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IB® · HL/SL

IB® Math Analysis & Approaches HL/SL

Step-by-step mastery of IB® Math AA: the five official syllabus topics with deep SL coverage and full HL extensions.

Start Unit 1 free. Topic 1: Number and Algebra is open to everyone, no account needed. Other topics are locked.

Topic 1: Number and Algebra

THE BIG PICTURE. Topic 1 builds the algebraic foundation for everything that follows. SL focuses on sequences and series, exponents and logarithms, and the binomial theorem. SL also covers scientific notation and an introduction to proof. HL extends to counting principles, complex numbers, formal proof (induction/contradiction/counter-example), systems of linear equations, and partial fractions. Strong fluency here pays dividends across all five topics: every other topic uses these tools.

SEQUENCES AND SERIES

Arithmetic: each term differs by a constant common difference dd:

  • un=u1+(n−1)du_n = u_1 + (n - 1) d
  • Sn=n2(2u1+(n−1)d)=n2(u1+un)S_n = \dfrac{n}{2}(2 u_1 + (n - 1) d) = \dfrac{n}{2}(u_1 + u_n)

Example. Find the 25th term and the sum of the first 25 terms of the arithmetic sequence 7,11,15,19,…7, 11, 15, 19, \ldots u1=7,d=4,u25=7+24⋅4=103.u_1 = 7,\quad d = 4,\quad u_{25} = 7 + 24 \cdot 4 = 103. S25=252(7+103)=25⋅1102=1,375.S_{25} = \dfrac{25}{2}(7 + 103) = \dfrac{25 \cdot 110}{2} = 1{,}375.

Geometric: each term multiplied by a constant common ratio rr:

  • un=u1⋅rn−1u_n = u_1 \cdot r^{n-1}
  • Sn=u1(1−rn)1−rS_n = \dfrac{u_1 (1 - r^n)}{1 - r} for r≠1r \neq 1
  • Sum to infinity (only when ∣r∣<1|r| < 1): S∞=u11−rS_\infty = \dfrac{u_1}{1 - r}

Example. Geometric sequence 4,12,36,…4, 12, 36, \ldots Find u8u_8 and S8S_8. u1=4,r=3,u8=4⋅37=4⋅2187=8,748.u_1 = 4,\quad r = 3,\quad u_8 = 4 \cdot 3^7 = 4 \cdot 2187 = 8{,}748. S8=4(1−38)1−3=4⋅(−6560)−2=13,120.S_8 = \dfrac{4(1 - 3^8)}{1 - 3} = \dfrac{4 \cdot (-6560)}{-2} = 13{,}120.

Example (sum to infinity). u1=9u_1 = 9 and r=1/3r = 1/3. Since ∣r∣<1|r| < 1, the sum to infinity exists. S∞=91−1/3=92/3=13.5.S_\infty = \dfrac{9}{1 - 1/3} = \dfrac{9}{2/3} = 13.5.

Sigma notation compresses long sums: ∑i=1nui=u1+u2+⋯+un\sum_{i=1}^n u_i = u_1 + u_2 + \cdots + u_n.

Compound interest is a geometric application: FV=PV(1+r100k)kn\text{FV} = \text{PV} \left(1 + \dfrac{r}{100k}\right)^{kn}, with kk compounding periods per year for nn years at nominal rate r%r\%. Annual depreciation uses the same form with negative rr.

Example. $3,000 invested at 6%6\% annual interest, compounded monthly for 5 years grows to the value below. FV=3000(1+61200)60=3000(1.005)60≈$4,046.55.\text{FV} = 3000\left(1 + \dfrac{6}{1200}\right)^{60} = 3000(1.005)^{60} \approx \$4{,}046.55.

Convergent geometric series When the absolute value of r is less than 1, the partial sums approach S infinity = u1/(1 − r). A positive ratio approaches from below; a negative ratio overshoots and undershoots on the way.

WORKED EXAMPLE: FINDING k IN A GEOMETRIC SEQUENCE (PAPER 1)

The first three terms of a geometric sequence are k+4k + 4, kk and k−3k - 3, where k≠0k \neq 0. Find kk, the common ratio, and the sum to infinity.

