Find $$ \int \sin x^2 \, dx $$ - HSC - SSCE Mathematics Extension 1 - Question 11 - 2015 - Paper 1

First, determine the slopes of the lines: The slope of the first line $m_1 = 2$ and for the second line, reorganizing gives $m_2 = 3$.

The formula for the angle $\theta$ between two lines is given by:

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

Substituting the slopes:

$\tan \theta = \left| \frac{2 - (-3)}{1 + 2 \cdot (-3)} \right| = \left| \frac{5}{1 - 6} \right| = \frac{5}{5} = 1$.

Thus, $\theta = \tan^{-1}(1) = \frac{\pi}{4}$ or $45^\circ$.

To solve the inequality:

1. Rearrange it: $\frac{4}{x + 3} - 1 \geq 0$

2. Combine the terms: $\frac{4 - (x + 3)}{x + 3} \geq 0$ $\frac{1 - x}{x + 3} \geq 0$.

3. Find the critical points: $x = 1$ and $x = -3$.

4. Test the intervals divided by these points:

   • For $x < -3$, the result is positive.
   • For $-3 < x < 1$, the result is negative.
   • For $x > 1$, the result is positive.

Thus, the solution set is $x \leq -3$ or $x \geq 1$.

To express $5 \cos x - 12 \sin x$ in the required form, we identify:

1. Calculate $A = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13$.
2. Determine $\tan \alpha = \frac{-12}{5} \Rightarrow \alpha = \tan^{-1}\left(-\frac{12}{5}\right)$.
3. Hence, we have: $5 \cos x - 12 \sin x = 13 \cos(x + \alpha)$.

Applying the substitution:

1. If $u = 2x - 1$, then $du = 2 \, dx$ or $dx = \frac{du}{2}$.
2. The limits of integration change as follows:
   • When $x = 1$, $u = 1$.
   • When $x = 2$, $u = 3$.
3. Substitute and evaluate: $\int_{1}^{3} \frac{\sqrt{2}}{u^2} \cdot \frac{du}{2} = \frac{\sqrt{2}}{2} \int_{1}^{3} u^{-2} \, du = \frac{\sqrt{2}}{2} [-u^{-1}]_{1}^{3}$ $= \frac{\sqrt{2}}{2} \left( -\frac{1}{3} + 1 \right) = \frac{\sqrt{2}}{2} \left( \frac{2}{3} \right) = \frac{\sqrt{2}}{3}$.

Since $A(x) = x - 3$ is a factor of $P(x)$, substituting $x = 3$ into the polynomial should yield 0:

$P(3) = (3)^3 - k(3)^2 + 5(3) + 12 = 27 - 9k + 15 + 12 = 54 - 9k = 0$.

This gives us:

$9k = 54 \Rightarrow k = 6$.

With $k = 6$, the polynomial becomes:

$P(x) = x^3 - 6x^2 + 5x + 12$.

Using the Rational Root Theorem and testing possible roots, we can find:

1. Testing $x = -3$: $P(-3) = (-3)^3 - 6(-3)^2 + 5(-3) + 12 = -27 - 54 - 15 + 12 = -84$ (not a root).
2. Testing $x = 1$: $P(1) = 1 - 6 + 5 + 12 = 12$ (not a root).
3. Testing $x = 2$: $P(2) = 8 - 24 + 10 + 12 = 6$ (not a root).
4. Testing $x = 3$: $P(3) = 27 - 54 + 15 + 12 = 0$ (a root).

Factoring out $(x - 3)$ gives:

$P(x) = (x - 3)(x^2 - 3x - 4)$.

Factoring the quadratic, we find:

$x^2 - 3x - 4 = (x - 4)(x + 1)$.

Thus, the zeros of $P(x)$ are $x = 3, 4, -1$.