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 graph of (y = |x + 1|):

• The vertex is at (-1, 0) and opens upwards.

To sketch the graph of (y = 3 - |x - 2|):

• The vertex is at (2, 1) and opens downwards.

The intercepts can be found at:

1. ( x = -1 ) for the first graph and ( x = 5 ) for the second graph.

• Both graphs intersect at the points (-1, 0) and (5, -1).
• Ensure both graphs are sketched accurately on the same axes, marking the intercepts clearly.

To solve the equation (|x + 1| + |x - 2| = 3):

1. Identify critical points from the expressions: (x = -1) and (x = 2).
2. Evaluate the intervals determined by these critical points: (-∞, -1), [-1, 2], and (2, ∞).
3. For (x < -1): Both expressions are negative, leading to: [-(x+1) - (x-2) = 3 \Rightarrow -2x + 1 = 3 \Rightarrow x = -1] (not valid as it lies outside).
4. For (-1 \leq x < 2): ( (x + 1) - (x - 2) = 3 \Rightarrow 3 = 3) (holds true for all values in this interval).
5. For (x \geq 2): ( (x + 1) + (x - 2) = 3 \Rightarrow 2x - 1 = 3 \Rightarrow x = 2) (valid).

Thus, the range of values for which the equation holds is ([-1, 2]).

To show that (h) satisfies the equation, we can determine the volumes of the solids formed by rotating the area about the x-axis:

1. The volume of the semicircle is given by:
   $V_s = \int_{-1}^{1} \sqrt{1 - x^2} dx = \pi/2.$
2. The volume of the solid formed from the line x = h is calculated next: $V_h = \pi \int_{-h}^{h} (h)^{2} dx = \pi h^{2} (1).$
3. Setting the ratio to 2:1, we will find that: $V_h = \frac{2}{3} V_s,$ leading to: rac{h^3 \pi}{3} = \frac{2\pi}{3} \Rightarrow 3h^3 – 9h + 2 = 0.
4. Therefore, (h) indeed satisfies the equation.

Using Newton's method for approximating roots:

1. We have our function ( f(h) = 3h^3 - 9h + 2 ) and its derivative ( f'(h) = 9h^2 - 9 ).
2. At the point ( h_0 = 0):
   • Calculate ( f(0) = 2 ) and ( f'(0) = -9 ).
3. Using Newton's method: $h_1 = h_0 - \frac{f(h_0)}{f'(h_0)} = 0 - \frac{2}{-9} = \frac{2}{9}.$
4. Therefore, the second approximation for h is ( h_1 = \frac{2}{9} ).

Given the displacement equation ( t = 4 - e^{-2x}):

1. Differentiate it to find the velocity:\n $\frac{dx}{dt} = 2e^{-2x}.$
2. Differentiate again to find acceleration: $\frac{d^2x}{dt^2} = -4e^{-2x} \frac{dx}{dt}.$
3. Substitute velocity into this equation: $\frac{d^2x}{dt^2} = -4e^{-2x} (2e^{-2x}) = -8e^{-4x}.$
4. Finally, the acceleration as a function of t is ( -8e^{-4x} ).

We can evaluate this limit using L'Hôpital's rule:

1. Start with the limit: $L = \lim_{x \to 0} \frac{1 - \cos 2\pi x}{x^2}.$
2. Applying L'Hôpital's rule, differentiate the numerator and denominator: ( \frac{d}{dx}(1 - \cos(2\pi x)) = 2\pi \sin(2\pi x)) and ( \frac{d}{dx}(x^2) = 2x.)
3. This gives: $L = \lim_{x \to 0} \frac{2\pi \sin(2\pi x)}{2x} = \lim_{x \to 0} \frac{\pi \sin(2\pi x)}{x}.$ \n4. Evaluating it yields ( L = \pi (2\pi) = 2\pi^2. )
4. Therefore, ( \lim_{x \to 0} \frac{1 - \cos 2\pi x}{x^2} = 2\pi^2.)