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

To sketch the region defined by $y \leq |2x + 3|$, we first need to find the line $y = 2x + 3$ and the line $y = -2x - 3$.

1. Identify the intersections with the x-axis:
   • For $y = 2x + 3$, set $y = 0$: $0 = 2x + 3 \Rightarrow x = -\frac{3}{2}$.
   • For $y = -2x - 3$, set $y = 0$: $0 = -2x - 3 \Rightarrow x = -\frac{3}{2}$ which gives a V-shaped graph.
2. The area below both lines where $y$ is less than or equal to these lines is the required region.

The function $y = \cos^{-1}(x)$ is defined for values of $x$ in the interval $[-1, 1]$. Therefore, for $y = \cos^{-1} \left( \frac{x}{4} \right)$:

• Domain: To satisfy $-1 \leq \frac{x}{4} \leq 1$, we find:

  • From $\frac{x}{4} \geq -1 \Rightarrow x \geq -4$.
  • From $\frac{x}{4} \leq 1 \Rightarrow x \leq 4$.
  • Thus, the domain is $[-4, 4]$.

• Range: For the range of $\cos^{-1}$, it is always $[0, \pi]$ regardless of the domain.

To solve this integral using the substitution $u = 2x^{2} + 1$, we compute:

• The derivative is $du = 4x \, dx$ or $dx = \frac{du}{4x}$.
• Therefore, we need to express $x$ in terms of $u$: $x = \sqrt{\frac{u - 1}{2}}$.
• Substitute this into the integral:

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

• This requires further simplification to evaluate.

Using the section formula, we can find the coordinates of point $B$. The formula for points dividing a segment in the ratio $m:n$ is given by:

$P(x, y) = \left( \frac{mx_{2} + nx_{1}}{m + n}, \frac{my_{2} + ny_{1}}{m + n} \right).$

Here,

• Let $A(-1, 8)$ and $B(x, y)$.
• The coordinates at point $P(1, y)$ are given. Plugging into the formulas, we have:

$1 = \frac{3x - 2}{5} \Rightarrow 5 = 3x - 2 \Rightarrow x = \frac{7}{3}$. For $y$:

$y = \frac{3y + 16}{5} \Rightarrow 5y = 3y + 16 \Rightarrow 2y = 16 \Rightarrow y = 8.$ Thus, the coordinates for balance point $B$ are $\left( \frac{7}{3}, 8 \right)$.

The slope of the first line is $m_{1} = 3$ and the second line is $m_{2} = m$. To find the acute angle $\theta$ between two lines:

$\tan(\theta) = \left| \frac{m_{2} - m_{1}}{1 + m_{1}m_{2}} \right|$.

For $\theta = 45^{\circ}$, we have: $\tan(45^{\circ}) = 1$. This gives:

$1 = \left| \frac{m - 3}{1 + 3m} \right|$.

This leads to two cases:

1. $\frac{m - 3}{1 + 3m} = 1$;
2. $\frac{m - 3}{1 + 3m} = -1$. Solving these will yield two values of $m$ respectively.