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

The converse of Statement A, which states 'If ( n^2 ) is even, then ( n ) is even', is:

Converse: 'If ( n ) is even, then ( n^2 ) is even.'

To show that the converse is true, let's assume ( n ) is even.

By definition of even numbers, there exists an integer ( k ) such that:

$n = 2k.$

Now, squaring ( n ):

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

Since ( 2k^2 ) is an integer, ( n^2 ) is also even, which proves that the converse of Statement A is true.

For the lines to be perpendicular, the dot product of their direction vectors must equal zero:

1. The direction vector of ( r_1 ) is ( \begin{pmatrix} 1 \ 0 \ 2 \end{pmatrix} ) and for ( r_2 ) it is ( \begin{pmatrix} p \ q \ -1 \end{pmatrix} ).

2. Set the dot product equation:

$1 \cdot p + 0 \cdot q + 2 \cdot (-1) = 0 \implies p - 2 = 0 \implies p = 2.$

3. To find ( q ), use the intersection condition by equating components:

From ( r_1 ):

• For ( x ): ( -2 + \lambda = 4 + \mu ) ( \implies \lambda - \mu = 6) (1)
• For ( y ): ( 1 = -2 + \mu ) (\implies \mu = 3) (2)
• For ( z ): ( 3 + 2\lambda = p - q )
  (\implies 2\lambda = 2 - q \implies \lambda = 1 - \frac{q}{2}) (3)

4. Substitute ( \mu ) in (1):
   • From equation (2): (\lambda - 3 = 6 \implies \lambda = 9). Then find ( q = 20 ). So,

$p = 2, \, q = 20.$

Base case: For ( n = 9 ), we have:

$\sqrt{9} = 3 \, \text{and} \, 2^9 = 512$ (3 < 512) so it's true for the base case.

Inductive step: Assume true for ( n = k ): ( \sqrt{k} \geq 2^k ).

We need to prove for ( n = k + 1 ):

$\sqrt{k + 1} \geq 2^{k + 1} = 2 \cdot 2^k.$

Using the assumption, we have:

\sqrt{k + 1} = \sqrt{k} \cdot \sqrt{1 + \frac{1}{k}} \geq 2^k \cdot \sqrt{1 + \frac{1}{9}} \. \textrm{(for the base case)}

This proves induction is satisfied. Therefore, by induction, the statement holds for all integers ( n \geq 9 ).

To show that the vectors add to zero, consider point ( H ) which is the intersection of diagonals in square ( ABCD ). By vector addition:

$HA + HB + HC + HD = 0.$

Since diagonal bisecting properties mean points distribute evenly around point H, combining these vectors leads back to point H, verifying the equation.

From part (i): We know that:

$GA + GB + GC + GD = 0$

Thus,

$\sum GH = 0 \implies 4GH = \lambda (HG + HS),$

showing that G is equidistant in vector representation.

From the previous part, where:

$4GH = \lambda HG + \lambda HS,$ In order to satisfy both sides, one can assert that:\n (\lambda = 4). Thus, the value of ( \lambda ) is 4.