Fig. 9 shows the curve \(y = f(x)\), where \(f(x) = e^{2x} + k e^{-2x}\) and \(k\) is a constant greater than 1.
The curve crosses the \(y\)-axis at P and has a turning point Q.
\includegraphics{figure_9}
- Find the \(y\)-coordinate of P in terms of \(k\). [1]
- Show that the \(x\)-coordinate of Q is \(\frac{1}{4}\ln k\), and find the \(y\)-coordinate in its simplest form. [5]
- Find, in terms of \(k\), the area of the region enclosed by the curve, the \(x\)-axis, the \(y\)-axis and the line \(x = \frac{1}{4}\ln k\). Give your answer in the form \(ak + b\). [4]
The function \(g(x)\) is defined by \(g(x) = f(x + \frac{1}{4}\ln k)\).
- Show that \(g(x) = \sqrt{k}(e^{2x} + e^{-2x})\). [3]
- Hence show that \(g(x)\) is an even function. [2]
- Deduce, with reasons, a geometrical property of the curve \(y = f(x)\). [3]