For the vectors $\mathbf{u} = \mathbf{i} - \mathbf{j}$ and $\mathbf{v} = 2\mathbf{i} + \mathbf{j}$, evaluate each of the following - HSC - SSCE Mathematics Extension 1 - Question 11 - 2022 - Paper 1

To compute the dot product $\mathbf{u} \cdot \mathbf{v}$:

$\mathbf{u} \cdot \mathbf{v} = (\mathbf{i} - \mathbf{j}) \cdot (2\mathbf{i} + \mathbf{j})$

This expands to:

$= 1\cdot2 + (-1)\cdot1 = 2 - 1 = 1.$

Using the substitution $u = x^2 + 4$, we have:

$du = 2x \; dx \implies dx = \frac{du}{2x}.$

When $x = 0$, $u = 4$; and when $x = 1$, $u = 5$.

This leads to:

$\int_{4}^{5} \frac{x}{\sqrt{u}} \cdot \frac{du}{2x} = \frac{1}{2} \int_{4}^{5} \frac{1}{\sqrt{u}} \; du$

Calculating the integral gives:

$= \frac{1}{2} \left[2\sqrt{u}\right]_{4}^{5} = \frac{1}{2} \left( 2\sqrt{5} - 2\sqrt{4} \right) = \sqrt{5} - 2.$

Using the binomial expansion, we know:

$\left( 1 - \frac{x}{2} \right)^{n} = \sum_{k=0}^{n} \binom{n}{k}\left(-\frac{x}{2}\right)^{k}.$

To find the coefficient of $x^2$:

$\binom{8}{2}\left(-\frac{1}{2}\right)^{2} = 28 \cdot \frac{1}{4} = 7.$

For the coefficient of $x^3$:

$\binom{8}{3}\left(-\frac{1}{2}\right)^{3} = 56 \cdot \left(-\frac{1}{8}\right) = -7.$

The vectors are perpendicular if:

$\mathbf{u} \cdot \mathbf{v} = 0.$

Thus:

$(\frac{a}{2})(\frac{a - 7}{4a - 1}) + (\frac{a - 7}{4a - 1})(\frac{a}{2}) = 0.$

This simplifies to:

$a(a - 7) = 0 \Rightarrow a = 0 \text{ or } a = 7.$

To express this in the required form:

Let $R = \sqrt{a^2 + b^2}$ where $a = -3$ and $b = \sqrt{3}$.

Calculating:

$R = \sqrt{(-3)^2 + (\sqrt{3})^2} = \sqrt{9 + 3} = \sqrt{12} = 2\sqrt{3}.$

Finding $\alpha$ using:

$\tan(\alpha) = \frac{b}{a} = \frac{\sqrt{3}}{-3}.$

Thus, we determine:

$\alpha = \tan^{-1}\left(-\frac{\sqrt{3}}{3}\right),$ which suggests $\alpha = \frac{5\pi}{6}$.

Therefore, the expression is:

$\sqrt{3}\sin(x) - 3\cos(x) = 2\sqrt{3}\sin\left(x + \frac{5\pi}{6}\right).$

To solve this, first rearrange:

$x \geq 5(2 - x) \\ x + 5x \geq 10 \\ 6x \geq 10 \\ x \geq \frac{5}{3}.$

Next, we consider the behavior of $\frac{x}{2 - x}$: It is defined for $x < 2$. Thus, the solution set is:

$x \in \left[\frac{5}{3}, 2\right).$