QCE Mathematical Methods

QCAA Maths Methods formula sheet: what is and is not on it (QCE)

Quick answer

The QCAA Mathematical Methods formula book that every QCE Maths Methods student is given is three printed pages. It prints mensuration, graph equations, logarithm laws, calculus rules, four trigonometry items and the statistics and probability rules, and it is provided in both external papers (and by your teacher in IA2 and IA3). It does not print the quadratic formula, the index laws, tan, radian or exact-value trigonometry or anything on the normal distribution, so those come from memory, and Paper 1 allows nothing else while Paper 2 adds an approved graphics and/or scientific calculator with no CAS.

The formulas below are reproduced from the June 2026 edition of the book (2025 course), with the meaning of each and a worked example added by AusGrader. Modified from the original: © State of Queensland (QCAA) 2025, licensed under CC BY 4.0, re-typeset here with explanation added. The original is the Mathematical Methods 2025 formula book on the QCAA website, and the copy in your exam is the one that counts. Check it against this page before you rely on a rule.

This guide describes the Mathematical Methods 2025 v1.2 syllabus, which Year 12 students sit under from 2026. The 2019 syllabus called the document it provided the "QCAA formula sheet", so older advice about the sheet describes the superseded course. The QCAA Maths Methods subject page holds the past papers, and the external exam guide covers the format and marking.

Do you get a formula sheet in the Methods exam?

Yes, in four assessments. The syllabus conditions say:

AssessmentWho provides the formula bookOther resources
External, Paper 1The QCAANone: no notes, calculators, technology or other resources
External, Paper 2The QCAAA handheld QCAA-approved graphics calculator (no CAS functionality) and/or a handheld QCAA-approved scientific calculator. No notes or other resources
IA2 (Unit 3 examination)Your teacherStudents are required to use technology; no notes
IA3 (Unit 4 examination)Your teacherStudents are required to use technology; no notes

The conditions we hold for IA1, the problem-solving and modelling task, do not mention the book. The two internal papers are covered in the IA2 and IA3 guide. The book itself says nothing about where it is issued or what calculators are allowed, so those rules come from the syllabus alone.

What is on the QCAA Maths Methods formula sheet?

The book prints formulas and no explanations. It has no key for its symbols: it does not say that ss is a slant height, that AA in a prism is the base area, or what zz, p^\hat{p} and nn stand for in the confidence interval. The "What it does" column below uses the usual meanings, and those words are ours, not QCAA's. The worked examples are also ours.

Mensuration (page 1)

FormulaWhat it doesWorked example
C=2πrC = 2\pi rCircumference of a circler=7r = 7 gives C=14πC = 14\pi
A=πr2A = \pi r^2Area of a circler=3r = 3 gives A=9πA = 9\pi
A=bhA = bhArea of a parallelogram (base times perpendicular height)b=8b = 8, h=5h = 5 gives A=40A = 40
A=12(a+b)hA = \frac{1}{2}(a + b)hArea of a trapezium (aa and bb are the parallel sides)a=6a = 6, b=10b = 10, h=4h = 4 gives 12(16)(4)=32\frac{1}{2}(16)(4) = 32
A=12bhA = \frac{1}{2}bhArea of a triangleb=10b = 10, h=6h = 6 gives A=30A = 30
S=πrs+πr2S = \pi r s + \pi r^2Total surface area of a cone (ss is the slant height)r=3r = 3, s=5s = 5 gives 15π+9π=24π15\pi + 9\pi = 24\pi
S=2πrh+2πr2S = 2\pi r h + 2\pi r^2Total surface area of a cylinderr=2r = 2, h=5h = 5 gives 20π+8π=28π20\pi + 8\pi = 28\pi
S=4πr2S = 4\pi r^2Surface area of a spherer=3r = 3 gives S=36πS = 36\pi
V=13πr2hV = \frac{1}{3}\pi r^2 hVolume of a coner=3r = 3, h=4h = 4 gives V=12πV = 12\pi (the same cone has s=5s = 5)
V=πr2hV = \pi r^2 hVolume of a cylinderr=2r = 2, h=5h = 5 gives V=20πV = 20\pi
V=AhV = AhVolume of a prism (AA is the area of the base)A=12A = 12, h=5h = 5 gives V=60V = 60
V=13AhV = \frac{1}{3}AhVolume of a pyramidA 6 by 6 square base (A=36A = 36) with h=5h = 5 gives V=60V = 60
V=43πr3V = \frac{4}{3}\pi r^3Volume of a spherer=3r = 3 gives V=36πV = 36\pi

Graph equations (page 1)

These are the standard forms. The syllabus asks you to recognise how the parameters change each graph, so the useful skill is reading features straight off the equation.

