TCE Mathematics Methods Level 4 deep dive: criteria 4 to 8 for the 2026 exam

Sketch $y = (x - 1)^2(x + 2)$.

$x$-intercepts: $x = 1$ (repeated, so the graph touches the axis and turns) and $x = -2$ (crosses). $y$-intercept: $(0 - 1)^2(0 + 2) = 2$. The leading term is $x^3$, so $y \to \infty$ as $x \to \infty$ and $y \to -\infty$ as $x \to -\infty$. The turning point at $(1, 0)$ is a local minimum, and there is a local maximum between $x = -2$ and $x = 1$ (at $x = -1$, where $y = 4$).

For $y = 2\sin\left(2\left(x - \dfrac{\pi}{4}\right)\right) + 1$: amplitude 2, period $\pi$, phase shift $\dfrac{\pi}{4}$ to the right, vertical shift 1, range $[-1, 3]$.

Solve $\cos(2x) = -\dfrac{1}{2}$ for $0 \le x \le \pi$. Let $\theta = 2x$ with $0 \le \theta \le 2\pi$. Cosine is negative in quadrants 2 and 3, with reference angle $\dfrac{\pi}{3}$: $\theta = \dfrac{2\pi}{3}$ or $\dfrac{4\pi}{3}$. So $x = \dfrac{\pi}{3}$ or $\dfrac{2\pi}{3}$.

Changing the domain for $\theta$ first is the step that stops you missing solutions.

Find the maximum and minimum values of $f(x) = x^3 - 3x$ on $-2 \le x \le 3$.

$f'(x) = 3x^2 - 3 = 0$ at $x = \pm 1$. $f(-1) = 2$ (local maximum) and $f(1) = -2$ (local minimum). End points: $f(-2) = -2$ and $f(3) = 18$.

The maximum value is 18, at the end point $x = 3$, not at the local maximum. The minimum value is $-2$, reached at both $x = -2$ and $x = 1$. Always check the end points on a closed interval.

Estimate $\displaystyle\int_0^2 e^x\,dx$ with 4 strips of width 0.5.

Heights: $e^0 = 1$, $e^{0.5} \approx 1.6487$, $e^1 \approx 2.7183$, $e^{1.5} \approx 4.4817$, $e^2 \approx 7.3891$.

Estimate $= \dfrac{0.5}{2}\left[1 + 2(1.6487 + 2.7183 + 4.4817) + 7.3891\right] \approx 6.521$.

Exact value: $e^2 - 1 \approx 6.389$. The trapezoidal rule overestimates here because $e^x$ is concave up, so each trapezium sits above the curve.

A machine produces faulty items with probability 0.04. In a batch of 50, let $X \sim \text{Bin}(50, 0.04)$.

$P(X = 0) = 0.96^{50} \approx 0.130$, so $P(X \ge 1) \approx 0.870$. $E(X) = 2$ and $\sigma = \sqrt{50(0.04)(0.96)} \approx 1.39$.

Separately, item masses are normally distributed with mean 250 g and standard deviation 5 g. $P(\text{mass} < 240) = P(Z < -2) \approx 0.023$: about 2.3 percent of items are under 240 g.