Flashcards · IB Diploma Programme Mathematics: Analysis and Approaches · Topic 1

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83 cards on Topic 1, subtopics 1.1 to 1.16. Pick a subtopic or study them all.

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

A number in the form a × 10ᵏ: what are the conditions on a and k?: 1 ≤ a < 10 and k is an integer. Calculator notation such as 2.4E−6 is not acceptable: write 2.4 × 10⁻⁶.
How do you multiply or divide two numbers in the form a × 10ᵏ?: Multiply (or divide) the a values and add (or subtract) the powers of 10. Then adjust so that 1 ≤ a < 10. Example: (6 × 10⁴)(4 × 10⁻⁹) = 24 × 10⁻⁵ = 2.4 × 10⁻⁴.
How do you add 4.3 × 10⁵ + 2.1 × 10⁴ without a calculator?: First write both with the same power of 10: 4.3 × 10⁵ + 0.21 × 10⁵ = 4.51 × 10⁵.
Write 0.000 407 in the form a × 10ᵏ: 4.07 × 10⁻⁴ (40.7 × 10⁻⁵ is not in the required form, because 40.7 is not less than 10.)

1.2 Arithmetic sequences and series

nth term of an arithmetic sequence: uₙ = u₁ + (n − 1)d u₁ is the first term and d is the common difference (d = u₂ − u₁).
Sum of the first n terms of an arithmetic sequence (two forms): Sₙ = (n/2)(2u₁ + (n − 1)d) Sₙ = (n/2)(u₁ + uₙ)
How do you evaluate a sigma sum such as Σ (2r + 1) from r = 5 to r = 20?: 1. Count the terms: top − bottom + 1 = 16 (not 15). 2. First term (r = 5) is 11; last term (r = 20) is 41. 3. It is arithmetic, so S = (16/2)(11 + 41) = 416.
An arithmetic sequence has u₃ = 11 and u₈ = 31. Find d and u₁.: Write both with the nth term formula and subtract: 5d = 20, so d = 4. Then u₁ = 11 − 2 × 4 = 3.
Why does simple interest give an arithmetic sequence?: The same amount is added each year. £2000 at 3% simple interest gains £60 a year: £2060, £2120, … After 10 years: 2000 + 10 × 60 = £2600.

1.3 Geometric sequences and series

nth term of a geometric sequence: uₙ = u₁rⁿ⁻¹ u₁ is the first term and r is the common ratio (r = u₂/u₁, not u₁/u₂).
Sum of the first n terms of a geometric sequence: Sₙ = u₁(rⁿ − 1)/(r − 1) = u₁(1 − rⁿ)/(1 − r), for r ≠ 1.
Find u₈ for the geometric sequence 3, −6, 12, …: r = −6/3 = −2, so u₈ = 3 × (−2)⁷ = −384.
How do you find the least n for which Sₙ exceeds a given value?: 1. Write the inequality using the Sₙ formula. 2. Solve it with logs (or a GDC table). 3. Round UP to the next integer and check Sₙ either side. Example: 5(3ⁿ − 1)/2 > 10 000 gives n > 7.55, so n = 8.
How do you test whether a sequence is arithmetic or geometric?: Arithmetic: constant difference, u₂ − u₁ = u₃ − u₂. Geometric: constant ratio, u₂/u₁ = u₃/u₂.

1.4 Financial applications of geometric sequences and series

Compound interest formula: FV = PV × (1 + r/(100k))^(kn) PV = amount invested, r = annual rate in %, k = compounding periods per year (1, 2, 4 or 12), n = number of years.
£2500 is invested at 4% a year, compounded half-yearly, for 5 years. Find its value.: 2% per period over 10 periods: 2500 × 1.02¹⁰ = £3047.49
Annual depreciation formula, and the value of an £18 000 car after 3 years at 20% a year: Value = initial value × (1 − r/100)ⁿ 18 000 × 0.8³ = £9216 (The multiplier is 0.8, not −0.2 or 0.2.)
Real value of a future amount after n years of inflation at i% a year: Real value = FV ÷ (1 + i/100)ⁿ It expresses the future amount in today’s money.
Compound interest on a GDC finance (TVM) solver: what do you enter?: N = number of periods (years × k), not years. I% = annual rate; P/Y = C/Y = k. PV negative (money paid in), PMT = 0. Write your inputs on the page.

