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

Given that $\alpha$ is a zero of $\rho(z)$, we have:

$\rho(\alpha) = 1 + \alpha^2 + \alpha^4 = 0.$

This can be rearranged to:

$\alpha^4 + \alpha^2 + 1 = 0.$

Letting $x = \alpha^2$, this gives:

$x^2 + x + 1 = 0.$

Using the quadratic formula, we find:

$x = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2}.$

The roots of $\alpha$, therefore, are complex. Use properties of complex numbers:

$\alpha^6 = \alpha^4 \cdot \alpha^2 = (-\alpha^2 - 1) \cdot \alpha^2 = -\alpha^4 - \alpha^2 = 1.$

To show that $\alpha^2$ is also a zero of $\rho(z)$, we can substitute $\alpha^2$ back into $\rho(z)$:

$\rho(\alpha^2) = 1 + (\alpha^2)^2 + (\alpha^2)^4 = 1 + \alpha^4 + \alpha^8.$

Using the earlier result from part (ii):

• Since $\alpha^4 + \alpha^2 + 1 = 0$, we know that $\alpha^8 = 1$ (because $\alpha^6 = 1$). Hence, we substitute it back:

$\rho(\alpha^2) = 1 + \alpha^4 + 1 = 0.$

Thus, $\alpha^2$ is a zero of $\rho(z)$.

To show that:

$I_n = \int_0^{\frac{\pi}{4}} \tan^{2n} \theta \, d\theta,$

we can use integration by parts. Let:

• $u = \tan^{2n - 1} \theta$ and $dv = \tan \theta \, d\theta$.

Then,

• $du = (2n - 1) \tan^{2n - 2} \theta \sec^2 \theta \, d\theta$ and $v = \int \tan \theta \, d\theta = -\ln(\cos \theta)$.

Applying integration by parts:

$I_n = - \left[ \tan^{2n - 1} \theta \ln(\cos \theta) \right]_0^{\frac{\pi}{4}} + \int_0^{\frac{\pi}{4}} \ln(\cos \theta) (2n - 1) \tan^{2n - 2} \theta \, \sec^2 \theta \, d\theta.$

This leads us to:

$I_n = \frac{1}{2n - 1} - I_{n - 1}.$

To calculate $I_3$, we apply the result from part (i):

We know: $I_3 = \frac{1}{2 \cdot 3 - 1} - I_2.$

Now we calculate $I_2$ as follows: $I_2 = \frac{1}{2 \cdot 2 - 1} - I_1.$

Assuming $I_1 = \frac{1}{1}$, we can proceed:

ightarrow I_2 = \frac{1}{3} - \frac{1}{1} = -\frac{2}{3},$$ Substituting back, we have: $$I_3 = \frac{1}{5} - (-\frac{2}{3}) = \frac{1}{5} + \frac{2}{3}.$$ Now finding a common denominator results: $$I_3 = \frac{3 + 10}{15} = \frac{13}{15}.$$

To show that:

$\omega^2 = \frac{g}{\ell \cos \alpha},$

we start by analyzing the forces acting on the particle $P$. The particle experiences:

• Tension $T$ directed along the string,
• Gravitational force $mg$ downwards.

Using Newton's second law in polar coordinates and resolving the forces in the vertical and horizontal directions, we observe that:

• The vertical component of the tension is:

ightarrow T \cos \alpha = mg.$$ - The horizontal component gives: $$T \sin \alpha = m\frac{v^2}{\ell}.$$ By rearranging these two equations, substituting for $T$ from the vertical component into the horizontal gives: $$\frac{mg}{\cos \alpha} \sin \alpha = m \frac{v^2}{\ell}.$$ This leads to: $$\omega^2 = \frac{g}{\ell \cos \alpha},$$ conclusively proving the required result.