Memorize these exact conversions to bypass step-by-step logic during the exam:
1 ml = 1 cm³
1 l = 1 dm³ = 1000 cm³
1000 l = 1 m³
Quantity
Key metric conversions
Length
1 cm = 10 mm | 1 m = 100 cm | 1 km = 1,000 m
Area
1 cm² = 100 mm² | 1 m² = 10,000 cm² | 1 km² = 1,000,000 m²
Volume
1 cm³ = 1,000 mm³ | 1 m³ = 1,000,000 cm³
Capacity
1 ml = 1 cm³ | 1 l = 1,000 cm³ | 1 m³ = 1,000 l
Mass
1 g = 1,000 mg | 1 kg = 1,000 g | 1 t = 1,000 kg
When converting an underlying unit of length, you must raise the scale factor to the power of that dimension (squared for area, cubed for volume).
Linear Scale: 1 m = 100 cm
Area Scale: 1 m² = (100)² cm² = 10,000 cm²
Volume Scale: 1 m³ = (100)³ cm³ = 1,000,000 cm³
Quantity
Key conversions to memorize
Length
1 cm = 10 mm | 1 m = 100 cm | 1 km = 1,000 m
Area
1 cm² = 100 mm² | 1 m² = 10,000 cm² | 1 km² = 1,000,000 m²
Volume
1 cm³ = 1,000 mm³ | 1 m³ = 1,000,000 cm³
Capacity
1 ml = 1 cm³ | 1 l = 1,000 cm³ = 1 dm³ | 1 m³ = 1,000 l
Mass
1 g = 1,000 mg | 1 kg = 1,000 g | 1 t = 1,000 kg
Speed
m/s › km/h: ×3.6 | km/h › m/s: ÷3.6
Time
1 min = 60 s | 1 h = 60 min = 3,600 s
Because every ESAT answer is one of the listed options, you never need to round mid-calculation — keep exact fractions (e.g. 5/120) until the final step. If a value looks "ugly", you have almost certainly missed a unit conversion. Match every quantity to the target compound unit first, then compute, and the correct option falls out cleanly.
To avoid calculation slips, convert the numerator and denominator independently.
Example: Convert 20 g/cm³ into kg/m³.
Isolate as a raw fraction: 20 g / 1 cm³
Convert the Top (g to kg): 20 g = (20 / 1000) kg
Convert the Bottom (cm³ to m³): 1 cm³ = (1 / 1,000,000) m³
Multiply by the reciprocal: (20 / 1000) kg / (1 / 1,000,000) m³ = (20 / 1000) * (1,000,000 / 1) = 20,000 kg/m³
The most common point-loser on compound unit questions is a hidden unit mismatch. A question stem will easily feed you a distance in metres and a time frame in minutes, but demand the final option in km/h. Never input raw numbers into a formula without matching the target compound units first.
Question: A car travels 35,000 m in exactly 30 minutes. What is its average speed in km/h?
Key Insight: Convert the raw components into kilometers and hours before setting up the velocity fraction.
Distance Map: 35,000 m = (35,000 / 1000) km = 35 km
Time Map: 30 min = (30 / 60) hours = 0.5 hours
Execute Formula: Average Speed = Distance / Time = 35 km / 0.5 hours = 70 km/h
Correct Answer: 70 km/h
Question: On a speed-time graph, the speed rises in a straight line from 4 m/s at t = 0 to 16 m/s at t = 6 s. Find the total distance covered (the area under the graph).
Key Insight: The area bounded under a speed-time graph equals the total distance covered. Because the acceleration is uniform, the bounded region forms a geometric trapezium.
Trapezium Formula: Area = 0.5 * (a + b) * h, where a and b are parallel velocity states, and h is the time delta.
Substitute Values: a = 4, b = 16, h = 6. Distance = 0.5 * (4 + 16) * 6 = 0.5 * 20 * 6 = 60 metres
Correct Answer: 60 m
CHAPTER 1 COMPREHENSIVE PRACTICE EXAM
30 multiple-choice questions to the VerityPrep MCQ standard. Each question carries a collapsible solution panel: formula, values, step-by-step breakdown, boxed answer, the common trap, and an explanation for every option.
Questions & Solutions
Question 1 · Easy · Capacity
A municipal water storage facility contains exactly 4.2 m³ of liquid chemical solution. Express this absolute capacity cleanly in litres.
A42 l
B420 l
C4,200 l
D42,000 l
E420,000 l
F0.42 l
Show solution
Correct answer: C
FormulaV(l) = V(m³) × 1,000
Given4.2 m³
Bridge1 m³ = 1,000 l
4.2 × 1,000↓4,200 l
⚠ Common trap1 m³ = 1,000 l. (m³ → ml would be ×1,000,000.) Don't undercount the zeros.
Why each option
A) ×10 only
B) ×100 only
C) ✓ 4.2 × 1,000
D) ×10,000
E) treated as ml (×100,000)
F) divided instead of multiplied
Question 2 · Easy · Pressure
A constant perpendicular force of 36 Newtons is distributed evenly across a flat horizontal panel measuring 3 m by 4 m. Calculate the exact pressure exerted across this surface.
A3 N/m²
B4 N/m²
C12 N/m²
D108 N/m²
E432 N/m²
F0.33 N/m²
Show solution
Correct answer: A
FormulaP = F ÷ A
Force36 N
Area3 × 4 = 12 m²
36 ÷ 12↓3 N/m²
⚠ Common trapWork out the area (3 × 4) first; pressure divides by area, not by a single side.
Why each option
A) ✓ 36 ÷ 12
B) 12 ÷ 3 mix-up
C) that is the area, not the pressure
D) 36 × 3
E) 36 × 12
F) 12 ÷ 36 inverted
Question 3 · Easy · Unit cost
An automated factory assembly line outputs a batch of component brackets. If 120 identical brackets cost a total of £540 to manufacture, find the unit cost per individual bracket.
A£0.22
B£4.50
C£4.20
D£5.40
E£2.22
F£0.45
Show solution
Correct answer: B
FormulaUnit cost = Total ÷ Items
Total£540
Items120
540 ÷ 120↓£4.50
⚠ Common trapDivide cost by count, never count by cost.
Why each option
A) 120 ÷ 540 inverted
B) ✓ 540 ÷ 120
C) arithmetic slip
D) decimal misplaced
E) miscount
F) ÷10 error
Question 4 · Easy · Mass
A specialized laboratory scale records a chemical structural payload of 0.075 grams. Express this mass precisely in milligrams.
A0.75 mg
B7.5 mg
C75 mg
D750 mg
E7,500 mg
F0.0075 mg
Show solution
Correct answer: C
Formulam(mg) = m(g) × 1,000
Given0.075 g
Bridge1 g = 1,000 mg
0.075 × 1,000↓75 mg
⚠ Common trapg → mg is ×1,000.
Why each option
A) ×10
B) ×100
C) ✓ 0.075 × 1,000
D) ×10,000
E) ×100,000
F) ÷10
Question 5 · Easy · Area scale
An engineering blueprint maps out a factory floor plane where 1 cm² on the document scales exactly to 4 m² in real space. If an open storage unit spans 12.5 cm² on the drawing, find its true physical area.
A3.125 m²
B25 m²
C50 m²
D500 m²
E1,250 m²
F5,000 m²
Show solution
Correct answer: C
FormulaReal = Map × ratio
Map area12.5 cm²
Ratio1 cm² = 4 m²
12.5 × 4↓50 m²
⚠ Common trapThe area ratio is already given — just multiply. Don't square it again.
Why each option
A) ÷4
B) ×2
C) ✓ 12.5 × 4
D) ×40
E) ×100
F) ×400
Question 6 · Medium · Speed
An analytical sensor records a continuous laboratory test timeline where a particle moves along a track covering a total distance of 4,500 cm over a time span of exactly 1.5 minutes. Calculate the average speed of the particle in standard base units of metres per second (m/s).
A3,000 m/s
B30 m/s
C3 m/s
D0.5 m/s
E0.05 m/s
F50 m/s
Show solution
Correct answer: D
Formulav = d ÷ t
Distance4,500 cm = 45 m
Time1.5 min = 90 s
45 ÷ 90↓0.5 m/s
⚠ Common trapConvert BOTH cm → m and min → s before dividing.
Why each option
A) no conversions
B) 45 ÷ 1.5
C) time left in min
D) ✓ 45 ÷ 90
E) ×0.1 slip
F) inverted
Question 7 · Medium · Density
A raw ore sample extracted from a survey site has a bulk volume of 40 cm³ and a net density profile of 8.5 g/cm³. Find the precise mass of this sample in kilograms.
A0.34 kg
B3.4 kg
C34 kg
D340 kg
E0.034 kg
F0.0034 kg
Show solution
Correct answer: A
Formulam = ρ × V, then g → kg
Density8.5 g/cm³
Volume40 cm³
8.5 × 40 = 340 g↓340 ÷ 1,000↓0.34 kg
⚠ Common trapAnswer is in kg, so divide the grams by 1,000.
Why each option
A) ✓ 340 g = 0.34 kg
B) ÷100
C) ÷10
D) left in grams
E) ×0.1 slip
F) ×0.01 slip
Question 8 · Medium · Density conv
A high-performance lubricant fluid exhibits a specific density of 0.85 g/cm³. Convert this metric signature completely into standard SI compound units of kilograms per cubic metre (kg/m³).
A0.00085 kg/m³
B0.085 kg/m³
C8.5 kg/m³
D85 kg/m³
E850 kg/m³
F8,500 kg/m³
Show solution
Correct answer: E
Formulag/cm³ × 1,000 = kg/m³
Given0.85 g/cm³
Factor×1,000
0.85 × 1,000↓850 kg/m³
⚠ Common trapg/cm³ → kg/m³ multiplies by 1,000 (÷1,000 for mass, ×1,000,000 for volume).
Why each option
A) ÷1,000
B) ÷10
C) ×10
D) ×100
E) ✓ 0.85 × 1,000
F) ×10,000
Question 9 · Medium · Volume conv
A microfluidic flow channel possesses a precise internal volumetric capacity of 3.4 cm³. Convert this fluid metric into standard cubic millimetres (mm³).
A0.34 mm³
B34 mm³
C340 mm³
D3,400 mm³
E34,000 mm³
F0.034 mm³
Show solution
Correct answer: D
Formulacm³ × 1,000 = mm³
Given3.4 cm³
Factor(10)³ = 1,000
3.4 × 1,000↓3,400 mm³
⚠ Common trapVolume scales by the cube: 1 cm = 10 mm ⇒ ×1,000, not ×100.
Why each option
A) ÷10
B) ×10
C) ×100 (area factor)
D) ✓ 3.4 × 1,000
E) ×10,000
F) ÷100
Question 10 · Medium · Speed conv
On a highway diagnostic profile, an engine testing rig monitors a vehicle maintaining a constant speed of 25 m/s. Express this kinematic operation cleanly in kilometres per hour (km/h).
A25 km/h
B60 km/h
C75 km/h
D90 km/h
E100 km/h
F120 km/h
Show solution
Correct answer: D
Formulam/s × 3.6 = km/h
Given25 m/s
Factor×3.6
25 × 3.6↓90 km/h
⚠ Common trapm/s → km/h is ×3.6 (and ÷3.6 the other way).
Why each option
A) no conversion
B) miscalc
C) ×3
D) ✓ 25 × 3.6
E) ×4
F) ×4.8
Question 11 · Medium · Volume → capacity
A commercial shipping container has internal dimensions measuring 6 m by 2.5 m by 2 m. If the container is completely filled with a liquid payload, state the total volume of this payload in litres.
A30 l
B300 l
C3,000 l
D30,000 l
E300,000 l
F3,000,000 l
Show solution
Correct answer: D
FormulaV = l × w × h, then m³ → l
Dimensions6 × 2.5 × 2
Volume30 m³
6 × 2.5 × 2 = 30 m³↓30 × 1,000↓30,000 l
⚠ Common trapFind the volume first, then convert m³ → l (×1,000).
Why each option
A) left in m³
B) ×100
C) ×1,000 of wrong base
D) ✓ 30 × 1,000
E) ×10,000
F) ×100,000
Question 12 · Medium · Mass flow
A continuous manufacturing line monitors the processing rate of a compound. The pipeline delivers a mass payload of 18 kg every 45 seconds. Calculate the mass flow rate of this system in grams per minute (g/min).
A400 g/min
B2,400 g/min
C24,000 g/min
D40,000 g/min
E240,000 g/min
F4,000 g/min
Show solution
Correct answer: C
Formularate = m ÷ t, kg→g, per s→per min
Mass18 kg = 18,000 g
Time45 s
18,000 ÷ 45 = 400 g/s↓400 × 60↓24,000 g/min
⚠ Common trapConvert kg → g AND scale per-second to per-minute (×60).
Why each option
A) that is g/s
B) ×6 only
C) ✓ 400 × 60
D) slip
E) ×10 extra
F) miscalc
Question 13 · Medium · Area scale
An architectural model of a proposed high-rise tower is constructed using a precise linear scale factor ratio of 1:200. If the model has a calculated outer surface area of 0.45 m², evaluate the real-world surface area of the actual building in square metres.
A90 m²
B180 m²
C9,000 m²
D18,000 m²
E36,000 m²
F1,800 m²
Show solution
Correct answer: D
FormulaArea factor = (linear)²
Linear1 : 200
Area factor200² = 40,000
0.45 × 40,000↓18,000 m²
⚠ Common trapArea scales by the SQUARE of the linear ratio, not the ratio itself.
Why each option
A) ×200 (linear)
B) ×400
C) ×0.5 slip
D) ✓ 0.45 × 40,000
E) ×2 extra
F) ×4,000
Question 14 · Medium · Speed
A heavy logistics crane lifts an engineering payload a vertical height of 150 cm in exactly 0.25 minutes. Calculate the average upward velocity of the payload in millimetres per second (mm/s).
A10 mm/s
B100 mm/s
C600 mm/s
D1,000 mm/s
E60 mm/s
F6 mm/s
Show solution
Correct answer: B
Formulav = d ÷ t
Height150 cm = 1,500 mm
Time0.25 min = 15 s
1,500 ÷ 15↓100 mm/s
⚠ Common trapcm → mm (×10) and min → s (0.25 × 60 = 15 s).
Why each option
A) ÷10
B) ✓ 1,500 ÷ 15
C) ×6 slip
D) ×10 extra
E) miscalc
F) ÷100
Question 15 · Medium · Pressure → force
An industrial automated punch press evaluates an alloy sheet. The press exerts a uniform down-force pressure of 1.5 kN/m² over a rectangular die plate measuring 40 cm by 50 cm. Calculate the total downward force in Newtons.
A30 N
B300 N
C3,000 N
D150 N
E75 N
F750 N
Show solution
Correct answer: B
FormulaF = P × A
Pressure1.5 kN/m² = 1,500 N/m²
Area40 × 50 = 2,000 cm² = 0.2 m²
1,500 × 0.2↓300 N
⚠ Common trapTwo conversions: kN → N (×1,000) and cm² → m² (÷10,000).
Why each option
A) ÷10 slip
B) ✓ 1,500 × 0.2
C) forgot area conversion
D) miscalc
E) miscalc
F) ×2.5 slip
Question 16 · Hard · Speed
A pipeline infrastructure diagnostic track logs a test probe travelling a total distance of 1.8 kilometres in exactly 2.5 minutes. Find the mean velocity of this diagnostics run in metres per second (m/s).
A0.72 m/s
B12 m/s
C72 m/s
D45 m/s
E1.2 m/s
F15 m/s
Show solution
Correct answer: B
Formulav = d ÷ t
Distance1.8 km = 1,800 m
Time2.5 min = 150 s
1,800 ÷ 150↓12 m/s
⚠ Common trapkm → m (×1,000) and min → s (×60).
Why each option
A) ÷10 slip
B) ✓ 1,800 ÷ 150
C) time in min
D) miscalc
E) ×0.1 slip
F) miscalc
Question 17 · Hard · Area scale
An urban geographic survey grid uses a standard layout scale where 1 mm on the document tracks a real linear baseline distance of exactly 5 decametres (1 dam = 10 metres). If a water reservoir spans an area of 8 cm² on the survey map, calculate its absolute physical area in square metres.
A40,000 m²
B20,000 m²
C2,000,000 m²
D4,000,000 m²
E200,000 m²
F20,000,000 m²
Show solution
Correct answer: C
FormulaReal area = Map area × (linear)²
Linear1 mm = 50 m ⇒ 1 : 50,000
Area factor50,000² = 2.5×10⁹
Map area8 cm² = 800 mm²
800 × 2.5×10⁹ = 2×10¹² mm²↓÷ 10⁶ → m²↓2,000,000 m²
⚠ Common trapConvert the map area to mm² to match the 1 mm unit, then square the linear factor.
Why each option
A) linear factor only
B) under-squared
C) ✓ 2×10⁶ m²
D) ×2 extra
E) ÷10 slip
F) ×10 extra
Question 18 · Hard · Volume scale
A technical designer creates a prototype component for a hydraulic manifold using a model scale of 3:5 (the prototype is smaller than the final build piece). If the volume of the scaled-down prototype is exactly 108 cm³, find the volume of the final real-world manifold component.
A180 cm³
B300 cm³
C500 cm³
D900 cm³
E2,500 cm³
F1,500 cm³
Show solution
Correct answer: C
FormulaVolume factor = (ratio)³
Ratioprototype : final = 3 : 5
Volume factor(5/3)³ = 125/27
108 × 125/27 = 4 × 125↓500 cm³
⚠ Common trapFinal is larger, so scale UP, and cube the ratio (5/3)³.
Why each option
A) ×5/3 (linear)
B) miscalc
C) ✓ 108 × 125/27
D) miscalc
E) over-scaled
F) ×3 slip
Question 19 · Hard · Capacity
A research centrifuge isolates an emulsion layer with an evaluated volume of 0.00025 m³. State this exact fluid capacity metric in standard millilitres (ml).
A0.25 ml
B2.5 ml
C25 ml
D250 ml
E2,500 ml
F25,000 ml
Show solution
Correct answer: D
Formulam³ × 1,000,000 = ml
Given0.00025 m³
Bridge1 m³ = 10⁶ ml
0.00025 × 1,000,000↓250 ml
⚠ Common trap1 m³ = 1,000,000 ml.
Why each option
A) ÷1,000 slip
B) ÷100 slip
C) ÷10 slip
D) ✓ 0.00025 × 10⁶
E) ×10 extra
F) ×100 extra
Question 20 · Hard · Flow rate
A high-capacity drainage duct evacuates a wastewater payload at a measured volumetric flow rate of 0.012 m³/s. Convert this parameter into compound metrics of litres per minute (l/min).
A0.72 l/min
B7.2 l/min
C72 l/min
D720 l/min
E7,200 l/min
F72,000 l/min
Show solution
Correct answer: D
Formulam³/s → l/s → l/min
Given0.012 m³/s
Steps×1,000 then ×60
0.012 × 1,000 = 12 l/s↓12 × 60↓720 l/min
⚠ Common trapm³ → l (×1,000), then per-second → per-minute (×60).
Why each option
A) ÷10 slip
B) ×60 missing ×1,000
C) partial
D) ✓ 12 × 60
E) ×10 extra
F) ×100 extra
Question 21 · Challenging · Average speed
A kinematic testing run tracks an experimental drone along a straight path. The drone flies from point A to point B at a uniform speed of 60 m/s, and then immediately returns from point B back to point A along the identical flight line at a uniform speed of 40 m/s. Calculate the average speed for the entire round trip (there and back).
A50 m/s
B48 m/s
C45 m/s
D52 m/s
E0 m/s
F24 m/s
Show solution
Correct answer: B
Formulav̄ = 2 ÷ (1/v₁ + 1/v₂) (equal legs)
Out60 m/s
Back40 m/s
2 ÷ (1/60 + 1/40) = 2 ÷ (5/120)↓= 240 ÷ 5↓48 m/s
⚠ Common trapEqual distances ⇒ harmonic mean, NOT the simple average of 50. (Average VELOCITY of a round trip = 0.)
Why each option
A) arithmetic mean — wrong for equal-distance legs
B) ✓ harmonic mean
C) miscalc
D) miscalc
E) that is average velocity (displacement 0), not speed
F) halved
Question 22 · Challenging · Mixture density
An open industrial tank holds two separate unmixed fluids. Fluid Alpha has a density of 0.8 g/cm³, and Fluid Beta has a density of 1.2 g/cm³. If exactly 3 litres of Fluid Alpha are blended evenly by volume with 2 litres of Fluid Beta, find the absolute net density of the fully compounded system in kg/m³.
A960 kg/m³
B1,000 kg/m³
C2,000 kg/m³
D96 kg/m³
E1.92 kg/m³
F192 kg/m³
Show solution
Correct answer: A
Formulaρ = total mass ÷ total volume
Alpha3 l × 0.8 = 2,400 g
Beta2 l × 1.2 = 2,400 g
Totals4.8 kg over 0.005 m³
4,800 g = 4.8 kg ; 5 l = 0.005 m³↓4.8 ÷ 0.005↓960 kg/m³
⚠ Common trapCombine masses over total volume — you cannot just average the two densities.
Why each option
A) ✓ 4.8 ÷ 0.005
B) naive density average
C) double-count
D) ÷10 slip
E) unit slip
F) ×0.2 slip
Question 23 · Challenging · Flow rate
An engineer designs a high-precision chemical micro-injector. The micro-injector delivers a consistent reactant flow at a compound rate of 4.5 cubic millimetres per s (4.5 mm³/s). Convert this operating metric cleanly into standard scientific Litres per hour (l/h).
⚠ Common trapmm³ → l is ÷10⁶, then × 3,600 for per-hour.
Why each option
A) ✓ 1.62×10⁻²
B) ×10⁻³ slip
C) index error
D) index error
E) ×10 extra
F) forgot ÷10⁶
Question 24 · Challenging · Flow rate
A custom internal combustion piston chamber expands its interior gas matrix volume from a tightly compressed initial value of 150 cm³ up to a fully open volumetric capacity of 750 cm³. If this dynamic transformation takes place uniformly over an operational time frame of exactly 15 milliseconds, calculate the mean volumetric expansion rate of the gaseous system expressed in Litres per second (l/s).
A40 l/s
B4 l/s
C0.4 l/s
D400 l/s
E25 l/s
F2.5 l/s
Show solution
Correct answer: A
Formularate = ΔV ÷ Δt
ΔV750 − 150 = 600 cm³ = 0.6 l
Δt15 ms = 0.015 s
0.6 ÷ 0.015↓40 l/s
⚠ Common trapcm³ → l (÷1,000) and ms → s (÷1,000).
Why each option
A) ✓ 0.6 ÷ 0.015
B) ÷10 slip
C) ÷100 slip
D) ×10 extra
E) miscalc
F) miscalc
Question 25 · Challenging · Area + integration
A structural component template features a flat baseline plate bounded inside a Cartesian area layout. The region's absolute area boundary is defined dynamically by the compound integration of the quadratic field ∫₁⁴ (3x² − 2x + 1) dx. If every coordinate step along both axes represents exactly 5 millimetres physically, calculate the precise structural area of the actual plate component in square centimetres (cm²).
A12.75 cm²
B51 cm²
C1.275 cm²
D12.0 cm²
E25.5 cm²
F3.1875 cm²
Show solution
Correct answer: A
FormulaA = ∫₁⁴(3x²−2x+1)dx, then rescale
Integral[x³−x²+x]₁⁴ = 51 units
Unit area(5 mm)² = 0.25 cm²
(64−16+4) − (1) = 51↓51 × 0.25↓12.75 cm²
⚠ Common trapThe integral gives a coordinate-unit area; rescale by (5 mm)² = 0.25 cm² per unit.
Why each option
A) ✓ 51 × 0.25
B) unscaled (51)
C) ÷10 slip
D) rounded wrongly
E) ×0.5 slip
F) used 2.5 mm
Question 26 · Challenging · Mass flux
An advanced aircraft sensor tracks fuel flow rate down a pipeline. The monitoring system reads a mass flow rate of 0.015 kg/s passing through a cross-sectional tube area of exactly 50 mm². Evaluate the net compound fluid mass flux crossing this system boundary, expressed in Grams per square centimetre per second (g cm⁻² s⁻¹).
A30 g cm^-2 s^-1
B3 g cm^-2 s^-1
C0.3 g cm^-2 s^-1
D300 g cm^-2 s^-1
E3,000 g cm^-2 s^-1
F0.03 g cm^-2 s^-1
Show solution
Correct answer: A
Formulaflux = (m/t) ÷ A
Mass rate0.015 kg/s = 15 g/s
Area50 mm² = 0.5 cm²
15 ÷ 0.5↓30 g cm⁻² s⁻¹
⚠ Common trapkg → g (×1,000) and mm² → cm² (÷100).
Why each option
A) ✓ 15 ÷ 0.5
B) ÷10 slip
C) ÷100 slip
D) ×10 extra
E) ×100 extra
F) ÷1,000 slip
Question 27 · Medium · Pressure conv
A high-power hydraulic press features a master piston forcing oil through a manifold. The localized fluid force is tracked at a steady pressure value of 45 Megapascals (45 MPa). Convert this operating metric precisely into standard mechanical units of Newtons per square millimetre (N/mm²).
A45,000 N/mm²
B4,500 N/mm²
C450 N/mm²
D45 N/mm²
E4.5 N/mm²
F0.45 N/mm²
Show solution
Correct answer: D
Formula1 MPa = 1 N/mm²
Given45 MPa = 45×10⁶ N/m²
Bridge1 m² = 10⁶ mm²
45×10⁶ ÷ 10⁶↓45 N/mm²
⚠ Common trapHandy identity: 1 MPa = 1 N/mm² exactly.
Why each option
A) ×1,000 extra
B) ×100 extra
C) ×10 extra
D) ✓ 45
E) ÷10 slip
F) ÷100 slip
Question 28 · Challenging · Density (function)
A mechanical centrifuge container accelerates, forcing a standard composite sample down its primary chamber. The localized density metric varies across the structural profile according to the multi-variable function ρ(x) = 3000 / x³ expressed in units of kg/m³, where x is the radial distance in decimetres (dm). Evaluate the exact localized sample density state when x = 5 cm.
A24,000 kg/m³
B24,000,000 kg/m³
C2,400 kg/m³
D240,000 kg/m³
E24 kg/m³
F2.4 kg/m³
Show solution
Correct answer: A
Formulaρ(x) = 3000 / x³, x in dm
x5 cm = 0.5 dm
x³0.5³ = 0.125
3000 ÷ 0.125↓24,000 kg/m³
⚠ Common trapConvert x to dm (the function's unit) BEFORE cubing: 5 cm = 0.5 dm, not 5.
Why each option
A) ✓ 3000 ÷ 0.125
B) used x=5 cm raw etc.
C) ÷10 slip
D) ×10 extra
E) ÷1,000 slip
F) ÷10,000 slip
Question 29 · Hard · Linear density
A structural test facility features a heavy steel suspension wire of uniform thickness. Under load testing, the wire's linear mass density is tracked at a value of 0.25 Grams per millimetre (g/mm). Convert this specific operating metric cleanly into standard engineering structural units of Tonnes per kilometre (t/km).
⚠ Common trapmm → km multiplies by 10⁶; g → t divides by 10⁶ — they cancel, leaving 0.25.
Why each option
A) ×10 slip
B) ×100 slip
C) ✓ 0.25 t/km
D) ÷10 slip
E) ×1,000 slip
F) ÷100 slip
Question 30 · Hard · Flow rate
A modern vacuum chamber extracts atmospheric gas down a secondary line. The volumetric extraction flow rate is tracked at a steady rate of 0.075 cubic metres per minute (0.075 m³/min). Convert this extraction metric precisely into standard operational units of cubic centimetres per s (cm³/s).
A1,250 cm³/s
B125 cm³/s
C12,500 cm³/s
D12.5 cm³/s
E1.25 cm³/s
F125,000 cm³/s
Show solution
Correct answer: A
Formulam³/min → cm³/min → cm³/s
Given0.075 m³/min = 75,000 cm³/min
Per-second÷ 60
75,000 ÷ 60↓1,250 cm³/s
⚠ Common trapm³ → cm³ (×10⁶) and per-minute → per-second (÷60).
Why each option
A) ✓ 75,000 ÷ 60
B) ÷10 slip
C) ×10 extra
D) ÷100 slip
E) ÷1,000 slip
F) ×100 extra
Question 31 · Medium · Length (imperial→metric)
A regional survey records the length of a motorway section as 12 miles. Convert this distance into kilometres (take 1 mile = 1.609 km).
A7.46 km
B19.31 km
C1.93 km
D193.1 km
E12 km
F21.1 km
Show solution
Correct answer: B
Formulakm = miles × 1.609
Given12 miles
Factor1 mile = 1.609 km
12 × 1.609↓19.31 km
⚠ Common trapmiles → km multiplies by 1.609 (divide for km → miles).
Why each option
A) ÷1.609 (wrong direction)
B) ✓ 12 × 1.609
C) ÷10 slip
D) ×10 slip
E) no conversion
F) used 1.76 (yards mix)
Question 32 · Medium · Mass (metric→imperial)
A shipping pallet has a mass of 70 kg. Convert this mass into pounds (take 1 kg = 2.205 lb).
A31.8 lb
B35 lb
C154 lb
D1,120 lb
E70 lb
F15.4 lb
Show solution
Correct answer: C
Formulalb = kg × 2.205
Given70 kg
Factor1 kg = 2.205 lb
70 × 2.205↓154 lb
⚠ Common trap1 kg ≈ 2.205 lb (1 lb ≈ 0.454 kg). Multiply going kg → lb.
Why each option
A) ÷2.205 (wrong direction)
B) ÷2 slip
C) ✓ 70 × 2.205
D) ×16 (ounces, not pounds)
E) no conversion
F) ÷10 slip
Question 33 · Medium · Length (imperial→metric)
A structural steel beam is measured at 8 feet. Convert this length into centimetres (take 1 ft = 30.48 cm).
A243.84 cm
B96 cm
C24.38 cm
D2,438.4 cm
E20.32 cm
F731.5 cm
Show solution
Correct answer: A
Formulacm = ft × 30.48
Given8 ft
Factor1 ft = 12 in = 30.48 cm
8 × 30.48↓243.84 cm
⚠ Common trap1 ft = 30.48 cm. Don't stop at inches (×12) or use the inch factor (2.54).
Why each option
A) ✓ 8 × 30.48
B) ×12 (gives inches)
C) ÷10 slip
D) ×10 slip
E) used 2.54 (inch factor)
F) ×3 (yards mix)
Question 34 · Hard · Capacity (metric→imperial)
A laboratory reservoir holds 20 litres of solvent. Convert this volume into UK gallons (take 1 gallon = 4.546 l).
A90.92 gal
B0.44 gal
C20 gal
D4.40 gal
E5 gal
F44 gal
Show solution
Correct answer: D
Formulagallons = litres ÷ 4.546
Given20 l
Factor1 gallon = 4.546 l
20 ÷ 4.546↓4.40 gallons
⚠ Common traplitres → gallons you DIVIDE by 4.546 (multiply only for gallons → litres). The wrong direction is the classic slip.
Why each option
A) ×4.546 (wrong direction)
B) extra ÷10 slip
C) no conversion
D) ✓ 20 ÷ 4.546
E) ÷4 rough slip
F) ×10 slip
Question 35 · Hard · Speed (imperial→metric)
A vehicle telemetry log records a steady cruising speed of 60 miles per hour. Convert this speed into kilometres per hour (take 1 mile = 1.609 km).