EquationWhat the parameters doWorked example
y=a(x−h)2+ky = a(x - h)^2 + kQuadratic in turning point form: turning point (h,k)(h, k)y=2(x−3)2+1y = 2(x - 3)^2 + 1 has turning point (3,1)(3, 1), opens up, and cuts the yy-axis at 2(9)+1=192(9) + 1 = 19
y=a(x−x1)(x−x2)y = a(x - x_1)(x - x_2)Quadratic in factorised form: xx-intercepts x1x_1 and x2x_2y=2(x−1)(x−5)y = 2(x - 1)(x - 5) cuts the xx-axis at 1 and 5. The turning point is at x=3x = 3, where y=2(2)(−2)=−8y = 2(2)(-2) = -8
y=a(x−h)3+ky = a(x - h)^3 + kCubic with a stationary point of inflection at (h,k)(h, k)y=−(x−2)3+4y = -(x - 2)^3 + 4 has its point of inflection at (2,4)(2, 4) and cuts the yy-axis at −(−8)+4=12-(-8) + 4 = 12
y=a(x−x1)(x−x2)(x−x3)y = a(x - x_1)(x - x_2)(x - x_3)Cubic in factorised form: xx-intercepts x1x_1, x2x_2, x3x_3y=(x+2)(x−1)(x−3)y = (x + 2)(x - 1)(x - 3) cuts the xx-axis at −2-2, 1 and 3, and the yy-axis at (2)(−1)(−3)=6(2)(-1)(-3) = 6
(x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2Circle with centre (h,k)(h, k) and radius rr(x−1)2+(y+2)2=9(x - 1)^2 + (y + 2)^2 = 9 has centre (1,−2)(1, -2) and radius 3
y=ax−h+ky = a\sqrt{x - h} + kSquare root graph starting at (h,k)(h, k)y=2x−1+3y = 2\sqrt{x - 1} + 3 starts at (1,3)(1, 3), so the domain is x≥1x \ge 1. At x=5x = 5, y=2(2)+3=7y = 2(2) + 3 = 7
y=ax−h+ky = \frac{a}{x - h} + kReciprocal graph with vertical asymptote x=hx = h and horizontal asymptote y=ky = ky=2x−3+1y = \frac{2}{x - 3} + 1 has asymptotes x=3x = 3 and y=1y = 1. At x=4x = 4, y=3y = 3
y=r(x−h)+k(r>0)y = r^{(x-h)} + k \quad (r > 0)Exponential graph with horizontal asymptote y=ky = ky=2(x−1)+3y = 2^{(x-1)} + 3 has asymptote y=3y = 3 and cuts the yy-axis at 2−1+3=3.52^{-1} + 3 = 3.5
y=log⁡a(x−h)+k(a>1)y = \log_a(x - h) + k \quad (a > 1)Logarithmic graph with vertical asymptote x=hx = hy=log⁡2(x−1)+3y = \log_2(x - 1) + 3 has asymptote x=1x = 1. At x=3x = 3, y=log⁡2(2)+3=4y = \log_2(2) + 3 = 4
y=asin⁡(b(x−h))+ky = a\sin\left(b(x - h)\right) + kSine graph: amplitude ∣a∣\lvert a \rvert, horizontal shift hh, midline y=ky = ky=3sin⁡(2(x−π4))+1y = 3\sin\left(2(x - \frac{\pi}{4})\right) + 1 has amplitude 3, midline y=1y = 1, maximum 4 and minimum −2-2
y=acos⁡(b(x−h))+ky = a\cos\left(b(x - h)\right) + kCosine graph with the same parametersy=5cos⁡(2x)−1y = 5\cos(2x) - 1 has maximum 4, minimum −6-6 and, because b=2b = 2, a period of π\pi

The book prints the equations but not the period rule. That the period is 2πb\frac{2\pi}{b} (for b>0b > 0) is something you need to know.

