1. Find $$\int \frac{1}{x^{2}+49} \: dx$$ - HSC - SSCE Mathematics Extension 1 - Question 1 - 2005 - Paper 1

This inequality describes a region below the V-shaped graph of $y = 2|x+3|$. To sketch:

1. Determine the vertex at the point where $x = -3$, giving $y = 0$.
2. The lines $y = 2x + 6$ (for $x+3 \geq 0$) and $y = -2x - 6$ (for $x+3 < 0$) are the arms of the V.
3. Shade the area below this graph, representing all points $(x, y)$ such that the inequality holds.

The function $y = \cos\left(\frac{x}{4}\right)$ is a cosine function:

• Domain: All real numbers, $(-\infty, \infty)$.
• Range: The output of the cosine function varies between -1 and 1, so the range is $[-1, 1]$.

Using the substitution:

1. Differentiate: $du = 4x \, dx \Rightarrow dx = \frac{du}{4x}$.
2. Solve for x in terms of u: $x = \sqrt{\frac{u-1}{2}}$.
3. Replace in integral:

$\int \frac{1}{u^{\frac{5}{2}}} \frac{du}{4\sqrt{\frac{u-1}{2}}}$

This can be simplified and solved using standard integral techniques. A cleaner method may involve reverting to $u$ integration first after substitution.

Using the section formula:

1. The coordinates of P can be calculated as: $P(x_P, y_P) = \left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right)$

2. Here, $(x_1, y_1) = (-1, 8)$, $(x_2, y_2) = (x, y)$, $\Rightarrow (1, y) = \left(\frac{3x - 2}{5}, \frac{3y + 16}{5}\right)$.

3. Setting these equal, we solve: $1 = \frac{3x - 2}{5} \\ \Rightarrow 3x - 2 = 5 \\ \Rightarrow 3x = 7 \\ x = \frac{7}{3}$ Then, for y: $y = \frac{3y + 16}{5} \Rightarrow 5y = 3y + 16 \Rightarrow 2y = 16 \Rightarrow y = 8$. Thus, point B coordinates are $\left( \frac{7}{3}, 8 \right)$.

To find the angle between these lines, we use the slope formula:

1. The slope of the first line is $m_1 = 3$.
2. The slope of the second line is $m_2 = m$.
3. The formula for the angle $\theta$ between two lines is given by:

$\tan(\theta) = \left|\frac{m_2 - m_1}{1 + m_1 m_2}\right|$.

4. Setting $\tan(45°) = 1$: $1 = \left|\frac{m - 3}{1 + 3m}\right|$.
5. Solving gives two scenarios:
   1. $m - 3 = 1 + 3m \Rightarrow 2m = 2 \Rightarrow m = 1$
   2. $m - 3 = -1 - 3m \Rightarrow 4m = 2 \Rightarrow m = \frac{1}{2}$.

Thus, the two possible values of m are $m = 1$ and $m = \frac{1}{2}$.