Solution (a): Absolute Maximum Value

Step 1 (Formula / Governing Law & Given Data): [0.5 Mark]

Given function: $f(x) = \cos x + \sin^2 x$ for $x \in [0, \pi]$.
To find the absolute maximum, we use the first derivative test to find critical points and evaluate the function at critical points and endpoints ($x=0, x=\pi$).

Step 2 (Step-by-Step Mathematical Working): [1.0 Mark]

Differentiating with respect to $x$:

$$f'(x) = -\sin x + 2\sin x \cos x = \sin x (2\cos x - 1)$$

Setting $f'(x) = 0$ for critical points:

$$\sin x = 0 \implies x = 0, \pi$$ $$2\cos x - 1 = 0 \implies \cos x = \frac{1}{2} \implies x = \frac{\pi}{3}$$

Evaluating $f(x)$ at critical points and endpoints:

• $f(0) = \cos(0) + \sin^2(0) = 1 + 0 = 1$
• $f(\pi) = \cos(\pi) + \sin^2(\pi) = -1 + 0 = -1$
• $f\left(\frac{\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) + \sin^2\left(\frac{\pi}{3}\right) = \frac{1}{2} + \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{2} + \frac{3}{4} = \frac{5}{4}$

Step 3 (Final Concluded Answer): [0.5 Mark]

Comparing the values $\{1, -1, 1.25\}$, the absolute maximum value is $\frac{5}{4}$.

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Solution (b): Rate of Change Proof

Step 1 (Formula / Governing Law & Given Data): [0.5 Mark]

Volume of hemisphere $V = \frac{2}{3}\pi r^3$. Surface area (curved) $S = 2\pi r^2$.
Given: $\frac{dV}{dt} = k$ (constant).

Step 2 (Step-by-Step Mathematical Working): [1.0 Mark]

Differentiating $V$ with respect to $t$:

$$\frac{dV}{dt} = 2\pi r^2 \frac{dr}{dt} = k \implies \frac{dr}{dt} = \frac{k}{2\pi r^2}$$

Now, differentiate $S$ with respect to $t$:

$$\frac{dS}{dt} = \frac{d}{dt}(2\pi r^2) = 4\pi r \frac{dr}{dt}$$

Substitute $\frac{dr}{dt}$ into the equation:

$$\frac{dS}{dt} = 4\pi r \left( \frac{k}{2\pi r^2} \right) = \frac{2k}{r}$$

Step 3 (Final Concluded Answer): [0.5 Mark]

Since $2k$ is a constant, $\frac{dS}{dt} \propto \frac{1}{r}$. Thus, the