1. (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 only when its argument is positive. Therefore, we need: $x - 3 > 0 \implies x > 3$ Thus, the domain of $f(x)$ is $(3, \infty)$.

To evaluate this limit, we can use the standard limit property that $\lim_{x \to 0} \frac{\sin x}{x} = 1$. We rewrite the limit as: $\lim_{x \to 0} \frac{\sin 2x}{x} = \lim_{x \to 0} \frac{\sin 2x}{2x} \cdot 2$ Applying the limit: $= 2 \cdot 1 = 2$

To solve the inequality:

1. Rearrange it: $\frac{x + 3}{2x} - 1 > 0$
2. Combine the fractions: $\frac{x + 3 - 2x}{2x} > 0 \implies \frac{-x + 3}{2x} > 0$
3. This requires that both the numerator and denominator are either positive or negative. We find the critical points:

• From $-x + 3 = 0 \implies x = 3$
• From $2x = 0 \implies x = 0$ (not included to avoid division by zero)

4. Test intervals:

• For $x < 0$: $\frac{-x + 3}{2x} < 0$
• For $0 < x < 3$: $\frac{-x + 3}{2x} > 0$
• For $x > 3$: $\frac{-x + 3}{2x} < 0$

5. From this, the solution is: $0 < x < 3$

To differentiate the function $y = x \cos^2 x$, we apply the product rule: $\frac{dy}{dx} = \frac{d}{dx}(x) \cdot \cos^2 x + x \cdot \frac{d}{dx}(\cos^2 x)$ Calculating each part:

1. $\frac{d}{dx}(x) = 1$
2. $\frac{d}{dx}(\cos^2 x) = 2\cos x \cdot (-\sin x) = -2\cos x \sin x$ Putting it all together: $\frac{dy}{dx} = \cos^2 x - 2x \cos x \sin x$

Using the substitution $u = x^3 + 1$, we differentiate: $du = 3x^2 dx \implies dx = \frac{du}{3x^2}$ Change the limits:

• When $x = 0, u = 1$
• When $x = 2, u = 9$ The integral becomes: $\int_1^9 e^{2(u - 1)} \cdot \frac{du}{3x^2}$ However, we relate $x^2$ in terms of $u$: $x^2 = (u - 1)^{\frac{2}{3}}$. This results in an intricate transformation. Alternatively, directly integral of $e^{2x^2 + 1}$ may need numerical methods if analytical methods become complex due to substitutions.