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

The function $f(x) = \ln(x - 3)$ is defined only when the argument of the logarithm is positive. Therefore, we require:

$x - 3 > 0$

This implies:

$x > 3$

Thus, the domain of $f(x)$ is:

$(3, \infty)$

To evaluate the limit, we can use L'Hôpital's rule, as both the numerator and denominator approach 0 as $x \to 0$:

$\lim_{x \to 0} \frac{\sin 2x}{x} = \lim_{x \to 0} \frac{d}{dx}(\sin 2x) \div \frac{d}{dx}(x) = \lim_{x \to 0} \frac{2 \cos 2x}{1}$

Evaluating this limit gives:

$2 \cos(0) = 2 \cdot 1 = 2$

First, we rewrite the inequality:

$\frac{x + 3}{2x} - 1 > 0$

This simplifies to:

$\frac{x + 3 - 2x}{2x} > 0 \implies \frac{3 - x}{2x} > 0$

To solve this, we identify critical points by setting the numerator and denominator to zero:

• $3 - x = 0 \Rightarrow x = 3$
• $2x = 0 \Rightarrow x = 0$

We test the intervals determined by these points:

• For $x < 0$: the fraction is negative.
• For $0 < x < 3$: the fraction is positive.
• For $x > 3$: the fraction is negative.

Thus, the solution set is:

$(0, 3)$

To differentiate the function using the product rule:

1. Let $u = x$ and $v = \cos^2 x$. Then,

   $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$

2. First, we need $\frac{du}{dx} = 1$ and for $v = \cos^2 x$, we use the chain rule: $\frac{dv}{dx} = 2 \cos x (-\sin x) = -2 \cos x \sin x$

3. Now substituting back: $\frac{d}{dx}(x \cos^2 x) = x(-2 \cos x \sin x) + \cos^2 x(1) = -2x \cos x \sin x + \cos^2 x$

Using the substitution $u = x^3 + 1$, then:

$du = 3x^2 \, dx \Rightarrow dx = \frac{du}{3x^2}$

Changing the limits:

• When $x = 0$, $u = 1$.
• When $x = 2$, $u = 9$.

Now, we express $x$ in terms of $u$:

$x = (u - 1)^{1/3} \Rightarrow dx = \frac{du}{3(u - 1)^{2/3}}$

Thus, we rewrite the integral as:

$\int_1^{9} e^{u^2 + 1} \frac{du}{3(u - 1)^{2/3}}$

This integral can be evaluated further depending on techniques, but it may not have a simple closed form.