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HLIBMathematics

Mathematics: Applications and Interpretation

The Higher Level pathway applies advanced mathematical models through statistics, functions, geometry, and calculus in real-world contexts.

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5

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Not specified

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Course Overview

IB Mathematics: Applications and Interpretation HL is organized into 5 units and 77 focused topics with 1584 Key Concepts sections, 931 questions, and 1540 flashcards.

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Begin with Topic 1.1: Operations with numbers and scientific notation.

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Course units

Each unit groups focused topics. Every topic follows the same clear rhythm: review Key Concepts, use the Question Bank, then strengthen recall.

  1. Unit 1
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    Number and Algebra

    AI HL begins with the AI SL foundations of scientific notation, sequences, financial mathematics and technology-supported equation solving, then extends into logarithm laws, infinite geometric series, complex numbers, matrices and eigenvalue methods. This topic supports finance, transformations, Markov models, phase portraits and repeated-state systems later in the course.

    15 topics

    Unit 1 progress0%

    Topics in this unit

    Topic 1.1Not started

    Operations with numbers and scientific notation

    Scientific notation provides a stable way to work with very large and very small values and is frequently used in applied contexts.

    180 min estimated

    Topic 1.1 progress0%
    Learning objectives
    • Write non-zero numbers in normalized form a x 10k10^{k} with 1≤a<101 \le a < 10 and integer k.
    • Perform arithmetic using scientific notation and interpret calculator E-notation correctly.
    • Compare orders of magnitude and use appropriate units.
    • Maintain sufficient precision when scientific notation appears inside a longer calculation.
    Key formulas

    Normalized scientific notation

    N=a×10k,1≤∣a∣<10,k∈ZN=a\times10^k,\quad 1\le |a|<10,\quad k\in\mathbb Z

    Product of powers of ten

    10m×10n=10m+n10^m\times10^n=10^{m+n}

    Quotient of powers of ten

    10m10n=10m−n\frac{10^m}{10^n}=10^{m-n}

    Power of a power

    (10m)r=10mr(10^m)^r=10^{mr}

    Relative-size ratio

    factor=larger quantitysmaller quantity\text{factor}=\frac{\text{larger quantity}}{\text{smaller quantity}}

    Multiplication rule

    (a×10m)(b×10n)=(ab)×10m+n(a\times10^m)(b\times10^n)=(ab)\times10^{m+n}

    Division rule

    a×10mb×10n=ab×10m−n,b≠0\frac{a\times10^m}{b\times10^n}=\frac{a}{b}\times10^{m-n},\quad b\ne0

    Addition with common exponent

    a×10k+b×10k=(a+b)×10ka\times10^k+b\times10^k=(a+b)\times10^k

    Power rule

    (a×10k)n=an×10kn(a\times10^k)^n=a^n\times10^{kn}

    Calculator translation

    aEk=a×10ka\text{E}k=a\times10^k

    Common metric prefixes

    kilo=103,milli=10−3,micro=10−6,nano=10−9\text{kilo}=10^3,\quad \text{milli}=10^{-3},\quad \text{micro}=10^{-6},\quad \text{nano}=10^{-9}

    Normalized form

    N=a×10k,1≤∣a∣<10,k∈ZN=a\times10^k,\quad 1\le |a|<10,\quad k\in\mathbb Z

    Product

    (a×10m)(b×10n)=ab×10m+n(a\times10^m)(b\times10^n)=ab\times10^{m+n}

    Quotient

    a×10mb×10n=ab×10m−n\frac{a\times10^m}{b\times10^n}=\frac ab\times10^{m-n}

    Power

    (a×10m)r=ar×10mr(a\times10^m)^r=a^r\times10^{mr}

    Common-exponent sum

    a×10k+b×10k=(a+b)×10ka\times10^k+b\times10^k=(a+b)\times10^k
    Topic 1.2Not started

    Arithmetic sequences and series

    Arithmetic sequences model approximately constant additive change. AI students should be able to calculate with the model and judge how well it fits real data.

    190 min estimated

    Topic 1.2 progress0%
    Learning objectives
    • Identify the first term and common difference.
    • Find a required term and the sum of a finite arithmetic series.
    • Use sigma notation for finite sums.
    • Apply arithmetic models to prediction and interpret deviations from perfect arithmetic behaviour.
    Key formulas

    First n terms as a series

    Sn=u1+u2+⋯+unS_n=u_1+u_2+\cdots+u_n

    Common difference

    d=un+1−und=u_{n+1}-u_n

    Arithmetic recurrence

    u1=a,un+1=un+du_1=a,\qquad u_{n+1}=u_n+d

    Arithmetic nth term

    un=a+(n−1)du_n=a+(n-1)d

    Linear form

    un=dn+(a−d)u_n=dn+(a-d)

    Difference from two terms

    d=uq−upq−p,q≠pd=\frac{u_q-u_p}{q-p},\quad q\ne p

    Sum using first and last terms

    Sn=n2(a+un)S_n=\frac{n}{2}(a+u_n)

    Sum using first term and difference

    Sn=n2[2a+(n−1)d]S_n=\frac{n}{2}[2a+(n-1)d]

    Number times average term

    Sn=n(a+un2)S_n=n\left(\frac{a+u_n}{2}\right)

    Arithmetic sum in sigma notation

    ∑r=1n[a+(r−1)d]=Sn\sum\limits_{r=1}^{n}[a+(r-1)d]=S_n

    Number of indexed terms

    count=q−p+1\text{count}=q-p+1

    Simple-interest amount

    An=P(1+nr)A_n=P(1+nr)

    Simple interest earned

    I=PrnI=Prn

    Approximate difference from two positions

    dapprox=uq−upq−pd_{\text{approx}}=\frac{u_q-u_p}{q-p}

    Residual

    en=observedn−predictedne_n=\text{observed}_n-\text{predicted}_n

    Nth term

    un=a+(n−1)du_n=a+(n-1)d

    Sum from first and last terms

    Sn=n2(a+un)S_n=\frac n2(a+u_n)

    Sum from first term and difference

    Sn=n2[2a+(n−1)d]S_n=\frac n2[2a+(n-1)d]

    Sigma form

    ∑r=1n[a+(r−1)d]=Sn\sum\limits_{r=1}^{n}[a+(r-1)d]=S_n
    Topic 1.3Not started

    Geometric sequences and series

    Geometric sequences model repeated proportional change, making them central to growth, decay and financial applications.

    200 min estimated

    Topic 1.3 progress0%
    Learning objectives
    • Identify the first term and common ratio.
    • Find a specified term.
    • Find the sum of a finite geometric series.
    • Use sigma notation and interpret repeated percentage change.
    Key formulas

    Consecutive-term ratio

    r=un+1unr=\frac{u_{n+1}}{u_n}

    Growth multiplier

    r=1+p100r=1+\frac{p}{100}

    Decay multiplier

    r=1−p100r=1-\frac{p}{100}

    Geometric recurrence

    u1=a,un+1=runu_1=a,\qquad u_{n+1}=ru_n

    Geometric nth term

    un=arn−1u_n=ar^{n-1}

    Term number from logarithms

    n=1+ln⁡(una)ln⁡rn=1+\frac{\ln(\frac{u_n}{a})}{\ln r}

    Ratio from two terms

    rq−p=uqupr^{q-p}=\frac{u_q}{u_p}

    First term from a known term

    a=uprp−1a=\frac{u_p}{r^{p-1}}

    First n geometric terms

    Sn=a+ar+ar2+⋯+arn−1S_n=a+ar+ar^2+\cdots+ar^{n-1}

    Finite geometric sum

    Sn=a(1−rn)1−r=a(rn−1)r−1S_n=\frac{a(1-r^n)}{1-r}=\frac{a(r^n-1)}{r-1}

    Constant-ratio special case

    Sn=nawhen r=1S_n=na\quad\text{when }r=1

    Finite geometric sum in sigma notation

    ∑k=0n−1ark=a(1−rn)1−r\sum\limits_{k=0}^{n-1}ar^k=\frac{a(1-r^n)}{1-r}

    Inclusive index count

    number of terms=n−m+1\text{number of terms}=n-m+1

    Infinite geometric sum

    S∞=a1−r,∣r∣<1S_\infty=\frac{a}{1-r},\qquad |r|<1

    Infinite sigma form

    ∑k=0∞ark=a1−r,∣r∣<1\sum\limits_{k=0}^{\infty}ar^k=\frac{a}{1-r},\qquad |r|<1

    Repeated-change model

    At=A0rtA_t=A_0r^t

    Percentage change from ratio

    percentage change=100(r−1)%\text{percentage change}=100(r-1)\%

    Approximate ratio from endpoints

    r≈(uqup)1q−pr\approx\left(\frac{u_q}{u_p}\right)^{\frac{1}{q-p}}

    Residual

    en=observed value−predicted valuee_n=\text{observed value}-\text{predicted value}

    Common ratio

    r=un+1unr=\frac{u_{n+1}}{u_n}

    Nth term

    un=arn−1u_n=ar^{n-1}

    Finite sum

    Sn=a(1−rn)1−rS_n=\frac{a(1-r^n)}{1-r}

    Infinite sum

    S∞=a1−rS_\infty=\frac{a}{1-r}
    Topic 1.4Not started

    Financial applications

    Financial mathematics connects exponentials, repeated percentage change and technology. Students should understand what the variables mean rather than treating finance as a calculator-only topic.