Logarithms (page 1)

FormulaWhat it doesWorked example
ax=b⇔x=log⁡a(b)a^x = b \Leftrightarrow x = \log_a(b)Converts between index and logarithm form53=1255^3 = 125 means log⁡5(125)=3\log_5(125) = 3
log⁡a(x)+log⁡a(y)=log⁡a(xy)\log_a(x) + \log_a(y) = \log_a(xy)Product lawlog⁡2(4)+log⁡2(8)=log⁡2(32)=5\log_2(4) + \log_2(8) = \log_2(32) = 5
log⁡a(x)−log⁡a(y)=log⁡a(xy)\log_a(x) - \log_a(y) = \log_a\left(\frac{x}{y}\right)Quotient lawlog⁡5(50)−log⁡5(2)=log⁡5(25)=2\log_5(50) - \log_5(2) = \log_5(25) = 2
log⁡a(xn)=nlog⁡a(x)\log_a(x^n) = n\log_a(x)Power lawlog⁡2(82)=2log⁡2(8)=6\log_2(8^2) = 2\log_2(8) = 6
log⁡a(x)=log⁡b(x)log⁡b(a)\log_a(x) = \frac{\log_b(x)}{\log_b(a)}Change of baselog⁡4(8)=log⁡2(8)log⁡2(4)=32\log_4(8) = \frac{\log_2(8)}{\log_2(4)} = \frac{3}{2}
log⁡a(1)=0\log_a(1) = 0Log of 1 is 0log⁡7(1)=0\log_7(1) = 0
log⁡a(a)=1\log_a(a) = 1Log of the base is 1log⁡7(7)=1\log_7(7) = 1

Put together, the laws solve equations by hand. To solve log⁡2(x)+log⁡2(x−2)=3\log_2(x) + \log_2(x - 2) = 3, combine the left side into log⁡2(x(x−2))=3\log_2\left(x(x - 2)\right) = 3, so x2−2x=8x^2 - 2x = 8 and (x−4)(x+2)=0(x - 4)(x + 2) = 0. Reject x=−2x = -2 because log⁡2(−2)\log_2(-2) is undefined, which leaves x=4x = 4.

Calculus (page 2)

