The diagram shows the graph of a function $f(x)$ - HSC - SSCE Mathematics Extension 2 - Question 12 - 2014 - Paper 1

Graph the function $y = \frac{1}{f(x)}$. Identify points where $f(x) = 0$ to find vertical asymptotes in the new graph, as these correspond to where the function approaches infinity. Also, indicate horizontal asymptotes based on the behavior of $f(x)$ as $x$ approaches infinity.

Using the assumption that $x = 2\cos\theta$ is a solution, substitute into the equation. Start from $x^3 - 3x = \sqrt{3}$, and derive the equation

$x^3 = (2\cos\theta)^3 = 8\cos^3\theta \quad \text{and} \quad -3x = -6\cos\theta$

Adding these gives the equation:

$8\cos^3\theta - 6\cos\theta = \sqrt{3}.$

Factoring out $2\cos\theta$, we find

$4\cos^3\theta - 3\cos\theta = \frac{\sqrt{3}}{2}.$

Thus, $\cos 3\theta = \frac{\sqrt{3}}{2}$.

We know that $\cos 3\theta = \frac{\sqrt{3}}{2}$ gives us angles $heta = \frac{\pi}{6} + k\frac{2\pi}{3}, k \in \mathbb{Z}$. Thus, substituting back, we have:

1. For $k=0$: $\theta = \frac{\pi}{6}$, then $x = 2\cos\left(\frac{\pi}{6}\right) = 2\frac{\sqrt{3}}{2} = \sqrt{3}$.
2. For $k=1$: $\theta = \frac{\pi}{2}$, then $x = 2\cos\left(\frac{\pi}{2}\right) = 0$.
3. For $k=2$: $\theta = \frac{7\pi}{6}$, then $x = 2\cos\left(\frac{7\pi}{6}\right) = 2\cdot -\frac{\sqrt{3}}{2} = -\sqrt{3}$.

Thus, the three solutions are $\sqrt{3}$, $0$, and $-\sqrt{3}$.

To find the slopes of the tangents, differentiate both curves with respect to $x$. For the first curve, using implicit differentiation:

1. For $x^2 - y^2 = 5$: $2x - 2y\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{x}{y}.$

2. For $xy = 6$: $y + x\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}.$

At point $P$, if slopes are $m_1$ and $m_2$, show that: $m_1 \cdot m_2 = -1$ this would confirm the tangents are perpendicular.

Evaluate:
$I_0 = \int_0^1 \frac{x^2}{x^2 + 1} \, dx + \int_0^1 \frac{2^0 x}{x^2 + 1} \, dx.$ The first integral simplifies to $[\frac{1}{2}x^2]_{0}^{1} = \frac{1}{2}$. The second integral is evaluated by substituting $u = x^2 + 1$, leading to a total value of $\frac{\pi}{4}$.

Using integration by parts or substitution for both integrals $I_n$ and $I_{n-1}$ deriving from the definitions, and showing the relationship using recursion gives us the required identity.

Using parts or substituting accordingly, we calculate the integral and confirm through earlier derived results; thus finding $\int_0^1 \frac{x^4}{x^2 + 1} \, dx$ can be achieved by relating to $I_1$. The final answer will be determined as required.