1.5 Laws of exponents and introduction to logarithms

Laws of exponents: aᵐ × aⁿ, aᵐ ÷ aⁿ, (aᵐ)ⁿ: aᵐ × aⁿ = aᵐ⁺ⁿ aᵐ ÷ aⁿ = aᵐ⁻ⁿ (aᵐ)ⁿ = aᵐⁿ
a⁰ and a⁻ⁿ (a ≠ 0): a⁰ = 1 and a⁻ⁿ = 1/aⁿ. So 2⁻³ = 1/8, which is not the same as −2³ = −8.
Definition of a logarithm: aˣ = b is equivalent to logₐ b = x, for a > 0, a ≠ 1, b > 0. A logarithm answers the question "what power?".
What do log x and ln x mean?: log x means log₁₀ x (common logarithm). ln x means logₑ x (natural logarithm), where e ≈ 2.718.
Solve 10ˣ = 50 and eˣ = 7 (calculator allowed): x = log 50 = 1.70 (3 s.f.) x = ln 7 = 1.95 (3 s.f.)
logₐ a and logₐ 1: logₐ a = 1 and logₐ 1 = 0, because a¹ = a and a⁰ = 1.

1.6 Simple deductive proof

Equation (=) vs identity (≡): An equation is true only for particular values: x² − 9 = 0 gives x = ±3. An identity is true for every value: x² − 9 ≡ (x − 3)(x + 3).
How do you lay out an LHS to RHS proof?: 1. Write "LHS =" and start from the left-hand side only. 2. Transform it one algebraic step per line. 3. Finish with "= RHS". Do not work on both sides at once or start from the result. Check with a numerical value.
Show that (2x + 1)² − (2x − 1)² ≡ 8x: LHS = (4x² + 4x + 1) − (4x² − 4x + 1) = 4x² + 4x + 1 − 4x² + 4x − 1 = 8x = RHS
Numerical proof vs algebraic proof: A numerical proof shows one case, such as 1/3 − 1/5 = 2/15. An algebraic proof shows the general case for all allowed values, such as 1/n − 1/(n + 2) ≡ 2/(n(n + 2)).

1.7 Laws of exponents and laws of logarithms

Rational exponents: a^(1/m) and a^(n/m): a^(1/m) = ᵐ√a and a^(n/m) = (ᵐ√a)ⁿ. Take the root first: 16^(3/4) = (⁴√16)³ = 2³ = 8.
Evaluate 8^(−2/3) without a calculator: 8^(2/3) = (∛8)² = 4, so 8^(−2/3) = 1/4.
Laws of logarithms: product, quotient and power: logₐ xy = logₐ x + logₐ y logₐ (x/y) = logₐ x − logₐ y logₐ xᵐ = m logₐ x (a, x, y > 0)
Change of base formula: logₐ x = (log_b x)/(log_b a) Example: log₃ 20 = ln 20 / ln 3 = 2.73 (3 s.f.).
How do you solve 5^(x+1) = 3^(2x)?: Take ln of both sides: (x + 1) ln 5 = 2x ln 3. Collect the x terms: x(2 ln 3 − ln 5) = ln 5. x = ln 5 / (2 ln 3 − ln 5) = 2.74 (3 s.f.)
Solve 9ˣ − 4(3ˣ) + 3 = 0: Let y = 3ˣ, so 9ˣ = y²: y² − 4y + 3 = 0, (y − 1)(y − 3) = 0. 3ˣ = 1 or 3ˣ = 3, so x = 0 or x = 1. (Reject any y ≤ 0, since 3ˣ > 0.)
Exam trap: can log(x + y) or log x × log y be simplified with the log laws?: No. log(x + y) is not log x + log y, and log x × log y is not log xy. Only log xy splits, into log x + log y.

1.8 Sum of infinite convergent geometric sequences

Sum to infinity of a geometric series, and when it exists: S∞ = u₁/(1 − r), only when |r| < 1 (a convergent series). State the condition when you use it.
Find S∞ for 81 + 27 + 9 + …: r = 1/3, so S∞ = 81/(1 − 1/3) = 81/(2/3) = 121.5
Write 0.888… as a fraction using a geometric series: 0.8 + 0.08 + 0.008 + …: u₁ = 0.8, r = 0.1 S∞ = 0.8/(1 − 0.1) = 8/9
For what values of x does the geometric series with r = x − 2 converge?: |x − 2| < 1, so −1 < x − 2 < 1, giving 1 < x < 3. The inequality is strict.

