The points A, B and C lie on a circle with centre O, as shown in the diagram - HSC - SSCE Mathematics Extension 1 - Question 12 - 2017 - Paper 1

To sketch the graphs:

• For the first graph, y = |x + 1|, the vertex is at (-1, 0), opening upwards.
• For the second graph, y = 3 - |x - 2|, the vertex is at (2, 1). It opens downwards. Make sure to label all axes and intercepts clearly for accuracy.

|x + 1| + |x - 2| equals 3 when evaluated over the relevant intervals derived from the sketch, revealing:

• For the interval where x is between -1 and 2, we find the valid x-values satisfying the equation, leading to the conclusion:

$-1 ≤ x ≤ 1$

In order to show that h satisfies this equation, we need to derive the volume of the solids of revolution and set them in the required ratio of 2:1. The integration yields the relation concerning h which simplifies to:

$3h³ - 9h + 2 = 0.$

Using Newton's method, we find:

1. Calculate f(h₁) and f'(h₁).
2. Substitute these into the Newton's method formula:

h_{n+1} = h_n - rac{f(h_n)}{f'(h_n)}.

Using h₁ = 0, compute the next approximation.

First, differentiate the equation ( t = 4 - e^{-2x} ) with respect to t to find velocity ( v = \frac{dx}{dt} ). Subsequently, derive acceleration by taking the second derivative, giving:

$a(t) = \frac{d^2x}{dt^2}.$ Use chain rule as applicable.

Using L'Hôpital's Rule or Taylor expansion:

$\lim_{x\to0} \frac{1 - \cos(2\pi x)}{x^2} = \lim_{x\to0} \frac{2\pi \sin(2\pi x)}{2x} = \lim_{x\to0} \frac{\pi \sin(2\pi x)}{x} = \pi \cdot 2\pi = 0.$