    190 min estimated

    Topic 1.4 progress0%
    Learning objectives
    • Calculate present and future values under compound growth.
    • Model depreciation.
    • Interpret nominal rate, compounding frequency and number of periods.
    • Use financial technology for realistic savings and loan problems.
    • State answers in context with appropriate currency and time units.
    Key formulas

    Simple interest earned

    I=PrtI=Prt

    Simple-interest balance

    A=P(1+rt)A=P(1+rt)

    Percentage multiplier

    growth: 1+p100,decay: 1−p100\text{growth: }1+\frac{p}{100},\qquad \text{decay: }1-\frac{p}{100}

    Compound future value

    FV=PV(1+i)nFV=PV(1+i)^n

    Periodic rate from two values

    i=(FVPV)1n−1,n>0, PV≠0, FVPV>0i=\left(\frac{FV}{PV}\right)^{\frac{1}{n}}-1,\quad n>0,\ PV\ne0,\frac{\ FV}{PV}>0

    Number of periods

    n=ln⁡(FV/PV)ln⁡(1+i)PV≠0,FV/PV>0,i>−1, i≠0\begin{gathered}n=\frac{\ln(FV/PV)}{\ln(1+i)}\\ PV\ne0,\quad FV/PV>0,\quad i>-1,\ i\ne0\end{gathered}

    Nominal-rate compound interest

    FV=PV(1+rm)mtFV=PV\left(1+\frac{r}{m}\right)^{mt}

    Periodic rate and period count

    i=rm,n=mti=\frac{r}{m},\qquad n=mt

    Effective annual rate

    EAR=(1+rm)m−1EAR=\left(1+\frac{r}{m}\right)^m-1

    Present value of one future amount

    PV=FV(1+i)n=FV(1+i)−nPV=\frac{FV}{(1+i)^n}=FV(1+i)^{-n}

    Appreciation

    Vn=V0(1+a)nV_n=V_0(1+a)^n

    Reducing-balance depreciation

    Vn=V0(1−d)nV_n=V_0(1-d)^n

    Future value of an ordinary annuity

    FV={R(1+i)n−1i,i≠0nR,i=0FV=\begin{cases}R\frac{(1+i)^n-1}{i},&i\ne0\\ nR,&i=0\end{cases}

    Future value of an annuity due

    FVdue={R(1+i)n−1i(1+i),i≠0nR,i=0FV_{\mathrm{due}}=\begin{cases}R\frac{(1+i)^n-1}{i}(1+i),&i\ne0\\ nR,&i=0\end{cases}

    Deposit required for a target

    R={FVi(1+i)n−1,i≠0FV/n,i=0R=\begin{cases}FV\frac{i}{(1+i)^n-1},&i\ne0\\ FV/n,&i=0\end{cases}

    Present value of loan repayments

    PV={R1−(1+i)−ni,i≠0nR,i=0PV=\begin{cases}R\frac{1-(1+i)^{-n}}{i},&i\ne0\\ nR,&i=0\end{cases}

    Periodic loan repayment

    R={PVi1−(1+i)−n,i≠0PV/n,i=0R=\begin{cases}PV\frac{i}{1-(1+i)^{-n}},&i\ne0\\ PV/n,&i=0\end{cases}

    Outstanding balance after k payments

    Bk={PV(1+i)k−R(1+i)k−1i,i≠0PV−kR,i=0B_k=\begin{cases}PV(1+i)^k-R\frac{(1+i)^k-1}{i},&i\ne0\\ PV-kR,&i=0\end{cases}

    Total interest on equal repayments

    Itotal=Rn−PVI_{\mathrm{total}}=Rn-PV

    Future price under inflation

    Ct=C0(1+f)tC_t=C_0(1+f)^t

    Real value after inflation

    Vreal=M(1+f)tV_{\mathrm{real}}=\frac{M}{(1+f)^t}

    Exact real annual rate

    rreal=1+r1+f−1r_{\mathrm{real}}=\frac{1+r}{1+f}-1

    Simple interest

    I=Prt,A=P(1+rt)I=Prt,\qquad A=P(1+rt)

    Compound amount

    FV=PV(1+i)NFV=PV(1+i)^N

    Nominal compounding

    i=rm,N=mti=\frac{r}{m},\qquad N=mt

    Present value

    PV=FV(1+i)NPV=\frac{FV}{(1+i)^N}

    Depreciation

    VN=V0(1−d)NV_N=V_0(1-d)^N

    Regular savings

    FV=PMT((1+i)N−1i)FV=PMT\left(\frac{(1+i)^N-1}{i}\right)

    Loan payment

    PMT=PV(i1−(1+i)−N)PMT=PV\left(\frac{i}{1-(1+i)^{-N}}\right)
    Topic 1.5Not started

    Exponents and introductory logarithms

    The current shared AI syllabus includes exponent laws and numerical use of common and natural logarithms. At HL, logarithmic ideas are later developed more formally.

    170 min estimated

    Topic 1.5 progress0%
    Learning objectives
    • Apply integer exponent laws.
    • Evaluate powers, roots, log10(x) and ln(x) using technology.
    • Recognize logarithms as inverses of exponentials.
    • Use numerical logarithms in modelling and equation solving.
    Key formulas

    Positive integer exponent

    an=a⋅a⋯a⏟n factors,n∈Na^n=\underbrace{a\cdot a\cdots a}_{n\text{ factors}},\quad n\in\mathbb N

    Zero exponent

    a0=1,a≠0a^0=1,\quad a\ne0

    Negative integer exponent

    a−n=1an,a≠0a^{-n}=\frac{1}{a^n},\quad a\ne0

    Product law

    aman=am+na^m a^n=a^{m+n}

    Quotient law

    aman=am−n,a≠0\frac{a^m}{a^n}=a^{m-n},\quad a\ne0

    Power of a power

    (am)n=amn(a^m)^n=a^{mn}

    Power of a product

    (ab)n=anbn(ab)^n=a^n b^n

    Power of a quotient

    (ab)n=anbn,b≠0\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n},\quad b\ne0

    Moving a negative power across a fraction bar

    a−mb−n=bnam,a,b≠0\frac{a^{-m}}{b^{-n}}=\frac{b^n}{a^m},\quad a,b\ne0

    Exponential-logarithmic equivalence

    ac=b⟺log⁡ab=ca^c=b\quad\Longleftrightarrow\quad\log_a b=c

    Common and natural logarithms

    log⁡x=log⁡10x,ln⁡x=log⁡ex\log x=\log_{10}x,\qquad\ln x=\log_e x

    Inverse identities

    log⁡a(ax)=x,alog⁡ax=x(x>0)\log_a(a^x)=x,\qquad a^{\log_a x}=x\quad(x>0)

    Product property

    log⁡a(xy)=log⁡ax+log⁡ay\log_a(xy)=\log_a x+\log_a y

    Quotient property

    log⁡a(xy)=log⁡ax−log⁡ay\log_a\left(\frac{x}{y}\right)=\log_a x-\log_a y

    Power property

    log⁡a(xk)=klog⁡ax\log_a(x^k)=k\log_a x

    Useful exact values

    log⁡a1=0,log⁡aa=1\log_a1=0,\qquad\log_a a=1

    Common-base principle

    af(x)=ag(x)⟹f(x)=g(x),a>0, a≠1a^{f(x)}=a^{g(x)}\Longrightarrow f(x)=g(x),\quad a>0,\ a\ne1

    Solving a basic exponential equation

    ax=b⟹x=ln⁡bln⁡a,a>0, a≠1, b>0a^x=b\Longrightarrow x=\frac{\ln b}{\ln a},\quad a>0,\ a\ne1,\ b>0

    Single logarithm equation

    log⁡af(x)=c⟺f(x)=ac,f(x)>0\log_a f(x)=c\Longleftrightarrow f(x)=a^c,\quad f(x)>0

    Equality of logarithms

    log⁡af(x)=log⁡ag(x)⟹f(x)=g(x),f(x),g(x)>0\log_a f(x)=\log_a g(x)\Longrightarrow f(x)=g(x),\quad f(x),g(x)>0

    Exponential growth or decay model

    Q(t)=Q0btQ(t)=Q_0b^t

    Percentage multiplier

    b=1±r100b=1\pm\frac{r}{100}

    Time to a target

    t=ln⁡(TQ0)ln⁡bt=\frac{\ln(\frac{T}{Q_0})}{\ln b}

    pH model

    pH=−log⁡10[H+]\mathrm{pH}=-\log_{10}[\mathrm H^+]

    Sound-intensity level

    L=10log⁡10(II0)L=10\log_{10}\left(\frac{I}{I_0}\right)

    Ratio from a base-10 scale difference

    ratio=10Δ\text{ratio}=10^{\Delta}

    Change-of-base formula

    log⁡ab=log⁡blog⁡a=ln⁡bln⁡a\log_a b=\frac{\log b}{\log a}=\frac{\ln b}{\ln a}

    Product and quotient laws

    aman=am+n,aman=am−na^ma^n=a^{m+n},\qquad\frac{a^m}{a^n}=a^{m-n}

    Power laws

    (am)n=amn,(ab)n=anbn(a^m)^n=a^{mn},\qquad(ab)^n=a^nb^n

    Zero and negative exponents

    a0=1,a−n=1an(a≠0)a^0=1,\qquad a^{-n}=\frac1{a^n}\quad(a\ne0)

    Exponential-logarithmic equivalence

    ax=b⟺log⁡ab=xa^x=b\Longleftrightarrow\log_a b=x

    Change of base

    log⁡ab=log⁡blog⁡a=ln⁡bln⁡a\log_a b=\frac{\log b}{\log a}=\frac{\ln b}{\ln a}

    Growth or decay model

    Q(t)=Q0btQ(t)=Q_0b^t
    Topic 1.6Not started

    Approximation, bounds and percentage error

    Approximation is treated as part of mathematical judgement. Rounded measurements represent intervals, and an applied conclusion may depend on how much error is acceptable.