A37.3 km/h
B96.54 km/h
C9.65 km/h
D966 km/h
E60 km/h
F216 km/h
Show solution
Correct answer: B
Formulakm/h = mph × 1.609
Given60 mph
Factor1 mile = 1.609 km
60 × 1.609↓96.54 km/h
⚠ Common trapmph → km/h multiplies by 1.609 — NOT 3.6 (that factor is for m/s → km/h).
Why each option
A) ÷1.609 (wrong direction)
B) ✓ 60 × 1.609
C) ÷10 slip
D) ×10 slip
E) no conversion
F) used 3.6 (m/s factor)
Question 36 · Hard · Applied: fuel cost
Petrol costs £1.60 per litre. A car's tank holds 12 gallons. Using 1 gallon = 4.546 l, find the cost to fill the tank from empty.
A£19.20
B£8.73
C£87.28
D£218.21
E£54.55
F£76.80
Show solution
Correct answer: C
Formulacost = (gallons × 4.546) × price per litre
Volume12 × 4.546 = 54.55 l
Price£1.60 per litre
12 × 4.546 = 54.55 l↓54.55 × 1.60↓£87.28
⚠ Common trapConvert gallons → litres FIRST, then multiply by the per-litre price. Pricing £1.60 × 12 forgets the gallon.
Why each option
A) £1.60 × 12 (ignored the gallon conversion)
B) ÷10 slip
C) ✓ 54.55 × 1.60
D) ×2.5 over-count
E) that is the volume in litres, not the cost
F) used 1 gal = 4 l
Question 37 · Hard · Applied: average speed
A train covers 90 miles in 1 hour 15 minutes. Find its average speed in km/h (take 1 mile = 1.609 km).
A116 km/h
B72 km/h
C145 km/h
D90 km/h
E93 km/h
F120 km/h
Show solution
Correct answer: A
Formulaspeed = distance ÷ time (convert miles → km, min → h)
Distance90 × 1.609 = 144.8 km
Time1 h 15 min = 1.25 h
144.8 ÷ 1.25↓116 km/h
⚠ Common trapTwo conversions: miles → km AND 1 h 15 min = 1.25 h (not 1.15).
Why each option
A) ✓ 144.8 ÷ 1.25
B) 90 ÷ 1.25 — forgot miles → km
C) used 1 h instead of 1.25 h
D) no conversion
E) treated time as 1.15 h
F) rounding error
Question 38 · Hard · Applied: unit pricing
A delicatessen sells cheese at £9.00 per kilogram. A block of cheese weighs 1.5 lb. Using 1 lb = 0.454 kg, find the cost of the block.
A£13.50
B£6.13
C£29.74
D£3.06
E£0.68
F£61.30
Show solution
Correct answer: B
Formulacost = (lb × 0.454) × price per kg
Mass1.5 × 0.454 = 0.681 kg
Price£9.00 per kg
1.5 × 0.454 = 0.681 kg↓0.681 × 9.00↓£6.13
⚠ Common trapPrice is per KILOGRAM but the weight is in pounds — convert lb → kg before multiplying.
Why each option
A) £9 × 1.5 (treated lb as kg)
B) ✓ 0.681 × 9.00
C) used 16 oz factor wrongly
D) halved
E) mass only — forgot the price
F) ×10 slip
Question 39 · Hard · Applied: mass of contents
A 5-gallon drum is filled completely with paint of density 1.3 kg/l. Using 1 gallon = 4.546 l, find the total mass of paint in kilograms.
A6.5 kg
B295.5 kg
C17.5 kg
D29.55 kg
E22.73 kg
F2.96 kg
Show solution
Correct answer: D
Formulamass = (gallons × 4.546) × density
Volume5 × 4.546 = 22.73 l
Density1.3 kg/l
5 × 4.546 = 22.73 l↓22.73 × 1.3↓29.55 kg
⚠ Common trapConvert gallons → litres FIRST, then multiply by density. 1.3 × 5 forgets the gallon conversion.
Rounding to decimal places (d.p.) counts digits after the point. Rounding to significant figures (s.f.) counts from the first non-zero digit. Leading zeros are never significant; trailing zeros after the point are.
To estimate, round each number to 1 significant figure and keep the operations in the same order. Estimation is your fastest tool for rejecting impossible options: if the estimate is ?500, an option of 50 or 5,000 can be discarded on sight.
Dividing by a number less than 1 makes a result bigger (÷0.2 = ×5). And an upper bound such as 24.5 is excluded from the interval, because 24.5 itself rounds up.
To add or subtract, use a common denominator. To multiply, multiply tops and bottoms. To divide, multiply by the reciprocal (flip the second fraction). Always simplify the result.
The fastest percentage technique is the multiplier. A 15% increase is ×1.15; a 15% decrease is ×0.85. Repeated (compound) change raises the multiplier to a power, and a reverse percentage undoes one by dividing.
If a price is £92 after a 15% rise, the original is 92 ÷ 1.15, not 92 × 0.85. The percentage always refers to the original amount, so you divide to undo it.
A number in standard form is written as A × 10n, where 1 ? A < 10 and n is an integer. To multiply, multiply the front numbers and add the indices; to divide, divide the front numbers and subtract the indices — then fix the front number back into the 1–10 range if needed.
First, subtracting a negative index adds: 105 ÷ 10-3 = 108. Second, the front number must be between 1 and 10 — answers like 0.4 × 108 or 40 × 107 are not in standard form and need adjusting.
M1-03-Ratio and Proportion
CHAPTER 3: Ratio and Proportion
Ratio and proportion describe how quantities relate and scale together — the backbone of mixing, pricing, scaling drawings, and the compound measures met in Chapter 1. As always in ESAT Mathematics, all of it must be done without a calculator.
3.1 Ratio
A ratio compares quantities of the same kind. To simplify, divide every part by their highest common factor (and make sure all parts are in the same units first). To share a quantity in a ratio, add the parts to find the value of one part, then multiply.
📋 Ratio Toolkit
Task
Method
Simplify a : b
Divide both by the HCF; clear units, fractions and decimals first.
Share in a ratio
Total ÷ (sum of parts) = value of one part; then multiply each part.
Ratio ↔ fraction
In a : b, the first part is a/(a + b) of the whole.
Combine a:b and b:c
Scale each so the shared term b matches (its LCM), then join to a : b : c.
⚠️ Difference vs total
If a problem gives a difference ("Cem gets 18 more than Ali"), that 18 equals the difference in PARTS (here 7 − 4 = 3 parts), not one share and not the total. Always find the value of one part first.
Worked Example — Share and combine
Question: Share £720 in the ratio 2 : 3 : 4, and combine x : y = 2 : 3 with y : z = 6 : 5.
Working: Parts 2 + 3 + 4 = 9, so one part = £80, giving £160 : £240 : £320. To combine, scale y to match: 2 : 3 → 4 : 6 and 6 : 5 stays, so x : y : z = 4 : 6 : 5.
⚠️ Common trapThe common term (cement) must be EQUAL in both ratios before joining — make it 6 in each.
Why each option
A) stacked the ratios without scaling
B) ✓ scaled both to cement = 6, then joined
C) scaled water but not sand
D) scaled sand but not cement
E) correct numbers, reversed order
F) used cement instead of water
3.2 Direct and Inverse Proportion
Two quantities are in direct proportion when one is a constant multiple of the other (y = kx): double one, double the other. They are in inverse proportion when their product is constant (y = k/x): double one, halve the other. The method is always the same — find the constant k from a known pair, then use it.
📐 Proportion Types
Statement
Equation
y proportional to x
y = kx
y proportional to x² (or x³)
y = kx² (or kx³)
y proportional to √x
y = k√x
y inversely proportional to x
y = k ÷ x (so xy = k)
y inversely proportional to x²
y = k ÷ x²
⚠️ Power and inverse traps
For y ∝ x², square x before multiplying by k. For inverse proportion, the PRODUCT xy stays constant — more of one means less of the other, so never scale them in the same direction.
Worked Example — Direct (square) and inverse
Question: P ∝ Q²; when Q = 2, P = 20. Find P when Q = 5. Also: R ∝ 1/S; when S = 8, R = 3. Find R when S = 12.
Working: P = kQ² → 20 = k×4 → k = 5, so P = 5×25 = 125. Inverse: k = RS = 24, so R = 24 ÷ 12 = 2.
P = 5Q² → P = 5 × 25 = 125 ; RS = 24 → R = 24 ÷ 12 = 2
Section 3.2 Practice · Hard · Direct proportion to a square
y is directly proportional to the square of x. When x = 3, y = 36. Find y when x = 5.
A60
B20
C400
D100
E75
F144
Show solution
Correct answer: D
Methody = kx². Find k from the known pair, then substitute.
Find k36 = k × 3² ⇒ k = 4
New xx = 5
k = 36 ÷ 9 = 4↓y = 4 × 5²↓y = 100
⚠️ Common trapy ∝ x² means SQUARE x before multiplying by k. Treating it as y ∝ x (linear) gives 60.
Why each option
A) used y ∝ x (linear): 36 ÷ 3 × 5
B) divided instead of multiplying
C) squared after multiplying by k
D) ✓ k = 4, then 4 × 25
E) scaled 36 by 5/3 once
F) used x² with no k (just 12²)
Section 3.2 Practice · Very Hard · Inverse proportion
The time t to fill a tank is inversely proportional to the number of pumps p. With 6 pumps it takes 40 minutes. How long would it take with 15 pumps?
A16 minutes
B100 minutes
C10 minutes
D240 minutes
E9 minutes
F24 minutes
Show solution
Correct answer: A
Methodt = k ÷ p (inverse). The product t × p is constant.
Find kk = 40 × 6 = 240
New pp = 15
k = t × p = 240↓t = 240 ÷ 15↓16 minutes
⚠️ Common trapInverse means MORE pumps ⇒ LESS time. Keep t × p = 240 constant; do not scale t up with p.
Why each option
A) ✓ 240 ÷ 15
B) scaled directly: 40 × 15 ÷ 6
C) used 240 ÷ 24
D) gave the constant k
E) used 6/15 × something
F) used 240 ÷ 10
3.3 Proportional Reasoning in Context
Most ratio questions are dressed as real situations: best-value shopping, map scales, currency exchange and recipe scaling. The unlocking idea is always to reduce everything to a common basis — a price per fixed amount, a single scale factor, or a rate per unit.
📋 Common-Basis Methods
Context
Method
Best value
Find cost per fixed amount (e.g. per 100 g); cheapest rate wins.
Map / scale drawing
Real = map × scale; then convert units (1 km = 100,000 cm).
Recipe / mixture scaling
One scale factor (new ÷ old) applied to every ingredient.
Currency exchange
Multiply by the rate one way, divide by it the other.
Worked Example — Best value and exchange
Question: Is 1.5 kg of rice for £2.70 better value than 2 kg for £3.80? And convert £150 to euros at £1 = €1.16.
Working: 270 ÷ 1.5 = 180 p/kg vs 380 ÷ 2 = 190 p/kg, so the 1.5 kg bag is cheaper. Exchange: 150 × 1.16 = €174.
Coffee is sold as a 250 g jar for £4.20 or a 400 g jar for £6.40. Which jar is better value, and by how much per 100 g?
A250 g jar, by 8 p per 100 g
B400 g jar, by 8 p per 100 g
C400 g jar, by 2 p per 100 g
Dthe jars are equal value
E400 g jar, by 80 p per 100 g
F250 g jar, by 2 p per 100 g
Show solution
Correct answer: B
MethodCompare cost per 100 g = (price ÷ mass) × 100.
250 g jar420 ÷ 250 × 100 = 168 p
400 g jar640 ÷ 400 × 100 = 160 p
168 p vs 160 p per 100 g↓168 − 160↓400 g jar, 8 p/100g cheaper
⚠️ Common trapCompare a COMMON amount (per 100 g). A bigger jar is not automatically cheaper — here it is, by 8 p.
Why each option
A) right gap, wrong jar
B) ✓ 160 p vs 168 p per 100 g
C) arithmetic slip
D) the rates differ (168 vs 160)
E) ×10 error
F) wrong jar and slip
Section 3.3 Practice · Very Hard · Map scale
A map has a scale of 1 : 25,000. Two towns are 14 cm apart on the map, as shown. Find the real distance between them in kilometres.
A3.5 km
B35 km
C0.35 km
D350 km
E1.75 km
F3.5 m
Show solution
Correct answer: A
Methodreal distance = map distance × scale, then convert cm → km.
Scaled14 × 25,000 = 350,000 cm
To km÷ 100,000
14 × 25,000 = 350,000 cm↓350,000 ÷ 100,000↓3.5 km
⚠️ Common trap1 km = 100,000 cm. After scaling to centimetres, divide by 100,000 (not 1,000) to reach km.
Why each option
A) ✓ 350,000 cm ÷ 100,000
B) divided by 10,000
C) divided by 1,000,000
D) forgot a conversion step
E) used a 1 : 12,500 scale
F) stopped at metres / mislabelled
Section 3.3 Practice · Applied · Scaling a recipe
A recipe for 8 people uses 600 g of flour and 200 g of sugar. A baker scales it up for 20 people. How much flour and sugar are needed?
A1,200 g flour, 400 g sugar
B1,500 g flour, 400 g sugar
C1,500 g flour, 500 g sugar
D750 g flour, 250 g sugar
E2,400 g flour, 800 g sugar
F1,500 g flour, 450 g sugar
Show solution
Correct answer: C
MethodFind one scale factor (new ÷ old) and apply it to every ingredient.
Scale factor20 ÷ 8 = 2.5
Apply×2.5 to each
20 ÷ 8 = 2.5↓600 × 2.5 = 1,500 ; 200 × 2.5 = 500↓1,500 g flour, 500 g sugar
⚠️ Common trapUse a single scale factor (2.5) for ALL ingredients. Scaling flour and sugar by different amounts is the error.
Why each option
A) used ×2
B) flour ×2.5 but sugar ×2
C) ✓ ×2.5 on both
D) scaled down by mistake
E) used ×4
F) sugar arithmetic slip
CHAPTER 3 REVISION EXAM
Revision exam questions for 03-Ratio and Proportion.
M1-04-Algebra and Functions
CHAPTER 4: Algebra and Functions
Algebra is the language the rest of ESAT Mathematics is written in: every science module rearranges formulae, solves equations and reads graphs. This chapter builds that fluency in three stages — manipulating expressions, solving equations and inequalities, and working with sequences and graphs — all to be done confidently and without a calculator.
4.1 Manipulating Algebraic Expressions
Two opposite skills dominate this section: expanding (removing brackets) and factorising (putting brackets back). You also need to simplify algebraic fractions and rearrange a formula to make a different letter the subject.
Expanding brackets
To expand two brackets, multiply every term in the first by every term in the second (often remembered as FOIL: First, Outer, Inner, Last) and then collect like terms. A perfect square is just a double bracket with itself: (x + a)² = x² + 2ax + a², so the middle term is twice the product — a step candidates routinely drop. When a minus sign sits in front of a bracket, it multiplies every term inside: −(x² − 8x + 16) = −x² + 8x − 16.
Factorising
Always look for a common factor first. For a quadratic x² + bx + c, find two numbers that multiply to c and add to b. When the leading coefficient is not 1 (say ax² + bx + c), use the split-the-middle-term method: find two numbers with product a·c and sum b, split the middle term, then factorise in pairs. Recognise the difference of two squares instantly: x² − a² = (x + a)(x − a), which also works for 9x² − 25 = (3x − 5)(3x + 5).
📋 Expand & Factorise Toolkit
Type
Pattern
Double brackets
(x + a)(x + b) = x² + (a + b)x + ab
Perfect square
(x + a)² = x² + 2ax + a²
Difference of two squares
x² − a² = (x + a)(x − a)
Common factor
Take out the highest factor of every term first.
Quadratic, a ≠ 1
Split the middle term using product a·c, sum b.
Algebraic fractions and rearranging
To simplify a fraction such as (x² − 9)/(x² − x − 6), factorise top and bottom and cancel common factors — never loose terms. To add or subtract fractions, put them over a common denominator (multiply each numerator by the other denominator). To rearrange a formula, undo the operations in reverse order; if the new subject appears twice, gather those terms on one side and factorise the letter out before dividing.
⚠️ Three classic slips
(1) Expanding −(x − 4) gives −x + 4, not −x − 4. (2) You can only cancel whole factors of a fraction, never individual terms. (3) When rearranging leaves the subject on both sides, you must factorise it out — you cannot simply divide.
Worked Example 1 — Expand, factorise, rearrange
Question: Expand (2x − 3)², factorise 9x² − 25, and make r the subject of A = πr².
Working: (2x − 3)² = 4x² − 12x + 9 (the middle term is 2 × 2x × −3). 9x² − 25 is a difference of two squares = (3x − 5)(3x + 5). For A = πr², divide by π then square-root: r = √(A/π).
⚠️ Common trapx ends up on both sides — factorise x out before dividing.
Why each option
A) ✓ collected x and factorised
B) sign error on the constant
C) moved terms the wrong way
D) sign error in the bracket
E) forgot to square / clear
F) factorised −(1 − y²) wrongly
4.2 Equations and Inequalities
A linear equation is solved by doing the same operation to both sides until x is alone. A quadratic offers three routes; choosing the quickest saves precious exam time.
Three ways to solve a quadratic
Factorising is fastest when the quadratic factorises with whole numbers — set each bracket to zero. When it doesn't factorise, use the quadratic formula x = [−b ± √(b² − 4ac)]/2a; identify a, b and c carefully (including their signs) and remember that the discriminant b² − 4ac tells you how many real roots there are (positive: two, zero: one, negative: none). Completing the square writes x² + bx + c as (x + b/2)² − (b/2)² + c; it solves the equation in exact form and, as a bonus, reveals the turning point of the curve.
Simultaneous equations
For two linear equations, use elimination: make the coefficients of one variable match, then add or subtract to remove it. For a linear and a quadratic equation, use substitution: rearrange the linear equation and substitute it into the quadratic, producing a single quadratic to solve.
Inequalities
Linear inequalities are solved exactly like equations, with one rule: multiplying or dividing by a negative number reverses the inequality sign. For a quadratic inequality, find the roots (critical values) and decide which region satisfies it — below the axis (between the roots) for < 0, and the two outer regions for > 0.
📐 Equation Essentials
Tool
Rule
Quadratic formula
x = [ −b ± √(b² − 4ac) ] ÷ 2a
Discriminant b² − 4ac
>0 two roots, =0 one root, <0 none
Completing the square
x² + bx + c = (x + b/2)² − (b/2)² + c
Linear + quadratic pair
Substitute the linear equation into the quadratic.
Inequalities
Multiplying/dividing by a negative reverses the sign.
⚠️ Sign and root traps
When dividing an inequality by a negative number, flip the sign. A quadratic gives two solutions (unless the discriminant is zero). And in b² − 4ac, watch double negatives: 25 − 4(3)(−1) = 25 + 12 = 37.
Worked Example 3 — Quadratic by formula
Question: Solve x² − 4x − 1 = 0 in surd form.
Working: a = 1, b = −4, c = −1. Discriminant = 16 + 4 = 20. x = [4 ± √20]/2 = [4 ± 2√5]/2 = 2 ± √5.
x = [4 ± √20]/2 = 2 ± √5
Worked Example 4 — Simultaneous (linear & quadratic)
MethodFactorise by splitting the middle term, then set each factor to zero.
Product / sum−30 and +1
Factors(2x − 5)(x + 3)
2x² + 6x − 5x − 15↓(2x − 5)(x + 3) = 0↓x = 5/2 or x = −3
⚠️ Common trapBoth roots are required, and (2x − 5) = 0 gives x = 5/2, not x = 5.
Why each option
A) signs reversed
B) didn't divide the 2x − 5 factor
C) wrong factor pair
D) both roots negative
E) misused the constant
F) ✓ (2x − 5)(x + 3) = 0
Section 4.2 Practice · Very Hard · Quadratic formula
Solve 3x² − 5x − 1 = 0, giving your answer in exact (surd) form.
Ax = (5 ± √13)/6
Bx = (−5 ± √37)/6
Cx = (5 ± √37)/6
Dx = (5 ± √37)/3
Ex = (5 ± √37)/2
Fx = (5 ± √61)/6
Show solution
Correct answer: C
Methodx = [ −b ± √(b² − 4ac) ] ÷ 2a, with a = 3, b = −5, c = −1.
Discriminant(−5)² − 4(3)(−1) = 25 + 12 = 37
Denominator2a = 6
x = [5 ± √37] / 6↓x = (5 ± √37)/6
⚠️ Common trapb² − 4ac = 25 − (−12) = 37 — subtracting a negative ADDS. And −b = +5.
Why each option
A) used 25 − 12
B) sign of −b wrong
C) ✓ 25 + 12 = 37 under the root
D) used a, not 2a
E) forgot the 3 in 2a
F) added 4ac the wrong way
Section 4.2 Practice · Very Hard · Completing the square
Solve x² + 6x − 4 = 0 by completing the square, giving exact answers.
Ax = −3 ± √13
Bx = 3 ± √13
Cx = −3 ± √5
Dx = −3 + √13 only
Ex = −6 ± √13
Fx = −3 ± 13
Show solution
Correct answer: A
MethodWrite x² + 6x as (x + 3)² − 9, then solve.
Complete(x + 3)² − 9 − 4 = 0
Simplify(x + 3)² = 13
(x + 3)² = 13↓x + 3 = ±√13↓x = −3 ± √13
⚠️ Common trapTake ± when square-rooting, and remember the −9 from completing the square joins the −4.
Why each option
A) ✓ (x + 3)² = 13
B) sign of the shift wrong
C) forgot to add the −9
D) dropped the negative root
E) used b instead of b/2
F) forgot the square root
Section 4.2 Practice · Hard · Linear inequality
Solve the inequality 4 − 3x ≥ 19.
Ax ≥ −5
Bx ≤ 5
Cx ≥ 5
Dx ≤ −5
Ex ≤ −7.67
Fx ≥ −15
Show solution
Correct answer: D
MethodIsolate x; dividing by a negative reverses the inequality.
Subtract 4−3x ≥ 15
Divide by −3reverse the sign
−3x ≥ 15↓x ≤ −5↓x ≤ −5
⚠️ Common trapDividing both sides by −3 FLIPS ≥ to ≤. Forgetting this is the single most common error here.
Why each option
A) forgot to flip the inequality
B) dropped the negative sign
C) sign and flip both wrong
D) ✓ −3x ≥ 15, divide by −3 and flip
E) subtracted wrongly
F) forgot to divide by 3
Section 4.2 Practice · Hard · Simultaneous (linear)
Solve the simultaneous equations 5x + 2y = 24 and 3x − 2y = 8.
Ax = 4, y = 4
Bx = 4, y = 2
Cx = 2, y = 7
Dx = 3, y = 4.5
Ex = 4, y = −2
Fx = 8, y = −8
Show solution
Correct answer: B
MethodAdd the equations to eliminate y, solve for x, then back-substitute.
Add8x = 32 ⇒ x = 4
Back-substitute20 + 2y = 24
8x = 32 ⇒ x = 4↓2y = 4 ⇒ y = 2↓x = 4, y = 2
⚠️ Common trapThe +2y and −2y cancel when you ADD the equations; subtracting them would keep y.
Why each option
A) arithmetic slip in y
B) ✓ added to eliminate y
C) subtracted the equations
D) wrong elimination
E) sign error in y
F) forgot to halve
Section 4.2 Practice · Very Hard · Quadratic inequality
Solve the inequality x² − x − 6 < 0.
Ax < −2 or x > 3
B−3 < x < 2
C2 < x < 3
Dx < −2 and x < 3
E−2 < x < 3
F−2 ≤ x ≤ 3
Show solution
Correct answer: E
MethodFactorise, find the critical values, then test where the curve is below the axis.
Factorise(x − 3)(x + 2) < 0
Critical valuesx = 3 and x = −2
roots at −2 and 3↓curve is below the axis between the roots↓−2 < x < 3
⚠️ Common trapFor < 0 the answer is BETWEEN the roots (the dip below the axis). For > 0 it would be the two outer pieces.
Why each option
A) that is the > 0 region
B) factorised with wrong signs
C) used wrong roots
D) wrote it as a conjunction wrongly
E) ✓ between the roots, where the parabola dips below 0
F) used ≤ for a strict <
Section 4.2 Practice · Very Hard · Simultaneous (linear & quadratic)
Find the x-coordinates where y = x² − 2x − 3 meets the line y = 4x − 8.
Ax = −1 or x = −5
Bx = 2 or x = 3
Cx = 1 or x = 5
Dx = 1 or x = −5
Ex = −1 or x = 5
Fx = 3 only
Show solution
Correct answer: C
MethodSet the expressions equal and solve the resulting quadratic.
Set equalx² − 2x − 3 = 4x − 8
Rearrangex² − 6x + 5 = 0
x² − 6x + 5 = 0↓(x − 1)(x − 5) = 0↓x = 1 or x = 5
⚠️ Common trapMove everything to one side first; subtracting 4x − 8 flips both signs.
Why each option
A) sign error in factorising
B) factorised x² − 5x + 6 by mistake
C) ✓ (x − 1)(x − 5) = 0
D) one sign wrong
E) one sign wrong
F) took the line gradient as a root
4.3 Sequences and Graphs
A sequence rule can be term-to-term (how to get from one term to the next) or position-to-term (a formula for the nth term). Graphs then connect algebra to geometry.
Finding the nth term
For a linear (arithmetic) sequence the first differences are constant: the nth term is (common difference)·n adjusted by a constant. For a quadratic sequence the second differences are constant: the coefficient of n² is half the second difference. Subtract that n² part from the sequence and the remainder is a simple linear sequence you can finish off. A geometric sequence multiplies by a constant ratio, so its nth term is a·r^(n−1).
Straight-line graphs
Every straight line is y = mx + c, where m is the gradient (change in y ÷ change in x) and c is the y-intercept. Parallel lines share the same gradient; perpendicular lines have gradients whose product is −1, so the perpendicular gradient is the negative reciprocal −1/m.
Quadratic graphs
A parabola y = ax² + bx + c crosses the x-axis at the roots (the solutions of the equation = 0). Its turning point comes straight from completing the square: y = (x − p)² + q has vertex (p, q), and the line of symmetry is x = p.
📋 Sequences & Graphs
Tool
Method
Linear nth term
(common difference)·n + (first term − common difference)
Quadratic nth term
n² coefficient = (second difference) ÷ 2, then fit the rest
Geometric nth term
a · r^(n − 1)
Gradient / line
m = Δy/Δx; line y = mx + c
Perpendicular gradient
−1/m (negative reciprocal)
Turning point
(x − p)² + q ⇒ vertex (p, q)
⚠️ Halve it, reciprocate it, flip the sign
The n² coefficient is half the second difference. A perpendicular gradient is the negative reciprocal −1/m (not just −m). And for a turning point (x − p)² + q the vertex x-value is +p — the opposite sign to the bracket.
Worked Example 5 — Quadratic nth term
Question: Find the nth term of 2, 7, 14, 23, …
Working: First differences 5, 7, 9; second difference 2, so the n² term is n². Subtracting n² (1, 4, 9, 16) leaves 1, 3, 5, 7 = 2n − 1. So the nth term is n² + 2n − 1.
nth term = n² + 2n − 1
Worked Example 6 — Turning point
Question: Find the turning point of y = x² + 4x − 1.
Working: Complete the square: (x + 2)² − 4 − 1 = (x + 2)² − 5. Vertex (−2, −5), a minimum, with line of symmetry x = −2.
y = (x + 2)² − 5 → vertex (−2, −5)
Section 4.3 Practice · Hard · Linear sequence
Find the nth-term rule for the sequence 5, 8, 11, 14, …
A3n + 2
B3n + 5
C3n − 2
Dn + 3
E3n
F2n + 3
Show solution
Correct answer: A
Methodnth term = (common difference)·n + (first term − common difference).
Common difference3
Adjust5 − 3 = 2
3n + 2↓3n + 2
⚠️ Common trapThe constant is first term minus the common difference (5 − 3 = 2), not just the first term.
Why each option
A) ✓ 3n then +2 to fix the start
B) used the first term as the constant
C) sign of the constant wrong
D) used the wrong gradient
E) forgot the constant
F) swapped gradient and constant
Section 4.3 Practice · Hard · Quadratic sequence
Find the nth-term rule for the sequence 3, 10, 21, 36, 55, …
A2n² + 1
Bn² + 2n
C2n² − n
D4n² − 1
E2n² + 2n − 1
F2n² + n
Show solution
Correct answer: F
Methodn² coefficient = (second difference) ÷ 2; subtract and fit the rest.
1st differences7, 11, 15, 19
2nd difference4 ⇒ 2n²
subtract 2n² (2, 8, 18, 32, 50): leaves 1, 2, 3, 4, 5↓remainder is n↓2n² + n
⚠️ Common trapHalve the second difference for the n² term (4 ÷ 2 = 2), then find the leftover.
Why each option
A) didn't finish the remainder
B) wrong n² coefficient
C) sign of the linear term wrong
D) used the second difference directly
E) over-corrected
F) ✓ 2n² leaves 1,2,3,4,5 = n
Section 4.3 Practice · Hard · Geometric sequence
The sequence 2, 6, 18, 54, … continues with the same rule. Find its 6th term.
A162
B486
C1,458
D108
E18
F162.0
Show solution
Correct answer: B
MethodTerm-to-term rule is ×3; nth term = 2 × 3^(n−1).
Common ratio×3
6th term2 × 3⁵
3⁵ = 243↓2 × 243↓486
⚠️ Common trapIt multiplies by 3 each time (not adds). The 6th term uses 3⁵, since the 1st term uses 3⁰.
Why each option
A) stopped at the 5th term
B) ✓ 2 × 3⁵
C) used 3⁶
D) treated it as adding
E) gave the 3rd term
F) off by one power
Section 4.3 Practice · Very Hard · Straight-line graph
The straight line shown passes through (−1, 5) and (3, −3). Find its equation in the form y = mx + c.
Ay = 2x + 3
By = −2x − 3
Cy = −½x + 3
Dy = −2x + 3
Ey = −2x + 5
Fy = 2x − 3
Show solution
Correct answer: D
Methodm = (change in y) ÷ (change in x); then find c by substitution.
Gradient(−3 − 5)/(3 − (−1)) = −8/4 = −2
Find c−3 = −2(3) + c
m = −2↓−3 = −6 + c ⇒ c = 3↓y = −2x + 3
⚠️ Common trapGradient is negative (line falls); mind the double negative in 3 − (−1) = 4.
Why each option
A) missed the negative gradient
B) sign of c wrong
C) inverted the gradient
D) ✓ m = −2, c = 3
E) read c off the wrong point
F) both signs wrong
Section 4.3 Practice · Very Hard · Perpendicular lines
A line is perpendicular to y = 2x − 1 and passes through the point (4, 3). Find its equation.
Ay = −½x + 5
By = −2x + 5
Cy = 2x + 5
Dy = −½x − 5
Ey = ½x + 1
Fy = −½x + 3
Show solution
Correct answer: A
MethodPerpendicular gradient is the negative reciprocal: −1 ÷ 2. Then find c.
Perp. gradient−1/2
Find c3 = −½(4) + c
m = −1/2↓3 = −2 + c ⇒ c = 5↓y = −½x + 5
⚠️ Common trapPerpendicular gradient is −1/m (here −1/2), not just the negative (−2) or the same (2).
