1. (a) Factorise $8x^3 + 27$ - HSC - SSCE Mathematics Extension 1 - Question 1 - 2009 - Paper 1

The function $f(x) = \text{ln}(x - 3)$ is defined only for values of $x$ such that $x - 3 > 0$. Thus, we need: $x > 3$ Therefore, the domain of $f(x)$ is: $\text{Domain: } (3, \, \infty)$

To find the limit, we can apply the limit property: $\lim_{x \to 0} \frac{\sin kx}{x} = k$ Therefore, for our limit: $\lim_{x \to 0} \frac{\sin 2x}{x} = 2$

First, we rewrite the inequality: $\frac{x + 3}{2x} - 1 > 0$ This simplifies to: $\frac{x + 3 - 2x}{2x} > 0$ So we have: $\frac{3 - x}{2x} > 0$ To solve this, we find the critical points by setting the numerator and denominator to zero:

• $3 - x = 0 \Rightarrow x = 3$
• $2x = 0 \Rightarrow x = 0$ Now we test intervals around these critical points. We find that the solution to the inequality is: $x < 0 \text{ or } x > 3$

To differentiate the function $y = x \cos^2 x$, we apply the product rule: $\frac{dy}{dx} = \frac{d}{dx}(x) \cdot \cos^2 x + x \cdot \frac{d}{dx}(\cos^2 x)$ Using the chain rule for the second term, we get: $\frac{dy}{dx} = \cos^2 x + x \cdot 2\cos x (-\sin x) = \cos^2 x - 2x \cos x \sin x$

Using the substitution $u = x^3 + 1$, we find: $du = 3x^2 \, dx \Rightarrow dx = \frac{du}{3x^2}$ We also need to change the limits of integration:

• When $x = 0$, $u = 0^3 + 1 = 1$
• When $x = 2$, $u = 2^3 + 1 = 9$

Now we rewrite the integral: $\int_{1}^{9} e^{2x^2 + 1} \frac{du}{3x^2}$ We would also need to express $x^2$ in terms of $u$. However, finding the exact solution requires further manipulation or numerical methods.