i

Test format & strategy

🗻 ACT International Subject Test — Math 2 Content Areas

Content Area	% of Test	Key Topics
Algebra II	~50%	Complex numbers, matrices & vectors, sequences & series, advanced equations/inequalities
Precalculus	~50%	Advanced functions, limits, trigonometry, polar coordinates, an introduction to statistics

ACT 2 Math Level 2 corresponds to the ACT International Subject Test (AIST) — Mathematics 2, the more advanced of ACT Inc.'s two international subject tests. Where Level 1 (AIST Math 1) covers first-year algebra and geometry, Level 2 assumes that material is mastered and moves into the topics that separate a solid Algebra II student from a precalculus-ready one: complex numbers, matrices, limits, and the polar/trigonometric machinery that leads into calculus.

What to expect

The real AIST Math 2 test is 50 multiple-choice questions in 60 minutes, split roughly evenly between Algebra II and precalculus content. A calculator is permitted, and a reference sheet of common formulas is provided in the test booklet — so this course leans on understanding why a formula works, not just memorizing it cold.

Prerequisite

ACT 2 Math Level 1 (AIST Math 1)

Level

Algebra II & Precalculus

Real exam format

50 MCQ · 60 minutes

Calculator

permitted throughout

▲ Rigor & Problem-Solving Strategy

AIST Math Level 2 questions test conceptual fluency rather than rote execution. Success requires two disciplines:

Algebraic Rigor: Show complete intermediate steps for every equation; avoid unverified shortcuts.
Verification Check: Plug found roots back into original rational and logarithmic expressions to eliminate extraneous roots immediately.

⚠ ACT Subject Test Traps & Calculator Hazards

No Formula Sheet: Unlike SAT Math, ACT Subject Tests do not provide a reference sheet. Double-angle identities, Law of Cosines, arithmetic/geometric series sum formulas, and the matrix inverse formula must be known cold.

Calculator Trap: Graphing calculators can obscure vertical asymptotes as lines or hide removable discontinuities (holes). Always analyze denominators algebraically before relying on graphs.

Pacing: 50 questions in 60 minutes allows exactly 72 seconds per problem. Checkpoint: 25 questions completed by the 30-minute mark. Never leave questions blank because there is no negative marking.

ACT 2 Math Level 2 domain weightings: Algebra II (~50%) and Precalculus (~50%).

A student spends 90 seconds on each of the first 20 questions of the 50-question AIST Math 2 exam. How much average time (in seconds) remains per question for the remaining 30 questions?

Total exam duration is

60\text{ minutes} = 3600\text{ seconds}

. The first 20 questions take

20 \times 90 = 1800\text{ seconds}

, leaving

3600 - 1800 = 1800\text{ seconds}

for the remaining 30 questions. The remaining time per question is

\dfrac{1800}{30} = 60\text{ seconds}

(1 minute per question). Pacing discipline early on prevents a frantic finish.

1

Complex Numbers

Every quadratic $ax^2+bx+c=0$ with real coefficients has solutions — as long as we're willing to accept square roots of negative numbers. Complex numbers make that promise precise: they extend $\mathbb{R}$ just enough that every polynomial equation gets a solution, using one new symbol, $i$ , and the ordinary rules of algebra everywhere else.

● Definition 1.1 — The imaginary unit and complex numbers

The imaginary unit $i$ satisfies $i^2=-1$ . A complex number is any expression $z=a+bi$ with $a,b\in\mathbb{R}$ : $a$ is the real part, written $\operatorname{Re}(z)$ , and $b$ is the imaginary part, written $\operatorname{Im}(z)$ . When $b=0$ , $z$ is an ordinary real number; when $a=0$ and $b\neq0$ , $z$ is called purely imaginary.

Complex numbers add and multiply exactly like binomials in $a+bx$ form — the only extra rule is $i^2=-1$ , which you use to simplify after multiplying out.

● Definition 1.2 — Addition and multiplication

For $z_1=a+bi$ and $z_2=c+di$ :

$z_1+z_2=(a+c)+(b+d)i$

$z_1 z_2=(ac-bd)+(ad+bc)i$

The multiplication rule is just FOIL followed by $i^2\to -1$ : $(a+bi)(c+di)=ac+adi+bci+bdi^2=(ac-bd)+(ad+bc)i$ .

Worked example 1.A — Add and multiply

Let $z_1=3+4i$ and $z_2=1-2i$ . Find $z_1+z_2$ and $z_1z_2$ .

$z_1+z_2=(3+1)+(4-2)i=4+2i.$

$z_1z_2=(3)(1)-(4)(-2)+\big[(3)(-2)+(4)(1)\big]i=(3+8)+(-6+4)i=11-2i.$

● Definition 1.3 — Conjugate and modulus

The complex conjugate of $z=a+bi$ is $\bar z=a-bi$ — same real part, opposite sign on the imaginary part. The modulus (or absolute value) of $z$ is $|z|=\sqrt{a^2+b^2},$ the distance from $z$ to the origin when $z$ is plotted as the point $(a,b)$ in the complex plane.

▲ Theorem 1.1 —

z\bar z=|z|^2

For every complex number $z=a+bi$ , $z\bar z=|z|^2.$ In particular $z\bar z$ is always a nonnegative real number.

Proof

Write $z=a+bi$ , so $\bar z=a-bi$ . Multiplying directly (Definition 1.2, with $c=a,\,d=-b$ ):

$z\bar z=(a+bi)(a-bi)=\big[a\cdot a-b\cdot(-b)\big]+\big[a\cdot(-b)+b\cdot a\big]i=(a^2+b^2)+0i=a^2+b^2.$

Since $|z|=\sqrt{a^2+b^2}$ , squaring both sides gives $|z|^2=a^2+b^2$ , which is exactly what the product $z\bar z$ equals. Hence $z\bar z=|z|^2$ , and this is nonnegative since it's a sum of two squares. $\blacksquare$

Theorem 1.1 is exactly what makes complex division possible: to divide by $z_2$ , multiply top and bottom by $\bar z_2$ so the denominator becomes the real number $|z_2|^2$ .