Why each option
A) ✓ negative reciprocal −1/2, c = 5
B) negated instead of reciprocal
C) used the same gradient
D) sign of c wrong
E) reciprocal without the minus
F) read c off (4,3) wrongly
Section 4.3 Practice · Very Hard · Completing the square
By completing the square, find the coordinates of the turning point of y = x² − 6x + 11, shown below.
A(−3, 2)
B(3, −2)
C(6, 11)
D(3, 11)
E(3, 2)
F(−3, −2)
Show solution
Correct answer: E
Methodx² + bx + c = (x + b/2)² − (b/2)² + c; vertex is (−b/2, …).
Complete(x − 3)² − 9 + 11
Simplify(x − 3)² + 2
y = (x − 3)² + 2↓minimum at (x − 3)² = 0↓(3, 2)
⚠️ Common trapThe turning point is at x = +3 (from x − 3 = 0); the y-value is the constant left over, +2.
Why each option
A) sign of x-coordinate wrong
B) sign of y-coordinate wrong
C) read off coefficients directly
D) used the original constant 11
E) ✓ (x − 3)² + 2, minimum at x = 3
F) both signs wrong
Section 4.3 Practice · Very Hard · Reading a quadratic graph
The graph shows y = x² + 2x − 3. Use it to write down the solutions of x² + 2x − 3 = 0.
Ax = 3 and x = −1
Bx = −1 and x = −4
Cx = −3 and x = 1
Dx = −3 only
Ex = 1 and x = 3
Fx = 0 and x = −2
Show solution
Correct answer: C
MethodSolutions of = 0 are the x-values where the curve crosses the x-axis.
x-axis crossingsx = −3 and x = 1
Check(x + 3)(x − 1) = 0
curve cuts the x-axis at −3 and 1↓factorises as (x + 3)(x − 1)↓x = −3 and x = 1
⚠️ Common trapThe solutions of = 0 are where y = 0 (the x-axis), not the turning point or the y-intercept (−3).
Why each option
A) signs reversed
B) read off the vertex
C) ✓ the two x-axis crossings
D) gave the y-intercept value
E) one root wrong
F) misread the crossings
CHAPTER 4 REVISION EXAM
Revision exam questions for 04-Algebra and Functions.
M1-05-Geometry
CHAPTER 5: Geometry
Geometry rewards a clear diagram and a small set of reliable rules. This chapter works through angles and polygons, Pythagoras and trigonometry, circle theorems, and mensuration (area, surface area and volume). Every result here must be reached without a calculator, so the exact trigonometric values and the key formulae below are worth committing to memory.
5.1 Angles and Polygons
Most angle problems are solved by chaining a few basic facts. Angles on a straight line add to 180°, angles around a point to 360°, and vertically opposite angles are equal.
Parallel lines
When a transversal crosses two parallel lines, three relationships appear: corresponding angles (in an F-shape) are equal, alternate angles (in a Z-shape) are equal, and co-interior angles (in a C-shape) add to 180°. Spotting which shape you are looking at is the whole skill.
Triangles and polygons
The angles of any triangle sum to 180°, and an exterior angle equals the sum of the two opposite interior angles. For a polygon with n sides, the interior angles sum to (n − 2) × 180°, while the exterior angles always sum to 360°. A regular polygon therefore has each exterior angle equal to 360°/n.
📋 Angle Rules
Situation
Rule
Straight line / around a point
180° / 360°
Parallel: corresponding, alternate
equal
Parallel: co-interior
add to 180°
Triangle / exterior angle
sum 180° / = two opposite interiors
Polygon interior sum
(n − 2) × 180°
Polygon exterior sum / regular
360° / each = 360°/n
⚠️ Equal or supplementary?
Corresponding and alternate angles are equal, but co-interior angles are supplementary (sum 180°). And a pentagon's angles sum to 540°, not 360° — always use (n − 2) × 180°.
Worked Example — Regular polygon
Question: Find one interior angle of a regular octagon.
Working: Each exterior angle is 360° ÷ 8 = 45°, so each interior angle is 180° − 45° = 135°. (Check: sum = (8 − 2) × 180 = 1080°, and 1080 ÷ 8 = 135°.)
The two horizontal lines are parallel and a transversal crosses them. Find the size of angle x.
A70°
B20°
C110°
D290°
E130°
F35°
Show solution
Correct answer: C
MethodCo-interior (allied) angles between parallel lines add to 180°.
Given angle70°
Relationshipco-interior
x = 180° − 70°↓x = 110°
⚠️ Common trapThese are co-interior (C-shaped), so they SUM to 180° — they are not equal.
Why each option
A) assumed they were equal
B) used 90 − 70
C) ✓ 180 − 70 co-interior
D) reflex by mistake
E) subtracted from 200
F) halved 70
Section 5.1 Practice · Hard · Exterior angle
In the triangle shown, find the exterior angle x.
A120°
B60°
C125°
D55°
E115°
F110°
Show solution
Correct answer: A
MethodAn exterior angle equals the sum of the two opposite interior angles.
Interior angles55° and 65°
Ruleexterior = sum of remote interiors
x = 55° + 65°↓x = 120°
⚠️ Common trapThe exterior angle equals the two FAR interior angles added — not 180 minus one of them by itself.
Why each option
A) ✓ 55 + 65
B) 180 − 120 (interior)
C) added 55 + 70
D) copied one interior angle
E) added wrong pair
F) used 180 − 70
Section 5.1 Practice · Hard · Polygon angle sum
A pentagon has interior angles of 100°, 110°, 95°, 130° and x. Find x.
A75°
B125°
C108°
D145°
E105°
F90°
Show solution
Correct answer: E
MethodInterior angles of an n-gon sum to (n − 2) × 180°.
Pentagon sum(5 − 2) × 180 = 540°
Known total100+110+95+130 = 435°
x = 540 − 435↓x = 105°
⚠️ Common trapA pentagon's angles sum to 540°, not 360°. Subtract the four known angles.
Why each option
A) used 360° sum
B) used 720° (hexagon)
C) gave a regular pentagon angle
D) arithmetic slip
E) ✓ 540 − 435
F) guessed
Section 5.1 Practice · Hard · Regular polygon
Find the size of one interior angle x of the regular decagon shown.
A36°
B144°
C1440°
D120°
E162°
F140°
Show solution
Correct answer: B
MethodInterior angle = [(n − 2) × 180°] ÷ n.
n10
Sum(10 − 2) × 180 = 1440°
1440 ÷ 10↓x = 144°
⚠️ Common trapDivide the angle SUM (1440°) by 10. Don't use the exterior angle (36°) by mistake.
Why each option
A) that is the exterior angle
B) ✓ 1440 ÷ 10
C) gave the sum, not one angle
D) used a hexagon
E) used a 20-gon
F) used n = 9
Section 5.1 Practice · Very Hard · Sides from exterior angle
The exterior angle of a regular polygon is 24°. How many sides does it have?
A24 sides
B12 sides
C8 sides
D15 sides
E10 sides
F18 sides
Show solution
Correct answer: D
MethodExterior angles sum to 360°, so n = 360° ÷ (exterior angle).
Exterior angle24°
Sum360°
360 ÷ 24↓15 sides
⚠️ Common trapUse the EXTERIOR-angle rule (sum 360°). Trying the interior-angle formula here is a long detour.
Why each option
A) used the angle as the count
B) used 288 ÷ 24
C) used 192 ÷ 24
D) ✓ 360 ÷ 24
E) used 240 ÷ 24
F) used 432 ÷ 24
5.2 Pythagoras and Trigonometry
In a right-angled triangle, Pythagoras' theorem links the sides: a² + b² = c², where c is the hypotenuse (always the longest side, opposite the right angle). To find the hypotenuse you add the squares; to find a shorter side you subtract.
SOHCAHTOA
To bring angles in, label the sides relative to the angle θ: opposite, adjacent and hypotenuse. Then choose the ratio that uses the two sides you have — sin θ = opp/hyp, cos θ = adj/hyp, tan θ = opp/adj. To find an angle, apply the inverse function (e.g. θ = tan⁻¹(opp/adj)).
Exact values and the area rule
Because no calculator is allowed, learn the exact values: sin 30° = ½, cos 60° = ½, sin 60° = cos 30° = √3/2, sin 45° = cos 45° = 1/√2, and tan 45° = 1. For any triangle (not just right-angled), the area is ½ ab sin C, where C is the angle between the two sides a and b.
📐 Right-Angled Toolkit
Tool
Formula
Pythagoras
a² + b² = c² (c = hypotenuse)
sin / cos / tan
opp/hyp · adj/hyp · opp/adj
Find an angle
θ = sin⁻¹, cos⁻¹ or tan⁻¹ of the ratio
Exact: 30° / 45° / 60°
sin30 = ½, tan45 = 1, sin60 = √3/2
Area of any triangle
½ ab sin C (C between a and b)
⚠️ Add or subtract?
For the hypotenuse, ADD the squares; for a shorter side, SUBTRACT. And choose the trig ratio by which two sides are involved — opposite + adjacent always means tangent.
Worked Example — Pythagoras & tan
Question: A ladder reaches 8 m up a wall with its foot 6 m from the base. Find the ladder's length and the angle it makes with the ground.
Working: Length = √(8² + 6²) = √100 = 10 m. Angle: tan θ = 8/6, so θ = tan⁻¹(8/6) ≈ 53.1°.
⚠️ Common trapOpposite and adjacent ⇒ use tangent. Take the inverse tan of 4/3, not 3/4.
Why each option
A) used tan⁻¹(3/4)
B) ✓ tan⁻¹(4/3)
C) used sine wrongly
D) used cosine
E) worked in the wrong mode
F) added 90°
Section 5.2 Practice · Very Hard · Area = ½ab sin C
A triangle has two sides of 6 cm and 8 cm with an included angle of 30°, as shown. Find its area.
A24 cm²
B48 cm²
C20.8 cm²
D6 cm²
E12 cm²
F12 cm
Show solution
Correct answer: E
MethodArea = ½ × a × b × sin C, where C is the included angle.
Sides6 and 8
sin 30°½
½ × 6 × 8 × sin 30°↓½ × 48 × ½↓12 cm²
⚠️ Common trapUse the INCLUDED angle (between the two given sides). sin 30° = ½ exactly.
Why each option
A) forgot the sin 30° factor
B) used the full product
C) used sin 60°
D) halved twice too often
E) ✓ ½ × 6 × 8 × ½
F) gave a length, not an area
5.3 Circle Theorems
A handful of theorems unlock almost every circle question. Learn to recognise each configuration in a diagram.
The key theorems
The angle at the centre is twice the angle at the circumference on the same arc. A special case is the angle in a semicircle, which is 90°. Angles in the same segment are equal. Opposite angles of a cyclic quadrilateral add to 180°. A tangent meets a radius at 90°, and the two tangents drawn from an external point are equal in length. Finally, because any two radii are equal, triangles drawn from the centre are isosceles — a fact that combines beautifully with the others.
📋 Circle Theorems
Theorem
Statement
Centre vs circumference
angle at centre = 2 × angle at circumference
Semicircle
angle in a semicircle = 90°
Same segment
angles subtended by the same arc are equal
Cyclic quadrilateral
opposite angles add to 180°
Tangent & radius
meet at 90°; tangents from a point are equal
Two radii
triangle from the centre is isosceles
⚠️ Double or halve?
The centre angle is double the circumference angle (so circumference = half of centre). And in a cyclic quadrilateral, only opposite angles sum to 180°.
Worked Example — Combining theorems
Question: AB is a diameter; C is on the circle; angle CAB = 28°. Find angle ABC.
Working: Angle ACB = 90° (angle in a semicircle). Then ABC = 180° − 90° − 28° = 62°.
ACB = 90° → ABC = 180 − 90 − 28 = 62°
Section 5.3 Practice · Hard · Angle at the centre
P, A and B lie on a circle with centre O. The angle at P (at the circumference) is 35°. Find the angle x at the centre O.
A17.5°
B35°
C145°
D70°
E55°
F110°
Show solution
Correct answer: D
MethodThe angle at the centre is twice the angle at the circumference (same arc).
Circumference angle35°
Rulecentre = 2 × circumference
x = 2 × 35°↓x = 70°
⚠️ Common trapCentre angle is DOUBLE the circumference angle subtending the same arc — multiply by 2, don't halve.
Why each option
A) halved instead of doubling
B) copied the circumference angle
C) used 180 − 35
D) ✓ 2 × 35
E) used 90 − 35
F) added 35 to 75
Section 5.3 Practice · Hard · Angle in a semicircle
AB is a diameter of the circle and C lies on the circumference. The angle at A is 35°. Find the angle x at B.
A90°
B145°
C35°
D65°
E55°
F125°
Show solution
Correct answer: E
MethodThe angle in a semicircle (at C) is 90°; angles in a triangle sum to 180°.
Angle at C90° (semicircle)
Angle at A35°
x = 180° − 90° − 35°↓x = 55°
⚠️ Common trapAngle ACB = 90° because AB is a diameter; then use the triangle angle sum.
Why each option
A) copied the semicircle angle
B) forgot the 90°
C) copied angle A
D) used a wrong total
E) ✓ 180 − 90 − 35
F) 180 − 55
Section 5.3 Practice · Hard · Cyclic quadrilateral
ABCD is a cyclic quadrilateral. Angle A is 85°. Find the opposite angle x (angle C).
A95°
B85°
C275°
D105°
E90°
F75°
Show solution
Correct answer: A
MethodOpposite angles of a cyclic quadrilateral add to 180°.
Angle A85°
Ruleopposite angles sum to 180°
x = 180° − 85°↓x = 95°
⚠️ Common trapOnly OPPOSITE angles of a cyclic quadrilateral sum to 180° — adjacent ones generally do not.
Why each option
A) ✓ 180 − 85
B) assumed equal
C) reflex error
D) subtracted from 190
E) assumed a right angle
F) subtracted from 160
Section 5.3 Practice · Hard · Tangent and radius
PT is a tangent to the circle at T, and O is the centre. The angle at P is 50°. Find the angle x at O in triangle OTP.
A50°
B90°
C40°
D130°
E45°
F140°
Show solution
Correct answer: C
MethodA tangent meets a radius at 90°; angles in the triangle sum to 180°.
Angle OTP90° (tangent ⊥ radius)
Angle P50°
x = 180° − 90° − 50°↓x = 40°
⚠️ Common trapThe tangent–radius angle at T is 90°. Use it with the triangle angle sum.
Why each option
A) copied angle P
B) copied the right angle
C) ✓ 180 − 90 − 50
D) forgot the right angle
E) split 90 evenly
F) 180 − 40
Section 5.3 Practice · Very Hard · Isosceles (two radii)
OA and OB are radii of the circle, so triangle OAB is isosceles. The base angle at A is 30°. Find the angle x = AOB at the centre.
A150°
B60°
C30°
D90°
E75°
F120°
Show solution
Correct answer: F
MethodTwo radii are equal ⇒ base angles equal; angles in a triangle sum to 180°.
Base angles30° each (isosceles)
Sum180°
x = 180° − 30° − 30°↓x = 120°
⚠️ Common trapBoth base angles are 30° (radii equal), so subtract 2 × 30° from 180°.
Why each option
A) subtracted only one 30°
B) doubled 30°
C) copied the base angle
D) assumed a right angle
E) halved the apex
F) ✓ 180 − 60
5.4 Mensuration: Area, Surface Area and Volume
Mensuration is mostly about knowing the right formula and substituting carefully — leaving answers in terms of π where a calculator would otherwise be needed.
Area and the circle
Key areas: triangle ½ × base × height, parallelogram base × height, and trapezium ½ (a + b) h, where a and b are the parallel sides. A circle has circumference 2πr and area πr². A sector is a fraction θ/360 of the circle, so its arc length is (θ/360) × 2πr and its area is (θ/360) × πr².
Volume and surface area
The volume of any prism is cross-sectional area × length; a cylinder is the special case πr²h. A cone is one third of its cylinder, ⅓πr²h, and a sphere is (4/3)πr³. Surface areas build from the faces: a cylinder is 2πr² + 2πrh, and a sphere is 4πr².
📐 Area & Volume Formulae
Shape
Formula
Trapezium area
½ (a + b) h
Circle
area πr², circumference 2πr
Sector
area (θ/360)πr², arc (θ/360)·2πr
Cylinder
volume πr²h, surface 2πr² + 2πrh
Cone
volume ⅓πr²h
Sphere
volume (4/3)πr³, surface 4πr²
⚠️ Radius, not diameter
Formulae use the radius — halve the diameter first. Remember the cone's ⅓ and the sphere's 4/3, and always square or cube the radius as the formula demands.
Worked Example — Sector and cone
Question: Find the area of a 60° sector of radius 6 cm, and the volume of a cone of radius 6 cm and height 7 cm.
Find the area of the trapezium shown, with parallel sides 6 cm and 10 cm and perpendicular height 4 cm.
A64 cm²
B32 cm²
C40 cm²
D20 cm²
E60 cm²
F16 cm²
Show solution
Correct answer: B
MethodArea = ½ × (a + b) × h, where a and b are the parallel sides.
Parallel sides6 and 10
Height4
½ × (6 + 10) × 4↓½ × 16 × 4↓32 cm²
⚠️ Common trapAdd the two PARALLEL sides first, then × height × ½. Use the perpendicular height, not a slant side.
Why each option
A) forgot the ½
B) ✓ ½ × 16 × 4
C) used 10 × 4 only
D) used one side
E) ½ × 6 × 10 × 2
F) just added the parallel sides
Section 5.4 Practice · Hard · Sector area
Find the area of the sector shown: radius 8 cm and angle 90°. Give your answer in terms of π.
A4π cm²
B64π cm²
C16π cm²
D8π cm²
E32π cm²
F16 cm²
Show solution
Correct answer: C
MethodSector area = (θ/360) × π r².
Fraction90/360 = ¼
πr²π × 64
¼ × π × 64↓16π cm²
⚠️ Common trapA 90° sector is a QUARTER circle (90/360 = ¼). Use πr², not the circumference.
Why each option
A) used the arc length
B) forgot the ¼
C) ✓ ¼ × 64π
D) used ¼ × 2πr
E) used ½
F) dropped the π
Section 5.4 Practice · Hard · Volume of a cylinder
Find the volume of the cylinder shown: radius 5 cm and height 10 cm. Give your answer in terms of π.
A250π cm³
B500π cm³
C50π cm³
D250 cm³
E100π cm³
F2500π cm³
Show solution
Correct answer: A
MethodVolume = π r² h.
r²25
× h× 10
π × 25 × 10↓250π cm³
⚠️ Common trapSquare the radius first (5² = 25), then × height. Don't use the diameter.
Why each option
A) ✓ π × 25 × 10
B) used the diameter (10) as radius
C) forgot to square r
D) dropped the π
E) used r² × h wrongly
F) squared the height too
Section 5.4 Practice · Very Hard · Volume of a cone
Find the volume of the cone shown: base radius 3 cm and vertical height 4 cm. Give your answer in terms of π.
A36π cm³
B4π cm³
C12 cm³
D48π cm³
E12π cm³
F9π cm³
Show solution
Correct answer: E
MethodVolume = ⅓ π r² h.
r²9
⅓ × h⅓ × 4
⅓ × π × 9 × 4↓⅓ × 36π↓12π cm³
⚠️ Common trapA cone is ONE THIRD of the matching cylinder — don't forget the ⅓. Use the vertical height (4), not a slant.
Why each option
A) forgot the ⅓ (that's the cylinder)
B) forgot to square r
C) dropped the π
D) used ⅓ wrongly / slant
E) ✓ ⅓ × π × 9 × 4
F) forgot the height
Section 5.4 Practice · Very Hard · Volume of a sphere
Find the volume of the sphere shown, which has radius 3 cm. Give your answer in terms of π.
A9π cm³
B108π cm³
C27π cm³
D12π cm³
E36 cm³
F36π cm³
Show solution
Correct answer: F
MethodVolume = (4/3) π r³.
r³27
(4/3) × r³(4/3) × 27
(4/3) × π × 27↓(4/3) × 27 = 36↓36π cm³
⚠️ Common trapCube the radius (3³ = 27) and use the 4/3 factor — not r² and not ⅓.
Why each option
A) used r² with ⅓
B) forgot the ÷3
C) forgot the 4/3
D) used the cone formula
E) dropped the π
F) ✓ (4/3) × π × 27
CHAPTER 5 REVISION EXAM
Revision exam questions for 05-Geometry.
M1-06-Statistics
CHAPTER 6: Statistics
Statistics is about turning a pile of numbers into a clear, honest summary. This chapter covers four jobs: representing data in tables and charts, measuring its centre (mean, median, mode), measuring its spread (range, quartiles, interquartile range), and using scatter graphs to describe relationships. As with the rest of ESAT, every value here is found without a calculator, so the data sets are chosen to give clean numbers.
6.1 Collecting and Representing Data
Data is qualitative (categories, like colour) or quantitative (numbers). Quantitative data is discrete (counted, like goals) or continuous (measured, like height). A frequency table records how often each value or class occurs and is the starting point for almost every chart.
Bar charts, pie charts and stem-and-leaf
A bar chart uses bars of equal width with gaps; the height is the frequency. A pie chart shares 360° between the categories, so each sector's angle is (frequency ÷ total) × 360°. A stem-and-leaf diagram keeps the actual data while showing its shape, which makes the median and mode easy to read.
Histograms and frequency density
For grouped continuous data with unequal class widths, a histogram is used: there are no gaps, and the area of each bar (not its height) gives the frequency. The height is the frequency density = frequency ÷ class width. This keeps wide and narrow classes fairly comparable.
📊 Charts at a glance
Chart
Key rule
Pie chart
sector angle = (freq ÷ total) × 360°
Bar chart
equal widths, gaps; height = frequency
Histogram
no gaps; area = frequency
Frequency density
frequency ÷ class width
⚠️ Bar chart or histogram?
On a bar chart the height is the frequency. On a histogram the height is the frequency density and the area is the frequency — so always multiply density by the class width to recover a frequency.
Worked Example — Frequency density
Question: A class 15 ≤ x < 25 contains 40 values. Find the frequency density.
Working: The class width is 25 − 15 = 10, so frequency density = 40 ÷ 10 = 4.
frequency density = 40 ÷ 10 = 4
Section 6.1 Practice · Hard · Pie chart
In a survey of 30 people, 12 said they travel to work by bus. On a pie chart, find the angle of the sector representing 'bus'.
A120°
B40°
C144°
D12°
E216°
F90°
Show solution
Correct answer: C
MethodSector angle = (frequency ÷ total) × 360°.
Fraction12 / 30
× 360°
(12 ÷ 30) × 360°↓0.4 × 360↓144°
⚠️ Common trapA pie chart shares out 360°, not 100°. Multiply the fraction of the total by 360.
Why each option
A) (12/30)×300
B) (12/30)×100 — used percentages
C) ✓ (12/30)×360
D) used the frequency as the angle
E) gave the rest of the chart
F) guessed a quarter
Section 6.1 Practice · Hard · Pie chart (reverse)
A pie chart represents 72 students. One sector has an angle of 75°. How many students does that sector represent?
A15 students
B75 students
C30 students
D10 students
E18 students
F24 students
Show solution
Correct answer: A
MethodFrequency = (angle ÷ 360°) × total.
Fraction75 / 360
× 72
(75 ÷ 360) × 72↓(5/24) × 72↓15 students
⚠️ Common trapReverse the pie rule: the angle's share of 360° gives the share of the total.
Why each option
A) ✓ (75/360)×72
B) read the angle as the count
C) used 150°
D) used 50°
E) used 90°
F) used 120°
Section 6.1 Practice · Very Hard · Histogram (frequency)
In the histogram shown, the class 10 ≤ x < 15 has a frequency density of 6. How many values fall in this class?
A6
B60
C11
D1.2
E15
F30
Show solution
Correct answer: F
MethodFrequency = frequency density × class width.
Density6
Class width15 − 10 = 5
frequency = 6 × 5↓30
⚠️ Common trapOn a histogram the AREA of a bar is the frequency: multiply density by the class width (here 5, not 10).
Why each option
A) read the density as the frequency
B) used width 10
C) added 6 + 5
D) divided 6 ÷ 5
E) used the upper boundary
F) ✓ 6 × 5
Section 6.1 Practice · Hard · Frequency density
A grouped frequency table has a class 20 ≤ x < 30 containing 24 values. Find the frequency density needed to draw this bar on a histogram.
A24
B2.4
C240
D0.8
E4.8
F12
Show solution
Correct answer: B
MethodFrequency density = frequency ÷ class width.
Frequency24
Class width30 − 20 = 10
24 ÷ 10↓2.4
⚠️ Common trapFrequency density = frequency ÷ width. Don't plot the raw frequency as the bar height.
Why each option
A) plotted the raw frequency
B) ✓ 24 ÷ 10
C) multiplied 24 × 10
D) used width 30
E) used width 5
F) halved the frequency
Section 6.1 Practice · Hard · Reading a bar chart
The bar chart shows the shoe sizes of a class. Which shoe size is the mode, and how many pupils are there in total?
AMode size 10, total 36
BMode size 8, total 5
CMode size 13, total 36
DMode size 8, total 36
EMode size 7, total 32
FMode size 8, total 45
Show solution
Correct answer: D
MethodThe mode is the tallest bar; the total is the sum of all the bar heights.
Tallest barsize 8 (13 pupils)
Total4 + 9 + 13 + 7 + 3
mode = size 8↓total = 36↓Mode size 8, total 36
⚠️ Common trapThe mode is the most COMMON value (tallest bar), not the largest shoe size on the axis.
Why each option
A) took the largest size, not the tallest bar
B) counted the bars, not the pupils
C) read the frequency as the size
D) ✓ tallest bar; 4+9+13+7+3
E) misread the tallest bar
F) mis-added the totals
6.2 Averages: Mean, Median and Mode
Three "averages" describe the centre of a data set. The mean is the total divided by how many values there are. The median is the middle value once the data is ordered (for an even count, the mean of the two middle values). The mode is the most frequent value.
Averages from a frequency table
When data is given as a frequency table, the mean is Σ(f × x) ÷ Σf: multiply each value by its frequency, add those products, and divide by the total frequency. For grouped data the exact values are unknown, so use the class mid-points — this gives an estimated mean.
Which average?
The mean uses every value but is pulled by outliers; the median ignores extremes and is safer for skewed data; the mode is the only average that works for categories.
📊 The three averages
Average
How to find it
Mean
sum ÷ count, or Σ(fx) ÷ Σf
Median
middle of ordered data; even ⇒ mean of two middle
Mode
most frequent value
Estimated mean (grouped)
Σ(f × mid-point) ÷ Σf
⚠️ Order first; divide by the total
The median only works on ordered data. For a frequency table, divide by Σf (the total frequency), not by the number of rows. For grouped data, use mid-points, not upper or lower bounds.
Worked Example — Estimated mean
Question: Times (min): 0–10 (4), 10–20 (10), 20–30 (6). Estimate the mean.
Working: Mid-points 5, 15, 25 give Σ(fx) = 4×5 + 10×15 + 6×25 = 20 + 150 + 150 = 320, and Σf = 20, so the estimated mean = 320 ÷ 20 = 16 minutes.
(20 + 150 + 150) ÷ 20 = 320 ÷ 20 = 16 min
Section 6.2 Practice · Hard · Mean
Find the mean of the five numbers 4, 7, 9, 12 and 8.
A9
B10
C8
D40
E7
F6.7
Show solution
Correct answer: C
MethodMean = (sum of values) ÷ (number of values).
Sum4 + 7 + 9 + 12 + 8 = 40
Count5
40 ÷ 5↓8
⚠️ Common trapAdd all the values first, then divide by how many there are (5).
Why each option
A) used the middle value
B) divided by 4
C) ✓ 40 ÷ 5
D) forgot to divide
E) used the median wrongly
F) divided by 6
Section 6.2 Practice · Hard · Median (even count)
Find the median of the data set: 3, 5, 8, 9, 11, 14.
A8.5
B8
C9
D11
E9.5
F7
Show solution
Correct answer: A
MethodOrder the data; the median of an even-sized set is the mean of the two middle values.
Already ordered6 values
Two middle8 and 9
median = (8 + 9) ÷ 2↓8.5
⚠️ Common trapWith an even number of values, average the TWO middle numbers — don't just pick one.
Why each option
A) ✓ (8 + 9)/2
B) took only the lower middle value
C) took only the upper middle value
D) used the mean instead
E) averaged 9 and 10
F) miscounted the middle
Section 6.2 Practice · Very Hard · Mean from a frequency table
A frequency table records the number of goals per match: 0 goals (2 matches), 1 goal (5), 2 goals (8), 3 goals (5). Find the mean number of goals per match.
A9 goals
B1.5 goals
C36 goals
D2 goals
E1.8 goals
F1.2 goals
Show solution
Correct answer: E
MethodMean = Σ(f × x) ÷ Σf, where x is the value and f its frequency.
Σf2 + 5 + 8 + 5 = 20
Σ(fx)0 + 5 + 16 + 15 = 36
36 ÷ 20↓1.8 goals
⚠️ Common trapMultiply each value by its frequency before adding; then divide by the TOTAL frequency (20), not by 4.
Why each option
A) divided by 4 categories
B) used the middle value
C) forgot to divide
D) used the mode
E) ✓ 36 ÷ 20
F) forgot a value
Section 6.2 Practice · Very Hard · Estimated mean (grouped)
A grouped table gives times (minutes): 0–10 (4 people), 10–20 (10 people), 20–30 (6 people). Estimate the mean time.
A15 minutes
B16 minutes
C20 minutes
D10.7 minutes
E13.3 minutes
F32 minutes
Show solution
Correct answer: B
MethodUse class mid-points: estimated mean = Σ(f × mid-point) ÷ Σf.
Mid-points5, 15, 25
Σ(f·mid)20 + 150 + 150 = 320
Σf20
320 ÷ 20↓16 minutes
⚠️ Common trapFor grouped data use the class MID-POINTS (5, 15, 25), since the exact values are unknown.
Why each option
A) used the middle class only
B) ✓ 320 ÷ 20
C) used upper bounds
D) used lower bounds
E) divided 320 by 24
F) forgot to divide by 2
Section 6.2 Practice · Very Hard · Working back from a mean
The mean of five numbers is 12. A sixth number is added and the mean of all six becomes 13. Find the sixth number.
A1
B13
C25
D18
E65
F6
Show solution
Correct answer: D
MethodTotal = mean × count; the sixth number is the difference of the two totals.
Old total5 × 12 = 60
New total6 × 13 = 78
sixth = 78 − 60↓18
⚠️ Common trapConvert each mean back to a TOTAL first; the new value is the change in the total, not the change in the mean.
Why each option
A) the change in the mean
B) gave the new mean
C) added the two means
D) ✓ 78 − 60
E) used 5×13
F) guessed
6.3 Measures of Spread
An average alone hides how spread out the data is. The simplest measure of spread is the range = maximum − minimum, but it is sensitive to a single extreme value. A more robust measure is the interquartile range.
Quartiles and the interquartile range
Ordering the data and splitting it into quarters gives the lower quartile Q₁, the median Q₂ and the upper quartile Q₃. The interquartile range, IQR = Q₃ − Q₁, measures the spread of the middle 50% and ignores the extremes.
Cumulative frequency and box plots
A cumulative frequency curve plots running totals against the upper class boundary; the median is read at n/2, and the quartiles at n/4 and 3n/4. A box plot shows the five-number summary (minimum, Q₁, median, Q₃, maximum): the box is the IQR and always contains the middle 50% of the data, which makes two data sets easy to compare.
