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

To find the dot product, we compute:

$\boldsymbol{u} \bullet \boldsymbol{v} = (\boldsymbol{i} - \boldsymbol{j}) \bullet (2\boldsymbol{i} + \boldsymbol{j})$

This expands to:

$= 1 \cdot 2 + (-1) \cdot 1 = 2 - 1 = 1$

Using the substitution $\boldsymbol{u} = x^2 + 4$, we have:

$du = 2x \, dx$ $dx = \frac{du}{2x}$

Thus substituting yields:

$\int \frac{\sqrt{x}}{\sqrt{u}} \cdot \frac{du}{2x} = \frac{1}{2} \int \frac{1}{\sqrt{u}} \, du$

Now, changing limits:

• When $x = 0$, $u = 4$.
• When $x = 1$, $u = 5$.

Calculation becomes:

$= \frac{1}{2} [2\sqrt{u}]_4^5 = \frac{1}{2} [2\sqrt{5} - 2\sqrt{4}] = \sqrt{5} - 2$

Using the binomial expansion:

$(1 - x)^n = \sum_{k=0}^{n} \binom{n}{k} (-x)^k$

For $(1 - \frac{x}{2})^8$, the coefficients are:

• Coefficient of $x^2$: (\binom{8}{2} (-\frac{1}{2})^2 = 28 \cdot \frac{1}{4} = 7)
• Coefficient of $x^3$: (\binom{8}{3} (-\frac{1}{2})^3 = 56 \cdot -\frac{1}{8} = -7)

The condition for perpendicular vectors is: $\boldsymbol{u} \bullet \boldsymbol{v} = 0$ Substituting the vectors: $\begin{pmatrix} \frac{a}{2} \\ 2 \end{pmatrix} \bullet \begin{pmatrix} \frac{a - 7}{4a - 1} \ 1 \end{pmatrix} = 0$ This leads to: $\frac{a}{2}\cdot \frac{a - 7}{4a - 1} + 2\cdot 1 = 0$ Solving the equation gives:

= a^2 - 15a + 4 = 0$$ Using the quadratic formula: $$a = \frac{15 \pm \sqrt{15^2 - 4 \cdot 1 \cdot 4}}{2} = 2 \quad \text{or} \quad 14$$

To express in the required form, calculate $R$ and $\alpha$:

First, find: $R = \sqrt{(\sqrt{3})^2 + (-3)^2} = \sqrt{3 + 9} = \sqrt{12} = 2\sqrt{3}$

Now, find $\alpha$: $\tan(\alpha) = \frac{-3}{\sqrt{3}} \Rightarrow \alpha = -\frac{\pi}{3}$ Therefore, we can write: $R\sin(x + \alpha) = 2\sqrt{3}\sin\left(x - \frac{\pi}{3}\right)$

To solve the inequality, we start by rearranging:

$x \leq 5(2 - x)$

Adding $5x$ to both sides: $x + 5x \leq 10$ $6x \leq 10$ This simplifies to: $x \leq \frac{10}{6} = \frac{5}{3}$ The domain also requires checking conditions: $2 - x > 0 \Rightarrow x < 2$ Thus combining gives: $x < 2 \text{ and } x \leq \frac{5}{3}$ The solution set is: $x < \frac{5}{3}$