ln\frac{5}{3}$, $\ln 1\frac{2}{3}$, $\ln 1.\dot{6}$ only but allow $k = \ldots$ No unrejected extra solutions |

## Way 3a — Squaring (sinh)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $4\sinh x - 1 = \cosh x \Rightarrow 16\sinh^2 x - 8\sinh x + 1 = \cosh^2 x \Rightarrow 16\sinh^2 x - 8\sinh x + 1 = 1 + \sinh^2 x$ | M1 | Squares (condone poor squaring) and uses correct hyperbolic identity to obtain quadratic equation in $\sinh x$ |
| $15\sinh^2 x - 8\sinh x = 0$ | M1 A1 | M1: Obtains 2TQ with no constant or 3TQ in $\sinh x$. A1: Correct 2TQ |
| $\sinh x = \frac{8}{15}$ | M1 | Solves 2TQ (with no constant) or 3TQ in $\sinh x$; must get correct non-zero root |
| $x = \operatorname{arsinh}\frac{8}{15} = \ln\!\left(\frac{8}{15}+\sqrt{\left(\frac{8}{15}\right)^2+1}\right)$ or $15e^{2x}-16e^x-15=0 \Rightarrow e^x = \frac{16\pm\sqrt{256+900}}{30}$ | A1 | Correct unsimplified expression for $x$ as a $\ln$. Must be exact |
| $x = \ln\frac{5}{3}$ | A1 | $\ln\frac{5}{3}$, $\ln 1\frac{2}{3}$, $\ln 1.\dot{6}$ only but allow $k = \ldots$ No unrejected extra solutions |

## Way 3b — Squaring (sech)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $4\tanh x = 1 + \operatorname{sech} x \Rightarrow 16\tanh^2 x = 1 + 2\operatorname{sech} x + \operatorname{sech}^2 x \Rightarrow 16(1-\operatorname{sech}^2 x) = 1 + 2\operatorname{sech} x + \operatorname{sech}^2 x$ | M1 | Squares (condone poor squaring) and uses correct hyperbolic identity to obtain quadratic in $\operatorname{sech} x$ |
| $17\operatorname{sech}^2 x + 2\operatorname{sech} x - 15 = 0$ | M1 A1 | M1: Obtains 2TQ (with no constant) or 3TQ in $\operatorname{sech} x$. A1: Correct 3TQ |
| $(17\operatorname{sech} x - 15)(\operatorname{sech} x + 1) = 0 \Rightarrow \operatorname{sech} x = \frac{15}{17}$ | M1 | Solves 2TQ with no constant or 3TQ in $\operatorname{sech} x$; must get correct non-zero root |
| $x = \operatorname{arcosh}\frac{17}{15} = \ln\!\left(\frac{17}{15}+\sqrt{\left(\frac{17}{15}\right)^2-1}\right)$ or $15e^{2x}-34e^x+15=0 \Rightarrow e^x = \frac{34\pm\sqrt{1156-900}}{30}$ | A1 | Correct unsimplified expression for $x$ as a $\ln$. Must be exact |
| $x = \ln\frac{5}{3}$ | A1 | $\ln\frac{5}{3}$, $\ln 1\frac{2}{3}$, $\ln 1.\dot{6}$ only but allow $k = \ldots$ No unrejected extra solutions |

# Question 3 (Way 3c - Squaring tanh):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $4\tanh x - 1 = \text{sech}\, x$ → $16\tanh^2 x - 8\tanh x + 1 = \text{sech}^2 x$ → $16\tanh^2 x - 8\tanh x + 1 = 1 - \tanh^2 x$ | M1 | Squares (condone poor squaring) and uses correct hyperbolic identity to obtain quadratic in $\tanh x$ |
| $17\tanh^2 x - 8\tanh x = 0$ | M1 A1 | M1: Obtains 2TQ with no constant or 3TQ in $\tanh x$; A1: Correct 2TQ |
| $\tanh x = \frac{8}{17}$ | M1 | Solves 2TQ with no constant or 3TQ in $\tanh x$. Apply usual rules. Must achieve one correct root |
| $x = \text{artanh}\,\frac{8}{17} = \frac{1}{2}\ln\!\left(\frac{1+\frac{8}{17}}{1-\frac{8}{17}}\right)$ or $9e^{2x}-25=0 \Rightarrow e^x = \frac{5}{3}$ | A1 | Correct unsimplified expression for $x$. Must be exact |
| $x = \ln\frac{5}{3}$ | A1 | Accept $\ln\frac{5}{3}$, $\ln 1\frac{2}{3}$, $\ln 1.\dot{6}$ only but allow $k=\ldots$; No unrejected extra solutions |

**Total: 6**

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# Question 3 (Way 3d - Squaring cosh):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $4\sinh x = 1 + \cosh x$ → $16\sinh^2 x = 1 + 2\cosh x + \cosh^2 x$ → $16\cosh^2 x - 16 = 1 + 2\cosh x + \cosh^2 x$ | M1 | Squares and uses correct hyperbolic identity to obtain quadratic in $\cosh x$ |
| $15\cosh^2 x - 2\cosh x - 17 = 0$ | M1 A1 | M1: Obtains 2TQ with no constant or 3TQ in $\cosh x$; A1: Correct 3TQ |
| $(15\cosh x - 17)(\cosh x + 1) = 0$ → $\cosh x = \frac{17}{15}$ | M1 | Solves 2TQ (no constant) or 3TQ in $\cosh x$. Apply usual rules |
| $x = \text{arcosh}\,\frac{17}{15} = \ln\!\left(\frac{17}{15}+\sqrt{\left(\frac{17}{15}\right)^2-1}\right)$ or $15e^{2x}-34e^x+15=0 \Rightarrow e^x = \frac{34\pm\sqrt{1156-900}}{30}$ | A1 | Correct unsimplified expression for $x$. Must be exact |
| $x = \ln\frac{5}{3}$ | A1 | Accept $\ln\frac{5}{3}$, $\ln 1\frac{2}{3}$, $\ln 1.\dot{6}$ only but allow $k=\ldots$; No unrejected extra solutions |

**Total: 6**

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