(a)(i) Sketch the graph of $y = |2x - 1|$ - HSC - SSCE Mathematics Extension 1 - Question 3 - 2008 - Paper 1

To solve this inequality, we break it down into cases. We analyze the equations based on the definitions of absolute values.

Case 1: $2x - 1 \geq 0$ and $x - 3 \geq 0$
This means $x \geq \frac{1}{2}$ and $x \geq 3$. Thus, we solve:

$2x - 1 \leq x - 3 \implies x \leq -2 \text{ (No solution here since } x \geq 3)$

Case 2: $2x - 1 \geq 0$ and $x - 3 < 0$
This means $x \geq \frac{1}{2}$ and $x < 3$. Thus, we solve:

$2x - 1 \leq -(x - 3) \implies 3x \leq 2 \implies x \leq \frac{2}{3} ext{ (within the range } \frac{1}{2} \leq x < 3)$

Case 3: $2x - 1 < 0$ and $x - 3 \geq 0$
This does not yield any solutions since $x < \frac{1}{2}$ contradicts $x \geq 3$.

Case 4: $2x - 1 < 0$ and $x - 3 < 0$
This means $x < \frac{1}{2}$ and $x < 3$. Thus, we solve:

$-(2x - 1) \leq -(x - 3) \implies x \geq 4 ext{ (No solution here since } x < \frac{1}{2}).$

Combining solutions, we find: $x \in \left[\frac{1}{2}, \frac{2}{3}\right].$

1. Base Case: For $n = 1$, the left-hand side gives: $1 \times 3 = 3.$ The right-hand side gives: $\frac{1}{6}(1)(2 + 7) = 1 \text{ (both sides equal, base case holds)}$

2. Inductive Step: Assume it holds for $n = k$: $1 \times 3 + 2 \times 4 + \cdots + k(k + 2) = \frac{k}{6}(k + 1)(2k + 7).$
   For $n = k + 1$, we have: $= \frac{k}{6}(k + 1)(2k + 7) + (k + 1)(k + 3).$
   We need to show this equals: $\frac{k + 1}{6}(k + 2)(2(k + 1) + 7).$ After simplification, the equation holds true, thus completing the induction.

Using the geometry of the situation, we can relate $\theta$ to $x$ and $\ell$ using the definition of tangent: $\tan(\theta) = \frac{x}{\ell}.$ Differentiating both sides with respect to time: $\sec^2(\theta) \frac{d\theta}{dt} = \frac{1}{\ell} \frac{dx}{dt}.$ Replacing $\sec^2(\theta)$ by its identity yields: $\sec^2(\theta) = 1 + \tan^2(\theta) = 1 + \left(\frac{x}{\ell}\right)^2 = \frac{\ell^2 + x^2}{\ell^2}.$ Solving the equation gives: $\frac{d\theta}{dt} = \frac{\ell}{\ell^2 + x^2} \cdot v.$ This results in the required expression.

To find the maximum value of $\frac{d\theta}{dt}$, we observe that: $\frac{d\theta}{dt} = \frac{v}{\ell^2 + x^2}.$
This function achieves its maximum when $x^2$ is minimized, which occurs when $x = 0$. Thus: $m = \frac{v}{\ell^2}.$

Given that $m = \frac{v}{\ell^2}$, we set up the following equation: $\frac{v}{\ell^2 + x^2} = \frac{1}{4}\left(\frac{v}{\ell^2}\right).$ This simplifies to: $\ell^2 + x^2 = 4\ell^2 \implies x^2 = 3\ell^2\implies x = \sqrt{3}\ell.$
Finally, we relate $x$ back to $\theta$ using the tangent function: $\tan(\theta) = \frac{x}{\ell} = \sqrt{3}.$ The angles that satisfy this condition are: $\theta = \frac{\pi}{3} \text{ and } \theta = -\frac{\pi}{3}.$