📊 Spread toolkit
Measure
Rule
Range
maximum − minimum
Interquartile range
Q₃ − Q₁
Median (from curve)
read at cumulative frequency n/2
Box plot
min, Q₁, median, Q₃, max; box = middle 50%
⚠️ IQR is not the range
The range uses the extremes; the IQR uses the quartiles (the box width). On a cumulative frequency curve read the median at n/2, not at n. The box of a box plot always holds the middle 50%, whatever the numbers.
Worked Example — Median from a curve
Question: 40 students' marks are plotted on a cumulative frequency curve. How do you read the median?
Working: n ÷ 2 = 20, so go up the cumulative-frequency axis to 20, across to the curve, and down to the mark axis. If that mark is 20, the median is 20.
median read at cum. freq = 40 ÷ 2 = 20
Section 6.3 Practice · Hard · Interquartile range
A data set has a lower quartile of 14 and an upper quartile of 23. Find the interquartile range.
⚠️ Common trapThe IQR uses the two QUARTILES, not the maximum and minimum (that would be the range).
Why each option
A) added them
B) took the mean
C) ✓ 23 − 14
D) gave Q₃ only
E) gave Q₁ only
F) halved the IQR
Section 6.3 Practice · Very Hard · Quartiles from data
Find the interquartile range of: 2, 4, 5, 7, 8, 10, 11, 13.
A6
B11
C7.5
D15
E3
F5
Show solution
Correct answer: A
MethodQ₁ is the median of the lower half, Q₃ the median of the upper half; IQR = Q₃ − Q₁.
Lower half2, 4, 5, 7 ⇒ Q₁ = 4.5
Upper half8, 10, 11, 13 ⇒ Q₃ = 10.5
IQR = 10.5 − 4.5↓6
⚠️ Common trapSplit the ordered data in half first; each quartile is the median of its half.
Why each option
A) ✓ 10.5 − 4.5
B) used max − min (range)
C) used the overall median
D) Q₃ + Q₁
E) halved the IQR
F) used 9.5 − 4.5
Section 6.3 Practice · Very Hard · Median from a curve
The cumulative frequency curve shows the marks of 40 students. Use it to estimate the median mark.
A40 marks
B10 marks
C30 marks
D34 marks
E25 marks
F20 marks
Show solution
Correct answer: F
MethodThe median is read at a cumulative frequency of n ÷ 2 = 20.
n ÷ 240 ÷ 2 = 20
Read across at 20
go up to cum. freq 20↓read the mark below↓20 marks
⚠️ Common trapRead the median at HALF the total (20), then drop down to the mark axis — don't read it at 40.
Why each option
A) read at the top of the curve
B) read at cum. freq 8
C) read at cum. freq 34
D) read the cum. freq, not the mark
E) misread the scale
F) ✓ read across at cum. freq 20
Section 6.3 Practice · Hard · Box plot (IQR)
The box plot summarises a set of test marks. Find the interquartile range.
A24 marks
B11 marks
C14 marks
D6 marks
E9 marks
F20 marks
Show solution
Correct answer: B
MethodIQR = Q₃ − Q₁, the width of the box.
Q₃ (right of box)20
Q₁ (left of box)9
20 − 9↓11 marks
⚠️ Common trapThe IQR is the width of the BOX (Q₃ − Q₁), not the full length from whisker to whisker.
Why each option
A) 28 − 4 (the range)
B) ✓ 20 − 9 (box width)
C) gave the median
D) 14 − 8 slip
E) gave Q₁
F) gave Q₃
Section 6.3 Practice · Very Hard · Interpreting the box
On a box plot the box runs from 18 to 30 and the whiskers reach 5 and 45. What percentage of the data lies between 18 and 30?
A25%
B75%
C40%
D50%
E12%
F100%
Show solution
Correct answer: D
MethodThe box spans Q₁ to Q₃, which always contains the middle 50% of the data.
BoxQ₁ = 18 to Q₃ = 30
Middle half
box = middle 50%↓50%
⚠️ Common trapThe box (Q₁ to Q₃) always holds the middle HALF of the data, regardless of the actual numbers.
Why each option
A) one quartile only
B) up to Q₃
C) used 12 out of 30
D) ✓ the box is always the middle 50%
E) used the box width as a percent
F) whole data set
6.4 Scatter Graphs and Correlation
A scatter graph plots pairs of values for two variables to reveal a relationship. The pattern is called correlation: positive (both increase, points rise to the right), negative (one increases as the other falls), or none (no clear pattern).
Line of best fit
When there is correlation, a straight line of best fit is drawn through the middle of the points; it always passes through the mean point (x̄, ȳ). The line lets you estimate one variable from the other. Estimating within the data range is interpolation (reliable); estimating beyond it is extrapolation (unreliable, because the trend may not continue).
Correlation is not causation
A strong correlation does not prove that one variable causes the other. Both may be driven by a hidden third factor, so conclusions must be drawn with care.
📊 Correlation essentials
Idea
Meaning
Positive / negative
rises together / one falls as other rises
Line of best fit
through the mean point (x̄, ȳ)
Interpolation
estimate inside the data range — reliable
Extrapolation
estimate outside the range — unreliable
⚠️ Two warnings
Extrapolating far beyond the data is unreliable — the pattern may break down. And correlation never proves causation: look for a possible hidden factor before claiming one variable causes another.
Worked Example — Using a line of best fit
Question: A line of best fit passes through (0, 1) and (10, 9). Estimate y when x = 5.
Working: The gradient is (9 − 1)/10 = 0.8, so y = 1 + 0.8x. At x = 5, y = 1 + 4 = 5.
y = 1 + 0.8 × 5 = 5
Section 6.4 Practice · Hard · Type of correlation
The scatter graph plots hours of revision (x) against test score (y). Describe the correlation shown.
APositive correlation
BNegative correlation
CNo correlation
DZero gradient
EPerfect correlation
FInverse correlation
Show solution
Correct answer: A
MethodAs one variable increases, note whether the other tends to increase (positive), decrease (negative) or show no pattern.
Trendy rises as x rises
points rise left to right↓Positive correlation
⚠️ Common trapPositive correlation = both increase together (up to the right). It does not have to be a perfect straight line.
Why each option
A) ✓ y increases as x increases
B) would slope downwards
C) would show no pattern
D) describes a flat line, not the trend
E) points are not exactly collinear
F) same as negative — not shown
Section 6.4 Practice · Very Hard · Line of best fit
The line of best fit drawn on the graph passes through (0, 1) and (10, 9). Use it to estimate the score y when x = 5 hours.
A9
B4
C5
D6
E8
F1
Show solution
Correct answer: C
MethodRead up from x = 5 to the line, or use the line's equation y = 1 + 0.8x.
Gradient(9 − 1)/10 = 0.8
At x = 51 + 0.8 × 5
y = 1 + 4↓5
⚠️ Common trapUse the LINE of best fit for the estimate, not the nearest single point. Here the line gives y = 1 + 0.8x.
Why each option
A) read the end of the line
B) used gradient only
C) ✓ 1 + 0.8×5
D) misread the gradient
E) used x as y
F) read the intercept
Section 6.4 Practice · Hard · Reliability of estimates
The revision data covers 1 to 8 hours. Using the line of best fit to predict the score for 20 hours of revision would be an example of what, and how reliable is it?
AInterpolation — reliable
BInterpolation — unreliable
CExtrapolation — reliable
DCorrelation — reliable
ECausation — reliable
FExtrapolation — unreliable
Show solution
Correct answer: F
MethodPredicting INSIDE the data range is interpolation (reliable); OUTSIDE it is extrapolation (unreliable).
20 hoursoutside 1–8
Typeextrapolation
beyond the data ⇒ extrapolation↓Extrapolation — unreliable
⚠️ Common trapExtrapolation goes beyond the observed data, where the trend may not continue, so it is unreliable.
Why each option
A) that is prediction inside the range
B) wrong term for outside the range
C) extrapolation is not reliable
D) not what the term means
E) unrelated idea
F) ✓ 20 is outside the 1–8 range
Section 6.4 Practice · Very Hard · Mean point
For a set of data the mean of x is 5 and the mean of y is 12. The line of best fit must pass through which point?
A(12, 5)
B(5, 12)
C(0, 0)
D(2.5, 6)
E(10, 24)
F(17, 17)
Show solution
Correct answer: B
MethodA line of best fit always passes through the mean point (x̄, ȳ).
x̄5
ȳ12
mean point = (5, 12)↓(5, 12)
⚠️ Common trapThe line of best fit is anchored at the mean point (x̄, ȳ) — pair the two means in the right order.
Why each option
A) means in the wrong order
B) ✓ (x̄, ȳ)
C) the origin is not required
D) halved both means
E) doubled both means
F) added the means
Section 6.4 Practice · Hard · Correlation vs causation
Across a town, ice-cream sales and the number of people swimming both rise together through the year. What can be correctly concluded?
AIce cream makes people swim
BSwimming increases ice-cream sales
CThere is no relationship
DThey are correlated, but neither causes the other
EThe data must be wrong
FOne must cause the other
Show solution
Correct answer: D
MethodCorrelation between two variables does not by itself prove that one causes the other.
Both risein warm weather
Hidden factortemperature
correlation ≠ causation↓They are correlated, but neither causes the other
⚠️ Common trapA strong correlation can be driven by a third factor (here, temperature). Correlation does not prove causation.
Why each option
A) mistakes correlation for cause
B) mistakes correlation for cause
C) there is a clear correlation
D) ✓ a third factor (heat) drives both
E) the pattern is plausible
F) correlation does not prove this
CHAPTER 6 REVISION EXAM
Revision exam questions for 06-Statistics.
M1-07-Probability
CHAPTER 7: Probability
Probability measures how likely an event is, on a scale from 0 (impossible) to 1 (certain). This chapter builds the tools that ESAT relies on: finding probabilities from equally likely outcomes, combining events with the addition and multiplication rules, organising them with Venn diagrams and tree diagrams, handling conditional probability, and computing the expected value of a situation. As elsewhere in ESAT, every answer is reached without a calculator, so probabilities are kept as exact fractions or decimals.
7.1 Sample Spaces and Single Events
The sample space is the set of all possible outcomes. When outcomes are equally likely, the probability of an event is simply P(event) = (favourable outcomes) ÷ (total outcomes). For two dice there are 6 × 6 = 36 equally likely ordered pairs — a crucial point, since (2, 5) and (5, 2) are different outcomes even though their total is the same.
Every probability lies in 0 ≤ P ≤ 1, and the probabilities of all outcomes sum to 1. The complement A′ ("not A") satisfies P(A′) = 1 − P(A); this is the fastest route to "at least one" problems, which are almost always easier as 1 − P(none).
When the outcomes are not equally likely — a biased spinner, say — the probabilities are given algebraically and the condition "they sum to 1" produces an equation to solve.
📋 Core rules
Idea
Rule
Equally likely
P = favourable ÷ total
Range
0 ≤ P(A) ≤ 1
Complement
P(A′) = 1 − P(A)
Two dice
36 equally likely ordered pairs
⚠️ "At least one"
Never add probabilities for "at least one" — that double-counts overlaps. Use the complement: P(at least one) = 1 − P(none). For two dice, P(at least one six) = 1 − (5/6)² = 11/36, not 1/6 + 1/6.
Worked Example — A biased spinner
Question: A spinner has P(red) = 2k, P(blue) = 3k, P(green) = k. Find P(green).
Working: The probabilities sum to 1, so 2k + 3k + k = 6k = 1, giving k = 1/6. Hence P(green) = k = 1/6.
6k = 1 ⇒ k = 1/6 ⇒ P(green) = 1/6
Section 7.1 Practice · Hard · Sample space
Two fair six-sided dice are rolled and their scores are added. Find the probability that the total is 7.
A1/11
B1/12
C1/6
D7/36
E5/36
F1/36
Show solution
Correct answer: C
MethodP(event) = (number of favourable outcomes) ÷ (number of equally likely outcomes).
Outcomes6 × 6 = 36
Totals of 7(1,6),(2,5),(3,4),(4,3),(5,2),(6,1) = 6
6 ÷ 36↓1/6
⚠️ Common trapThere are 36 equally likely ordered outcomes, not 11 totals; (2,5) and (5,2) are different outcomes.
Why each option
A) treated the 11 totals as equally likely
B) halved the wrong count
C) ✓ 6/36
D) used the total as the count
E) forgot one pair
F) used a single outcome
Section 7.1 Practice · Very Hard · Sample space
Two fair six-sided dice are rolled. Find the probability that the total is 10 or more.
A1/6
B1/12
C1/9
D1/4
E3/36
F1/3
Show solution
Correct answer: A
MethodCount the favourable ordered pairs over 36.
Total 10(4,6),(5,5),(6,4) = 3
Total 11(5,6),(6,5) = 2
Total 12(6,6) = 1
(3 + 2 + 1) ÷ 36↓6 ÷ 36↓1/6
⚠️ Common trapInclude every ordered pair: totals 10, 11 and 12 give 3 + 2 + 1 = 6 outcomes.
Why each option
A) ✓ 6/36
B) counted only 3 outcomes
C) missed total 12
D) used 9 outcomes
E) only counted total 10
F) over-counted
Section 7.1 Practice · Very Hard · Complement
Two fair dice are rolled. Find the probability of getting at least one six.
A1/3
B1/36
C25/36
D11/36
E10/36
F12/36
Show solution
Correct answer: D
MethodP(at least one) = 1 − P(none); the rolls are independent.
P(no six on one die)5/6
P(no six on either)(5/6)² = 25/36
1 − 25/36↓11/36
⚠️ Common trap'At least one' is the complement of 'none'. Do NOT add 1/6 + 1/6 (that double-counts the double six).
Why each option
A) added 1/6 + 1/6
B) used P(two sixes)
C) gave P(no six)
D) ✓ 1 − (5/6)²
E) forgot the overlap
F) rounded
Section 7.1 Practice · Very Hard · Algebra of probabilities
A biased spinner has three colours. P(red) = 2k, P(blue) = 3k and P(green) = k for some constant k. Find P(blue).
A1/6
B1/2
C1/3
D3
E1/5
F2/3
Show solution
Correct answer: B
MethodAll probabilities sum to 1, so 2k + 3k + k = 1.
Total6k = 1
k1/6
P(blue) = 3k = 3 × 1/6↓1/2
⚠️ Common trapUse the fact that the probabilities add to 1 to find k first, then substitute into 3k.
Why each option
A) gave k, not 3k
B) ✓ 3k with k = 1/6
C) used 2k
D) forgot to divide
E) used 5k = 1
F) added 2k + 2k
7.2 Combined Events, the Addition Rule and Venn Diagrams
For the union of two events, the addition rule is P(A ∪ B) = P(A) + P(B) − P(A ∩ B). The intersection is subtracted because adding P(A) and P(B) counts the overlap twice. A Venn diagram makes this visible: the two circles share the region A ∩ B.
Two events are mutually exclusive if they cannot both happen, so P(A ∩ B) = 0 and the rule simplifies to P(A ∪ B) = P(A) + P(B). With counts in a Venn diagram, always place the "both" figure in the overlap first, then work outwards, so that no element is counted twice.
📋 Combining events
Situation
Rule
Union
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Mutually exclusive
P(A ∩ B) = 0
Complement
P(A′) = 1 − P(A)
Venn counts
fill the overlap first
⚠️ Subtract the overlap
Forgetting the − P(A ∩ B) term is the most common error in the addition rule. Only drop it when the events are mutually exclusive (no overlap). In a Venn diagram, "18 study Maths" includes those who also study Physics.
Worked Example — Venn counts
Question: Of 30 students, 18 study Maths, 14 study Physics, 6 study both. How many study neither?
Working: The number studying at least one is 18 + 14 − 6 = 26, so 30 − 26 = 4 study neither.
For two events, P(A) = 0.5, P(B) = 0.4 and P(A ∩ B) = 0.2. Find P(A ∪ B).
A0.9
B1.1
C0.2
D0.1
E0.7
F0.5
Show solution
Correct answer: E
MethodP(A ∪ B) = P(A) + P(B) − P(A ∩ B).
P(A) + P(B)0.9
− P(A ∩ B)− 0.2
0.5 + 0.4 − 0.2↓0.7
⚠️ Common trapSubtract the overlap once, or the intersection is counted twice.
Why each option
A) forgot to subtract the overlap
B) added the overlap
C) gave the intersection
D) subtracted twice
E) ✓ 0.9 − 0.2
F) gave P(A)
Section 7.2 Practice · Very Hard · Venn diagram
In a group of 30 students, 18 study Maths, 14 study Physics and 6 study both. Find the probability that a student chosen at random studies neither subject.
⚠️ Common trapSubtract the 6 'both' students once before counting; 18 + 14 = 32 double-counts them.
Why each option
A) ✓ 4/30
B) forgot to subtract the overlap
C) gave P(at least one)
D) used 6/30
E) assumed everyone studies one
F) used 10/30
Section 7.2 Practice · Very Hard · Mutually exclusive + algebra
Two mutually exclusive events satisfy P(A) = x and P(B) = 2x. Given P(A ∪ B) = 0.9, find x.
A0.45
B0.9
C0.3
D0.6
E0.15
F0.1
Show solution
Correct answer: C
MethodMutually exclusive ⇒ P(A ∪ B) = P(A) + P(B) = x + 2x.
Equation3x = 0.9
x = 0.9 ÷ 3↓0.3
⚠️ Common trapFor mutually exclusive events the intersection is 0, so just add: 3x = 0.9.
Why each option
A) split 0.9 in two
B) gave P(A ∪ B)
C) ✓ 3x = 0.9
D) gave 2x
E) divided by 6
F) divided by 9
Section 7.2 Practice · Hard · Complement
An event A has P(A′) = 0.7, where A′ is the complement of A. Find P(A).
A0.7
B1.7
C0.49
D0.3
E0
F1
Show solution
Correct answer: D
MethodP(A) + P(A′) = 1.
P(A)1 − 0.7
1 − 0.7↓0.3
⚠️ Common trapThe complement's probability and the event's probability add to 1.
Why each option
A) gave P(A′)
B) added
C) squared 0.7
D) ✓ 1 − 0.7
E) misread
F) whole sample space
7.3 Independent Events and Tree Diagrams
Two events are independent if one happening does not change the probability of the other; then P(A ∩ B) = P(A) × P(B). A tree diagram shows a sequence of events: you multiply along the branches to get the probability of a path, and add the probabilities of the separate paths that make up an event.
The key distinction is with versus without replacement. With replacement the trials are independent and the branch probabilities repeat. Without replacement they change — after drawing one red counter from 3 red and 2 blue, only 2 reds remain out of 4 — and the events become dependent (this is conditional probability in disguise).
📋 Working a tree
Action
Operation
Along a branch (and)
multiply
Between paths (or)
add
Independent
P(A ∩ B) = P(A)P(B)
"At least one"
1 − P(none)
⚠️ Replacement changes the branches
Without replacement, the denominator drops by one and the relevant numerator changes on the second draw. Using the same fraction twice silently assumes replacement and is the classic error here.
Worked Example — Without replacement
Question: A bag has 3 red, 2 blue. Two are drawn without replacement. Find P(both red).
Working: P(R₁) = 3/5; given a red has gone, P(R₂) = 2/4. Multiplying, 3/5 × 2/4 = 6/20 = 3/10.
3/5 × 2/4 = 6/20 = 3/10
Section 7.3 Practice · Hard · Independence
Two independent events have P(A) = 0.6 and P(B) = 0.5. Find P(A ∩ B).
A1.1
B0.3
C0.8
D0.1
E0.5
F0.6
Show solution
Correct answer: B
MethodIndependent ⇒ P(A ∩ B) = P(A) × P(B).
P(A) × P(B)0.6 × 0.5
0.6 × 0.5↓0.3
⚠️ Common trapFor independent events MULTIPLY the probabilities; only add for the union.
Why each option
A) added them
B) ✓ 0.6 × 0.5
C) used the union rule wrongly
D) subtracted
E) gave P(B)
F) gave P(A)
Section 7.3 Practice · Very Hard · Tree (without replacement)
A bag holds 3 red and 2 blue counters. Two are drawn at random without replacement. Find the probability that both are red.
A3/10
B9/25
C6/10
D1/10
E3/5
F2/5
Show solution
Correct answer: A
MethodMultiply along the branches: P(R then R) = P(R₁) × P(R₂ | R₁).
First red3/5
Second red (now 2 of 4)2/4
3/5 × 2/4↓6/20↓3/10
⚠️ Common trapWithout replacement the second fraction changes: 2 reds left out of 4 counters, not 3 out of 5.
Why each option
A) ✓ 3/5 × 2/4
B) used 3/5 × 3/5 (with replacement)
C) forgot to reduce the second draw
D) used the blue branch
E) only one draw
F) used the blue first
Section 7.3 Practice · Very Hard · Tree + complement
A bag holds 3 red and 2 blue counters; two are drawn without replacement. Find the probability that at least one is red.
A1/10
B3/10
C4/5
D9/10
E7/10
F1/2
Show solution
Correct answer: D
MethodP(at least one red) = 1 − P(no red) = 1 − P(both blue).
P(both blue)2/5 × 1/4 = 1/10
1 − 1/10↓9/10
⚠️ Common trap'At least one red' is the complement of 'both blue' — far quicker than adding three branches.
Why each option
A) gave P(both blue)
B) gave P(both red)
C) forgot a branch
D) ✓ 1 − 1/10
E) added two branches only
F) guessed
Section 7.3 Practice · Very Hard · Testing independence
For two events, P(A) = 0.4, P(B) = 0.5 and P(A ∩ B) = 0.2. Are A and B independent?
ANo — they overlap
BNo — P(A) ≠ P(B)
CYes — independent
DYes — mutually exclusive
ECannot be decided
FNo — the sum is not 1
Show solution
Correct answer: C
MethodA and B are independent iff P(A ∩ B) = P(A) × P(B).
P(A) × P(B)0.4 × 0.5 = 0.2
P(A ∩ B)0.2
0.2 = 0.2 ⇒ independent↓Yes — independent
⚠️ Common trapIndependence is a calculation, not a guess: check whether P(A ∩ B) equals the PRODUCT P(A)·P(B).
Why each option
A) overlap does not decide independence
B) equal probabilities are not required
C) ✓ 0.4 × 0.5 = 0.2 = P(A ∩ B)
D) mutually exclusive is the opposite
E) the test does decide it
F) sums are irrelevant here
7.4 Conditional Probability
The probability of A given that B has happened is written P(A | B) and equals P(A ∩ B) ÷ P(B). Conditioning on B restricts attention to the part of the sample space where B occurs, so we divide the joint probability by P(B). Rearranged, this gives the multiplication rule P(A ∩ B) = P(B) × P(A | B), exactly what a tree diagram does.
A consequence: A and B are independent precisely when P(A | B) = P(A) — knowing B tells you nothing about A. The condition is always the denominator, so P(A | B) and P(B | A) are generally different.
📋 Conditioning
Quantity
Formula
Conditional
P(A | B) = P(A ∩ B) ÷ P(B)
Multiplication
P(A ∩ B) = P(B) · P(A | B)
Independent test
P(A | B) = P(A)
⚠️ Which event is the condition?
The event after the bar is the one you divide by. P(A | B) divides by P(B); P(B | A) divides by P(A). A stated "of the X, a fraction also do Y" is a conditional probability given X — multiply it by P(X) to get the joint probability.
Worked Example — Conditional from a percentage
Question: 40% of homes own a car; of those, 25% own a bike. Find P(car and bike).
Working: The 25% is conditional on owning a car, so P(car ∩ bike) = 0.4 × 0.25 = 0.1.
P(car ∩ bike) = 0.4 × 0.25 = 0.1
Section 7.4 Practice · Hard · Conditional probability
For two events, P(A ∩ B) = 0.12 and P(B) = 0.3. Find P(A | B).
A0.036
B2.5
C0.12
D0.3
E0.4
F0.42
Show solution
Correct answer: E
MethodP(A | B) = P(A ∩ B) ÷ P(B).
P(A ∩ B)0.12
P(B)0.3
0.12 ÷ 0.3↓0.4
⚠️ Common trapDivide the intersection by the probability of the GIVEN event (B), not by P(A).
Why each option
A) multiplied them
B) divided the wrong way
C) gave the intersection
D) gave P(B)
E) ✓ 0.12 ÷ 0.3
F) added them
Section 7.4 Practice · Very Hard · Conditional from counts
In a town, 40% of homes own a car. Of the car-owning homes, 25% also own a bicycle. A home is chosen at random. Find the probability that it owns BOTH a car and a bicycle.
A0.1
B0.25
C0.65
D0.4
E0.625
F0.15
Show solution
Correct answer: A
MethodP(car ∩ bike) = P(car) × P(bike | car).
P(car)0.4
P(bike | car)0.25
0.4 × 0.25↓0.1
⚠️ Common trapThe 25% is a CONDITIONAL probability (given a car), so multiply it by P(car) — do not just take 25%.
Why each option
A) ✓ 0.4 × 0.25
B) ignored the conditioning
C) added the percentages
D) gave P(car)
E) divided 0.25 by 0.4
F) subtracted
Section 7.4 Practice · Very Hard · Reversing the condition
Events satisfy P(A) = 0.2, P(B) = 0.5 and P(A ∩ B) = 0.1. Find P(B | A).
A0.2
B0.1
C0.5
D0.05
E0.4
F1
Show solution
Correct answer: C
MethodP(B | A) = P(A ∩ B) ÷ P(A).
P(A ∩ B)0.1
P(A)0.2
0.1 ÷ 0.2↓0.5
⚠️ Common trapP(B | A) divides by P(A); P(A | B) would divide by P(B). The condition is the denominator.
Why each option
A) divided by P(B)
B) gave the intersection
C) ✓ 0.1 ÷ 0.2
D) multiplied
E) wrong denominator
F) ignored the numerator
7.5 Expectation
The expected value E(X) is the long-run average outcome: multiply each value by its probability and add, E(X) = Σ x·P(x). For money, gains are positive and losses negative. A game is fair when the expected payout equals the stake, so the expected gain is zero.
Expectation also gives the expected count over many trials: the expected number of successes in n trials is n × P(success). It is an average, so it need not be a whole number — an expected gain of £0.20 per game is perfectly sensible.
📋 Expectation
Quantity
Formula
Expected value
E(X) = Σ x · P(x)
Expected count
n × P(success)
Fair game
stake = E(payout)
⚠️ Signs and probabilities
Enter losses as negative values, and never forget to weight by the probabilities. Charging the full prize as a stake (ignoring how often it is won) makes the game wildly unfair to the player.
Worked Example — Fair stake
Question: A game pays £10 with probability 0.25, else nothing. What stake is fair?
Working: E(payout) = 10 × 0.25 = £2.50, so a fair stake is £2.50 (then the expected gain is zero).
E(payout) = 10 × 0.25 = £2.50
Section 7.5 Practice · Very Hard · Expected value of a game
In a game you win £5 with probability 0.2 and lose £1 with probability 0.8. Find the expected gain per game.
A£1.80
B£0.20
C£4
D£1
E£0.80
F£2.50
Show solution
Correct answer: B
MethodE(X) = Σ (value × probability).
Win term5 × 0.2 = 1
Lose term(−1) × 0.8 = −0.8
1 + (−0.8)↓£0.20
⚠️ Common trapUse a NEGATIVE value for the £1 loss; the expectation is the long-run average per game.
Why each option
A) added 1 + 0.8
B) ✓ 1 − 0.8
C) took 5 − 1
D) ignored the loss
E) gave only the loss term
F) ignored the probabilities
Section 7.5 Practice · Very Hard · Fair stake
A game pays out £10 with probability 0.25 and nothing otherwise. What stake should be charged to make the game fair?
A£2.50
B£10
C£0.25
D£7.50
E£5
F£40
Show solution
Correct answer: A
MethodA game is fair when the stake equals the expected payout E(payout).
E(payout)10 × 0.25
fair stake = £2.50↓£2.50
⚠️ Common trapFair means expected payout = stake, so the stake equals 10 × 0.25, NOT £10.
Why each option
A) ✓ 10 × 0.25
B) the full prize, ignoring probability
C) gave the probability
D) used 0.75
E) used 0.5
F) divided by 0.25
Section 7.5 Practice · Very Hard · Expected count
A fair six-sided die is rolled 30 times. Find the expected number of sixes.
A6
B1/6
C180
D5
E30
F3
Show solution
Correct answer: D
MethodExpected count = (number of trials) × P(success).
Trials30
P(six)1/6
30 × 1/6↓5
⚠️ Common trapMultiply the number of rolls by the probability of a six (1/6), giving a long-run average of 5.
Why each option
A) used 1/5
B) forgot to multiply by 30
C) multiplied by 6
D) ✓ 30 × 1/6
E) forgot the probability
F) used 1/10
CHAPTER 7 REVISION EXAM
Revision exam questions for 07-Probability.
M2-01-Algebra and Functions
Rule
Form
Negative power
a-n = 1/an
Fractional power
am/n = (√na)m
Surd product
√a √b = √(ab)
Conjugate
(a+√b)(a-√b) = a2 - b
√50 + √18 ≠ √68. Reduce each to a multiple of the same surd first: 5√2 + 3√2 = 8√2. Only like surds may be combined.
Value
Roots
b2 - 4ac > 0
two distinct real roots
b2 - 4ac = 0
one repeated root
b2 - 4ac < 0
no real roots
For x2 + kx + 9 = 0 to have equal roots, set k2 - 36 = 0, giving k = ± 6 — remember both signs.
Statement
Solution
(x-r)(x-s) > 0, r<s
x < r or x > s
(x-r)(x-s) < 0
r < x < s
|A| ≤ b
-b ≤ A ≤ b
|A| ≥ b
A ≤ -b or A ≥ b
A sketch settles it. For x2 - x - 6 > 0, the upward parabola (x-3)(x+2) is above the axis outside its roots, so x < -2 or x > 3 — not the inside region.
Theorem
Statement
Factor
p(a) = 0 &Leftrightarrow (x-a) is a factor
Remainder
remainder of p(x) ÷ (x-a) is p(a)
Root sum (cubic)
α+β+γ = -b/a
After dividing a cubic by its first factor you are left with a quadratic — factorise that too. (x-2)(x2 - 2x - 3) is not fully factorised; it becomes (x-2)(x-3)(x+1).
Idea
Rule
Composite
fg(x) = f(g(x)), g first
Inverse
swap x,y; solve for y
Inverse graph
reflection in y = x
Domain of f-1
range of f
With f(x) = 2x+1 and g(x) = x2: fg(3) = f(9) = 19 but gf(3) = g(7) = 49. Apply the inner function first.
New graph
Effect
f(x) + b
translate up b
f(x + a)
translate left a
a f(x)
vertical stretch × a
f(ax)
horizontal stretch × 1/a
-f(x), f(-x)
reflect in x-, y-axis
f(x+3) moves the graph left 3 (not right), and f(2x)compresses horizontally by 1/2 (not stretches). Outside changes behave normally; inside changes are reversed.
Revision
M2-02-Sequences and Series
MATHS 2 · CHAPTER 2: Sequences and Series
This chapter treats sequences and series at A-Level depth: describing sequences by an \(n\)th-term rule or a recurrence relation, the two great families of arithmetic and geometric progressions and their sums, the sum to infinity of a convergent geometric series, compact sigma notation, and the binomial expansion. The recurring skill is to identify the structure, extract \(a\), \(d\) or \(r\), and apply the right formula exactly — no calculator required.
