AEMAM Semester 1 Exam Cheatsheet

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Exam Structure Seen Repeatedly in Papers

SectionReadingWorkingTypical marks/weighting in ingested papers
Section One (Calculator-free)5 min50 min~52 marks (~35%)
Section Two (Calculator-assumed)10 min100 min~98 marks (~65%)

High-Frequency Question Patterns (from ingested question snippets)

Pattern familyCount in extracted snippets*Typical prompt style
Functions/relations184domain/range, function test, graph interpretation
Circle/geometry134circle equations, radius/centre, triangle geometry
Trig identities/equations99simplify/evaluate/solve trig expressions
Probability/sets71P(AB)P(A\cup B), conditional probability, independence
Line equations/coordinates60perpendicular/parallel line, midpoint, distance
Modelling/context60time-height models, applied probability contexts
Quadratic skills44turning point, factor form, min/max
Asymptotes/rational25hyperbola graph/asymptotes/intersections
Counting/binomial9combinations, expansion, coefficient meaning

*Counts include repeated exam+solution variants, so use as a priority signal only.


Core Concepts + Formula Bank

1. Functions, relations, domain/range

  • Function test: each xx maps to one yy (vertical line test on graphs).
  • Mappings: One-to-One and Many-to-One are functions. One-to-Many is not a function.
  • Domain: allowed xx-values.
  • Range: resulting yy-values.
  • Common exam move: state domain/range from graph/features first, then solve.

2. Lines and coordinate geometry

  • Gradient from y=mx+cy=mx+c: mm.
  • Angle of Inclination: Gradient m=tanθm = \tan \theta (θ\theta is the angle with the positive xx-axis).
  • Perpendicular gradients: m1m2=1m_1m_2=-1.
  • Point-slope form: yy1=m(xx1)y-y_1=m(x-x_1).
  • Midpoint: (x1+x22,y1+y22)\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right), distance by Pythagoras.

3. Quadratics

  • Turning point form: y=a(xh)2+ky=a(x-h)^2+k (vertex (h,k)(h,k)).
  • Symmetry line: x=hx=h.
  • Min/max on restricted domain: check vertex + domain endpoints.
  • Quadratic Formula: x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
  • Discriminant (Δ=b24ac\Delta = b^2 - 4ac): Δ>0\Delta > 0 (2 distinct real roots), Δ=0\Delta = 0 (1 repeated real root/tangent to x-axis), Δ<0\Delta < 0 (0 real roots).

4. Trigonometry

  • Degree-radian: θrad=θπ/180\theta_{\text{rad}}=\theta^\circ\pi/180, θ=θrad180/π\theta^\circ=\theta_{\text{rad}}180/\pi.
  • Identity used repeatedly: cos(A+B)=cosAcosBsinAsinB\cos(A+B)=\cos A\cos B-\sin A\sin B.
  • Trig Graphs (y=asin(b(xh))+ky = a \sin(b(x-h)) + k): Amplitude = a|a|, Period = 2πb\frac{2\pi}{|b|} (for sin/cos\sin/\cos) or πb\frac{\pi}{|b|} (for tan\tan), Phase shift = hh (right).
  • Solve trig equations by:
    1. isolate trig term,
    2. find reference angles,
    3. apply interval limits.

5. Probability

  • Addition rule: P(AB)=P(A)+P(B)P(AB)P(A\cup B)=P(A)+P(B)-P(A\cap B).
  • Conditional: P(AB)=P(AB)P(B)P(A\mid B)=\frac{P(A\cap B)}{P(B)}.
  • Independence test: P(AB)=P(A)P(B)P(A\cap B)=P(A)P(B).
  • Complementary / Not A: P(A)=1P(A)P(A') = 1 - P(A).
  • Mutually Exclusive: P(AB)=0P(A\cap B) = 0.