1.9 The binomial theorem

The binomial theorem for n ∈ ℕ: (a + b)ⁿ = Σ ⁿCᵣ aⁿ⁻ʳ bʳ, summed from r = 0 to n. General term: ⁿCᵣ aⁿ⁻ʳ bʳ.
ⁿCᵣ: formula and link to Pascal’s triangle: ⁿCᵣ = n!/(r!(n − r)!) It is entry r of row n of Pascal’s triangle, counting from r = 0. Example: ⁵C₂ = 120/12 = 10.
How do you find a specific term in a binomial expansion?: 1. Write the general term ⁿCᵣ aⁿ⁻ʳ bʳ with a and b in full, signs included. 2. Simplify the power of x as an expression in r. 3. Set it equal to the power you want and solve for r. 4. Substitute r and evaluate.
Find the term independent of x in (x² − 2/x)⁶: General term: ⁶Cᵣ (x²)⁶⁻ʳ (−2/x)ʳ = ⁶Cᵣ (−2)ʳ x^(12 − 3r). 12 − 3r = 0 gives r = 4: 15 × 16 = 240.
Find the coefficient of x² in (1 + 2x)⁶: ⁶C₂ × 2² = 15 × 4 = 60 (The coefficient includes the power of the number in the bracket, not just ⁶C₂.)

1.10 Counting principles and extension of the binomial theorem (AHL only)

Permutation vs combination, with formulasHL: Permutation: an ordered selection. ⁿPᵣ = n!/(n − r)! Combination: an unordered selection. ⁿCᵣ = n!/(r!(n − r)!) n different objects in a line: n! ways (0! = 1).
Choose a president and a secretary from 5 people, vs choose 2 representatives from 5HL: President and secretary (order matters): ⁵P₂ = 20. Two representatives (order does not matter): ⁵C₂ = 10.
4 different red cards and 3 different blue cards are put in a row with the blue cards together. How many arrangements?HL: Glue the blues into one block: arrange 5 items in 5! = 120 ways, then order the blues in 3! = 6 ways. 120 × 6 = 720
Extended binomial theorem for n ∈ ℚHL: (1 + x)ⁿ = 1 + nx + (n(n − 1)/2!)x² + (n(n − 1)(n − 2)/3!)x³ + … Infinite unless n ∈ ℕ; valid for |x| < 1.
Expand √(9 − x) as far as the x² term and state when it is validHL: √(9 − x) = 3(1 − x/9)^(1/2) = 3(1 − x/18 − x²/648 + …) = 3 − x/6 − x²/216 + … Valid for |x/9| < 1, so |x| < 9.
How do you count "at least one" selections without double-counting?HL: Either add the cases with exactly 1, exactly 2, … or use total − the cases you do not want (for example, total − none).

1.11 Partial fractions (AHL only)

Partial fractions: the form for two distinct linear factorsHL: (px + q)/((ax + b)(cx + d)) ≡ A/(ax + b) + B/(cx + d) The numerator must be of lower degree than the denominator.
How do you find A and B in partial fractions?HL: 1. Factorise the denominator. 2. Multiply both sides by it. 3. Substitute each root of the denominator, or equate coefficients. 4. Check with one value of x. It is an identity, so any x works.
Express 5/((x − 2)(x + 3)) in partial fractionsHL: 5 ≡ A(x + 3) + B(x − 2). x = 2 gives A = 1; x = −3 gives B = −1. 1/(x − 2) − 1/(x + 3)
Express 3x/((x + 1)(x − 2)) in partial fractionsHL: 3x ≡ A(x − 2) + B(x + 1). x = −1 gives A = 1; x = 2 gives B = 2. 1/(x + 1) + 2/(x − 2)

1.12 Complex numbers: Cartesian form and the complex plane (AHL only)

iHL: The number with i² = −1. Powers cycle: i, −1, −i, 1.
ConjugateHL: z* = a − bi. Reflection of z in the real axis.
ModulusHL: |z| = √(a² + b²). Distance from the origin.
EqualityHL: a + bi = c + di means a = c and b = d.
How do you divide complex numbers in Cartesian form?HL: Multiply top and bottom by the conjugate of the denominator. The denominator becomes c² + d², a real number. Then split into real and imaginary parts.
Re z and Im z for z = 2 − 7iHL: Re z = 2 and Im z = −7 (not −7i). Both are real numbers.
Solve z² − 6z + 13 = 0HL: z = (6 ± √(36 − 52))/2 = (6 ± √(−16))/2 = 3 ± 2i