FormulaWhat it doesWorked example
f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}Derivative from first principlesf(x)=x2f(x) = x^2: (x+h)2−x2h=2xh+h2h=2x+h\frac{(x+h)^2 - x^2}{h} = \frac{2xh + h^2}{h} = 2x + h, which tends to 2x2x
∫abf(x) dx≈limit of sums ∑if(xi) δxi\int_a^b f(x)\,dx \approx \text{limit of sums } \sum_i f(x_i)\,\delta x_iThe definite integral as a limit of sums of strip areasFour right-endpoint strips of width 0.5 under y=x2y = x^2 from 0 to 2: 0.5(0.25+1+2.25+4)=3.750.5(0.25 + 1 + 2.25 + 4) = 3.75. More strips would close in on the exact area
∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a)Fundamental theorem of calculus, where FF is an antiderivative of ff∫132x dx=[x2]13=9−1=8\int_1^3 2x\,dx = \left[x^2\right]_1^3 = 9 - 1 = 8
ddx(f(x)+g(x))=ddxf(x)+ddxg(x)\frac{d}{dx}\left(f(x) + g(x)\right) = \frac{d}{dx}f(x) + \frac{d}{dx}g(x)Differentiate a sum term by termddx(x3+sin⁡(x))=3x2+cos⁡(x)\frac{d}{dx}(x^3 + \sin(x)) = 3x^2 + \cos(x)
∫(f(x)+g(x))dx=∫f(x) dx+∫g(x) dx\int \left(f(x) + g(x)\right) dx = \int f(x)\,dx + \int g(x)\,dxIntegrate a sum term by term∫(x2+ex) dx=x33+ex+c\int (x^2 + e^x)\,dx = \frac{x^3}{3} + e^x + c
∫k f(x) dx=k∫f(x) dx\int k\,f(x)\,dx = k\int f(x)\,dxA constant factor comes outside the integral∫6x2 dx=6⋅x33=2x3+c\int 6x^2\,dx = 6 \cdot \frac{x^3}{3} = 2x^3 + c
ddxxn=nxn−1\frac{d}{dx}x^n = nx^{n-1}Power rule for derivativesddxx−2=−2x−3\frac{d}{dx}x^{-2} = -2x^{-3}
∫xn dx=xn+1n+1+c(n≠−1)\int x^n\,dx = \frac{x^{n+1}}{n + 1} + c \quad (n \neq -1)Power rule for integrals∫x3 dx=x44+c\int x^3\,dx = \frac{x^4}{4} + c
ddxex=ex\frac{d}{dx}e^x = e^x and ddxef(x)=f′(x) ef(x)\frac{d}{dx}e^{f(x)} = f'(x)\,e^{f(x)}Derivative of an exponentialddxe3x2=6x e3x2\frac{d}{dx}e^{3x^2} = 6x\,e^{3x^2}
∫ex dx=ex+c\int e^x\,dx = e^x + cIntegral of exe^x∫01ex dx=e−1\int_0^1 e^x\,dx = e - 1
ddxln⁡(x)=1x\frac{d}{dx}\ln(x) = \frac{1}{x} and ddxln⁡(f(x))=f′(x)f(x)\frac{d}{dx}\ln\left(f(x)\right) = \frac{f'(x)}{f(x)}Derivative of a natural logarithmddxln⁡(x2+1)=2xx2+1\frac{d}{dx}\ln(x^2 + 1) = \frac{2x}{x^2 + 1}
∫1x dx=ln⁡(x)+c(x>0)\int \frac{1}{x}\,dx = \ln(x) + c \quad (x > 0)Integral of 1x\frac{1}{x}∫1e1x dx=ln⁡(e)−ln⁡(1)=1\int_1^e \frac{1}{x}\,dx = \ln(e) - \ln(1) = 1
ddxsin⁡(x)=cos⁡(x)\frac{d}{dx}\sin(x) = \cos(x) and ddxsin⁡(f(x))=f′(x)cos⁡(f(x))\frac{d}{dx}\sin\left(f(x)\right) = f'(x)\cos\left(f(x)\right)Derivative of sineddxsin⁡(x2)=2xcos⁡(x2)\frac{d}{dx}\sin(x^2) = 2x\cos(x^2)
∫sin⁡(x) dx=−cos⁡(x)+c\int \sin(x)\,dx = -\cos(x) + cIntegral of sine∫0πsin⁡(x) dx=[−cos⁡(x)]0π=1−(−1)=2\int_0^{\pi} \sin(x)\,dx = \left[-\cos(x)\right]_0^{\pi} = 1 - (-1) = 2
ddxcos⁡(x)=−sin⁡(x)\frac{d}{dx}\cos(x) = -\sin(x) and ddxcos⁡(f(x))=−f′(x)sin⁡(f(x))\frac{d}{dx}\cos\left(f(x)\right) = -f'(x)\sin\left(f(x)\right)Derivative of cosineddxcos⁡(3x)=−3sin⁡(3x)\frac{d}{dx}\cos(3x) = -3\sin(3x)
∫cos⁡(x) dx=sin⁡(x)+c\int \cos(x)\,dx = \sin(x) + cIntegral of cosine∫0π/2cos⁡(x) dx=sin⁡(π2)−sin⁡(0)=1\int_0^{\pi/2} \cos(x)\,dx = \sin\left(\frac{\pi}{2}\right) - \sin(0) = 1
If y=f(u)y = f(u) and u=g(x)u = g(x) then dydx=dydu×dudx\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}Chain ruley=(3x+1)4y = (3x + 1)^4: with u=3x+1u = 3x + 1, dydx=4u3×3=12(3x+1)3\frac{dy}{dx} = 4u^3 \times 3 = 12(3x + 1)^3
ddx(uv)=udvdx+vdudx\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}Product ruleu=x2u = x^2, v=exv = e^x: x2ex+2x ex=x ex(x+2)x^2e^x + 2x\,e^x = x\,e^x(x + 2)
ddx(uv)=vdudx−udvdxv2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}Quotient ruleu=sin⁡(x)u = \sin(x), v=xv = x: xcos⁡(x)−sin⁡(x)x2\frac{x\cos(x) - \sin(x)}{x^2}

The trapezoidal rule, with nn strips of equal width ww, is too long for the table:

∫abf(x) dx≈w2[f(x0)+2(f(x1)+f(x2)+f(x3)+…+f(xn−1))+f(xn)],where w=b−an\int_a^b f(x)\,dx \approx \frac{w}{2}\left[f(x_0) + 2\left(f(x_1) + f(x_2) + f(x_3) + \ldots + f(x_{n-1})\right) + f(x_n)\right], \quad \text{where } w = \frac{b - a}{n}

Worked example: for ∫02x2 dx\int_0^2 x^2\,dx with n=4n = 4, w=0.5w = 0.5 and the values are 0,0.25,1,2.25,40, 0.25, 1, 2.25, 4. So 0.52[0+2(3.5)+4]=2.75\frac{0.5}{2}\left[0 + 2(3.5) + 4\right] = 2.75 (the exact value is 83\frac{8}{3}).

The book prints the derivative of a function inside an exponential, a natural logarithm, a sine and a cosine, but only the basic integrals of exe^x, 1x\frac{1}{x}, sin⁡(x)\sin(x) and cos⁡(x)\cos(x). The trapezoidal rule in the book has no plus sign before the dots in the middle bracket; the version above adds it.