  • Use the constant ratio. kk+4=k−3k\dfrac{k}{k + 4} = \dfrac{k - 3}{k}, so k2=(k+4)(k−3)k^2 = (k + 4)(k - 3). (M1)
  • Solve. k2=k2+k−12k^2 = k^2 + k - 12, so k=12k = 12. (A1) The terms are 16,12,916, 12, 9.
  • Ratio. r=1216=0.75r = \dfrac{12}{16} = 0.75. (A1)
  • Check the condition, then sum. ∣r∣=0.75<1|r| = 0.75 < 1, so the sum to infinity exists: S∞=161−0.75=64S_\infty = \dfrac{16}{1 - 0.75} = 64. (A1)

The left panel of the figure plots SnS_n for this sequence: S10=16(1−0.7510)0.25≈60.4S_{10} = \dfrac{16(1 - 0.75^{10})}{0.25} \approx 60.4, already close to 64. Always state the ∣r∣<1|r| < 1 check in words: examiners look for it.

WORKED EXAMPLE: COMPOUND INTEREST WITH A GDC (PAPER 2)

An investor deposits $5,000 in an account paying a nominal annual rate of 4.5%4.5\%, compounded monthly. (a) Find the value after 10 years. (b) Find the number of complete years until the value first exceeds $10,000.

(a) Here k=12k = 12 and n=10n = 10: FV=5000(1+4.51200)120=5000(1.00375)120≈7834.96.\text{FV} = 5000\left(1 + \dfrac{4.5}{1200}\right)^{120} = 5000(1.00375)^{120} \approx 7834.96. The value is about $7,834.96 (money answers go to 2 decimal places).

(b) Solve 5000(1.00375)12n>100005000(1.00375)^{12n} > 10000, that is (1.00375)12n>2(1.00375)^{12n} > 2: 12n>ln⁡2ln⁡1.00375≈185.2  ⇒  n>15.43.12n > \dfrac{\ln 2}{\ln 1.00375} \approx 185.2 \;\Rightarrow\; n > 15.43. Check with the GDC table: after 15 years the value is about $9,807.78; after 16 years it is about $10,258.34. Answer: 16 years. The question says complete years, so round up, not to the nearest whole number. The TVM solver on a GDC gives the same result with N = number of months.

EXPONENTS AND LOGARITHMS

Exponent laws: am⋅an=am+na^m \cdot a^n = a^{m+n}, aman=am−n\dfrac{a^m}{a^n} = a^{m-n}, (am)n=amn(a^m)^n = a^{mn}, a0=1a^0 = 1, a−n=1ana^{-n} = \dfrac{1}{a^n}, a1/n=ana^{1/n} = \sqrt[n]{a}.

Definition of logarithm: ax=b  ⇔  x=log⁡aba^x = b \;\Leftrightarrow\; x = \log_a b (with a>0a > 0, b>0b > 0, a≠1a \neq 1).

Logarithm laws mirror exponent laws

  • log⁡a(xy)=log⁡ax+log⁡ay\log_a (xy) = \log_a x + \log_a y
  • log⁡axy=log⁡ax−log⁡ay\log_a \dfrac{x}{y} = \log_a x - \log_a y
  • log⁡axm=mlog⁡ax\log_a x^m = m \log_a x
  • log⁡aa=1\log_a a = 1, log⁡a1=0\log_a 1 = 0
  • Change of base: log⁡ax=log⁡bxlog⁡ba\log_a x = \dfrac{\log_b x}{\log_b a}
  • Natural log: ln⁡x=log⁡ex\ln x = \log_e x, with e≈2.71828e \approx 2.71828

Solving exponential equations: take logs of both sides; isolate the variable. Always check that the argument is positive.

