i

Test format & strategy

🗻 EST II Mathematics Level 2 — Content Areas

Content Area	Key Topics
Advanced Algebra	Complex numbers, sequences & series, advanced equations
Functions & Graphs	Advanced functions, transformations, function composition
Matrices & Vectors	Matrices, determinants, vector algebra
Introductory Calculus	Limits, an introduction to differential equations
Probability	Counting principles, probability distributions

EST II — the individual subject test taken alongside EST I by students applying to practical colleges (Medicine, Engineering, Pharmacy, and similar) — has two separate math components. Math Level 1 covers algebra, geometry, trigonometry, and statistics at the level EST I already builds toward. Math Level 2, covered here, goes further: complex numbers, matrices and vectors, sequences and series, and an introduction to calculus — the material that distinguishes students aiming for the most competitive, math-heavy programs.

What to expect

EST II Math Level 2 is a timed, multiple-choice exam; a calculator is permitted, and a formula reference sheet is printed in the test booklet. Because the same booklet formula sheet covers most standard identities, the highest-value study time is spent on applying formulas correctly under time pressure, not re-memorizing ones you'll be handed anyway.

Prerequisite

EST 2 Math Level 1

Level

Advanced algebra & introductory calculus

Required for

Practical-college applicants (Medicine, Engineering, ...)

Calculator

permitted, formula sheet provided

▲ EST II Math Level 2 Strategic Approach

EST II Math Level 2 questions reward quick conceptual recognition. Key priorities include:

Algebraic Checks: Verify roots in original logarithmic and rational expressions to eliminate extraneous answers.
Geometric Visuals: Sketch vectors, complex plane points, and function graphs to verify signs and quadrants before calculating.

⚠ EST II Timing Pitfalls & Calculator Hazards

Pacing Strategy: The test has 50 questions in 60 minutes (72 seconds/question). Do not spend more than 2 minutes on any single question on the first pass; flag it and return.

Calculator Precision: Avoid rounding intermediate values during multi-step matrix or trigonometry problems, which causes final rounding errors that lead to distractor choices.

No Guessing Penalty: EST II does not deduct points for incorrect answers; always ensure every question is answered before time expires.

EST II Mathematics Level 2 approximate content distribution across the major syllabus domains.

A student answers 35 questions in 42 minutes. To finish all 50 questions on the exam, how many seconds per question must the student maintain for the remaining questions?

The exam has

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

total. 42 minutes is

42 \times 60 = 2520\text{ seconds}

. The remaining time is

3600 - 2520 = 1080\text{ seconds}

. The remaining number of questions is

50 - 35 = 15

. The allowed pacing is

\dfrac{1080\text{ seconds}}{15\text{ questions}} = 72\text{ seconds per question}

.

1

Complex Numbers

A quadratic equation with a negative discriminant has no real solutions — the parabola never crosses the $x$ -axis. Complex numbers resolve this: by defining a number whose square is $-1$ , every quadratic (in fact, every polynomial) gets exactly as many roots as its degree.

● Definition 1.1 — The imaginary unit and complex numbers

The imaginary unit $i$ is defined by $i^2=-1$ . A complex number $z=a+bi$ pairs a real part $a$ with an imaginary part $b$ (both real numbers). Two complex numbers are equal exactly when their real parts match and their imaginary parts match.

● Definition 1.2 — Arithmetic with complex numbers

Complex numbers add, subtract, and multiply using ordinary algebra, replacing $i^2$ with $-1$ wherever it appears. For $z_1=a+bi$ , $z_2=c+di$ :

$z_1\pm z_2=(a\pm c)+(b\pm d)i, \qquad z_1z_2=(ac-bd)+(ad+bc)i.$

Worked example 1.A — Sum and product

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

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

$z_1z_2=(2)(-1)-(-3)(4)+\big[(2)(4)+(-3)(-1)\big]i=(-2+12)+(8+3)i=10+11i.$

The most common way a Level 2 question meets complex numbers is by asking you to solve a quadratic that the discriminant test says has no real roots. The quadratic formula still works — it just hands back a complex answer.