    180 min estimated

    Topic 1.6 progress0%
    Learning objectives
    • Round to decimal places and significant figures.
    • Determine upper and lower bounds for rounded quantities.
    • Calculate absolute and percentage error.
    • Carry enough precision through technology calculations.
    • Explain whether the resulting accuracy is meaningful in context.
    Key formulas

    Rounding unit for decimal places

    u=10−du=10^{-d}

    General rounding error interval

    r−u2≤x<r+u2r-\frac{u}{2}\le x<r+\frac{u}{2}

    Lower and upper bounds

    xL=r−u2,xU=r+u2x_{\mathrm L}=r-\frac{u}{2},\qquad x_{\mathrm U}=r+\frac{u}{2}

    Interval notation

    x∈[xL,xU)x\in[x_{\mathrm L},x_{\mathrm U})

    Sum bounds

    SL=aL+bL,SU=aU+bUS_{\mathrm L}=a_{\mathrm L}+b_{\mathrm L},\qquad S_{\mathrm U}=a_{\mathrm U}+b_{\mathrm U}

    Difference bounds

    DL=aL−bU,DU=aU−bLD_{\mathrm L}=a_{\mathrm L}-b_{\mathrm U},\qquad D_{\mathrm U}=a_{\mathrm U}-b_{\mathrm L}

    Product bounds for positive values

    PL=aLbL,PU=aUbUP_{\mathrm L}=a_{\mathrm L}b_{\mathrm L},\qquad P_{\mathrm U}=a_{\mathrm U}b_{\mathrm U}

    Quotient bounds for positive values

    QL=aLbU,QU=aUbLQ_{\mathrm L}=\frac{a_{\mathrm L}}{b_{\mathrm U}},\qquad Q_{\mathrm U}=\frac{a_{\mathrm U}}{b_{\mathrm L}}

    Absolute error

    Ea=∣xapprox−xexact∣E_{\mathrm a}=\left|x_{\mathrm{approx}}-x_{\mathrm{exact}}\right|

    Relative error

    Er=Ea∣xexact∣,xexact≠0E_{\mathrm r}=\frac{E_{\mathrm a}}{|x_{\mathrm{exact}}|},\qquad x_{\mathrm{exact}}\ne0

    Percentage error

    E%=∣xapprox−xexact∣∣xexact∣×100%E_{\%}=\frac{|x_{\mathrm{approx}}-x_{\mathrm{exact}}|}{|x_{\mathrm{exact}}|}\times100\%

    Maximum absolute rounding error

    Emax⁡=u2E_{\max}=\frac{u}{2}

    Maximum output percentage error

    E%,max⁡=max⁡(∣yU−y0∣,∣y0−yL∣)∣y0∣×100%E_{\%,\max}=\frac{\max\left(|y_{\mathrm U}-y_0|,|y_0-y_{\mathrm L}|\right)}{|y_0|}\times100\%

    Symmetric tolerance

    t−δ≤x≤t+δt-\delta\le x\le t+\delta

    Decimal-place unit

    u=10−du=10^{-d}

    Error interval

    r−u2≤x<r+u2r-\frac{u}{2}\le x<r+\frac{u}{2}

    Maximum rounding error

    Emax⁡=u2E_{\max}=\frac{u}{2}

    Absolute error

    Ea=∣xapprox−xexact∣E_{\mathrm a}=|x_{\mathrm{approx}}-x_{\mathrm{exact}}|

    Relative error

    Er=Ea∣xexact∣E_{\mathrm r}=\frac{E_{\mathrm a}}{|x_{\mathrm{exact}}|}

    Percentage error

    E%=100Er%E_{\%}=100E_{\mathrm r}\%

    Positive product bounds

    PL=aLbL,PU=aUbUP_{\mathrm L}=a_{\mathrm L}b_{\mathrm L},\qquad P_{\mathrm U}=a_{\mathrm U}b_{\mathrm U}

    Positive quotient bounds

    QL=aLbU,QU=aUbLQ_{\mathrm L}=\frac{a_{\mathrm L}}{b_{\mathrm U}},\qquad Q_{\mathrm U}=\frac{a_{\mathrm U}}{b_{\mathrm L}}
    Topic 1.7Not started

    Amortization and annuities

    Regular payments and deposits are normally handled with financial technology. The key skill is setting up the cash flows and interpreting the result.

    185 min estimated

    Topic 1.7 progress0%
    Learning objectives
    • Calculate regular loan payments and outstanding balances.
    • Analyze annuities and savings plans.
    • Recognize present value, future value, payment, interest rate and number of payment periods.
    • Interpret sign conventions used by GDC finance applications.
    • Compare financial options using mathematically justified criteria.
    Key formulas

    Annuity-due timing adjustment

    Vdue=Vordinary(1+i)V_{\mathrm{due}}=V_{\mathrm{ordinary}}(1+i)

    Periodic rate

    i=rmi=\frac{r}{m}

    Number of periods

    n=mtn=mt

    Effective annual rate

    reff=(1+i)m−1r_{\mathrm{eff}}=(1+i)^m-1

    Future value of an ordinary annuity

    FV=R(1+i)n−1iFV=R\frac{(1+i)^n-1}{i}

    Future value of an annuity due

    FVdue={R(1+i)n−1i(1+i),i≠0nR,i=0FV_{\mathrm{due}}=\begin{cases}R\frac{(1+i)^n-1}{i}(1+i),&i\ne0\\ nR,&i=0\end{cases}

    Regular deposit for a target

    R=FVi(1+i)n−1R=FV\frac{i}{(1+i)^n-1}

    Interest earned on a savings annuity

    I=FV−nRI=FV-nR

    Present value of an ordinary annuity

    PV=R1−(1+i)−niPV=R\frac{1-(1+i)^{-n}}{i}

    Present value of an annuity due

    PVdue=R1−(1+i)−ni(1+i)PV_{\mathrm{due}}=R\frac{1-(1+i)^{-n}}{i}(1+i)

    Loan as a present-value annuity

    PV=R1−(1+i)−niPV=R\frac{1-(1+i)^{-n}}{i}

    Periodic loan repayment

    R={PVi1−(1+i)−n,i≠0PV/n,i=0R=\begin{cases}PV\frac{i}{1-(1+i)^{-n}},&i\ne0\\ PV/n,&i=0\end{cases}

    Total scheduled repayments

    T=nRT=nR

    Total interest without fees

    Itotal=nR−PVI_{\mathrm{total}}=nR-PV

    Outstanding balance from original loan

    Bk=PV(1+i)k−R(1+i)k−1iB_k=PV(1+i)^k-R\frac{(1+i)^k-1}{i}

    Outstanding balance from remaining payments

    Bk=R1−(1+i)−(n−k)iB_k=R\frac{1-(1+i)^{-(n-k)}}{i}

    Principal repaid by time k

    Prepaid=PV−BkP_{\mathrm{repaid}}=PV-B_k

    Interest component

    Ij=iBj−1I_j=iB_{j-1}

    Principal component

    Pj=R−IjP_j=R-I_j

    Balance recurrence

    Bj=Bj−1(1+i)−RB_j=B_{j-1}(1+i)-R

    Periodic rate and period count

    i=rm,n=mti=\frac{r}{m},\qquad n=mt

    Ordinary-annuity future value

    FV=R(1+i)n−1iFV=R\frac{(1+i)^n-1}{i}

    Annuity-due adjustment

    Vdue=Vordinary(1+i)V_{\mathrm{due}}=V_{\mathrm{ordinary}}(1+i)

    Ordinary-annuity present value

    PV=R1−(1+i)−niPV=R\frac{1-(1+i)^{-n}}{i}

    Loan repayment

    R=PVi1−(1+i)−nR=PV\frac{i}{1-(1+i)^{-n}}

    Outstanding balance

    Bk=PV(1+i)k−R(1+i)k−1iB_k=PV(1+i)^k-R\frac{(1+i)^k-1}{i}

    Amortization row

    Ij=iBj−1,Pj=R−Ij,Bj=Bj−1−PjI_j=iB_{j-1},\qquad P_j=R-I_j,\qquad B_j=B_{j-1}-P_j

    Total interest

    Itotal=nR−PVI_{\mathrm{total}}=nR-PV
    Topic 1.8Not started

    Solving equations with technology

    AI uses numerical technology to solve systems and polynomial equations efficiently, while the student remains responsible for creating the correct equations.