Worked example 1.B — Dividing complex numbers

Simplify $\dfrac{5+i}{2-3i}$ .

Multiply numerator and denominator by the conjugate of the denominator, $2+3i$ :

$\frac{5+i}{2-3i}\cdot\frac{2+3i}{2+3i}=\frac{(5+i)(2+3i)}{(2-3i)(2+3i)}.$

Numerator: $(5)(2)-(1)(3)+\big[(5)(3)+(1)(2)\big]i=10-3+17i=7+17i$ . Denominator, by Theorem 1.1: $|2-3i|^2=2^2+3^2=13$ . So

$\frac{5+i}{2-3i}=\frac{7+17i}{13}=\frac{7}{13}+\frac{17}{13}i.$

⊙ Explorer · z, z̄, and z·z̄ = |z|²interactive

real part a: 3.0

imaginary part b: 4.0

|z|5.000

z̄3.00 − 4.00i

z · z̄ (= |z|²)25.000

The indigo arrow is z; the amber arrow is its conjugate z̄, always the mirror image of z across the real axis. z·z̄ always comes out real and nonnegative — Theorem 1.1 in action.

Quick reference

Powers of i

i^1=i,\ i^2=-1,\ i^3=-i,\ i^4=1

, then repeats

Sum of conjugates

z+\bar z=2\operatorname{Re}(z)

, always real

Difference of conjugates

z-\bar z=2i\operatorname{Im}(z)

, always purely imaginary

Geometric representation of a complex number z = a + bi in the Argand plane, showing real part a, imaginary part b, and modulus |z| = √(a² + b²).

⚠ Exam Traps in Complex Arithmetic

Modulus with 'i': A frequent error is writing

|a+bi| = \sqrt{a^2 + (bi)^2} = \sqrt{a^2 - b^2}

. The imaginary unit

i

is never included inside the modulus radicand:

|a+bi| = \sqrt{a^2 + b^2}

.

Powers of i: For large exponents

i^n

, evaluate

n \pmod 4

:

n \equiv 0 \implies 1

,

n \equiv 1 \implies i

,

n \equiv 2 \implies -1

,

n \equiv 3 \implies -i

. For negative powers,

i^{-1} = -i

.

Division by Binomials: Never invert terms individually (

1/(2+3i) \neq 1/2 + 1/(3i)

). Always multiply numerator and denominator by the conjugate

2-3i

.

Given

z = \dfrac{4 + 7i}{1 - 2i}

, express

z

in standard form

a + bi

and compute

|z|^2

.

Multiply numerator and denominator by the conjugate

1 + 2i

:

z = \frac{(4+7i)(1+2i)}{(1-2i)(1+2i)} = \frac{4(1) - 7(2) + [4(2) + 7(1)]i}{1^2 + (-2)^2} = \frac{4 - 14 + (8+7)i}{5} = \frac{-10 + 15i}{5} = -2 + 3i.

Then

|z|^2 = (-2)^2 + 3^2 = 4 + 9 = 13

. (Alternatively, by modulus properties:

|z|^2 = \dfrac{|4+7i|^2}{|1-2i|^2} = \dfrac{16+49}{1+4} = \dfrac{65}{5} = 13

).

📝 Chapter Quiz

Questions Level

2

Matrices & Vectors

A matrix is just a rectangular table of numbers — but that simple idea packs an entire system of linear equations into one object you can add, scale, and multiply. This chapter builds the algebra of $2\times2$ matrices from scratch, then uses it to solve linear systems a genuinely new way, and closes with the vector tools (magnitude, dot product, angle) AIST Math 2 pairs with it.

● Definition 2.1 — Matrix, addition, and scalar multiplication

An $m\times n$ matrix is a rectangular array of numbers with $m$ rows and $n$ columns, written $A=[a_{ij}]$ where $a_{ij}$ is the entry in row $i$ , column $j$ . Two matrices of the same dimensions add entrywise, and a scalar $k$ multiplies every entry:

$(A+B)_{ij}=a_{ij}+b_{ij}, \qquad (kA)_{ij}=k\,a_{ij}.$

Worked example 2.A — Add and scale

Let $A=\begin{bmatrix}3&-2\\5&1\end{bmatrix}$ and $B=\begin{bmatrix}-4&6\\2&-3\end{bmatrix}$ . Find $A+B$ and $3A$ .

$A+B=\begin{bmatrix}3+(-4)&-2+6\\5+2&1+(-3)\end{bmatrix}=\begin{bmatrix}-1&4\\7&-2\end{bmatrix}.$

$3A=\begin{bmatrix}9&-6\\15&3\end{bmatrix}.$

● Definition 2.2 — Matrix multiplication

If $A$ is $m\times n$ and $B$ is $n\times p$ (the inner dimensions must match), the product $AB$ is the $m\times p$ matrix whose $(i,j)$ entry is the dot product of row $i$ of $A$ with column $j$ of $B$ :

$(AB)_{ij}=\sum_{k=1}^{n} a_{ik}b_{kj}.$

Matrix multiplication is not commutative in general — $AB\neq BA$ even when both products are defined.

Worked example 2.B — Multiply two matrices, and a matrix by a vector

Let $A=\begin{bmatrix}2&1\\0&3\end{bmatrix}$ , $B=\begin{bmatrix}4&-1\\2&5\end{bmatrix}$ , and $\mathbf{v}=\begin{bmatrix}5\\-2\end{bmatrix}$ .