Trigonometry (page 2)

Only four items.

FormulaWhat it doesWorked example
c2=a2+b2−2abcos⁡(C)c^2 = a^2 + b^2 - 2ab\cos(C)Cosine rule: find a side from two sides and the included angle, or an angle from three sidesa=5a = 5, b=8b = 8, C=60∘C = 60^\circ: c2=25+64−80cos⁡(60∘)=49c^2 = 25 + 64 - 80\cos(60^\circ) = 49, so c=7c = 7
asin⁡(A)=bsin⁡(B)=csin⁡(C)\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}Sine rule: sides over the sines of their opposite angles. The syllabus requires the ambiguous caseA=30∘A = 30^\circ, a=4a = 4, B=45∘B = 45^\circ: b=4sin⁡(45∘)sin⁡(30∘)=42b = \frac{4\sin(45^\circ)}{\sin(30^\circ)} = 4\sqrt{2}
area=12bcsin⁡(A)\text{area} = \frac{1}{2}bc\sin(A)Area of a triangle from two sides and the included angleb=6b = 6, c=8c = 8, A=30∘A = 30^\circ: 12(6)(8)(12)=12\frac{1}{2}(6)(8)\left(\frac{1}{2}\right) = 12
sin⁡2(A)+cos⁡2(A)=1\sin^2(A) + \cos^2(A) = 1Pythagorean identityIf sin⁡(A)=35\sin(A) = \frac{3}{5} with AA acute, cos⁡2(A)=1−925=1625\cos^2(A) = 1 - \frac{9}{25} = \frac{16}{25}, so cos⁡(A)=45\cos(A) = \frac{4}{5}

Statistics and probability (page 3)

FormulaWhat it doesWorked example
(x+y)n=xn+(n1)xn−1y+…+(nr)xn−ryr+…+yn(x + y)^n = x^n + \binom{n}{1}x^{n-1}y + \ldots + \binom{n}{r}x^{n-r}y^r + \ldots + y^nBinomial theorem(2x−1)3(2x - 1)^3: the coefficients are 1, 3, 3, 1, so 8x3−12x2+6x−18x^3 - 12x^2 + 6x - 1
P(X=r)=(nr)pr(1−p)n−rP(X = r) = \binom{n}{r}p^r(1 - p)^{n-r}Binomial probability of rr successes in nn trialsn=4n = 4, p=12p = \frac{1}{2}, r=2r = 2: 6×116=386 \times \frac{1}{16} = \frac{3}{8}
E(X)=μ=∑pixiE(X) = \mu = \sum p_i x_iMean of a discrete random variablex=1,2,3x = 1, 2, 3 with p=0.2,0.5,0.3p = 0.2, 0.5, 0.3: 0.2+1.0+0.9=2.10.2 + 1.0 + 0.9 = 2.1
Var(X)=∑pi(xi−μ)2\mathrm{Var}(X) = \sum p_i (x_i - \mu)^2Variance of a discrete random variableSame table: 0.2(1.1)2+0.5(0.1)2+0.3(0.9)2=0.242+0.005+0.243=0.490.2(1.1)^2 + 0.5(0.1)^2 + 0.3(0.9)^2 = 0.242 + 0.005 + 0.243 = 0.49
Var(X)\sqrt{\mathrm{Var}(X)}Standard deviation0.49=0.7\sqrt{0.49} = 0.7
E(X)=μ=∫−∞∞x p(x) dxE(X) = \mu = \int_{-\infty}^{\infty} x\,p(x)\,dxMean of a continuous random variable, with p(x)p(x) its probability density functionp(x)=2xp(x) = 2x on 0≤x≤10 \le x \le 1 and 0 elsewhere: ∫012x2 dx=23\int_0^1 2x^2\,dx = \frac{2}{3}
Var(X)=σ2=∫−∞∞(x−μ)2 p(x) dx\mathrm{Var}(X) = \sigma^2 = \int_{-\infty}^{\infty} (x - \mu)^2\,p(x)\,dxVariance of a continuous random variableSame p(x)p(x): ∫01(x−23)22x dx=12−89+49=118\int_0^1 \left(x - \frac{2}{3}\right)^2 2x\,dx = \frac{1}{2} - \frac{8}{9} + \frac{4}{9} = \frac{1}{18}
Bernoulli: mean pp, variance p(1−p)p(1 - p)A single trial with probability of success ppp=0.3p = 0.3: mean 0.3, variance 0.3×0.7=0.210.3 \times 0.7 = 0.21
Binomial: mean npnp, variance np(1−p)np(1 - p)Number of successes in nn independent trialsn=50n = 50, p=0.2p = 0.2: mean 10, variance 50×0.2×0.8=850 \times 0.2 \times 0.8 = 8
Sample proportion: mean pp, standard deviation p(1−p)n\sqrt{\dfrac{p(1 - p)}{n}}Centre and spread of the sample proportion p^\hat{p}p=0.5p = 0.5, n=100n = 100: 0.25100=0.05\sqrt{\dfrac{0.25}{100}} = 0.05
(p^−zp^(1−p^)n, p^+zp^(1−p^)n)\left(\hat{p} - z\sqrt{\dfrac{\hat{p}(1 - \hat{p})}{n}},\ \hat{p} + z\sqrt{\dfrac{\hat{p}(1 - \hat{p})}{n}}\right)Approximate confidence interval for the population proportion ppp^=0.5\hat{p} = 0.5, n=100n = 100, z=1.96z = 1.96: the margin is 1.96×0.05=0.0981.96 \times 0.05 = 0.098, so the interval is (0.402,0.598)(0.402, 0.598)
zp^(1−p^)nz\sqrt{\dfrac{\hat{p}(1 - \hat{p})}{n}}Approximate margin of errorSame values: 0.0980.098. Quadrupling nn halves it, because nn sits under the square root
P(A‾)=1−P(A)P(\overline{A}) = 1 - P(A)ComplementP(A)=0.35P(A) = 0.35 gives P(A‾)=0.65P(\overline{A}) = 0.65
P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)General addition rule0.5+0.4−0.2=0.70.5 + 0.4 - 0.2 = 0.7
P(A∩B)=P(A)P(B)P(A \cap B) = P(A)P(B)Independent eventsP(A)=0.3P(A) = 0.3, P(B)=0.5P(B) = 0.5, independent: 0.150.15
P(A∩B)=P(A∣B)P(B)P(A \cap B) = P(A \mid B)P(B)Conditional probability, multiplication formP(B)=0.5P(B) = 0.5, P(A∣B)=0.6P(A \mid B) = 0.6: P(A∩B)=0.3P(A \cap B) = 0.3

