a) The direction field for a differential equation is given on page 1 of the Question 12 Writing Booklet - HSC - SSCE Mathematics Extension 1 - Question 12 - 2021 - Paper 1

To solve the differential equation $\frac{dT}{dt} = k(T - 25)$:

1. Rearranging gives: $\frac{dT}{T - 25} = k \, dt$.
2. Integrate both sides: [ \int \frac{dT}{T - 25} = \int k , dt ] which gives [ \ln|T - 25| = kt + C ].
3. Exponentiating yields: [ |T - 25| = e^{kt+C} ].
4. Since the initial condition T(0) = 5°C: [ |5 - 25| = e^C \Rightarrow C = \ln 20 ].
5. Thus, the equation becomes: [ T - 25 = 20e^{kt} \Rightarrow T = 25 - 20e^{-kt} ].
6. At t = 8, T = 10°C: [ 10 = 25 - 20e^{-8k} \Rightarrow 20e^{-8k} = 15 \Rightarrow e^{-8k} = \frac{3}{4}.]
7. Solve for k: [ -8k = \ln \frac{3}{4} \Rightarrow k = -\frac{\ln \frac{3}{4}}{8}.]
8. Now, substitute k back in to find t when T = 20°C: [ 20 = 25 - 20e^{-kt} \Rightarrow 20e^{-kt} = 5 \Rightarrow e^{-kt} = \frac{1}{4}.]
9. Therefore, solve for t: [ -kt = \ln \frac{1}{4} \Rightarrow t = -\frac{8 \ln \frac{1}{4}}{\ln \frac{3}{4}}.]
10. This yields t ≈ 9.04 minutes, rounded to the nearest minute gives t = 9 minutes.

The graph of T as a function of time t should be sketched as follows:

1. The graph starts at T = 5°C when t = 0.
2. It should show an increasing curve that approaches 25°C (the room temperature).
3. Indicate the point (8, 10) on your graph where T = 10°C.
4. Mark the point (9, 20) where the temperature reaches 20°C.
5. The behavior of the graph should be asymptotic to 25°C, indicating that it approaches but never quite reaches this temperature.

Assume the statement is true for n = k: $S_k = \frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \cdots + \frac{1}{k(k + 1)(k + 2)} = \frac{1}{2(k)(k + 2)}.$ To prove this for n = k + 1:

1. Adding the next term: $S_{k + 1} = S_k + \frac{1}{(k + 1)(k + 2)(k + 3)}.$
2. Substitute $S_k$ into the equation: $S_{k + 1} = \frac{1}{2(k)(k + 2)} + \frac{1}{(k + 1)(k + 2)(k + 3)}.$
3. Find a common denominator, which is $2(k)(k + 2)(k + 1)(k + 3)$.
4. Simplifying shows: $S_{k + 1} = \frac{k + 3}{2(k + 1)(k + 3)} = \frac{1}{2(k + 1)(k + 3)}.$
5. Therefore, the hypothesis holds for k + 1, completing the induction proof.

To sketch the graph of $f(x) = 4 - \left(1 - \frac{x}{2}\right)^{2}$:

1. Identify intercepts:
   • y-intercept: when x = 0, $f(0) = 4 - (1 - 0)^{2} = 3$.
   • x-intercept: set f(x) = 0: $4 - \left(1 - \frac{x}{2}\right)^{2} = 0 \Rightarrow (1 - \frac{x}{2})^{2} = 4 \Rightarrow 1 - \frac{x}{2} = \pm 2$ gives x = -4 or 6.
2. The graph is a downward-opening parabola, peaking at (2, 4).
3. Sketch the curve through the intercepts and vertex, ensuring it is symmetric about the line x = 2.

1. Start with $y = 4 - \left(1 - \frac{x}{2}\right)^{2}$.
2. Rearrange to find x: $y = 4 - z^{2}$ where $z = 1 - \frac{x}{2}$.
3. Therefore, $z^{2} = 4 - y \Rightarrow z = \sqrt{4 - y} \text{ (only using positive since f(x) is decreasing)}$.
4. Substitute back to find x: $1 - \frac{x}{2} = \sqrt{4 - y} \Rightarrow \frac{x}{2} = 1 - \sqrt{4 - y} \Rightarrow x = 2(1 - \sqrt{4 - y}).$
5. Thus, the inverse function is given by: $f^{-1}(y) = 2 - 2\sqrt{4 - y}.$
6. The domain of the inverse is the range of the original function, which can be noted as y ∈ [-∞, 4].

To sketch the graph of the inverse function:

1. Start with the inverse equation $y = 2 - 2\sqrt{4 - x}$.
2. Identify the x-intercept (when y = 0): $0 = 2 - 2\sqrt{4 - x} \Rightarrow \sqrt{4 - x} = 1 \Rightarrow x = 3.$
3. Establish points for the graph:
   • For x = 0, $y = 2 - 2\sqrt{4} = -2.$
4. As x approaches 4, y approaches 0, confirming an intercept at (3, 0).
5. The graph is decreasing, symmetric to the line y = x, confirming that the original function and its inverse mirror each other across this line.