The curve is given by the equation y = 2x(3−x). a) State the coordinates of the point B. b) Show that the value of k is (5)/(2). c) Find an equation of the straight line passing through D and E.

This equation gives two solutions: $2x=0 \implies x=0$ (which corresponds to point O) or $3-x=0 \implies x=3$.
So, point B has an x-coordinate of 3.
The coordinates of point B are $(3, 0)$.

b) Show that the value of $k$ is $\frac{5}{2}$.
The straight line crosses the y-axis at point E with coordinates $(0, 5)$. It meets the curve at points C and D. The coordinates of point D are $(k, k)$.
Since point D $(k, k)$ lies on the curve, its coordinates must satisfy the curve's equation $y = 2x(3-x)$.

$$k = 6k - 2k^2$$ $$2k^2 - 5k = 0$$

Factor out $k$:

$$k(2k - 5) = 0$$

This gives two possible values for $k$: $k=0$ or $2k-5=0$.
If $k=0$, then point D would be $(0,0)$, which is point O. From the diagram, D is clearly not O.
Therefore, $2k-5=0 \implies 2k=5 \implies k=\frac{5}{2}$.
Thus, the value of $k$ is $\frac{5}{2}$.

c) Find an equation of the straight line passing through D and E.
Point E is $(0, 5)$. Point D is $(k, k) = (\frac{5}{2}, \frac{5}{2})$.

$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\frac{5}{2} - 5}{\frac{5}{2} - 0}$$ $$m = \frac{\frac{5}{2} - \frac{10}{2}}{\frac{5}{2}} = \frac{-\frac{5}{2}}{\frac{5}{2}} = -1$$

Since point E is $(0, 5)$, the y-intercept $c$ is 5.
Using $m=-1$ and $c=5$:

$$y = -1x + 5$$

The equation of the straight line is $y = -x + 5$.

d) Show that the x-coordinates of the point C is 1.
Point C is an intersection of the curve $y = 2x(3-x)$ and the straight line $y = -x + 5$.

$$2x(3-x) = -x + 5$$ $$6x - 2x^2 = -x + 5$$

We can factorize the quadratic equation:

$$2x^2 - 2x - 5x + 5 = 0$$ $$2x(x-1) - 5(x-1) = 0$$ $$(2x-5)(x-1) = 0$$

This gives two solutions for $x$: $2x-5=0 \implies x=\frac{5}{2}$ or $x-1=0 \implies x=1$.
The x-coordinate $x=\frac{5}{2}$ corresponds to point D.
Therefore, the x-coordinate of point C is 1.

e) Calculate the area of the region R bounded by the curve.
The region R is bounded by the curve $y = 2x(3-x)$ and the straight line $y = -x + 5$. The region is between the x-coordinates of C and D.
The x-coordinate of C is 1. The x-coordinate of D is $\frac{5}{2}$.
The area R is given by the integral of the upper function minus the lower function, from $x=1$ to $x=\frac{5}{2}$.
The upper function is the curve $y_c = 2x(3-x) = 6x - 2x^2$.
The lower function is the line $y_l = -x + 5$.

$$\text{Area R} = \int_{1}^{\frac{5}{2}} [(6x - 2x^2) - (-x + 5)]\,dx$$ $$\text{Area R} = \int_{1}^{\frac{5}{2}} (6x - 2x^2 + x - 5)\,dx$$ $$\text{Area R} = \int_{1}^{\frac{5}{2}} (-2x^2 + 7x - 5)\,dx$$

First, evaluate at $x=\frac{5}{2}$:

$$F\left(\frac{5}{2}\right) = -\frac{2}{3}\left(\frac{5}{2}\right)^3 + \frac{7}{2}\left(\frac{5}{2}\right)^2 - 5\left(\frac{5}{2}\right)$$ $$F\left(\frac{5}{2}\right) = -\frac{2}{3}\left(\frac{125}{8}\right) + \frac{7}{2}\left(\frac{25}{4}\right) - \frac{25}{2}$$ $$F\left(\frac{5}{2}\right) = -\frac{125}{12} + \frac{175}{8} - \frac{25}{2}$$

Find a common denominator, which is 24:

$$F\left(\frac{5}{2}\right) = -\frac{125 \times 2}{24} + \frac{175 \times 3}{24} - \frac{25 \times 12}{24}$$ $$F\left(\frac{5}{2}\right) = -\frac{250}{24} + \frac{525}{24} - \frac{300}{24}$$ $$F\left(\frac{5}{2}\right) = \frac{-250 + 525 - 300}{24} = \frac{275 - 300}{24} = -\frac{25}{24}$$

Next, evaluate at $x=1$:

$$F(1) = -\frac{2(1)^3}{3} + \frac{7(1)^2}{2} - 5(1)$$ $$F(1) = -\frac{2}{3} + \frac{7}{2} - 5$$

Find a common denominator, which is 6:

$$F(1) = -\frac{2 \times 2}{6} + \frac{7 \times 3}{6} - \frac{5 \times 6}{6}$$ $$F(1) = -\frac{4}{6} + \frac{21}{6} - \frac{30}{6}$$ $$F(1) = \frac{-4 + 21 - 30}{6} = \frac{17 - 30}{6} = -\frac{13}{6}$$

Finally, calculate the area:

$$\text{Area R} = F\left(\frac{5}{2}\right) - F(1)$$ $$\text{Area R} = -\frac{25}{24} - \left(-\frac{13}{6}\right)$$ $$\text{Area R} = -\frac{25}{24} + \frac{13}{6}$$

Convert $\frac{13}{6}$ to have a denominator of 24:

$$\text{Area R} = -\frac{25}{24} + \frac{13 \times 4}{6 \times 4}$$ $$\text{Area R} = -\frac{25}{24} + \frac{52}{24}$$ $$\text{Area R} = \frac{52 - 25}{24} = \frac{27}{24}$$

Simplify the fraction:

$$\text{Area R} = \frac{9}{8}$$

a) The coordinates of point B are $\boxed{(3, 0)}$.

b) The value of $k$ is $\boxed{\frac{5}{2}}$.

c) The equation of the straight line is $\boxed{y = -x + 5}$.

d) The x-coordinate of point C is $\boxed{1}$.

e) The area of the region R is $\boxed{\frac{9}{8} \text{ units}^2}$.