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

Starting with the differential equation: $\frac{dT}{dt} = k(T - 25)$ Substituting for T = 10°C at t = 8 minutes: $\int_{5}^{10} \frac{1}{T - 25} dT = k \int_{0}^{t} dt$ This gives us the relationship to find the time when T = 20°C. Solving the equation for t: $t = \frac{1}{k} \ln\left(\frac{20 - 25}{5 - 25}\right)$ Calculate the value and round it to the nearest minute.

The graph should illustrate how the water temperature T approaches the room temperature of 25°C over time. Start at T = 5°C and show an increasing curve that levels off as it nears T = 25°C. Mark the points where T equals 10°C and 20°C to indicate rapid growth before the curve stabilizes.

Base case (n=1): $\frac{1}{1 \cdot 2 \cdot 3} = \frac{1}{6}$ Right side: $\frac{1}{2(2)(3)} = \frac{1}{12},$ which holds. Assume for n=k, prove for n=k+1: $1/(1 \cdot 2 \cdot 3) + ... + 1/[k(k + 1)(k + 2)] + 1/[k+1)(k + 2)(k + 3)] = 1/[2(k + 2)(k + 3)]$ With algebra, combine fractions and simplify to prove the statement holds.

The function is defined as: f(x) = 4 - (1 - x^2)^2. Identify the intercepts: For x-intercepts, set f(x) = 0 and solve for x. For y-intercept, substitute x = 0. Graph the function within the domain (-∞, 2), ensuring to mark the intercepts clearly and show the general shape of the graph, which should display symmetry around the y-axis.

To find the inverse, switch x and y: $x = 4 - (1 - y^2)^2$ Solve for y in terms of x: Rearranging, find y in the appropriate ranges. State the domain of f^{-1}(x) based on the original function's range.

Plot the inverse function using the points from the original function's graph. Ensure proper reflection over the line y = x, noting where the function is strictly increasing or decreasing, and mark any critical points found during the inverse calculation.