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, start by isolating $x$:

1. Multiply both sides by $(x + 1)$ (noting that we need to consider the sign of $x + 1$):

   If $x + 1 > 0$, $4 < 3(x + 1)$ $4 < 3x + 3$ $3x > 1$ $x > \frac{1}{3}$

   If $x + 1 < 0$ (i.e., $x < -1$), then the inequality sign flips:

   4 > 3x + 3\ 1 > 3x\ x < \frac{1}{3}$$

The solution is: $x > \frac{1}{3}$ for $x > -1$. Thus, $x \in \left(\frac{1}{3}, \infty\right)$.

The coordinates of point P dividing the segment AB externally in the ratio 5:2 can be calculated using the formula for external division.

For points A $(x_1, y_1) = (3, -1)$ and B $(x_2, y_2) = (9, 2)$, the coordinates of point P $(x, y)$ are given by: $x = \frac{mx_2 - nx_1}{m - n},$ $y = \frac{my_2 - ny_1}{m - n},$ where $m=5$ and $n=2$.

Substituting the values: $x = \frac{5 \cdot 9 - 2 \cdot 3}{5 - 2} = \frac{45 - 6}{3} = 13,$ $y = \frac{5 \cdot 2 - 2 \cdot (-1)}{5 - 2} = \frac{10 + 2}{3} = 4.$

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

To evaluate this integral, we can recognize that it is related to the arcsine function. Using the substitution $x = 2\sin(\theta)$, we have: $dx = 2\cos(\theta) d\theta, \quad \sqrt{4 - x^2} = 2\cos(\theta).$

The limits change as follows: When $x = 0, \theta = 0$ and when $x = 1, \theta = \arcsin(\frac{1}{2}) = \frac{\pi}{6}.$

Thus, the integral becomes: $\int_0^{\frac{\pi}{6}} \frac{2 \cos(\theta)}{2 \cos(\theta)} d\theta = \int_0^{\frac{\pi}{6}} d\theta = \left[\theta\right]_0^{\frac{\pi}{6}} = \frac{\pi}{6}.$

Using the substitution $u = x - 3$, we have $du = dx$, and the limits change accordingly: when $x = 3$, $u = 0$ and when $x = 4$, $u = 1$. The integral transforms as follows: $\int_0^1 \sqrt{u} du.$ The evaluation of this integral gives: $\int_0^1 u^{\frac{1}{2}} du = \left[\frac{u^{\frac{3}{2}}}{\frac{3}{2}}\right]_0^1 = \frac{2}{3}[u^{\frac{3}{2}}]_0^1 = \frac{2}{3}(1 - 0) = \frac{2}{3}.$

Thus, the final result is $\frac{2}{3}$.