Flashcards · IB Diploma Programme Mathematics: Applications and Interpretation · Topic 1

Number and algebra flashcards

73 cards on Topic 1, subtopics 1.1 to 1.15. Pick a subtopic or study them all.

Study mode needs JavaScript. Every card is listed below, so you can still cover the answers and test yourself.

All 73 cards

For a quick read-through or printing. Cover the right-hand side to test yourself. Cards marked HL are HL content only.

1.1 Operations with numbers in the form a × 10^k

Standard form: a × 10ᵏ, where 1 ≤ a < 10 and k is an integer.
Your GDC shows 6.1E−4. How do you write it in an answer?: 6.1 × 10⁻⁴. Never write calculator notation such as 6.1E−4 as a final answer.
Calculate (6 × 10⁵) × (3 × 10⁻²) in standard form: Multiply the numbers and add the powers: 18 × 10³ = 1.8 × 10⁴.
Why estimate before accepting a GDC answer such as (4.8 × 10⁶) ÷ (1.6 × 10⁻³)?: To catch keying errors. 4.8 ÷ 1.6 = 3 and 10⁶ ÷ 10⁻³ = 10⁹, so the answer should be about 3 × 10⁹, not 3 × 10³.

1.2 Arithmetic sequences and series

Arithmetic sequence: nth term formula: uₙ = u₁ + (n − 1)d u₁ is the first term, d the common difference, n the term number.
Arithmetic series: sum of the first n terms: Sₙ = (n/2)(2u₁ + (n − 1)d) = (n/2)(u₁ + uₙ)
How do you find d and u₁ when you are given u₅ and u₁₂?: 1. Subtract: u₁₂ − u₅ = 7d, so find d. 2. Back-substitute: u₁ = u₅ − 4d. 3. Use the nth term or sum formula.
What does Σ (from r = 1 to n) of an expression in r mean?: Substitute r = 1, 2, …, n into the expression and add the results. Example: Σ (5r + 2) from r = 1 to 16 has u₁ = 7, u₁₆ = 82, so the sum is 8(7 + 82) = 712.
Find S₁₂ for an arithmetic sequence with u₁ = 9 and d = −2: S₁₂ = 6(2 × 9 + 11 × (−2)) = 6 × (−4) = −24.
Simple interest vs compound interest: which is arithmetic and which is geometric?: Simple interest adds the same amount each year: arithmetic. Compound interest adds the same percentage each year: geometric.

1.3 Geometric sequences and series

Geometric sequence: nth term formula: uₙ = u₁rⁿ⁻¹ u₁ is the first term, r the common ratio.
Geometric series: sum of the first n terms: Sₙ = u₁(rⁿ − 1)/(r − 1), for r ≠ 1.
A salary starts at £30 000 and rises 4% a year. Which power of 1.04 gives the salary in year 10?: 1.04⁹: u₁₀ = 30 000 × 1.04⁹ = £42 699.35. Using 1.04¹⁰ puts the answer one year out.
Find S₆ for the geometric sequence with u₁ = 3 and r = 2: S₆ = 3(2⁶ − 1)/(2 − 1) = 3 × 63 = 189.

1.4 Financial applications of geometric sequences and series (compound interest, depreciation)

Compound interest formula, and what each letter means: FV = PV × (1 + r/(100k))^(kn) PV present value, FV future value, r the annual rate (%), k compounding periods per year (1, 2, 4 or 12), n the number of years.
Depreciation by i% per year: formula: V = V₀ × (1 − i/100)ⁿ Example: £18 000 depreciating 15% a year for 3 years: 18 000 × 0.85³ = £11 054.25.
Real value of an amount after n years of inflation at i% per year: Real value = FV ÷ (1 + i/100)ⁿ. Example: €5627.54 after 4 years of 2% inflation is worth 5627.54 ÷ 1.02⁴ = €5198.98 in today's money.
TVM solver settings for compound interest on a lump sum: N = years × k, I% = annual rate, PV = −amount invested (money out is negative), PMT = 0, P/Y = C/Y = k. Solve for FV.
Interest is compounded monthly for 5 years. What is N in the TVM solver?: N = 12 × 5 = 60 periods, not 5.

1.5 Laws of exponents with integer exponents; introduction to logarithms

Laws of exponents (integer powers): aᵐ × aⁿ = aᵐ⁺ⁿ aᵐ ÷ aⁿ = aᵐ⁻ⁿ (aᵐ)ⁿ = aᵐⁿ a⁻ⁿ = 1/aⁿ
What is a logarithm? State the link with exponents: aˣ = b is equivalent to x = log_a b (for b > 0). A logarithm answers ‘what power?’. At SL the base is 10 (log) or e (ln), evaluated with technology.
Solve 10ˣ = 350: x = log 350 = 2.54 (3 s.f.).
Simplify (2x⁻²)³: 2³x⁻⁶ = 8x⁻⁶ = 8/x⁶.

