The following diagram shows the graph of $y = g(x)$ - HSC - SSCE Mathematics Extension 2 - Question 3 - 2008 - Paper 1

To sketch $y = \frac{1}{g(x)}$, identify where $g(x)$ intersects the x-axis (the asymptotes) and exactly where it approaches $0$. For sections where $g(x) > 0$, take the reciprocal of $g(x)$ to find the corresponding values on this graph. As $g(x)$ approaches $0$, $y$ approaches infinity.

For $x \geq 1$, simply plot $g(x)$ from the original graph. For $x < 1$, calculate $f(x)$ using the defined piecewise condition by substituting $x$ values less than 1 into the expression $f(x) = \frac{g(2-x)}{g(x)}$. The overall graph should connect smoothly at $x = 1$.

Begin by considering the function $p(z) = 1 + z^2 + z^3$. To find real zeros, analyze the behavior of $p(z)$ by inspecting the sum of positive terms. Since $1$, $z^2$, and $z^3$ are all non-negative for any real $z$, it follows that $p(z) > 0$ for all real $z$, implying no real zeros exist.

Since eta is a zero of $p(z)$, use the property of $p(z)$ to express conditions on eta. Notably, verify if eta^3 = -1 leads to eta^6 = 1. This concludes that all the sixth roots of unity, eta^6 = 1, hold.

Substituting eta^2 into $p(z)$: Calculate $p(\beta^2) = 1 + (\beta^2)^2 + (\beta^2)^3$ and simplify. Using previous results, show that $p(\beta^2) = 0$, thereby establishing that eta^2 is indeed a zero.

Start by integrating the expression for $I_n$. Utilize integration by parts, letting $u = \tan^{2n-1}\theta$ and $dv = \tan\theta \, d\theta$. This produces relationships between $I_n$ and $I_{n-1}$ as derived.

Using the established recurrence relation, calculate down from $I_3$ using the relationship. Solve from $I_1$ through $I_2$ until reaching $I_3$, executing the integration to reveal the values recursively.

Analyze the forces acting on the particle. The vertical component of tension combined with the gravitational force leads to a relation involving $\omega$. Set up the equation from the components of these forces, consequently isolating and solving for $\omega^2$.