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

The function $f(x) = ext{ln}(x - 3)$ is defined when the argument of the logarithm is positive.

Thus, we require:

$x - 3 > 0 \\ ext{or} \\ x > 3$

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

$x ext{ such that } x > 3$

To evaluate the limit $\text{lim}_{x \to 0} \frac{\sin 2x}{x}$, we can use the substitution $t = 2x$.

Then as $x \to 0$, $t \to 0$ and we can rewrite the limit as:

$\text{lim}_{t \to 0} \frac{\sin t}{t/2} = 2 \cdot \text{lim}_{t \to 0} \frac{\sin t}{t}$

Using the known limit $\text{lim}_{t \to 0} \frac{\sin t}{t} = 1$, we find:

$= 2 \cdot 1 = 2$

To solve the inequality $\frac{x + 3}{2x} > 1$, we first multiply each side by $2x$, ensuring $x \neq 0$ and adjusting for the sign:

$x + 3 > 2x \\ \Rightarrow 3 > x$

Next, we consider when $2x > 0$, which occurs when $x > 0$. Thus, we also consider the domain of the solution.

Combining, we have:

• From $x > 0$, and $3 > x$, we find:

$0 < x < 3$

To differentiate the function $y = x \cos^2 x$, we will use the product rule. Denote:

• $u = x$,
• $v = \cos^2 x$.

The derivative using the product rule is:

$y' = u'v + uv'$

First, we compute:

• $u' = 1$
• To find $v'$, we use the chain rule:

$v' = 2\cos x (-\sin x) = -2 \cos x \sin x$

Thus, substituting back, we have:

$y' = 1 \cdot \cos^2 x + x (-2 \cos x \sin x) = \cos^2 x - 2x \cos x \sin x$

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

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

Now we need to express the differential:

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

Rewriting the integral:

$\int_{1}^{9} e^{u - 1} \frac{du}{3 \sqrt[3]{u - 1}^2}$

Next, we can integrate:

$\int e^{u - 1} \, du = e^{u - 1}$

Now substituting back would yield the evaluated limits, but a thorough integration is required for the final answer: