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

For $y = \frac{1}{f(x)}$, we need to find the reciprocal of the function. Identify all intercepts of $f(x)$; these points will become asymptotes for the new graph. As $f(x)$ approaches infinity, $y$ will approach 0. Plot the new behavior according to the original asymptotes, ensuring to reflect the behavior of $f(x)$ in the positive and negative x-directive.

Starting with the identity, substitute $x = 2\cos \theta$ into the equation $x^3 - 3x = \sqrt{3}$. This yields:

$\cos 3\theta = 4\cos^3 \theta - 3\cos \theta$

When $x = 2\cos \theta = \sqrt{3}$, we derive:

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

This follows from evaluating the cosine function using the known relationships for specific angles.

To find the real solutions, we equate the derived expression to known cosine values:

1. Solve for $\theta$ when $\cos 3\theta = \frac{\sqrt{3}}{2}$. The angles that satisfy this are:

   • $3\theta = \frac{\pi}{6}$
   • $3\theta = \frac{11\pi}{6}$
   • Additional cycle solutions up to $2\pi$.

2. Solve for $heta$: $\theta = \frac{\pi}{18}, \frac{11\pi}{18}, \frac{\pi}{6} + k\frac{2\pi}{3} \text{ for integers } k$

3. Find corresponding values for $x$ using $x = 2\cos(\theta)$.

To show that the tangents at point $P(x_0, y_0)$ are perpendicular, calculate the derivatives of each curve:

1. For $x^2 - y^2 = 5$ use implicit differentiation: $\frac{dy}{dx} = \frac{x}{y}$

2. For $xy = 6$ apply implicit differentiation: $\frac{dy}{dx} = -\frac{y}{x}$

3. At point $P$, substitute values into both derivatives to find slopes.

4. Verify that the product of the slopes equals $-1$, indicating perpendicular tangents.

Calculate the integral: $I_0 = \int_0^1 \frac{x^0}{x^2 + 0} \, dx = \int_0^1 \frac{1}{x^2} \, dx$ Evaluate as: $I_0 = \lim_{t \to 0} \left[ \tan^{-1}(t) \right]_0^1 = \frac{\pi}{4}$

Use integration by parts to relate $I_n$ and $I_{n-1}$, leading to: $I_n = \int_0^1 \frac{x^{2n}}{x^2 + 2n} \, dx$ Upon differentiation and using limits, demonstrate that: $\int_0^1 \frac{x^{2n+1}}{x^2 + 2n} \, dx + \int_0^1 \frac{x^{2n-1}}{x^2 + 2(n-1)} \, dx = \frac{1}{2n - 1}$

Identify the relation between $I_n$ and values for specific $n$: $\int_0^1 \frac{x^4}{x^2 + 1} \, dx = I_2$ Using the already established relationship, compute: $I_2 = \int_0^1 \frac{x^4}{x^2 + 1} \, dx = \frac{1}{2}$ where substitutions lead to $I_1 + I_0$ completing the evaluation.