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

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

Calculating the components:

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

Hence, ( \mathbf{u} \cdot \mathbf{v} = 1 ).

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

1. The differential: [ du = 2x , dx \implies dx = \frac{du}{2x} ] Hence, the integral changes to: [ \int \frac{x}{\sqrt{u}} \frac{du}{2x} = \frac{1}{2} \int u^{-1/2} , du ]

2. Now we need to evaluate the limits when ( x = 0 ) (gives ( u = 4 )) and ( x = 1 ) (gives ( u = 5 )): [ \int_0^1 \frac{x}{\sqrt{x^2 + 4}} , dx = \frac{1}{2} \left[ 2\sqrt{u} \right]_4^5 = \frac{1}{2} (2\sqrt{5} - 2\sqrt{4}) ] This simplifies to: [ = \sqrt{5} - 2 ]

Using the binomial expansion:

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

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

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

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

To find the values of ( a ) such that vectors are perpendicular, we set:

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

This means:

[ \left( \frac{a}{2} \right) \cdot \left( \frac{a - 7}{4a - 1} \right) = 0 ]

We can simplify this to:

[ a(a - 7) = 0 \quad \text{or} \quad (4a - 1) \neq 0 ]

The solutions to this are: [ a = 0 \quad \text{or} \quad a = 7. ]

Additionally, we check that ( 4a - 1 ) is not zero to avoid division by zero; ( a \neq \frac{1}{4}. )

To express ( \sqrt{3}\sin(x) - 3\cos(x) ) in that form:

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

2. Calculate ( \alpha ) using: [ \tan(\alpha) = \frac{-3}{\sqrt{3}} \implies \alpha = \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3} ]

Thus, we find: [ \sqrt{3}\sin(x) - 3\cos(x) = 2\sqrt{3} \sin\left(x - \frac{\pi}{3}\right) ]

To solve the inequality:

1. Multiply both sides by ( 2 - x ) (noting the sign based on the value of ( x )): [ x \geq 5(2 - x)\n ] This leads to: [ x + 5x \geq 10 \implies 6x \geq 10 \implies x \geq \frac{5}{3} ]

2. Identify critical values and intervals using a sign chart to be comprehensive: The critical points will give intervals to test whether they satisfy the original inequality.

3. Final interval solution: [ x \geq \frac{5}{3} \quad ( 2 - x > 0)\implies x < 2 \quad \therefore \frac{5}{3} \leq x < 2\text{ is the solution set.} ]