In the confidence interval, zz is the quantile of the standard normal distribution that matches the confidence level. The book does not say so. The value 1.96 for 95% confidence is standard, and the syllabus says quantiles are found with technology, so a technology-free paper gives you no way to find zz yourself.

What is not on the formula sheet?

We compared the book line by line with the Units 1 to 4 subject matter in the 2025 syllabus. The syllabus asks for each of these, and the book does not print it.

Not printedWhat the syllabus asks forUnit
The quadratic formula and the discriminantSolve quadratics with the quadratic formula (exact and approximate answers) and use the discriminant to count solutions1
Surd rulesSimplify square roots with perfect square factors, rationalise denominators and use the four operations on surds1
Index lawsNegative and fractional indices, radicals as fractional indices, scientific notation2
The meaning of (nr)\binom{n}{r}The book uses the symbol in two formulas and prints no factorial or combination definition1, 3
Radian measure and exact valuesRadians and degrees, arc lengths and sector areas, exact values of cos⁡\cos, sin⁡\sin and tan⁡\tan at multiples of π6\frac{\pi}{6} and π4\frac{\pi}{4}1
tan⁡\tanThe unit circle definition of tan⁡(θ)\tan(\theta) and the graph of y=tan⁡(x)y = \tan(x)1
Period of a sine or cosine graphThe effect of aa, bb, hh and kk on y=asin⁡(b(x−h))+ky = a\sin\left(b(x - h)\right) + k1
Straight line formulasThe syllabus calls a link to linear functions assumed knowledge for tangents and normals2
The definition of independenceIndependence of AA from BB, defined as P(A∣B)=P(A)P(A \mid B) = P(A). The book prints only P(A∩B)=P(A∣B)P(B)P(A \cap B) = P(A \mid B)P(B) and P(A∩B)=P(A)P(B)P(A \cap B) = P(A)P(B), so to find P(A∣B)P(A \mid B) you divide the first by P(B)P(B) yourself1
Integrals of the form ∫f(ax+b) dx\int f(ax + b)\,dxDetermine indefinite integrals of this form. The book has only the basic integrals of exe^x, 1x\frac{1}{x}, sin⁡(x)\sin(x) and cos⁡(x)\cos(x)3
Area between two curvesCalculate the area between curves. The book has the fundamental theorem for one function4
Motion relationshipsDisplacement from velocity or acceleration with initial values, and velocity as the rate of change of displacement2, 3, 4
Probability density function propertiesProbabilities of a continuous random variable as integrals, and the cumulative distribution function. The book uses p(x)p(x) without defining it4
The normal distributionThe notation X∼N(μ,σ2)X \sim N(\mu, \sigma^2), standardised zz-scores, probabilities and quantiles. The book has no density, no standardisation formula and no table, and the syllabus says tables are not required4
The standardised sample proportionp^−pp^(1−p^)/n\dfrac{\hat{p} - p}{\sqrt{\hat{p}(1 - \hat{p})/n}} being approximately standard normal for large samples4
Definitions of zz, p^\hat{p} and nnThe book uses them in the interval and margin of error without a key4

