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

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

Let ( u = x^2 + 4 ). Then ( du = 2x , dx ) or ( dx = \frac{1}{2} \frac{du}{x} ).

When ( x = 0, u = 4 ) and when ( x = 1, u = 5 ). Thus,

[ \int_0^1 \frac{x}{\sqrt{x^2 + 4}} , dx = \int_4^5 \frac{1}{2 \sqrt{u}} , du = \left[ \sqrt{u} \right]_4^5 = \sqrt{5} - \sqrt{4} = \sqrt{5} - 2 .

Using the binomial theorem, coefficients can be found as:

[ \text{Coefficient of } x^2 = \binom{8}{2} \left( -\frac{1}{2} \right)^2 = \frac{8 \times 7}{2} \times \frac{1}{4} = 14. ]

[ \text{Coefficient of } x^3 = \binom{8}{3} \left( -\frac{1}{2} \right)^3 = \frac{8 \times 7 \times 6}{6} \times \left( -\frac{1}{8} \right) = -7. ]

To show that ( \mathbf{u} \cdot \mathbf{v} = 0 ), we set:

( \mathbf{u} = \left( \frac{a}{2} \right) \quad \mathbf{v} = \left( \frac{a-7}{4a-1} \right) )

( \mathbf{u} \cdot \mathbf{v} = \left( \frac{a}{2} \right) \cdot \left( \frac{a-7}{4a-1} \right) = 0 )

This gives: ( a(a-7) - 2(4a-1) = 0 ) leading to ( a^2 - 7a + 8 = 0 ), which factors to ( (a-2)(a-4) = 0 ), thus ( a = 2 ) or ( a = 4. )

We rewrite the expression:

( R \sin(x + \alpha) = R \sin x \cos \alpha + R \cos x \sin \alpha ). Matching coefficients gives the system:

[ R \cos \alpha = -3 , (1) \quad R \sin \alpha = \sqrt{3} , (2) ]

From (1) & (2):

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

To find ( \alpha ): ( \tan \alpha = \frac{\sqrt{3}}{-3} ) leads to ( \alpha = \tan^{-1}(-\frac{1}{\sqrt{3}}) + \pi = \frac{5\pi}{6} ).

First, rearranging,

( x \geq 5(2 - x) \quad \Rightarrow \quad x + 5x \geq 10 \quad \Rightarrow \quad 6x \geq 10 \quad \Rightarrow \quad x \geq \frac{5}{3} )

And also ( 2 - x \neq 0 \Rightarrow x \neq 2 ).

Adding this gives: ( x \geq \frac{5}{3}, x \neq 2. )