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

To solve the inequality $|2x - 1| \leq |x - 3|$, we break it into cases based on the critical points encountered in both absolute values:

1. Determine the critical points: $2x - 1 = 0$ gives $x = \frac{1}{2}$, and $x - 3 = 0$ gives $x = 3$. Thus, we analyze intervals determined by these points: ((-\infty, \frac{1}{2})$,$[\frac{1}{2}, 3]$, and$(3, \infty)$.

2. Interval 1: For $x < \frac{1}{2}$, The inequality to solve is $-(2x - 1) \leq -(x - 3)$ which simplifies to $2x - 1 \geq x - 3$ leading to $x \geq -2$.

3. Interval 2: For $x \in [\frac{1}{2}, 3]$, Here, $2x - 1 \leq x - 3$ gives $x \leq -2$. This interval yields no solutions.

4. Interval 3: For $x > 3$, We have $2x - 1 \leq 2x - 6$, leading to $1 \leq -6$, which is false.

5. The results yield a non-overlapping region suggesting $x \in [-2, \frac{1}{2}]$. Thus, the solution to the inequality is $x \in [-2, \frac{1}{2}]$.

To prove the equation $1 \times 3 + 2 \times 4 + 3 \times 5 + \cdots + n(n + 2) = \frac{n}{6}(n + 1)(2n + 7)$ by induction:

1. Base case: For $n = 1$, we check: LHS = $1 \times 3 = 3$ and RHS = $\frac{1}{6}(2)(9) = 3$, thus both sides are equal.

2. Inductive Step: Assume for $n = k$, the equation holds: $1 \times 3 + 2 \times 4 + \ldots + k(k + 2) = \frac{k}{6}(k + 1)(2k + 7)$

3. We need to show it holds for $n = k + 1$: LHS = $1 \times 3 + 2 \times 4 + \ldots + k(k + 2) + (k + 1)(k + 3)$ Substituting the inductive hypothesis: $= \frac{k}{6}(k + 1)(2k + 7) + (k + 1)(k + 3)$

4. Combine like terms to manipulate the expression, factor out $(k + 1)$. After simplification, show that it equals the RHS for $n = k + 1$. This completes the proof.

To demonstrate that $\frac{d\theta}{dt} = \frac{v}{\ell^2 + x^2}$, apply the following:

1. Using the relationship of the triangle formed, we know that: $\tan(\theta) = \frac{x}{\ell}$ Taking the derivative with respect to time $t$ yields: $\sec^2(\theta) \frac{d\theta}{dt} = \frac{1}{\ell}\frac{dx}{dt} = \frac{v}{\ell}$

2. Given that $\sec^2(\theta) = 1 + \tan^2(\theta) = 1 + \frac{x^2}{\ell^2}$, substitute this in to get: $\frac{d\theta}{dt} = \frac{v}{\ell \sec^2(\theta)} = \frac{v}{\ell} \cdot \frac{\ell^2}{\ell^2 + x^2}$

3. Simplifying this gives the required expression, $\frac{d\theta}{dt} = \frac{v}{\ell^2 + x^2}$.

To determine the maximum value $m$ of $\frac{d\theta}{dt}$, first note:

1. The derived formula is $\frac{d\theta}{dt} = \frac{v}{\ell^2 + x^2}$. The maximum occurs when $x$ is minimized.

2. The minimum value of $x$ happens when the race car is closest to $O$, ideally at $x = 0$. Thus: $m = \frac{v}{\ell^2 + 0^2} = \frac{v}{\ell^2}$

3. Therefore, we conclude that $m = \frac{v}{\ell^2}$.

To find the two values of $\theta$ such that $\frac{d\theta}{dt} = \frac{m}{4}$:

1. Substitute $m$ into the equation: $\frac{v}{\ell^2 + x^2} = \frac{1}{4}\frac{v}{\ell^2}$ This leads to: $\ell^2 + x^2 = 4\ell^2$ implying that $x^2 = 3\ell^2$, hence: $x = \sqrt{3}\ell$ and condition can yield $x = -\sqrt{3}\ell$ when considering symmetry.

2. Using the aforementioned $x$ in $\tan(\theta) = \frac{x}{\ell}$ gives: $\tan(\theta) = \frac{\sqrt{3}}{1} \Rightarrow \theta = \frac{\pi}{3}$ which also gives $\tan(\theta) = -\frac{\sqrt{3}}{1}$ yielding: $\theta = -\frac{\pi}{3}$.

3. Thus, the two values of $\theta$ are: $\theta = \frac{\pi}{3},\ -\frac{\pi}{3}$.