Question 1 (12 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 1 - 2006 - Paper 1

To solve this, we apply the substitution:

1. Let $u = x^2 + 8$, then $du = 2x \, dx$. Hence, $dx = \frac{du}{2x}$.
2. Rewrite the integral in terms of $u$: $\int x \sqrt{u - 4} \frac{du}{2x} = \frac{1}{2} \int \sqrt{u - 4} \, du.$
3. To integrate, use the formula: $\int \sqrt{u - a} \, du = \frac{2}{3}(u - a)^{\frac{3}{2}} + C.$
4. Thus: $\frac{1}{2} \cdot \frac{2}{3}(u - 4)^{\frac{3}{2}} + C = \frac{1}{3}(x^2 + 4)^{\frac{3}{2}} + C.$

For this limit, we apply the limit property:

$\lim_{x \to 0} \frac{\sin kx}{x} = k.$ Setting $k = 5$, we have: $\lim_{x \to 0} \frac{\sin 5x}{3x} = \frac{5}{3}.$ Therefore, the answer is $\frac{5}{3}$.

Using the identity for the sum of cubes: $a^3 + b^3 = (a + b)(a^2 - ab + b^2),$ where $a = \sin \theta and b = \cos \theta$:

1. The expression becomes: $\frac{(\sin \theta + \cos \theta)(\sin^2 \theta - \sin \theta \cos \theta + \cos^2 \theta)}{\sin \theta + \cos \theta} - 1$
2. Simplifying yields: $\sin^2 \theta - \sin \theta \cos \theta + \cos^2 \theta - 1 = -\sin \theta \cos \theta.$ Thus, the final answer is: $-\sin \theta \cos \theta.$

To determine the values of $b$, we need to find where the line and the curve intersect:

1. Set $12x + b = x^3$.
2. Rearranging gives: $x^3 - 12x - b = 0.$
3. For tangency, the above cubic must have a double root. Therefore, calculate the derivative: $f'(x) = 3x^2 - 12.$ Set this equal to 0 to find the double root: $x^2 = 4 \Rightarrow x = 2 \text{ or } x = -2.$
4. Substitute $x = 2$ into the original equation: $y = 12(2) + b = 8 \Rightarrow b = -16.$
5. Substitute $x = -2$: $y = 12(-2) + b = -8 \Rightarrow b = 24.$ The values of $b$ are thus $-16$ and $24.$