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

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

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

To solve this integral:

1. Make the substitution $u = x^{2} + 4$, then $du = 2x \, dx$.

2. Change limits:

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

3. Update the integral: $\int_{4}^{5} \frac{1}{2\sqrt{u}} du$

4. Integrate: $= \frac{1}{2} \left[ 2\sqrt{u} \right]_{4}^{5}$

5. Evaluate: $= \left[ \sqrt{5} - 2 \right]$

Using the binomial theorem for the expansion:

1. The general term in the expansion: $T_k = \binom{n}{k} a^{n-k} b^k$ where $a = 1$, $b = -\frac{x}{2}$, and $n=8$.

2. For $x^{2}$ (where $k=2$): $T_2 = \binom{8}{2} \left(-\frac{x}{2}\right)^{2} = \binom{8}{2} \frac{x^{2}}{4} = 28 \cdot \frac{1}{4} = 7$

3. For $x^{3}$ (where $k=3$): $T_3 = \binom{8}{3} \left(-\frac{x}{2}\right)^{3} = \binom{8}{3} \frac{-x^{3}}{8} = -56 \cdot \frac{1}{8} = -7$

To find $a$ where the vectors are perpendicular:

1. The condition is $\mathbf{u} \cdot \mathbf{v} = 0$: $\left(\frac{a}{2}\right)\left(\frac{a - 7}{4a - 1}\right) + (2)\cdot 0 = 0$

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

To express in the desired form:

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

2. Find $\alpha$ using: $\tan \alpha = \frac{b}{a} = \frac{-3}{\sqrt{3}} = -\sqrt{3}$

3. Determine $\alpha$ in the context of the unit circle: $\alpha = -\frac{\pi}{3}$

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

To solve the inequality:

1. Rewrite it: $x \geq 5(2 - x)$

2. Expand and simplify: $x \geq 10 - 5x$ $6x \geq 10 \Rightarrow x \geq \frac{5}{3}$

3. Consider the critical point $x = 2$ (where the denominator is zero):

   • Test intervals around $x = 2$ to see where the expression holds true to find: $x \leq 2 \ \, \text{and} \ x \geq \frac{5}{3}$