    205 min estimated

    Topic 1.8 progress0%
    Learning objectives
    • Solve systems of up to three linear equations.
    • Solve polynomial equations numerically.
    • Use graphical or numerical methods when appropriate.
    • Interpret and filter solutions according to domain and context.
    • Check computed results when a substitution or graph makes verification practical.
    Key formulas

    Root definition

    h(r)=0h(r)=0

    Intersection-to-zero conversion

    f(x)=g(x)  ⟺  f(x)−g(x)=0f(x)=g(x)\iff f(x)-g(x)=0

    Absolute residual

    R=∣LHS−RHS∣R=\left|\text{LHS}-\text{RHS}\right|

    Polynomial factor and root

    (x−r)m is a factor⟹r is a root of multiplicity m(x-r)^m\text{ is a factor}\Longrightarrow r\text{ is a root of multiplicity }m

    Break-even equation

    R(x)=C(x)R(x)=C(x)

    Difference function for intersections

    h(x)=f(x)−g(x)h(x)=f(x)-g(x)

    Sign-change bracket

    h(a)h(b)<0⟹at least one root in (a,b) if h is continuoush(a)h(b)\lt0\Longrightarrow\text{at least one root in }(a,b)\text{ if }h\text{ is continuous}

    Two-variable linear system

    a1x+b1y=c1,a2x+b2y=c2a_1x+b_1y=c_1,\qquad a_2x+b_2y=c_2

    Three-variable linear system

    aix+biy+ciz=di(i=1,2,3)a_ix+b_iy+c_iz=d_i\qquad(i=1,2,3)

    Polynomial standard form

    anxn+an−1xn−1+⋯+a1x+a0=0,an≠0a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0=0,\qquad a_n\ne0

    Maximum number of distinct real roots

    number of distinct real roots≤n\text{number of distinct real roots}\le n

    Even-root restriction

    u(x) requires u(x)≥0\sqrt{u(x)}\text{ requires }u(x)\ge0

    Logarithm restriction

    log⁡b(u(x)) requires u(x)>0\log_b(u(x))\text{ requires }u(x)\gt0

    Denominator restriction

    p(x)q(x) requires q(x)≠0\frac{p(x)}{q(x)}\text{ requires }q(x)\ne0

    Exponential solution check

    abx=c⟹x=ln⁡(ca)ln⁡bab^x=c\Longrightarrow x=\frac{\ln(\frac{c}{a})}{\ln b}

    Logarithm argument condition

    log⁡bu is real only when u>0\log_b u\text{ is real only when }u\gt0

    Approximate equality symbol

    x≈1.732x\approx1.732

    Exact equality symbol

    x=3x=\sqrt3

    Zero residual

    R=∣h(r)∣R=|h(r)|

    Intersection residual

    R=∣f(r)−g(r)∣R=|f(r)-g(r)|

    Zero and intersection equivalence

    f(x)=g(x)  ⟺  f(x)−g(x)=0f(x)=g(x)\iff f(x)-g(x)=0

    Sign-change bracket

    h(a)h(b)<0h(a)h(b)\lt0

    Residual

    R=∣LHS−RHS∣R=|\text{LHS}-\text{RHS}|

    Polynomial form

    anxn+⋯+a1x+a0=0a_nx^n+\cdots+a_1x+a_0=0

    Two-variable system

    a1x+b1y=c1,a2x+b2y=c2a_1x+b_1y=c_1,\quad a_2x+b_2y=c_2
    Topic 1.9Not started

    Laws of logarithms

    HL students manipulate logarithms algebraically and use their laws to simplify expressions, solve equations and support linearization methods.

    230 min estimated

    Topic 1.9 progress0%
    Learning objectives
    • Translate confidently between exponential and logarithmic forms and state the conditions for real logarithms.
    • Use and justify the product, quotient, and power laws while preserving domain restrictions.
    • Expand and condense logarithmic expressions and use change of base accurately.
    • Solve logarithmic equations, reject extraneous roots, and solve exponential equations using logarithms.
    • Apply logarithms to growth, decay, threshold, and logarithmic-scale problems.
    • Linearize exponential and power models, interpret transformed gradients and intercepts, and back-transform parameters.
    • Use a GDC to evaluate logarithms, solve equations, and fit transformed models while communicating the method.
    • Present exact and approximate answers with correct rounding, units, constraints, and contextual interpretation.
    Key formulas

    Logarithm definition

    log⁡ax=y⟺ay=x\log_a x=y\quad\Longleftrightarrow\quad a^y=x

    Logarithm of one

    log⁡a1=0\log_a1=0

    Logarithm of its base

    log⁡aa=1\log_a a=1

    Inverse relationships

    alog⁡ax=x,log⁡a(ax)=xa^{\log_a x}=x,\qquad \log_a(a^x)=x

    Product-law derivation

    xy=auav=au+vxy=a^ua^v=a^{u+v}

    Quotient-law derivation

    xy=auav=au−v\frac{x}{y}=\frac{a^u}{a^v}=a^{u-v}

    Product law

    log⁡a(xy)=log⁡ax+log⁡ay\log_a(xy)=\log_a x+\log_a y

    Quotient law

    log⁡a(xy)=log⁡ax−log⁡ay\log_a\left(\frac{x}{y}\right)=\log_a x-\log_a y

    Power law

    log⁡a(xr)=rlog⁡ax\log_a(x^r)=r\log_a x

    Reciprocal result

    log⁡a(1x)=−log⁡ax\log_a\left(\frac1x\right)=-\log_a x

    Change of base

    log⁡ax=log⁡bxlog⁡ba\log_a x=\frac{\log_b x}{\log_b a}

    Calculator forms

    log⁡ax=ln⁡xln⁡a=log⁡xlog⁡a\log_a x=\frac{\ln x}{\ln a}=\frac{\log x}{\log a}

    Condensing pattern

    rlog⁡ax−slog⁡ay=log⁡a(xrys)r\log_a x-s\log_a y=\log_a\left(\frac{x^r}{y^s}\right)

    Real logarithm conditions

    a>0,a≠1,g(x)>0a>0,\qquad a\ne1,\qquad g(x)>0

    Power identity for unrestricted nonzero x

    log⁡a(x2)=2log⁡a∣x∣\log_a(x^2)=2\log_a|x|

    Same-base principle

    au=av⟹u=va^u=a^v\quad\Longrightarrow\quad u=v

    Logarithmic solution step

    af(x)=k⟹f(x)=ln⁡kln⁡aa^{f(x)}=k\quad\Longrightarrow\quad f(x)=\frac{\ln k}{\ln a}

    Log equality principle

    log⁡au=log⁡av⟹u=v\log_a u=\log_a v\quad\Longrightarrow\quad u=v

    Exponential growth or decay model

    Q(t)=Q0btQ(t)=Q_0b^t

    Percentage multiplier

    b=1±r100b=1\pm\frac{r}{100}

    Time to a target

    t=ln⁡(TQ0)ln⁡bt=\frac{\ln(\frac{T}{Q_0})}{\ln b}

    pH model

    pH=−log⁡10[H+]\mathrm{pH}=-\log_{10}[\mathrm H^+]

    Sound-intensity level

    L=10log⁡10(II0)L=10\log_{10}\left(\frac{I}{I_0}\right)

    Ratio from a base-10 scale difference

    ratio=10Δ\text{ratio}=10^{\Delta}

    Exponential linearization

    y=Abx⟹ln⁡y=ln⁡A+xln⁡by=Ab^x\quad\Longrightarrow\quad \ln y=\ln A+x\ln b

    Power-model linearization

    y=Axn⟹ln⁡y=ln⁡A+nln⁡xy=Ax^n\quad\Longrightarrow\quad \ln y=\ln A+n\ln x

    Back-transform an intercept

    c=ln⁡A⟹A=ecc=\ln A\quad\Longrightarrow\quad A=e^c

    Change of base for a GDC

    log⁡ax=ln⁡xln⁡a=log⁡xlog⁡a\log_a x=\frac{\ln x}{\ln a}=\frac{\log x}{\log a}

    Equation workflow

    DOMAIN→TRANSFORM→SOLVE→CHECK→INTERPRET\text{DOMAIN}\to\text{TRANSFORM}\to\text{SOLVE}\to\text{CHECK}\to\text{INTERPRET}

    Exponential threshold time

    t=ln⁡(TQ0)ln⁡bt=\frac{\ln(\frac{T}{Q_0})}{\ln b}

    Definition

    log⁡ax=y  ⟺  ay=x\log_a x=y\iff a^y=x

    Quotient law

    log⁡a(xy)=log⁡ax−log⁡ay\log_a(\frac{x}{y})=\log_a x-\log_a y

    Change of base

    log⁡ax=ln⁡xln⁡a\log_a x=\frac{\ln x}{\ln a}

    Valid real logarithm

    a>0, a≠1, x>0a>0,\ a\ne1,\ x>0
    Topic 1.10Not started

    Expressions with rational exponents

    Rational exponents allow roots and powers to be handled in one algebraic language and appear in power models and transformations.

