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

1. The graph of ( y = |x + 1| ) will have a V-shape with the vertex at ( (-1,0) ).
2. The graph of ( y = 3 - |x - 2| ) will also have a V-shape with the vertex at ( (2, 3) ) and intercepts at ( (1, 2) ) and ( (3, 0) ).
3. Ensure the x-values are clearly marked and the y-intercepts are labeled correctly.

To find the range, we solve the equation:

1. For ( x < -1 ): ( -(x + 1) + (2 - x) = 3 ) leads to no solution.
2. For ( -1 \leq x < 2 ): ( (x + 1) + (2 - x) = 3 ), leads to the range ( x = -1 ).
3. For ( x \geq 2 ): ( (x + 1) + (x - 2) = 3 ) translates to ( 2x - 1 = 3 ), yielding ( x = 2 ).

Thus, the range of ( x ) is ( -1 \leq x ext{ or } x = 2. )

To demonstrate this,

1. The volume of the first solid formed by the semicircle is given by: $V_1 = \pi \int_0^h (\sqrt{1-x^2})^2 dx = \pi \int_0^h (1-x^2) dx = \pi \left[ x - \frac{x^3}{3} \right]_0^h = \pi \left( h - \frac{h^3}{3} \right).$
2. The volume of the second solid formed by the remaining area is: $V_2 = \pi \int_h^1 (\sqrt{1-x^2})^2 dx = \pi \left( \frac{\pi}{2} - V_1 \right).$
3. Setting the volumes ratio to 2:1 will lead to: $\frac{V_1}{V_2} = \frac{1}{2} \Rightarrow \text{ultimately giving the equation } 3h^3 - 9h + 2 = 0.$

To apply Newton's method:

1. Compute the function: $f(h) = 3h^3 - 9h + 2$ with ( f(0) = 2. )
2. Find the derivative: $f'(h) = 9h^2 - 9.$
3. Calculate: $h_2 = h_1 - \frac{f(h_1)}{f'(h_1)} = 0 - \frac{2}{-9} = \frac{2}{9}.$
4. Therefore, the approximation for ( h ) would be rendered as ( h_2 = \frac{2}{9}. )

To find the acceleration, we first differentiate the displacement with respect to time:

1. ( x = 4 - e^{-2t} )
2. Taking the first derivative: $\frac{dx}{dt} = 2e^{-2t}$
3. The acceleration, which is the second derivative: $\frac{d^2x}{dt^2} = -4e^{-2t}.$ \nHence, the acceleration of the particle as a function of ( t ) is given by ( -4e^{-2t}. )

To evaluate this limit, we can use the substitution:

1. We know that as ( x \to 0 ), ( 1 - \cos(2\pi x) \approx \frac{(2\pi x)^2}{2} ) using the Taylor series expansion.
2. Therefore: $\lim_{x \to 0} \frac{1 - \cos 2\pi x}{x^2} = \lim_{x \to 0} \frac{2\pi^2x^2}{2x^2} = \pi^2.$
3. Thus, the limit evaluates to ( \pi^2. )