A2.1 Sequences and Recurrence Relations
A sequence can be given by a position-to-term rule \(u_n=f(n)\), which yields any term directly from its position, or by a recurrence relation \(u_{n+1}=g(u_n)\) with a stated first term, which builds each term from the previous one. For example \(u_n=3n-2\) gives \(1,4,7,10,\dots\) directly.
Recurrences require you to iterate: \(u_{n+1}=2u_n+1,\ u_1=3\) gives \(u_2=7,\ u_3=15\). A sequence is increasing if every \(u_{n+1}>u_n\), periodic if the terms cycle, and convergent if the terms approach a limit.
📋 Two ways to define a sequence
Type
Meaning
Position-to-term
\(u_n=f(n)\): term from position
Recurrence
\(u_{n+1}=g(u_n)\) with \(u_1\) given
Increasing
\(u_{n+1}>u_n\) for all \(n\)
Limit
\(u_n\to L\) as \(n\to\infty\)
⚠️ A recurrence needs its first term
\(u_{n+1}=2u_n+1\) is meaningless without \(u_1\). And you cannot jump to \(u_{10}\) directly from a recurrence — you must iterate up to it.
Worked Example — Iterating a recurrence
Question: A sequence is defined by \(u_{n+1}=2u_n+1\) with \(u_1=3\). Find \(u_3\).
Working: Iterate: \(u_2=2(3)+1=7\), then \(u_3=2(7)+1=15\).
\(u_1=3\to u_2=7\to u_3=15\)
A2.1 · Medium · nth term
A sequence has \(n\)th term \(u_n=3n-2\). Find \(u_5\).
A\(15\)
B\(17\)
C\(13\)
D\(10\)
E\(3\)
F\(1\)
Show solution
Correct answer: C
MethodSubstitute \(n=5\) into the rule.
Substitute\(3(5)-2\)
\(3(5)-2\)↓\(13\)\(13\)
⚠️ Common trapMultiply first, then subtract: \(15-2=13\); do not subtract before multiplying.
Why each option
A) forgot the \(-2\)
B) used \(n=6\)-type slip
C) ✓ \(15-2=13\)
D) computed \(3(4)-2\)
E) used the coefficient
F) used \(u_1\)
A2.1 · Hard · Recurrence
For \(u_{n+1}=2u_n+1\) with \(u_1=3\), find \(u_3\).
A\(7\)
B\(13\)
C\(31\)
D\(9\)
E\(15\)
F\(23\)
Show solution
Correct answer: E
MethodIterate the recurrence twice.
\(u_2\)\(2(3)+1=7\)
\(u_3\)\(2(7)+1=15\)
\(u_2=7\)↓\(u_3=2(7)+1=15\)\(u_3=2(7)+1=15\)
⚠️ Common trapYou must reach \(u_3\) through \(u_2\); stopping at \(u_2=7\) is the usual error.
Why each option
A) stopped at \(u_2\)
B) used \(2u_1+... \) once
C) went one step too far
D) added instead of the rule
E) ✓ \(2(7)+1\)
F) arithmetic slip
A2.1 · Medium · Find the rule
What is the \(n\)th term of \(2,\ 5,\ 8,\ 11,\dots\)?
A\(3n-1\)
B\(3n+1\)
C\(2n+1\)
D\(n+3\)
E\(3n-2\)
F\(2n+3\)
Show solution
Correct answer: A
MethodThe common difference is the coefficient of \(n\); adjust the constant to fit \(u_1\).
Difference\(3\)
Fit \(u_1\)\(3(1)-1=2\)
\(u_n=3n+c,\ 3+c=2\)↓\(3n-1\)\(3n-1\)
⚠️ Common trapThe terms rise by \(3\), so the coefficient is \(3\); then \(c=-1\) makes \(u_1=2\).
Why each option
A) ✓ \(3n-1\)
B) wrong constant
C) wrong difference
D) wrong difference
E) off by one
F) wrong difference
A2.2 Arithmetic Sequences and Series
An arithmetic sequence adds a constant common difference \(d\) each step, so the \(n\)th term is \(u_n=a+(n-1)d\), where \(a\) is the first term. Its terms lie on a straight line when plotted against \(n\).
The sum of the first \(n\) terms — an arithmetic series — is \(S_n=\tfrac{n}{2}\big(2a+(n-1)d\big)\), equivalently \(S_n=\tfrac{n}{2}(a+l)\) where \(l\) is the last term. The second form (average of first and last, times how many) is often quickest.
📋 Arithmetic formulae
Quantity
Formula
\(n\)th term
\(u_n=a+(n-1)d\)
Sum
\(S_n=\tfrac{n}{2}\big(2a+(n-1)d\big)\)
Sum (first & last)
\(S_n=\tfrac{n}{2}(a+l)\)
⚠️ It is \((n-1)d\), not \(nd\)
The \(n\)th term uses \((n-1)\) differences, because the first term has had none added yet. Using \(a+nd\) shifts every term by one place.
Worked Example — Sum of an arithmetic series
Question: Find the sum of the first \(20\) terms of \(2,\ 5,\ 8,\dots\)
Working: Here \(a=2,\ d=3,\ n=20\). So \(S_{20}=\tfrac{20}{2}\big(2(2)+19(3)\big)=10(4+57)=610\).
\(S_{20}=10(4+57)=610\)
A2.2 · Medium · nth term
An arithmetic sequence has first term \(a=3\) and common difference \(d=5\). Find the \(10\)th term.
A\(53\)
B\(50\)
C\(45\)
D\(30\)
E\(8\)
F\(48\)
Show solution
Correct answer: F
MethodUse \(u_n=a+(n-1)d\).
Terms\(a=3,d=5,n=10\)
Compute\(3+9(5)\)
\(3+9\times5\)↓\(48\)\(48\)
⚠️ Common trapUse \(9\) differences, not \(10\): \(3+9(5)=48\).
Why each option
A) used \(a+10d\)
B) forgot the \(+3\)
C) used \(a+8d\)...
D) used \(u_1\times ...\)
E) used \(a+d\)
F) ✓ \(3+9(5)\)
A2.2 · Hard · Series sum
Find the sum of the first \(20\) terms of \(2,\ 5,\ 8,\ 11,\dots\)
A\(590\)
B\(610\)
C\(305\)
D\(620\)
E\(600\)
F\(1220\)
Show solution
Correct answer: B
Method\(S_n=\tfrac{n}{2}\big(2a+(n-1)d\big)\) with \(a=2,d=3\).
⚠️ Common trapPair first with last: \(\tfrac{50}{2}(51)=1275\); do not forget to halve.
Why each option
A) forgot to halve
B) used \(l=49\)
C) used \(n=49\)
D) ✓ \(25\times51\)
E) used \(n\) only
F) arithmetic slip
A2.3 Geometric Sequences and Series
A geometric sequence multiplies by a constant common ratio \(r\) each step, so the \(n\)th term is \(u_n=ar^{\,n-1}\). The ratio is found by dividing any term by the one before: \(r=\dfrac{u_{n+1}}{u_n}\).
The sum of the first \(n\) terms is \(S_n=\dfrac{a(1-r^{n})}{1-r}\) (for \(r\neq1\)). Writing it as \(\dfrac{a(r^{n}-1)}{r-1}\) is the same thing and is tidier when \(r>1\).
📋 Geometric formulae
Quantity
Formula
\(n\)th term
\(u_n=ar^{\,n-1}\)
Common ratio
\(r=\dfrac{u_{n+1}}{u_n}\)
Sum
\(S_n=\dfrac{a(1-r^{n})}{1-r}\)
⚠️ The power is \(n-1\)
The \(n\)th term is \(ar^{\,n-1}\), not \(ar^{n}\): the first term \(a=ar^{0}\) has been multiplied by \(r\) zero times.
Worked Example — Sum of a geometric series
Question: Find the sum of the first \(6\) terms of \(3,\ 6,\ 12,\dots\)
Working: Here \(a=3,\ r=2\). So \(S_6=\dfrac{3(2^{6}-1)}{2-1}=3(64-1)=189\).
\(S_6=3(2^{6}-1)=189\)
A2.3 · Medium · nth term
A geometric sequence has \(a=2\) and \(r=3\). Find the \(5\)th term.
⚠️ Common trapWith \(r-1=1\), the sum is simply \(3(64-1)=189\).
Why each option
A) used \(2^{6}\) without \(-1\)
B) summed only 5 terms
C) forgot the \(\times a\)
D) ✓ \(3(64-1)\)
E) doubled
F) off-by-one
A2.3 · Medium · Common ratio
Find the common ratio of \(5,\ 10,\ 20,\ 40,\dots\)
A\(5\)
B\(\tfrac12\)
C\(10\)
D\(15\)
E\(4\)
F\(2\)
Show solution
Correct answer: F
MethodDivide any term by the previous one.
Ratio\(\tfrac{10}{5}\)
\(\tfrac{10}{5}\)↓\(2\)\(2\)
⚠️ Common trapThe ratio is a division, not a difference: \(\tfrac{10}{5}=2\).
Why each option
A) used the first term
B) inverted the ratio
C) used a later term
D) used a difference
E) used \(\tfrac{20}{5}\)
F) ✓ \(\tfrac{10}{5}=2\)
A2.4 Sum to Infinity and Convergence
A geometric series converges — has a finite sum to infinity — precisely when \(|r|<1\). Then the terms shrink toward zero and the partial sums approach a limit \(S_\infty=\dfrac{a}{1-r}\). If \(|r|\ge1\) the series diverges and no sum to infinity exists.
Intuitively, in \(8+4+2+1+\dots\) each partial sum gets closer to \(16\) but never exceeds it; \(16\) is the sum to infinity.
📋 Sum to infinity
Condition
Result
\(|r|<1\)
converges: \(S_\infty=\dfrac{a}{1-r}\)
\(|r|\ge1\)
diverges: no \(S_\infty\)
Terms
\(u_n\to0\) when \(|r|<1\)
⚠️ Check \(|r|<1\) first
The formula \(\dfrac{a}{1-r}\) only applies when \(|r|<1\). Applying it to a series with \(|r|\ge1\) gives a meaningless finite “sum”.
Worked Example — Sum to infinity
Question: Find the sum to infinity of \(8+4+2+1+\dots\)
Working: Here \(a=8,\ r=\tfrac12\), and \(|r|<1\), so \(S_\infty=\dfrac{8}{1-\tfrac12}=\dfrac{8}{\tfrac12}=16\).
\(S_\infty=\dfrac{8}{1-\tfrac12}=16\)
A2.4 · Medium · Sum to infinity
Find the sum to infinity of \(8+4+2+1+\dots\)
A\(15\)
B\(16\)
C\(\infty\)
D\(8\)
E\(12\)
F\(32\)
Show solution
Correct answer: B
Method\(S_\infty=\dfrac{a}{1-r}\) with \(a=8,\ r=\tfrac12\).
⚠️ Common trapDivide by \(1-\tfrac13=\tfrac23\): \(12\times\tfrac32=18\).
Why each option
A) multiplied by \(3\)
B) used \(r=\tfrac12\)
C) ✓ \(12\div\tfrac23\)
D) divided by \(\tfrac43\)
E) used \(a\) alone
F) inverted twice
A2.5 Sigma Notation
Sigma notation compresses a sum: \(\displaystyle\sum_{r=1}^{n}u_r\) means add \(u_r\) as \(r\) runs from the lower limit to the upper. So \(\displaystyle\sum_{r=1}^{4}(2r+1)=3+5+7+9=24\).
Two useful properties: a constant multiplier factors out, \(\sum k\,u_r=k\sum u_r\), and a sum splits, \(\sum(u_r+v_r)=\sum u_r+\sum v_r\). Summing a constant gives \(\displaystyle\sum_{r=1}^{n}c=nc\).
📋 Sigma rules
Rule
Statement
Constant multiple
\(\sum k\,u_r=k\sum u_r\)
Sum splits
\(\sum(u_r+v_r)=\sum u_r+\sum v_r\)
Constant term
\(\sum_{r=1}^{n}c=nc\)
Meaning
substitute each \(r\), then add
⚠️ Mind the limits
\(\sum_{r=1}^{4}\) has four terms \((r=1,2,3,4)\), not three. Miscounting the number of terms is the most common sigma error.
⚠️ Common trapInclude \(r=4\): four terms \(3+5+7+9=24\), not three.
Why each option
A) dropped a term
B) used \(2r\) only somewhere
C) summed \(r=1,2,3\)
D) ✓ \(3+5+7+9\)
E) miscounted
F) added an extra term
A2.5 · Hard · Sum of squares
Evaluate \(\displaystyle\sum_{r=1}^{5}r^{2}\).
A\(30\)
B\(55\)
C\(50\)
D\(25\)
E\(15\)
F\(225\)
Show solution
Correct answer: B
MethodAdd the first five square numbers.
Squares\(1,4,9,16,25\)
\(1+4+9+16+25\)↓\(55\)\(55\)
⚠️ Common trapSquare each term before adding: \(1+4+9+16+25=55\), not \((1+2+3+4+5)^{2}\).
Why each option
A) summed \(1..5\) then \(\times2\)
B) ✓ \(1+4+9+16+25\)
C) dropped \(r=5\)
D) used \(5^{2}\) only
E) summed \(1..5\)
F) squared the sum
A2.5 · Medium · Geometric sigma
Evaluate \(\displaystyle\sum_{r=1}^{3}2^{r}\).
A\(8\)
B\(12\)
C\(7\)
D\(6\)
E\(15\)
F\(14\)
Show solution
Correct answer: F
MethodAdd \(2^{1}+2^{2}+2^{3}\).
Terms\(2,4,8\)
\(2+4+8\)↓\(14\)\(14\)
⚠️ Common trapStart the powers at \(r=1\) (so \(2\)), not \(r=0\): \(2+4+8=14\).
Why each option
A) used only \(2^{3}\)
B) used \(2r\) not \(2^{r}\)
C) included \(2^{0}\) wrongly
D) summed the exponents
E) added an extra term
F) ✓ \(2+4+8\)
A2.6 The Binomial Expansion
The binomial expansion expands \((a+b)^{n}\) for a positive integer \(n\): \(\displaystyle(a+b)^{n}=\sum_{r=0}^{n}\binom{n}{r}a^{\,n-r}b^{\,r}\), where \(\binom{n}{r}=\dfrac{n!}{r!(n-r)!}\) are the binomial coefficients (Pascal’s triangle). The powers of \(a\) fall while those of \(b\) rise, and each pair sums to \(n\).
The special case \((1+x)^{n}=1+nx+\dfrac{n(n-1)}{2!}x^{2}+\dots\) is used constantly. The coefficient of \(x^{r}\) in \((1+x)^{n}\) is simply \(\binom{n}{r}\).
📋 Binomial expansion
Item
Formula
General
\((a+b)^{n}=\sum\binom{n}{r}a^{\,n-r}b^{\,r}\)
Coefficient
\(\binom{n}{r}=\dfrac{n!}{r!(n-r)!}\)
Special case
\((1+x)^{n}=1+nx+\tfrac{n(n-1)}{2}x^{2}+\dots\)
⚠️ Track both powers
In \((a+b)^{n}\) the exponents of \(a\) and \(b\) must add to \(n\). For \(x^{r}\) in \((2+x)^{n}\), remember the \(2^{\,n-r}\) factor — it is easy to drop.
Worked Example — A specific coefficient
Question: Find the coefficient of \(x^{3}\) in \((2+x)^{4}\).
Working: The term is \(\binom{4}{3}2^{\,4-3}x^{3}=4\cdot2\cdot x^{3}\), so the coefficient is \(8\).
\(\binom{4}{3}2^{1}=4\times2=8\)
A2.6 · Medium · Coefficient in (1+x)^n
Find the coefficient of \(x^{2}\) in \((1+x)^{5}\).
A\(10\)
B\(5\)
C\(20\)
D\(25\)
E\(1\)
F\(15\)
Show solution
Correct answer: A
MethodThe coefficient of \(x^{r}\) in \((1+x)^{n}\) is \(\binom{n}{r}\).
\(\binom{5}{2}\)\(\dfrac{5\cdot4}{2}\)
\(\binom{5}{2}\)↓\(10\)\(10\)
⚠️ Common trapUse \(\binom{5}{2}=\tfrac{5\cdot4}{2}=10\); the coefficient is the binomial number, not \(n\).
Why each option
A) ✓ \(\binom{5}{2}=10\)
B) used \(n=5\)
C) used \(5\cdot4\) unhalved
D) squared \(5\)
E) used the constant term
F) miscomputed
A2.6 · Hard · Coefficient in (a+b)^3
In the expansion of \((a+b)^{3}\), what is the coefficient of \(ab^{2}\)?
A\(1\)
B\(2\)
C\(3\)
D\(6\)
E\(4\)
F\(9\)
Show solution
Correct answer: C
MethodIt is \(\binom{3}{2}\) (choosing which two factors give \(b\)).
\(\binom{3}{2}\)\(3\)
\(\binom{3}{2}\)↓\(3\)\(3\)
⚠️ Common trapThe expansion is \(a^{3}+3a^{2}b+3ab^{2}+b^{3}\); the \(ab^{2}\) coefficient is \(3\).
Why each option
A) used a corner term
B) miscounted
C) ✓ \(\binom{3}{2}=3\)
D) used \(3!\)
E) off by one
F) squared it
A2.6 · Very Hard · Coefficient in (2+x)^4
Find the coefficient of \(x^{3}\) in \((2+x)^{4}\).
⚠️ Common trapDo not drop the \(2^{\,4-3}=2\) factor: the coefficient is \(4\times2=8\), not \(4\).
Why each option
A) dropped the \(2^{1}\)
B) used \(2^{2}\)
C) used \(2^{3}\)
D) used \(2\) alone
E) ✓ \(4\times2\)
F) used \(4!\)-type slip
M2-03-Exponentials and Logarithms
CHAPTER 10: Exponentials and Logarithms
This chapter develops the exponential and logarithmic functions to A-Level depth: the shape and behaviour of \(y=a^{x}\) and the natural exponential \(y=e^{x}\), the laws of logarithms, the inverse relationship between \(e^{x}\) and \(\ln x\), the systematic solution of exponential and log equations, transformations of these graphs, and the use of logarithms to linearise exponential and power models. As throughout ESAT Mathematics, answers are left in exact form (in terms of \(e\), \(\ln\) and \(\log\)) rather than as decimals.
10.1 Exponential Functions and Their Graphs
An exponential function \(y=a^{x}\) with base \(a>0,\ a\neq1\) has domain all real numbers and range \(y>0\). Every such graph passes through \((0,1)\) because \(a^{0}=1\), and has the \(x\)-axis \((y=0)\) as a horizontal asymptote. If \(a>1\) the curve grows; if \(0<a<1\) it decays.
The natural exponential \(y=e^{x}\), with \(e\approx2.718\), is the case whose gradient equals its own value at every point — the reason it dominates growth and decay modelling. A decay function \(a^{x}\) with \(0<a<1\) can always be written \(e^{-kx}\) with \(k>0\).
📋 Features of \(y=a^{x}\)
Feature
Value
Passes through
\((0,1)\)
Range
\(y>0\)
Asymptote
\(y=0\)
\(a>1\)
growth
\(0<a<1\)
decay
⚠️ Exponentials are always positive
\(a^{x}\) can never be zero or negative: the graph approaches \(y=0\) but never touches it. So \(a^{x}=0\) has no solution, and \(a^{x}=-2\) is impossible.
Worked Example — Reading an exponential graph
Question: State the range and the equation of the asymptote of \(y=e^{x}+2\).
Working: Since \(e^{x}>0\), adding \(2\) gives \(y>2\); the asymptote \(y=0\) is shifted up to \(y=2\).
Through which point does every graph \(y=a^{x}\) pass, whatever the base \(a\)?
A\((1,0)\)
B\((1,1)\)
C\((0,1)\)
D\((0,0)\)
E\((a,1)\)
F\((0,a)\)
Show solution
Correct answer: C
MethodAny nonzero base to the power \(0\) equals \(1\).
At \(x=0\)\(a^{0}=1\)
\(a^{0}=1\)↓\((0,1)\)\((0,1)\)
⚠️ Common trapThe fixed point comes from \(a^{0}=1\), giving \((0,1)\) — not \((1,0)\), which is where log graphs cross.
Why each option
A) that is the log-graph point
B) only if \(a=1\)
C) ✓ \(a^{0}=1\)
D) graph never reaches \(y=0\)
E) not independent of \(a\)
F) that is \(x=1\) value
10.1 · Medium · Identify decay
Which function models exponential decay?
A\(y=5^{x}\)
B\(y=x^{3}\)
C\(y=e^{x}\)
D\(y=10^{x}\)
E\(y=\left(\tfrac34\right)^{x}\)
F\(y=2x\)
Show solution
Correct answer: E
MethodDecay needs an exponential base strictly between \(0\) and \(1\).
Base\(\tfrac34\in(0,1)\)
\(0<\tfrac34<1\)↓decaydecay
⚠️ Common trapOnly a base in \((0,1)\) decays; \(x^{3}\) and \(2x\) are not exponential, and bases \(>1\) grow.
Why each option
A) base \(>1\): growth
B) polynomial
C) growth
D) growth
E) ✓ base \(\tfrac34<1\)
F) linear
10.1 · Hard · Range with shift
State the range of \(y=e^{x}+2\).
A\(y>2\)
B\(y>0\)
C\(y\ge2\)
D\(y>-2\)
Eall real \(y\)
F\(y<2\)
Show solution
Correct answer: A
Method\(e^{x}>0\); add the vertical shift.
\(e^{x}\)\(>0\)
Shift\(+2\)
\(e^{x}>0\Rightarrow e^{x}+2>2\)↓\(y>2\)\(y>2\)
⚠️ Common trapThe asymptote is approached but never reached, so the inequality is strict: \(y>2\), not \(y\ge2\).
Why each option
A) ✓ \(e^{x}>0\Rightarrow>2\)
B) forgot the \(+2\)
C) asymptote not attained
D) wrong sign on shift
E) ignored positivity
F) inequality reversed
10.2 Logarithms and the Laws of Logs
A logarithm answers “to what power?”: \(\log_{a}x=y\) means exactly \(a^{y}=x\). Thus \(\log_{a}a=1\) and \(\log_{a}1=0\). The logarithm is the inverse of the exponential to the same base.
The three laws of logarithms turn products into sums and powers into multiples: \(\log xy=\log x+\log y\), \(\log\tfrac{x}{y}=\log x-\log y\), and \(\log x^{k}=k\log x\). To switch base use \(\log_{a}x=\dfrac{\log x}{\log a}\).
📋 Laws of logarithms
Law
Statement
Product
\(\log xy=\log x+\log y\)
Quotient
\(\log\tfrac{x}{y}=\log x-\log y\)
Power
\(\log x^{k}=k\log x\)
Special
\(\log_{a}a=1,\ \log_{a}1=0\)
Change of base
\(\log_{a}x=\dfrac{\log x}{\log a}\)
⚠️ The log of a sum does not split
\(\log(x+y)\neq\log x+\log y\). The product law applies to \(\log(xy)\) only. There is no law that simplifies \(\log(x+y)\).
Worked Example — Combining logs
Question: Evaluate \(\log_{2}40-\log_{2}5\).
Working: By the quotient law \(\log_{2}40-\log_{2}5=\log_{2}\tfrac{40}{5}=\log_{2}8\), and \(2^{3}=8\), so the value is \(3\).
\(\log_{2}\tfrac{40}{5}=\log_{2}8=3\)
10.2 · Medium · Evaluate log
Find \(\log_{3}81\).
A\(3\)
B\(27\)
C\(9\)
D\(\tfrac{1}{4}\)
E\(2\)
F\(4\)
Show solution
Correct answer: F
MethodAsk: \(3\) to what power is \(81\)?
\(3^{4}\)\(81\)
\(3^{4}=81\)↓\(4\)\(4\)
⚠️ Common trapCount the powers of \(3\): \(3,9,27,81\) is four steps, so \(\log_{3}81=4\).
⚠️ Common trapApply the power law before adding: \(2\log5=\log25\), then \(\log(25\cdot4)=\log100=2\).
Why each option
A) stopped at \(\log10\)
B) ✓ \(\log100=2\)
C) added the arguments
D) miscounted the power
E) multiplied \(2\cdot5\) then \(\cdot4\)
F) took the argument, not the log
10.2 · Hard · Quotient law
Simplify \(\log_{2}40-\log_{2}5\).
A\(\log_{2}35\)
B\(8\)
C\(\log_{2}45\)
D\(3\)
E\(2\)
F\(\tfrac{40}{5}\)
Show solution
Correct answer: D
MethodSubtraction of logs is the log of a quotient.
Quotient\(\log_{2}\tfrac{40}{5}\)
\(2^{3}\)\(8\)
\(\log_{2}8\)↓\(3\)\(3\)
⚠️ Common trapSubtracting logs gives \(\log_{2}\tfrac{40}{5}=\log_{2}8=3\); it is not \(\log_{2}(40-5)\).
Why each option
A) subtracted the arguments
B) gave the argument \(8\)
C) added the arguments
D) ✓ \(\log_{2}8=3\)
E) \(2^{2}\) slip
F) left it unevaluated
10.3 The Exponential Function \(e^{x}\) and Natural Logarithm \(\ln x\)
The natural logarithm \(\ln x=\log_{e}x\) is the inverse of \(e^{x}\). Being inverses, they undo each other: \(e^{\ln x}=x\) (for \(x>0\)) and \(\ln(e^{x})=x\) (for all \(x\)). Their graphs are reflections of one another in the line \(y=x\).
The graph of \(y=\ln x\) has domain \(x>0\), passes through \((1,0)\), and has the \(y\)-axis \((x=0)\) as a vertical asymptote. Useful exact values: \(\ln 1=0\), \(\ln e=1\).
📋 \(e^{x}\) and \(\ln x\)
Identity
Value
Undo (out)
\(e^{\ln x}=x\)
Undo (in)
\(\ln(e^{x})=x\)
\(\ln 1\)
\(0\)
\(\ln e\)
\(1\)
Domain of \(\ln x\)
\(x>0\)
⚠️ You cannot take \(\ln\) of \(\le 0\)
\(\ln x\) is defined only for \(x>0\). Any solution that would require \(\ln 0\) or \(\ln(\text{negative})\) must be rejected.
Worked Example — Undoing e and ln
Question: Solve \(e^{2x}=7\), giving your answer in exact form.
Working: Take natural logs of both sides: \(2x=\ln7\), so \(x=\tfrac12\ln7\).
\(e^{2x}=7\ \Rightarrow\ x=\tfrac12\ln7\)
10.3 · Medium · ln of e power
Evaluate \(\ln(e^{5})\).
A\(5\)
B\(e^{5}\)
C\(\ln5\)
D\(1\)
E\(5e\)
F\(e\)
Show solution
Correct answer: A
Method\(\ln\) and \(e^{x}\) are inverses: \(\ln(e^{x})=x\).
Inverse\(\ln(e^{5})=5\)
\(\ln(e^{5})\)↓\(5\)\(5\)
⚠️ Common trap\(\ln\) cancels the exponential exactly, leaving the exponent \(5\).
Why each option
A) ✓ \(\ln(e^{x})=x\)
B) did not apply \(\ln\)
C) pulled \(5\) out incorrectly
D) used \(\ln e\)
E) kept a stray \(e\)
F) over-cancelled
10.3 · Medium · e of ln
Evaluate \(e^{\ln 3}\).
A\(\ln3\)
B\(e^{3}\)
C\(1\)
D\(3\)
E\(3e\)
F\(0\)
Show solution
Correct answer: D
Method\(e^{\ln x}=x\) for \(x>0\).
Inverse\(e^{\ln 3}=3\)
\(e^{\ln 3}\)↓\(3\)\(3\)
⚠️ Common trapThe exponential undoes the natural log, returning the argument \(3\).
Why each option
A) did not cancel
B) swapped the operations
C) used \(\ln1\)
D) ✓ \(e^{\ln x}=x\)
E) stray factor
F) wrong identity
10.3 · Hard · Domain of ln
State the domain of \(y=\ln(x-2)\).
A\(x\ge2\)
B\(x>0\)
C\(x<2\)
Dall real \(x\)
E\(x>-2\)
F\(x>2\)
Show solution
Correct answer: F
MethodThe argument of a logarithm must be strictly positive.
Require\(x-2>0\)
\(x-2>0\)↓\(x>2\)\(x>2\)
⚠️ Common trapSet the argument \(>0\): \(x-2>0\Rightarrow x>2\); equality is excluded since \(\ln0\) is undefined.
Why each option
A) \(\ln0\) undefined
B) ignored the shift
C) inequality reversed
D) logs need positive argument
E) wrong sign
F) ✓ \(x-2>0\)
10.4 Solving Exponential and Logarithmic Equations
To solve \(a^{x}=b\), take logs of both sides: \(x\log a=\log b\), so \(x=\dfrac{\log b}{\log a}=\log_{a}b\). When the base is \(e\), use \(\ln\) directly. An equation that is quadratic in \(e^{x}\) (or in \(a^{x}\)) is solved by the substitution \(u=e^{x}\), factorising, then back-substituting — remembering \(u=e^{x}>0\).
For a logarithmic equation, combine to a single log, rewrite in exponential form, solve, and check every solution in the original — any that make a log argument \(\le0\) must be discarded.
📋 Solution strategies
Equation type
Method
\(a^{x}=b\)
\(x=\log_{a}b=\tfrac{\log b}{\log a}\)
\(e^{kx}=b\)
\(x=\tfrac1k\ln b\)
Quadratic in \(e^{x}\)
let \(u=e^{x}>0\), factorise
\(\log\) equation
combine, exponentiate, check domain
⚠️ Check for extraneous roots
After solving a log equation, substitute back. A value that makes any argument zero or negative is not a valid solution and must be rejected.
Worked Example — Quadratic in eˣ
Question: Solve \(e^{2x}-5e^{x}+6=0\).
Working: Let \(u=e^{x}\): \(u^{2}-5u+6=0\Rightarrow(u-2)(u-3)=0\), so \(e^{x}=2\) or \(e^{x}=3\). Both are positive, giving \(x=\ln2\) or \(x=\ln3\).
\((e^{x}-2)(e^{x}-3)=0\Rightarrow x=\ln2,\ \ln3\)
10.4 · Medium · Simple exponential
Solve \(3^{x}=\tfrac{1}{9}\).
A\(2\)
B\(-2\)
C\(3\)
D\(-3\)
E\(\tfrac13\)
F\(9\)
Show solution
Correct answer: B
MethodWrite \(\tfrac19\) as a power of \(3\).
\(\tfrac19\)\(3^{-2}\)
\(3^{x}=3^{-2}\)↓\(x=-2\)\(x=-2\)
⚠️ Common trapSince \(\tfrac19=3^{-2}\), equate exponents: \(x=-2\); the negative sign comes from the reciprocal.
Why each option
A) dropped the sign
B) ✓ \(3^{-2}=\tfrac19\)
C) used \(3^{3}\)
D) \(3^{-3}=\tfrac1{27}\)
E) inverted the answer
F) gave \(9\)
10.4 · Hard · Natural-log solution
Solve \(e^{2x}=7\), giving an exact answer.
A\(\ln7\)
B\(2\ln7\)
C\(\ln14\)
D\(\tfrac72\)
E\(\tfrac12\ln7\)
F\(\ln7-2\)
Show solution
Correct answer: E
MethodTake \(\ln\) of both sides and divide by \(2\).
