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Exponential, logarithmic and trigonometric functions: QCE Math Methods Unit 2 Year 11

Syllabus dot point

“Exponential, logarithmic and trigonometric functions (including their graphs and transformations), and applications to growth and decay and periodic phenomena”

9 min read

Quick answer

Unit 2 extends function families to exponential (growth and decay), logarithmic (inverse of exponential), and trigonometric (unit circle, exact values, graphs of sin⁡\sin, cos⁡\cos and tan⁡\tan with their transformations); solving trig equations uses principal solutions plus symmetry / periodicity.

Jump to a section
  1. What this dot point is asking
  2. Exponential functions
  3. Logarithmic functions
  4. Trigonometric functions
  5. The CAST diagram and exact values
  6. Solving trig equations
  7. Applications
  8. Exam-style questions
Key fact

In the 2025 syllabus, trigonometric functions are taught in Unit 1 (Topic 4), while exponential and logarithmic functions are Unit 2 topics. This overview page stays in Unit 2 so its link does not change.

What this dot point is asking

QCAA wants Year 11 students to extend their understanding of exponential, logarithmic and trigonometric functions, including their graphs and transformations, and to apply them to growth-and-decay and periodic phenomena. These function families are the modelling backbone of the course and reappear with calculus in Year 12.

Exponential functions

An exponential function y=axy = a^x (with a>0a > 0, a≠1a \neq 1) is always positive, has horizontal asymptote y=0y = 0, and passes through (0,1)(0, 1). When a>1a > 1 it increases (growth); when 0<a<10 < a < 1 it decreases (decay). Writing the base as ee gives the natural exponential y=y0ekty = y_0 e^{kt}, whose constant kk directly measures the proportional rate of change.

Exponential growth uses y=y0ekty = y_0 e^{kt} with k>0k > 0, and the doubling time is t=ln⁡2kt = \dfrac{\ln 2}{k}. Exponential decay uses y=y0e−kty = y_0 e^{-kt} with k>0k > 0, and the half-life is t=ln⁡2kt = \dfrac{\ln 2}{k}. The doubling-time and half-life formulas both come from solving 2=ekt2 = e^{kt} and 12=e−kt\tfrac{1}{2} = e^{-kt} respectively.

Logarithmic functions

y=log⁡a(x)y = \log_a(x) inverse of y=axy = a^x. Defined for x>0x > 0, vertical asymptote x=0x = 0, xx-intercept (1,0)(1, 0).

Natural log: ln⁡x=log⁡ex\ln x = \log_e x.

Trigonometric functions

Unit circle. Point at angle θ\theta: (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta).

Exact values at 0,π/6,π/4,π/3,π/2,π,3π/2,2π0, \pi/6, \pi/4, \pi/3, \pi/2, \pi, 3\pi/2, 2\pi.

Graphs.

  • sin⁡(x)\sin(x): wave, amplitude 1, period 2π2\pi.
  • cos⁡(x)\cos(x): shifted sin, yy-intercept 1.
  • tan⁡(x)\tan(x): period π\pi, asymptotes at π/2+πk\pi/2 + \pi k.

Identities. sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1. Even/odd: sin⁡(−θ)=−sin⁡θ\sin(-\theta) = -\sin\theta, cos⁡(−θ)=cos⁡θ\cos(-\theta) = \cos\theta.

Transformations. y=Asin⁡(B(x−C))+Dy = A\sin(B(x - C)) + D: amplitude ∣A∣|A|, period 2π/∣B∣2\pi/|B|, phase shift CC, vertical shift DD.

The CAST diagram and exact values

Knowing the signs of the three ratios in each quadrant (the CAST rule: all positive in the first, sine in the second, tangent in the third, cosine in the fourth) lets you place every solution without a calculator. Combined with the exact values at π6\tfrac{\pi}{6}, π4\tfrac{\pi}{4} and π3\tfrac{\pi}{3} (for example sin⁡π6=12\sin\tfrac{\pi}{6} = \tfrac{1}{2}, cos⁡π4=12\cos\tfrac{\pi}{4} = \tfrac{1}{\sqrt 2}, tan⁡π3=3\tan\tfrac{\pi}{3} = \sqrt 3), this is the foundation of calculator-free trigonometry in Paper 1. The reference angle (the acute angle to the xx-axis) gives the size of the solution, and the quadrant gives its sign and position.