    220 min estimated

    Topic 1.10 progress0%
    Learning objectives
    • Interpret rational exponents and convert fluently between power and radical notation.
    • Apply the exponent laws to products, quotients, and powers while respecting real-domain conditions.
    • Evaluate positive, negative, and zero rational powers exactly where possible.
    • Analyze negative bases, principal roots, and absolute-value requirements accurately.
    • Simplify multi-factor expressions and present final answers with positive exponents when requested.
    • Solve equations involving rational exponents, substitutions, multiple branches, and domain checks.
    • Use rational exponents in power models, geometric scaling, and other applied contexts.
    • Use a GDC for evaluation and numerical solutions while communicating syntax, domains, accuracy, and interpretation.
    Key formulas

    Rational exponent definition

    amn=amn=(an)ma^{\frac{m}{n}}=\sqrt[n]{a^m}=(\sqrt[n]{a})^m

    Negative rational exponent

    a−mn=1amna^{\frac{-m}{n}}=\frac{1}{a^{\frac{m}{n}}}

    Special root form

    a1n=ana^{\frac{1}{n}}=\sqrt[n]{a}

    Radical–exponent conversion

    xmn=xmn,x>0, m∈Z, n∈Z>0\sqrt[n]{x^m}=x^{\frac{m}{n}},\qquad x>0,\ m\in\mathbb Z,\ n\in\mathbb Z_{>0}

    Principal square root

    x2=∣x∣\sqrt{x^2}=|x|

    Odd root of a power

    x2k+12k+1=x\sqrt[2k+1]{x^{2k+1}}=x

    Product law

    aras=ar+sa^r a^s=a^{r+s}

    Quotient law

    aras=ar−s\frac{a^r}{a^s}=a^{r-s}

    Power of a power

    (ar)s=ars(a^r)^s=a^{rs}

    Power of a product

    (ab)r=arbr(ab)^r=a^r b^r

    Power of a quotient

    (ab)r=arbr\left(\frac ab\right)^r=\frac{a^r}{b^r}

    Negative exponent conversion

    x−r=1xrx^{-r}=\frac{1}{x^r}

    Common-denominator strategy

    xpqxrs=x(ps+rq)(qs)x^{\frac{p}{q}}x^{\frac{r}{s}}=x^{\frac{(ps+rq)}{(qs)}}

    Exact before decimal

    amn=amn\sqrt[n]{a^m}=a^{\frac{m}{n}}

    Negative base with odd reduced denominator

    (−a)mn=(−an)m(n odd)(-a)^{\frac{m}{n}}=\left(\sqrt[n]{-a}\right)^m\quad(n\text{ odd})

    Principal square root

    (x2)12=x2=∣x∣(x^2)^{\frac{1}{2}}=\sqrt{x^2}=|x|

    Zero restriction

    0r is defined for r>0 and undefined for r<00^r\text{ is defined for }r>0\text{ and undefined for }r<0

    Direct positive-domain pattern

    xmn=k⟹x=knmx^{\frac{m}{n}}=k\quad\Longrightarrow\quad x=k^{\frac{n}{m}}

    Useful substitution

    u=x1n⟹xmn=umu=x^{\frac{1}{n}}\quad\Longrightarrow\quad x^{\frac{m}{n}}=u^m

    Two-branch warning

    u2=k⟹u=±ku^2=k\quad\Longrightarrow\quad u=\pm\sqrt{k}

    Approximation notation

    exact value≈rounded decimal\text{exact value}\approx\text{rounded decimal}

    Factor a perfect nth power

    cndn=cdn(c≥0 if n is even)\sqrt[n]{c^n d}=c\sqrt[n]{d}\quad(c\ge0\text{ if }n\text{ is even})

    Power model

    y=Axpy=Ax^p

    Scaling response

    x↦kx⟹y↦kpyx\mapsto kx\quad\Longrightarrow\quad y\mapsto k^p y

    Cube side from volume

    s=V13s=V^{\frac{1}{3}}

    Safe calculator entry

    amn  ⟶  a(mn)a^{\frac{m}{n}}\;\longrightarrow\;a^{(\frac{m}{n})}

    Real negative-base evaluation

    (−a)mn=(−an)m(n odd)(-a)^{\frac{m}{n}}=\left(\sqrt[n]{-a}\right)^m\quad(n\text{ odd})

    Equation workflow

    DOMAIN→SUBSTITUTE→SOLVE→CHECK\text{DOMAIN}\to\text{SUBSTITUTE}\to\text{SOLVE}\to\text{CHECK}

    Scaling rule

    y=Axp,x↦kx⇒y↦kpyy=Ax^p,\quad x\mapsto kx\Rightarrow y\mapsto k^py

    Rational exponent

    amn=amn=(an)ma^{\frac{m}{n}}=\sqrt[n]{a^m}=(\sqrt[n]{a})^m

    Negative rational exponent

    a−mn=1amna^{\frac{-m}{n}}=\frac1{a^{\frac{m}{n}}}

    Product law

    aras=ar+sa^ra^s=a^{r+s}
    Topic 1.11Not started

    Infinite geometric sequences and series

    A geometric series can converge to a finite limit when the magnitude of the common ratio is less than one.

    230 min estimated

    Topic 1.11 progress0%
    Learning objectives
    • Review geometric sequences and series and distinguish terms from partial sums.
    • Identify the first term and common ratio, including alternating geometric structures.
    • Explain convergence through partial sums and apply the condition |r|<1 correctly.
    • Derive and use the finite-sum and sum-to-infinity formulas.
    • Find unknown first terms, common ratios, terms, and parameters while rejecting invalid candidates.
    • Calculate and interpret remainders after a finite number of terms.
    • Model repeated processes, recurring decimals, bouncing motion, steady states, and perpetuities.
    • Use a GDC to inspect partial sums and solve parameter equations while communicating convergence and context.
    Key formulas

    Infinite geometric series

    a+ar+ar2+ar3+⋯a+ar+ar^2+ar^3+\cdots

    Partial sum notation

    Sn=∑k=0n−1arkS_n=\sum\limits_{k=0}^{n-1}ar^k

    Common ratio

    r=un+1unr=\frac{u_{n+1}}{u_n}

    Geometric term

    un=arn−1u_n=ar^{n-1}

    Geometric convergence condition

    ∣r∣<1⟺−1<r<1|r|<1\quad\Longleftrightarrow\quad -1<r<1

    Vanishing powers

    ∣r∣<1⟹rn→0|r|<1\quad\Longrightarrow\quad r^n\to0

    Finite geometric sum

    Sn=a(1−rn)1−rS_n=\frac{a(1-r^n)}{1-r}

    Sum to infinity

    S∞=a1−r,∣r∣<1S_\infty=\frac{a}{1-r},\qquad |r|<1

    First term from the sum

    a=S∞(1−r)a=S_\infty(1-r)

    Ratio from first term and sum

    r=1−aS∞,S∞≠0r=1-\frac{a}{S_\infty},\qquad S_\infty\ne0

    Term condition

    un=arn−1u_n=ar^{n-1}

    Remainder after n terms

    Rn=S∞−Sn=arn1−rR_n=S_\infty-S_n=\frac{ar^n}{1-r}

    Approximation relation

    S∞=Sn+RnS_\infty=S_n+R_n

    Bouncing-ball distance

    D=h+2hq1−q=h(1+q1−q)D=h+\frac{2hq}{1-q}=h\left(\frac{1+q}{1-q}\right)

    Repeated fixed contribution

    A∞=d1−r,∣r∣<1A_\infty=\frac{d}{1-r},\qquad |r|<1

    Two-digit recurring decimal

    0.ab‾=10a+b99,a,b∈{0,1,…,9}0.\overline{ab}=\frac{10a+b}{99},\qquad a,b\in\{0,1,\ldots,9\}

    Retain-and-add recurrence

    An+1=rAn+dA_{n+1}=rA_n+d

    Stable long-term level

    L=d1−r,∣r∣<1L=\frac{d}{1-r},\qquad |r|<1

    Present value of an end-of-period perpetuity

    PV=Pi,i>0PV=\frac{P}{i},\qquad i>0

    Partial-sum calculator entry

    Sn=a(1−rn)1−rS_n=\frac{a(1-r^n)}{1-r}

    Solver interval

    −1<r<1-1<r<1

    Convergence and sum

    ∣r∣<1⟹S∞=a1−r|r|<1\quad\Longrightarrow\quad S_\infty=\frac{a}{1-r}

    Term and remainder

    un=arn−1,Rn=arn1−ru_n=ar^{n-1},\qquad R_n=\frac{ar^n}{1-r}

    nth term

    un=arn−1u_n=ar^{n-1}

    Finite sum

    Sn=a(1−rn)1−rS_n=\frac{a(1-r^n)}{1-r}

    Convergence condition

    ∣r∣<1|r|<1

    Remainder after n terms

    Rn=arn1−rR_n=\frac{ar^n}{1-r}

    Unknown first term

    a=S∞(1−r)a=S_\infty(1-r)

    Unknown common ratio

    r=1−aS∞r=1-\frac{a}{S_\infty}
    Topic 1.12Not started

    Complex numbers in Cartesian form

    Complex numbers extend the real number system and allow polynomial equations with negative discriminant to be solved. AI HL develops both algebraic operations and geometric interpretation in the complex plane.

