(a) The graph of \(f\left( x \right) = \sqrt {4 - {x^2}} \) is a semicircle with radius 2 units. Hence, the area bounded by the graph of \(f\) and the x-axis is \(\frac{1}{2}{\rm{\pi }}{r^2} = \frac{1}{2}{\rm{\pi }}{2^2} = 2{\rm{\pi }}\;\;{\rm{unit}}{{\rm{s}}^2}\).(b) (i) For any value of \(p\) the graph of \(g\left( x \right) = \sqrt {\frac{x + 2}{p}} \) will contain the point \(\left( { - \,2,0} \right)\). Below are graphs of g for \(p = 1\) and \(p = 9\). Thus, the point of intersection where the x-coordinate is negative is always \(\left( { - \,2,0} \right)\).(ii) \(\sqrt {4 - {x^2}} = \sqrt {\frac{{x + 2}}{p}} \;\;\; \Rightarrow \;\;\;4 - {x^2} = \frac{{x + 2}}{p}\;\;\; \Rightarrow \;\;\;4p - p{x^2} = x + 2\)\(p{x^2} + x + 2 - 4p = 0\)