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The functions f and g are defined by

\begin{aligned} & f(x)=\frac{\mathrm{e}^{x}+\mathrm{e}^{-x}}{2}, x \in \mathbb{R} \\ & g(x)=\frac{\mathrm{e}^{x}-\mathrm{e}^{-x}}{2}, x \in \mathbb{R} \end{aligned}

Let $t(x)=\frac{g(x)}{f(x)}$ .

Hence show that $t^{\prime}(x)>0$ for $x \in \mathbb{R}$ .

METHOD 1
$[f(x)]^{2}>[g(x)]^{2}$ (or equivalent)
$[f(x)]^{2}>0$
hence $t^{\prime}(x)>0, x \in \mathbb{R}$
Note: Award as above for use of either $f(x)=\frac{\mathrm{e}^{x}+\mathrm{e}^{-x}}{2}$ and $g(x)=\frac{\mathrm{e}^{x}-\mathrm{e}^{-x}}{2}$ or $\mathrm{e}^{x}+\mathrm{e}^{-x}$ and $\mathrm{e}^{x}-\mathrm{e}^{-x}$ .
METHOD 2

\begin{array}{lr} {[f(x)]^{2}-[g(x)]^{2}=1 \text { (or equivalent) }} & \text { M1A1 } \\ {[f(x)]^{2}>0} & \text { R1 } \\ \text { hence } t^{\prime}(x)>0, x \in \mathbb{R} & \text { AG } \end{array}

Note: Award as above for use of either $f(x)=\frac{\mathrm{e}^{x}+\mathrm{e}^{-x}}{2}$ and $g(x)=\frac{\mathrm{e}^{x}-\mathrm{e}^{-x}}{2}$ or $\mathrm{e}^{x}+\mathrm{e}^{-x}$ and $\mathrm{e}^{x}-\mathrm{e}^{-x}$ .

METHOD 3

IB Maths AA HL Sl 5 2 Increasing and Decreasing Functions Topic Practice

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IB Maths AA HL Sl 5 2 Increasing and Decreasing Functions Topic Practice

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Question 1