Fig. 9 shows the curves \(y = \text{f}(x)\) and \(y = \text{g}(x)\). The function \(y = \text{f}(x)\) is given by
$$\text{f}(x) = \ln \left( \frac{2x}{1+x} \right), \quad x > 0.$$
The curve \(y = \text{f}(x)\) crosses the \(x\)-axis at P, and the line \(x = 2\) at Q.
\includegraphics{figure_9}
- Verify that the \(x\)-coordinate of P is 1.
Find the exact \(y\)-coordinate of Q. [2]
- Find the gradient of the curve at P. [Hint: use \(\frac{a}{b} = \ln a - \ln b\).] [4]
The function \(\text{g}(x)\) is given by
$$\text{g}(x) = \frac{e^x}{2-e^x}, \quad x < \ln 2.$$
The curve \(y = \text{g}(x)\) crosses the \(y\)-axis at the point R.
- Show that \(\text{g}(x)\) is the inverse function of \(\text{f}(x)\).
Write down the gradient of \(y = \text{g}(x)\) at R. [5]
- Show, using the substitution \(u = 2 - e^x\) or otherwise, that \(\int_0^{\ln \frac{4}{3}} \text{g}(x) dx = \ln \frac{3}{2}\).
Using this result, show that the exact area of the shaded region shown in Fig. 9 is \(\ln \frac{32}{27}\).
[Hint: consider its reflection in \(y = x\).] [7]