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

First, we simplify (\sqrt{4 + 8} = \sqrt{12} = 2\sqrt{3}).

Now, substituting (u = x^2 + 8) also requires finding (du = 2x , dx) or (dx = \frac{du}{2x}). Thus we have:

$$\int x (2\sqrt{3}) \, dx = 2\sqrt{3} \int x \, dx = 2\sqrt{3} \cdot \frac{x^2}{2} + C = \sqrt{3}x^2 + C.$$

Using the fact that (\lim_{x \to 0} \frac{\sin kx}{kx} = 1), we can write:

$$\lim_{x \to 0} \frac{\sin 5x}{3x} = \frac{5}{3} \cdot \lim_{x \to 0} \frac{\sin 5x}{5x} = \frac{5}{3} \cdot 1 = \frac{5}{3}.$$

The sum of cubes can be factored as:

$$\sin^3 \theta + \cos^3 \theta = (\sin \theta + \cos \theta)(\sin^2 \theta - \sin \theta \cos \theta + \cos^2 \theta).$$

Substituting this into the expression gives:

$$\frac{(\sin \theta + \cos \theta)(\sin^2 \theta - \sin \theta \cos \theta + \cos^2 \theta)}{\sin \theta + \cos \theta} - 1 = \sin^2 \theta - \sin \theta \cos \theta + \cos^2 \theta - 1.$$

Since (\sin^2 \theta + \cos^2 \theta = 1), we find:

$$\sin^2 \theta + \cos^2 \theta - 1 = 0.$$

To determine the values of (b), we first set the equations equal:

$$12x + b = x^3.$$

This implies:\n(x^3 - 12x - b = 0). For the line to be tangent, this cubic equation must have a double root. Finding the derivative:

$$\frac{d}{dx}(x^3 - 12x - b) = 3x^2 - 12.$$

Setting this equal to zero gives us the critical points:

$$x^2 = 4 \, \Rightarrow \, x = \pm 2.$$

We now substitute (x = 2) back into the original equation to find (b):

$$12(2) + b = 2^3 \Rightarrow 24 + b = 8 \Rightarrow b = 8 - 24 = -16.$$

Repeating for (x = -2):

$$12(-2) + b = (-2)^3 \Rightarrow -24 + b = -8 \Rightarrow b = -8 + 24 = 16.$$

Thus, the values of (b) for tangency are (b = -16) and (b = 16).