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

For the dot product ( \mathbf{u} \cdot \mathbf{v} ):

[ \mathbf{u} \cdot \mathbf{v} = (\mathbf{i} - \mathbf{j}) \cdot (2\mathbf{i} + \mathbf{j}) ] [ = 1 \cdot 2 + (-1) \cdot 1 ] [ = 2 - 1 = 1 ]

Using the substitution ( u = x^{2} + 4 ), we have ( du = 2x , dx ) which gives us:

[ dx = \frac{du}{2x} \quad \text{and} \quad x = \sqrt{u - 4} ]

The limits change as follows:

• When ( x = 0, u = 4 )
• When ( x = 1, u = 5 )

Substituting these into the integral:

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

Using the binomial expansion:

[ \left( 1 - \frac{x}{2} \right)^{8} = \sum_{k=0}^{8} \binom{8}{k} \left( -\frac{x}{2} \right)^{k} ]

The coefficient of ( x^{2} ): [ \text{Coefficient} = \binom{8}{2} \cdot \left( -\frac{1}{2} \right)^{2} = 28 \cdot \frac{1}{4} = 7 ]

The coefficient of ( x^{3} ): [ \text{Coefficient} = \binom{8}{3} \cdot \left( -\frac{1}{2} \right)^{3} = 56 \cdot \left( -\frac{1}{8} \right) = -7 ]

Setting ( \mathbf{u} \cdot \mathbf{v} = 0 ), we have:

[ \left( \frac{a}{2}, \frac{a-7}{4a-1} \right) \cdot \left( \frac{a}{2}, \frac{a-7}{4a-1} \right) = 0 ] [ \Rightarrow a(a-7) - 2 \cdot (4a - 1) = 0 ] [ \Rightarrow a^{2} - 7a + 8a - 2 = 0 ] [ a^{2} - 2a = 0 \Rightarrow a(a - 2) = 0 ] [ \Rightarrow a = 0 \text{ or } a = 2 ]

Rewriting in the form ( R \sin(x + \alpha) ):

1. Find ( R ): [ R = \sqrt{(\sqrt{3})^{2} + (-3)^{2}} = \sqrt{3 + 9} = \sqrt{12} = 2\sqrt{3} ]

2. Find ( \alpha ): [ \tan(\alpha) = \frac{-3}{\sqrt{3}} = -\sqrt{3} \quad \Rightarrow \alpha = \frac{4\pi}{3} ]

So, ( \sqrt{3} \sin(x) - 3 \cos(x) = 2\sqrt{3} \sin(x + \frac{4\pi}{3}) )

To solve the inequality: [ x \leq 5(2 - x) ] [ \Rightarrow x + 5x \leq 10 ] [ 6x \leq 10 \Rightarrow x \leq \frac{5}{3} ]

Check critical points for discontinuities:

• Denominator cannot be zero: ( 2 - x \neq 0 \Rightarrow x \neq 2 )
• Thus, the solution is ( x \leq \frac{5}{3} \text{ and } x \neq 2 ), which simplifies to ( x < 2 ).