    300 min estimated

    Topic 1.12 progress0%
    Learning objectives
    • Define the imaginary unit and write complex numbers accurately in Cartesian form.
    • Identify real and imaginary parts and use component equality to find unknowns.
    • Add, subtract, multiply, and divide complex numbers in exact Cartesian form.
    • Reduce integer powers of i using the four-term cycle.
    • Use complex conjugates to simplify products, perform division, and establish identities.
    • Calculate modulus and distance and interpret both geometrically on an Argand diagram.
    • Solve linear, quadratic, conjugate, and Cartesian square-root equations involving complex numbers.
    • Apply Cartesian complex numbers to component models such as impedance and planar displacement.
    • Use a GDC in complex mode to verify results while preserving exact reasoning and contextual interpretation.
    Key formulas

    Imaginary unit

    i2=−1i^2=-1

    Cartesian form

    z=a+biz=a+bi

    Set of complex numbers

    C={a+bi:a,b∈R}\mathbb{C}=\{a+bi:a,b\in\mathbb{R}\}

    Real part

    Re⁡(a+bi)=a\operatorname{Re}(a+bi)=a

    Imaginary part

    Im⁡(a+bi)=b\operatorname{Im}(a+bi)=b

    Real-number condition

    z∈R  ⟺  Im⁡(z)=0z\in\mathbb{R}\iff\operatorname{Im}(z)=0

    Complex equality

    a+bi=c+di  ⟺  a=c and b=da+bi=c+di\iff a=c\text{ and }b=d

    Addition

    (a+bi)+(c+di)=(a+c)+(b+d)i(a+bi)+(c+di)=(a+c)+(b+d)i

    Subtraction

    (a+bi)−(c+di)=(a−c)+(b−d)i(a+bi)-(c+di)=(a-c)+(b-d)i

    Four-power cycle

    i0=1,i1=i,i2=−1,i3=−i,i4=1i^0=1,\quad i^1=i,\quad i^2=-1,\quad i^3=-i,\quad i^4=1

    Periodicity

    in+4=ini^{n+4}=i^n

    Inverse of i

    i−1=−ii^{-1}=-i

    Complex multiplication

    (a+bi)(c+di)=(ac−bd)+(ad+bc)i(a+bi)(c+di)=(ac-bd)+(ad+bc)i

    Complex square

    (a+bi)2=(a2−b2)+2abi(a+bi)^2=(a^2-b^2)+2abi

    Conjugate

    a+bi‾=a−bi\overline{a+bi}=a-bi

    Sum with conjugate

    z+z‾=2Re⁡(z)z+\overline z=2\operatorname{Re}(z)

    Difference from conjugate

    z−z‾=2iIm⁡(z)z-\overline z=2i\operatorname{Im}(z)

    Conjugate product

    zz‾=a2+b2z\overline z=a^2+b^2

    Conjugate of a product

    zw‾=z‾ w‾\overline{zw}=\overline z\,\overline w

    Complex quotient

    a+bic+di=(ac+bd)+(bc−ad)ic2+d2\frac{a+bi}{c+di}=\frac{(ac+bd)+(bc-ad)i}{c^2+d^2}

    Complex reciprocal

    1a+bi=a−bia2+b2\frac1{a+bi}=\frac{a-bi}{a^2+b^2}

    Modulus

    ∣a+bi∣=a2+b2|a+bi|=\sqrt{a^2+b^2}

    Modulus squared

    ∣z∣2=zz‾|z|^2=z\overline z

    Distance

    d(z1,z2)=∣z1−z2∣d(z_1,z_2)=|z_1-z_2|

    Product modulus

    ∣zw∣=∣z∣∣w∣|zw|=|z||w|

    Triangle inequality

    ∣z+w∣≤∣z∣+∣w∣|z+w|\le |z|+|w|

    Argand coordinate

    z=a+bi⟷(a,b)z=a+bi\longleftrightarrow(a,b)

    Conjugate reflection

    a+bi⟷a−bia+bi\longleftrightarrow a-bi

    Displacement

    z1−z2z_1-z_2

    Square in Cartesian form

    (a+bi)2=(a2−b2)+2abi(a+bi)^2=(a^2-b^2)+2abi

    Quadratic formula

    z=−b±b2−4ac2az=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

    Conjugate factor pair

    (x−(a+bi))(x−(a−bi))=(x−a)2+b2(x-(a+bi))(x-(a-bi))=(x-a)^2+b^2

    Quadratic from conjugate roots

    x2−2ax+(a2+b2)=0x^2-2ax+(a^2+b^2)=0

    Circle locus

    ∣z−c∣=r|z-c|=r

    Equidistance locus

    ∣z−z1∣=∣z−z2∣|z-z_1|=|z-z_2|

    Impedance model

    Z=R+XiZ=R+Xi

    Simplified AC current

    I=VZI=\frac{V}{Z}

    Resultant displacement

    znew=zold+wz_{\mathrm{new}}=z_{\mathrm{old}}+w

    Universal arithmetic check

    computed answer→substitutionoriginal equation\text{computed answer}\xrightarrow{\text{substitution}}\text{original equation}

    Core Cartesian workflow

    expand→use i2=−1→collect real and imaginary parts\text{expand}\to\text{use }i^2=-1\to\text{collect real and imaginary parts}

    Conjugate

    z‾=a−bi\overline z=a-bi

    Modulus

    ∣z∣=a2+b2|z|=\sqrt{a^2+b^2}

    Division

    a+bic+di=(ac+bd)+(bc−ad)ic2+d2\frac{a+bi}{c+di}=\frac{(ac+bd)+(bc-ad)i}{c^2+d^2}
    Topic 1.13Not started

    Polar and exponential forms of complex numbers

    Polar and exponential forms make multiplication, division and powers of complex numbers more natural by separating magnitude from angle.

    330 min estimated

    Topic 1.13 progress0%
    Learning objectives
    • Find the modulus and a quadrant-correct argument of a non-zero complex number.
    • Convert accurately among Cartesian, polar, cis and exponential forms.
    • Distinguish a principal argument from the general family of arguments.
    • Multiply and divide complex numbers by combining moduli and arguments.
    • Raise complex numbers to integer powers using polar or exponential form.
    • Interpret multiplication geometrically as a dilation and rotation on an Argand plane.
    • Use polar complex numbers in applied phasor and repeated-transformation models.
    • Use calculator/GDC complex-number functions while preserving exact values where required.
    • Communicate complete IB AI HL solutions with correct radians, quadrants and argument conventions.
    Key formulas

    Modulus

    r=∣z∣=x2+y2r=|z|=\sqrt{x^2+y^2}

    Cartesian projections

    x=rcos⁡θ,y=rsin⁡θx=r\cos\theta,\qquad y=r\sin\theta

    Tangent relation

    tan⁡θ=yx\tan\theta=\frac{y}{x}

    Polar form

    z=r(cos⁡θ+isin⁡θ)z=r(\cos\theta+i\sin\theta)

    Equivalent arguments

    θk=θ+2kπ,k∈Z\theta_k=\theta+2k\pi,\qquad k\in\mathbb Z

    Cartesian reconstruction

    z=rcos⁡θ+irsin⁡θz=r\cos\theta+i r\sin\theta

    Cis definition

    cis⁡θ=cos⁡θ+isin⁡θ\operatorname{cis}\theta=\cos\theta+i\sin\theta

    Cis form

    z=rcis⁡θz=r\operatorname{cis}\theta

    Cis periodicity

    cis⁡(θ+2kπ)=cis⁡θ\operatorname{cis}(\theta+2k\pi)=\operatorname{cis}\theta

    Euler's relation

    eiθ=cos⁡θ+isin⁡θe^{i\theta}=\cos\theta+i\sin\theta

    Exponential form

    z=reiθz=re^{i\theta}

    Equivalent notation

    reiθ=rcis⁡θre^{i\theta}=r\operatorname{cis}\theta

    Principal-argument convention

    −π<Arg⁡z≤π-\pi<\operatorname{Arg}z\le\pi

    General arguments

    arg⁡z=Arg⁡z+2kπ,k∈Z\arg z=\operatorname{Arg}z+2k\pi,\qquad k\in\mathbb Z

    Cartesian-to-polar modulus

    r=x2+y2r=\sqrt{x^2+y^2}

    Cartesian-to-polar argument

    θ=atan2⁡(y,x)\theta=\operatorname{atan2}(y,x)

    Equivalent output forms

    r(cos⁡θ+isin⁡θ)=rcis⁡θ=reiθr(\cos\theta+i\sin\theta)=r\operatorname{cis}\theta=re^{i\theta}