6. Counting & Set Theory

  • Arrangement / Permutation (PP): Order matters     nPr=n!(nr)!\implies {}^nP_r=\frac{n!}{(n-r)!}.
  • Selection / Combination (CC): Order doesn’t matter     (nr)=n!r!(nr)!\implies \binom{n}{r}=\frac{n!}{r!(n-r)!}.
  • Multiplication Principle: Multiply independent event outcomes (e.g. choosing shirt AND pants).
  • Subsets & Binomial Terms: A set of nn elements has 2n2^n subsets. General binomial term in (a+b)n(a+b)^n is (nr)anrbr\binom{n}{r} a^{n-r} b^r.
  • Sets: ABA \cap B (AND / intersection shape), ABA \cup B (OR / union shape), AA' (NOT / complement).

7. Circular/triangle geometry

  • Sine rule: asinA=bsinB=csinC\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} (Ambiguous case: Given two sides and a non-included angle (SSA\text{SSA}), the unknown angle can be θ\theta or 180θ180^\circ - \theta).
  • Cosine rule: c2=a2+b22abcosCc^2=a^2+b^2-2ab\cos C.
  • Arcs & Sectors (radians): Arc length L=rθL=r\theta, sector area A=12r2θA=\frac12 r^2\theta.
  • Triangle area: A=12absinCA=\frac12ab\sin C.
  • Circular Segment area: A=12r2(θsinθ)A=\frac12 r^2(\theta - \sin\theta).
  • Circles: Center (h,k)(h,k), radius r    (xh)2+(yk)2=r2r \implies (x-h)^2 + (y-k)^2 = r^2.

8. Polynomials & Division

  • Remainder Theorem: If a polynomial P(x)P(x) is divided by (xa)(x-a), the remainder is P(a)P(a).
  • Factor Theorem: (xa)(x-a) is a factor of P(x)P(x) if and only if P(a)=0P(a) = 0.
  • Synthetic Division / Equating Coefficients: Use to completely factorise cubics once one root is found.

9. General Transformations (y=af(b(xh))+ky=af(b(x-h))+k)

  • Parent functions: Hyperbola y=1/xy=1/x (asymptotes x=0,y=0x=0, y=0), Square root y=xy=\sqrt{x}, Circle x2+y2=r2x^2+y^2=r^2.
  • aa: Vertical dilation by factor aa. If negative, reflection in the xx-axis.
  • bb: Horizontal dilation by factor 1b\frac{1}{b}. If negative, reflection in the yy-axis.
  • hh: Horizontal translation (h>0h>0 moves right).
  • kk: Vertical translation (k>0k>0 moves up).

10. Abstract Algebra & Proportion

  • Sum of Cubes: a3+b3=(a+b)(a2ab+b2)a^3 + b^3 = (a + b)(a^2 - ab + b^2).
  • Difference of Cubes: a3b3=(ab)(a2+ab+b2)a^3 - b^3 = (a - b)(a^2 + ab + b^2).
  • Direct Proportion: yx    y=kxy \propto x \implies y = kx.
  • Inverse Proportion: y1x    y=kxy \propto \frac{1}{x} \implies y = \frac{k}{x}.

Blueprint Solving Methods (Pattern → Method)

If you see…Do this immediately
“perpendicular line through point”get given slope → invert/negate for perpendicular slope → point-slope form
“radian/degree conversion”apply conversion factor first; avoid calculator rounding until end
trig equation with intervalisolate trig ratio → solve base angles → list all interval-valid solutions
hyperbola/rational with asymptotesrewrite to expose denominator zero and horizontal trend
“min/max on interval”convert to turning-point form, then test both endpoints
P(AB)P(A\cup B), P(AB)P(A\mid B), independencewrite the correct rule first, then substitute values
combinations selection constraintssplit into cases (e.g., 4+2, 5+1, 6+0), sum each case
model h(t)=a+bcos(kt)h(t)=a+b\cos(kt)baseline aa, amplitude $

Marking Insights Seen in Solutions

  • Solutions repeatedly allocate marks for method milestones (“Specific behaviours”), not just the last line.
  • In exam instructions across papers: for multi-mark parts, unsupported final answers lose marks.
  • Practical rule: every part should show:
    1. formula/rule selected,
    2. substitution,
    3. simplified result with units/context.