Three kinds of item sit in that list. Some are formulas you must memorise (quadratic formula, exact values, index laws, surd rules). Some are methods, not formulas (area between curves, motion problems, the second derivative test), so nothing on any sheet could replace them. And some are done with technology in Paper 2 (normal probabilities and quantiles).

The book also has nothing on sequences and series, finance, matrices, vectors, complex numbers, integration by substitution or by parts, or hypothesis testing. None of those appear in the Methods Units 1 to 4 subject matter either, so their absence is not a gap. They belong to other subjects.

What is the difference between Paper 1 and Paper 2?

Same formula book, same 90 minutes working time and 5 minutes perusal, 25% each. Technology is the only variable the syllabus changes.

  • Paper 1 is technology-free. Anything that needs a decimal has to be left exact or estimated by hand. That makes these rows the ones to practise by hand: the log laws, the power, chain, product and quotient rules, the trapezoidal rule with a few strips, the cosine rule with an angle that has an exact cosine, binomial coefficients, and the confidence interval with friendly numbers.
  • Paper 2 is technology-active, with the calculator rules in the table above. The syllabus says normal probabilities and quantiles are found using technology, and binomial means and variances are calculated using technology and algebraic methods.

Check your calculator model against QCAA's approved lists before the day. QCAA's 2026 memo (044/26) says the lists were updated, and that the instructions for resetting or clearing approved graphics calculators before and after the external exam changed. Teachers are asked to review those instructions for the models their students use. The lists themselves are on the QCAA website, so read the current copy, not a screenshot from a friend.

What do technology-free questions ask you to do by hand?

The formula book does not do the technology-free work for you. QCAA's 2023 to 2025 reports (those cohorts sat the 2019 course, with its own formula sheet) describe Paper 1 questions that were won or lost on a skill, not a lookup:

  • Exact answers (2023, Paper 1, Question 19). The effective response simplified 10ln⁡2ln⁡(2x)−10ln⁡2ln⁡(x/16)\frac{10}{\ln 2}\ln(2x) - \frac{10}{\ln 2}\ln(x/16) with log laws to 10ln⁡2ln⁡(32)=50\frac{10}{\ln 2}\ln(32) = 50 exactly, with no calculator.
  • Estimation (2024, Paper 1, Question 18). The question needed the cosine rule to connect the sides to one given angle. The report says many students could progress part of the way but could not convert the progressing calculation to a reasonable approximation. The effective excerpt reached h=624/25≈5h = \sqrt{624/25} \approx 5 and compared it with the 4 m requirement.
  • Algebra before calculus (2024, Paper 1, Question 13b). Many students could not integrate correctly because they made errors expanding a squared bracket and simplifying.
  • Confidence intervals (2025, Paper 1, Question 12). The report says many students struggle to determine confidence intervals in technology-free situations.
  • Logarithmic equations (2025, Paper 1, Question 15). The effective response applied the power law log⁡b(an)=nlog⁡b(a)\log_b(a^n) = n\log_b(a), recognised the quadratic a2=2a+8a^2 = 2a + 8 with a=log⁡2(x)a = \log_2(x), solved for both roots and checked one by substitution.

Where do you practise using it?

The question bank holds marked past-paper questions from Paper 1 and Paper 2 for 2020 to 2025, 116 questions on each paper across the six years. Do a Paper 1 set with only the book open and no calculator, the way QCAA sets the conditions, and note every time you reach for a rule that is not printed. That list is your memory work, and the table above is where to start it.

To see which topics the papers test most and where students lose marks, read the most tested topics and hardest topics guides. Everything else about the subject sits on the Mathematical Methods study guide, and other subjects are on the QCE hub.

Frequently asked questions

Do you get a formula sheet in QCAA Mathematical Methods Paper 1?