Example. Solve log⁡5(x+3)+log⁡5(x−1)=1\log_5(x + 3) + \log_5(x - 1) = 1. log⁡5[(x+3)(x−1)]=1  ⇒  (x+3)(x−1)=5.\log_5[(x+3)(x-1)] = 1 \;\Rightarrow\; (x+3)(x-1) = 5. x2+2x−3=5  ⇒  (x+4)(x−2)=0.x^2 + 2x - 3 = 5 \;\Rightarrow\; (x+4)(x-2) = 0. Domain requires x>1x > 1. Reject x=−4x = -4. Answer: x=2x = 2.

Example. Solve 7x=307^x = 30. x=log⁡730=ln⁡30ln⁡7≈1.748.x = \log_7 30 = \dfrac{\ln 30}{\ln 7} \approx 1.748.

BINOMIAL THEOREM

(a+b)n=∑r=0n(nr)an−rbr(a + b)^n = \sum_{r=0}^n \binom{n}{r} a^{n-r} b^r, where (nr)=n!r!(n−r)!\binom{n}{r} = \dfrac{n!}{r!(n-r)!}.

Pascal's triangle Row n gives the binomial coefficients nCr. Row 7 supplies every coefficient of (a + b) to the power 7; the highlighted 35 is the one used for the x to the 4 term of (3 − 2x) to the power 7.

  • Pascal's triangle generates binomial coefficients for small nn.
  • The (r+1)(r+1)th term is (nr)an−rbr\binom{n}{r} a^{n-r} b^r: useful for "find the term containing xkx^k" questions.
  • HL extension: the theorem extends to negative and rational nn for ∣b/a∣<1|b/a| < 1: (1+x)n=1+nx+n(n−1)2!x2+⋯(1 + x)^n = 1 + nx + \dfrac{n(n-1)}{2!} x^2 + \cdots

Example. Find the coefficient of x4x^4 in (3−2x)7(3 - 2x)^7. Term=(74)(3)3(−2x)4=35⋅27⋅16⋅x4=15,120 x4.\text{Term} = \binom{7}{4}(3)^3(-2x)^4 = 35 \cdot 27 \cdot 16 \cdot x^4 = 15{,}120\, x^4. Coefficient is 15,12015{,}120.

Example. Find the constant term in (2x+1x2)9\left(2x + \dfrac{1}{x^2}\right)^9. General term: (9r)(2x)9−r(x−2)r=(9r)29−rx9−3r\binom{9}{r}(2x)^{9-r}(x^{-2})^r = \binom{9}{r} 2^{9-r} x^{9 - 3r}. Constant requires 9−3r=0⇒r=39 - 3r = 0 \Rightarrow r = 3. (93)26=84⋅64=5,376.\binom{9}{3} 2^6 = 84 \cdot 64 = 5{,}376.

WORKED EXAMPLE: AN EXPONENTIAL EQUATION IN DISGUISE (PAPER 1)

Solve 22x+1−5⋅2x+2=02^{2x + 1} - 5 \cdot 2^x + 2 = 0.

  • Spot the quadratic. 22x+1=2⋅(2x)22^{2x + 1} = 2 \cdot (2^x)^2. Let y=2xy = 2^x (so y>0y > 0): the equation becomes 2y2−5y+2=02y^2 - 5y + 2 = 0. (M1)
  • Factorise. (2y−1)(y−2)=0(2y - 1)(y - 2) = 0, so y=12y = \tfrac{1}{2} or y=2y = 2. (A1)
  • Undo the substitution. 2x=12⇒x=−12^x = \tfrac{1}{2} \Rightarrow x = -1; 2x=2⇒x=12^x = 2 \Rightarrow x = 1. (A1)(A1)

Both values are valid because both yy-values are positive. If one root had been negative, you would reject it, since 2x>02^x > 0 for every real xx.

PRACTICE: WHICH TOOL FITS THE QUESTION?

