Find two numbers that multiply to 3 × 2 = 6 and add to −7: that's −6 and −1
Split middle term: 3x² − 6x − x + 2 = 0
Factor: 3x(x − 2) − 1(x − 2) = 0
(3x − 1)(x − 2) = 0, so x = 1/3 or x = 2
⚠ Common Error
When completing the square for 2x² + 8x + 3, students forget to factor out the 2 first. You MUST write 2(x² + 4x) + 3 before completing the square inside the bracket.
Practice Problems
Q1: Solve x² − 5x + 6 = 0
(x − 2)(x − 3) = 0, so x = 2 or x = 3
Q2: Use the discriminant to show that x² + 3x + 5 = 0 has no real roots.
Example: Find S∞ for the series 8 + 4 + 2 + 1 + ...
a = 8, r = 4/8 = 1/2. Since |1/2| < 1, sum to infinity exists.
S∞ = 8 / (1 − 1/2) = 8 / (1/2) = 16
Example: The 3rd term of a GP is 18 and the 6th term is 486. Find a and r.
ar² = 18 and ar⁵ = 486
Divide: ar⁵/ar² = 486/18 → r³ = 27 → r = 3
From ar² = 18: a(9) = 18 → a = 2
⚠ Common Error
The sum to infinity only converges when |r| < 1. Always check this condition first! If |r| ≥ 1 the series diverges and S∞ does not exist.
📐 Binomial Expansion (Positive Integer)
Binomial Theorem
(a + b)ⁿ = Σ ⁿCᵣ an−r br ⁿCᵣ = n! / (r!(n−r)!)
Example: Expand (2 + x)⁴
Coefficients from Pascal's row 4: 1, 4, 6, 4, 1
1(2)⁴ + 4(2)³(x) + 6(2)²(x²) + 4(2)(x³) + 1(x⁴)
= 16 + 32x + 24x² + 8x³ + x⁴
Practice Problems
Q1: Find the 8th term of the arithmetic sequence 2, 5, 8, ...
u₈ = 2 + 7(3) = 23
Q2: A geometric series has a = 5 and r = 3. Find S₆.
Example: Find the equation of the tangent to y = x³ − 2x at x = 1
dy/dx = 3x² − 2. At x = 1: gradient = 3 − 2 = 1
At x = 1: y = 1 − 2 = −1. Point is (1, −1)
Tangent: y − (−1) = 1(x − 1) → y = x − 2
Practice Problems
Q1: Find the gradient of y = 3x² − 4x + 1 at x = 2
dy/dx = 6x − 4. At x = 2: gradient = 8
Q2: Find the stationary point of y = x² − 8x + 3 and state if it is max or min.
dy/dx = 2x − 8 = 0 → x = 4, y = 16 − 32 + 3 = −13. d²y/dx² = 2 > 0 → minimum at (4, −13).
Q3: Find the equation of the normal to y = x² at x = 3.
dy/dx = 2x. At x=3: gradient = 6, point (3,9). Normal gradient = −1/6. Equation: y − 9 = −(1/6)(x − 3) → 6y + x = 57
▶
6. Integration (AS)
AS
📐 Basic Integration
Integration Rule
∫ xn dx = xn+1/(n+1) + C (n ≠ −1) "Add 1 to the power, divide by the new power, add C"
Example: Find ∫ (3x² + 4x − 1) dx
∫3x² dx = 3x³/3 = x³
∫4x dx = 4x²/2 = 2x²
∫−1 dx = −x
Answer: x³ + 2x² − x + C
📐 Area Under a Curve
Method: Area Under a Curve
Set up the definite integral: ∫ₐᵇ f(x) dx
Integrate the function
Substitute upper limit, then subtract lower limit: [F(b)] − [F(a)]
If the area is below the x-axis, the integral gives a negative value — take the modulus
Example: Find the area under y = x² + 1 from x = 0 to x = 3
∫₀³ (x² + 1) dx = [x³/3 + x]₀³
Upper: (27/3 + 3) = 9 + 3 = 12
Lower: (0 + 0) = 0
Area = 12 − 0 = 12 square units
⚠ Common Error
Don't forget +C for indefinite integrals! For definite integrals, the C cancels so it's not needed. Also, always check if any part of the curve is below the x-axis — you need to split the integral and take absolute values for those parts.
