Fig. 8 shows parts of the curves \(y = f(x)\) and \(y = g(x)\), where \(f(x) = \tan x\) and \(g(x) = 1 + f(x - \frac{1}{4}\pi)\).
\includegraphics{figure_8}
- Describe a sequence of two transformations which maps the curve \(y = f(x)\) to the curve \(y = g(x)\). [4]
It can be shown that \(g(x) = \frac{2\sin x}{\sin x + \cos x}\).
- Show that \(g'(x) = \frac{2}{(\sin x + \cos x)^2}\). Hence verify that the gradient of \(y = g(x)\) at the point \((\frac{1}{4}\pi, 1)\) is the same as that of \(y = f(x)\) at the origin. [7]
- By writing \(\tan x = \frac{\sin x}{\cos x}\) and using the substitution \(u = \cos x\), show that \(\int_0^{\frac{1}{4}\pi} f(x)dx = \int_{\frac{1}{\sqrt{2}}}^1 \frac{1}{u}du\).
Evaluate this integral exactly. [4]
- Hence find the exact area of the region enclosed by the curve \(y = g(x)\), the \(x\)-axis and the lines \(x = \frac{1}{4}\pi\) and \(x = \frac{1}{2}\pi\). [2]