The question says...Reach forFirst step
"Find the 20th term" of 5,8,11,…5, 8, 11, \ldotsArithmetic unu_nd=3d = 3, u20=5+19(3)=62u_{20} = 5 + 19(3) = 62
"Sum of all terms" of 24,12,6,…24, 12, 6, \ldotsS∞S_\inftycheck the size of r=0.5r = 0.5 first
"Solve 3x=203^x = 20"Logarithmsx=ln⁡20/ln⁡3≈2.73x = \ln 20 / \ln 3 \approx 2.73
"Coefficient of x3x^3" in (2+x)6(2 + x)^6Binomial term(63)23=160\binom{6}{3} 2^3 = 160
"Value after nn years, compounded quarterly"FV\text{FV} formula or TVMk=4k = 4, total periods 4n4n
"Smallest nn so that Sn>1000S_n > 1000"GDC table or inequalityround up to the next integer

COUNTING PRINCIPLES (HL)

  • Permutations (order matters): nPr=n!(n−r)!^n P_r = \dfrac{n!}{(n-r)!}
  • Combinations (order doesn't matter): nCr=(nr)=n!r!(n−r)!^n C_r = \binom{n}{r} = \dfrac{n!}{r!(n-r)!}
  • Inclusion–exclusion for overlapping sets.

Example. A class of 12 students forms a 4-person debate team with a designated captain. How many ways? Choose 4 from 12: (124)=495\binom{12}{4} = 495. Then pick captain from those 4: 44 ways. Total=495⋅4=1,980.\text{Total} = 495 \cdot 4 = 1{,}980. Equivalently, 12P1⋅(113)=12⋅165=1,980^{12}P_1 \cdot \binom{11}{3} = 12 \cdot 165 = 1{,}980.

COMPLEX NUMBERS (HL)

Cartesian form: z=a+biz = a + bi with i=−1i = \sqrt{-1}, so i2=−1i^2 = -1. Re(z)=a\text{Re}(z) = a, Im(z)=b\text{Im}(z) = b.

Modulus: ∣z∣=a2+b2|z| = \sqrt{a^2 + b^2}. Argument: arg⁡(z)=θ\arg(z) = \theta where tan⁡θ=b/a\tan \theta = b/a (always check the quadrant).

Polar (modulus-argument) form: z=r(cos⁡θ+isin⁡θ)=r cis θz = r(\cos\theta + i\sin\theta) = r\,\text{cis}\,\theta.

Exponential (Euler) form: z=reiθz = r e^{i\theta}, leveraging Euler's identity eiθ=cos⁡θ+isin⁡θe^{i\theta} = \cos\theta + i\sin\theta.

Example. Convert z=−2+2iz = -2 + 2i to polar and exponential form. ∣z∣=4+4=22.|z| = \sqrt{4 + 4} = 2\sqrt{2}. zz is in quadrant II; reference angle tan⁡−1(2/2)=π/4\tan^{-1}(2/2) = \pi/4, so arg⁡z=π−π/4=3π/4\arg z = \pi - \pi/4 = 3\pi/4. z=22(cos⁡(3π/4)+isin⁡(3π/4))=22 ei⋅3π/4.z = 2\sqrt{2}\bigl(\cos(3\pi/4) + i\sin(3\pi/4)\bigr) = 2\sqrt{2}\, e^{i \cdot 3\pi/4}.

De Moivre's Theorem: zn=rn(cos⁡nθ+isin⁡nθ)=rneinθz^n = r^n (\cos n\theta + i \sin n\theta) = r^n e^{i n\theta}. Used for powers and nnth roots of complex numbers; the nn roots of zz are equally spaced by 2πn\dfrac{2\pi}{n} on a circle of radius ∣z∣1/n|z|^{1/n}.

Example. Compute (3+i)6(\sqrt{3} + i)^6. First convert: r=2r = 2, θ=π/6\theta = \pi/6. (3+i)6=26ei⋅6⋅π/6=64eiπ=−64.(\sqrt{3} + i)^6 = 2^6 e^{i \cdot 6 \cdot \pi/6} = 64 e^{i\pi} = -64.

