Find $$\int \sin x^2 \, dx.$$ (b) Calculate the size of the acute angle between the lines $$y = 2x + 5$$ and $$y = 4 - 3x$$ - HSC - SSCE Mathematics Extension 1 - Question 11 - 2015 - Paper 1

The slopes of the lines are:
For $y = 2x + 5$, the slope $m_1 = 2$.
For $y = 4 - 3x$, the slope $m_2 = -3$.
Using the formula for the angle between two lines:
$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$,
we can calculate:
$\tan \theta = \left| \frac{2 - (-3)}{1 + 2(-3)} \right| = \left| \frac{5}{-5} \right| = 1$.
Therefore, $\theta = 45^\circ$.

To solve the inequality, we rearrange it:
$\frac{4}{x + 3} - 1 \geq 0$,
which simplifies to
$\frac{4 - (x + 3)}{x + 3} \geq 0$,
or
$\frac{1 - x}{x + 3} \geq 0$.
The critical points are at $x = 1$ and $x = -3$.
We analyze the intervals:

1. For $x < -3$: the fraction is positive.
2. For $-3 < x < 1$: the fraction is negative.
3. For $x > 1$: the fraction is negative.
   Hence, the solution set is: $(-\infty, -3) \cup [1, \infty).$

To express the equation $5\cos x - 12\sin x$ in the desired form, we set:
$A = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = 13$.
Also, we calculate $\tan \alpha = \frac{-12}{5}$.
Thus, $\alpha = \tan^{-1}\left(\frac{-12}{5}\right)$.
This gives us the expression:
$5\cos x - 12\sin x = 13\cos(x + \alpha)$.

Using the substitution:
When $u = 2x - 1$, then $du = 2dx$ or $dx = \frac{du}{2}$.
Also, when $x=1$, $u=1$ and when $x=2$, $u=3$.
Thus, the integral becomes:
$\int_{1}^{3} \frac{\frac{u + 1}{2}}{u^3} \frac{du}{2} = \frac{1}{4}\int_{1}^{3} \frac{u + 1}{u^3}$.
This can be separated as:
$\frac{1}{4}\int_{1}^{3} \left(\frac{1}{u^2} + \frac{1}{u^3}\right) du$.
Evaluating this gives:
$= \frac{1}{4}\left[-\frac{1}{u} - \frac{1}{2u^2}\right]_{1}^{3} = \frac{1}{4}\left[-\frac{1}{3} - \frac{1}{18} + 1 + \frac{1}{2}\right] = \frac{1}{4}\left[-\frac{1}{3} -\frac{1}{18} + \frac{9}{18} + \frac{3}{18}\right]$.

Since $P(x)$ is divisible by $A(x)$, we have $P(3) = 0$.
Substituting $x = 3$ into $P(x)$ gives:
$P(3) = 3^3 - k(3^2) + 5(3) + 12 = 27 - 9k + 15 + 12 = 54 - 9k$.
Setting this equal to zero yields:
$54 - 9k = 0 \Rightarrow 9k = 54 \Rightarrow k = 6.$

With $k = 6$, we have:
$P(x) = x^3 - 6x^2 + 5x + 12$.
Using synthetic or polynomial division by $x - 3$ (since $3$ is a root), we get:
$P(x) = (x - 3)(x^2 - 3x - 4)$.
Factoring further, $x^2 - 3x - 4 = (x - 4)(x + 1)$.
Thus, the zeros are:
$x = 3, x = 4, x = -1.$