For the vectors egin{align*} oldsymbol{u} &= oldsymbol{i} - oldsymbol{j}, \\ oldsymbol{v} &= 2oldsymbol{i} + oldsymbol{j} \\ ext{evaluate each of the following.} \\ oldsymbol{(i)} \\ oldsymbol{u + 3v} \\ oldsymbol{(ii)} \\ oldsymbol{u ullet v} \\ ext{(b) Find the exact value of} \\ oldsymbol{rac{1}{2} extstyleigg( extstylerac{x}{oldsymbol{ oot{2}}{x^{2}+4}} igg)} \\ ext{using the substitution} \\ oldsymbol{u = x^{2} + 4} - HSC - SSCE Mathematics Extension 1 - Question 11 - 2022 - Paper 1

Next, compute the dot product u • v:

oldsymbol{u ullet v} = (1)(2) + (-1)(1) = 2 - 1 = 1.

To solve this integral, use the substitution (u = x^2 + 4), which gives (du = 2x dx). Thus, we have:

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

Next, change the limits of integration. When (x = 0), (u = 4) and when (x = 1), (u = 5).

The integral then becomes:

$\int_0^1 \frac{x}{\sqrt{x^2 + 4}} dx = \frac{1}{2} \int_4^5 \frac{1}{\sqrt{u}} du = \frac{1}{2} [2\sqrt{u}]_4^5 = [\sqrt{5} - \sqrt{4}] = \sqrt{5} - 2.$

So the exact value is (\sqrt{5} - 2).

Using the binomial theorem:

$(a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k,$

set (a = 1), (b = -\frac{x}{2}), and (n = 8).

The coefficient of (x^2) is:

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

The coefficient of (x^3) is:

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

Thus, the coefficients are 7 and -7 for (x^2) and (x^3) respectively.

For the vectors (u) and (v) to be perpendicular, their dot product must equal zero:

$\begin{aligned} u \bullet v &= 0 \Rightarrow \frac{(a)}{2} \cdot \frac{(a - 7)}{4a - 1} = 0 \end{aligned}$

This leads to:

$a(a - 7) - 2(a + 7) = 0 \Rightarrow a^2 - 7a + 2 = 0.$

We can solve this quadratic for (a) using the quadratic formula:

$a = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} = \frac{7 \pm \sqrt{49 - 8}}{2} = \frac{7 \pm \sqrt{41}}{2}.$

To express the given function in the desired form, first identify R and α:

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

and then,

$tan(α) = \frac{-3}{\sqrt{3}} = -\sqrt{3}$ implies that (α = -\frac{\pi}{3} + 2k\pi, \forall k \in \mathbb{Z}.$$

Thus,

$\sqrt{3}sin(x) - 3cos(x) = 2\sqrt{3}sin(x - \frac{\pi}{3}).$

To solve this inequality:

1. Rewrite it:

$x \leq 5(2 - x)$

2. Rearrange:

$x + 5x \leq 10$

3. Combine like terms:

$6x \leq 10 \Rightarrow x \leq \frac{5}{3}.$

4. Identify critical points by setting the denominator to zero:

$2 - x = 0 \Rightarrow x = 2.$

5. Consider test intervals (e.g., (-∞, 2), (2, 5/3), and (5/3, ∞)).
6. Collect valid intervals, which yield the final answer.