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

Polar (modulus-argument) form and Euler form of zHL: z = r(cos θ + i sin θ) = r cis θ = re^(iθ), where r = |z| and θ = arg z. Back to Cartesian: a = r cos θ, b = r sin θ.
ArgumentHL: arg z, the angle from the positive real axis. These notes use −π < arg z ≤ π unless a question states another range.
Multiplying and dividing in polar formHL: r₁ cis θ₁ × r₂ cis θ₂ = r₁r₂ cis(θ₁ + θ₂) r₁ cis θ₁ ÷ r₂ cis θ₂ = (r₁/r₂) cis(θ₁ − θ₂)
How do you find arg z without a quadrant error?HL: 1. Sketch z on the Argand diagram. 2. Reference angle α from tan α = |b|/|a|. 3. First quadrant θ = α; second θ = π − α; third θ = −(π − α); fourth θ = −α.
Write 1 − i in Euler formHL: |1 − i| = √2 and arg(1 − i) = −π/4 (fourth quadrant), so 1 − i = √2 e^(−iπ/4).
Geometric effect of multiplying a complex number by r cis θ, and by iHL: Multiplying by r cis θ scales by r and rotates anticlockwise by θ about the origin. Multiplying by i is a rotation of π/2 anticlockwise; the modulus is unchanged.

1.14 Complex conjugate roots and De Moivre’s theorem (AHL only)

Conjugate root theoremHL: If a polynomial has real coefficients and a + bi is a root, then a − bi is also a root. The pair gives the real quadratic factor z² − 2az + a² + b².
De Moivre’s theoremHL: (r cis θ)ⁿ = rⁿ cis nθ, that is (r(cos θ + i sin θ))ⁿ = rⁿ(cos nθ + i sin nθ).
Find (1 + √3 i)⁶HL: 1 + √3 i = 2 cis(π/3), so (1 + √3 i)⁶ = 2⁶ cis(6π/3) = 64 cis 2π = 64.
How do you solve zⁿ = w?HL: 1. Write w = R cis(φ + 2kπ). 2. z = R^(1/n) cis((φ + 2kπ)/n), for k = 0, 1, …, n − 1. The n roots form a regular n-gon centred at the origin, spaced 2π/n apart.
The three cube roots of 1HL: cis(2kπ/3) for k = 0, 1, 2: 1, −1/2 + (√3/2)i, −1/2 − (√3/2)i
A cubic with real coefficients and leading coefficient 1 has roots 1 and 2i. Find it.HL: −2i is also a root, giving the factor z² + 4. (z − 1)(z² + 4) = z³ − z² + 4z − 4

1.15 Proof by mathematical induction, contradiction and counterexample (AHL only)

Layout of a proof by mathematical inductionHL: Basis: show P(1) is true. Assumption: assume P(k) is true for some k ∈ ℤ⁺. Step: show P(k + 1) is true, using P(k). Conclusion: P(1) is true and P(k) ⇒ P(k + 1), so P(n) is true for all n ∈ ℤ⁺.
Layout of a proof by contradiction (for example, √7 is irrational)HL: 1. "Assume, for a contradiction, that √7 = p/q, where p, q ∈ ℤ, q ≠ 0, with no common factor." 2. Rearrange to a statement about integers. 3. Show p and q share a factor (or another impossibility). 4. State the contradiction, so the original statement is true.
Disprove: "if p is prime then 2p + 1 is prime"HL: Counterexample: p = 7 is prime, but 2(7) + 1 = 15 = 3 × 5, which is not prime. Show the working that breaks the statement.
Induction for divisibility: the key rewrite for 9ⁿ − 1HL: Write 9ᵏ⁺¹ − 1 = 9(9ᵏ − 1) + 8, then use the assumption that 9ᵏ − 1 is divisible by 8 (so both terms are multiples of 8).
Does checking n = 1, 2 and 3 prove a statement for all n?HL: No. Examples never prove a "for all" statement; the inductive step does the work. One counterexample is enough to disprove one.

1.16 Solutions of systems of linear equations (AHL only)

How do you solve three linear equations in x, y, z by elimination?HL: 1. Use equation (1) to remove x from (2) and (3). 2. Use the new pair to remove y. 3. Solve the last line for z and back-substitute. On Papers 2 and 3 a GDC solver can do the arithmetic.
After elimination the last line is cz = d. What does it tell you?HL: c ≠ 0: a unique solution. c = 0, d ≠ 0: no solution (inconsistent). c = 0, d = 0: infinitely many solutions; let z = λ and back-substitute.
Find the general solution of x + y + z = 2, y − z = 1HL: Let z = λ. Then y = 1 + λ and x = 2 − y − z = 1 − 2λ. x = 1 − 2λ, y = 1 + λ, z = λ, λ ∈ ℝ
Solve x + y = 3, 2x + 2y = 7HL: Doubling the first gives 2x + 2y = 6, which contradicts 2x + 2y = 7. The system is inconsistent: no solution.