Example. Find the four 4th roots of 16i16 i. Write 16i=16 ei(π/2+2πk)16 i = 16\, e^{i(\pi/2 + 2\pi k)} for integer kk. 4th roots: zk=161/4 ei(π/8+πk/2)=2 ei(π/8+πk/2)z_k = 16^{1/4}\, e^{i(\pi/8 + \pi k/2)} = 2\, e^{i(\pi/8 + \pi k/2)} for k=0,1,2,3k = 0, 1, 2, 3. Four points equally spaced on a circle of radius 2, starting at angle π/8\pi/8.

Conjugate z∗=a−biz^* = a - bi. Properties: z+z∗=2az + z^* = 2a (real), z⋅z∗=∣z∣2z \cdot z^* = |z|^2 (real, non-negative). Used to rationalise complex denominators.

Fundamental theorem of algebra: every polynomial of degree nn with real or complex coefficients has exactly nn roots (counting multiplicity) in C\mathbb{C}. Real polynomial roots come in conjugate pairs: if p+qip + qi is a root, so is p−qip - qi.

The Argand diagram (HL) Modulus is distance from the origin and argument is the anticlockwise angle from the positive real axis, so always check the quadrant. The n roots of a complex number sit on a circle, equally spaced by 2 pi over n.

WORKED EXAMPLE: USING A CONJUGATE ROOT (HL, PAPER 1)

Given that 2−i2 - i is a root of z3−5z2+9z−5=0z^3 - 5z^2 + 9z - 5 = 0, find the other two roots.

  • Conjugate pair. The coefficients are real, so 2+i2 + i is also a root. (R1)
  • Build the quadratic factor. (z−(2−i))(z−(2+i))=(z−2)2−i2=z2−4z+5(z - (2 - i))(z - (2 + i)) = (z - 2)^2 - i^2 = z^2 - 4z + 5. (M1)(A1)
  • Divide or compare coefficients. z3−5z2+9z−5=(z2−4z+5)(z−1)z^3 - 5z^2 + 9z - 5 = (z^2 - 4z + 5)(z - 1): the constant terms give 5×(−1)=−55 \times (-1) = -5. (M1)
  • Answer. The roots are 2−i2 - i, 2+i2 + i and 11. (A1)

Check with sum and product of roots. The sum is (2−i)+(2+i)+1=5(2 - i) + (2 + i) + 1 = 5, matching −a2a3=−−51=5-\dfrac{a_2}{a_3} = -\dfrac{-5}{1} = 5. The product is (2−i)(2+i)(1)=5(2 - i)(2 + i)(1) = 5, matching (−1)3a0a3=−−51=5(-1)^3\dfrac{a_0}{a_3} = -\dfrac{-5}{1} = 5.

PROOF: SL FOUNDATION, HL EXTENSION

SL students are introduced to the formal concept of proof: what a proof is, and simple deductive proof (numerical and algebraic; laying out a clear LHS → RHS argument with correct equality/identity notation). HL extends this to the three formal methods below: induction, contradiction, and counter-example.

Methods (direct/deductive at SL; induction, contradiction, counter-example are HL):

  • Direct (deductive) proof: assume the hypothesis, deduce the conclusion through valid algebraic steps.
  • Proof by contradiction: assume the negation of what you want to prove, derive a contradiction. Classic example: 2\sqrt{2} is irrational.
  • Proof by induction: for statements indexed by n∈Z+n \in \mathbb{Z}^+:

1. Base case: verify P(1)P(1) is true. 2. Inductive step: assume P(k)P(k) is true; prove P(k+1)P(k+1) follows. 3. Conclusion: by induction, P(n)P(n) is true for all n∈Z+n \in \mathbb{Z}^+.

  • Disproof by counter-example: one specific case where the statement fails is enough to disprove a universal claim.

Induction is heavily tested on summation formulas, divisibility, inequalities, and De Moivre's theorem (whose induction proof is explicitly on the HL syllabus). Matrices are not part of Math AA.

WORKED EXAMPLE: A SIMPLE DEDUCTIVE PROOF (SL)

Prove that the difference between the squares of two integers that differ by 2 is always a multiple of 4.