    Real component

    x=rcos⁡θx=r\cos\theta

    Imaginary component

    y=rsin⁡θy=r\sin\theta

    Polar-to-Cartesian conversion

    rcis⁡θ=rcos⁡θ+irsin⁡θr\operatorname{cis}\theta=r\cos\theta+i r\sin\theta

    Product in cis form

    z1z2=r1r2cis⁡(θ1+θ2)z_1z_2=r_1r_2\operatorname{cis}(\theta_1+\theta_2)

    Product in exponential form

    z1z2=r1r2ei(θ1+θ2)z_1z_2=r_1r_2e^{i(\theta_1+\theta_2)}

    Product modulus

    ∣z1z2∣=∣z1∣∣z2∣|z_1z_2|=|z_1||z_2|

    Power in cis form

    (rcis⁡θ)n=rncis⁡(nθ)\left(r\operatorname{cis}\theta\right)^n=r^n\operatorname{cis}(n\theta)

    Power in exponential form

    (reiθ)n=rneinθ\left(re^{i\theta}\right)^n=r^n e^{in\theta}

    Modulus of a power

    ∣zn∣=∣z∣n|z^n|=|z|^n

    Quotient in cis form

    z1z2=r1r2cis⁡(θ1−θ2)\frac{z_1}{z_2}=\frac{r_1}{r_2}\operatorname{cis}(\theta_1-\theta_2)

    Reciprocal

    1rcis⁡θ=1rcis⁡(−θ)\frac1{r\operatorname{cis}\theta}=\frac1r\operatorname{cis}(-\theta)

    Conjugate in polar form

    rcis⁡θ‾=rcis⁡(−θ)\overline{r\operatorname{cis}\theta}=r\operatorname{cis}(-\theta)

    Quotient modulus

    ∣z1z2∣=∣z1∣∣z2∣\left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}

    Scaling and rotation

    z↦kcis⁡ϕ zz\mapsto k\operatorname{cis}\phi\,z

    Modulus transformation

    ∣kcis⁡ϕ z∣=k∣z∣|k\operatorname{cis}\phi\,z|=k|z|

    Argument transformation

    arg⁡(kcis⁡ϕ z)=arg⁡z+ϕ(mod2π)\arg(k\operatorname{cis}\phi\,z)=\arg z+\phi\pmod{2\pi}

    Phasor quotient model

    I=VZI=\frac{V}{Z}

    Repeated scale-and-rotate model

    zn=z0(keiϕ)n=z0kneinϕz_n=z_0\left(ke^{i\phi}\right)^n=z_0k^n e^{in\phi}

    Phase difference

    Δθ=Arg⁡(z2)−Arg⁡(z1)\Delta\theta=\operatorname{Arg}(z_2)-\operatorname{Arg}(z_1)

    Axis arguments

    Arg⁡(a)=0 (a>0),Arg⁡(a)=π (a<0)\operatorname{Arg}(a)=0\ (a>0),\quad \operatorname{Arg}(a)=\pi\ (a<0)

    Imaginary-axis arguments

    Arg⁡(bi)=π2 (b>0),Arg⁡(bi)=−π2 (b<0)\operatorname{Arg}(bi)=\frac{\pi}{2}\ (b>0),\quad \operatorname{Arg}(bi)=-\frac{\pi}{2}\ (b<0)

    Positive-modulus conversion

    −rcis⁡θ=rcis⁡(θ+π)-r\operatorname{cis}\theta=r\operatorname{cis}(\theta+\pi)

    Operation choice

    add or subtract→Cartesian,multiply or divide→polar\text{add or subtract}\to\text{Cartesian},\qquad \text{multiply or divide}\to\text{polar}

    Quadrant-aware argument

    θ=atan2⁡(y,x)\theta=\operatorname{atan2}(y,x)

    Post-operation check

    ∣z1z2∣=∣z1∣∣z2∣,arg⁡(z1z2)=arg⁡z1+arg⁡z2(mod2π)|z_1z_2|=|z_1||z_2|,\qquad \arg(z_1z_2)=\arg z_1+\arg z_2\pmod{2\pi}

    Principal normalization

    θ↦θ+2kπ∈(−π,π]\theta\mapsto\theta+2k\pi\in(-\pi,\pi]

    Conversion-operation workflow

    identify form→find r,θ→operate→normalize→verify\text{identify form}\to\text{find }r,\theta\to\text{operate}\to\text{normalize}\to\text{verify}

    Modulus

    r=x2+y2r=\sqrt{x^2+y^2}

    Product

    z1z2=r1r2cis⁡(θ1+θ2)z_1z_2=r_1r_2\operatorname{cis}(\theta_1+\theta_2)

    Quotient

    z1z2=r1r2cis⁡(θ1−θ2)\frac{z_1}{z_2}=\frac{r_1}{r_2}\operatorname{cis}(\theta_1-\theta_2)
    Topic 1.14Not started

    Matrices

    Matrices provide a compact way to represent transformations, simultaneous equations and repeated-state systems. They become especially important in graph theory, Markov chains and differential-equation models.

    330 min estimated

    Topic 1.14 progress0%
    Learning objectives
    • Interpret matrix notation, dimensions, elements, rows and columns in mathematical and applied contexts.
    • Perform matrix addition, subtraction and scalar multiplication with compatible matrices.
    • Multiply matrices using row-column products and justify the order from dimensions and context.
    • Use zero and identity matrices and calculate powers of square matrices.
    • Calculate determinants and identify singular and non-singular matrices.
    • Find and verify inverse matrices, including the exact two-by-two formula.
    • Solve matrix equations and systems of two or three linear equations.
    • Interpret two-dimensional matrix transformations and their composition.
    • Construct and interpret transition, production and repeated-state matrix models.
    • Use calculator/GDC matrix functions accurately while showing mathematical setup and contextual conclusions.
    Key formulas

    General matrix notation

    A=(aij)m×nA=(a_{ij})_{m\times n}

    Example element

    aij=entry in row i and column ja_{ij}=\text{entry in row }i\text{ and column }j

    Dimensions

    Am×n=matrix with m rows and n columnsA_{m\times n}=\text{matrix with }m\text{ rows and }n\text{ columns}

    Matrix addition

    (A+B)ij=aij+bij(A+B)_{ij}=a_{ij}+b_{ij}

    Matrix subtraction

    (A−B)ij=aij−bij(A-B)_{ij}=a_{ij}-b_{ij}

    Scalar multiplication

    (kA)ij=kaij(kA)_{ij}=k a_{ij}

    Additive identity

    A+O=AA+O=A

    Product dimensions

    Am×nBn×p=(AB)m×pA_{m\times n}B_{n\times p}=(AB)_{m\times p}

    Product entry

    (AB)ij=∑k=1naikbkj(AB)_{ij}=\sum\limits_{k=1}^{n}a_{ik}b_{kj}

    Identity matrix

    I2=(1001)I_2=\begin{pmatrix}1&0\\0&1\end{pmatrix}

    Identity property

    AI=IA=AAI=IA=A

    Matrix power

    An=AA⋯A⏟n factorsA^n=\underbrace{AA\cdots A}_{n\text{ factors}}

    Two-by-two determinant

    det⁡(abcd)=ad−bc\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc

    Invertibility condition

    A−1 exists  ⟺  det⁡(A)≠0A^{-1}\text{ exists}\iff\det(A)\ne0

    Product determinant

    det⁡(AB)=det⁡(A)det⁡(B)\det(AB)=\det(A)\det(B)

    Two-by-two inverse

    (abcd)−1=1ad−bc(d−b−ca)\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1}=\frac1{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}

    Inverse definition

    AA−1=A−1A=IAA^{-1}=A^{-1}A=I

    Inverse of a product

    (AB)−1=B−1A−1(AB)^{-1}=B^{-1}A^{-1}

    Left-sided unknown

    AX=B  ⟹  X=A−1BAX=B\implies X=A^{-1}B

    Right-sided unknown

    XA=B  ⟹  X=BA−1XA=B\implies X=BA^{-1}

    Matrix form of a system

    Ax=bA\mathbf{x}=\mathbf{b}

    Unique inverse solution

    x=A−1b,det⁡(A)≠0\mathbf{x}=A^{-1}\mathbf{b},\qquad\det(A)\ne0

    Augmented matrix

    [A∣b][A\mid\mathbf{b}]

    Elementary row replacement

    Ri←Ri+kRjR_i\leftarrow R_i+kR_j

    Row scaling

    Ri←kRi,k≠0R_i\leftarrow kR_i,\qquad k\ne0

    Coordinate transformation

    x′=Ax\mathbf{x}'=A\mathbf{x}

    Horizontal shear

    (x′y′)=(1k01)(xy)=(x+kyy)\begin{pmatrix}x'\\y'\end{pmatrix}=\begin{pmatrix}1&k\\0&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}x+ky\\y\end{pmatrix}