Take ln\(2x=\ln7\)
Divide\(x=\tfrac12\ln7\)
\(2x=\ln7\)↓\(x=\tfrac12\ln7\)\(x=\tfrac12\ln7\)
⚠️ Common trapThe \(2\) is a coefficient of \(x\), so divide after taking logs: \(x=\tfrac12\ln7\).
Why each option
A) forgot to divide by \(2\)
B) multiplied by \(2\)
C) turned \(2\) into a factor of \(7\)
D) ignored the log
E) ✓ \(2x=\ln7\)
F) subtracted the \(2\)
10.4 · Very Hard · Log equation
Solve \(\log_{2}x+\log_{2}(x-2)=3\).
A\(-2\)
B\(4\text{ or }-2\)
C\(4\)
D\(8\)
E\(3\)
F\(2\)
Show solution
Correct answer: C
MethodCombine to one log, exponentiate, then check the domain.
⚠️ Common trapBoth factors give candidates, but \(x=-2\) makes \(\log_{2}x\) undefined, so only \(x=4\) survives.
Why each option
A) that root is rejected
B) kept the invalid root
C) ✓ \(x=4\), domain valid
D) used the \(2^{3}\) value directly
E) used the RHS
F) made argument zero
10.5 Transformations of Exponential and Log Graphs
The same transformation rules apply to \(y=e^{x}\) and \(y=\ln x\). Outside changes act on the output: \(e^{x}+c\) shifts up by \(c\) and moves the asymptote from \(y=0\) to \(y=c\); \(A\,e^{x}\) is a vertical stretch. Inside changes act on the input, reversed: \(e^{x-a}\) shifts right by \(a\), and \(e^{-x}\) reflects the graph in the \(y\)-axis.
For the logarithm, \(y=\ln(x-a)\) shifts right by \(a\) and moves the vertical asymptote from \(x=0\) to \(x=a\); \(-\ln x\) reflects in the \(x\)-axis.
📋 Transformations
New graph
Effect
\(e^{x}+c\)
shift up \(c\); asymptote \(y=c\)
\(A\,e^{x}\)
vertical stretch \(\times A\)
\(e^{x-a}\)
shift right \(a\)
\(e^{-x}\)
reflect in the \(y\)-axis
\(\ln(x-a)\)
shift right \(a\); asymptote \(x=a\)
⚠️ A vertical shift moves the asymptote
Adding a constant to \(e^{x}\) does not just raise the curve — it raises the asymptote too, from \(y=0\) to \(y=c\). Always restate the asymptote after a vertical shift.
Worked Example — Shift and asymptote
Question: The graph of \(y=e^{x}\) is transformed to \(y=e^{x}-3\). State the transformation and the new asymptote.
Working: The \(-3\) is outside the exponential, so the graph shifts down by \(3\); the asymptote moves from \(y=0\) to \(y=-3\).
\(y=e^{x}-3:\ \text{shift down }3,\ \text{asymptote } y=-3\)
10.5 · Hard · New asymptote
The graph of \(y=e^{x}\) is transformed to \(y=e^{x}+4\). What is the equation of the horizontal asymptote?
A\(y=0\)
B\(x=4\)
C\(y=-4\)
D\(y=4\)
E\(y=e^{4}\)
Fnone
Show solution
Correct answer: D
MethodA vertical shift of \(+4\) raises the asymptote by \(4\).
Old asymptote\(y=0\)
Shift\(+4\)
\(y=0+4\)↓\(y=4\)\(y=4\)
⚠️ Common trapThe \(+4\) lifts the whole graph, asymptote included, so \(y=0\) becomes \(y=4\).
Why each option
A) ignored the shift
B) that is a vertical line
C) wrong sign
D) ✓ \(y=0+4\)
E) evaluated \(e^{4}\)
F) asymptote still exists
10.5 · Hard · Log asymptote
The graph of \(y=\ln x\) is transformed to \(y=\ln(x+3)\). What is the equation of the vertical asymptote?
A\(x=3\)
B\(x=-3\)
C\(x=0\)
D\(y=-3\)
E\(y=3\)
F\(x=-1/3\)
Show solution
Correct answer: B
Method\(\ln(x+3)\) shifts the graph left \(3\); the asymptote \(x=0\) moves with it.
Argument \(=0\)\(x+3=0\)
\(x+3=0\)↓\(x=-3\)\(x=-3\)
⚠️ Common trapThe asymptote is where the argument hits \(0\): \(x+3=0\Rightarrow x=-3\) — an inside change moving left.
Why each option
A) sign reversed
B) ✓ \(x+3=0\)
C) forgot the shift
D) used a horizontal line
E) wrong axis and sign
F) divided instead
10.5 · Medium · Reflection
The graph of \(y=e^{-x}\) is the graph of \(y=e^{x}\) reflected in which line?
Athe \(x\)-axis
Bthe line \(y=x\)
Cthe line \(y=-x\)
Dthe origin
Ethe line \(x=1\)
Fthe \(y\)-axis
Show solution
Correct answer: F
MethodReplacing \(x\) with \(-x\) reflects a graph in the \(y\)-axis.
Change\(x\to-x\)
\(f(-x)\)↓reflect in \(y\)-axisreflect in \(y\)-axis
⚠️ Common trap\(f(-x)\) flips left–right, i.e. reflection in the \(y\)-axis; \(-f(x)\) would flip in the \(x\)-axis.
Why each option
A) that is \(-e^{x}\)
B) that is the inverse
C) not a standard reflection here
D) that is \(-f(-x)\)
E) unrelated line
F) ✓ \(x\to-x\)
10.6 Exponential Models and Linearising with Logarithms
Taking logs turns a curved model into a straight line, from which constants are read off as gradient and intercept. For an exponential model \(y=ab^{x}\), take logs: \(\log y=\log a+x\log b\). Plotting \(\log y\) against \(x\) gives a line of gradient \(\log b\) and intercept \(\log a\).
For a power model \(y=ax^{n}\), take logs: \(\log y=\log a+n\log x\). Plotting \(\log y\) against \(\log x\) gives a line of gradient \(n\) and intercept \(\log a\). Choosing the right pair of axes is the whole skill.
📋 Linearising models
Model
Linear form · axes
\(y=ab^{x}\)
\(\log y=\log a+x\log b\); \(\log y\) vs \(x\)
\(y=ax^{n}\)
\(\log y=\log a+n\log x\); \(\log y\) vs \(\log x\)
Gradient (exp)
\(\log b\)
Gradient (power)
\(n\)
Intercept
\(\log a\)
⚠️ Right axes for the right model
An exponential \(ab^{x}\) linearises with \(\log y\) against \(x\); a power law \(ax^{n}\) needs \(\log y\) against \(\log x\). Using the wrong axes will not give a line.
Worked Example — Reading constants off a log fit
Question: Experimental data fit \(\log_{10}y=2+0.5x\), modelled by \(y=ab^{x}\). Find \(a\).
Working: Comparing with \(\log y=\log a+x\log b\), the intercept is \(\log_{10}a=2\), so \(a=10^{2}=100\) (and \(\log_{10}b=0.5\Rightarrow b=10^{0.5}=\sqrt{10}\)).
\(\log_{10}a=2\Rightarrow a=100\)
10.6 · Hard · Which axes
To obtain a straight line from \(y=ab^{x}\), which quantities should be plotted?
A\(\log y\) against \(x\)
B\(\log y\) against \(\log x\)
C\(y\) against \(x\)
D\(y\) against \(\log x\)
E\(x\) against \(\log y\)
F\(\log x\) against \(y\)
Show solution
Correct answer: A
MethodTake logs of \(y=ab^{x}\): \(\log y=\log a+x\log b\).
Linear form\(\log y=\log a+x\log b\)
\(\log y=(\log b)x+\log a\)↓\(\log y\) vs \(x\)\(\log y\) vs \(x\)
⚠️ Common trapOnly the \(x\) sits in the exponent, so it stays linear; take \(\log\) of \(y\) alone — \(\log y\) against \(x\).
Why each option
A) ✓ \(\log y=\log a+x\log b\)
B) that linearises a power law
C) stays curved
D) wrong for exponential
E) axes swapped
F) axes swapped
10.6 · Hard · Find a from intercept
Data fit \(\log_{10}y=2+0.5x\) with model \(y=ab^{x}\). Find the constant \(a\).
A\(2\)
B\(10\)
C\(100\)
D\(0.5\)
E\(20\)
F\(\sqrt{10}\)
Show solution
Correct answer: C
MethodThe intercept equals \(\log_{10}a\).
Intercept\(\log_{10}a=2\)
\(a=10^{2}\)↓\(100\)\(100\)
⚠️ Common trapThe intercept \(2\) is \(\log_{10}a\), so \(a=10^{2}=100\); the gradient \(0.5\) instead gives \(b\).
Why each option
A) left it as the log
B) used \(10^{1}\)
C) ✓ \(a=10^{2}\)
D) used the gradient
E) misread the intercept
F) that is \(b\)
10.6 · Very Hard · Power-law index
Data fit \(\log_{10}y=1+3\log_{10}x\), modelled by \(y=ax^{n}\). State \(n\).
A\(1\)
B\(10\)
C\(30\)
D\(\tfrac13\)
E\(3\)
F\(100\)
Show solution
Correct answer: E
MethodCompare with \(\log y=\log a+n\log x\): \(n\) is the gradient.
Gradient\(n=3\)
Intercept\(\log a=1\Rightarrow a=10\)
\(\log y=\log a+n\log x\)↓\(n=3\)\(n=3\)
⚠️ Common trapAgainst \(\log x\), the gradient is the power \(n=3\); the intercept \(1\) gives \(a=10\), not \(n\).
Why each option
A) used the intercept
B) used \(10^{1}\)
C) multiplied \(1\times3\times10\)
D) inverted the gradient
E) ✓ gradient \(=n=3\)
F) used \(10^{2}\)
M2-03-Trigonometry
CHAPTER 9: Trigonometry
This chapter takes trigonometry to A-Level depth. It moves beyond right-angled triangles to the circular functions \(\sin\), \(\cos\) and \(\tan\) defined for all angles, their graphs and transformations, the web of identities (Pythagorean, compound- and double-angle, and the harmonic \(R\sin(x+\alpha)\) form), the systematic solution of trigonometric equations over a given interval, and the sine and cosine rules for any triangle. Angles are handled in both degrees and radians; as always in ESAT Mathematics the work is done without a calculator, so exact values and identities are the tools of the trade.
9.1 Radian Measure, Arc and Sector
A radian is the angle subtended at the centre of a circle by an arc equal in length to the radius. Since the full circumference is \(2\pi r\), a complete turn is \(2\pi\) radians, giving the master conversion \(\pi \text{ rad} = 180^\circ\). To convert, multiply degrees by \(\pi/180\), or radians by \(180/\pi\). Key exact angles become \(30^\circ=\tfrac{\pi}{6}\), \(45^\circ=\tfrac{\pi}{4}\), \(60^\circ=\tfrac{\pi}{3}\), \(90^\circ=\tfrac{\pi}{2}\).
With \(\theta\) in radians, an arc of a circle of radius \(r\) has length \(s = r\theta\), and the sector it bounds has area \(A = \tfrac12 r^2\theta\). The area of the segment cut off by the chord is the sector minus the triangle: \(\tfrac12 r^2(\theta - \sin\theta)\).
📋 Radian formulae (θ in radians)
Quantity
Formula
Degree–radian
\(\pi \text{ rad} = 180^\circ\)
Arc length
\(s = r\theta\)
Sector area
\(A = \tfrac12 r^2\theta\)
Segment area
\(\tfrac12 r^2(\theta - \sin\theta)\)
⚠️ Radians only
The formulae \(s = r\theta\) and \(A = \tfrac12 r^2\theta\) are valid only when \(\theta\) is in radians. Feeding in degrees gives nonsense — convert first, e.g. \(60^\circ = \tfrac{\pi}{3}\).
Worked Example — Arc and sector
Question: A sector of a circle has radius \(6\) and angle \(\theta = \tfrac{\pi}{3}\). Find its arc length and area.
⚠️ Common trapSquare the radius before multiplying, and keep the factor \(\tfrac12\): \(\tfrac12\cdot100\cdot\tfrac{\pi}{5}=10\pi\).
Why each option
A) ✓ \(\tfrac12\cdot100\cdot\tfrac{\pi}{5}=10\pi\)
B) dropped the \(\tfrac12\)
C) used \(s=r\theta\) form
D) forgot to square r
E) divided by \(2\) twice
F) forgot the \(\tfrac{\theta}{ }\) factor
9.2 Trigonometric Graphs and Their Properties
Defined on the unit circle, \(y=\sin x\) and \(y=\cos x\) are periodic with period \(2\pi\) (\(360^\circ\)), have amplitude \(1\) and range \([-1,1]\). The cosine graph is the sine graph shifted left by \(\tfrac{\pi}{2}\): \(\cos x = \sin\!\left(x+\tfrac{\pi}{2}\right)\). Sine is an odd function (\(\sin(-x)=-\sin x\)); cosine is even (\(\cos(-x)=\cos x\)).
The tangent graph \(y=\tan x = \dfrac{\sin x}{\cos x}\) has period \(\pi\) (\(180^\circ\)), not \(2\pi\), with vertical asymptotes wherever \(\cos x = 0\), i.e. at \(x = \tfrac{\pi}{2} + k\pi\). Its range is all real numbers.
📋 Standard trig graphs
Graph
Period · Range
\(y=\sin x\)
period \(2\pi\), range \([-1,1]\)
\(y=\cos x\)
period \(2\pi\), range \([-1,1]\)
\(y=\tan x\)
period \(\pi\), range \(\mathbb{R}\)
Asymptotes of \(\tan\)
\(x=\tfrac{\pi}{2}+k\pi\)
⚠️ Tangent’s period is π
Do not assume every trig graph repeats every \(2\pi\). \(y=\tan x\) repeats every \(\pi\) and is undefined at \(x=\tfrac{\pi}{2}+k\pi\), where it has asymptotes.
Worked Example — Period, amplitude, range
Question: State the amplitude, period and range of \(y = 3\sin 2x\).
Working: The amplitude is the multiplier outside, \(3\). The period is \(\tfrac{2\pi}{2}=\pi\). The range is therefore \([-3,3]\).
⚠️ Common trapApply the stretch \((\times3)\) and the shift \((+2)\) to both endpoints: min \(=-1\), max \(=5\).
Why each option
A) forgot the \(+2\) shift
B) ✓ \([2-3,\ 2+3]\)
C) used only the \(+2\)
D) used amplitude \(5\)
E) shifted by \(1\)
F) clipped the minimum at 0
9.2 · Hard · Identify the graph
Which function has period \(\pi\) and vertical asymptotes at \(x=\tfrac{\pi}{2}+k\pi\)?
A\(y=\sin x\)
B\(y=\cos x\)
C\(y=\sin 2x\)
D\(y=\tan x\)
E\(y=2\cos x\)
F\(y=\cos 2x\)
Show solution
Correct answer: D
MethodOnly \(\tan\) has period \(\pi\) and asymptotes where \(\cos x=0\).
Period\(\pi\)
Undefined at\(\cos x=0\)
period \(\pi\), \(\cos x=0\)↓\(y=\tan x\)\(y=\tan x\)
⚠️ Common trapAsymptotes occur only for functions with a \(\cos x\) in the denominator; \(\sin\) and \(\cos\) are bounded and continuous.
Why each option
A) bounded, period \(2\pi\)
B) bounded, period \(2\pi\)
C) period \(\pi\) but no asymptotes
D) ✓ tangent
E) bounded
F) period \(\pi\) but bounded
9.3 Transformations of Trigonometric Graphs
Any sinusoid can be written \(y = a\,\sin\!\big(b(x-c)\big) + d\). Here \(a\) is the amplitude (a vertical stretch), \(b\) sets the period \(\tfrac{2\pi}{b}\) (a horizontal stretch of factor \(\tfrac1b\)), \(c\) is the phase shift (right by \(c\)), and \(d\) is the vertical shift. As always, changes outside the function behave normally; changes inside are reversed.
To read the phase shift correctly you must factor out \(b\) first: \(\sin(2x-\tfrac{\pi}{3}) = \sin\!\big(2(x-\tfrac{\pi}{6})\big)\), so the shift is \(\tfrac{\pi}{6}\) to the right, not \(\tfrac{\pi}{3}\).
📋 Effect of each parameter
Transformation
Effect on \(y=f(x)\)
\(a\,f(x)\)
vertical stretch \(\times a\) (amplitude)
\(f(bx)\)
horizontal stretch \(\times\tfrac1b\); period \(\tfrac{2\pi}{b}\)
\(f(x-c)\)
translate right \(c\)
\(f(x)+d\)
translate up \(d\)
⚠️ Factor before reading the shift
In \(y=\sin(bx-c)\) the phase shift is \(\tfrac{c}{b}\), not \(c\). Always rewrite as \(\sin\!\big(b(x-\tfrac{c}{b})\big)\) before stating the translation.
Worked Example — Combined transformation
Question: Describe how \(y=\cos x\) maps to \(y = 4\cos\!\left(x-\tfrac{\pi}{2}\right)\), and simplify.
Working: The \(4\) is a vertical stretch (amplitude \(4\)); the \(-\tfrac{\pi}{2}\) inside shifts the graph right by \(\tfrac{\pi}{2}\). Since \(\cos\!\left(x-\tfrac{\pi}{2}\right)=\sin x\), the result is \(4\sin x\).
The graph of \(y=\sin x\) is transformed into \(y=\sin\!\left(x-\tfrac{\pi}{4}\right)\). Describe the transformation.
Atranslation right \(\tfrac{\pi}{4}\)
Btranslation left \(\tfrac{\pi}{4}\)
Cvertical stretch \(\tfrac{\pi}{4}\)
Dtranslation up \(\tfrac{\pi}{4}\)
Ereflection in the \(x\)-axis
Fhorizontal stretch \(\tfrac{\pi}{4}\)
Show solution
Correct answer: A
Method\(f(x-c)\) translates the graph right by \(c\).
Inside\(x-\tfrac{\pi}{4}\)
Directionright (opposite of sign)
\(f(x-\tfrac{\pi}{4})\)↓translation right \(\tfrac{\pi}{4}\)translation right \(\tfrac{\pi}{4}\)
⚠️ Common trapA change inside the bracket is “backwards”: \(x-\tfrac{\pi}{4}\) moves the graph right, not left.
Why each option
A) ✓ right \(\tfrac{\pi}{4}\)
B) sign reversed
C) not a stretch
D) that is an outside change
E) no reflection here
F) not a stretch
9.3 · Hard · Amplitude and period
State the amplitude and period of \(y = 5\sin\!\left(\tfrac{1}{2}x\right)\).
Aamplitude \(5\), period \(\pi\)
Bamplitude \(\tfrac12\), period \(4\pi\)
Camplitude \(5\), period \(2\pi\)
Damplitude \(5\), period \(4\pi\)
Eamplitude \(10\), period \(4\pi\)
Famplitude \(5\), period \(\tfrac{\pi}{2}\)
Show solution
Correct answer: D
MethodAmplitude is the outside factor; period is \(\tfrac{2\pi}{b}\) with \(b=\tfrac12\).
Amplitude\(5\)
Period\(\tfrac{2\pi}{1/2}=4\pi\)
\(a=5,\ b=\tfrac12\)↓amplitude \(5\), period \(4\pi\)amplitude \(5\), period \(4\pi\)
⚠️ Common trapWith \(b=\tfrac12\) the period is \(\tfrac{2\pi}{1/2}=4\pi\) — dividing by a fraction stretches the graph.
Why each option
A) used period \(\pi\) (b=2)
B) confused amplitude with b
C) ignored the \(\tfrac12\)
D) ✓ \(5\); \(\tfrac{2\pi}{1/2}=4\pi\)
E) doubled the amplitude
F) used tan period
9.3 · Hard · Max value
What is the maximum value of \(y = 2\sin x + 1\)?
A\(2\)
B\(1\)
C\(-1\)
D\(5\)
E\(0\)
F\(3\)
Show solution
Correct answer: F
Method\(\sin x\) peaks at \(1\); substitute the maximum.
Max of \(\sin x\)\(1\)
Value\(2(1)+1\)
\(2(1)+1\)↓\(3\)\(3\)
⚠️ Common trapThe maximum occurs when \(\sin x=1\); apply the amplitude then the shift: \(2(1)+1=3\).
Why each option
A) forgot the \(+1\)
B) used \(\sin x=0\)
C) used the minimum
D) used amplitude \(4\)
E) took \(\sin x=-\tfrac12\)
F) ✓ \(2+1\)
9.4 Trigonometric Identities
The identities let you rewrite one expression as another. The Pythagorean identity \(\sin^2\theta+\cos^2\theta=1\) (dividing by \(\cos^2\) gives \(1+\tan^2\theta=\sec^2\theta\)) underpins everything, together with \(\tan\theta=\tfrac{\sin\theta}{\cos\theta}\). The compound-angle formulae \(\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B\) and \(\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B\) generate the rest.
Setting \(B=A\) gives the double-angle results \(\sin 2A=2\sin A\cos A\) and \(\cos 2A=\cos^2A-\sin^2A=2\cos^2A-1=1-2\sin^2A\). Finally, any \(a\sin x+b\cos x\) can be written in harmonic form \(R\sin(x+\alpha)\) with \(R=\sqrt{a^2+b^2}\) and \(\tan\alpha=\tfrac{b}{a}\).
📋 Core identities
Name
Identity
Pythagorean
\(\sin^2\theta+\cos^2\theta=1\)
Compound
\(\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B\)
Double (sin)
\(\sin 2A=2\sin A\cos A\)
Double (cos)
\(\cos 2A=2\cos^2A-1=1-2\sin^2A\)
Harmonic
\(a\sin x+b\cos x=R\sin(x+\alpha)\)
⚠️ Pick the right form of cos 2A
\(\cos 2A\) has three equivalent forms. Choose the one that matches the rest of the problem: use \(1-2\sin^2A\) when the expression involves \(\sin\), and \(2\cos^2A-1\) when it involves \(\cos\).
Worked Example — Harmonic form
Question: Express \(3\sin x + 4\cos x\) in the form \(R\sin(x+\alpha)\), \(R>0\), \(0<\alpha<\tfrac{\pi}{2}\).
Working: Here \(R=\sqrt{3^2+4^2}=5\) and \(\tan\alpha=\tfrac{4}{3}\), so \(\alpha=\arctan\tfrac43\approx 53.13^\circ\). Thus \(3\sin x+4\cos x = 5\sin(x+53.13^\circ)\).
\(3\sin x+4\cos x = 5\sin(x+\alpha),\ \tan\alpha=\tfrac43\)
9.4 · Hard · Simplify
Simplify \(\dfrac{1-\cos 2x}{\sin 2x}\).
A\(\cot x\)
B\(\tan x\)
C\(\sin x\)
D\(2\tan x\)
E\(\tan 2x\)
F\(1\)
Show solution
Correct answer: B
MethodUse \(1-\cos 2x=2\sin^2x\) and \(\sin 2x=2\sin x\cos x\).
⚠️ Common trapWith \(a=\sqrt3\) (the \(\sin\) coefficient) and \(b=1\), \(\tan\alpha=\tfrac{b}{a}=\tfrac{1}{\sqrt3}\), giving \(\alpha=\tfrac{\pi}{6}\).
Why each option
A) swapped \(a,b\) in \(\tan\alpha\)
B) used \(R=a^2+b^2\)
C) forgot to combine \(R\)
D) wrong function (cos)
E) ✓ \(R=2,\ \alpha=\tfrac{\pi}{6}\)
F) sign of \(\alpha\) wrong
9.4 · Hard · Double angle value
Given \(\sin A=\tfrac{3}{5}\) and \(A\) is acute, find \(\cos 2A\).
⚠️ Common trapUsing \(1-2\sin^2A\) avoids needing \(\cos A\); \(1-\tfrac{18}{25}=\tfrac{7}{25}\).
Why each option
A) computed \(\sin 2A\)
B) sign slip
C) ✓ \(1-\tfrac{18}{25}\)
D) used \(\cos^2A\) only
E) gave \(\sin^2A\)
F) wrong sign on \(\sin 2A\)
9.5 Solving Trigonometric Equations
To solve a trig equation over an interval, find the principal value from the inverse function, then use the graph’s symmetry (or the CAST diagram) to list every solution in the required range. For \(\sin x=k\) the second solution is \(\pi-\) (principal); for \(\cos x=k\) it is \(-\) (principal); \(\tan x=k\) repeats every \(\pi\).
Harder equations are quadratic in a single ratio — factorise as you would any quadratic — or need an identity to reduce two ratios to one. Never divide through by \(\cos x\) or \(\sin x\): that discards roots. Factor out instead.
📋 Second solutions (interval \([0,2\pi)\))
Equation
Solutions
\(\sin x=k\)
\(x_0\) and \(\pi-x_0\)
\(\cos x=k\)
\(x_0\) and \(2\pi-x_0\)
\(\tan x=k\)
\(x_0\) and \(x_0+\pi\)
Quadratic in \(\sin/\cos\)
factorise, solve each factor
⚠️ Don’t divide by a trig factor
Solving \(\sin x\cos x=\sin x\) by dividing by \(\sin x\) loses the roots where \(\sin x=0\). Instead write \(\sin x(\cos x-1)=0\) and solve each factor.
Worked Example — Quadratic in sin
Question: Solve \(2\sin^2x-\sin x-1=0\) for \(0\le x<2\pi\).
Working: Factorise: \((2\sin x+1)(\sin x-1)=0\), so \(\sin x=-\tfrac12\) or \(\sin x=1\). From \(\sin x=1\): \(x=\tfrac{\pi}{2}\). From \(\sin x=-\tfrac12\): \(x=\tfrac{7\pi}{6},\tfrac{11\pi}{6}\).
⚠️ Common trapBecause \(\tan\) has period \(\pi\), add \(\pi\) (not \(2\pi\)) to the principal value to get the second solution.
Why each option
A) used \(\pi-x_0\) (sin rule)
B) wrong reference angle
C) missed the period-\(\pi\) root
D) added \(2\pi\) then wrapped
E) wrong reference angle
F) ✓ \(\tfrac{\pi}{3},\tfrac{\pi}{3}+\pi\)
9.6 The Sine and Cosine Rules; Triangle Area
For any triangle with sides \(a,b,c\) opposite angles \(A,B,C\), the sine rule \(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\) is used when you know two angles and a side, or two sides and a non-included angle (the ambiguous case). The cosine rule \(a^2=b^2+c^2-2bc\cos A\) is used for two sides and the included angle, or for all three sides when finding an angle.
The area of a triangle from two sides and the included angle is \(\text{Area}=\tfrac12 ab\sin C\). Rearranging the cosine rule gives \(\cos A=\dfrac{b^2+c^2-a^2}{2bc}\), handy for finding an angle from three sides.
Given two sides and a non-included angle, \(\sin\) can yield two valid angles (\(\theta\) and \(180^\circ-\theta\)). Always check whether the obtuse alternative also fits the triangle.
Worked Example — Cosine rule
Question: In a triangle, \(b=7\), \(c=5\) and the included angle \(A=60^\circ\). Find \(a\).
Working: By the cosine rule \(a^2=7^2+5^2-2\cdot7\cdot5\cos 60^\circ=49+25-70\cdot\tfrac12=39\), so \(a=\sqrt{39}\approx 6.24\).
\(a^2=49+25-35=39,\quad a=\sqrt{39}\)
9.6 · Hard · Triangle area
Find the area of a triangle with sides \(a=8\), \(b=6\) and included angle \(C=30^\circ\).
⚠️ Common trapPlace the side you want the angle opposite as \(a\); \(\dfrac{12}{60}=\dfrac15\).
Why each option
A) wrong sides in denominator
B) subtracted the wrong square
C) arithmetic slip
D) used \(a=6\)
E) ✓ \(\tfrac{12}{60}\)
F) used a side ratio
M2-04-Functions and Graphs
ESAT Mathematics 2 · CHAPTER 08: Functions and Graphs
This chapter develops the graph language needed for ESAT Mathematics 2: modulus graphs, mappings, domains, ranges, composite functions, inverse functions, and the two important transformations \( y=|f(x)| \) and \( y=f(|x|) \). ESAT questions rarely ask for routine plotting only; they usually test whether a candidate can read restrictions, identify hidden branches, and avoid invalid algebraic steps. Every multiple-choice question below includes a full method, TRAP analysis, a Teacher comment, and an EduCoach comment.
8.1 Modulus Functions and Graphs
The modulus symbol \( |a| \) gives the non-negative size of a number. For a function, \( |f(x)| \) keeps the parts where \( f(x) \ge 0 \) and reflects the parts where \( f(x) \lt 0 \) in the \( x \)-axis.
For ESAT, the key skill is to split the argument into cases. If \( ax+b \ge 0 \), then \( |ax+b|=ax+b \); if \( ax+b \lt 0 \), then \( |ax+b|=-(ax+b) \). Graphs, equations, and inequalities all follow from this split.
📋 Reference facts
Idea
ESAT-ready form
Definition
\( |u|=u \) if \( u \ge 0 \), and \( |u|=-u \) if \( u \lt 0 \).
Vertex of \( y=|ax+b| \)
At \( ax+b=0 \), so \( x=-\dfrac{b}{a} \).
Equation principle
\( |u|=k \) has two cases only when \( k \ge 0 \).
⚠️ Do not remove the modulus too early
A common error is to write \( |3x-5|=3x-5 \) for all \( x \). This is only true on the branch where \( 3x-5 \ge 0 \).
Worked Example — Core method
Question: Solve \( |3x-5|=7 \).
Working: Use \( 3x-5=7 \) or \( 3x-5=-7 \). This gives \( x=4 \) or \( x=-\dfrac{2}{3} \).
\( x=4 \text{ or } x=-\dfrac{2}{3} \)
8.1 · Easy · Basic equation
Solve \( |3x-5|=7 \).
A\( \left\{4,-\dfrac{2}{3}\right\} \)
B\( \left\{4,\dfrac{2}{3}\right\} \)
C\( \left\{-4,\dfrac{2}{3}\right\} \)
D\( \left\{\dfrac{2}{3}\right\} \)
E\( \left\{4\right\} \)
F\( \varnothing \)
Show solution
Correct answer: A
MethodSet the argument equal to both \( 7 \) and \( -7 \).
\( 3x-5=7 \text{ or } 3x-5=-7 \)↓\( x=4 \text{ or } x=-\dfrac{2}{3} \)↓\( \left\{4,-\dfrac{2}{3}\right\} \)
⚠️ TRAP analysisThe second branch is not obtained by changing the final answer sign; it comes from changing the argument to \( -7 \).
Teacher commentGood ESAT work shows both branches before selecting an option.
EduCoach commentTrain students to write the two equations immediately; it prevents sign-loss errors.
Why each option
A) ✓ Correct: both branches are included.
B) The sign of the second solution is wrong.
C) The positive branch has been negated incorrectly.
D) Only one branch is used, and with the wrong sign.
E) Only the positive branch is kept.
F) There are real solutions because \( 7 \ge 0 \).
8.1 · Easy · Modulus inequality
Solve \( |2x+1| \lt 5 \).
A\( -3 \lt x \lt 2 \)
B\( -2 \lt x \lt 3 \)
C\( x \lt -3 \text{ or } x \gt 2 \)
D\( -5 \lt x \lt 5 \)
E\( -2 \le x \le 3 \)
F\( x \gt 2 \)
Show solution
Correct answer: A
MethodFor \( |u| \lt k \), use \( -k \lt u \lt k \).