▲ Theorem 1.1 — Complex roots of a quadratic

For $ax^2+bx+c=0$ with real $a,b,c$ ( $a\neq0$ ) and discriminant $\Delta=b^2-4ac<0$ , the two roots are the complex conjugate pair

$x=-\frac{b}{2a}\pm i\,\frac{\sqrt{4ac-b^2}}{2a}.$

Proof

The quadratic formula gives $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ regardless of the sign of the discriminant. When $\Delta=b^2-4ac<0$ , write $b^2-4ac=-(4ac-b^2)$ , where $4ac-b^2>0$ since $\Delta<0$ . Then

$\sqrt{b^2-4ac}=\sqrt{-(4ac-b^2)}=\sqrt{4ac-b^2}\cdot\sqrt{-1}=i\sqrt{4ac-b^2},$

using $i=\sqrt{-1}$ from Definition 1.1. Substituting back:

$x=\frac{-b\pm i\sqrt{4ac-b^2}}{2a}=-\frac{b}{2a}\pm i\,\frac{\sqrt{4ac-b^2}}{2a}.$

The two roots share the same real part $-\tfrac{b}{2a}$ and have opposite-sign imaginary parts — exactly the definition of a conjugate pair. $\blacksquare$

Worked example 1.B — Solving a quadratic with complex roots

Solve $x^2+4x+13=0$ .

Here $a=1,\,b=4,\,c=13$ , so $\Delta=4^2-4(1)(13)=16-52=-36<0$ — no real roots, so we expect a conjugate pair by Theorem 1.1.

$x=\frac{-4\pm\sqrt{-36}}{2}=\frac{-4\pm6i}{2}=-2\pm3i.$

Check (for $x=-2+3i$ ): $(-2+3i)^2=4-12i+9i^2=4-12i-9=-5-12i$ , and $4(-2+3i)=-8+12i$ , so $x^2+4x+13=(-5-12i)+(-8+12i)+13=0. \;\checkmark$

Modulus — distance from the origin

Definition

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

Example

For

z=-2+3i

(Example 1.B's root),

|z|=\sqrt{(-2)^2+3^2}=\sqrt{13}

Conjugate pairs

A conjugate pair

a\pm bi

always has equal modulus for both roots

Complex number representation in the Argand plane: z = -2 + 3i forming a right triangle with real component -2, imaginary component 3, and modulus |z| = √13.

⚠ Complex Numbers Exam Traps

Negative Radicand Multiplication: Never write

\sqrt{-a}\sqrt{-b} = \sqrt{(-a)(-b)} = \sqrt{ab}

. You must first factor out

i

:

i\sqrt{a} \times i\sqrt{b} = i^2 \sqrt{ab} = -\sqrt{ab}

.

Sign Error in Modulus: In

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

, squaring the negative imaginary part yields a positive value:

|3 - 4i| = \sqrt{9 + 16} = 5

, not

\sqrt{9 - 16}

.

Cyclicity of Powers: Remember

i^1 = i

,

i^2 = -1

,

i^3 = -i

,

i^4 = 1

. Always divide large exponents by 4 and evaluate the remainder.

Evaluate the expression

\dfrac{(3 + 2i)^2}{i}

and write the result in standard form

a + bi

.

1. Expand the numerator:

(3+2i)^2 = 3^2 + 2(3)(2i) + (2i)^2 = 9 + 12i - 4 = 5 + 12i

.
2. Divide by

i

by multiplying numerator and denominator by

-i

(or using

1/i = -i

):

\frac{5 + 12i}{i} = (5 + 12i)(-i) = -5i - 12i^2 = -5i + 12 = 12 - 5i.

In standard form

a + bi

, the answer is

12 - 5i

.

📝 Chapter Quiz

Questions Level

2

Sequences & Series

This chapter builds the two sequence families EST II Math Level 2 tests — arithmetic (constant difference) and geometric (constant ratio) — and the surprising idea that an infinite sum can still equal a finite number, as long as the terms shrink fast enough.

● Definition 2.1 — Arithmetic sequence

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

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

Worked example 2.A — Find a term

An arithmetic sequence has $a_1=7$ and $d=5$ . Find $a_{12}$ .