Use this for partial-mark maximization.


Exam Strategy Layer

Time structure

  • Section One: reading 5 + working 50.
  • Section Two: reading 10 + working 100.

Fast pattern-recognition workflow

  1. Circle trigger words (“perpendicular”, “independent”, “amplitude”, “at least one”, “asymptote”).
  2. Map to one blueprint method (table above).
  3. Write first method line immediately (locks in method marks).

Partial-mark tactics

  • If stuck, still write the governing formula.
  • For graph-heavy questions, state key features (asymptotes/intercepts/turning point) even before full sketch.
  • In probability/combinatorics, define the case split explicitly before arithmetic.

Special Consideration Mode: “I did not prepare. I have 2 days.”

Last-resort strategy (Replication > Understanding)

  1. Memorize 10 templates only (line, trig solve, trig identity, rational asymptote, quadratic min/max, P(AB)P(A\cup B), conditional probability, combinations case split, binomial coefficient meaning, cosine model threshold time).
  2. Drill the Master Reference Exam below twice:
    • Pass 1: copy method lines exactly.
    • Pass 2: reproduce from memory with timing.
  3. Exam-day rule: if a question matches a template, replicate the structure first, then compute.

This is a pattern-matching system under pressure, optimized for marks.


What to Put on Your Notes Sheet

(2 double-sided A4, calculator section only, formula sheet also available)

Sheet 1 (front): Trigger → First line

  • Perpendicular line: m=1/mm_\perp=-1/m, then yy1=m(xx1)y-y_1=m_\perp(x-x_1)
  • Trig solve template with interval line-by-line
  • P(AB)P(A\cup B), P(AB)P(A\mid B), independence check template
  • Hyperbola checklist: vertical asymptote from denominator =0=0, horizontal asymptote from end behaviour

Sheet 1 (back): Exact-value + conversion block

  • Degree↔radian conversions
  • cos(A±B)\cos(A\pm B), sin(A±B)\sin(A\pm B)
  • Exact values table:
    θ\theta00π6\frac{\pi}{6} (3030^\circ)π4\frac{\pi}{4} (4545^\circ)π3\frac{\pi}{3} (6060^\circ)π2\frac{\pi}{2} (9090^\circ)
    sinθ\sin\theta001/21/21/21/\sqrt{2}3/2\sqrt{3}/211
    cosθ\cos\theta113/2\sqrt{3}/21/21/\sqrt{2}1/21/200
    tanθ\tan\theta001/31/\sqrt{3}113\sqrt{3}Undefined

Sheet 2 (front): Counting + modelling block

  • (nr)\binom{n}{r} setup template
  • “At least one” = total − forbidden cases
  • Binomial expansion skeleton (a+b)6(a+b)^6
  • h(t)=a+bcos(kt)h(t)=a+b\cos(kt): min/max/time threshold solving steps

Sheet 2 (back): Mini worked prototypes

  • One full line-equation example
  • One trig-equation interval example
  • One probability conditional example
  • One combinations case-split example

Keep this side dense and procedural: method lines over theory paragraphs.


Master Reference Exam (Print on Notes)

1. Questions Section

Q1 (Source: content/SEM1 2025 YR11 METH U1 S1)

  1. Through (2,7)(2,-7), find the line perpendicular to y=14x+1y=-\frac14x+1.
  2. Solve (x7)2=3(x-7)^2=3.

Q2 (Source: content/SEM1 2025 YR11 METH U1 S1)

  1. Convert 33^\circ to radians.
  2. Convert 11π18\frac{11\pi}{18} to degrees.
  3. Solve 2sin(2x)2=02\sin(2x)-\sqrt2=0, for 0xπ0\le x\le\pi.

Q3 (Source: content/SEM1 2025 YR11 METH U1 S1)

  1. For f(t)=5+3cos(2t)f(t)=5+3\cos(2t), state amplitude and period.
  2. Evaluate cos ⁣(2π3+π4)\cos\!\left(\frac{2\pi}{3}+\frac{\pi}{4}\right).

