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

The dot product ( \mathbf{u} \cdot \mathbf{v} ) is calculated as follows:

1. ( \mathbf{u} \cdot \mathbf{v} = (\mathbf{i} - \mathbf{j}) \cdot (2\mathbf{i} + \mathbf{j}) ).
2. This expands to ( (1)(2) + (-1)(1) = 2 - 1 = 1 ).

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

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

1. Calculate ( du = 2x , dx ) which implies ( dx = \frac{du}{2x} ).
2. Adjust the limits: when ( x = 0 ), ( u = 4 ); and when ( x = 1 ), ( u = 5 ).
3. The integral becomes: [ \int_4^5 \frac{x}{\sqrt{u}} \cdot \frac{du}{2x} = \frac{1}{2} \int_4^5 u^{-\frac{1}{2}} , du. ]
4. Solving this, [ \frac{1}{2} \cdot 2(u^{\frac{1}{2}})\Big|_4^5 = \sqrt{5} - 2. ]

Using the binomial expansion:

1. The general term is given by ( T_k = \binom{n}{k} a^{n-k} b^k ), where ( n = 8, a = 1, b = -\frac{x}{2} ).
2. Coefficients for ( x^2 ) and ( x^3 ):
   • For ( x^2 ): ( \binom{8}{2} \left(-\frac{1}{2}\right)^2 = 28 \cdot \frac{1}{4} = 7 ).
   • For ( x^3 ): ( \binom{8}{3} \left(-\frac{1}{2}\right)^3 = 56 \cdot -\frac{1}{8} = -7 ).

For ( \mathbf{u} ) and ( \mathbf{v} ) to be perpendicular, their dot product must be zero:

1. ( \mathbf{u} \cdot \mathbf{v} = 0 ) implies: [ \left( \frac{a}{2} \right) \cdot \left( \frac{a - 7}{4a - 1} \right) = 0. ]
2. This leads to two equations:
   • ( \frac{a}{2} = 0 \rightarrow a = 0 )
   • or ( \frac{a - 7}{4a - 1} = 0 \rightarrow a - 7 = 0 \rightarrow a = 7. )

Thus, the possible values of ( a ) are ( 0 ) or ( 7 ).

1. First, identify ( R = \sqrt{a^2 + b^2} ) where ( a = -3 ) (coefficient of ( \cos(x) )) and ( b = \sqrt{3} ) (coefficient of ( \sin(x) )). So, [ R = \sqrt{(-3)^2 + (\sqrt{3})^2} = \sqrt{9 + 3} = \sqrt{12} = 2\sqrt{3}. ]

2. Next, find ( \tan(\alpha) = \frac{b}{a} = \frac{\sqrt{3}}{-3} \rightarrow \alpha = \tan^{-1}(\frac{\sqrt{3}}{-3}) \rightarrow \alpha = -\frac{ p{\pi}}{3} ).

3. Hence, ( \sqrt{3}\sin(x) - 3\cos(x) = 2\sqrt{3}\sin\left( x - \frac{\pi}{3} \right) ).

1. Rearranging gives ( x \leq 5(2 - x) ) leading to ( x \leq 10 - 5x ).
2. This simplifies to ( 6x \leq 10 \rightarrow x \leq \frac{5}{3} ).
3. Also consider the denominator: ( 2 - x > 0 \rightarrow x < 2. )
4. Thus, the solution is ( x \leq \frac{5}{3} ) and ( x < 2 ) providing the interval ( x < 2 ).