$a_{12}=7+5(12-1)=7+55=62.$

● Definition 2.2 — Arithmetic series

The sum of an arithmetic sequence's first $n$ terms is

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

Worked example 2.B — Sum an arithmetic series

Find the sum of the first $12$ terms of the sequence from Example 2.A ( $a_1=7,\,d=5$ ).

$S_{12}=\frac{12}{2}\big(2(7)+11(5)\big)=6(14+55)=6(69)=414.$

● Definition 2.3 — Geometric sequence

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

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

Worked example 2.C — Find a geometric term

A geometric sequence has $a_1=4$ and $r=3$ . Find $a_6$ .

$a_6=4\cdot3^{5}=4\cdot243=972.$

▲ Theorem 2.1 — 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 $S_n=a_1+a_1r+a_1r^2+\cdots+a_1r^{n-1}$ . Multiply by $r$ : $rS_n=a_1r+a_1r^2+\cdots+a_1r^{n}$ . Subtracting the first sum from the second cancels every middle term, leaving $rS_n-S_n=a_1r^{n}-a_1$ , so $S_n(r-1)=a_1(r^n-1)$ . Since $r\neq1$ , divide by $r-1$ : $S_n=\dfrac{a_1(r^n-1)}{r-1}$ . $\blacksquare$

Worked example 2.D — Sum a geometric series

Find $S_6$ for the sequence from Example 2.C ( $a_1=4,\,r=3$ ).

$S_6=\frac{4(3^6-1)}{3-1}=\frac{4(729-1)}{2}=\frac{4(728)}{2}=1456.$

▲ Theorem 2.2 — Infinite geometric series

If $|r|<1$ , the infinite geometric series converges to

$S_\infty=\frac{a_1}{1-r}.$ If $|r|\geq1$ (and $a_1\neq0$ ), the series diverges.

Proof

By Theorem 2.1, $S_n=\dfrac{a_1}{1-r}(1-r^n)$ . When $|r|<1$ , repeated multiplication by a factor smaller than $1$ in absolute value forces $r^n\to0$ as $n\to\infty$ , so $S_n\to\dfrac{a_1}{1-r}$ . When $|r|\geq1$ , $r^n$ never shrinks toward $0$ , so $S_n$ never settles on a finite value. $\blacksquare$

Worked example 2.E — Sum an infinite geometric series

Find $54+18+6+2+\cdots$ .

Here $a_1=54$ and $r=\tfrac13$ (each term is a third of the last), and $|r|=\tfrac13<1$ , so $S_\infty=\frac{54}{1-\tfrac13}=\frac{54}{\tfrac23}=81.$

Quick reference

Sigma notation

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

Geometric sequence decay with first term a_1 = 54 and common ratio r = 1/3, with terms rapidly approaching 0.

⚠ Sequences & Series Traps

Common Ratio Sign: In alternating geometric series like

12 - 6 + 3 - \cdots

, the common ratio is negative:

r = -1/2

. Writing

r = 1/2

alters the denominator

1 - r

from

1 - (-1/2) = 3/2

to

1/2

, resulting in an answer off by a factor of 3.

Number of Terms in Arithmetic Series: Never assume the number of terms

n

is the last number written. Always compute

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

.

Divergent Geometric Series: If

|r| \geq 1

, the infinite sum

S_\infty

does not exist. Do not blindly use

\dfrac{a_1}{1-r}

on divergent series.

An arithmetic sequence has third term

a_3 = 11

and eighth term

a_8 = 31

. Find the common difference

d

, the first term

a_1

, and the sum of the first 20 terms

S_{20}

.

1. Express the terms:

a_3 = a_1 + 2d = 11

and

a_8 = a_1 + 7d = 31

.
2. Subtract the equations:

(a_1 + 7d) - (a_1 + 2d) = 5d = 20 \implies d = 4

.
3. Find

a_1

:

a_1 + 2(4) = 11 \implies a_1 = 3

.
4. Compute

S_{20}

:

S_{20} = \frac{20}{2}[2(3) + (20-1)(4)] = 10[6 + 19(4)] = 10[6 + 76] = 10(82) = 820.

📝 Chapter Quiz

Questions Level

3

Functions & Graphs

This chapter is about reading and reshaping graphs algebraically: what a formula tells you about symmetry before you ever plot a point, how shifting/stretching/reflecting a base graph changes its equation, and how to pin down a function's domain from its formula alone.

● Definition 3.1 — Function transformations

Starting from a base function $f(x)$ , the transformed function $g(x)=a\,f(x-h)+k$ applies, in this order: a horizontal shift by $h$ (right if $h>0$ ), a vertical stretch/compression and reflection by $a$ (a reflection over the $x$ -axis if $a<0$ ), and a vertical shift by $k$ .

Worked example 3.A — Apply a transformation

Let $f(x)=x^2$ and $g(x)=2f(x-3)+1=2(x-3)^2+1$ . Find $g(5)$ , and describe the transformation from $f$ to $g$ in words.

$g(5)=2(5-3)^2+1=2(4)+1=9.$

$g$ is $f$ shifted right $3$ , stretched vertically by a factor of $2$ , and shifted up $1$ .

⊙ Explorer · Transforming y = f(x)interactive

stretch/reflect a: 1.0

horizontal shift h: 0

vertical shift k: 0

y = 1.0(x-0)² + 0

The grey curve is the base f(x) = x²; the indigo curve is g(x) = a·f(x−h)+k, exactly Definition 3.1's transformation order: shift right h, stretch/reflect by a, then shift up k.

● Definition 3.2 — Even and odd functions

A function is even if $f(-x)=f(x)$ for every $x$ in its domain (graph symmetric about the $y$ -axis). A function is odd if $f(-x)=-f(x)$ for every $x$ (graph symmetric about the origin — a $180^\circ$ rotation of itself). Most functions are neither.

▲ Theorem 3.1 — Even/odd powers determine even/odd polynomials

A polynomial made up only of even-degree terms (including a constant) is an even function; a polynomial made up only of odd-degree terms is an odd function.

Proof

If every term has even degree, write $f(x)=\sum_k a_k x^{2k}$ . Then $f(-x)=\sum_k a_k(-x)^{2k}=\sum_k a_k\big((-1)^{2k}x^{2k}\big)=\sum_k a_k x^{2k}=f(x)$ since $(-1)^{2k}=1$ for every integer $k$ . If every term has odd degree, write $f(x)=\sum_k a_k x^{2k+1}$ ; then $(-1)^{2k+1}=-1$ , so $f(-x)=\sum_k a_k(-1)x^{2k+1}=-f(x)$ . $\blacksquare$

Worked example 3.B — Classify even, odd, or neither

Classify $f(x)=x^4-3x^2+2$ and $h(x)=x^3-4x$ .

$f$ has only even-degree terms ( $x^4,x^2,$ and the constant $2=2x^0$ ), so by Theorem 3.1 it's even. Check: $f(2)=16-12+2=6$ and $f(-2)=16-12+2=6$ ✓.

$h$ has only odd-degree terms ( $x^3,\,x$ ), so it's odd. Check: $h(1)=1-4=-3$ and $h(-1)=-1+4=3=-h(1)$ ✓.

● Definition 3.3 — Domain restrictions

Two formula types automatically restrict a function's domain: a square root $\sqrt{g(x)}$ requires $g(x)\geq0$ , and a rational expression $\dfrac{1}{g(x)}$ requires $g(x)\neq0$ .

Worked example 3.C — Find a domain

Find the domain of $f(x)=\sqrt{x-5}$ and of $g(x)=\dfrac{1}{x^2-9}$ .

For $f$ : need $x-5\geq0$ , so the domain is $x\geq5$ .

For $g$ : need $x^2-9\neq0$ , i.e. $x\neq\pm3$ . Domain: all reals except $3$ and $-3$ .

● Definition 3.4 — Piecewise functions

A piecewise function is defined by different formulas on different parts of its domain. To evaluate it at a specific input, first identify which piece's condition that input satisfies, then apply only that piece's formula.

Worked example 3.D — Evaluate a piecewise function

Let $p(x)=\begin{cases}-x+1,&x<0\\x^2-2,&x\geq0\end{cases}$ . Find $p(-3)$ and $p(3)$ .

Since $-3<0$ , use the first piece: $p(-3)=-(-3)+1=4$ .

Since $3\geq0$ , use the second piece: $p(3)=3^2-2=7$ .

Quick reference

Transformation order

a\,f(x-h)+k

: shift right

h

, stretch/reflect by

a

, shift up

k

Even function

f(-x)=f(x)

, symmetric about the

y

-axis

Odd function

f(-x)=-f(x)

, symmetric about the origin

Graph of transformed absolute value function f(x) = 2|x - 2| - 3 with vertex at (2, -3) and vertical stretch factor 2.

⚠ Transformations & Domain Traps

Horizontal Shift Sign:

f(x - h)

shifts the graph right by

h

units, not left. For example,

y = (x - 3)^2

has its vertex at

x = +3

.

Radical Domain: For

\sqrt{g(x)}

, the domain requires

g(x) \geq 0

. For

g(x) = 16 - x^2

,

16 - x^2 \geq 0 \implies -4 \leq x \leq 4

. Forgetting that the solution is a closed bounded interval is a common trap.

Inverse Function Reflection: The inverse

f^{-1}(x)

reflects across the line

y = x

, which swaps the domain and range:

\operatorname{dom}(f^{-1}) = \operatorname{range}(f)

.

Given

f(x) = \dfrac{2x + 5}{x - 3}

for

x \neq 3

, find the inverse function

f^{-1}(x)

and state its domain.

1. Set

y = \dfrac{2x + 5}{x - 3}

and swap

x

and

y

:

x = \frac{2y + 5}{y - 3}.

2. Multiply by

y - 3

:

x(y - 3) = 2y + 5 \implies xy - 3x = 2y + 5

.
3. Collect

y

terms:

xy - 2y = 3x + 5 \implies y(x - 2) = 3x + 5

.
4. Divide by

x - 2

:

f^{-1}(x) = \frac{3x + 5}{x - 2}.

The domain of

f^{-1}(x)

excludes

x = 2

:

(-\infty, 2) \cup (2, \infty)

. (This matches the range of

f

, which has horizontal asymptote

y = 2

).

📝 Chapter Quiz

Questions Level

4

Matrices & Vectors

A matrix packs a rectangular table of numbers into one object you can add, scale, and multiply — and that algebra gives a genuinely new way to solve a linear system. This chapter builds $2\times2$ matrix algebra from scratch, then pairs it with the vector tools (magnitude, dot product, angle) EST II Math Level 2 tests alongside it.

● Definition 4.1 — Matrix, addition, and scalar multiplication

An $m\times n$ matrix is a rectangular array of numbers with $m$ rows and $n$ columns. 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 4.A — Add and scale

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

$A+B=\begin{bmatrix}2&4\\4&-1\end{bmatrix}, \qquad 2A=\begin{bmatrix}8&-2\\6&4\end{bmatrix}.$

● Definition 4.2 — Matrix multiplication

If $A$ is $m\times n$ and $B$ is $n\times p$ , 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.

Worked example 4.B — Multiply two matrices, and a matrix by a vector

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

$AB=\begin{bmatrix}(3)(1)+(-1)(0)&(3)(2)+(-1)(-3)\\(2)(1)+(4)(0)&(2)(2)+(4)(-3)\end{bmatrix}=\begin{bmatrix}3&9\\2&-8\end{bmatrix}.$

$A\mathbf{v}=\begin{bmatrix}(3)(3)+(-1)(-4)\\(2)(3)+(4)(-4)\end{bmatrix}=\begin{bmatrix}13\\-10\end{bmatrix}.$

● Definition 4.3 — Determinant and the identity matrix

For $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$ , the determinant is $\det A=ad-bc$ . The identity matrix $I=\begin{bmatrix}1&0\\0&1\end{bmatrix}$ satisfies $IA=AI=A$ .

▲ Theorem 4.1 — The

2\times2

inverse formula

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

$A^{-1}=\frac{1}{\det A}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}.$ If $\det A=0$ , no inverse exists.