Let the integers be n−1n - 1 and n+1n + 1, where n∈Zn \in \mathbb{Z}. LHS=(n+1)2−(n−1)2=(n2+2n+1)−(n2−2n+1)=4n.\text{LHS} = (n + 1)^2 - (n - 1)^2 = (n^2 + 2n + 1) - (n^2 - 2n + 1) = 4n. Since nn is an integer, 4n4n is a multiple of 4. The layout matters: start from one side, use == between each step, and finish with a sentence that states what has been proved.

WORKED EXAMPLE: PROOF BY INDUCTION (HL, PAPER 1)

Prove that ∑r=1nr(r+1)=n(n+1)(n+2)3\displaystyle\sum_{r=1}^{n} r(r + 1) = \dfrac{n(n + 1)(n + 2)}{3} for all n∈Z+n \in \mathbb{Z}^+.

Base case. For n=1n = 1: LHS =1×2=2= 1 \times 2 = 2 and RHS =1×2×33=2= \dfrac{1 \times 2 \times 3}{3} = 2. True for n=1n = 1. (A1)

Assumption. Assume true for n=kn = k: ∑r=1kr(r+1)=k(k+1)(k+2)3\displaystyle\sum_{r=1}^{k} r(r + 1) = \dfrac{k(k + 1)(k + 2)}{3}. (M1)

Inductive step. For n=k+1n = k + 1, add the (k+1)(k + 1)th term to the assumed sum. ∑r=1k+1r(r+1)=k(k+1)(k+2)3+(k+1)(k+2)=(k+1)(k+2)(k+3)3,\sum_{r=1}^{k+1} r(r + 1) = \dfrac{k(k + 1)(k + 2)}{3} + (k + 1)(k + 2) = \dfrac{(k + 1)(k + 2)(k + 3)}{3}, after taking out the common factor (k+1)(k+2)3\dfrac{(k + 1)(k + 2)}{3}. This is the formula with n=k+1n = k + 1. (M1)(A1)

Conclusion. The result is true for n=1n = 1, and if it is true for n=kn = k then it is true for n=k+1n = k + 1. Therefore, by mathematical induction, it is true for all n∈Z+n \in \mathbb{Z}^+. (R1)

The conclusion mark is lost more often than any other: it must mention both the base case and the implication "true for kk implies true for k+1k + 1". A quick numerical check (n=10n = 10 gives 440440 both ways) catches algebra slips before you write the proof.

PARTIAL FRACTIONS (HL)

In Math AA HL, partial fractions are limited to a denominator with a maximum of two distinct linear factors and numerator degree less than denominator degree:

  • 1(x−a)(x−b)=Ax−a+Bx−b\dfrac{1}{(x - a)(x - b)} = \dfrac{A}{x - a} + \dfrac{B}{x - b} (find A,BA, B by substitution or cover-up).

(Repeated linear factors such as 1(x−a)2\dfrac{1}{(x-a)^2}, and irreducible quadratic factors, are beyond required Math AA HL: useful enrichment only.)

Example. Decompose 4x+1(x−2)(x+3)\dfrac{4x + 1}{(x - 2)(x + 3)}. Set 4x+1(x−2)(x+3)=Ax−2+Bx+3\dfrac{4x + 1}{(x - 2)(x + 3)} = \dfrac{A}{x - 2} + \dfrac{B}{x + 3}, so 4x+1=A(x+3)+B(x−2)4x + 1 = A(x + 3) + B(x - 2). Let x=2x = 2: 9=5A⇒A=9/59 = 5A \Rightarrow A = 9/5. Let x=−3x = -3: −11=−5B⇒B=11/5-11 = -5B \Rightarrow B = 11/5. 4x+1(x−2)(x+3)=9/5x−2+11/5x+3.\dfrac{4x + 1}{(x - 2)(x + 3)} = \dfrac{9/5}{x - 2} + \dfrac{11/5}{x + 3}.

Useful for integration and Maclaurin series.

