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 for when the temperature T reaches 20°C:

1. Start with the differential equation: $\frac{dT}{dt} = k(T - 25)$
2. Separate variables and integrate: $\int \frac{1}{T-25} dT = k \int dt$
3. Solve for T using the initial condition T(0) = 5.
4. Set T = 20°C and solve for t. The specific calculations will yield t approximately equal to 4.

The graph of T as a function of t is a simple curve:

1. The graph begins at T = 5°C when t = 0.
2. It approaches T = 25°C as t increases, displaying exponential behavior.
3. Sketch a curve that starts at (0, 5) and gradually moves toward T = 25°C without crossing this line.

To prove by induction:

1. Base case: For n = 1, calculate LHS = 1/(2×3) = 1/6; RHS = 1/(1(1+2)) = 1/3 (holds).
2. Assume true for n = k: LHS = 1/(2×3) + ... + 1/(k(k+1)) = 1/k(k+2).
3. For n = k+1: Add the next term to LHS; simplify to show LHS = 1/((k+1)(k+2)). Induction confirms it is true for all n.

To sketch y = f(x):

1. Identify intercepts:
   • Set f(x) = 0 to find x-intercepts.
   • Set x = 0 to find the y-intercept (f(0) = 4).
2. Draw the curve starting from y = 4, drooping downwards with concave shapes.

To find the inverse:

1. Replace f(x) with y: y = 4 - (1 - (x/2))^2.
2. Solve for x in terms of y: x = 2(4 - y)^{1/2}.
3. Domain is restricted to (-∞, 4].

The graph of y = f^(-1)(x):

1. Reflect the original graph over the line y = x.
2. Mark key points to identify the trend of the inverse function.