For the vectors $ extbf{u} = extbf{i} - extbf{j}$ and $ extbf{v} = 2 extbf{i} + extbf{j}$, evaluate each of the following: a) (i) $ extbf{u} + 3 extbf{v}$ (ii) $ extbf{u} ullet extbf{v}$ b) Find the exact value of \( \int_{0}^{1} \frac{x}{\sqrt{x^{2}+4}} \, dx \) using the substitution \( u = x^{2}+4 \) - HSC - SSCE Mathematics Extension 1 - Question 11 - 2022 - Paper 1

$\textbf{u} \bullet \textbf{v} = (\textbf{i} - \textbf{j}) \bullet (2\textbf{i} + \textbf{j}) = 1 \cdot 2 + (-1) \cdot 1 = 1.$

Let $u = x^{2} + 4$, then $du = 2x \, dx$. Changing the limits when $x=0$, $u=4$ and when $x=1$, $u=5$. Hence,

$\int_{0}^{1} \frac{x}{\sqrt{x^{2}+4}} \, dx = \int_{4}^{5} \frac{1}{2\sqrt{u}} \, du = \left[\sqrt{u}\right]_{4}^{5} = \sqrt{5} - 2.$

Using the binomial expansion, we have:

$\left( 1 - \frac{x}{2} \right)^{8} = \sum_{k=0}^{8} \binom{8}{k} (-1)^{k} \left(\frac{x}{2}\right)^{k}.$ For $x^{2}$, use $k=2$: $\binom{8}{2} \left( -\frac{1}{2} \right)^{2} = 28 \cdot \frac{1}{4} = 7.$ For $x^{3}$, use $k=3$: $\binom{8}{3} \left( -\frac{1}{2} \right)^{3} = 56 \cdot -\frac{1}{8} = -7.$

Given the vectors $\textbf{u} = \begin{pmatrix} \frac{a}{2} \\ 2 \end{pmatrix}$ and $\textbf{y} = \begin{pmatrix} \frac{a-7}{4a-1} \\ 1 \end{pmatrix}$ are perpendicular if $\textbf{u} \bullet \textbf{y} = 0$:

$\frac{a}{2} \cdot \frac{a-7}{4a-1} + 2 \cdot 1 = 0$ Solving this leads to the quadratic in terms of $a$, which gives possible values of $a = -2$ or $a = 7.$

To express this form, let $R = \sqrt{\sqrt{3}^{2} + (-3)^{2}} = \sqrt{3 + 9} = 2\sqrt{3}$. Now let $\tan(\alpha) = \frac{-3}{\sqrt{3}}$. From here, we find: $\alpha = \tan^{-1}\left(-\sqrt{3}\right) = \frac{5\pi}{3}.$

Cross-multiplying provides: $\Rightarrow x \geq 10 - 5x$ $\Rightarrow 6x \geq 10$', leading to $x \geq \frac{5}{3}$. We then examine the critical points to find valid solutions.