Decide which form suits the operation: rectangular a + jb for adding and subtracting, polar r∠θ for multiplying, dividing, powers and roots.
Multiply out with j² = −1; divide by multiplying the top and bottom by the conjugate of the denominator.
Convert with r = √(a² + b²) and θ from a sketch of the point (check the quadrant), and back with a = r cos θ, b = r sin θ.
For powers use De Moivre, (r∠θ)ⁿ = rⁿ∠nθ; for zⁿ = R∠φ list all n roots R1/n∠((φ + 360°k)/n).
Differential equations
Exponential growth/decay
dy/dt = ky → y = y0ekt
Half-life
k = ln 2 / t½
Newton's law of cooling
T(t) = Ta + (T0 − Ta)e−kt
Integrating factor
y′ + p(x)y = q(x): μ = e∫p dx, (μy)′ = μq
Mixing tank
dm/dt = r·c − (r/V)·m
Matrices & determinants
2×2 determinant
det = ad − bc
2×2 inverse
A⁻¹ = (1/det)·[d −b; −c a]
Product
(AB)ij = Σk AikBkj
Eigenvalues
det(A − λI) = 0
Cramer’s rule
xi = det(Ai)/det(A)
Vectors: dot & cross product
Magnitude
|a| = √(ax² + ay² + az²)
Unit vector
û = a/|a|
Components from an angle
Fx = F·cos θ, Fy = F·sin θ
Dot product
a·b = axbx + ayby + azbz = |a||b|cos θ
Scalar projection of a onto b
a·b/|b|
Work
W = F·d
Cross product
a × b = (aybz − azby, azbx − axbz, axby − aybx)
Areas
parallelogram |a × b|, triangle ½|a × b|
Moment about O
MO = r × F, r = A − O
Volume of a parallelepiped
V = |a·(b × c)|
Method
Write every vector in components; for a force given by a magnitude and an angle, use F·cos θ and F·sin θ.
For lengths and directions use the magnitude √(x² + y² + z²) and the unit vector a/|a|.
Use the dot product for angles, projections, perpendicularity and work: a·b = |a||b|cos θ.
Use the cross product for perpendicular vectors, areas and moments: write the three components carefully, keeping the order a × b.
Probability
Complement
P(not A) = 1 − P(A)
Addition rule
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Independent events
P(A ∩ B) = P(A)·P(B)
Conditional probability
P(A|B) = P(A ∩ B)/P(B)
Binomial
P(X = k) = C(n,k)·pk(1 − p)n−k
Descriptive statistics
Mean
x̄ = Σx/n
Sample standard deviation
s = √(Σ(x − x̄)²/(n − 1))
Median
middle value of the sorted data
z-score
z = (x − μ)/σ
Normal distribution
68% within ±1σ, 95% within ±2σ
Normal distribution
z-score
z = (x − μ)/σ
Below x
P(X < x) = Φ(z)
Above x
P(X > x) = 1 − Φ(z)
Between
Φ(z2) − Φ(z1)
Percentile
x = μ + z·σ
Method
Convert the value to a z-score: z = (x − μ)/σ.
Look up Φ(z), the area to the left of z.
For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
For a percentile, work backwards: find z for the area, then x = μ + z·σ.
2Practice problems
1.LimitsMedium
Find the limit L = limx→∞ (1 + 3/x)2x. Give a decimal value.
f(x) = (1 + 3/x)2xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + x)/sin(2x), with x in radians.
f(x) = ln(1 + x)/sin(2x)x → 0
L =
3.DerivativesMedium
Let f(x) = x2·e−x. Find f′(0.5).
f(x) = x2·e−xx = 0.5
f′(0.5) =
4.DerivativesHard
The tangent line to f(x) = 0.5x2 + x − 5 at x = 3 crosses the y-axis at y = b. Find b.
f(x) = 0.5x2 + x − 5x = 3
b =
5.IntegralsMedium
Evaluate I = ∫₀2 x·sin x dx (x in radians).
