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 the dot product:

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

Using the dot product formula:

[ = 1 \cdot 2 + (-1) \cdot 1 = 2 - 1 = 1 ]

Using the substitution ( u = x^2 + 4 ), we have:

[ du = 2x , dx \quad \Rightarrow \quad dx = \frac{du}{2x} ]

Thus, the integral becomes:

[ \int \frac{x}{\sqrt{u}} \cdot \frac{1}{2x} , du = \frac{1}{2} \int u^{-1/2} , du ]

Integrating from the limits: when ( x = 0 ), ( u = 4 ) and when ( x = 1 ), ( u = 5 ).

Evaluating:

[ = \frac{1}{2} [2u^{1/2}]{4}^{5} = [u^{1/2}]{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 ]

For ( x^2 ):

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

For ( x^3 ):

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

To find when the vectors are perpendicular, we set:

[ \mathbf{u} \cdot \mathbf{v} = 0 ]

This gives:

[ \frac{a}{2} \cdot \frac{a - 7}{4a - 1} = 0 \quad \Rightarrow \quad a = -2 \text{ or } a = 7 ]

To express in that form, we note:

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

And find ( \alpha ):

[ \tan(\alpha) = \frac{-3}{\sqrt{3}} = -\sqrt{3} \quad \Rightarrow \quad \alpha = -\frac{\pi}{3} ]

Thus, we have:

[ \sqrt{3} \sin(x) - 3 \cos(x) = 2\sqrt{3} \sin\left(x - \frac{\pi}{3}\right)]

Rearranging gives:

[ \frac{x}{2 - x} - 5 \leq 0 \quad \Rightarrow \quad \frac{x - 5(2 - x)}{2 - x} \leq 0 ]

This simplifies to:

[ \frac{6x - 10}{2 - x} \leq 0 \quad \Rightarrow \text{Critical points are } x = 5 \text{ and } x = 2 ]

From the test intervals, we find the solution:

[ 2 < x \leq 5 ]