SYSTEMS OF LINEAR EQUATIONS (HL)

Solve systems of up to three linear equations in three unknowns, algebraically (elimination/substitution, row reduction) and with technology. Classify the outcome:

  • Unique solution: the lines/planes meet at a single point.
  • No solution: the system is inconsistent (e.g., parallel planes; equations reduce to 0=c0 = c, c≠0c \neq 0).
  • Infinitely many solutions: dependent equations (a line or plane of solutions); express the solution set with a parameter.

This links directly to intersections of lines and planes in 3D vector geometry (Topic 3).

WORKED EXAMPLE: CLASSIFYING A 3 BY 3 SYSTEM (HL)

Consider the system x+y+z=6x + y + z = 6,   2x−y+z=3\;2x - y + z = 3,   x+2y+kz=c\;x + 2y + kz = c.

(a) For k=1k = 1 and c=8c = 8, solve the system. Row reduction (or a GDC) gives the unique solution x=1x = 1, y=2y = 2, z=3z = 3. Check in the third equation: 1+4+3=81 + 4 + 3 = 8.

(b) Find the value of kk for which there is no unique solution. The coefficient determinant is 4−3k4 - 3k, so there is no unique solution when k=43k = \dfrac{4}{3}. Without determinants: eliminate xx and yy and look for a row that becomes 0z=…0z = \ldots

(c) With k=43k = \dfrac{4}{3}, the first two equations give x=3−2λx = 3 - 2\lambda, y=3−λy = 3 - \lambda, z=3λz = 3\lambda. Substituting into the third: (3−2λ)+2(3−λ)+4λ=9(3 - 2\lambda) + 2(3 - \lambda) + 4\lambda = 9 for every λ\lambda.

  • If c=9c = 9: infinitely many solutions, the line (x,y,z)=(3,3,0)+λ(−2,−1,3)(x, y, z) = (3, 3, 0) + \lambda(-2, -1, 3). Geometrically the three planes meet in a common line.
  • If c≠9c \neq 9: no solution (inconsistent). The third plane is parallel to the line where the first two meet.

SCIENTIFIC NOTATION (SL)

Numbers in the form a×10ka \times 10^k where 1≤a<101 \le a < 10 and k∈Zk \in \mathbb{Z}. Perform operations and present results in proper standard form. Calculator/display notation such as "5.2E30" or "5.2 E 30" is NOT acceptable in a final written answer: rewrite it as 5.2×10305.2 \times 10^{30}.

EXAM CONNECTIONS.

  • Paper 1 (no GDC): algebraic manipulation of logs, geometric sums, induction proofs (HL), De Moivre (HL).
  • Paper 2 (with GDC): finance/compound-interest problems, larger numerical sums, complex roots via GDC.
  • Paper 3 (HL only): extended investigations frequently use induction, counting, or complex-number identities.

Key Terms

Arithmetic Sequence

Sequence with constant common difference dd. un=u1+(n−1)du_n = u_1 + (n-1) d; Sn=n2(u1+un)S_n = \dfrac{n}{2}(u_1 + u_n).

Geometric Sequence

Sequence with constant common ratio rr. un=u1rn−1u_n = u_1 r^{n-1}; Sn=u1(1−rn)1−rS_n = \dfrac{u_1(1 - r^n)}{1 - r}. Infinite sum exists only when ∣r∣<1|r| < 1: S∞=u11−rS_\infty = \dfrac{u_1}{1 - r}.

Sigma Notation

∑i=1nui\sum_{i=1}^n u_i denotes u1+u2+⋯+unu_1 + u_2 + \cdots + u_n. The index ii runs from the lower to upper bound.

Compound Interest

FV=PV(1+r100k)kn\text{FV} = \text{PV}\left(1 + \dfrac{r}{100k}\right)^{kn}, with kk compounding periods per year for nn years at annual rate r%r\%.

Logarithm Laws

log⁡(AB)=log⁡A+log⁡B\log(AB) = \log A + \log B; log⁡(A/B)=log⁡A−log⁡B\log(A/B) = \log A - \log B; log⁡An=nlog⁡A\log A^n = n \log A; log⁡aa=1\log_a a = 1; log⁡a1=0\log_a 1 = 0.