Solving trig equations

Find principal solutions, then use symmetry/periodicity for all in range.

For sin⁡x=k\sin x = k: principal x1=arcsin⁡(k)x_1 = \arcsin(k); second x2=π−x1x_2 = \pi - x_1.

For cos⁡x=k\cos x = k: principal x1=arccos⁡(k)x_1 = \arccos(k); second x2=−x1x_2 = -x_1 or 2π−x12\pi - x_1.

For tan⁡x=k\tan x = k: principal x1=arctan⁡(k)x_1 = \arctan(k); add multiples of π\pi.

Applications

Exponentials model anything changing by a constant factor per period: compound interest, population growth and decay, and carbon dating (where the known half-life pins down the decay constant). Trigonometric functions model anything that repeats: tides, sound waves, daylight hours and oscillating systems. In a model y=Asin⁡(B(x−C))+Dy = A\sin(B(x - C)) + D, the amplitude AA is half the peak-to-trough range, DD is the midline (average value), the period 2πB\tfrac{2\pi}{B} is the time for one cycle, and CC is the horizontal shift, so reading these off a worded context is the core modelling skill.

Solving a cosine equation over an interval

Find the principal solution

Solve cos⁡x=−12\cos x = -\tfrac{1}{2} for x∈[0,2π]x \in [0, 2\pi]. The reference angle with cosine 12\tfrac{1}{2} is π3\dfrac{\pi}{3}.

Use the quadrants where cosine is negative

Cosine is negative in the second and third quadrants, so the solutions are x=π−π3=2π3x = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3} and x=π+π3=4π3x = \pi + \dfrac{\pi}{3} = \dfrac{4\pi}{3}.

State the solutions

x=2π3orx=4π3.x = \frac{2\pi}{3} \quad \text{or} \quad x = \frac{4\pi}{3}.

Markers reward the reference angle, the correct quadrants for a negative cosine, and both solutions in the interval.

Common errors
Calculator in degree mode
QCE Methods works in radians; set the calculator accordingly.
Missing solutions
sin⁡x=12\sin x = \tfrac{1}{2} has two solutions per period; always find all solutions in the stated interval using symmetry.
Reciprocal versus inverse
sin⁡−1\sin^{-1} denotes the inverse function (arcsine), not the reciprocal 1sin⁡\tfrac{1}{\sin}.

Exam-style questions

Questions in the style of QCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice3 marks
Paper 1 (technique). Solve sin⁡x=0.5\sin x = 0.5 for x∈[0,2π]x \in [0, 2\pi].
Show worked answer →

Principal solution: sin⁡−1(0.5)=π6\sin^{-1}(0.5) = \dfrac{\pi}{6}. By the symmetry of sine, the second solution in the interval is π−π6=5π6\pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}.

No further solutions lie in [0,2π][0, 2\pi], so x=π6x = \dfrac{\pi}{6} or x=5π6x = \dfrac{5\pi}{6}.

Markers reward both solutions and the symmetry reasoning.

Practice4 marks
Paper 2 (complex familiar). The temperature in a greenhouse is modelled by T(t)=18+6sin⁡ ⁣(πt12)T(t) = 18 + 6\sin\!\left(\dfrac{\pi t}{12}\right), where tt is hours after midnight. (a) State the maximum temperature and the time it first occurs. (b) Determine the temperature at t=4t = 4.
Show worked answer →

(a) The maximum of sin⁡\sin is 11, so Tmax⁡=18+6=24∘T_{\max} = 18 + 6 = 24^\circC, occurring when πt12=π2\dfrac{\pi t}{12} = \dfrac{\pi}{2}, that is t=6t = 6 (6 am).

(b) T(4)=18+6sin⁡ ⁣(π3)=18+6×32=18+33≈23.2∘T(4) = 18 + 6\sin\!\left(\dfrac{\pi}{3}\right) = 18 + 6 \times \dfrac{\sqrt 3}{2} = 18 + 3\sqrt 3 \approx 23.2^\circC.

Markers reward the amplitude-plus-midline maximum, the time from the sine argument, and the exact-value evaluation.

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Sources & how we know this

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