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:

1. Start by rewriting it: $\frac{4}{x+1} - 3 < 0$ This is equivalent to: $\frac{4 - 3(x + 1)}{x + 1} < 0$ Simplifying yields: $\frac{4 - 3x - 3}{x + 1} < 0$ $\frac{1 - 3x}{x + 1} < 0$

2. The critical points occur when the numerator or denominator is zero:

   • $1 - 3x = 0 \Rightarrow x = \frac{1}{3}$
   • $x + 1 = 0 \Rightarrow x = -1$

3. Test intervals determined by $-1$ and $\frac{1}{3}$ to find where the inequality holds true. Testing points in each interval:

   • For $x < -1$: both numerator and denominator negative, positive.
   • For $-1 < x < \frac{1}{3}$: numerator positive and denominator positive, negative.
   • For $x > \frac{1}{3}$: numerator negative and denominator positive, negative.

4. The solution set where the inequality holds is: $(-1, \frac{1}{3}) \cup (\frac{1}{3}, \infty)$.

Using the external division formula for points (x,y):

Given points A(3, -1) and B(9, 2), we use the ratio 5:2:

The coordinates P can be calculated as follows:

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

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

Thus, the point P is (13, 4).

This integral represents the area under the curve. It can be evaluated using a trigonometric substitution.

Let: $x = 2 \sin \theta \, , \ dx = 2 \cos \theta \, d\theta$ Accordingly, the limits become:

• When $x = 0$, $\theta = 0$.
• When $x = 1$, $\theta = \arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6}$.

Now substituting:

$\int_{0}^{1} \frac{dx}{\sqrt{4 - x^2}} = \int_{0}^{\frac{\pi}{6}} \frac{2 \cos \theta \, d\theta}{\sqrt{4 - 4 \sin^2 \theta}} = \int_{0}^{\frac{\pi}{6}} \cos \theta \, d\theta$

Evaluating gives: $= [\sin \theta]_{0}^{\frac{\pi}{6}} = \sin(\frac{\pi}{6}) - \sin(0) = \frac{1}{2} - 0 = \frac{1}{2}$.

Using the substitution: $u = x - 3 \Rightarrow x = u + 3 \, , \ dx = du$

Changing the limits:

• When $x = 3$, $u = 0$.
• When $x = 4$, $u = 1$.

Thus, the integral becomes:

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

Evaluating term-by-term:

1. $\int u^{3/2} \, du = \frac{2}{5} u^{5/2}$
2. $\int 3u^{1/2} \, du = 2u^{3/2}$

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