$AB=\begin{bmatrix}(2)(4)+(1)(2)&(2)(-1)+(1)(5)\\(0)(4)+(3)(2)&(0)(-1)+(3)(5)\end{bmatrix}=\begin{bmatrix}10&3\\6&15\end{bmatrix}.$

$A\mathbf{v}=\begin{bmatrix}(2)(5)+(1)(-2)\\(0)(5)+(3)(-2)\end{bmatrix}=\begin{bmatrix}8\\-6\end{bmatrix}.$

● Definition 2.3 — Determinant and the identity matrix

For a $2\times2$ matrix $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$ , the determinant is the scalar $\det A=ad-bc.$ The $2\times2$ identity matrix $I=\begin{bmatrix}1&0\\0&1\end{bmatrix}$ satisfies $IA=AI=A$ for every $2\times2$ matrix $A$ — it's the multiplicative identity.

▲ Theorem 2.1 — The

2\times2

inverse formula

If $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$ and $\det A=ad-bc\neq0$ , then $A$ is invertible, with

$A^{-1}=\frac{1}{\det A}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}.$

If $\det A=0$ , no inverse exists.

Proof

Let $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$ and $d_0=ad-bc\neq0$ . Multiply $A$ by the claimed inverse $\tfrac{1}{d_0}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$ (Definition 2.2):

$A\cdot\frac{1}{d_0}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}=\frac{1}{d_0}\begin{bmatrix}ad-bc&-ab+ba\\cd-dc&-cb+da\end{bmatrix}=\frac{1}{d_0}\begin{bmatrix}d_0&0\\0&d_0\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}=I.$

The same computation in the other order gives $\tfrac{1}{d_0}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}\cdot A=I$ as well, so this matrix is a genuine two-sided inverse. When $\det A=0$ , suppose toward contradiction some matrix $B$ satisfied $AB=I$ ; a short computation (or the fact that $\det(AB)=\det A\cdot\det B$ ) would force $\det A\cdot\det B=\det I=1$ , impossible since $\det A=0$ makes the left side $0$ . So no inverse exists. $\blacksquare$

Worked example 2.C — Invert a matrix and solve a system with it

Solve $\begin{cases}2x+3y=16\\x-y=-2\end{cases}$ by writing it as $A\mathbf{x}=\mathbf{b}$ and inverting $A$ .

Here $A=\begin{bmatrix}2&3\\1&-1\end{bmatrix}$ , $\mathbf{b}=\begin{bmatrix}16\\-2\end{bmatrix}$ . $\det A=(2)(-1)-(3)(1)=-5$ , so by Theorem 2.1:

$A^{-1}=\frac{1}{-5}\begin{bmatrix}-1&-3\\-1&2\end{bmatrix}=\begin{bmatrix}\tfrac{1}{5}&\tfrac{3}{5}\\[2pt]\tfrac{1}{5}&-\tfrac{2}{5}\end{bmatrix}.$

Then $\mathbf{x}=A^{-1}\mathbf{b}$ : $x=\tfrac{1}{5}(16)+\tfrac{3}{5}(-2)=\tfrac{16-6}{5}=2, \qquad y=\tfrac{1}{5}(16)+\left(-\tfrac{2}{5}\right)(-2)=\tfrac{16+4}{5}=4.$ Check: $2(2)+3(4)=4+12=16$ ✓ and $2-4=-2$ ✓.

⊙ Explorer · Determinant as signed areainteractive

a: 2.0

b: 3.0

c: 1.0

d: -1.0

det A = ad − bc-5.000

A is...invertible

A⁻¹ (rows)(0.200, 0.600), (0.200, -0.400)

The indigo arrow is column 1 of A (a, c); the amber arrow is column 2 (b, d). The shaded parallelogram they span has area |det A| — amber-tinted when det A < 0, indigo when det A > 0. By Theorem 2.1, A has an inverse exactly when det A ≠ 0.

● Definition 2.4 — Vectors: magnitude and dot product

A vector $\mathbf{u}=\langle u_1,u_2\rangle$ in the plane has magnitude $|\mathbf{u}|=\sqrt{u_1^2+u_2^2}$ . The dot product of $\mathbf{u}=\langle u_1,u_2\rangle$ and $\mathbf{v}=\langle v_1,v_2\rangle$ is the scalar $\mathbf{u}\cdot\mathbf{v}=u_1v_1+u_2v_2.$

▲ Theorem 2.2 — Angle between two vectors

For nonzero vectors $\mathbf{u},\mathbf{v}$ with angle $\theta$ between them, $\cos\theta=\frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}|\,|\mathbf{v}|}.$

Proof

Place $\mathbf{u}$ and $\mathbf{v}$ tail-to-tail; the vector from the tip of $\mathbf{u}$ to the tip of $\mathbf{v}$ is $\mathbf{v}-\mathbf{u}$ , and the three sides $|\mathbf{u}|,|\mathbf{v}|,|\mathbf{v}-\mathbf{u}|$ form a triangle with included angle $\theta$ at the shared tail. The Law of Cosines gives $|\mathbf{v}-\mathbf{u}|^2=|\mathbf{u}|^2+|\mathbf{v}|^2-2|\mathbf{u}||\mathbf{v}|\cos\theta.$ Expand the left side using $|\mathbf{v}-\mathbf{u}|^2=(\mathbf{v}-\mathbf{u})\cdot(\mathbf{v}-\mathbf{u})=|\mathbf{v}|^2-2\,\mathbf{u}\cdot\mathbf{v}+|\mathbf{u}|^2$ (dot product distributes like ordinary multiplication). Setting the two expressions for $|\mathbf{v}-\mathbf{u}|^2$ equal and cancelling $|\mathbf{u}|^2+|\mathbf{v}|^2$ from both sides leaves $-2\,\mathbf{u}\cdot\mathbf{v}=-2|\mathbf{u}||\mathbf{v}|\cos\theta$ , i.e. $\cos\theta=\dfrac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}||\mathbf{v}|}$ . $\blacksquare$

Worked example 2.D — Find the angle between two vectors

Find the angle between $\mathbf{u}=\langle4,3\rangle$ and $\mathbf{v}=\langle-1,2\rangle$ .

