(b)

1. Sketch the graph of $y = -3\sin x$ for $0 \le x \le 2\pi$

The graph of $y = -3\sin x$ is an inverted sine wave with amplitude 3 and period $2\pi$.
Key points:

• $x = 0, y = 0$
• $x = \frac{\pi}{2}, y = -3$
• $x = \pi, y = 0$
• $x = \frac{3\pi}{2}, y = 3$
• $x = 2\pi, y = 0$
  Draw a smooth curve through these points.

2. Sketch the graph of $y = 1 - \frac{2x}{\pi}$ for $0 \le x \le 2\pi$

The graph of $y = 1 - \frac{2x}{\pi}$ is a straight line.
Find the endpoints:

• When $x = 0$, $y = 1 - 0 = 1$. Point is $(0, 1)$.
• When $x = 2\pi$, $y = 1 - \frac{2(2\pi)}{\pi} = 1 - 4 = -3$. Point is $(2\pi, -3)$.
  Draw a straight line connecting $(0, 1)$ and $(2\pi, -3)$.

1. Rearrange the given equation to match the equations of the graphs sketched in part (i).

$2x = \pi + 3\pi \sin x$
Divide both sides by $\pi$:
$\frac{2x}{\pi} = 1 + 3\sin x$
Rearrange to isolate $-3\sin x$:
$-3\sin x = 1 - \frac{2x}{\pi}$

2. Determine the number of intersections between the two graphs.

The solutions to the equation correspond to the points of intersection between the curve $y = -3\sin x$ and the line $y = 1 - \frac{2x}{\pi}$.
Analyzing the intervals:

• At $x = 0$: curve $y = 0$, line $y = 1$ (curve < line)
• At $x = \frac{\pi}{2}$: curve $y = -3$, line $y = 0$ (curve < line)
• At $x = \pi$: curve $y = 0$, line $y = -1$ (curve > line)
• For $\pi < x \le 2\pi$: curve $y \ge 0$, line $y \le -1$ (curve > line)
  Since the continuous curve crosses the line exactly once between $x = \frac{\pi}{2}$ and $x = \pi$, there is 1 point of intersection.

1. Deduce that line $AB$ is perpendicular to line $BC$.

Since $AD \parallel BC$ and $\angle BAD = 90^\circ$, the interior angle $\angle ABC$ must also be $180^\circ - 90^\circ = 90^\circ$.
Therefore, $AB \perp BC$.

2. Express the gradients of $AB$ and $BC$ in terms of $k$ and set up an equation.

Gradient of $AB = \frac{8 - 4}{k - (-2)} = \frac{4}{k + 2}$
Gradient of $BC = \frac{4 - 8}{6 - k} = \frac{-4}{6 - k}$
Since $AB \perp BC$, the product of their gradients is $-1$:
$\left(\frac{4}{k + 2}\right) \left(\frac{-4}{6 - k}\right) = -1$

3. Solve the equation to show that $k = 2$.

$\frac{-16}{(k + 2)(6 - k)} = -1$
$16 = (k + 2)(6 - k)$
$16 = 6k - k^2 + 12 - 2k$
$16 = -k^2 + 4k + 12$
$k^2 - 4k + 4 = 0$
$(k - 2)^2 = 0$
$k = 2 \text{ (shown)}$