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

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

[ = 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} ]

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

Then,

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

Calculating gives:

[ = \frac{1}{2}[2 \sqrt{u}]_4^5 = \frac{1}{2}(2\sqrt{5} - 4) = \sqrt{5} - 2 ]

Using the binomial theorem:

[ \left( a + b \right)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k ]

Setting ( a = 1 ), ( b = -\frac{x}{2} ), and ( n = 8 ), we find the coefficients:

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 ]

To check for perpendicularity, we use the condition: ( \mathbf{u} \cdot \mathbf{y} = 0 ).

So, [ \begin{pmatrix} a \ 2 \end{pmatrix} \cdot \begin{pmatrix} a - 7 \ 4a - 1 \end{pmatrix} = a(a - 7) + 2(4a - 1) = 0 ]

This simplifies to: [ a^2 - 7a + 8a - 2 = 0 \quad \Rightarrow \quad a^2 + a - 2 = 0 ]

Factoring gives: [ (a - 2)(a + 1) = 0 ]

Thus, the possible values of ( a ) are: ( a = 2 ) or ( a = -1 ).

To express in the desired form, first calculate ( R ):

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

Next, find ( \alpha ):

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

Thus, we can write:

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

To solve this inequality, begin by setting:

[ x , (2 - x) \leq 10 ]

This simplifies to: [ 2x - x^2 - 10 \leq 0 \quad \Rightarrow \quad -x^2 + 2x - 10 \leq 0 ]

Factoring gives: [ -(x - 5)(x + 2) \leq 0 ]

The critical points are ( x = -2 ) and ( x = 5 ). The sign chart shows:

• The solution is valid for ( -2 \leq x \leq 5 ).