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How are exponential, logarithmic and trigonometric functions extended in QCE Math Methods Unit 2?

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

A focused answer to the QCE Math Methods Unit 2 subject-matter point on extended functions. Exponential growth and decay models, logarithmic functions, trigonometric functions (unit circle, exact values, graphs and transformations), and applications.

Generated by Claude Opus 4.89 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  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

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, a1a \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=ln2kt = \dfrac{\ln 2}{k}. Exponential decay uses y=y0ekty = y_0 e^{-kt} with k>0k > 0, and the half-life is t=ln2kt = \dfrac{\ln 2}{k}. The doubling-time and half-life formulas both come from solving 2=ekt2 = e^{kt} and 12=ekt\tfrac{1}{2} = e^{-kt} respectively.

Logarithmic functions

y=loga(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: lnx=logex\ln x = \log_e x.

Trigonometric functions

Unit circle. Point at angle θ\theta: (cosθ,sinθ)(\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. sin2θ+cos2θ=1\sin^2\theta + \cos^2\theta = 1. Even/odd: sin(θ)=sinθ\sin(-\theta) = -\sin\theta, cos(θ)=cosθ\cos(-\theta) = \cos\theta.

Transformations. y=Asin(B(xC))+Dy = A\sin(B(x - C)) + D: amplitude A|A|, period 2π/B2\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 sinπ6=12\sin\tfrac{\pi}{6} = \tfrac{1}{2}, cosπ4=12\cos\tfrac{\pi}{4} = \tfrac{1}{\sqrt 2}, tanπ3=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 sinx=k\sin x = k: principal x1=arcsin(k)x_1 = \arcsin(k); second x2=πx1x_2 = \pi - x_1.

For cosx=k\cos x = k: principal x1=arccos(k)x_1 = \arccos(k); second x2=x1x_2 = -x_1 or 2πx12\pi - x_1.

For tanx=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(xC))+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.

Exam-style practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

QCAA 20223 marksPaper 1 (technique). Solve sinx=0.5\sin x = 0.5 for x[0,2π]x \in [0, 2\pi].
Show worked answer →

Principal solution: sin1(0.5)=π6\sin^{-1}(0.5) = \dfrac{\pi}{6}. By the symmetry of sine, the second solution in the interval is ππ6=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.

QCAA 20234 marksPaper 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=24T_{\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+3323.2T(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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