Change of Base

log⁡ax=log⁡bxlog⁡ba\log_a x = \dfrac{\log_b x}{\log_b a}. Lets you convert between logarithm bases (e.g., to common log or natural log on the GDC).

Binomial Theorem

(a+b)n=∑r=0n(nr)an−rbr(a + b)^n = \sum_{r=0}^n \binom{n}{r} a^{n-r} b^r, with (nr)=n!r!(n−r)!\binom{n}{r} = \dfrac{n!}{r!(n-r)!}. The (r+1)(r+1)th term contains brb^r.

Permutation

nPr=n!(n−r)!^n P_r = \dfrac{n!}{(n-r)!}. Number of ordered arrangements of rr items chosen from nn.

Combination

nCr=(nr)=n!r!(n−r)!^n C_r = \binom{n}{r} = \dfrac{n!}{r!(n-r)!}. Number of unordered selections of rr items from nn.

Imaginary Unit (HL)

i=−1i = \sqrt{-1}, so i2=−1i^2 = -1. Powers cycle: i1=ii^1 = i, i2=−1i^2 = -1, i3=−ii^3 = -i, i4=1i^4 = 1.

Modulus / Argument (HL)

For z=a+biz = a + bi: ∣z∣=a2+b2|z| = \sqrt{a^2 + b^2}; arg⁡(z)=θ\arg(z) = \theta with tan⁡θ=b/a\tan\theta = b/a (check quadrant). Principal argument −π<θ≤π-\pi < \theta \leq \pi.

De Moivre's Theorem (HL)

[r(cos⁡θ+isin⁡θ)]n=rn(cos⁡nθ+isin⁡nθ)[r(\cos\theta + i\sin\theta)]^n = r^n(\cos n\theta + i\sin n\theta). Equivalently (reiθ)n=rneinθ(re^{i\theta})^n = r^n e^{in\theta}.

Complex Conjugate (HL)

If z=a+biz = a + bi, then z∗=a−biz^* = a - bi. z⋅z∗=∣z∣2z \cdot z^* = |z|^2. Real polynomials' complex roots come in conjugate pairs.

Deductive Proof (SL)

SL foundation: a logical argument from given assumptions to a conclusion via valid steps (numerical/algebraic), often laid out LHS → RHS with correct equality/identity notation. HL extends to induction, contradiction, and counter-example.

Proof by Induction (HL)

Prove P(n)P(n) for all n∈Z+n \in \mathbb{Z}^+: (1) base case P(1)P(1); (2) inductive step: assume P(k)P(k), deduce P(k+1)P(k+1); (3) conclude by induction.

Proof by Contradiction (HL)

Assume the negation of the claim is true; derive a logical contradiction; conclude the original claim is true.

Partial Fractions (HL)

Decompose a rational function into simpler fractions. AA HL is limited to a denominator with a MAXIMUM of two distinct linear factors and numerator degree < denominator degree: 1(x−a)(x−b)=Ax−a+Bx−b\dfrac{1}{(x-a)(x-b)} = \dfrac{A}{x-a} + \dfrac{B}{x-b}. Repeated/quadratic factors are enrichment only.

Exam Tips

  • Geometric series with ∣r∣≥1|r| \geq 1: there is NO infinite sum. Always check before applying S∞=u11−rS_\infty = \dfrac{u_1}{1-r}.
  • Logarithm domain: arguments must be strictly positive. After solving, reject any solution that produces log⁡\log of zero or a negative number.
  • Binomial term selection: identify aa, bb, nn first. The (r+1)(r+1)th term contains brb^r. For "find the term in xkx^k", solve for rr from the exponent of xx.
  • Complex argument quadrant (HL): arctan⁡\arctan alone won't tell you which quadrant. Sketch the Argand diagram and adjust to get the correct principal argument.
  • Induction proof structure (HL): state base, state assumption, manipulate to reach P(k+1)P(k+1), conclude. Examiners look for the explicit framing: write the three steps clearly.
  • Show all algebraic working in Paper 1 (no GDC): every line earns potential method marks even if a later step contains an error.

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