## Question 7(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \arccos(\text{sech } x)$, differentiates e.g. $\cos y = \text{sech } x \Rightarrow -\sin y \frac{dy}{dx} = -\text{sech } x \tanh x$ or $-\sin y = -\text{sech } x \tanh x \frac{dx}{dy}$ or $\cos y = (\cosh x)^{-1} \Rightarrow -\sin y \frac{dy}{dx} = -(\cosh x)^{-2}\sinh x$ | M1 | Differentiates to obtain equation in $\frac{dy}{dx}$ or $\frac{dx}{dy}$ of correct form; condone coefficient sign errors only |
| $\frac{dy}{dx} = \frac{\text{sech } x \tanh x}{\tanh x}$ or $\sqrt{1-\text{sech}^2 x}\frac{dy}{dx} = \text{sech } x \tanh x \Rightarrow \tanh x \frac{dy}{dx} = \text{sech } x \tanh x$ or $\sqrt{1-\text{sech}^2 x}\frac{dy}{dx} = \frac{\sinh x}{\cosh^2 x} \Rightarrow \tanh x \frac{dy}{dx} = \frac{\sinh x}{\cosh^2 x}$ | dM1 | Uses correct identities to obtain equation in $\frac{dy}{dx}$ or $\frac{dx}{dy}$ in terms of $x$ only with no roots; accept $\sqrt{\tanh^2 x}$ as "no roots" |
| $\Rightarrow \frac{dy}{dx} = \text{sech } x$ | A1* | Fully correct proof; must see equation in $\frac{dy}{dx}$ or $\frac{dx}{dy}$ with exactly two different hyperbolic functions and no roots before given answer. Withhold for clear errors e.g. $\frac{d}{dx}(\arccos x) = +\frac{1}{\sqrt{1-x^2}}$ or $\frac{d}{dx}(\text{sech } x) = +\text{sech } x \tanh x$; allow slips if recovered |

**(3 marks)**

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## Question 7(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d}{dx}(\coth x) = -\text{cosech}^2 x$ or equivalent forms e.g. $\frac{\sinh^2 x - \cosh^2 x}{\sinh^2 x}$ or $\frac{-\text{sech}^2 x}{\tanh^2 x}$ or $1 - \coth^2 x$ or exponential equivalents | B1 | Correct derivative of $\coth x$ in any form; allow recovery if $-\text{cosec}^2 x$ written when $-\text{cosech}^2 x$ clearly implied |
| $\text{sech } x - \text{cosech}^2 x = 0 \Rightarrow \text{sech } x = \text{cosech}^2 x \Rightarrow \frac{1}{\cosh x} = \frac{1}{\sinh^2 x}$ leading to $a\cosh^2 x + b\cosh x + c = 0$ or $a\text{sech}^2 x + b\text{sech } x + c = 0$, **or** substitutes exponential forms to obtain $Ae^{4x}+Be^{3x}+Ce^{2x}+De^x+E=0$ | M1 | Sets $f'(x)=0$ **and** uses correct identities to obtain 3TQ in $\cosh x$ or $\text{sech } x$, **or** substitutes correct exponential forms to get 5-term quartic in $e^x$ |
| $\cosh^2 x - \cosh x - 1 = 0$ or $\text{sech}^2 x + \text{sech } x - 1 = 0$ **or** $e^{4x} - 2e^{3x} - 2e^{2x} - 2e^x + 1 = 0$ | A1 | Correct quadratic or correct quartic equation |
| $\cosh x = \frac{-(-1)\pm\sqrt{(-1)^2-4(1)(-1)}}{2(1)} = \frac{1+\sqrt{5}}{2}$ or $\left(\text{sech } x + \frac{1}{2}\right)^2 - \frac{1}{4} - 1 = 0 \Rightarrow \text{sech } x = \frac{-1+\sqrt{5}}{2}$ | dM1 | Solves quadratic in $\text{sech } x$ or $\cosh x$; must obtain real and exact value $>1$ (or between 0 and 1 if sech used); reject invalid values |
| $x = \text{arcosh}\!\left(\frac{1+\sqrt{5}}{2}\right) = \ln\!\left(\frac{1+\sqrt{5}}{2}+\sqrt{\left(\frac{1+\sqrt{5}}{2}\right)^2-1}\right)$ **or** $\frac{e^x+e^{-x}}{2}=\frac{1+\sqrt{5}}{2}\Rightarrow e^{2x}-(1+\sqrt{5})e^x+1=0 \Rightarrow e^x = \frac{1+\sqrt{5}+\sqrt{(1+\sqrt{5})^2-4}}{2}$ | ddM1 | Uses correct logarithmic form or exponentials to find $x$ as a ln of exact value; exponential definition must be correct and quadratic solving subject to usual rules |
| $x = \ln\!\left(\frac{1}{2}(1+\sqrt{5})+\sqrt{\frac{1}{2}(1+\sqrt{5})}\right)$ or accept $x = \ln\!\left(\frac{1+\sqrt{5}}{2}+\sqrt{\frac{1+\sqrt{5}}{2}}\right)$ | A1 | Correct answer — any exact simplified equivalent; note $x=\ln\frac{1}{2}(1+\sqrt{5})+\sqrt{\frac{1}{2}(1+\sqrt{5})}$ scores A0 |

**(6 marks)**

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