The point P(1, a) lies on the curve with equation $y = (x + 1)^{2}(2 - x)$ - Edexcel - A-Level Maths Pure - Question 10 - 2009 - Paper 1

The curve $y = (x + 1)^{2}(2 - x)$ is a cubic function. It has a minimum at x = -1 and its maximum intersection point can be calculated and sketched as follows:

1. Identify intercepts:
   • For the y-intercept (when x = 0):
     $y = (0 + 1)^{2}(2 - 0) = 1^{2}(2) = 2.$
   • For x-intercept (setting y=0):
     $0 = (x + 1)^{2}(2 - x) => (x + 1)^{2} = 0 ext{ or } (2 - x) = 0$ yielding x = -1 or x = 2.
2. Plot these critical points on the graph (0, 2), (-1, 0), and (2, 0).
3. Sketch the curve through these points, ensuring it displays a minimum at (-1, 0) and intersects the axes at the calculated points.

The curve $y = \frac{2}{x}$ is a hyperbola. It has the following characteristics:

1. Asymptotes at both axes.
2. Intersects the y-axis at (0, 2).
3. Intersects the x-axis at (2, 0), although x cannot actually be zero.
4. Sketch this hyperbola in each quadrant avoiding the axes.
5. Clearly denote the shape for both curves, indicating where they meet the axes and their attributes on the graph.

To find the number of real solutions, observe the intersections of the curves from part (b).

1. The first curve has at least two intersection points in the positive quadrant (roots) given its behavior.
2. Since the second curve approaches infinity but never touches the x-axis, the two curves, therefore, intersect at two points.
   This results in exactly two solutions to the equation.