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 calculate $\mathbf{u} \cdot \mathbf{v}$:

1. Use the dot product: $\mathbf{u} \cdot \mathbf{v} = (\mathbf{i} - \mathbf{j}) \cdot (2\mathbf{i} + \mathbf{j})$

2. Compute the dot product: $= 1 \cdot 2 + (-1) \cdot 1$ $= 2 - 1 = 1$

To solve the integral:

1. Use substitution: Let $u = x^2 + 4$, then $du = 2x \, dx$.

2. Change limits:

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

3. Rewrite the integral: $\int_0^1 \frac{\sqrt{x}}{\sqrt{x^2 + 4}} \, dx = \int_4^5 \frac{\sqrt{u - 4}}{\sqrt{u}} \cdot \frac{1}{2} \, du$

4. Simplify and solve: $= \frac{1}{2} \int_4^5 \frac{\sqrt{u - 4}}{\sqrt{u}} \, du$ $= \left[ \sqrt{5} - 2 \right]$

To find the coefficients:

1. Use the binomial expansion: $\left(1 - \frac{x}{2}\right)^8 = \sum_{k=0}^{8} \binom{8}{k} \left(-\frac{x}{2}\right)^k$

2. For $k=2$ (coefficient of $x^2$): $\binom{8}{2}\left(-\frac{1}{2}\right)^2 = 28 \cdot \frac{1}{4} = 7$

3. For $k=3$ (coefficient of $x^3$): $\binom{8}{3}\left(-\frac{1}{2}\right)^3 = 56 \cdot -\frac{1}{8} = -7$

To determine $a$:

1. Set up the equation for perpendicularity: $\mathbf{u} \cdot \mathbf{v} = 0$ $\Rightarrow \left(\frac{a}{2}\right) \left(\frac{a}{2}\right) + \left(\frac{a-7}{4a-1}\right) \left(\frac{a-7}{4a-1}\right) = 0$

2. Solve the resulting polynomial: $a(a-7) = 0$ $\Rightarrow a = 2 \text { or } a = -2$

To express in the required form:

1. Identify $R$ and $\alpha$: $R = \sqrt{(\sqrt{3})^2 + (-3)^2} = \sqrt{3 + 9} = 2\sqrt{3}$

2. Find $\tan(\alpha)$: $\tan(\alpha) = \frac{-3}{\sqrt{3}} = -\sqrt{3}$ $\Rightarrow \alpha = -\frac{\pi}{3}$

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

To solve the inequality:

1. Rearrange and clear the fraction: $x \geq 5(2 - x)$ $\Rightarrow x + 5x \geq 10$ $\Rightarrow 6x \geq 10$ $\Rightarrow x \geq \frac{5}{3}$

2. However, ensure the denominator is valid: $2 - x > 0 \Rightarrow x < 2$

3. Combine the conditions: $\frac{5}{3} \leq x < 2$