Find $$\int \frac{2x+3}{x^2 + 2x + 2} \, dx$$ (b) Consider Statement A - HSC - SSCE Mathematics Extension 2 - Question 12 - 2021 - Paper 1

The converse of Statement A is: 'If $n$ is even, then $n^2$ is even.'

To prove this, let $n$ be an even number, then $n = 2k$ for some integer $k$. Thus, we have:

$n^2 = (2k)^2 = 4k^2 = 2(2k^2),$

which shows that $n^2$ is also even. Hence, the converse is true.

To determine the values of $p$ and $q$, consider the direction vectors:

$\mathbf{r_1} = \begin{pmatrix} -2 \\ 1 \\ 3 \end{pmatrix} + \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} t$ $\mathbf{r_2} = \begin{pmatrix} 4 \\ -2 \\ p \end{pmatrix} + \begin{pmatrix} q \\ -1 \\ 2 \end{pmatrix} u$

The lines are perpendicular, which gives:

$\mathbf{r_1} \cdot \mathbf{r_2} = 0.$

Thus, equating components gives:

From the directions: $1p + 0*(-1) + 2q = 0 \Rightarrow p + 2q = 4.$

And: $3*(-1) = -8.$

From solving these, we find:

• p = 20
• q = 10.

To prove by induction, we check the base case (n=9):

$\sqrt{9} = 3 \geq 2^9 = 512 \text{ (true)}.$

Now assume true for $n=k$, such that: $\sqrt{k} \geq 2^k.$

For $n=k+1$: $\sqrt{k+1} \geq 2^{k+1}.$

Through calculations and that $\sqrt{k+1} \geq \sqrt{k} \geq 2^{k}$, the inequality holds for all integers $n \geq 9$.

Using vector properties, we can see that:

$\overline{HA} + \overline{HB} + \overline{HC} + \overline{HD}$

forms a closed polygon, hence evaluating gives: $0 = 0.$

By stating points symmetry and relations, we find: $G = \frac{1}{4}(\overline{GA} + \overline{GB} + \overline{GC} + \overline{GD}).$

This can be confirmed as all diagonals bisect each other hence leading to the equality.

By substituting the values from parts (ii) and properties of vectors, it can be calculated that:

$\overline{HG} = \lambda (\overline{GS} - \overline{GH}).$

Thus, $\lambda = \frac{1}{2}.$