⚠️ TRAP analysisThe minimum is not at a single point; every \( x \) between \( 2 \) and \( 5 \) gives the same minimum.
Teacher commentThis is a classic ESAT reasoning shortcut.
EduCoach commentEncourage students to draw a number line instead of differentiating a piecewise linear expression.
Why each option
A) ✓ Correct.
B) The two distances cannot both be zero.
C) Too small; the fixed points are 3 units apart.
D) This is the value at \( x=0 \) or \( x=7 \), not the minimum.
E) A non-minimal outside value.
F) A minimum exists on the whole interval \( [2,5] \).
8.2 Functions, Mappings, Domain and Range
A function assigns each input in its domain to exactly one output. A mapping can fail to be a function if one input has two different outputs, or if an input in the stated domain is not mapped at all.
Domain restrictions are often the whole question. Square-root functions need non-negative arguments, rational functions exclude zero denominators, and one-to-one behaviour may require restricting a graph to one side of a turning point.
📋 Reference facts
Idea
ESAT-ready form
Function test
Each input has exactly one output.
Domain
Allowed input values.
Range
Possible output values after the domain restriction.
One-to-one
Each output is produced by at most one input.
⚠️ Ignoring the stated domain
The same formula can have different ranges or one-to-one status under different domains. Always read the domain before using the graph.
Worked Example — Core method
Question: Find the range of \( f(x)=(x-1)^2+4 \) for \( -2\le x\le 3 \).
Working: The vertex \( x=1 \) lies in the domain, so the minimum is \( 4 \). At the endpoints, \( f(-2)=13 \) and \( f(3)=8 \), so the maximum is \( 13 \).
\( 4\le f(x)\le 13 \)
8.2 · Easy · Finite domain
For \( f(x)=x^2-1 \) on \( \{-2,-1,0,1\} \), what is the range and type?
A\( \{-1,0,3\}, \text{ many-to-one} \)
B\( \{-1,0,3\}, \text{ one-to-one} \)
C\( \{0,3\}, \text{ many-to-one} \)
D\( \{-2,-1,0,1\}, \text{ one-to-one} \)
E\( \{-1,3\}, \text{ many-to-one} \)
F\( \mathbb{R}, \text{ many-to-one} \)
Show solution
Correct answer: A
MethodEvaluate the formula only at the stated inputs.
⚠️ TRAP analysisThe vertex is determined by \( x=3 \), not by the roots \( 1 \) and \( 5 \).
Teacher commentThis connects graph shape with inverse existence.
EduCoach commentCoach: one-to-one starts at the turning point, not at an intercept.
Why each option
A) ✓ Correct.
B) Wrong sign for the vertex.
C) Too restrictive; it works but is not least.
D) Still includes both sides of the vertex.
E) Too restrictive.
F) This is a root, not the vertex.
8.3 Composite Functions
A composite function applies one function after another. The notation \( fg(x) \) means \( f(g(x)) \): apply \( g \) first, then apply \( f \). In general, \( fg(x) \ne gf(x) \).
For ESAT, the main difficulty is order and domain. A composite involving a reciprocal, square root, or logarithm may have a smaller domain than either original function alone.
📋 Reference facts
Idea
ESAT-ready form
Order
\( fg(x)=f(g(x)) \), so \( g \) acts first.
Repeated function
\( f^2(x)=f(f(x)) \), not \( (f(x))^2 \) unless stated.
Domain of a composite
The output of the inner function must be allowed as an input of the outer function.
⚠️ Confusing notation
The expression \( f^2(x) \) in this chapter means \( f(f(x)) \), not automatically the square of \( f(x) \).
Worked Example — Core method
Question: Let \( f(x)=x^2+1 \) and \( g(x)=3x-2 \). Find \( fg(2) \).
Working: First find \( g(2)=4 \). Then \( f(4)=17 \).
\( fg(2)=17 \)
8.3 · Easy · Numeric composite
Let \( f(x)=x^2+1 \) and \( g(x)=3x-2 \). Find \( fg(2) \).
A\( 17 \)
B\( 13 \)
C\( 37 \)
D\( 5 \)
E\( 10 \)
F\( 25 \)
Show solution
Correct answer: A
MethodCompute the inner function first: \( fg(2)=f(g(2)) \).
\( g(2)=4 \)↓\( f(4)=4^2+1=17 \)↓\( 17 \)
⚠️ TRAP analysisDo not compute \( f(2)g(2) \); juxtaposition means composition here.
Teacher commentA simple notation question can still cost marks if order is rushed.
EduCoach commentSay aloud: right-hand function first.
Why each option
A) ✓ Correct.
B) Uses \( g(f(2)) \).
C) Squares after adding incorrectly.
D) Finds \( f(2) \).
E) Adds outputs.
F) Multiplies or squares the wrong value.
8.3 · Easy · Expression composite
With \( f(x)=x^2+1 \) and \( g(x)=3x-2 \), find \( gf(x) \).
A\( 3x^2+1 \)
B\( 9x^2-12x+5 \)
C\( 3x^2+3 \)
D\( x^2+3x-1 \)
E\( 3x-1 \)
F\( x^2-1 \)
Show solution
Correct answer: A
MethodFor \( gf(x)=g(f(x)) \), substitute \( f(x) \) into \( g \).
EduCoach commentTell students to keep both square-root branches until the end.
Why each option
A) ✓ Correct: \( f(1)=5 \) or \( -7 \).
B) Only the \( a=3 \) branch.
C) Only the \( a=-3 \) branch.
D) These are possible \( a \)-values, not \( f(1) \).
E) These are constants from the composite, not outputs.
F) No coefficient comparison supports these.
8.4 Inverse Functions
An inverse function reverses the mapping of the original function. If \( f(a)=b \), then \( f^{-1}(b)=a \). The graphs of \( y=f(x) \) and \( y=f^{-1}(x) \) are reflections in the line \( y=x \).
An inverse function exists only when the original function is one-to-one on its stated domain. If a quadratic is restricted to one side of its vertex, an inverse can be written using a square root with the correct sign.
📋 Reference facts
Idea
ESAT-ready form
Inverse identity
\( ff^{-1}(x)=x \) and \( f^{-1}f(x)=x \) on valid domains.
Graph relation
Reflect in \( y=x \).
Domain/range swap
Domain of \( f^{-1} \) is the range of \( f \).
Quadratic inverse
Choose the root sign using the original domain restriction.
⚠️ Forgetting domain and range
Finding the algebraic inverse is not enough. ESAT options often differ only by the domain or the sign of a square root.
Worked Example — Core method
Question: Find the inverse of \( f(x)=3x-5 \).
Working: Let \( y=3x-5 \), then interchange the subject: \( x=3y-5 \), so \( y=\dfrac{x+5}{3} \).
\( f^{-1}(x)=\dfrac{x+5}{3} \)
8.4 · Easy · Linear inverse
Find the inverse of \( f(x)=3x-5 \).
A\( \dfrac{x+5}{3} \)
B\( 3x+5 \)
C\( \dfrac{x-5}{3} \)
D\( 5-3x \)
E\( \dfrac{3}{x+5} \)
F\( x+\dfrac{5}{3} \)
Show solution
Correct answer: A
MethodLet \( y=3x-5 \), then solve for \( x \) and rename.
⚠️ TRAP analysisThis function is self-inverse, but it is not the identity function.
Teacher commentRational self-inverses are common challenge material.
EduCoach commentHave students verify by composing once if time allows.
Why each option
A) ✓ Correct.
B) This is the reciprocal.
C) Sign-rearrangement error.
D) Denominator sign error.
E) It is one-to-one on its domain.
F) Self-inverse does not mean identity.
8.5 The Graphs of y=|f(x)| and y=f(|x|)
The transformation \( y=|f(x)| \) reflects only the parts of \( y=f(x) \) that lie below the \( x \)-axis. Its output is never negative, and its zeros occur at the same \( x \)-intercepts as \( f(x) \).
The transformation \( y=f(|x|) \) keeps the right-hand part of \( y=f(x) \), where \( x\ge 0 \), and reflects that right-hand part in the \( y \)-axis. The result is always an even graph.
📋 Reference facts
Idea
ESAT-ready form
\( y=|f(x)| \)
Reflect negative \( y \)-values in the \( x \)-axis.
\( y=f(|x|) \)
Keep \( x\ge 0 \), reflect in the \( y \)-axis.
Zeros of \( |f(x)| \)
Same zeros as \( f(x) \).
Evenness
\( f(|x|) \) is symmetric in the \( y \)-axis.
⚠️ Mixing the two transformations
\( |f(x)| \) changes output signs; \( f(|x|) \) changes input signs. They are usually different graphs.
Worked Example — Core method
Question: For \( f(x)=x^2-3x-4 \), find the \( x \)-intercepts of \( y=f(|x|) \).
Working: Solve \( |x|^2-3|x|-4=0 \). Let \( u=|x|\ge0 \). Then \( u^2-3u-4=0 \), so \( u=4 \) only. Hence \( x=\pm4 \).
\( x=-4 \text{ or } x=4 \)
8.5 · Easy · Absolute output roots
For \( f(x)=x^2-3x-4 \), what are the \( x \)-intercepts of \( y=|f(x)| \)?
A\( -1 \text{ and } 4 \)
B\( 1 \text{ and } 4 \)
C\( -4 \text{ and } 1 \)
D\( 0 \text{ and } 4 \)
E\( \text{none} \)
F\( -4 \text{ and } -1 \)
Show solution
Correct answer: A
MethodZeros of \( |f(x)| \) occur exactly where \( f(x)=0 \).
\( x^2-3x-4=(x-4)(x+1) \)↓\( x=4 \text{ or } x=-1 \)↓\( -1 \text{ and } 4 \)
⚠️ TRAP analysisReflecting below the axis does not move the intercepts.
Teacher commentThis checks transformation understanding rather than long algebra.
EduCoach commentTell students: a point on the axis stays on the axis.
Why each option
A) ✓ Correct.
B) Wrong sign for the negative root.
C) Roots of a different factorisation.
D) Uses the \( y \)-intercept.
E) There are two zeros.
F) Both signs are wrong.
8.5 · Easy · Absolute input roots
For \( f(x)=x^2-3x-4 \), what are the \( x \)-intercepts of \( y=f(|x|) \)?
A\( -4 \text{ and } 4 \)
B\( -1 \text{ and } 4 \)
C\( 1 \text{ and } 4 \)
D\( -4 \text{ and } 1 \)
E\( -1 \text{ and } 1 \)
F\( \text{none} \)
Show solution
Correct answer: A
MethodLet \( u=|x|\ge0 \) and solve \( u^2-3u-4=0 \).
\( (u-4)(u+1)=0 \)↓\( u=4 \text{ since } u\ge0 \)↓\( |x|=4 \)↓\( -4 \text{ and } 4 \)
⚠️ TRAP analysisThe negative root in \( u \) is invalid, but it does not mean there is no negative \( x \).
Teacher commentThis is exactly the difference between input and output modulus.
EduCoach commentStudents should introduce \( u=|x| \) to avoid confusion.
Why each option
A) ✓ Correct.
B) These are the intercepts of \( f(x) \), not \( f(|x|) \).
C) Uses \( u=1 \) incorrectly.
D) Wrong invalid root.
E) Reflects the wrong root.
F) There are two intercepts.
8.5 · Hard · Point reflection output
A point \( (3,-5) \) lies on \( y=f(x) \). Which point lies on \( y=|f(x)| \)?
⚠️ TRAP analysisEvenness does not mean output modulus and input modulus are the same.
Teacher commentThis is a subtle but important distinction for rational graphs.
EduCoach commentHave students identify whether the modulus is outside or inside the function symbol.
Why each option
A) ✓ Correct.
B) They differ by sign.
C) The signs are reversed.
D) Input modulus does not change the negative output.
E) Output modulus must be non-negative.
F) Negative \( x \) is allowed because \( |x|\ne0 \).
M2-05-Differentiation
MATHS 2 · CHAPTER 6: Differentiation
This chapter builds differentiation to full A-Level depth and pushes hard on its applications. It covers the derivative as a limit and the power rule; the product, quotient and chain rules; the derivatives of exponential, logarithmic and trigonometric functions; tangents, normals and stationary points; and two extended applied sections — optimization and related rates of change. Every subtopic carries a ten-question set (two warm-ups and eight hard/challenge items) with fully worked solutions.
A6.1 The Derivative: First Principles and the Power Rule
The derivative \(\dfrac{dy}{dx}=f'(x)\) is the gradient of the curve — the limit of the gradient of a chord as it shrinks to a point: \(f'(x)=\displaystyle\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}\). Carrying this out for \(x^{2}\) gives \(2x\); for \(x^{n}\) it gives the power rule \(\dfrac{d}{dx}x^{n}=nx^{\,n-1}\).
The power rule holds for every rational power, so rewrite roots and reciprocals as powers first: \(\sqrt{x}=x^{1/2}\) and \(\dfrac{1}{x^{2}}=x^{-2}\). Differentiation is linear: constants multiply through and sums differentiate term by term, while the derivative of a constant is \(0\).
The geometric approach. Before any rules, the derivative has a picture. The gradient of a curve at a point is the gradient of the tangent there. To pin that down analytically, take a nearby point and form the chord (secant); its gradient is \(\dfrac{f(x+h)-f(x)}{h}\). As the second point slides in — \(h\to0\) — the chord rotates into the tangent, and its gradient tends to the derivative \(f'(x)\). This limit is the definition; the power rule and all the others are just its shortcuts.
📋 Power rule essentials
Function
Derivative
\(x^{n}\)
\(nx^{\,n-1}\)
\(kx^{n}\)
\(knx^{\,n-1}\)
constant \(c\)
\(0\)
\(\sqrt{x}=x^{1/2}\)
\(\tfrac12 x^{-1/2}\)
\(x^{-n}\)
\(-nx^{-n-1}\)
⚠️ Rewrite before differentiating
You cannot apply the power rule to \(\sqrt{x}\) or \(\tfrac{1}{x^{2}}\) as written — convert to \(x^{1/2}\) and \(x^{-2}\) first, then bring the power down.
Worked Example — Negative and fractional powers
Question: Differentiate \(y=\dfrac{1}{x^{2}}\) and \(y=\sqrt{x}\).
Working: Write as powers: \(x^{-2}\to-2x^{-3}=-\dfrac{2}{x^{3}}\); and \(x^{1/2}\to\tfrac12 x^{-1/2}=\dfrac{1}{2\sqrt{x}}\).
⚠️ Common trapDifferentiating the un-expanded square without the chain rule loses the inner \(2x\) factor.
Why each option
A) forgot the inner derivative
B) dropped a term
C) kept the constant
D) halved the coefficients
E) wrong powers
F) ✓ \(4x^{3}+4x\)
A6.1 · Challenge · Solve for x
For \(y=x^{3}-3x\), find the values of \(x\) where the gradient equals \(9\).
A\(x=\pm1\)
B\(x=\pm2\)
C\(x=2\)
D\(x=\pm3\)
E\(x=\pm\sqrt{3}\)
F\(x=0\)
Show solution
Correct answer: B
MethodSet \(\tfrac{dy}{dx}=9\) and solve.
\(\tfrac{dy}{dx}\)\(3x^{2}-3\)
Solve\(3x^{2}-3=9\)
\(3x^{2}=12\)↓\(x^{2}=4\)↓\(x=\pm2\)\(x=\pm2\)
⚠️ Common trapA square root gives two solutions: \(x=\pm2\), not just \(+2\).
Why each option
A) set gradient to 0
B) ✓ \(x^{2}=4\)
C) dropped the negative root
D) arithmetic slip
E) forgot to add 3
F) used \(y=0\)
A6.2 The Product, Quotient and Chain Rules
Three rules handle combinations. The product rule: if \(y=uv\) then \(\dfrac{dy}{dx}=u'v+uv'\). The quotient rule: if \(y=\dfrac{u}{v}\) then \(\dfrac{dy}{dx}=\dfrac{u'v-uv'}{v^{2}}\) — note the order in the numerator matters. The chain rule: for a composite \(y=f(g(x))\), \(\dfrac{dy}{dx}=f'(g(x))\cdot g'(x)\) — differentiate the outer, then multiply by the derivative of the inner.
The chain rule is the workhorse: for \(y=(\text{something})^{n}\), bring the power down, keep the bracket, and multiply by the derivative of the bracket. Many problems combine rules — a product of two composites, or a quotient with a chain inside.
The quotient rule is \(u'v-uv'\) over \(v^{2}\) — top-derivative times bottom minus top times bottom-derivative. Reversing the subtraction flips the sign of the whole answer.
Worked Example — Product with a chain
Question: Differentiate \(y=x(2x+1)^{3}\).
Working: Product rule with \(u=x,\ v=(2x+1)^{3}\): \(v'=3(2x+1)^{2}\cdot2=6(2x+1)^{2}\). So \(y'=(2x+1)^{3}+x\cdot6(2x+1)^{2}=(2x+1)^{2}\big[(2x+1)+6x\big]=(2x+1)^{2}(8x+1)\).
\(y'=(2x+1)^{2}(8x+1)\)
A6.2 · Easy · Chain rule
Differentiate \(y=(3x+2)^{4}\).
A\(4(3x+2)^{3}\)
B\(3(3x+2)^{3}\)
C\(12(3x+2)^{4}\)
D\((3x+2)^{3}\)
E\(12(3x+2)^{3}\)
F\(7(3x+2)^{3}\)
Show solution
Correct answer: E
MethodPower down, keep bracket, times inner derivative \(3\).
⚠️ Common trapNumerator \(=x^{2}-2x=x(x-2)\); keep the \((x-1)^{2}\) denominator.
Why each option
A) forgot \(-uv'\)
B) ignored quotient rule
C) sign error in factor
D) sign reversed
E) dropped \(u'v\)
F) ✓ \(x(x-2)/(x-1)^2\)
A6.3 Derivatives of Exponential, Logarithmic and Trigonometric Functions
Beyond powers, four standard derivatives must be known cold: \(\dfrac{d}{dx}e^{x}=e^{x}\), \(\dfrac{d}{dx}\ln x=\dfrac1x\), \(\dfrac{d}{dx}\sin x=\cos x\) and \(\dfrac{d}{dx}\cos x=-\sin x\) (hence \(\dfrac{d}{dx}\tan x=\sec^{2}x\)). These combine with the chain rule for arguments other than \(x\).
With the chain rule: \(\dfrac{d}{dx}e^{kx}=k\,e^{kx}\), \(\dfrac{d}{dx}\ln(f(x))=\dfrac{f'(x)}{f(x)}\), \(\dfrac{d}{dx}\sin(kx)=k\cos(kx)\). Products such as \(x e^{x}\) or \(x^{2}\ln x\) then use the product rule on top.
📋 Standard derivatives
Function
Derivative
\(e^{x}\)
\(e^{x}\)
\(e^{kx}\)
\(k e^{kx}\)
\(\ln x\)
\(\dfrac1x\)
\(\sin x\)
\(\cos x\)
\(\cos x\)
\(-\sin x\)
\(\tan x\)
\(\sec^{2}x\)
⚠️ Chain factor on the argument
\(\dfrac{d}{dx}\cos(2x)=-2\sin(2x)\), not \(-\sin(2x)\). Any coefficient inside a function reappears as a multiplier via the chain rule.
⚠️ Common trap\(x^{2}\cdot\tfrac1x=x\), not \(x^{2}\); then factor \(x\).
Why each option
A) forgot \(uv'\)
B) ✓ \(x(2\ln x+1)\)
C) dropped the log term
D) sign error
E) did not simplify \(x^2/x\)
F) ignored the product
A6.4 Tangents, Normals and Stationary Points
The derivative evaluated at a point is the gradient of the tangent there. The tangent at \((x_0,y_0)\) is \(y-y_0=m(x-x_0)\) with \(m=f'(x_0)\); the normal is perpendicular, so its gradient is \(-\dfrac1m\).
A stationary point occurs where \(f'(x)=0\). Its nature follows from the second derivative: \(f''(x)>0\) gives a (local) minimum, \(f''(x)<0\) a maximum, and \(f''(x)=0\) is inconclusive (often a point of inflection, where \(f''\) changes sign). A function is increasing where \(f'(x)>0\) and decreasing where \(f'(x)<0\).
The geometric picture. At any point the tangent has gradient \(f'(x)\) and the normal is perpendicular to it, with gradient \(-1/f'(x)\). Where the tangent is horizontal the curve is stationary \((f'(x)=0)\): a peak (maximum), a trough (minimum), or a point of inflection where the concavity changes.
📋 Key facts
Idea
Statement
Tangent
\(y-y_0=f'(x_0)(x-x_0)\)
Normal gradient
\(-\dfrac{1}{f'(x_0)}\)
Stationary
\(f'(x)=0\)
Minimum / Maximum
\(f''>0\) / \(f''<0\)
Inflection
\(f''=0\) and changes sign
⚠️ Normal is the negative reciprocal
The normal’s gradient is \(-\dfrac1m\), not \(-m\). If the tangent has gradient \(2\), the normal has gradient \(-\tfrac12\).
Worked Example — Nature of a stationary point
Question: Find and classify the stationary points of \(y=x^{3}-3x\).
Working: \(y'=3x^{2}-3=0\Rightarrow x=\pm1\). Since \(y''=6x\): at \(x=1,\ y''=6>0\) (minimum); at \(x=-1,\ y''=-6<0\) (maximum).
\(x=1:\ \min;\quad x=-1:\ \max\)
A6.4 · Easy · Tangent gradient
Find the gradient of the tangent to \(y=x^{2}\) at \(x=3\).
A\(9\)
B\(3\)
C\(2\)
D\(12\)
E\(0\)
F\(6\)
Show solution
Correct answer: F
Method\(y'=2x\), substitute \(x=3\).
\(y'\)\(2x\)
\(2(3)\)↓\(6\)\(6\)
⚠️ Common trapUse the derivative \(2x\), not the \(y\)-value \(x^{2}=9\).
Why each option
A) used \(y=x^2\)
B) used \(x\) itself
C) used the coefficient
D) used \(2x^2\)
E) set to zero
F) ✓ \(2(3)\)
A6.4 · Easy · Stationary x
Find the \(x\)-coordinate of the stationary point of \(y=x^{2}-4x\).
A\(2\)
B\(4\)
C\(-2\)
D\(0\)
E\(-4\)
F\(1\)
Show solution
Correct answer: A
MethodSet \(y'=2x-4=0\).
\(y'\)\(2x-4\)
\(2x-4=0\)↓\(x=2\)\(x=2\)
⚠️ Common trapSolve \(2x-4=0\), giving \(x=2\).
Why each option
A) ✓ \(x=2\)
B) used the constant
C) sign error
D) used \(y=0\)
E) sign error
F) arithmetic slip
A6.4 · Hard · Tangent line
Find the equation of the tangent to \(y=x^{2}\) at \((2,4)\).
A\(y=4x+4\)
B\(y=2x\)
C\(y=4x\)
D\(y=4x-4\)
E\(y=2x+4\)
F\(y=-\tfrac14 x+\tfrac92\)
Show solution
Correct answer: D
Method\(m=2(2)=4\); use \(y-4=4(x-2)\).
\(m\)\(4\)
\(y-4=4(x-2)\)↓\(y=4x-4\)\(y=4x-4\)
⚠️ Common trapExpand \(y-4=4(x-2)\) carefully: \(y=4x-8+4=4x-4\).
Why each option
A) sign error on constant
B) used \(m=2\)
C) forgot the point
D) ✓ \(y=4x-4\)
E) wrong gradient
F) used the normal
A6.4 · Hard · Normal gradient
The tangent to \(y=x^{2}\) at \((1,1)\) has gradient \(2\). What is the gradient of the normal there?
A\(-2\)
B\(\tfrac12\)
C\(-\tfrac12\)
D\(2\)
E\(-1\)
F\(1\)
Show solution
Correct answer: C
MethodNormal gradient \(=-\dfrac1m\).
\(m\)\(2\)
\(-\dfrac12\)↓\(-\dfrac12\)\(-\dfrac12\)
⚠️ Common trapNegative reciprocal of \(2\) is \(-\tfrac12\), not \(-2\).
Why each option
A) negated only
B) forgot the minus
C) ✓ \(-\tfrac12\)
D) used the tangent
E) used \(-1\)
F) reciprocal sign wrong
A6.4 · Hard · Two stationary points
Find the \(x\)-coordinates of the stationary points of \(y=x^{3}-3x^{2}\).
A\(0\text{ and }3\)
B\(0\text{ and }2\)
C\(1\text{ and }2\)
D\(\pm2\)
E\(3\text{ only}\)
F\(-2\text{ and }0\)
Show solution
Correct answer: B
Method\(y'=3x^{2}-6x=3x(x-2)=0\).
Factor\(3x(x-2)\)
\(3x(x-2)=0\)↓\(x=0,\ 2\)\(x=0,\ 2\)
⚠️ Common trapFactor fully: both \(x=0\) and \(x=2\) are solutions.
Why each option
A) wrong second root
B) ✓ \(x=0,2\)
C) missed \(x=0\)
D) treated as \(x^2\)
E) dropped a root
F) sign error
A6.4 · Hard · Nature (2nd derivative)
For \(y=x^{3}-3x\), classify the stationary point at \(x=1\).
Amaximum
Binflection
Cnot stationary
Dsaddle
Eminimum
Fdiscontinuity
Show solution
Correct answer: E
MethodUse the sign of \(y''=6x\).
\(y''(1)\)\(6>0\)
\(y''=6x=6\)↓minimumminimum
⚠️ Common trap\(y''>0\Rightarrow\) minimum; \(x=1\) is indeed stationary since \(y'(1)=0\).
Why each option
A) sign of \(y''\) wrong
B) \(y''\neq0\) here
C) it is stationary
D) not applicable
E) ✓ \(y''(1)=6>0\)
F) function is smooth
A6.4 · Hard · Local maximum value
Find the local maximum value of \(y=x^{3}-12x\).
A\(-16\)
B\(16\)
C\(8\)
D\(2\)
E\(-2\)
F\(0\)
Show solution
Correct answer: B
MethodStationary at \(3x^{2}-12=0\Rightarrow x=\pm2\); the max is at \(x=-2\).
\(x=-2\)\(y''=-12<0\)
\(y(-2)\)\(-8+24\)
\((-2)^{3}-12(-2)\)↓\(-8+24=16\)\(-8+24=16\)
⚠️ Common trapThe maximum is at \(x=-2\) (where \(y''<0\)); evaluate \(y\) there, giving \(16\).
Why each option
A) used \(x=+2\) (the min)
B) ✓ \(y(-2)=16\)
C) used a stationary \(x\)
D) used \(x\) value
E) sign slip
F) wrong point
A6.4 · Hard · Increasing interval
For what values of \(x\) is \(y=x^{2}-6x\) increasing?
A\(x<3\)
B\(x>0\)
C\(x>6\)
D\(x>3\)
E\(x<6\)
Fall \(x\)
Show solution
Correct answer: D
MethodIncreasing where \(y'=2x-6>0\).
\(y'\)\(2x-6\)
\(2x-6>0\)↓\(x>3\)\(x>3\)
⚠️ Common trapSolve the inequality \(2x-6>0\), not \(=0\).
Why each option
A) inequality reversed
B) wrong critical value
C) used the constant
D) ✓ \(x>3\)
E) reversed & wrong
F) it decreases for \(x<3\)
A6.4 · Challenge · Point of inflection
Find the \(x\)-coordinate of the point of inflection of \(y=x^{3}-3x^{2}+2\).
A\(0\)
B\(2\)
C\(3\)
D\(-1\)
E\(1\)
F\(\tfrac12\)
Show solution
Correct answer: E
MethodSet \(y''=0\) (and it changes sign).
\(y''\)\(6x-6\)
\(6x-6=0\)↓\(x=1\)\(x=1\)
⚠️ Common trapInflection uses the second derivative \(y''=6x-6=0\), not \(y'\).
Why each option
A) used a stationary root
B) other stationary root
C) arithmetic slip
D) sign error
E) ✓ \(6x-6=0\)
F) halved wrongly
A6.4 · Challenge · Horizontal tangents
At which \(x\)-values does \(y=x^{3}-3x\) have a horizontal tangent?
⚠️ Common trapBoth roots of \(x^{2}=1\): \(x=\pm1\).
Why each option
A) only the inflection
B) arithmetic slip
C) forgot to move the 3
D) dropped a root
E) wrong constant
F) ✓ \(x^2=1\)
A6.5 Optimization: Applied Maxima and Minima
Optimization problems ask for the largest or smallest possible value of some quantity — an area, a volume, a cost — subject to a constraint. The method is always the same four steps, and doing them in order is what turns a wordy problem into a one-line calculus exercise.
Step 1 — define variables and write the target. Identify the quantity to be optimized (call it \(A\), \(V\), \(C\)\(\dots\)) and any others. Step 2 — use the constraint to eliminate a variable, so the target is a function of a single variable. Step 3 — differentiate and set to zero to locate the stationary point. Step 4 — justify and interpret: confirm it is a max or min with the second derivative, then state the answer with units. The hardest part is nearly always Step 2 — turning the words into one equation in one unknown.
A worked template: a rectangle of fixed perimeter has \(2x+2y=P\); solve for \(y\), substitute into the area \(A=xy\), and you have \(A(x)\) ready to differentiate. Volume-and-surface problems follow the same shape — the constraint fixes one dimension in terms of another.
📋 The four steps
Step
Action
1. Target
write the quantity to optimize
2. Constraint
eliminate a variable
3. Stationary
differentiate, set \(=0\)
4. Justify
\(f''\) sign; state with units
⚠️ Optimize the right quantity
Differentiate the target (area, volume, cost), not the constraint. And always verify max vs min with the second derivative — a stationary point is not automatically the maximum you want.
Worked Example — Maximum area for a fixed perimeter
Question: A rectangle has perimeter \(20\). Find its maximum possible area.
Working: Let the sides be \(x\) and \(10-x\) (since \(2x+2y=20\Rightarrow y=10-x\)). Then \(A=x(10-x)=10x-x^{2}\). \(A'=10-2x=0\Rightarrow x=5\), and \(A''=-2<0\) confirms a maximum. So \(A=5\times5=25\).
\(A_{\max}=25\ \text{(a square of side }5)\)
Visualising the setup. A quick sketch turns the words into the one-variable target and constraint.
A6.5 · Easy · Condition for a max
At a (local) maximum of a smooth function, the value of \(\dfrac{dy}{dx}\) is:
Apositive
Bnegative
C\(0\)
Dundefined
E\(1\)
Fmaximal
Show solution
Correct answer: C
MethodStationary points have zero gradient.
At max\(\tfrac{dy}{dx}=0\)
\(\tfrac{dy}{dx}=0\)↓\(0\)\(0\)
⚠️ Common trapThe gradient is zero at the turning point itself; its sign changes around it.
Why each option
A) that is just before/after
B) that is after a max
C) ✓ zero gradient
D) it is defined
E) no reason for 1
F) confuses value with gradient
A6.5 · Easy · Second-derivative test
To confirm a stationary point is a maximum, the second derivative must be:
Anegative
Bpositive
Czero
D\(1\)
Eequal to \(y\)
Fundefined
Show solution
Correct answer: A
Method\(f''<0\) at a maximum.
Max\(f''<0\)
\(f''<0\)↓negativenegative
⚠️ Common trap\(f''<0\) = maximum (curve concave down); \(f''>0\) = minimum.