Yes. The syllabus says the QCAA provides the QCAA Mathematical Methods formula book for both external papers. Paper 1 is technology-free and allows nothing else: no notes, calculators, technology or other resources. In IA2 and IA3 the teacher must provide the same formula book.

Is the formula sheet the same in Paper 1 and Paper 2?

The syllabus names one document, the QCAA Mathematical Methods formula book, as provided for both papers. What differs between the papers is the technology: none in Paper 1, and in Paper 2 a handheld QCAA-approved graphics calculator (no CAS functionality) and/or a handheld QCAA-approved scientific calculator.

What is on the QCAA Maths Methods formula book?

Three printed pages. Page 1 has mensuration, graph equations and logarithm laws. Page 2 has calculus (first principles, the fundamental theorem, derivative and integral rules, the chain, product and quotient rules and the trapezoidal rule) and four trigonometry items. Page 3 has the binomial theorem and binomial probability, random-variable means and variances, sample proportions, the confidence interval and margin of error, and the basic probability rules.

Is the quadratic formula on the Maths Methods formula sheet?

No. The 2025 syllabus asks you to solve quadratics with the quadratic formula and to use the discriminant, and the book prints neither. It also leaves out the index laws, surd rules, radian and exact-value trigonometry, tan, integrals of the f(ax + b) type and anything on the normal distribution, so these come from memory.

Is the normal distribution on the Maths Methods formula sheet?

No. The book has no normal density, no table and no standardisation formula, and it does not define z. The syllabus says normal probabilities and quantiles are calculated with technology and that distribution tables are not required.

Is it called a formula sheet or a formula book?

The 2025 syllabus says formula book. The 2019 syllabus said formula sheet. Students and search engines still use formula sheet, so both names point at the same QCAA document for the current course.

Can I use a CAS calculator in Mathematical Methods Paper 2?

No. The syllabus allows a handheld QCAA-approved graphics calculator with no CAS functionality, and/or a handheld QCAA-approved scientific calculator. QCAA publishes an approved list for each type, and its 2026 memo (044/26) says the instructions for resetting or clearing approved graphics calculators before and after the external exam were updated.

Which Paper 1 skills did QCAA's subject reports flag?

Both cohorts sat the 2019 course. In the 2024 report: algebraic and arithmetic fluency (expanding brackets, simplifying, working with fractions and decimals) and estimating a calculated value in a technology-free two- or three-dimensional problem. In the 2025 report: determining confidence intervals in a technology-free question (Paper 1, Question 12) and solving logarithmic equations that need several logarithmic rules before a quadratic.

Sources

  • Mathematical Methods 2025 formula book, QCAA, June 2026, pp. 1-3. Every formula reproduced on this page, and the list of what the book does not print. The cover is dated June 2026 and the licence line reads © State of Queensland (QCAA) 2025.
  • Mathematical Methods 2025 v1.2 General senior syllabus, QCAA, October 2024, pp. 15-30, 35-45. Units 1 to 4 subject matter compared with the book to find what is not printed, and the IA2, IA3 and external assessment conditions: who provides the formula book, what technology Paper 1 and Paper 2 allow.
  • Mathematical Methods 2019 v1.2 General Senior Syllabus, QCAA, July 2018. Superseded version, cited only for the earlier name of the document, the QCAA formula sheet.
  • Mathematical Methods subject report, 2025 cohort, QCAA, January 2026, pp. 28-39. Technology-free confidence-interval and logarithmic-equation findings, and the effective-response excerpts.
  • Mathematical Methods subject report, 2024 cohort, QCAA, January 2025, pp. 25-30. Technology-free algebra and estimation findings (Paper 1, Questions 13b and 18), and the cosine rule question excerpt (Paper 1, Question 18).
  • Mathematical Methods subject report, 2023 cohort, QCAA, January 2024, pp. 37-38. The technology-free logarithm excerpt (Paper 1, Question 19).
  • Memo 044/26: Scientific and graphics calculator lists (updated 2026) now available, QCAA, August 2026, pp. 1-2. Existence of the approved scientific and graphics lists for external assessment, and the updated reset or clear instructions for graphics calculators.
  • AusGrader question bank, QCE Mathematical Methods, AusGrader. Paper 1 and Paper 2 for 2020 to 2025 are held, 116 questions on each paper across the six years (examPapers coverage).

Syllabus and assessment material referenced in this guide is © State of Queensland (Queensland Curriculum and Assessment Authority), licensed under CC BY 4.0. See our QCAA licensing notice. AusGrader is an independent study tool and is not affiliated with, endorsed by, or operated by the QCAA.

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