Practice Problems
Q1: Find ∫ (6x² − 2x + 3) dx
2x³ − x² + 3x + C
Q2: Evaluate ∫₁² (4x³) dx
[x⁴]₁² = 16 − 1 = 15
Q3: Find the area enclosed between y = x² and y = 4.
Understanding functions, their domains, ranges, compositions and inverses.
📐 Domain, Range & Types
Key Definitions
Domain: the set of allowed input values (x)
Range: the set of possible output values (y = f(x)) One-to-one: each output has exactly one input | Many-to-one: different inputs can give the same output
Method: Finding Domain & Range
Domain: look for restrictions — can't divide by zero, can't square-root negatives, ln needs positive input
Range: consider the graph — what y-values are actually reached?
For quadratics: completing the square gives the vertex, which tells you the min/max of the range
Example: Find the domain and range of f(x) = 1/(x − 3)
Domain: x − 3 ≠ 0, so x ≠ 3. Domain: x ∈ ℝ, x ≠ 3
Range: 1/(x − 3) can take any value except 0 (the curve never crosses y = 0)
Range: f(x) ∈ ℝ, f(x) ≠ 0
📐 Composite Functions
Composite Functions
fg(x) = f(g(x)) — apply g first, then f Domain of fg: x must be in domain of g, AND g(x) must be in domain of f
Example: If f(x) = 2x + 1 and g(x) = x², find fg(3) and gf(x)
⚠ Common Error
For (a + bx)n where a ≠ 1: you MUST factor out an first! Write an(1 + bx/a)n then expand (1 + bx/a)n. The validity is |bx/a| < 1, i.e., |x| < |a/b|.
Practice Problems
Q1: Expand (1 − 2x)−1 up to x³ and state the validity.
⚠ Common Error
When using sec²θ = 1 + tan²θ to solve equations, always substitute to get one trig function only. Don't forget to find ALL solutions in the given range using the appropriate quadrant rule.
📐 Addition Formulae
Addition Formulae
sin(A ± B) = sinA cosB ± cosA sinB
cos(A ± B) = cosA cosB ∓ sinA sinB
tan(A ± B) = (tanA ± tanB) / (1 ∓ tanA tanB)
Substitute back for x (or change limits if definite)
Example: Find ∫ 2x(x² + 1)⁴ dx
Let u = x² + 1, then du/dx = 2x, so du = 2x dx
Integral becomes ∫ u⁴ du
= u⁵/5 + C = (x² + 1)⁵/5 + C
Example: Evaluate ∫₀¹ x/√(1 + x²) dx using substitution
Let u = 1 + x². du = 2x dx, so x dx = du/2
Change limits: x = 0 → u = 1; x = 1 → u = 2
∫₁² u⁻¹/² × ½ du = ½[2u¹/²]₁² = [√u]₁²
= √2 − √1 = √2 − 1 ≈ 0.414
⚠ Common Error
For definite integrals by substitution, either: (a) change the limits to u-values and DON'T substitute back, or (b) keep original limits and substitute back to x. Never mix the two approaches!
⚠ Common Error
Newton-Raphson can fail if: (1) f'(xₙ) = 0 (division by zero), (2) the starting point is too far from the root, or (3) the function has a turning point near the root. Always check convergence.
Practice Problems
Q1: Show that ex = 3x has a root between x = 1 and x = 2.
Magnitude: |a| = √(a₁² + a₂² + a₃²)
Scalar product: a · b = a₁b₁ + a₂b₂ + a₃b₃ = |a||b|cosθ
Angle: cosθ = (a · b) / (|a| |b|)
Example: Find the angle between a = (2, 1, −1) and b = (1, −1, 2)
a · b = (2)(1) + (1)(−1) + (−1)(2) = 2 − 1 − 2 = −1
|a| = √(4 + 1 + 1) = √6. |b| = √(1 + 1 + 4) = √6
cosθ = −1 / (√6 × √6) = −1/6
θ = cos⁻¹(−1/6) = 99.6° (1 d.p.)