Q4 (Source: content/SEM1 2025 YR11 METH U1 S1) Given y=c2xy=\frac{c}{2-x} and x+y+3=0x+y+3=0, and the hyperbola crosses the yy-axis at y=3y=3:

  1. Find the line slope.
  2. Find cc.
  3. State both asymptotes.
  4. Find the xx-coordinates of intersections.

Q5 (Source: content/SEM1 2025 YR11 METH U1 S1) Square has vertices (0,a+6)(0,a+6) and (a,0)(a,0), with 7a4-7\le a\le4:

  1. Find area when a=7a=-7.
  2. Show A=2a2+12a+36A=2a^2+12a+36.
  3. Find turning point of A(a)A(a).
  4. Find min and max area on the given domain.

Q6 (Source: Rossmoyne Methods 11 2024 Sem 1 CA) Given P(A)=0.3,  P(B)=0.5P(A)=0.3,\;P(B)=0.5, find P(AB)P(A\cup B) when:

  1. A,BA,B mutually exclusive.
  2. P(AB)=0.8P(A\mid B)=0.8.
  3. A,BA,B independent.

Q7 (Source: Rossmoyne Methods 11 2024 Sem 1 CA) Choose 6 dancers from 7 Year 11 and 6 Year 12 students:

  1. equal numbers from each year.
  2. at least one from each year.
  3. exactly one choreographer included (one of two choreographers, not both).
  4. more Year 11 than Year 12.
  5. Expand (a+b)6(a+b)^6.
  6. Interpret coefficient of a2b4a^2b^4 in context.

Q8 (Source: Rossmoyne Methods 11 2024 Sem 1 CA)

  1. Using cos36=45,sin36=35,cos55=47,sin55=56\cos36^\circ=\frac45,\sin36^\circ=\frac35,\cos55^\circ=\frac47,\sin55^\circ=\frac56, approximate cos19\cos19^\circ.
  2. Peter sees Ray 450 m away on bearing 110110^\circT. Mary is 200 m south of Peter. Find Ray’s bearing to Mary.

Q9 (Source: Rossmoyne Methods 11 2024 Sem 1 CA) Malaria test: prevalence 2%2\%, sensitivity 87%87\%, specificity 94%94\%.

  1. Find P(malaria and test positive)P(\text{malaria and test positive}).
  2. Find P(malariatest positive)P(\text{malaria}\mid\text{test positive}).

Q10 (Source: Rossmoyne Methods 11 2024 Sem 1 CA)

h(t)=5.4+1.7cos ⁣(πt6),0t24h(t)=5.4+1.7\cos\!\left(\frac{\pi t}{6}\right),\quad 0\le t\le24

tt is hours after 5 am.

  1. Find clearance at 5:00 am and 8:45 am.
  2. First minimum time and value.
  3. Latest morning time for clearance 4.6\ge 4.6 m.

Q11 (Source: Rossmoyne Methods 11 2024 Sem 1 CA) Voucher table (total 400):

Event1 person2 people3 people4 people
Basketball50352540
Football27172145
Soccer48204230

Find probabilities that a random voucher:

  1. is soccer + admits 2.
  2. is basketball.
  3. admits no more than 3.
  4. admits 1 person or is basketball.
  5. admits an odd number given soccer is excluded.

2. Full Worked Solutions Section

Q1

  1. Given line slope is 14-\frac14, so perpendicular slope m=4m=4.
    y+7=4(x2)y=4x15y+7=4(x-2)\Rightarrow y=4x-15.
    y=4x15y=4x-15
  2. (x7)2=3x7=±3x=7±3(x-7)^2=3\Rightarrow x-7=\pm\sqrt3\Rightarrow x=7\pm\sqrt3.
    x=7±3x=7\pm\sqrt3

