The polynomial $p(x) = x^3 - ax + b$ has a remainder of 2 when divided by $(x - 1)$ and a remainder of 5 when divided by $(x + 2)$ - HSC - SSCE Mathematics Extension 1 - Question 2 - 2009 - Paper 1

To express $3 \sin x + 4 \cos x$ in the required form, we find:

1. $A = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$
2. To find $\alpha$, we use: $\tan \alpha = \frac{4}{3}$ Therefore: $\alpha = \tan^{-1}\left(\frac{4}{3}\right)$

So we can rewrite: $3 \sin x + 4 \cos x = 5 \sin\left(x + \alpha\right)$

Now we solve: $5 \sin\left(x + \alpha\right) = 5$ This implies: $\sin \left(x + \alpha\right) = 1$ The general solution for $\sin$ is given by: $x + \alpha = \frac{\pi}{2} + 2k\pi \; (k \in \mathbb{Z})$ Hence: $x = \frac{\pi}{2} - \alpha + 2k\pi$

For $0 \leq x \leq 2\pi$: Calculate:

1. When $k=0$, $x = \frac{\pi}{2} - \tan^{-1}\left(\frac{4}{3}\right)$
2. When $k=1$, this value will exceed $2\pi$. Check for any required cases and adjust for the range to find the specific answers.

Taking the derivative of the function $y = \frac{x^2}{4}$:

1. The derivative is: $\frac{dy}{dx} = \frac{x}{2}$
2. At point $P(2, r^2)$, we find: $\frac{dy}{dx} = \frac{2}{2} = 1$
   Therefore, the slope $m = 1$.
3. The tangent line at $P$ is given by: $y - r^2 = m(x - 2)$ Simplifying gives us: $y = x - 2 + r^2 = x - r^2$ Thus: $y = mx - r^2$

For point $Q(4, 4)$:

1. Calculate the derivative at $Q$: $\frac{dy}{dx} = \frac{4}{2} = 2$ Here, the slope $m = 2$.

2. The tangent equation at $Q$ is: $y - 4 = 2(x - 4)$ Then: $y = 2x - 8 + 4 = 2x - 4$

3. To find point $R$, set the two tangent equations equal: $x - r^2 = 2x - 4$ Rearranging gives us: $x - 2 = r^2$ Thus, express $R$ as: $R(x, x - 2)$

To find the Cartesian equation of point $R$:

1. Substitute $r^2$ in terms of $x$: $y = x - 2$ Thus, the locus of $R$ is represented as: $y = x - 2$
2. This gives us a linear equation which describes the path of point $R$ as it moves along the trajectory formed by points $P$ and $Q$.