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} ):

1. Use the formula: ( \mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 ).
2. Here, ( u_1 = 1, u_2 = -1 ) and ( v_1 = 2, v_2 = 1 ).
3. Therefore: [ \mathbf{u} \cdot \mathbf{v} = 1 \times 2 + (-1) \times 1 = 2 - 1 = 1. ]

To solve the integral:

1. Set the substitution: ( u = x^2 + 4 ). Then, differentiate: ( du = 2x , dx ), yielding ( dx = \frac{du}{2x} ).

2. Change the limits of integration: When ( x = 0 ), ( u = 4 ) and 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{du}{2\sqrt{u - 4}} = \frac{1}{2} \int 1 , du = \frac{u}{2} \bigg|_{4}^{5} = \frac{5 - 4}{2} = \frac{1}{2}. ]

To find the coefficients:

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

2. For ( n = 8 ),\ identify the terms for ( k = 2 ) and ( k = 3 ):

   • Coefficient of ( x^2 ): [ \binom{8}{2} \left(-\frac{1}{2}\right)^2 = 28 \cdot \frac{1}{4} = 7. ]
   • Coefficient of ( x^3 ): [ \binom{8}{3} \left(-\frac{1}{2}\right)^3 = 56 \cdot \left(-\frac{1}{8}\right) = -7. ]

For ( \mathbf{u} ) and ( \mathbf{v} ) to be perpendicular:

1. The dot product must equal zero: [ \mathbf{u} \cdot \mathbf{v} = 0. ]

2. Thus: [ \left( \frac{a}{2} \right) \cdot \left( \frac{a - 7}{4a - 1} \right) = 0. ]
   This implies ( a(a - 7) = 0 ), yielding ( a = 0 ) or ( a = 7 ).

To express in desired form:

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

2. Determine ( \alpha ): Use ( \tan \alpha = \frac{\sqrt{3}}{-3} ):

   • Thus, ( \alpha = \pi - \frac{\pi}{3} = \frac{2\pi}{3} ).

3. Express the original equation: [ \sqrt{3}\sin(x) - 3\cos(x) = 2\sqrt{3}\sin\left(x + \frac{2\pi}{3}\right). ]

To solve the inequality:

1. Rearrange: [ x \geq 5(2 - x) \Rightarrow x + 5x \geq 10 \Rightarrow 6x \geq 10 \Rightarrow x \geq \frac{5}{3}. ]

2. Check for the vertical asymptote at ( x = 2 ).

3. The solution set: ( x \geq \frac{5}{3} ) and ( x \neq 2).