Q2

  1. 3×π180=π603^\circ\times\frac{\pi}{180}=\frac{\pi}{60}.
    π60\frac{\pi}{60} rad
  2. 11π18×180π=110\frac{11\pi}{18}\times\frac{180}{\pi}=110^\circ.
    110110^\circ
  3. 2sin(2x)2=0sin(2x)=222\sin(2x)-\sqrt2=0\Rightarrow \sin(2x)=\frac{\sqrt2}{2}.
    2x=π4,3π42x=\frac\pi4,\frac{3\pi}4 in [0,2π]x=π8,3π8[0,2\pi]\Rightarrow x=\frac\pi8,\frac{3\pi}8.
    x=π8,3π8x=\frac\pi8,\frac{3\pi}8

Q3

  1. For f(t)=5+3cos(2t)f(t)=5+3\cos(2t): amplitude =3=3, period =2π2=π=\frac{2\pi}{2}=\pi.
    Amplitude 33, period π\pi

$$
\cos\!\left(\frac{2\pi}{3}+\frac{\pi}{4}\right)
=\cos\frac{2\pi}{3}\cos\frac\pi4-\sin\frac{2\pi}{3}\sin\frac\pi4
=-\frac12\frac{\sqrt2}{2}-\frac{\sqrt3}{2}\frac{\sqrt2}{2}
=-\frac{\sqrt2+\sqrt6}{4}.
$$

<span class="highlight">$-\frac{\sqrt2+\sqrt6}{4}$</span>

Q4

  1. x+y+3=0y=x3x+y+3=0\Rightarrow y=-x-3, slope =1=-1.
    1-1
  2. At x=0,  y=3x=0,\;y=3: 3=c20c=63=\frac{c}{2-0}\Rightarrow c=6.
    c=6c=6
  3. For y=62xy=\frac{6}{2-x}: horizontal asymptote y=0y=0, vertical asymptote x=2x=2.
    y=0,  x=2y=0,\;x=2
  4. Intersections: 62x=x3\frac{6}{2-x}=-x-3.
    6=(2x)(x3)=x2+x6x2+x12=0(x+4)(x3)=06=(2-x)(-x-3)=x^2+x-6\Rightarrow x^2+x-12=0\Rightarrow(x+4)(x-3)=0.
    x=4,  x=3x=-4,\;x=3

Q5

  1. At a=7a=-7, vertices (0,1)(0,-1) and (7,0)(-7,0): side l=72+12=50l=\sqrt{7^2+1^2}=\sqrt{50}.
    Area A=l2=50A=l^2=50.
    5050
  2. l2=a2+(a+6)2A=2a2+12a+36l^2=a^2+(a+6)^2\Rightarrow A=2a^2+12a+36.
    A=2a2+12a+36A=2a^2+12a+36
  3. A=2(a2+6a+18)=2(a+3)2+18A=2(a^2+6a+18)=2(a+3)^2+18\Rightarrow turning point (3,18)(-3,18).
    (3,18)(-3,18)
  4. Minimum at vertex: 1818.
    Endpoints: A(7)=50,  A(4)=116A(-7)=50,\;A(4)=116\Rightarrow maximum 116116.
    Minimum 1818, maximum 116116

Q6

  1. Mutually exclusive P(AB)=0\Rightarrow P(A\cap B)=0:
    P(AB)=0.3+0.5=0.8P(A\cup B)=0.3+0.5=0.8.
    0.80.8
  2. P(AB)=0.8P(AB)=0.8×0.5=0.4P(A\mid B)=0.8\Rightarrow P(A\cap B)=0.8\times0.5=0.4.
    P(AB)=0.3+0.50.4=0.4P(A\cup B)=0.3+0.5-0.4=0.4.
    0.40.4
  3. Independent P(AB)=0.3×0.5=0.15\Rightarrow P(A\cap B)=0.3\times0.5=0.15.
    P(AB)=0.3+0.50.15=0.65P(A\cup B)=0.3+0.5-0.15=0.65.
    0.650.65

Q7

  1. Equal split 3+33+3: (73)(63)=700\binom73\binom63=700.
    700700
  2. At least one from each year: (136)(70)(66)(76)(60)=1708\binom{13}6-\binom70\binom66-\binom76\binom60=1708.
    17081708
  3. Exactly one choreographer: (21)(115)=924\binom21\binom{11}5=924.
    924924
  4. More Year 11:

\binom74\binom62+\binom75\binom61+\binom76\binom60 =525+126+7=658.

