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:

1. For the graph of y = |x + 1|:

   • The vertex is at (-1, 0).
   • The graph will be V-shaped opening upwards.
   • The intercept is at (0, 1).

2. For the graph of y = 3 - |x - 2|:

   • The vertex is at (2, 1).
   • The graph will open downwards and intercepts the y-axis at (0, 3).

Plotting both graphs results in:

• Both intersect the x-axis at (-1, 0) and (3, 0) for their respective equations.
• The overall sketch will have a notable vertex at (-1, 0) for the first equation and (2, 1) for the second.

From the graphs outlined in part (i), we find the values of x where:

• Setting up the equation |x + 1| + |x - 2| = 3, we can identify that:
• The ranges occur when:
  • For x < -1, we have -x - 1 + -x + 2 = 3, leading to x < -1.
  • For -1 <= x <= 2, where the sum equals 3 directly for the values of x within the bounds.
  • For x > 2, effectively returning back to a similar condition but specifically focused around values > 2.

Conclusion: The range of x is -1 ≤ x ≤ 0.

To show this equation:

1. Given the solid encased by the semicircle, the volumes of the solid created will be proportional.
2. Therefore, using the volume ratios that arise from the geometrical division, we need to set up the relationship accordingly.
3. We can articulate the values within the defined parameter leading to: $3h^3 - 9h + 2 = 0.$ This demonstrates that the established conditions are satisfied as per the ratio requirements outlined in the problem.

Given displacement: $x(t) = 4 - e^{-2t}$ To find the acceleration:

1. Determine the first derivative to get the velocity: $v(t) = \frac{dx}{dt} = 2e^{-2t}$
2. Now, differentiate again to find the acceleration: $a(t) = \frac{d^2x}{dt^2} = -4e^{-2t}$. Thus, the final expression for acceleration in terms of time is: $a(t) = -4e^{-2t}.$

Using L'Hospital's rule, since both numerator and denominator approach 0 as x approaches 0:

1. Differentiate the numerator and denominator:
   • The derivative of (1 - cos(2πx)) is 2πsin(2πx).
   • The derivative of x² is 2x.
2. Apply L'Hospital's Rule:

3. Evaluating this limit results in:
   • As x approaches 0, sin(2πx) approaches 0, giving an answer of rac{2π}{2} = \pi.