1. (a) Indicate the region on the number plane satisfied by $y \geq |x| + 1$ - HSC - SSCE Mathematics Extension 1 - Question 1 - 2004 - Paper 1

To solve the inequality $\frac{4}{x + 1} < 3$, first multiply both sides by $(x + 1)$, ensuring to consider the sign of $(x + 1)$:

• For $x + 1 > 0$ (or $x > -1$), we have $4 < 3(x + 1)$, which simplifies to $4 < 3x + 3$, leading to $3x > 1$, so $x > \frac{1}{3}$.
• For $x + 1 < 0$ (or $x < -1$), multiply by -1 to reverse the inequality: $4 > 3(x + 1)$ leads to $4 > 3x + 3$, or $x < \frac{1}{3}$ (this is already true for $x < -1$).

Combining these results, the solution is $x > -1$ and $x > \frac{1}{3}$, hence $x > \frac{1}{3}$.

Let the coordinates of point P be $(x, y)$. The coordinates of the external division can be calculated using the section formula:

$$P = \left( \frac{m_2 x_1 - m_1 x_2}{m_2 - m_1}, \frac{m_2 y_1 - m_1 y_2}{m_2 - m_1} \right)$$

where $A(3, -1)$ and $B(9, 2)$, with $m_1 = 5$ and $m_2 = 2$.

Substituting in:

$$P = \left( \frac{2 \cdot 3 - 5 \cdot 9}{2 - 5}, \frac{2 \cdot (-1) - 5 \cdot 2}{2 - 5} \right) = \left( \frac{6 - 45}{-3}, \frac{-2 - 10}{-3} \right) = \left( \frac{-39}{-3}, \frac{-12}{-3} \right) = (13, 4).$$

Thus, the coordinates of point P are (13, 4).

To find the integral $\int_{0}^{1} \frac{dx}{\sqrt{4 - x^2}}$, use the substitution $x = 2\sin{\theta}$, then $dx = 2\cos{\theta} d\theta$ and the limits change from $x = 0 \to \theta = 0$ and $x = 1 \to \theta = \arcsin{\frac{1}{2}} = \frac{\pi}{6}$. The integral becomes:

$$\int_{0}^{\frac{\pi}{6}} \frac{2\cos{\theta}}{\sqrt{4 - 4\sin^2{\theta}}} \cdot 2\cos{\theta} d\theta = 2\int_{0}^{\frac{\pi}{6}} d\theta = 2\left(\frac{\pi}{6} - 0\right) = \frac{\pi}{3}.$$

Using the substitution $u = x - 3$, then $x = u + 3$ and $dx = du$. The limits change to $x = 3 \to u = 0$ and $x = 4 \to u = 1$. The integral becomes:

$$\int_{0}^{1} (u + 3) \sqrt{u} \, du = \int_{0}^{1} (u^{3/2} + 3u^{1/2}) \, du.$$

Integrating gives:

$$\left[ \frac{u^{5/2}}{\frac{5}{2}} + \frac{3u^{3/2}}{\frac{3}{2}} \right]_{0}^{1} = \left[ \frac{2}{5}u^{5/2} + 2u^{3/2} \right]_{0}^{1} = \left(\frac{2}{5} + 2\right) - 0 = \frac{2}{5} + \frac{10}{5} = \frac{12}{5}.$$

Thus, the value of the integral is $\frac{12}{5}$.