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

For $y = \frac{1}{f(x)}$, identify the x-values where $f(x) = 0$, as these will create vertical asymptotes in the graph. Plot the reciprocal values for positive and negative ranges of $x$ for which $f(x)$ is not zero. Clearly label any horizontal asymptotes and intercepts if they exist.

Given that $x = 2\cos\theta$, substitute into the equation:
[ x^3 - 3x = \sqrt{3} ].
This forms the equation as follows:
[ (2\cos\theta)^3 - 3(2\cos\theta) = \sqrt{3} ]
[ 8\cos^3\theta - 6\cos\theta = \sqrt{3} ]
Rearranging gives:
[ 8\cos^3\theta - 6\cos\theta - \sqrt{3} = 0 ].
Now applying the identity $4\cos^3 \theta - 3\cos \theta = \cos 3\theta$, we replace and find:
[ \cos 3\theta = \frac{\sqrt{3}}{2} ].

From $\cos 3\theta = \frac{\sqrt{3}}{2}$, we know that the general solutions for $3\theta$ are given by:
[ 3\theta = \frac{\pi}{6} + 2k\pi \quad \text{and} \quad 3\theta = \frac{11\pi}{6} + 2k\pi \text{ for } k \in \mathbb{Z} ].
Dividing by 3 provides:
[ \theta = \frac{\pi}{18} + \frac{2k\pi}{3} \quad \text{and} \quad \theta = \frac{11\pi}{18} + \frac{2k\pi}{3}. ]
Determining $x = 2\cos \theta$, we evaluate for $k = 0, 1, 2$ to find the three distinct solutions in the interval $[-2, 2]$.

To prove that the tangents to the curves $x^2 - y^2 = 5$ and $xy = 6$ at point $P$ are perpendicular:

1. Find rac{dy}{dx} for both curves using implicit differentiation:

• For $x^2 - y^2 = 5$:
  [ 2x - 2y\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{x}{y}. ]
• For $xy = 6$:
  Applying product rule gives:
  [ x\frac{dy}{dx} + y = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}. ]

2. The product of the slopes should equal $-1$ to confirm perpendicularity: [ m_1 \cdot m_2 = -1 ext{ where } m_1 = \frac{dy}{dx} ext{ for the first curve and } m_2 = -\frac{y}{x}. ]
   Evaluating at point $P(x_0, y_0)$ will complete the proof.

To find $I_0$, evaluate the integral:
[ I_0 = \int_0^1 \frac{2}{x^2 + 0} , dx = 2 \int_0^1 \frac{1}{x^2 + 1} , dx. ]
This integral conforms to the standard form:
[ \frac{1}{1+x^2} ] which integrates to:
[ I_0 = 2 \left[ \tan^{-1}(x) \right]_0^1 = 2 \left( \frac{\pi}{4} - 0 \right) = \frac{\pi}{2}. ]
Thus, confirming:
[ I_0 = \frac{\pi}{4}. ]

Using integration by parts for $I_n$, let:
[ u = \frac{2}{x^2 + 2n} \text{ and } dv = dx ].
Differentiating and integrating gives:
[ du = -\frac{4x}{(x^2 + 2n)^2} dx \text{ and } v = x. ]
Applying parts yields:
[ I_n = uv - \int v , du = I_{n-1}. ]
Recursive substitution leads to:
[ I_n = \frac{1}{2n - 1}. ]

To find [ \int_0^1 \frac{x^4}{x^2 + 1} , dx = \int_0^1 (x^2 - 1 + \frac{1}{x^2 + 1}) , dx. ]
Calculating each part results in:

1. $\int_0^1 (x^2 - 1) \, dx$ evaluates to $-\frac{1}{3}$,
2. $\int_0^1 \frac{1}{x^2 + 1} \, dx$ evaluates to $\frac{\pi}{4}$.
   Altogether gives: [ -\frac{1}{3} + \frac{\pi}{4} = \frac{\pi}{4} - \frac{1}{3}. ]