∫₀2 x·sin x dx
I =
6.IntegralsHard
The region between the line y = 2x and the parabola y = 2x² is rotated about the x-axis. Find the volume V of the solid.
y = 2xy = 2x²
V =
7.Complex numbersMedium
Given z₁ = 5∠70° and z₂ = 1.5∠−95°, find the imaginary part of w = z₁/z₂.
z₁ = 5∠70°z₂ = 1.5∠−95°
Im(w) =
8.Complex numbersHard
Solve z² − 5z + 8.5 = 0. The roots are a ± jb; find b > 0.
p = −5q = 8.5
b =
9.Differential equationsMedium
Solve y″ + 4y′ + 4y = 0 with y(0) = 2 and y′(0) = 0. Find y(1.5).
y″ + 4y′ + 4y = 0y(0) = 2y′(0) = 0
y(1.5) =
10.Differential equationsHard
Solve y″ + 4y′ + 13y = 0 with y(0) = 1 and y′(0) = 0 (released from rest). Find y(0.75); angles in radians.
y″ + 4y′ + 13y = 0y(0) = 1y′(0) = 0
y(0.75) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−2 1 −4; 0 −5 −3; 5 0 −3]
det A =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [0 −4; −3 1]
λ =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (4, −4, −2) and b = (−3, k, −4) are perpendicular.
a = (4, −4, −2)b = (−3, k, −4)
k =
14.Vectors: dot & cross productHard
For a = (2, 3, 2) and b = (2, −5, 0), find the x-component (a × b)x of the cross product a × b.
a = (2, 3, 2)b = (2, −5, 0)
(a × b)x =
15.ProbabilityMedium
A fair die is rolled 5 times. What is the probability P of getting at least one six?
n = 5
P =
16.ProbabilityHard
A game costs €1 to play. You win €5 with probability 0.1 and €10 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €1€5: p = 0.1€10: p = 0.01
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 22, 23, 28, 46, 53, 56, 51.
n = 7
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 12. What proportion P of values is below 94?
μ = 70σ = 12x = 94
P =
19.Normal distributionMedium
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 246.25 and 259.25 g? Give P as a decimal.
μ = 250σ = 5x1 = 246.25x2 = 259.25
P =
20.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded bottom 70% is to the left of it. Find the cut-off x.
μ = 100 pointsσ = 15 points70%
x = points
3Solutions
1.Limits
Answer: L = 403.4
Limits: first try direct substitution. If it gives 0/0, simplify first: factor and cancel the common factor, or multiply by the conjugate. As x → ∞, divide numerator and denominator by the highest power of x. L'Hôpital's rule: if f/g gives 0/0 or ∞/∞, then lim f/g = lim f′/g′, differentiating the numerator and the denominator separately (not with the quotient rule).
Take the logarithm
ln L = limx→∞ 2·x·ln(1 + 3/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 2·3 = 6
Undo the logarithm
L = e6 = 403.4
Answer
L = 403.4
2.Limits
Answer: L = 0.5
Limits: first try direct substitution. If it gives 0/0, simplify first: factor and cancel the common factor, or multiply by the conjugate. As x → ∞, divide numerator and denominator by the highest power of x. L'Hôpital's rule: if f/g gives 0/0 or ∞/∞, then lim f/g = lim f′/g′, differentiating the numerator and the denominator separately (not with the quotient rule).
L'Hôpital's rule
L = limx→0 [1/(1 + x)]/[2·cos 2x]
Substitute x = 0
L = 1/2 = 0.5
Answer
L = 0.5
3.Derivatives
Answer: f′(0.5) = 0.4549
Differentiation: the derivative f′(x) is the slope of the graph at x. Power rule: (xⁿ)′ = n·xⁿ⁻¹. Product rule: (uv)′ = u′v + uv′. Chain rule: (f(g(x)))′ = f′(g(x))·g′(x). Where f′(x) = 0 the graph has a horizontal tangent: a maximum, minimum or saddle point.
Product rule
f′(x) = u′v + uv′ = e−x·(2x + (−1)·x2)
Substitute
f′(0.5) = 0.6065·(1 + (−0.25)) = 0.4549
Answer
f′(0.5) = 0.4549
4.Derivatives
Answer: b = −9.5
Differentiation: the derivative f′(x) is the slope of the graph at x. Power rule: (xⁿ)′ = n·xⁿ⁻¹. Product rule: (uv)′ = u′v + uv′. Chain rule: (f(g(x)))′ = f′(g(x))·g′(x). Where f′(x) = 0 the graph has a horizontal tangent: a maximum, minimum or saddle point.
