TCE Specialist Mathematics exam 2026Exam: Mon 9 Nov · TASC timetable

Memorise these two readings. A single modulus equal to a constant gives a circle. Two moduli set equal to each other gives a straight line, the perpendicular bisector of the two centres. Spotting which one you have is the whole battle.

When $f(x) = (mx + c) + \dfrac{r}{x - a}$ with $r > 0$, solving $f'(x) = 0$ gives $x = a \pm \sqrt{r}$. The two critical points are equidistant from the vertical asymptote, one a local maximum and one a local minimum, and the curve hugs the oblique asymptote far from $x = a$.

Multiply the moduli and add the arguments; divide the moduli and subtract the arguments.

At a simple zero of $f$ the reciprocal's two branches go to opposite infinities ($+\infty$ on the side where $f > 0$, $-\infty$ where $f < 0$). At a double zero of $f$, where $f$ keeps the same sign, both branches of the reciprocal go to the same infinity.

Multiply through by the integrating factor $\mu(x) = e^{\int P(x)\, dx}$. The left side then becomes $\tfrac{d}{dx}\big(\mu y\big)$, so

For a sufficiently large sample size, the distribution of the sample mean $\bar{X}$ is approximately normal,

regardless of the shape of the original population distribution.