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

We start with the expression:

$3 ext{sin} x + 4 ext{cos} x$

To express this in the desired form, we need to identify $A$ and $heta$:

1. Calculate $A$: $A = ext{sqrt}(3^2 + 4^2) = ext{sqrt}(9 + 16) = ext{sqrt}(25) = 5$

2. Find $heta$: an heta = rac{4}{3} ext{, hence } heta = an^{-1} rac{4}{3}

Thus, we express it as: $3 ext{sin} x + 4 ext{cos} x = 5 ext{sin}(x + heta)$, with 0 ext{ } heta ext{ } rac{ au}{2}.

From the previous part, we have: $5 ext{sin}(x + heta) = 5$ Dividing by 5 gives: $ext{sin}(x + heta) = 1$ The solutions for $ext{sin}(x + heta) = 1$ are: x + heta = rac{ au}{2} + k au ext{ where } k ext{ is any integer} Thus we can solve for $x$: x = rac{ au}{2} - heta + k au

Considering the range $[0, 2 au]$, finding $x$ results in:

1. For $k=0$, compute: x = rac{ au}{2} - an^{-1}rac{4}{3}
2. For $k=1$, compute: x = rac{3 au}{2} - an^{-1}rac{4}{3}

Evaluating these gives specific solutions. All values must then be rounded to two decimal places.

The tangent line to the parabola at point $P(2, r^2)$ can be derived using the derivative. The slope of the tangent line at $(x^2 = 4y)$ can be found as follows:

1. The equation of the parabola is: y = rac{x^2}{4} Differentiating gives: rac{dy}{dx} = rac{x}{2}

2. At $P(2, r^2)$, the slope is: m = rac{2}{2} = 1

3. Using the point-slope form for the equation of the line yields: [y - r^2 = m(x - 2)] Substituting $m = 1$ gives: $y - r^2 = (x - 2)$ Rearranging results in: $y = mx - r^2$, showing the tangent at $P$.

1. Following a similar process, find the slope at $Q(4, 4^2)$: m = rac{4}{2} = 2 Thus, the equation of the line at $Q$ in point-slope form is: $y - 16 = 2(x - 4)$ Simplifying results in: $y = 2x - 8$.

2. To find point $R$, we need the intersection of the tangents: Set: $mx - r^2 = 2x - 8$ Thus: $(1 - 2)x = r^2 - 8$ Solving gives: Coordinates of $R$ can be presented as $(x_R, y_R)$ in terms of $t$.

To find the locus of point $R$, it involves deriving the relationship between $x$ and $y$ based on the earlier equations obtained in terms of $t$. Using the known coordinates of $P$ and $Q$, we can relate both:

1. Establish a relation between both tangents and equalize their parameters.
2. This results in an equation that describes $y$ as a function of $x$. The explicit form will lead to the Cartesian relationship needed.