$\mathbf{u}\cdot\mathbf{v}=(4)(-1)+(3)(2)=-4+6=2. \qquad |\mathbf{u}|=\sqrt{16+9}=5, \qquad |\mathbf{v}|=\sqrt{1+4}=\sqrt5.$

$\cos\theta=\frac{2}{5\sqrt5}\approx0.1789 \quad\Rightarrow\quad \theta=\cos^{-1}(0.1789)\approx79.7^\circ.$

Quick reference

Determinant test

A

invertible

\iff \det A\neq0

2×2 inverse

swap diagonal, negate off-diagonal, divide by

\det A

Perpendicular vectors

\mathbf{u}\cdot\mathbf{v}=0 \iff \mathbf{u}\perp\mathbf{v}

(for nonzero vectors)

Geometric vector triangle representation: vectors u and v placed tail-to-tail, with included angle θ and difference vector v - u.

⚠ Matrix & Vector Exam Traps

Matrix Multiplication is Non-Commutative: In general,

AB \neq BA

. For dimensions

m \times n

and

n \times p

, the inner dimensions must match to multiply. Swapping them may not even be defined.

Inverse Invertibility Condition: A matrix

A = \begin{pmatrix}a & b\\ c & d\end{pmatrix}

is invertible if and only if

\det(A) = ad - bc \neq 0

. If

\det(A) = 0

, the system has either no unique solution or infinitely many solutions.

Dot Product vs Vector: The dot product

\mathbf{u}\cdot\mathbf{v}

is a scalar (a single real number), not a vector. The vectors are orthogonal if and only if

\mathbf{u}\cdot\mathbf{v} = 0

.

Let

A = \begin{pmatrix}3 & 1\\ 5 & 2\end{pmatrix}

. Find

A^{-1}

and use it to solve the linear system:

3x + y = 7

5x + 2y = 12

1. Compute

\det(A) = (3)(2) - (1)(5) = 6 - 5 = 1

.
2. Form

A^{-1}

: swap main diagonal entries, negate off-diagonal entries, and divide by

\det(A) = 1

:

A^{-1} = \begin{pmatrix}2 & -1\\ -5 & 3\end{pmatrix}.

3. Multiply

A^{-1}\begin{pmatrix}7\\ 12\end{pmatrix}

:

\begin{pmatrix}x\\ y\end{pmatrix} = \begin{pmatrix}2 & -1\\ -5 & 3\end{pmatrix}\begin{pmatrix}7\\ 12\end{pmatrix} = \begin{pmatrix}2(7) + (-1)(12)\\ -5(7) + 3(12)\end{pmatrix} = \begin{pmatrix}14 - 12\\ -35 + 36\end{pmatrix} = \begin{pmatrix}2\\ 1\end{pmatrix}.

Thus

x = 2

and

y = 1

.

📝 Chapter Quiz

Questions Level

3

Sequences & Series

A sequence is just an ordered list of numbers generated by a rule; a series is what you get when you add the terms of a sequence together. AIST Math 2 tests two specific families — arithmetic (constant difference) and geometric (constant ratio) — and one genuinely infinite idea: a sum with infinitely many terms that still adds up to a finite number.

● Definition 3.1 — Arithmetic sequence

A sequence $a_1,a_2,a_3,\dots$ is arithmetic if consecutive terms differ by a constant $d$ , the common difference: $a_{n+1}=a_n+d$ . Its explicit (closed-form) formula is

$a_n=a_1+(n-1)d.$

Worked example 3.A — Find a term

An arithmetic sequence has $a_1=5$ and $d=3$ . Find $a_{10}$ .

$a_{10}=5+3(10-1)=5+27=32.$

● Definition 3.2 — Arithmetic series

An arithmetic series is the sum of an arithmetic sequence's first $n$ terms. Pairing the first term with the last, the second with the second-to-last, and so on (each pair sums to $a_1+a_n$ ) gives

$S_n=\frac{n}{2}\big(a_1+a_n\big)=\frac{n}{2}\big(2a_1+(n-1)d\big).$

Worked example 3.B — Sum an arithmetic series

Find the sum of the first $10$ terms of the sequence from Example 3.A ( $a_1=5,\,d=3$ ).

$S_{10}=\frac{10}{2}\big(2(5)+9(3)\big)=5(10+27)=5(37)=185.$

● Definition 3.3 — Geometric sequence

A sequence is geometric if consecutive terms share a constant ratio $r\neq0$ : $a_{n+1}=r\,a_n$ . Its explicit formula is

$a_n=a_1\,r^{\,n-1}.$

Worked example 3.C — Find a geometric term

A geometric sequence has $a_1=3$ and $r=2$ . Find $a_7$ .

$a_7=3\cdot2^{6}=3\cdot64=192.$

▲ Theorem 3.1 — Finite geometric series

For $r\neq1$ , the sum of the first $n$ terms of a geometric sequence is

$S_n=\frac{a_1(r^n-1)}{r-1}.$

Proof

Write out $S_n=a_1+a_1r+a_1r^2+\cdots+a_1r^{n-1}$ . Multiply both sides by $r$ : $rS_n=a_1r+a_1r^2+\cdots+a_1r^{n-1}+a_1r^{n}.$ Subtract the first equation from the second — every middle term cancels (this is why the trick works), leaving $rS_n-S_n=a_1r^{n}-a_1.$ Factor: $S_n(r-1)=a_1(r^n-1)$ , and since $r\neq1$ we can divide by $r-1$ to get $S_n=\dfrac{a_1(r^n-1)}{r-1}$ . $\blacksquare$

Worked example 3.D — Sum a geometric series

Find $S_7$ for the sequence from Example 3.C ( $a_1=3,\,r=2$ ).

