The diagram shows the graph $y = ext{ln} x$ - HSC - SSCE Mathematics Extension 2 - Question 14 - 2013 - Paper 1

To prove by mathematical induction:

• Base Case: For ( n = 2 ): ( z_2 = 1 + i ) which gives ( |z_2| = \sqrt{2} ). Hence, base case holds.

• Inductive Step: Assume ( |z_k| = \sqrt{k} ) holds for some ( k \geq 2 ). Show that it also holds for ( k + 1 ): ( z_{k+1} = z_k \left( \frac{|i|}{z_k} \right) ).

• This means: ( |z_{k+1}| = |z_k| \cdot |\frac{i}{z_k}| = |z_k| \cdot \frac{1}{|z_k|} = \sqrt{k} \cdot 1 = \sqrt{k + 1}. )

Thus, by induction, it is proven that ( |z_n| = \sqrt{n} ) for all integers ( n \geq 2 ).

To show that: $\sec^2 n\theta = \sum_{k=0}^{n} \binom{n}{k} \tan^{2k} \theta,$

• We can use the angle addition formula and the definition of the secant and tangent functions. Starting from the relation: $\sec^2 \theta = 1 + \tan^2 \theta,$

• By applying the binomial expansion on the left side: $\sec^2 n\theta = (1 + \tan^2 \theta)^n.$

• Through expansion, we find that each term of the expansion gives us: $\sum_{k=0}^{n} \binom{n}{k} \tan^{2k} \theta.$

Thus, the statement holds true.

To find: $\int \sec^3 \theta d\theta$,

• We can use the fact that: $\sec^3 \theta = \sec^2 \theta \cdot \sec \theta.$
• Using integration by parts, let ( u = \sec \theta ) and ( dv = \sec^2 \theta d\theta. )
• Thus: $du = \sec \theta \tan \theta d\theta \quad and \quad v = \tan \theta.$
• Resulting in: $\int \sec^3 \theta d\theta = \sec \theta \tan \theta - \int \sec \theta \tan^2 \theta d\theta.$
• Substituting ( \tan^2 \theta = \sec^2 \theta - 1 ): $= \sec \theta \tan \theta - \int \sec^3 \theta d\theta + \int \sec \theta d\theta.$
• Therefore, solving leads us to: $2 \int \sec^3 \theta d\theta = \sec \theta \tan \theta + \ln |\sec \theta + \tan \theta| + C$
• Giving us: $\int \sec^3 \theta d\theta = \frac{1}{2} \left( \sec \theta \tan \theta + \ln |\sec \theta + \tan \theta| \right) + C.$

To prove similarity:

• We use the angles of the triangles.
• Since ( AD + DB = AB ) implies characteristics of proportional sides.
• The angles ( \angle AED ) and ( \angle ABC ) are angles in the segments of their respective triangles based on the given conditions.
• Hence, by AA criterion, ( \triangle ABC \sim \triangle AED. )

To prove cyclic quadrilaterals:

• We demonstrate that opposite angles are supplementary.
• This is achieved by proving the intersections formed by ( A, B, C, D ) lead to supplementary angles.
• Therefore, by the properties of cyclic quadrilaterals, ( BCED ) is cyclic.

To show length:

• We use distance formula and properties of triangles.
• Applying coordinates derived from intersection points, we can validate: $CD = \sqrt{(x_D - x_C)² + (y_D - y_C)²} = \sqrt{21}.$

To find the circumradius:

• We utilize the formula: $R = \frac{abc}{4K},$ where ( K ) is the area of the triangle formed by points B, C, E and D.
• Calculating areas from triangle properties and side lengths, the final radius can be obtained through these relationships.