Point and slope
f(3) = 2.5
f′(x) = x + 1 → f′(3) = 4
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 2.5 − 4·3 = −9.5
Answer
b = −9.5
5.Integrals
Answer: I = 1.742
Integration by parts: ∫u·dv = u·v − ∫v·du, from the product rule. Choose u as the factor that gets simpler when you differentiate it (a power of x, or ln x), and dv as the factor you can integrate (eˣ, sin x, cos x, or x when u = ln x). For a definite integral, evaluate u·v at both limits: ∫ₐᵇ u dv = [u·v]ₐᵇ − ∫ₐᵇ v du. A factor x² needs two rounds.
Choose u and dv
u = x, du = dx; dv = sin x dx, v = −cos x
Antiderivative
∫x sin x dx = −x cos x + ∫cos x dx = −x cos x + sin x
Limits
(−2·cos 2 + sin 2) − (0 + 0) = 1.742
Answer
I = 1.742
6.Integrals
Answer: V = 1.676
Volumes of revolution: rotating the region under y = f(x) from x = a to x = b about the x-axis makes a solid whose cross-section at x is a disc of radius f(x). Adding the discs gives V = π∫ₐᵇ f(x)² dx. When the region lies between an outer curve R(x) and an inner curve r(x), each slice is a washer: V = π∫ₐᵇ (R(x)² − r(x)²) dx. Square the radii before you subtract them.
Intersection points
2x = 2x² → x = 0, x = 1
Washer method
V = π∫₀1 ((2x)² − (2x²)²) dx
Integrate
V = π(4·1³/3 − 4·1⁵/5) = π·0.5333 = 1.676
Answer
V = 1.676
7.Complex numbers
Answer: Im(w) = 0.8627
Polar form: z = r∠θ = r(cos θ + j sin θ) = r·ejθ, with r = |z| and θ = arg z. Back to rectangular form: a = r cos θ, b = r sin θ. The argument needs the quadrant: θ = atan(b/a) only when a > 0; for a < 0 add or subtract 180°. In polar form multiplication is easy: multiply the moduli and add the arguments; for division divide the moduli and subtract the arguments.
Modulus and argument of w
|w| = 5 / 1.5 = 3.333
θ = 70° − (−95)° = 165°
Rectangular form
Re(w) = 3.333·cos(165°) = −3.22
Im(w) = 3.333·sin(165°) = 0.8627
Im(w) = 0.8627
Answer
Im(w) = 0.8627
8.Complex numbers
Answer: b = 1.5
De Moivre and roots: (r∠θ)ⁿ = rⁿ∠nθ. The equation zⁿ = R∠φ has exactly n roots, all with modulus R1/n, at the arguments (φ + 360°·k)/n for k = 0, 1, …, n − 1: they lie evenly spaced on a circle, 360°/n apart. A quadratic z² + pz + q = 0 with p² < 4q has the complex roots z = −p/2 ± j√(q − p²/4); a polynomial with real coefficients always has its complex roots in conjugate pairs a ± jb.
Discriminant
p² − 4q = 25 − 34 = −9 < 0
Quadratic formula
z = (5 ± j√9)/2 = 2.5 + j1.5, 2.5 − j1.5
Answer
b = 1.5
Answer
b = 1.5
9.Differential equations
Answer: y(1.5) = 0.3983
Second-order linear equations: for y″ + b·y′ + c·y = 0 try y = ert: this gives the characteristic equation r² + b·r + c = 0. Two real roots r₁ ≠ r₂ give y = A·er₁t + B·er₂t (overdamped); a double root r gives y = (A + Bt)·ert (critically damped); complex roots −α ± jω give y = e−αt(A cos ωt + B sin ωt) (underdamped, oscillating). The initial values y(0) and y′(0) fix A and B. Written as y″ + 2ζωny′ + ωn²y = 0, ωn is the natural frequency and ζ the damping ratio: ζ < 1 oscillates. With a constant input, y″ + by′ + cy = c·K, the solution is the steady value K plus the free response.