⚠ Common Error
If a · b = 0, the vectors are perpendicular. Don't forget this key fact! Also remember: the scalar product gives a number, not a vector.
Example: Points A and B have position vectors a = 2i + j − 3k and b = 4i − j + k. Find the midpoint M and the distance AB.
AB = b − a = (4−2)i + (−1−1)j + (1−(−3))k = 2i − 2j + 4k
|AB| = √(4 + 4 + 16) = √24 = 2√6
M = ½(a + b) = ½(6i + 0j − 2k) = 3i − k. Midpoint is (3, 0, −1).
Practice Problems
Q1: Find |a| where a = 3i + 4j.
|a| = √(9 + 16) = √25 = 5
Q2: Show that a = (2, −1, 3) and b = (1, 5, 1) are perpendicular.
a · b = 2 − 5 + 3 = 0. Since a · b = 0, the vectors are perpendicular. ✓
Q3: A = (1, 2, 3), B = (3, 0, −1). Find a unit vector in the direction of AB.
AB = (2, −2, −4). |AB| = √(4+4+16) = √24 = 2√6. Unit vector = (1/√6, −1/√6, −2/√6)
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14. Proof
A2
Methods of mathematical proof required at A2 level.
📐 Proof by Contradiction
Method: Proof by Contradiction
Assume the OPPOSITE of what you want to prove
Use logical steps to derive a contradiction (something impossible)
Conclude that the original assumption must be false
Therefore the original statement is true
Example: Prove that √2 is irrational
Assume √2 IS rational. Then √2 = a/b where a, b are integers with no common factors.
Squaring: 2 = a²/b², so a² = 2b². Therefore a² is even, so a must be even. Write a = 2k.
(2k)² = 2b² → 4k² = 2b² → b² = 2k². So b² is even, meaning b is also even.
Both a and b are even — they share factor 2. This contradicts our assumption that a/b has no common factors. Therefore √2 is irrational. ∎
Example: Prove there are infinitely many prime numbers
Assume there are finitely many primes: p₁, p₂, ..., pₙ
Consider N = p₁ × p₂ × ... × pₙ + 1
N is not divisible by any of p₁, p₂, ..., pₙ (always remainder 1). So either N is prime, or N has a prime factor not in our list.
Either way, there's a prime not in the list — contradiction. Therefore there are infinitely many primes. ∎
📐 Disproof by Counter-Example
Method: Disproof by Counter-Example
To disprove a statement, you only need ONE example where it fails
Find a specific value that makes the statement false
State the counter-example and show why the statement fails
This only works to DISPROVE. One example is NOT enough to prove something is always true.
Example: Disprove: "If n is a prime number, then 2ⁿ − 1 is also prime"
Try n = 11 (prime): 2¹¹ − 1 = 2047
2047 = 23 × 89, which is NOT prime
Counter-example: n = 11 gives 2ⁿ − 1 = 2047 = 23 × 89. Statement is false. ✗
Practice Problems
Q1: Disprove: "n² + n + 41 is prime for all positive integers n".
Q2: Prove by contradiction that if n² is even, then n is even.
Assume n is odd. Then n = 2k+1, so n² = 4k²+4k+1 = 2(2k²+2k)+1, which is odd. Contradicts n² being even. So n must be even. ∎
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14. Kinematics
AS
📐 SUVAT Equations
SUVAT Equations (constant acceleration)
v = u + at
s = ut + ½at²
s = vt − ½at²
v² = u² + 2as
s = ½(u + v)t
Velocity-Time & Displacement-Time Graphs
v-t graph: gradient = acceleration | area under = displacement
s-t graph: gradient = velocity | curve → changing velocity
Method: Solving SUVAT Problems
List what you know: s, u, v, a, t
Identify what you need to find
Choose the equation that contains all known variables plus the unknown
Substitute and solve
For vertical motion: use g = 9.8 m/s² (positive downwards usually)
Define a positive direction at the start. Be consistent with signs!
Example: A ball is thrown upwards at 15 m/s. Find the max height and time to return.
Taking upwards as positive: u = 15, v = 0 (at max height), a = −9.8
v² = u² + 2as → 0 = 225 + 2(−9.8)s → s = 225/19.6 = 11.5 m
Time up: v = u + at → 0 = 15 − 9.8t → t = 15/9.8 = 1.53 s
Time to return = 2 × 1.53 = 3.06 s (by symmetry)
Example: A car accelerates uniformly from rest to 20 m/s in 5 s, then travels at constant speed for 10 s, then decelerates to rest in 5 s. Find the total distance.
Phase 1 (0-5s): s₁ = ½(0 + 20)(5) = 50 m (triangle on v-t graph)
1st: Object remains at rest or constant velocity unless acted on by a force
2nd: F = ma (resultant force = mass × acceleration)
3rd: Every action has an equal and opposite reaction
Weight: W = mg | Friction: F ≤ μR
Method: F = ma Problems
Draw a clear free body diagram
Resolve forces parallel and perpendicular to motion
Apply F = ma in the direction of motion
If equilibrium: resultant force = 0
For connected particles: consider each body separately or the whole system
Example: A 5 kg box is pushed along a rough floor by a 30 N force. μ = 0.2. Find the acceleration.
Weight = mg = 5 × 9.8 = 49 N. Normal reaction R = 49 N (no vertical acceleration)
Friction = μR = 0.2 × 49 = 9.8 N
Resultant = 30 − 9.8 = 20.2 N
F = ma → 20.2 = 5a → a = 4.04 m/s²
📐 Inclined Planes
Forces on an Inclined Plane (angle α)
Component down the slope: mg sinα
Component normal to slope: mg cosα
Normal reaction: R = mg cosα (no acceleration perpendicular to slope)
Example: A 4 kg box slides down a smooth slope at 30° to the horizontal. Find the acceleration.
Force down slope = mg sin30° = 4 × 9.8 × 0.5 = 19.6 N
Smooth → no friction. F = ma: 19.6 = 4a
a = 4.9 m/s² (= g sin30°)
Example: A 6 kg box is on a rough slope at 25° (μ = 0.3). Find whether it slides and, if so, its acceleration.
Component down slope = mg sin25° = 6(9.8)(0.4226) = 24.85 N
R = mg cos25° = 6(9.8)(0.9063) = 53.29 N. Max friction = μR = 0.3 × 53.29 = 15.99 N
24.85 > 15.99, so the box slides. Resultant = 24.85 − 15.99 = 8.86 N
F = ma: 8.86 = 6a → a = 1.48 m/s² down the slope
⚠ Common Error
On inclined planes, the component of weight down the slope is mg sinθ, NOT mg cosθ. Remember: sin goes with the slope direction, cos goes with the normal direction.
Practice Problems
Q1: A 10 kg box is on a smooth slope at 30°. Find its acceleration.
a = g sin30° = 9.8 × 0.5 = 4.9 m/s²
Q2: A 5 kg block is held on a rough slope (30°, μ = 0.4) by a horizontal force P. Find P for equilibrium.
Resolve along slope: P cos30° + F = mg sin30°. Perpendicular: R = mg cos30° + P sin30°. F = μR. Substituting and solving gives P ≈ 11.5 N.
Q3: Two particles (3 kg and 5 kg) are connected by a string over a pulley. Find the acceleration and tension.
Whole system: 5g − 3g = 8a → a = 2g/8 = 2.45 m/s². For 3 kg: T − 3g = 3a → T = 3(9.8 + 2.45) = 36.75 N.
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16. Moments
AS
📐 Moments & Equilibrium
Moments
Moment = Force × Perpendicular distance from pivot Equilibrium: Sum of clockwise moments = Sum of anticlockwise moments
Example: A uniform rod AB of length 4 m and weight 60 N rests on supports at A and at C, 3 m from A. Find the reactions at A and C.
Weight acts at midpoint (2 m from A). Let reactions be Rₐ and R꜀.
Taking moments about A: R꜀ × 3 = 60 × 2 → R꜀ = 40 N
Example: A non-uniform rod AB (4 m, 80 N) has its centre of gravity 1.5 m from A. It's supported at P (1 m from A) and Q (3.5 m from A). Find the reactions.
Weight acts at centre of gravity: 1.5 m from A. Let reactions be Rₚ and R_Q.
Moments about P: R_Q × 2.5 = 80 × 0.5 (weight is 0.5 m from P, on same side as Q)
⚠ Common Error
For a uniform rod, the weight acts at the midpoint. For a non-uniform rod, the weight acts at the centre of gravity — this will be specified in the question. Always take moments about one of the unknown forces to eliminate it.
Practice Problems
Q1: A 3 m uniform plank weighing 120 N is supported at its ends. A 200 N weight is placed 1 m from one end. Find the reactions.
Moments about A: R_B × 3 = 120(1.5) + 200(1) = 380. R_B = 126.7 N. R_A = 320 − 126.7 = 193.3 N.
Q2: A 2 m light rod has a 40 N weight at one end and a 60 N weight at the other. Find the position of the support for equilibrium.
Let support be x from the 40 N end. Moments about support: 40x = 60(2 − x) → 40x = 120 − 60x → 100x = 120 → x = 1.2 m from the 40 N end.
Q3: A uniform beam of length 6 m and weight 200 N rests on two supports, one at each end. A child weighing 400 N sits 2 m from one end. Find both reaction forces.
Moments about A: R_B × 6 = 200(3) + 400(2) = 1400. R_B = 233.3 N. R_A = 600 − 233.3 = 366.7 N.
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17. Projectiles
A2
📐 Projectile Motion
Projectile Equations (angle θ, speed u)
Horizontal: x = (u cosθ)t [no acceleration]
Vertical: y = (u sinθ)t − ½gt²
Time of flight: T = 2u sinθ / g
Max height: H = u²sin²θ / 2g
Range: R = u²sin2θ / g
Example: A ball is launched at 20 m/s at 30° above horizontal. Find the range and max height.
u = 20, θ = 30°, g = 9.8
Max height: H = 20² × sin²30° / (2 × 9.8) = 400 × 0.25 / 19.6 = 5.10 m
Range: R = 20² × sin60° / 9.8 = 400 × 0.866 / 9.8 = 35.3 m
Method: Projectile Problems
Split velocity into components: Horizontal = u cosθ, Vertical = u sinθ
Horizontal: constant velocity (no acceleration), use s = vt
Vertical: use SUVAT with a = −g (taking up as positive)
At maximum height, vertical velocity = 0
Time of flight: time for vertical displacement to return to 0
Example: A ball is kicked horizontally at 12 m/s from the top of a 15 m cliff. Find where it lands.
Vertically (down positive): s = 15, u = 0, a = 9.8. Using s = ½at²: 15 = ½(9.8)t²
t² = 30/9.8 = 3.061 → t = 1.75 s
Horizontal distance = 12 × 1.75 = 21.0 m from base of cliff
⚠ Common Error
Horizontal acceleration is ZERO (not g). Only the vertical component is affected by gravity. Don't mix up horizontal and vertical components!
Practice Problems
Q1: A projectile is fired at 25 m/s at 60° above the horizontal. Find the time of flight.
T = 2u sinθ / g = 2(25)sin60° / 9.8 = 50(0.866)/9.8 = 4.42 s
Q2: Find the range of a projectile fired at 30 m/s at 45°.
R = u²sin2θ / g = 900 × sin90° / 9.8 = 900/9.8 = 91.8 m (maximum range occurs at 45°)
Q3: A ball is thrown horizontally at 8 m/s from a height of 5 m. Find its speed when it hits the ground.
Momentum: p = mv
Impulse: I = Ft = mv − mu (change in momentum)
Conservation: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Coefficient of restitution: e = (v₂ − v₁)/(u₁ − u₂) [separation/approach speed]
Example: A 2 kg ball at 6 m/s hits a stationary 3 kg ball. e = 0.4. Find velocities after collision.
⚠ Common Error
Direction matters! When objects rebound, their velocity changes sign. Impulse = change in momentum = mv − mu, and you must be consistent with your positive direction throughout.
Practice Problems
Q1: A 3 kg object moving at 4 m/s collides with a stationary 2 kg object. They stick together. Find their common velocity.
Conservation: 3(4) + 2(0) = 5v → v = 12/5 = 2.4 m/s
Q2: A force acts on a 2 kg ball for 0.3 s, changing its velocity from 5 m/s to 11 m/s. Find the force.
Impulse = mv − mu = 2(11) − 2(5) = 12 Ns. F = I/t = 12/0.3 = 40 N
Q3: Two particles (2 kg at 5 m/s, 3 kg at 2 m/s) travel in the same direction and collide. e = 0.5. Find final velocities.
Q1: A disc rotates at 120 rpm. Find the angular speed in rad/s.
ω = 2π × 120/60 = 4π = 12.6 rad/s
Q2: Find the centripetal force on a 3 kg object moving in a circle of radius 2 m at 5 m/s.
F = mv²/r = 3 × 25/2 = 37.5 N
Q3: A satellite orbits at radius r. Show that v = √(gr²/r) if gravity provides the centripetal force mg = mv²/r.
mg = mv²/r → g = v²/r → v² = gr → v = √(gr). For a 200 km orbit (r ≈ 6571 km), v ≈ √(9.8 × 6.571 × 10⁶) ≈ 7.9 km/s.
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21. Variable Acceleration & Calculus Kinematics
A2
When acceleration is not constant, we use calculus instead of SUVAT.
📐 Calculus Methods for Kinematics
Calculus Kinematics
v = ds/dt (velocity = derivative of displacement)
a = dv/dt = d²s/dt² (acceleration = derivative of velocity)
s = ∫ v dt (displacement = integral of velocity)
v = ∫ a dt (velocity = integral of acceleration)
Method: Variable Acceleration Problems
If given s(t) or v(t): differentiate to find velocity or acceleration
If given a(t): integrate to find velocity (don't forget +C!)
Use initial conditions (e.g., "at t = 0, v = 5") to find the constant C
Maximum/minimum displacement: set v = 0 and solve for t
Object at rest: v = 0. Returning to start: s = 0
Example: A particle moves with s = 2t³ − 9t² + 12t. Find when the particle is at rest and its acceleration at that instant.
a = dv/dt = 12t − 18. At t = 1: a = −6 m/s². At t = 2: a = 6 m/s²
Example: A particle has acceleration a = 4 − 2t. At t = 0, v = 3 and s = 0. Find s when t = 3.
v = ∫(4 − 2t) dt = 4t − t² + C₁. At t = 0, v = 3: C₁ = 3
v = 4t − t² + 3
s = ∫(4t − t² + 3) dt = 2t² − t³/3 + 3t + C₂. At t = 0, s = 0: C₂ = 0
At t = 3: s = 2(9) − 27/3 + 9 = 18 − 9 + 9 = 18 m
⚠ Common Error
When integrating to find v or s, you MUST include the constant of integration (+C) and use the given initial conditions to find it. Forgetting the constant loses marks every time.
Practice Problems
Q1: s = t³ − 6t² + 9t + 1. Find the times when v = 0 and the displacement at each.
v = 3t² − 12t + 9 = 3(t² − 4t + 3) = 3(t−1)(t−3) = 0. t = 1: s = 1−6+9+1 = 5. t = 3: s = 27−54+27+1 = 1.
Q2: a = 6t. At t = 0, v = 2 and s = 1. Find s when t = 2.
v = ∫6t dt = 3t² + C. v(0)=2 → C=2. v = 3t²+2. s = ∫(3t²+2)dt = t³+2t+D. s(0)=1 → D=1. s(2) = 8+4+1 = 13 m.
Q3: v = 8 − 4t. Find the total distance travelled from t = 0 to t = 4 s.
v = 0 at t = 2. Distance = |∫₀² (8−4t)dt| + |∫₂⁴ (8−4t)dt| = |[8t−2t²]₀²| + |[8t−2t²]₂⁴| = |8| + |−8| = 16 m