SACE Stage 2 Mathematical Methods deep dive: all six topics for the 2026 exam

Differentiate $f(x) = x^2\sin(3x)$.

Product rule with $u = x^2$ and $v = \sin(3x)$, where the chain rule gives $v' = 3\cos(3x)$:

A closed cylindrical can must hold $500\text{ cm}^3$. Find the radius that minimises its surface area.

Volume: $\pi r^2h = 500$, so $h = \dfrac{500}{\pi r^2}$. Surface area: $A = 2\pi r^2 + 2\pi rh = 2\pi r^2 + \dfrac{1000}{r}$.

$\dfrac{dA}{dr} = 4\pi r - \dfrac{1000}{r^2} = 0$ gives $r^3 = \dfrac{250}{\pi}$, so $r \approx 4.30$ cm.

$\dfrac{d^2A}{dr^2} = 4\pi + \dfrac{2000}{r^3} > 0$, confirming a minimum. At this radius $h \approx 8.60$ cm, twice the radius: the most efficient can is as tall as it is wide.

A multiple-choice quiz has 8 questions, each with 4 options. A student guesses every answer. Let $X$ be the number correct, so $X \sim \text{Bin}(8, 0.25)$.

$P(X = 2) = \binom{8}{2}(0.25)^2(0.75)^6 \approx 0.3115$.

$\mu = 8(0.25) = 2$ and $\sigma = \sqrt{8(0.25)(0.75)} \approx 1.22$. So a score of 6 or more (more than three standard deviations above the mean) would be very unusual for a guesser.

Find the area between $y = x^2 - 4x$ and the $x$-axis for $0 \le x \le 5$.

The curve crosses the axis at $x = 0$ and $x = 4$; it is below the axis on $0 < x < 4$.

$\displaystyle\int_0^4 (x^2 - 4x)\,dx = \left[\frac{x^3}{3} - 2x^2\right]_0^4 = \frac{64}{3} - 32 = -\frac{32}{3}$

$\displaystyle\int_4^5 (x^2 - 4x)\,dx = \left(\frac{125}{3} - 50\right) - \left(-\frac{32}{3}\right) = \frac{7}{3}$

Area $= \dfrac{32}{3} + \dfrac{7}{3} = 13$ square units. Integrating straight from 0 to 5 would give $-\dfrac{25}{3}$, the signed area.

A bacterial culture grows according to $N = 200e^{0.35t}$, with $t$ in hours. When does it reach 5000?

$e^{0.35t} = 25$, so $t = \dfrac{\ln 25}{0.35} \approx 9.20$ hours.

The rate of growth is $\dfrac{dN}{dt} = 70e^{0.35t} = 0.35N$, so at the moment $N = 5000$ the culture is growing at 1750 bacteria per hour.

Ella scored 78 in a Chemistry test (mean 70, standard deviation 5) and 82 in a Biology test (mean 76, standard deviation 8). In which test did she perform better relative to the cohort?

Chemistry: $z = \dfrac{78 - 70}{5} = 1.6$. Biology: $z = \dfrac{82 - 76}{8} = 0.75$.

Her Chemistry result is further above the mean in standard deviation terms, so it is the relatively better performance, even though the raw mark is lower.

Checkout times at a supermarket have mean 50 seconds and standard deviation 12 seconds. For random samples of 36 customers, $\text{SD}(\bar{X}) = \dfrac{12}{\sqrt{36}} = 2$ seconds, so $P(\bar{X} > 53) = P(Z > 1.5) \approx 0.0668$.

To estimate the mean checkout time to within 2 seconds with 95 percent confidence (taking $\sigma = 10$ from a pilot study): $1.96 \times \dfrac{10}{\sqrt{n}} \le 2$ gives $n \ge 96.04$, so at least 97 customers are needed.