Double root
(r + 2)² = 0 → r = −2
Constants
A = y(0) = 2
y′(0) = B − 2A → B = 0 + 4 = 4
Evaluate
y(1.5) = (2 + 4·1.5)·e−3 = 0.3983
Answer
y(1.5) = 0.3983
10.Differential equations
Answer: y(0.75) = −0.02442
Second-order linear equations: for y″ + b·y′ + c·y = 0 try y = ert: this gives the characteristic equation r² + b·r + c = 0. Two real roots r₁ ≠ r₂ give y = A·er₁t + B·er₂t (overdamped); a double root r gives y = (A + Bt)·ert (critically damped); complex roots −α ± jω give y = e−αt(A cos ωt + B sin ωt) (underdamped, oscillating). The initial values y(0) and y′(0) fix A and B. Written as y″ + 2ζωny′ + ωn²y = 0, ωn is the natural frequency and ζ the damping ratio: ζ < 1 oscillates. With a constant input, y″ + by′ + cy = c·K, the solution is the steady value K plus the free response.
Matrices: the determinant of a 2×2 matrix is ad − bc. For 3×3, expand along a row: each entry times the determinant of the 2×2 matrix left after removing its row and column, with alternating signs + − +. Eigenvalues λ solve det(A − λI) = 0; for 2×2 that is λ² − (trace)·λ + det = 0.
2×2 minors of the first row
M₁₁ = 15, M₁₂ = 15, M₁₃ = 25
Cofactor expansion
det A = (−2)·15 − 1·15 + (−4)·25 = −145
Answer
det A = −145
12.Matrices & determinants
Answer: λ = 4
Matrices: the determinant of a 2×2 matrix is ad − bc. For 3×3, expand along a row: each entry times the determinant of the 2×2 matrix left after removing its row and column, with alternating signs + − +. Eigenvalues λ solve det(A − λI) = 0; for 2×2 that is λ² − (trace)·λ + det = 0.
Characteristic equation
λ² − (1)λ + (−12) = 0
Solve
λ = (1 ± √(49))/2 → λ = −3, 4
λ = 4
Answer
λ = 4
13.Vectors: dot & cross product
Answer: k = −1
Dot product:a·b = axbx + ayby + azbz = |a|·|b|·cos θ, a scalar. It gives the angle between two vectors, cos θ = a·b/(|a|·|b|), and it is zero when they are perpendicular. The scalar projection of a onto b is a·b/|b|, and a constant force F moving through a displacement d does work W = F·d.
Dot product equals zero
a·b = 4·(−3) + (−4)·k + (−2)·(−4) = 0
−4 + (−4)·k = 0
Solve for k
k = −(−4)/(−4) = −1
Answer
k = −1
14.Vectors: dot & cross product
Answer: (a × b)x = 10
Cross product:a × b = (aybz − azby, azbx − axbz, axby − aybx) is perpendicular to both vectors, and |a × b| = |a|·|b|·sin θ is the area of the parallelogram they span (the triangle has half of it). Order matters: b × a = −(a × b). The moment of a force F applied at point A about point O is MO = r × F, with r = A − O.
Cross product, component by component
(a × b)x = aybz − azby = 3·0 − 2·(−5) = 10
(a × b)y = azbx − axbz = 2·2 − 2·0 = 4
(a × b)z = axby − aybx = 2·(−5) − 3·2 = −16
Requested component
(a × b)x = 10
Answer
(a × b)x = 10
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: P = 0.5981
Probability: with equally likely outcomes, P(A) = favourable/total. For independent events multiply: P(A and B) = P(A)·P(B). The complement often helps: P(at least one) = 1 − P(none). Binomial: in n independent trials with success probability p, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ.
Complement
P(no six) = (5/6)5 = 0.4019
P = 1 − 0.4019 = 0.5981
Answer
P = 0.5981
16.Probability
Answer: E(G) = −0.4 €
Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.
Net outcomes
G: 4 (p = 0.1), 9 (p = 0.01), −1 (p = 0.89)
Expected value
E(G) = 4·0.1 + 9·0.01 − 1·0.89 = −0.4 €
Answer
E(G) = −0.4 €
A negative value means you lose money on average.
17.Descriptive statistics
Answer: s = 14.94
Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.
Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.
Standardise
z = 2
Standard normal table
Φ(2) = 0.9772
P = 0.9772
Answer
P = 0.9772
19.Normal distribution
Answer: P = 0.7412
Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.
z-scores
z₁ = (246.25 − 250)/5 = −0.75
z₂ = (259.25 − 250)/5 = 1.85
Area between
P = Φ(1.85) − Φ(−0.75) = 0.9678 − 0.2266 = 0.7412
Answer
P = 0.7412
20.Normal distribution
Answer: x = 107.8 points
From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.