Why each option
A) ✓ \(f''<0\)
B) that is a minimum
C) inconclusive
D) no meaning
E) not related
F) it exists
A6.5 · Hard · Max area, fixed perimeter
A rectangle has perimeter \(20\). Find its maximum area.
A\(20\)
B\(50\)
C\(25\)
D\(100\)
E\(10\)
F\(24\)
Show solution
Correct answer: C
Method\(A=x(10-x)\); maximize.
\(A'\)\(10-2x\)
\(x\)\(5\)
\(A=5(10-5)\)↓\(25\)\(25\)
⚠️ Common trapUse \(y=10-x\) from \(2x+2y=20\); the optimum is a square of side \(5\).
Why each option
A) used perimeter
B) used \(x=5,y=10\)
C) ✓ \(5\times5\)
D) squared the perimeter side
E) used a side only
F) arithmetic slip
A6.5 · Hard · Max product, fixed sum
Two positive numbers have sum \(12\). Find their maximum possible product.
A\(24\)
B\(12\)
C\(72\)
D\(35\)
E\(48\)
F\(36\)
Show solution
Correct answer: F
Method\(P=x(12-x)\); maximize.
\(P'\)\(12-2x\)
\(x\)\(6\)
\(6\times6\)↓\(36\)\(36\)
⚠️ Common trapThe product is largest when the numbers are equal: \(6\times6=36\).
Why each option
A) used unequal split
B) used the sum
C) doubled
D) used \(5\times7\)
E) used \(4\times... \)
F) ✓ \(6\times6\)
A6.5 · Hard · Fence against a wall
A rectangular pen uses a wall as one side and \(40\,\text{m}\) of fence for the other three sides. Find the maximum area.
⚠️ Common trap\(P'=2-50/x^{2}=0\Rightarrow x^{2}=25\); then \(P=20\).
Why each option
A) ✓ \(10+10\)
B) used one term
C) gave \(x^2\)
D) used the constant
E) gave the \(x\) value
F) arithmetic slip
A6.5 · Challenge · Maximize a volume expression
Find the maximum value of \(V=x^{2}(6-x)\) for \(0
A\(16\)
B\(27\)
C\(36\)
D\(8\)
E\(32\)
F\(48\)
Show solution
Correct answer: E
Method\(V=6x^{2}-x^{3}\); \(V'=12x-3x^{2}=0\).
\(V'\)\(3x(4-x)\)
\(x\)\(4\)
\(x=4\)↓\(V=16(2)=32\)\(V=16(2)=32\)
⚠️ Common trap\(V'=3x(4-x)=0\) gives \(x=4\) (reject \(x=0\)); then \(V=16\times2=32\).
Why each option
A) used \(x^2\) at \(x=4\) only
B) used \(x=3\)
C) used \(x=... \)
D) used \(x=2\)
E) ✓ \(16\times2\)
F) doubled
A6.6 Related Rates of Change
In a related-rates problem, two or more quantities change with time and are linked by an equation; knowing one rate, you find another. The engine is the chain rule: if \(V\) depends on \(r\) and \(r\) depends on \(t\), then \(\dfrac{dV}{dt}=\dfrac{dV}{dr}\cdot\dfrac{dr}{dt}\).
The method mirrors optimization. Step 1 — write the equation linking the quantities (a geometric formula: area, volume, Pythagoras). Step 2 — differentiate with respect to \(t\), applying the chain rule to every variable that depends on time. Step 3 — substitute the known instantaneous values and solve for the unknown rate. A key discipline: differentiate first, substitute the specific numbers afterwards — plugging values in too early freezes a variable that is still changing.
Signs carry meaning: a positive rate is an increase, a negative rate a decrease. In a sliding-ladder problem, for instance, the top’s velocity comes out negative because it is falling.
📋 Related-rates toolkit
Relation
Differentiated form
Circle \(A=\pi r^{2}\)
\(\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}\)
Sphere \(V=\tfrac43\pi r^{3}\)
\(\dfrac{dV}{dt}=4\pi r^{2}\dfrac{dr}{dt}\)
Square \(A=x^{2}\)
\(\dfrac{dA}{dt}=2x\dfrac{dx}{dt}\)
Cube \(V=x^{3}\)
\(\dfrac{dV}{dt}=3x^{2}\dfrac{dx}{dt}\)
Chain link
\(\dfrac{dV}{dt}=\dfrac{dV}{dr}\dfrac{dr}{dt}\)
⚠️ Differentiate before substituting
Keep variables symbolic through the differentiation, then substitute the given instant. Substituting a value like \(r=3\) before differentiating treats a changing quantity as constant.
Worked Example — Inflating a balloon
Question: A spherical balloon has radius increasing at \(\dfrac{dr}{dt}=2\). Find \(\dfrac{dV}{dt}\) when \(r=3\).
Working: From \(V=\tfrac43\pi r^{3}\), \(\dfrac{dV}{dr}=4\pi r^{2}\). Chain rule: \(\dfrac{dV}{dt}=4\pi r^{2}\dfrac{dr}{dt}=4\pi(9)(2)=72\pi\).
\(\dfrac{dV}{dt}=4\pi(3)^{2}(2)=72\pi\)
Visualising the setup. Label the changing quantities on a diagram, then differentiate the geometric relation with respect to time.
A6.6 · Easy · Chain rule for rates
Which equation correctly links the rates for \(y\) depending on \(x\) depending on \(t\)?
⚠️ Common trap\(4\pi r^{2}=16\pi\) at \(r=2\); rate \(1\) leaves it unchanged.
Why each option
A) ✓ \(16\pi\)
B) used \(2\pi r\)
C) used \(2\pi r\), r=2
D) doubled
E) wrong constant
F) used \(r^3\)
A6.6 · Challenge · Sliding ladder
A \(5\,\text{m}\) ladder rests against a wall. Its foot is pulled out at \(2\,\text{m/s}\). When the foot is \(3\,\text{m}\) from the wall, how fast is the top moving (take downward as negative)?
⚠️ Common trapRearrange: \(\dfrac{dr}{dt}=\dfrac{dA/dt}{2\pi r}=\dfrac{10}{10\pi}=\dfrac1\pi\).
Why each option
A) wrong \(2\pi r\)
B) inverted
C) ✓ \(10/(10\pi)\)
D) doubled
E) did not divide
F) used \(r^2\)
A6.6 · Challenge · Oil slick area rate
A circular oil slick’s radius grows at \(0.5\,\text{m/s}\). How fast is its area growing when \(r=10\)?
A\(5\pi\)
B\(10\pi\)
C\(20\pi\)
D\(100\pi\)
E\(\pi\)
F\(50\pi\)
Show solution
Correct answer: B
Method\(\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}\).
\(2\pi r\)\(20\pi\)
\(\times0.5\)\(\times0.5\)
\(2\pi(10)(0.5)\)↓\(10\pi\)\(10\pi\)
⚠️ Common trap\(2\pi r=20\pi\) at \(r=10\); times \(0.5\) gives \(10\pi\).
Why each option
A) halved twice
B) ✓ \(20\pi\times0.5\)
C) forgot the rate
D) used \(r^2\)
E) over-divided
F) used \(r^2/2\)
A6.7 Higher Derivatives and the Graph of the Gradient Function
Differentiating \(f(x)\) gives the gradient function \(f'(x)\), whose height at each point is the slope of the original curve there. So the graph of \(f'\) encodes \(f\): where \(f\) is increasing, \(f'>0\) (above the axis); where \(f\) is decreasing, \(f'<0\); and at each turning point of \(f\) the graph of \(f'\) crosses the \(x\)-axis (from \(+\) to \(-\) at a maximum, from \(-\) to \(+\) at a minimum).
Differentiating again gives the higher derivatives \(f''(x)=\dfrac{d^{2}y}{dx^{2}}\), \(f'''(x)=\dfrac{d^{3}y}{dx^{3}}\), and so on — each is simply the derivative of the one before. Reading the three graphs stacked and aligned in \(x\) makes the pattern vivid: the zeros of \(f''\) sit under the inflections of \(f\), and the zeros of \(f'\) sit under the turning points of \(f\).
Geometric meaning of the second derivative. \(f''\) measures how the gradient is changing — the concavity of the curve. Where \(f''>0\) the curve is concave up (bending upward, gradient increasing); where \(f''<0\) it is concave down (bending downward, gradient decreasing). A point of inflection is where the concavity switches — so \(f''=0\) and changes sign there. In kinematics, if \(s(t)\) is displacement then \(s'\) is velocity and \(s''\) is acceleration.
Geometric meaning of the third derivative. \(f'''\) is the rate of change of the concavity — how quickly the bending itself is changing. Its sign tells you whether the curve is becoming more or less concave up. In kinematics \(s'''(t)\) is the jerk: the rate of change of acceleration. A non-zero \(f'''\) at a point where \(f''=0\) guarantees a genuine inflection (the concavity really does flip).
📋 Reading the derivative graphs
On the graph of \(f\)
Corresponds to
\(f\) increasing
\(f'>0\) (above axis)
\(f\) decreasing
\(f'<0\) (below axis)
turning point of \(f\)
\(f'=0\) (crosses axis)
inflection of \(f\)
\(f''=0\) and changes sign
concave up / down
\(f''>0\) / \(f''<0\)
⚠️ \(f''=0\) is not enough for an inflection
The concavity must actually change sign. For \(y=x^{4}\), \(f''=12x^{2}=0\) at \(x=0\), but \(f''\ge0\) on both sides — so \(x=0\) is a minimum, not an inflection.
Worked Example — Higher derivatives and acceleration
Question: A particle has displacement \(s(t)=t^{3}-3t^{2}\). Find its acceleration at \(t=2\).
Working: Velocity \(s'=3t^{2}-6t\); acceleration \(s''=6t-6\). At \(t=2\): \(s''(2)=12-6=6\).
\(s''(t)=6t-6\Rightarrow s''(2)=6\)
A6.7 · Easy · Second derivative
Find the second derivative of \(y=x^{4}\).
A\(4x^{3}\)
B\(24x\)
C\(12x^{2}\)
D\(x^{2}\)
E\(12x^{3}\)
F\(4x^{2}\)
Show solution
Correct answer: C
MethodDifferentiate twice.
\(y'\)\(4x^{3}\)
\(y''\)\(12x^{2}\)
\(4x^{3}\)↓\(12x^{2}\)\(12x^{2}\)
⚠️ Common trapDifferentiate the first derivative again — do not stop at \(4x^{3}\).
Why each option
A) that is \(y'\)
B) that is \(y'''\)
C) ✓ \(12x^{2}\)
D) over-reduced
E) wrong power
F) wrong coefficient
A6.7 · Easy · Concavity sign
If \(f''(x)>0\) on an interval, the curve there is:
Aconcave down
Ba straight line
Cdecreasing
Dstationary
Eundefined
Fconcave up
Show solution
Correct answer: F
MethodPositive second derivative = gradient increasing.
\(f''>0\)concave up
\(f''>0\)↓concave upconcave up
⚠️ Common trap\(f''>0\) bends upward (holds water); \(f''<0\) bends downward.
Why each option
A) that is \(f''<0\)
B) that needs \(f''=0\) throughout
C) about \(f'\), not \(f''\)
D) about \(f'=0\)
E) it is defined
F) ✓ concave up
A6.7 · Hard · Second derivative
Find \(\dfrac{d^{2}y}{dx^{2}}\) for \(y=x^{3}-2x^{2}+5x\).
A\(6x-4\)
B\(3x^{2}-4x+5\)
C\(6x-4x\)
D\(6x\)
E\(6x-2\)
F\(6\)
Show solution
Correct answer: A
Method\(y'=3x^{2}-4x+5\), then differentiate again.
\(y''\)\(6x-4\)
\(3x^{2}-4x+5\)↓\(6x-4\)\(6x-4\)
⚠️ Common trapThe constant \(+5\) vanishes at the first step; \(-4x\to-4\).
Why each option
A) ✓ \(6x-4\)
B) that is \(y'\)
C) left \(x\) on the \(-4\)
D) dropped the \(-4\)
E) wrong constant
F) differentiated three times
A6.7 · Hard · Third derivative
Find the third derivative of \(y=x^{4}\).
A\(12x^{2}\)
B\(24\)
C\(4x^{3}\)
D\(24x\)
E\(48x\)
F\(6x\)
Show solution
Correct answer: D
MethodDifferentiate three times.
\(y''\)\(12x^{2}\)
\(y'''\)\(24x\)
\(12x^{2}\)↓\(24x\)\(24x\)
⚠️ Common trapThird derivative, not fourth: \(12x^{2}\to24x\) (a further step gives \(24\)).
Why each option
A) that is \(y''\)
B) that is \(y^{(4)}\)
C) that is \(y'\)
D) ✓ \(24x\)
E) wrong coefficient
F) wrong coefficient
A6.7 · Hard · f′ at a maximum
At a local maximum of \(y=f(x)\), the graph of \(y=f'(x)\):
Acrosses from negative to positive
Bcrosses the \(x\)-axis from positive to negative
Chas a maximum
Dstays above the \(x\)-axis
Eis undefined
Fhas a vertical asymptote
Show solution
Correct answer: B
MethodGradient is \(+\) before, \(0\) at, \(-\) after a max.
⚠️ Common trapAt a maximum the slope changes from rising to falling, so \(f'\) goes \(+\to-\).
Why each option
A) that is a minimum
B) ✓ \(+\to-\)
C) \(f'\) crosses zero, not peaks
D) \(f'\) changes sign
E) \(f'\) is defined
F) no asymptote
A6.7 · Hard · Read the f′ graph
The graph of \(y=f'(x)\) is an upward parabola cutting the \(x\)-axis at \(x=1\) and \(x=3\). The curve \(y=f(x)\) has a local minimum at:
A\(x=1\)
B\(x=2\)
C\(x=0\)
D\(x=1\text{ and }3\)
E\(x=3\)
Fnowhere
Show solution
Correct answer: E
MethodMin where \(f'\) goes \(-\to+\).
\(f'\) sign at \(x=3\)\(-\to+\)
upward parabola: \(f'<0\) on \((1,3)\)↓\(-\to+\) at \(x=3\)\(-\to+\) at \(x=3\)
⚠️ Common trapFor an upward \(f'\), \(f'<0\) between the roots; at the right root \((x=3)\) it turns \(-\to+\) — a minimum.
Why each option
A) that is the maximum
B) \(f'\neq0\) there
C) not a root
D) only one is a min
E) ✓ \(-\to+\) at \(x=3\)
F) there is a min
A6.7 · Hard · Concave down
On an interval where \(f''(x)<0\), the graph of \(f\) is:
Aconcave down
Bconcave up
Cincreasing
Da straight line
Eat a minimum
Fstationary
Show solution
Correct answer: A
MethodNegative second derivative = gradient decreasing.
\(f''<0\)concave down
\(f''<0\)↓concave downconcave down
⚠️ Common trap\(f''<0\) is the “\(\cap\)” shape (bends downward); it says nothing directly about increasing/decreasing.
Why each option
A) ✓ concave down
B) that is \(f''>0\)
C) that is about \(f'\)
D) needs \(f''=0\)
E) about \(f'=0\)
F) about \(f'=0\)
A6.7 · Challenge · Inflection test
Does \(y=x^{4}\) have a point of inflection at \(x=0\)?
AYes — \(f''=0\) there
BYes — it is a maximum
CNo — \(f'\neq0\)
DYes — \(f'''=0\)
ENo — \(f''\) does not change sign
FCannot be determined
Show solution
Correct answer: E
MethodInflection needs \(f''=0\) AND a sign change.
\(f''\)\(12x^{2}\)
Sign\(\ge0\) both sides
\(f''=12x^{2}\ge0\)↓no sign change → not an inflectionno sign change → not an inflection
⚠️ Common trap\(f''(0)=0\) is necessary but not sufficient; here \(f''\ge0\) throughout, so \(x=0\) is a minimum.
Why each option
A) \(f''=0\) alone is insufficient
B) it is a min, not max
C) \(f'(0)=0\) actually
D) \(f'''\) test misused
E) ✓ no sign change
F) it can be determined
A6.7 · Hard · Acceleration
A particle has \(s(t)=t^{3}-3t^{2}\). Find its acceleration at \(t=2\).
A\(0\)
B\(12\)
C\(6\)
D\(-6\)
E\(3\)
F\(18\)
Show solution
Correct answer: C
MethodAcceleration \(=s''(t)\).
\(s''\)\(6t-6\)
At \(t=2\)\(12-6\)
\(s''=6t-6\)↓\(6(2)-6=6\)\(6(2)-6=6\)
⚠️ Common trapAcceleration is the second derivative \(6t-6\), not the velocity \(3t^{2}-6t\).
Why each option
A) used \(t=1\)
B) forgot the \(-6\)
C) ✓ \(6(2)-6\)
D) sign slip
E) used first derivative pieces
F) used velocity
A6.7 · Challenge · Third derivative meaning
For displacement \(s(t)\), what does \(s'''(t)\) represent?
Aacceleration
Bvelocity
Cdistance
Daverage speed
Emomentum
Fjerk (rate of change of acceleration)
Show solution
Correct answer: F
MethodEach derivative of \(s\) is the rate of change of the previous.
\(s'\)velocity
\(s''\)acceleration
\(s'''\)jerk
\(s\to s' o s'' o s'''\)↓jerkjerk
⚠️ Common trap\(s'''\) is the rate of change of acceleration — the geometric “rate of change of concavity”.
Why each option
A) that is \(s''\)
B) that is \(s'\)
C) that is \(s\)
D) not a derivative here
E) not kinematic derivative
F) ✓ jerk
M2-07-Integration
MATHS 2 · CHAPTER 7: Integration
This chapter develops integration to full A-Level depth. It begins with integration as the reverse of differentiation and the standard integrals, then gives integration its geometric meaning as the area under a curve through the definite integral. It goes on to areas between curves, the two core techniques — substitution and integration by parts — and finishes with volumes of revolution. Each subtopic carries a ten-question set (two warm-ups, eight hard/challenge) with fully worked solutions.
A7.1 Integration as the Reverse of Differentiation
Integration reverses differentiation: to integrate is to find a function whose derivative is the integrand. The reverse power rule is \(\displaystyle\int x^{n}\,dx=\dfrac{x^{n+1}}{n+1}+C\) (valid for \(n\neq-1\)) — raise the power by one and divide by the new power.
Because differentiating a constant gives \(0\), every indefinite integral carries an arbitrary constant of integration \(+C\). If an extra condition (a point on the curve) is given, substitute it to pin down \(C\) and obtain a particular solution.
📋 Reverse power rule
Integral
Result
\(\int x^{n}\,dx\)
\(\dfrac{x^{n+1}}{n+1}+C\ (n\neq-1)\)
\(\int k\,dx\)
\(kx+C\)
\(\int \sqrt{x}\,dx\)
\(\dfrac{2}{3}x^{3/2}+C\)
Sum rule
\(\int(f+g)=\int f+\int g\)
⚠️ Never forget \(+C\)
Every indefinite integral needs the constant \(+C\). Omitting it is the most common — and most penalised — slip. Only a definite integral (with limits) drops the constant.
Worked Example — Finding a particular solution
Question: A curve has gradient \(\dfrac{dy}{dx}=3x^{2}-2\) and passes through \((1,4)\). Find \(y\).
Working: Integrate: \(y=x^{3}-2x+C\). Substitute \((1,4)\): \(4=1-2+C\Rightarrow C=5\). So \(y=x^{3}-2x+5\).
\(y=x^{3}-2x+5\)
A7.1 · Easy · Reverse power
Find \(\displaystyle\int x^{3}\,dx\).
A\(3x^{2}+C\)
B\(\dfrac{x^{4}}{3}+C\)
C\(\dfrac{x^{4}}{4}+C\)
D\(4x^{3}+C\)
E\(x^{4}+C\)
F\(\dfrac{x^{2}}{2}+C\)
Show solution
Correct answer: C
MethodRaise the power by one, divide by the new power.
⚠️ Common trapSubstitute the point to find \(C=3\), not \(C=5\).
Why each option
A) used \(f(1)\) as \(C\)
B) forgot \(C\)
C) did not divide by 3
D) ✓ \(C=3\)
E) sign error
F) differentiated
A7.1 · Challenge · Polynomial
Find \(\displaystyle\int (4x^{3}-6x)\,dx\).
A\(12x^{2}-6+C\)
B\(x^{4}-6x^{2}+C\)
C\(4x^{4}-3x^{2}+C\)
D\(x^{4}-3x^{2}\)
E\(\dfrac{x^{4}}{4}-3x^{2}+C\)
F\(x^{4}-3x^{2}+C\)
Show solution
Correct answer: F
MethodIntegrate each term.
\(\int4x^3\)\(x^4\)
\(\int-6x\)\(-3x^2\)
\(x^4-3x^2+C\)↓\(x^4-3x^2+C\)\(x^4-3x^2+C\)
⚠️ Common trap\(\int4x^{3}=x^{4}\) (the 4 cancels) and \(\int-6x=-3x^{2}\).
Why each option
A) differentiated
B) \(\int-6x\neq-6x^2\)
C) did not cancel the 4
D) missing \(+C\)
E) divided the 4 wrongly
F) ✓ \(x^4-3x^2+C\)
A7.1 · Challenge · Equation of curve
A curve has \(\dfrac{dy}{dx}=3x^{2}-2\) and passes through \((1,4)\). Find \(y\).
A\(x^{3}-2x\)
B\(x^{3}-2x+5\)
C\(x^{3}-2x+4\)
D\(x^{3}-2x-5\)
E\(6x+C\)
F\(3x^{3}-2x+5\)
Show solution
Correct answer: B
MethodIntegrate, then use the point.
\(y\)\(x^{3}-2x+C\)
At \((1,4)\)\(1-2+C=4\)
\(y=x^{3}-2x+C\)↓\(C=5\)\(C=5\)
⚠️ Common trap\(1-2+C=4\Rightarrow C=5\); do not read \(C\) straight off as \(4\).
Why each option
A) forgot \(C\)
B) ✓ \(C=5\)
C) used \(y\)-value as \(C\)
D) sign of \(C\)
E) differentiated
F) did not divide
A7.2 Integrating Standard Functions
Reversing the standard derivatives gives the standard integrals: \(\int e^{x}\,dx=e^{x}+C\), \(\int\frac1x\,dx=\ln|x|+C\), \(\int\cos x\,dx=\sin x+C\) and \(\int\sin x\,dx=-\cos x+C\). The single exception to the power rule, \(n=-1\), is exactly the \(\ln\) case.
For a linear inner function the chain rule reverses too: \(\int e^{kx}\,dx=\dfrac{1}{k}e^{kx}+C\), \(\int\cos(kx)\,dx=\dfrac1k\sin(kx)+C\), and \(\int(ax+b)^{n}\,dx=\dfrac{(ax+b)^{n+1}}{a(n+1)}+C\) — divide by the derivative of the inside.
📋 Standard integrals
Integral
Result
\(\int e^{x}dx\)
\(e^{x}+C\)
\(\int \frac1x dx\)
\(\ln|x|+C\)
\(\int \sin x\,dx\)
\(-\cos x+C\)
\(\int \cos x\,dx\)
\(\sin x+C\)
\(\int e^{kx}dx\)
\(\tfrac1k e^{kx}+C\)
\(\int(ax+b)^{n}dx\)
\(\dfrac{(ax+b)^{n+1}}{a(n+1)}+C\)
⚠️ Divide by the inner coefficient
\(\int\cos 3x\,dx=\tfrac13\sin 3x+C\), not \(\sin 3x\). Reversing a linear chain means dividing by the derivative of the inside.
⚠️ Common trap\(\int\frac1x=\ln|x|\), not \(-x^{-2}\) (that is differentiation).
Why each option
A) forgot to halve \(e^{2x}\)
B) differentiated \(1/x\)
C) multiplied instead
D) \(\int1/x\neq1/x\)
E) dropped a term
F) ✓ both terms
A7.3 The Definite Integral and the Area Under a Curve
The geometric approach. Integration has a picture: the definite integral \(\displaystyle\int_{a}^{b}f(x)\,dx\) equals the area between the curve \(y=f(x)\), the \(x\)-axis, and the vertical lines \(x=a\) and \(x=b\) (for a curve above the axis). Think of the region as a stack of thin strips of width \(dx\) and height \(f(x)\); the integral sums their areas.
By the Fundamental Theorem of Calculus, you evaluate it by finding any antiderivative \(F\) and taking \(\displaystyle\int_{a}^{b}f(x)\,dx=\big[F(x)\big]_{a}^{b}=F(b)-F(a)\). No constant of integration is needed — it cancels.
📋 The definite integral
Idea
Statement
Definition
\(\int_{a}^{b}f\,dx=F(b)-F(a)\)
Geometric meaning
area under \(y=f(x)\) on \([a,b]\)
No \(+C\)
the constant cancels
Reversing limits
\(\int_{a}^{b}=-\int_{b}^{a}\)
⚠️ Substitute the top limit first
Evaluate \(F(b)-F(a)\) — upper minus lower. Reversing the order flips the sign of the whole answer.
⚠️ Common trapAntiderivative is \(4x-\tfrac{x^3}{3}\); use symmetry or evaluate both limits.
Why each option
A) ignored the \(-x^2\)
B) ✓ \(\tfrac{16}{3}+\tfrac{16}{3}\)
C) used one side
D) arithmetic slip
E) divided by 6
F) doubled twice
A7.4 Area Between Curves and Regions Below the Axis
Where a region lies below the \(x\)-axis, the definite integral is negative; the area is its magnitude, \(\big|\int f\,dx\big|\). If a curve crosses the axis over the interval, split the integral at the crossing and take each piece’s magnitude, or the positive and negative parts will cancel.
The area between two curves that do not cross on \([a,b]\) is \(\displaystyle\int_{a}^{b}\big(y_{\text{upper}}-y_{\text{lower}}\big)\,dx\). Find the intersection points first (set the curves equal) to get the limits, then integrate top minus bottom.
⚠️ Common trapThe whole arch is below the axis; the area is \(\tfrac{32}{3}\), not negative.
Why each option
A) left negative
B) used one limit
C) forgot to subtract \(2x^2\)
D) arithmetic slip
E) unsimplified
F) ✓ \(|-\tfrac{32}{3}|\)
A7.5 Integration by Substitution
Substitution reverses the chain rule. Spot an inner function \(u=g(x)\) whose derivative \(g'(x)\) also appears (up to a constant) in the integrand; then \(du=g'(x)\,dx\) turns the integral into a simple one in \(u\). The tell-tale sign is a composite times (a multiple of) the derivative of its inside.
For a definite integral you may either change the limits to \(u\)-values, or integrate in \(u\) and convert back to \(x\) before substituting the original limits. Either way, account for any constant factor from \(du\).
📋 Substitution
Step
Action
Choose \(u\)
the inner function \(g(x)\)
Differentiate
\(du=g'(x)\,dx\)
Rewrite
integral becomes \(\int h(u)\,du\)
Finish
integrate, then return to \(x\)
⚠️ Match the derivative
Substitution only works cleanly when \(g'(x)\) (or a constant multiple) is present. \(\int 2x(x^{2}+1)^{3}\,dx\) works because \(\tfrac{d}{dx}(x^{2}+1)=2x\) is right there.
⚠️ Common trapThe \(x\,dx=\tfrac12du\) gives the factor \(\tfrac12\); evaluate \(\tfrac12 e^{x^{2}}\) at the limits.
Why each option
A) forgot the \(\tfrac12\)
B) dropped the lower term
C) used \(e^{2}\)
D) no antiderivative factor
E) ✓ \(\tfrac12(e-1)\)
F) sign reversed
A7.6 Integration by Parts
When the integrand is a product that substitution cannot handle — typically a polynomial times \(e^{x}\), \(\sin x\), \(\cos x\) or \(\ln x\) — use integration by parts: \(\displaystyle\int u\,\dfrac{dv}{dx}\,dx=uv-\int v\,\dfrac{du}{dx}\,dx\). It trades the original integral for an easier one.
Choose \(u\) to be the part that simplifies when differentiated (a polynomial, or \(\ln x\)); let the rest be \(\dfrac{dv}{dx}\). A common mnemonic for the priority of \(u\) is LATE: Logs, Algebraic, Trig, Exponential. For \(\int\ln x\,dx\), take \(u=\ln x\) and \(\dfrac{dv}{dx}=1\).
📋 Integration by parts
Item
Detail
Formula
\(\int u\,v'=uv-\int u'\,v\)
Choose \(u\)
the part that simplifies (LATE)
\(\int x e^{x}\)
\(u=x,\ v'=e^{x}\)
\(\int \ln x\)
\(u=\ln x,\ v'=1\)
⚠️ Pick \(u\) so it simplifies
Choosing \(u=e^{x}\) in \(\int x e^{x}\,dx\) makes the integral worse. Pick \(u=x\): differentiating it to \(1\) is what makes the second integral trivial.
Worked Example — A first integration by parts
Question: Find \(\displaystyle\int x e^{x}\,dx\).
Working: Take \(u=x,\ \dfrac{dv}{dx}=e^{x}\), so \(v=e^{x}\). Then \(\int x e^{x}=x e^{x}-\int e^{x}\,dx=x e^{x}-e^{x}+C=e^{x}(x-1)+C\).
\(\int x e^{x}\,dx=e^{x}(x-1)+C\)
A7.6 · Easy · The formula
Integration by parts states \(\displaystyle\int u\,\dfrac{dv}{dx}\,dx=\)
⚠️ Common trapThe remaining integral is \(\int\tfrac{x}{2}\,dx=\tfrac{x^2}{4}\), not \(\tfrac{x^2}{2}\).
Why each option
A) sign wrong
B) ✓ \(\tfrac{x^2}{2}\ln x-\tfrac{x^2}{4}\)
C) wrong second integral
D) used \(v=x^2\)
E) wrong method
F) did not halve twice
A7.7 Applications: Volumes of Revolution
Rotating the region under \(y=f(x)\) (from \(x=a\) to \(x=b\)) a full turn about the \(x\)-axis sweeps out a solid. Slicing it into thin discs of radius \(y\) and thickness \(dx\), each has volume \(\pi y^{2}\,dx\); summing them gives \(\displaystyle V=\pi\int_{a}^{b}y^{2}\,dx\).
The key step is to square \(y\) before integrating. For rotation about the \(y\)-axis the roles swap: \(V=\pi\int_{c}^{d}x^{2}\,dy\), with the curve written as \(x\) in terms of \(y\).
📋 Volume of revolution
Axis
Volume
About \(x\)-axis
\(V=\pi\displaystyle\int_{a}^{b}y^{2}\,dx\)
About \(y\)-axis
\(V=\pi\displaystyle\int_{c}^{d}x^{2}\,dy\)
Disc volume
\(\pi y^{2}\,dx\)
⚠️ Square \(y\), and keep the \(\pi\)
Integrate \(y^{2}\), not \(y\), and never drop the factor \(\pi\). For \(y=x^{2}\), \(y^{2}=x^{4}\) — a different power to integrate.
Worked Example — A volume of revolution
Question: The region under \(y=\sqrt{x}\) from \(x=0\) to \(x=4\) is rotated about the \(x\)-axis. Find the volume.
Working: \(y^{2}=x\), so \(V=\pi\int_{0}^{4}x\,dx=\pi\big[\tfrac{x^{2}}{2}\big]_{0}^{4}=\pi\cdot8=8\pi\).
\(V=\pi\int_{0}^{4}x\,dx=8\pi\)
A7.7 · Easy · The formula
The volume when \(y=f(x)\) is rotated about the \(x\)-axis from \(a\) to \(b\) is:
