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, we start with ( y = |x + 1| ):

• It is a V-shaped graph with vertex at (-1, 0).
• It intersects the x-axis at ( x = -1 ).

Next, for ( y = 3 - |x - 2| ):

• It also has a vertex at (2, 1).
• Intercepts can be found by setting ( y = 0 ):

$$3 - |x - 2| = 0 \Rightarrow |x - 2| = 3$$

Which gives us ( x = 5 ) and ( x = -1 ).

The correct sketch shows the vertices and intercepts clearly.

We need to find intersections where the sum of the two functions equals three.

Calculating the points where:

1. ( x < -1: |x + 1| = -x - 1; |x - 2| = -x + 2 ) gives: $-x - 1 - x + 2 = 3 \Rightarrow -2x + 1 = 3 \Rightarrow -2x = 2 \Rightarrow x = -1.$
2. ( -1 \leq x < 2: |x + 1| = x + 1; |x - 2| = -x + 2 ) gives: $x + 1 - x + 2 = 3 \Rightarrow 3 = 3.$
3. For ( x \geq 2: |x + 1| = x + 1; |x - 2| = x - 2 ) gives: $x + 1 + x - 2 = 3 \Rightarrow 2x - 1 = 3 \Rightarrow 2x = 4 \Rightarrow x = 2.$

Thus, the range of values is ( x \in [-1, 2]. )

To find ( h ) such that the volumes of the two solids are in the ratio 2:1, we express the volumes of each solid as:

• Volume of solid 1: $V_1 = \pi \int_{-1}^{h} (1 - x^2) \, dx \Rightarrow V_1 = \pi \left[ x - \frac{x^3}{3} \right]_{-1}^{h}.$

• Volume of solid 2: $V_2 = \pi \int_{h}^{1} (1 - x^2) \, dx \Rightarrow V_2 = \pi \left[ x - \frac{x^3}{3} \right]_{h}^{1}.$

Setting up the ratio of the volumes leads to: $\frac{V_1}{V_2} = \frac{2}{1}.$

This relationship simplifies to derive the polynomial equation ( 3h^3 - 9h + 2 = 0. )

Using Newton's method, we start with the first approximation:\n( h_1 = 0 ). Calculating:

• ( f(h) = 3h^3 - 9h + 2 )
• ( f'(h) = 9h^2 - 9 ) ( f(0) = 2, f'(0) = -9. )

Applying Newton's formula: $h_2 = h_1 - \frac{f(h_1)}{f'(h_1)} \Rightarrow h_2 = 0 - \frac{2}{-9} = \frac{2}{9}.$

Therefore, the second approximation for ( h ) is ( h = \frac{2}{9}. )

To find the acceleration of the particle, we start with the displacement function given as: $t = 4 - e^{-2x}.$

Differentiating with respect to time, $\frac{dx}{dt} = 2e^{-2x}.$

To find acceleration: $a = \frac{d^2x}{dt^2} = \frac{d}{dt}(\frac{dx}{dt}) = \frac{d}{dt}(2e^{-2x}).$ Using the chain rule: $a = \frac{d(2e^{-2x})}{dx} \cdot \frac{dx}{dt} = -4e^{-2x} \cdot 2e^{-2x}.$

Thus the acceleration function simplifies to: $a(x) = -8e^{-4x}.$

Using L'Hôpital's rule, since this is an indeterminate form (0/0): Taking derivatives: $\frac{d}{dx}[1 - \cos 2\pi x] = 2\pi \sin 2\pi x \text{ and } \frac{d}{dx}[x^2] = 2x.$ Thus: $\lim_{x \to 0} \frac{2\pi \sin 2\pi x}{2x} = \pi \lim_{x \to 0} \frac{\sin 2\pi x}{x} = \pi \times 2\pi = 2\pi^2.$

Therefore, the final result is ( 2\pi^2. )