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 $\frac{4}{x + 1} < 3$, we first rewrite it without the fraction by multiplying both sides by $x + 1$, keeping in mind that $x + 1 > 0$ (for this inequality to hold):

$4 < 3(x + 1)$

This simplifies to:

$4 < 3x + 3$

Subtracting 3 from both sides gives:

$1 < 3x$

Dividing both sides by 3 yields:

$\frac{1}{3} < x$

Thus, the solution is $x > \frac{1}{3}$.

The coordinates of point P that divides the segment AB externally in the ratio 5:2 can be found using the formula for external division. If A is $(3, -1)$ and B is $(9, 2)$, the coordinates of P are given by:

$P_x = \frac{m x_2 - n x_1}{m - n}, \, P_y = \frac{m y_2 - n y_1}{m - n}$

where $(x_1, y_1) = (3, -1)$, $(x_2, y_2) = (9, 2)$, $m = 5$, and $n = 2$. Thus:

$P_x = \frac{5 \cdot 9 - 2 \cdot 3}{5 - 2} = \frac{45 - 6}{3} = \frac{39}{3} = 13$

$P_y = \frac{5 \cdot 2 - 2 \cdot (-1)}{5 - 2} = \frac{10 + 2}{3} = \frac{12}{3} = 4$

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

To evaluate the integral $\int_0^1 \frac{dx}{\sqrt{4 - x^2}}$, we can apply a trigonometric substitution. Letting $x = 2 \sin(\theta)$ implies that $dx = 2 \cos(\theta) \, d\theta$. When $x = 0$, $\theta = 0$ and when $x = 1$, $\theta = \sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}$. Thus, the limits change from 0 to $\frac{\pi}{6}$.

Substituting into the integral gives:

$$$$

Using the substitution $u = x - 3$ leads to $x = u + 3$ and thus $dx = du$. The limits change as follows: when $x = 3$, $u = 0$; when $x = 4$, $u = 1$. Therefore, we can rewrite the integral:

$\int_3^4 x \sqrt{x - 3} \, dx = \int_0^1 (u + 3) \sqrt{u} \, du.$

Expanding this gives:

$\int_0^1 (u \sqrt{u} + 3 \sqrt{u}) \, du = \int_0^1 (u^{3/2} + 3u^{1/2}) \, du.$

Now integrate term by term:

$= \left[ \frac{2}{5} u^{5/2} + 3 \cdot \frac{2}{3} u^{3/2} \right]_0^1 = \left[ \frac{2}{5} (1) + 2(1) \right] - 0 = \frac{2}{5} + 2 = \frac{2}{5} + \frac{10}{5} = \frac{12}{5}.$

Thus, the final answer is $\frac{12}{5}$.