1.6 Approximation: decimal places, significant figures, upper/lower bounds, percentage error, estimation

A mass is 8.6 kg to 1 d.p. State its bounds: 8.55 ≤ m < 8.65. The upper bound uses <, not ≤ and not 8.649…
Percentage error formula: ε = |(v_A − v_E)/v_E| × 100% v_A is the approximate value, v_E the exact value. Divide by the exact value.
Find the percentage error when π is approximated by 3.14: |3.14 − π|/π × 100% = 0.0507% (3 s.f.).
How do you find the maximum percentage error from a rounded measurement?: 1. Write the bounds of the measurement. 2. Calculate the quantity with the measured value and with each bound. 3. Find the percentage error with each bound as the exact value. 4. The larger one is the maximum percentage error.
y = 2600 to the nearest 100. State its bounds: 2550 ≤ y < 2650.

1.7 Amortization and annuities using technology

Amortization and annuity: Amortization: repaying a loan with equal regular payments. Annuity: equal regular payments into or out of an account. In exams, payments are made at the end of each period.
TVM solver settings to find a loan's regular payment: N = total number of payments, I% = annual rate, PV = +loan (you receive it), FV = 0, P/Y = C/Y = payments per year. Solve PMT; it comes out negative.
How do you find the total interest paid on a loan?: Total interest = (number of payments × payment) − amount borrowed. Example: 48 payments of £352.28 on a £15 000 loan: 16 909.44 − 15 000 = £1909.44.
How do you find the balance outstanding on a loan after m payments?: In the TVM solver set N = m and PMT = −payment, keep PV = +loan, I% and P/Y = C/Y, and solve for FV.
TVM solver trap: the signs of PV and PMT: Money you receive is positive and money you pay out is negative, so PV and PMT must have opposite signs. Same signs give a wrong answer.

1.8 Use technology to solve systems of linear equations and polynomial equations

Zeros and roots: The x-values where f(x) = 0 are the zeros of the function f and the roots of the equation f(x) = 0.
How do you solve a system of linear equations or a polynomial equation in AI?: 1. Write the equations in full (method mark). 2. Use the GDC simultaneous-equation or polynomial solver. 3. Reject roots outside the context's domain (for example a negative length).
Solve 2x³ − 5x² − 4x + 3 = 0, where x is a length: GDC polynomial solver: x = −1, 0.5 or 3. As a length cannot be negative, reject x = −1, leaving x = 0.5 or x = 3.
What do exam systems of linear equations look like at SL?: Up to 3 equations in 3 unknowns, always with a unique solution, solved with technology. Write the system down before quoting the answer.

1.9 Laws of logarithms (AHL only)

The three laws of logarithmsHL: Product: log_a xy = log_a x + log_a y Quotient: log_a (x/y) = log_a x − log_a y Power: log_a xᵐ = m log_a x (a, x, y > 0)
Evaluate 2 log 5 + log 8 − log 2HL: log (5² × 8 ÷ 2) = log 100 = 2.
How do you solve an equation with logarithms on both sides?HL: 1. Use the laws to collect each side into a single log. 2. Remove the log: log A = log B gives A = B; log A = k gives A = 10ᵏ. 3. Solve. 4. Substitute back and reject any root that makes a log argument zero or negative.
Log trap: can log (x + 3) be split into log x + log 3?HL: No. Only a product splits: log (xy) = log x + log y. There is no rule for log (x + 3), and log (xy) is not log x × log y.
Simplify ln (e³/√e)HL: ln e^(3 − 1/2) = 5/2.

1.10 Simplifying expressions involving rational exponents (AHL only)

Rational exponentHL: a^(m/n) = (ⁿ√a)ᵐ, and a^(−p) = 1/aᵖ.
Evaluate 243^(2/5) and 64^(−2/3)HL: 243^(2/5): the fifth root of 243 is 3, then 3² = 9. 64^(−2/3) = 1/(cube root of 64)² = 1/4² = 1/16.
Simplify (4x^(1/2))³ ÷ 2xHL: 64x^(3/2) ÷ 2x = 32x^(1/2).
Why take the root first when evaluating 81^(3/4) by hand?HL: The numbers stay small: the fourth root of 81 is 3, then 3³ = 27. Cubing 81 first gives a large number and invites slips.

1.11 The sum of infinite geometric sequences (AHL only)

Sum to infinity of a geometric series, and when it existsHL: S∞ = u₁/(1 − r). It exists only when |r| < 1. State this condition when asked to justify that the sum exists.
Find the sum to infinity of 12 + 4 + 4/3 + …HL: r = 4/12 = 1/3, and |r| < 1, so S∞ = 12/(1 − 1/3) = 12 ÷ (2/3) = 18.
A ball is dropped from height h and each bounce reaches a fraction r of the previous height. Total distance travelled?HL: h + 2 × hr/(1 − r). The first drop counts once; every rebound counts twice (up and down).
Write 0.272727… as a fraction using a geometric seriesHL: 0.27 + 0.0027 + … has u₁ = 0.27 and r = 0.01, so S∞ = 0.27/(1 − 0.01) = 27/99 = 3/11.
For which x does 1 + 2x + 4x² + … have a sum to infinity?HL: r = 2x, so you need |2x| < 1, which gives −1/2 < x < 1/2.