$S_7=\frac{3(2^7-1)}{2-1}=3(128-1)=3(127)=381.$

▲ Theorem 3.2 — Infinite geometric series

If $|r|<1$ , the infinite geometric series $a_1+a_1r+a_1r^2+\cdots$ converges to

$S_\infty=\frac{a_1}{1-r}.$ If $|r|\geq1$ (and $a_1\neq0$ ), the series diverges — it has no finite sum.

Proof

By Theorem 3.1, $S_n=\dfrac{a_1(r^n-1)}{r-1}=\dfrac{a_1}{1-r}\big(1-r^n\big)$ (multiplying numerator and denominator by $-1$ ). When $|r|<1$ , repeated multiplication by a number smaller than $1$ in absolute value shrinks $r^n\to0$ as $n\to\infty$ (e.g. $r=\tfrac12$ gives $\tfrac12,\tfrac14,\tfrac18,\dots\to0$ ). So $S_n\to\dfrac{a_1}{1-r}(1-0)=\dfrac{a_1}{1-r}$ , which is exactly $S_\infty$ . When $|r|\geq1$ , $r^n$ does not shrink to $0$ (it stays constant at $|a_1|$ or grows without bound), so $S_n$ never settles on a finite value. $\blacksquare$

Worked example 3.E — Sum an infinite geometric series

Find $8+4+2+1+\cdots$ .

Here $a_1=8$ and $r=\tfrac12$ (each term is half the last), and $|r|=\tfrac12<1$ , so Theorem 3.2 applies: $S_\infty=\frac{8}{1-\tfrac12}=\frac{8}{\tfrac12}=16.$

Quick reference

Sigma notation

\displaystyle\sum_{k=1}^{n}a_k

means

a_1+a_2+\cdots+a_n

Arithmetic

a_n=a_1+(n-1)d,\quad S_n=\tfrac{n}{2}(a_1+a_n)

Geometric

a_n=a_1r^{n-1},\quad S_n=\tfrac{a_1(r^n-1)}{r-1},\quad S_\infty=\tfrac{a_1}{1-r}\ (|r|<1)

Successive terms of a converging geometric sequence a_n = 16 · (1/2)^(n-1), halving at each step and converging to 0.

⚠ Sequences & Series Exam Traps

Convergence Restriction: The infinite geometric sum formula

S_\infty = \dfrac{a_1}{1-r}

is only valid when

|r| < 1

. If

|r| \geq 1

, the series diverges and has no finite sum.

Term Index Off-by-One: Remember the exponent is

n-1

, not

n

:

a_n = a_1 r^{n-1}

and

a_n = a_1 + (n-1)d

. A common error is writing

a_5 = a_1 r^5

instead of

a_1 r^4

.

Arithmetic Term Count: When using

S_n = \dfrac{n}{2}(a_1 + a_n)

, ensure

n

is determined correctly via

n = \dfrac{a_n - a_1}{d} + 1

.

The second term of a geometric sequence is

12

and the fifth term is

\dfrac{32}{9}

. Find the common ratio

r

, the first term

a_1

, and the sum of the infinite series

S_\infty

.

1. Express the terms:

a_2 = a_1 r = 12

and

a_5 = a_1 r^4 = \dfrac{32}{9}

.
2. Divide

a_5

by

a_2

:

\frac{a_1 r^4}{a_1 r} = r^3 = \frac{32/9}{12} = \frac{32}{108} = \frac{8}{27}.

Taking the real cube root gives

r = \dfrac{2}{3}

.
3. Find

a_1

:

a_1 (2/3) = 12 \implies a_1 = 12 \times \dfrac{3}{2} = 18

.
4. Since

|r| = 2/3 < 1

, the infinite sum exists:

S_\infty = \frac{a_1}{1 - r} = \frac{18}{1 - 2/3} = \frac{18}{1/3} = 54.

📝 Chapter Quiz

Questions Level

4

Advanced Functions & Equations

This chapter covers two closely related ideas AIST Math 2 tests together: building new functions from old ones (composition, inverses), and solving equation types that go beyond linear/quadratic — rational, polynomial, exponential, and logarithmic.

● Definition 4.1 — Function composition

The composition of $f$ and $g$ , written $(f\circ g)(x)$ or $f(g(x))$ , means: apply $g$ first, then apply $f$ to the result. In general $f(g(x))\neq g(f(x))$ — order matters.

Worked example 4.A — Compose functions

Let $f(x)=2x+3$ and $g(x)=x^2-1$ . Find $f(g(2))$ and $g(f(1))$ .

$g(2)=2^2-1=3, \qquad f(g(2))=f(3)=2(3)+3=9.$

$f(1)=2(1)+3=5, \qquad g(f(1))=g(5)=5^2-1=24.$

Different order, different answer — exactly as the definition warns.

● Definition 4.2 — Inverse function

A function $f$ is one-to-one if it never sends two different inputs to the same output. Only one-to-one functions have an inverse $f^{-1}$ , the function that undoes $f$ : $f\big(f^{-1}(x)\big)=x \quad\text{and}\quad f^{-1}\big(f(x)\big)=x.$ To find $f^{-1}$ algebraically: write $y=f(x)$ , swap $x$ and $y$ , then solve for $y$ .

Worked example 4.B — Find and verify an inverse

Find $f^{-1}(x)$ for $f(x)=2x+3$ , and check it.

Write $y=2x+3$ ; swap: $x=2y+3$ ; solve: $y=\dfrac{x-3}{2}$ . So $f^{-1}(x)=\dfrac{x-3}{2}$ .

Check at $x=7$ : $f^{-1}(7)=\dfrac{7-3}{2}=2$ , and $f(2)=2(2)+3=7$ ✓ — back where we started.

● Definition 4.3 — Rational equations

A rational equation has a variable in a denominator. Solve by clearing denominators (multiply both sides by them), then solve the resulting polynomial equation — but always check each solution in the original equation, since a value that makes a denominator zero is an extraneous solution and must be rejected even if it satisfies the cleared version.

