Find $$\int \sin x \, dx.$$ Calculate the size of the acute angle between the lines $$y = 2x + 5$$ and $$y = 4 - 3x.$$ Solve the inequality $$\frac{4}{x + 3} \geq 1.$$ Express $$5 \cos x - 12 \sin x$$ in the form $$A \cos(x + \alpha), \text{ where } 0 \leq \alpha \leq \frac{\pi}{2}.$$ Use the substitution $$u = 2x - 1$$ to evaluate $$\int_{1}^{2} \frac{x}{(2x - 1)^3} \, dx.$$ Consider the polynomials $$P(x) = x^3 - kx^2 + 5x + 12$$ and $$A(x) = x - 3.$$ (i) Given that $$P(x)$$ is divisible by $$A(x)$$, show that $$k = 6.$$ (ii) Find all the zeros of $$P(x)$$ when $$k = 6.$$ - HSC - SSCE Mathematics Extension 1 - Question 11 - 2015 - Paper 1

First, we find the slopes of both lines:

• For the line $y = 2x + 5$, the slope $m_1 = 2$.
• For the line $y = 4 - 3x$, rearranging gives $y = -3x + 4$, so the slope $m_2 = -3$.

Next, we use the formula for the angle between two lines: $\tan \theta = \frac{|m_1 - m_2|}{1 + m_1 m_2}.$ Substituting the values gives: $\tan \theta = \frac{|2 - (-3)|}{1 + (2)(-3)} = \frac{5}{-5} = -1.$

Thus, the acute angle $\theta$ can be found using: $\theta = \arctan(1) = \frac{\pi}{4}$ radians or $45^\circ$.

To solve this inequality, we first set up the equation: $\frac{4}{x + 3} - 1 \geq 0.$ This simplifies to: $\frac{4 - (x + 3)}{x + 3} \geq 0$ or $\frac{1 - x}{x + 3} \geq 0.$

The critical points are found by setting the numerator and denominator to zero:

1. $$$$

ightarrow x = 1$2.$x + 3 = 0 ightarrow x = -3$$

We create a sign chart and test intervals:

• For ($-\infty, -3$)
• For ($-3, 1$)
• For ($1, \infty$)

From the tests, the solution set is: $-3 < x \leq 1.$

We represent $5 \cos x - 12 \sin x$ as: $A \cos(x + \alpha) = A(\cos x \cos \alpha - \sin x \sin \alpha).$ By comparison, we have:

• $A \cos \alpha = 5$
• $-A \sin \alpha = -12.$

Thus, we can find $A$ as follows: $A = \sqrt{(5)^2 + (12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13.$

To find $\alpha$, we solve: $\cos \alpha = \frac{5}{13}, \sin \alpha = \frac{12}{13}.$

Therefore, we can express: $5 \cos x - 12 \sin x = 13 \cos(x + \alpha), \text{ where } \tan \alpha = \frac{12}{5}$.

Using the substitution, we first convert the limits:

• When $x = 1$, $u = 2(1) - 1 = 1$.
• When $x = 2$, $u = 2(2) - 1 = 3$.

Next, we express $x$ in terms of $u$: $x = \frac{u + 1}{2}.$

The integral becomes: $\int_{1}^{3} \frac{\frac{u + 1}{2}}{u^3} \cdot \frac{1}{2} \, du = \frac{1}{4} \int_{1}^{3} \frac{u + 1}{u^3} \, du$.

Separating the integral: $= \frac{1}{4} \left(\int_{1}^{3} \frac{1}{u^2} \, du + \int_{1}^{3} u^{-2} \, du \right).$

Calculating gives: $= \frac{1}{4} \left(-\frac{1}{u}\bigg|_{1}^{3} + \frac{1}{u}\bigg|_{1}^{3}\right) = \frac{1}{4} \left(-\frac{1}{3} + 1 + 1 - \frac{1}{1}\right) = \frac{1}{4} \left(\frac{5}{3} - 0\right) = \frac{5}{12}.$

To show that $k = 6$ given that $P(x)$ is divisible by $A(x)$, we apply the Factor Theorem: Since $A(3) = 0,$ it implies: $P(3) = 3^3 - 6(3)^2 + 5(3) + 12 = 0.$

Solving the equation results in the condition: simplifying:

ightarrow 54 - 6k = 0 ightarrow k = 9.$$ Thus, we conclude that $$k = 6$$ is incorrect, based on the calculations above. Next, for the zeros when $$k = 6$$, we substitute back into $$P(x):$$ This leads us to factor the cubic polynomial resulting from the above. The possible zeros can be found through synthetic division or factor theorem where necessary.