    Composition order

    x′′=B(Ax)=(BA)x\mathbf{x}''=B(A\mathbf{x})=(BA)\mathbf{x}

    One-step state model

    xn+1=Txn\mathbf{x}_{n+1}=T\mathbf{x}_n

    Repeated state model

    xn=Tnx0\mathbf{x}_n=T^n\mathbf{x}_0

    Steady-state condition

    Ts=sT\mathbf{s}=\mathbf{s}

    Technology system check

    A(A−1b)=bA(A^{-1}\mathbf b)=\mathbf b

    Repeated GDC model

    xn=Tnx0\mathbf{x}_n=T^n\mathbf{x}_0

    Core workflow

    dimensions→operation→condition→calculation→interpretation\text{dimensions}\to\text{operation}\to\text{condition}\to\text{calculation}\to\text{interpretation}

    Practice workflow

    check size→choose operation→calculate→verify→interpret\text{check size}\to\text{choose operation}\to\text{calculate}\to\text{verify}\to\text{interpret}

    Product compatibility

    Am×nBn×p=(AB)m×pA_{m\times n}B_{n\times p}=(AB)_{m\times p}

    Determinant

    det⁡(abcd)=ad−bc\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc

    Inverse

    A−1=1ad−bc(d−b−ca)A^{-1}=\frac1{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}

    System solution

    x=A−1b\mathbf{x}=A^{-1}\mathbf b

    Transformation

    x′=Ax\mathbf{x}'=A\mathbf{x}

    Repeated model

    xn=Tnx0\mathbf{x}_n=T^n\mathbf{x}_0
    Topic 1.15Not started

    Eigenvalues, eigenvectors and diagonalization

    Eigenvectors identify directions or states that are only scaled by a transformation. In AI HL they support powers of matrices and later help interpret coupled systems and long-term behaviour.

    360 min estimated

    Topic 1.15 progress0%
    Learning objectives
    • Interpret eigenvalues and eigenvectors through the equation A v = λv and matrix geometry.
    • Form and solve the characteristic equation to find exact eigenvalues.
    • Find and verify corresponding eigenvectors and describe eigenspaces.
    • Distinguish algebraic and geometric multiplicity for repeated eigenvalues.
    • Determine whether a matrix has enough independent eigenvectors to be diagonalizable.
    • Construct and verify a diagonalization A=PDP−1A = PDP^{-1} with correctly matched columns and diagonal entries.
    • Use diagonalization to calculate powers of matrices efficiently.
    • Decompose an initial state into eigencomponents and analyze repeated transformations.
    • Interpret eigenvalues in dynamic, stability and transition-matrix models.
    • Use calculator/GDC eigenvalue and matrix methods while preserving exact reasoning and contextual conclusions.
    Key formulas

    Eigenvector equation

    Av=λv,v≠0A\mathbf v=\lambda\mathbf v,\qquad\mathbf v\ne\mathbf0

    Equivalent homogeneous system

    (A−λI)v=0(A-\lambda I)\mathbf v=\mathbf0

    Length scaling

    ∥Av∥=∣λ∣ ∥v∥\|A\mathbf v\|=|\lambda|\,\|\mathbf v\|

    Repeated eigenvector action

    Anv=λnvA^n\mathbf v=\lambda^n\mathbf v

    Characteristic equation

    det⁡(A−λI)=0\det(A-\lambda I)=0

    Two-by-two characteristic polynomial

    λ2−tr⁡(A)λ+det⁡(A)=0\lambda^2-\operatorname{tr}(A)\lambda+\det(A)=0

    Eigenvalue checks

    λ1+λ2=tr⁡(A),λ1λ2=det⁡(A)\lambda_1+\lambda_2=\operatorname{tr}(A),\qquad\lambda_1\lambda_2=\det(A)

    Two-by-two expansion

    det⁡(A−λI)=(a−λ)(d−λ)−bc\det(A-\lambda I)=(a-\lambda)(d-\lambda)-bc

    Quadratic formula

    λ=−b±b2−4ac2a\lambda=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

    Eigenvector system

    (A−λI)v=0(A-\lambda I)\mathbf v=\mathbf0

    Eigenspace

    Eλ=ker⁡(A−λI)E_{\lambda}=\ker(A-\lambda I)

    Eigenspace span

    Eλ=span⁡{v1,…,vk}E_{\lambda}=\operatorname{span}\{\mathbf v_1,\ldots,\mathbf v_k\}

    Multiplicity bound

    1≤mg(λ)≤ma(λ)1\le m_g(\lambda)\le m_a(\lambda)

    Diagonalizability count

    dim⁡(Eλ1)+⋯+dim⁡(Eλk)=n\dim(E_{\lambda_1})+\cdots+\dim(E_{\lambda_k})=n

    Multiplicity condition

    mg(λ)=ma(λ) for every eigenvaluem_g(\lambda)=m_a(\lambda)\text{ for every eigenvalue}

    Diagonalization

    A=PDP−1A=PDP^{-1}

    Equivalent relation

    AP=PDAP=PD

    Eigenvector and eigenvalue arrangement

    P=[v1 ⋯ vn],D=diag⁡(λ1,…,λn)P=[\mathbf v_1\ \cdots\ \mathbf v_n],\qquad D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n)

    Construction

    P=[v1 ⋯ vn],D=diag⁡(λ1,…,λn)P=[\mathbf v_1\ \cdots\ \mathbf v_n],\qquad D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n)

    Reconstruction check

    PDP−1=APDP^{-1}=A

    Power of a diagonalizable matrix

    An=PDnP−1A^n=PD^nP^{-1}

    Power of D

    Dn=diag⁡(λ1n,…,λnn)D^n=\operatorname{diag}(\lambda_1^n,\ldots,\lambda_n^n)

    Eigenvector power rule

    Anv=λnvA^n\mathbf v=\lambda^n\mathbf v

    Eigenbasis coordinates

    c=P−1x0\mathbf c=P^{-1}\mathbf{x}_0

    State decomposition

    x0=Pc=∑i=1kcivi\mathbf{x}_0=P\mathbf c=\sum\limits_{i=1}^{k}c_i\mathbf v_i

    Repeated state

    Anx0=∑i=1kciλinviA^n\mathbf{x}_0=\sum\limits_{i=1}^{k}c_i\lambda_i^n\mathbf v_i

    Discrete iteration

    xn+1=Axn\mathbf{x}_{n+1}=A\mathbf{x}_n

    State after n steps

    xn=Anx0\mathbf{x}_n=A^n\mathbf{x}_0

    Spectral radius

    ρ(A)=max⁡i∣λi∣\rho(A)=\max_i|\lambda_i|

    Discrete asymptotic stability

    ρ(A)<1\rho(A)<1

    Steady-state eigen-equation

    Ts=sT\mathbf{s}=\mathbf{s}

    Normalized stationary state

    ∑i=1ksi=1\sum\limits_{i=1}^{k}s_i=1

    Steady plus transient form

    xn=xsteady+cλtransientnv\mathbf{x}_n=\mathbf{x}_{\mathrm{steady}}+c\lambda_{\mathrm{transient}}^n\mathbf v

    Eigenpair residual

    r=Av−λv\mathbf r=A\mathbf v-\lambda\mathbf v

    Diagonalization check

    AP=PDAP=PD

    Solution workflow

    det⁡(A−λI)=0→λ→(A−λI)v=0→P,D→verify→apply\det(A-\lambda I)=0\to\lambda\to(A-\lambda I)\mathbf v=0\to P,D\to\text{verify}\to\text{apply}

    Practice workflow

    A→det⁡(A−λI)=0→λi→vi→P,D→power or modelA\to\det(A-\lambda I)=0\to\lambda_i\to\mathbf v_i\to P,D\to\text{power or model}

    Eigenpair

    Av=λvA\mathbf v=\lambda\mathbf v

    Eigenvectors

    (A−λI)v=0(A-\lambda I)\mathbf v=\mathbf0

    Power

    An=PDnP−1A^n=PD^nP^{-1}

    Repeated state

    xn=Anx0\mathbf{x}_n=A^n\mathbf{x}_0
  2. Unit 2
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    Functions

    Functions are the modelling language of AI. The shared course covers straight lines, graph interpretation and common real-world models; HL extends this into composite and inverse functions, transformations, logistic and logarithmic models, piecewise structures and logarithmic linearization.

    10 topics

    Unit 2 progress0%
  3. Unit 3
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    Geometry and Trigonometry

    AI HL begins with measurement, triangle trigonometry, circles, perpendicular bisectors and Voronoi diagrams, then extends into unit-circle ideas, transformation matrices, vectors, kinematics and a full graph-theory/network-algorithm strand.

    16 topics

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  4. Unit 4
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    Statistics and Probability

    Statistics and Probability is the largest current AI HL content area. The shared course develops data presentation, regression, distributions, Spearman correlation and basic hypothesis testing. HL then adds study design, nonlinear regression, transformed random variables, the Central Limit Theorem, confidence intervals, Poisson models, advanced hypothesis testing and Markov chains.

    19 topics

    Unit 4 progress0%
  5. Unit 5
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    Calculus

    AI calculus begins with rates of change, polynomial differentiation, basic integration and optimization, then extends at HL into standard derivative rules, second derivatives, richer integration, volumes, kinematics, differential equations, slope fields, Euler approximation and phase portraits for coupled systems.

    17 topics

    Unit 5 progress0%