Worked example 4.C — Solve a rational equation

Solve $\dfrac{x+1}{x-2}=3$ (note $x\neq2$ ).

Multiply both sides by $x-2$ : $x+1=3(x-2)=3x-6$ . So $7=2x$ , $x=\dfrac{7}{2}$ . Since $\tfrac72\neq2$ , it's valid. Check: $\dfrac{\tfrac72+1}{\tfrac72-2}=\dfrac{\tfrac92}{\tfrac32}=3$ ✓.

▲ Theorem 4.1 — Factor Theorem

For a polynomial $f(x)$ and a number $c$ : $f(c)=0 \quad\iff\quad (x-c)\text{ is a factor of }f(x).$

Proof

( $\Leftarrow$ ) If $(x-c)$ is a factor, then $f(x)=(x-c)q(x)$ for some polynomial $q$ , so $f(c)=(c-c)q(c)=0$ .

( $\Rightarrow$ ) Polynomial division always lets us write $f(x)=(x-c)q(x)+r$ for some constant remainder $r$ (dividing by a linear factor leaves a constant remainder). Evaluate at $x=c$ : $f(c)=(c-c)q(c)+r=0+r=r$ . So if $f(c)=0$ , then $r=0$ , which means $f(x)=(x-c)q(x)$ exactly — $(x-c)$ is a factor. $\blacksquare$

Worked example 4.D — Factor a cubic completely

Factor $f(x)=x^3-4x^2+x+6$ completely.

Test small integer candidates: $f(-1)=(-1)^3-4(1)+(-1)+6=-1-4-1+6=0$ . By Theorem 4.1, $(x+1)$ is a factor. Dividing gives $f(x)=(x+1)(x^2-5x+6)$ , and the quadratic factors further: $x^2-5x+6=(x-2)(x-3)$ . So

$f(x)=(x+1)(x-2)(x-3).$

Check: expanding $(x+1)(x-2)(x-3)$ does give back $x^3-4x^2+x+6$ , and indeed $f(2)=8-16+2+6=0$ and $f(3)=27-36+3+6=0$ confirm the other two roots.

● Definition 4.4 — Exponential and logarithmic equations

An exponential equation has the variable in an exponent; solve by matching bases ( $b^{m}=b^{n}\Rightarrow m=n$ ) when possible. A logarithmic equation has the variable inside a log; the key identity is $\log_b x+\log_b y=\log_b(xy)$ , and every solution must be checked against the log's domain ( $x>0$ ).

Worked example 4.E — Solve an exponential equation

Solve $2^{x+1}=32$ .

Write $32=2^5$ , so $2^{x+1}=2^5$ . Matching exponents: $x+1=5$ , so $x=4$ .

Worked example 4.F — Solve a logarithmic equation, checking domain

Solve $\log_2 x+\log_2(x-2)=3$ .

Combine: $\log_2\big(x(x-2)\big)=3$ , so $x(x-2)=2^3=8$ , i.e. $x^2-2x-8=0$ , which factors as $(x-4)(x+2)=0$ , giving $x=4$ or $x=-2$ .

Check domain: $\log_2 x$ requires $x>0$ , so $x=-2$ is rejected (extraneous). Only $x=4$ is valid — check: $\log_2 4+\log_2 2=2+1=3$ ✓.

Quick reference

Inverse recipe

swap

x

and

y

, then solve for

y

Rational equations

always re-check for extraneous (denominator-zero) roots

Log equations

always re-check the domain

x>0

inside every log

Rational function f(x) = (2x + 1)/(x - 1), illustrating the vertical asymptote at x = 1 and horizontal asymptote at y = 2.

⚠ Advanced Functions & Domain Traps

Extraneous Solutions: Multiplying rational equations by clearing denominators can introduce false roots where the original denominator is zero. Always check candidates in the original expression.

Logarithm Domain Violation: In

\log_b(g(x))

, the argument must satisfy

g(x) > 0

. Never test roots in combined log forms (like

\log(x^2)

); you must test them in the original separate logs.

Composition Order:

(f \circ g)(x) = f(g(x))

is generally not equal to

(g \circ f)(x) = g(f(x))

. Also remember that the domain of

f \circ g

requires

x \in \operatorname{dom}(g)

and

g(x) \in \operatorname{dom}(f)

.

Solve the rational equation for

x

:

\frac{x}{x-2} + \frac{1}{x-4} = \frac{2}{(x-2)(x-4)}

1. Multiply both sides by the common denominator

(x-2)(x-4)

, noting domain restrictions

x \neq 2

and

x \neq 4

:

x(x-4) + 1(x-2) = 2

x^2 - 4x + x - 2 = 2 \implies x^2 - 3x - 4 = 0.

2. Factor the quadratic:

(x - 4)(x + 1) = 0

, yielding potential solutions

x = 4

and

x = -1

.
3. Check domain restrictions:

x = 4

makes the denominator

x-4 = 0

, so

x = 4

is extraneous.
4. Therefore, the only valid solution is

x = -1

.

📝 Chapter Quiz

Questions Level

5

Trigonometry & Polar Coordinates

AIST Math 2 pushes trigonometry past right triangles into the Law of Sines and Cosines (any triangle, any shape), double-angle identities (getting $\sin2x$ or $\cos2x$ from $\sin x$ and $\cos x$ alone), and a second coordinate system — polar coordinates, which locate a point by distance and direction instead of by $x$ and $y$ .

● Definition 5.1 — Law of Sines

In any triangle with angles $A,B,C$ and opposite sides $a,b,c$ :

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}.$

Use it when you know an angle-side pair plus one more angle or side.

Worked example 5.A — Law of Sines

In $\triangle ABC$ , $A=40^\circ$ , $a=12$ , $B=65^\circ$ . Find $b$ to the nearest tenth.