1.12 Complex numbers: Cartesian form, the complex plane, quadratic equations with complex roots (AHL only)

Cartesian form of a complex number, Re and ImHL: z = a + bi, where i² = −1. Re(z) = a, Im(z) = b. Note Im(4 − 7i) = −7, not −7i.
ConjugateHL: z* = a − bi (reflection in the real axis).
How do you divide complex numbers by hand?HL: Multiply top and bottom by the conjugate of the denominator, expand with i² = −1, then split into a + bi. Example: (3 + i)/(1 − i) = (3 + i)(1 + i)/2 = (2 + 4i)/2 = 1 + 2i.
Solve x² + 2x + 10 = 0HL: x = (−2 ± √(4 − 40))/2 = (−2 ± √(−36))/2 = −1 ± 3i.
What does a negative discriminant tell you about a quadratic?HL: b² − 4ac < 0: the parabola does not meet the x-axis and the roots are a complex conjugate pair, a ± bi.

1.13 Complex numbers: modulus-argument (polar) and exponential (Euler) forms (AHL only)

Modulus of z = a + biHL: |z| = √(a² + b²). It is a length, never negative.
Argument of zHL: arg z = θ, the angle from the positive real axis. The Marlbridge notes use radians, −π < θ ≤ π. Check the quadrant from a sketch; do not read arctan(b/a) straight off the GDC.
Polar and exponential (Euler) formsHL: Polar: z = r(cos θ + i sin θ) = r cis θ. Exponential: z = re^(iθ). Same r and θ, different notation.
Multiplying and dividing in exponential formHL: r₁e^(iθ₁) × r₂e^(iθ₂) = r₁r₂e^(i(θ₁ + θ₂)) r₁e^(iθ₁) ÷ r₂e^(iθ₂) = (r₁/r₂)e^(i(θ₁ − θ₂)) Multiply moduli and add arguments; divide moduli and subtract arguments.
Write 2e^(iπ/6) in Cartesian formHL: 2(cos π/6 + i sin π/6) = 2(√3/2 + i/2) = √3 + i.
How do you add two sinusoids of the same frequency with complex numbers?HL: 1. Write A cos(ωt + α) as Ae^(iα). 2. Add the complex numbers. 3. The modulus is the new amplitude and the argument the new phase. 4. Answer as R cos(ωt + β). Do not add the amplitudes directly.

1.14 Matrices: definition, algebra, multiplication, determinants, inverses, solving linear systems (AHL only)

Order of a matrix, and the order of a productHL: m × n means m rows and n columns (rows first). (m × n)(n × p) gives m × p: the inner numbers must match.
Is matrix multiplication commutative?HL: No. AB ≠ BA in general, and a product may exist one way only. It is associative, (AB)C = A(BC), and distributive, A(B + C) = AB + AC.
Determinant and inverse of the 2 × 2 matrix with rows (a, b) and (c, d)HL: det A = ad − bc. A⁻¹ = (1/(ad − bc)) × rows (d, −b) and (−c, a): swap a and d, change the signs of b and c, divide by the determinant. Only if ad − bc ≠ 0.
Singular matrixHL: A square matrix with determinant 0. It has no inverse.
How do you solve a system of linear equations with matrices?HL: Write it as Ax = b, check det A ≠ 0, then x = A⁻¹b (A⁻¹ on the left, never bA⁻¹). Use technology for 3 × 3 systems. Example: 5x + 3y = 19, 3x + 2y = 12 gives x = 2, y = 3.
Identity matrix IHL: A square matrix with 1s on the leading diagonal and 0s elsewhere. AI = IA = A.

1.15 Eigenvalues, eigenvectors, characteristic polynomial and diagonalization of 2×2 matrices (AHL only)

Eigenvector and eigenvalue of a matrix MHL: An eigenvector is a non-zero vector v with Mv = λv. The scalar λ is its eigenvalue. Any non-zero multiple of v is also an eigenvector.
Characteristic polynomial of a 2 × 2 matrix MHL: det(M − λI). For M with rows (a, b), (c, d) it is (a − λ)(d − λ) − bc. Its roots are the eigenvalues.
Find the eigenvalues and eigenvectors of M with rows (2, 1) and (1, 2)HL: (2 − λ)² − 1 = λ² − 4λ + 3 = (λ − 1)(λ − 3), so λ = 1 or 3. λ = 3: −x + y = 0, v = (1, 1). λ = 1: x + y = 0, v = (1, −1).
Diagonalization: how do you find a formula for Mⁿ?HL: M = PDP⁻¹, so Mⁿ = PDⁿP⁻¹. P has the eigenvectors as columns; D has the matching eigenvalues on its diagonal in the same order. Dⁿ: raise each diagonal entry to n. Check n = 1 gives M.
In xₙ = Tⁿx₀, the eigenvalues of T are 1 and λ₂ with |λ₂| < 1. What happens in the long term?HL: Write x₀ = αv₁ + βv₂. Then xₙ = α(1)ⁿv₁ + β(λ₂)ⁿv₂. The λ₂ⁿ term shrinks to 0, so xₙ approaches αv₁.