<span class="highlight">$658$</span> 5. $$ (a+b)^6=a^6+6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6.

(a+b)6(a+b)^6 expansion shown above 6. Coefficient of a2b4a^2b^4 is 1515: 15 ways to allocate 2 dancers to group aa and 4 to group bb.
15 allocations

Q8

\cos19^\circ=\cos(55^\circ-36^\circ) =\cos55^\circ\cos36^\circ+\sin55^\circ\sin36^\circ =\frac47\cdot\frac45+\frac56\cdot\frac35 =\frac{67}{70}\approx0.9571.

<span class="highlight">$\cos19^\circ\approx0.9571$</span> 2. Triangle $PMR$: $PR=450,\;PM=200,\;\angle RPM=70^\circ$. $$ MR^2=200^2+450^2-2(200)(450)\cos70^\circ\Rightarrow MR\approx425.366.

Then cosine rule gives angle at RR: θ26.22\theta\approx26.22^\circ.
Bearing from RR to MM: 36070θ263.78360^\circ-70^\circ-\theta\approx263.78^\circ.
264T (nearest degree)264^\circ\text{T (nearest degree)}

Q9

  1. P(M+)=0.02×0.87=0.0174P(M\cap +)=0.02\times0.87=0.0174.
    0.01740.0174
  2. False positive =0.98×0.06=0.0588=0.98\times0.06=0.0588.
    P(+)=0.0174+0.0588=0.0762P(+)=0.0174+0.0588=0.0762.

P(M\mid +)=\frac{0.0174}{0.0762}=0.2283.

<span class="highlight">$0.2283$</span> ### Q10 1. $h(0)=5.4+1.7=7.1$ m. For 8:45 am, $t=3.75$: $h(3.75)=5.4+1.7\cos\!\left(\frac{\pi(3.75)}6\right)\approx4.75$ m. <span class="highlight">$7.1$ m and $4.75$ m</span> 2. Minimum when $\cos\left(\frac{\pi t}{6}\right)=-1\Rightarrow t=6$. Time $=11{:}00$ am, clearance $=5.4-1.7=3.7$ m. <span class="highlight">$11{:}00$ am, $3.7$ m</span> 3. Solve $5.4+1.7\cos\!\left(\frac{\pi t}{6}\right)=4.6\Rightarrow t\approx3.9357$ h after 5 am. $5{:}00+3{:}56\approx8{:}56$ am. <span class="highlight">$8{:}56$ am (nearest minute)</span> ### Q11 1. Soccer + 2 people: $\frac{20}{400}=0.05$. <span class="highlight">$0.05$</span> 2. Basketball total: $\frac{50+35+25+40}{400}=\frac{150}{400}=\frac38=0.375$. <span class="highlight">$\frac38$</span> 3. No more than 3 people: $\frac{400-(40+45+30)}{400}=\frac{285}{400}=\frac{57}{80}=0.7125$. <span class="highlight">$\frac{57}{80}$</span> 4. One person or basketball: $$ \frac{(50+27+48)+(50+35+25+40)-50}{400} =\frac{225}{400}=\frac9{16}=0.5625.

916\frac9{16} 5. Odd number given soccer excluded:
Non-soccer total =400(48+20+42+30)=260=400-(48+20+42+30)=260.
Odd-count non-soccer =(50+25)+(27+21)=123=(50+25)+(27+21)=123.
$$ P=\frac{123}{260}\approx0.473.

<span class="highlight">$\frac{123}{260}\approx0.473$</span> --- ## Final Use Rule Before each exam question, identify the template first. Then copy the matching method structure from this sheet. That is the fastest path to <span class="highlight">replicable marks under pressure</span>.