$\frac{12}{\sin40^\circ}=\frac{b}{\sin65^\circ} \quad\Rightarrow\quad b=\frac{12\sin65^\circ}{\sin40^\circ}\approx\frac{12(0.9063)}{0.6428}\approx16.9.$

● Definition 5.2 — Law of Cosines

In any triangle, with $C$ the angle between sides $a$ and $b$ (opposite side $c$ ):

$c^2=a^2+b^2-2ab\cos C.$

Use it when you know two sides and the included angle (SAS), or all three sides (SSS, solving for an angle).

Worked example 5.B — Law of Cosines

In $\triangle ABC$ , $a=8$ , $b=5$ , and the included angle $C=60^\circ$ . Find $c$ .

$c^2=8^2+5^2-2(8)(5)\cos60^\circ=64+25-80\left(\tfrac12\right)=89-40=49.$

$c=\sqrt{49}=7.$

▲ Theorem 5.1 — Double-angle identities

For any angle $x$ :

$\sin2x=2\sin x\cos x, \qquad \cos2x=\cos^2x-\sin^2x.$

Proof

Both follow from the angle-sum identities $\sin(a+b)=\sin a\cos b+\cos a\sin b$ and $\cos(a+b)=\cos a\cos b-\sin a\sin b$ , setting $a=b=x$ :

$\sin2x=\sin(x+x)=\sin x\cos x+\cos x\sin x=2\sin x\cos x.$

$\cos2x=\cos(x+x)=\cos x\cos x-\sin x\sin x=\cos^2x-\sin^2x. \quad\blacksquare$

Worked example 5.C — Apply the double-angle identities

If $\sin x=\tfrac35$ and $\cos x=\tfrac45$ (angle $x$ in Quadrant I), find $\sin2x$ and $\cos2x$ .

$\sin2x=2\left(\tfrac35\right)\left(\tfrac45\right)=\tfrac{24}{25}.$

$\cos2x=\left(\tfrac45\right)^2-\left(\tfrac35\right)^2=\tfrac{16}{25}-\tfrac{9}{25}=\tfrac{7}{25}.$

⊙ Explorer · Double angle on the unit circleinteractive

angle θ: 30°

cos θ0.866

sin θ0.500

cos 2θ0.500

sin 2θ0.866

The indigo point is (cos θ, sin θ); the amber point is (cos 2θ, sin 2θ). Watch sin 2θ and cos 2θ track exactly what Theorem 5.1 predicts as θ moves — including past 90°, where sin 2θ can turn negative before sin θ does.

● Definition 5.3 — Polar coordinates

A point's polar coordinates $(r,\theta)$ give its distance $r$ from the origin and the angle $\theta$ (measured counterclockwise from the positive $x$ -axis). Converting between polar and rectangular $(x,y)$ :

$x=r\cos\theta,\quad y=r\sin\theta, \qquad r=\sqrt{x^2+y^2},\quad \tan\theta=\frac{y}{x}\ (x\neq0).$

Worked example 5.D — Polar to rectangular

Convert $(r,\theta)=(4,120^\circ)$ to rectangular coordinates.

$x=4\cos120^\circ=4\left(-\tfrac12\right)=-2, \qquad y=4\sin120^\circ=4\left(\tfrac{\sqrt3}{2}\right)=2\sqrt3.$

So the point is $(-2,\,2\sqrt3)$ .

Worked example 5.E — Rectangular to polar

Convert $(x,y)=(-3,\,3\sqrt3)$ to polar coordinates.

$r=\sqrt{(-3)^2+(3\sqrt3)^2}=\sqrt{9+27}=\sqrt{36}=6.$

Since $x<0,\,y>0$ , the point is in Quadrant II. The reference angle is $\tan^{-1}\!\left(\dfrac{3\sqrt3}{3}\right)=\tan^{-1}(\sqrt3)=60^\circ$ , so $\theta=180^\circ-60^\circ=120^\circ$ . Polar form: $(6,\,120^\circ)$ .

Quick reference

Law of Sines — use when

AAS or ASA (angle-angle-side)

Law of Cosines — use when

SAS or SSS (no matching angle-side pair)

Quadrant angle signs

QI:

(+,+)

, QII:

(-,+)

, QIII:

(-,-)

, QIV:

(+,-)

Oblique triangle with sides a, b, c and opposite angles A, B, C used in the Law of Sines and Law of Cosines.

⚠ Trigonometry & Polar Exam Traps

Ambiguous Case (SSA): When given two sides and a non-included angle (SSA), the Law of Sines

\sin B = \dfrac{b\sin A}{a}

can yield 0, 1, or 2 solutions. Always check whether the supplementary angle

180^\circ - B

allows a valid triangle (i.e.

A + (180^\circ - B) < 180^\circ

).

Negative r in Polar Coordinates:

(-r, \theta)

represents the same point as

(r, \theta + 180^\circ)

or

(r, \theta + \pi)

. The point is plotted along the ray opposite to

\theta

.

Calculator Angle Mode: Never use degree mode when problems involve arc length

s = r\theta

, radian polar equations, or trigonometric limits. Double check radian vs degree indicators on each question.

In triangle

ABC

,

a = 7

,

b = 5

, and

\cos C = \dfrac{1}{5}

. Find the exact length of side

c

and determine

\sin^2 C

.

1. Apply the Law of Cosines:

c^2 = a^2 + b^2 - 2ab\cos C

c^2 = 7^2 + 5^2 - 2(7)(5)\left(\frac{1}{5}\right) = 49 + 25 - 14 = 60.

Thus

c = \sqrt{60} = 2\sqrt{15}

.
2. Using the Pythagorean identity:

\sin^2 C = 1 - \cos^2 C = 1 - \left(\frac{1}{5}\right)^2 = 1 - \frac{1}{25} = \frac{24}{25}.

📝 Chapter Quiz

Questions Level

6

Limits

A limit asks: as $x$ gets closer and closer to some value, what does $f(x)$ get closer and closer to? This is precalculus' bridge into calculus — the same machinery that eventually defines the derivative and the integral, here used to make sense of behavior a function's formula doesn't show directly.

● Definition 6.1 — Limit

We write $\lim_{x\to a}f(x)=L$ to mean: as $x$ approaches $a$ (from either side, without necessarily equaling $a$ ), $f(x)$ approaches $L$ . When $f$ is a polynomial or rational function and $a$ is in its domain, the limit is found simply by direct substitution: $\lim_{x\to a}f(x)=f(a)$ .

Worked example 6.A — Direct substitution

Find $\displaystyle\lim_{x\to3}\big(x^2-2x+1\big)$ .

Since this polynomial is defined everywhere, substitute directly: $\lim_{x\to3}\big(x^2-2x+1\big)=3^2-2(3)+1=9-6+1=4.$

● Definition 6.2 — Indeterminate form

If direct substitution gives $\tfrac{0}{0}$ , the limit isn't automatically $0$ , undefined, or anything else — it's an indeterminate form, meaning more work is needed. For rational functions, the usual fix is to factor and cancel the common factor causing the zero, then substitute again.

Worked example 6.B — Factor and cancel

Find $\displaystyle\lim_{x\to2}\dfrac{x^2-4}{x-2}$ .

Direct substitution gives $\tfrac{4-4}{2-2}=\tfrac00$ — indeterminate. Factor the numerator: $\frac{x^2-4}{x-2}=\frac{(x-2)(x+2)}{x-2}=x+2 \quad(x\neq2).$ The canceled form agrees with the original everywhere except exactly at $x=2$ — which is fine, since a limit only cares about values near $x=2$ , never at it. Now substitute: $\lim_{x\to2}\frac{x^2-4}{x-2}=\lim_{x\to2}(x+2)=2+2=4.$

● Definition 6.3 — Limits at infinity

$\displaystyle\lim_{x\to\infty}f(x)$ describes what $f(x)$ approaches as $x$ grows without bound. For a rational function $\dfrac{p(x)}{q(x)}$ , the result depends only on comparing the degrees of $p$ and $q$ .

▲ Theorem 6.1 — Limits at infinity for rational functions

For $f(x)=\dfrac{p(x)}{q(x)}$ with $\deg p=m$ , $\deg q=n$ , leading coefficients $a$ (of $p$ ) and $b$ (of $q$ ):

If \( m
If $m=n$ : $\displaystyle\lim_{x\to\infty}f(x)=\dfrac{a}{b}$ (ratio of leading coefficients).
If $m>n$ : the limit does not exist as a finite number — $f(x)\to\pm\infty$ .

Proof

Divide numerator and denominator by $x^n$ (the denominator's highest power): every term of the form $\dfrac{c\,x^k}{x^n}=c\,x^{k-n}$ with $ kn $, a positive power of $x$ survives in the numerator after dividing by $x^n$ , and that power grows without bound. $\blacksquare$

Worked example 6.C — Limits at infinity, all three cases

Evaluate each as $x\to\infty$ :

$\text{(i) } \frac{5x+1}{2x^2+3}\to0 \quad(\deg\,\text{num}=1<\deg\,\text{den}=2)$

$\text{(ii) } \frac{3x^2+2x}{x^2-5}\to\frac{3}{1}=3 \quad(\text{equal degrees, ratio of leading coefficients})$

$\text{(iii) } \frac{2x^3+1}{x^2}\to\infty \quad(\deg\,\text{num}=3>\deg\,\text{den}=2)$

● Definition 6.4 — Continuity

A function $f$ is continuous at $x=a$ if all three of these hold: $f(a)$ is defined, $\displaystyle\lim_{x\to a}f(x)$ exists, and $\displaystyle\lim_{x\to a}f(x)=f(a)$ . Informally: the graph has no hole, jump, or break at $x=a$ .

Quick reference

0/0 on substitution

indeterminate — factor and cancel, then re-substitute

Limits at infinity

compare degrees: lower

\to0

, equal

\to

ratio of leading coeffs, higher

\to\infty

Horizontal asymptote

y=L

where

L=\lim_{x\to\infty}f(x)

, when that limit is finite

Rational function with removable discontinuity: f(x) = (x² - 1)/(x - 1) = x + 1 for x ≠ 1, displaying an open hole at (1, 2).

⚠ Indeterminate Forms & Limit Traps

0/0 is NOT 0 or Undefined: When direct substitution gives

\dfrac{0}{0}

, this is an indeterminate form. It signals that algebraic work is required (factoring, rationalizing, or simplifying) to determine the true limit value.

Hole vs Value: A limit

\lim_{x\to a}f(x)

describes the behavior of

f(x)

as

x

gets arbitrarily close to

a

, completely independent of whether

f(a)

is defined.

One-Sided Agreement: For the two-sided limit

\lim_{x\to a}f(x)

to exist, the left-hand limit

\lim_{x\to a^-}f(x)

and right-hand limit

\lim_{x\to a^+}f(x)

must both exist and be equal.

Evaluate the limit:

\lim_{x\to 4} \frac{\sqrt{x} - 2}{x - 4}

1. Direct substitution yields

\dfrac{\sqrt{4}-2}{4-4} = \dfrac{0}{0}

, an indeterminate form.
2. Multiply numerator and denominator by the conjugate

\sqrt{x} + 2

:

\lim_{x\to 4}\frac{(\sqrt{x}-2)(\sqrt{x}+2)}{(x-4)(\sqrt{x}+2)} = \lim_{x\to 4}\frac{x - 4}{(x-4)(\sqrt{x}+2)}.

3. Cancel the common factor

(x-4)

for

x \neq 4

:

\lim_{x\to 4}\frac{1}{\sqrt{x}+2} = \frac{1}{\sqrt{4}+2} = \frac{1}{2+2} = \frac{1}{4}.

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