Proof

Let $d_0=ad-bc\neq0$ . Multiplying $A$ by $\tfrac{1}{d_0}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$ gives $\frac{1}{d_0}\begin{bmatrix}ad-bc&-ab+ba\\cd-dc&-cb+da\end{bmatrix}=\frac{1}{d_0}\begin{bmatrix}d_0&0\\0&d_0\end{bmatrix}=I,$ and the same computation in the other order also gives $I$ , so this is a genuine two-sided inverse. If $\det A=0$ and some $B$ satisfied $AB=I$ , then $\det A\cdot\det B=\det(AB)=\det I=1$ — impossible since the left side is $0$ . So no inverse exists. $\blacksquare$

Worked example 4.C — Invert a matrix and solve a system with it

Solve $\begin{cases}4x+y=14\\3x+2y=13\end{cases}$ using $\mathbf{x}=A^{-1}\mathbf{b}$ .

$A=\begin{bmatrix}4&1\\3&2\end{bmatrix}$ , $\det A=(4)(2)-(1)(3)=5$ , so $A^{-1}=\frac{1}{5}\begin{bmatrix}2&-1\\-3&4\end{bmatrix}.$

$x=\tfrac{1}{5}\big(2(14)-1(13)\big)=\tfrac{15}{5}=3, \qquad y=\tfrac{1}{5}\big(-3(14)+4(13)\big)=\tfrac{10}{5}=2.$ Check: $4(3)+2=14$ ✓ and $3(3)+2(2)=13$ ✓.

● Definition 4.4 — Vectors: magnitude and dot product

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

▲ Theorem 4.2 — Angle between two vectors

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

Proof

Placing $\mathbf{u},\mathbf{v}$ tail-to-tail, the Law of Cosines on the triangle with sides $|\mathbf{u}|,|\mathbf{v}|,|\mathbf{v}-\mathbf{u}|$ gives $|\mathbf{v}-\mathbf{u}|^2=|\mathbf{u}|^2+|\mathbf{v}|^2-2|\mathbf{u}||\mathbf{v}|\cos\theta$ . Expanding the left side as $(\mathbf{v}-\mathbf{u})\cdot(\mathbf{v}-\mathbf{u})=|\mathbf{v}|^2-2\mathbf{u}\cdot\mathbf{v}+|\mathbf{u}|^2$ and equating the two expressions cancels $|\mathbf{u}|^2+|\mathbf{v}|^2$ , leaving $-2\mathbf{u}\cdot\mathbf{v}=-2|\mathbf{u}||\mathbf{v}|\cos\theta$ , i.e. $\cos\theta=\dfrac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}||\mathbf{v}|}$ . $\blacksquare$

Worked example 4.D — Find the angle between two vectors

Find the angle between $\mathbf{u}=\langle5,-2\rangle$ and $\mathbf{v}=\langle3,4\rangle$ .

$\mathbf{u}\cdot\mathbf{v}=(5)(3)+(-2)(4)=7, \qquad |\mathbf{u}|=\sqrt{29},\qquad |\mathbf{v}|=5.$

$\cos\theta=\frac{7}{5\sqrt{29}}\approx0.2600 \quad\Rightarrow\quad \theta\approx74.9^\circ.$

⊙ Explorer · Angle between two vectorsinteractive

u direction: 37°

u length: 5.0

v direction: 127°

v length: 5.5

u · v0.000

angle between u and v90.0°

The default 90° apart is not an accident: u · v = 0 exactly when the vectors are perpendicular, whatever their lengths — Theorem 4.2 in action.

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}

Geometric representation of vector addition and dot product geometry in two dimensions.

⚠ Matrix & Vector Pitfalls

Matrix Multiplication Dimension Mismatch: An

m \times k

matrix can only multiply a

k \times n

matrix; the result is

m \times n

. Multiplication is strictly non-commutative (

AB \neq BA

).

Inverse Formula Sign Errors: In

A^{-1} = \dfrac{1}{ad - bc}\begin{pmatrix}d & -b\\ -c & a\end{pmatrix}

, note that the main diagonal entries

a

and

d

swap positions without sign change, while off-diagonal entries

b

and

c

change signs without swapping.

Perpendicular vs Parallel Vectors:

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

means the vectors are perpendicular (orthogonal). If

\mathbf{u} = k\mathbf{v}

, they are parallel.

Let

A = \begin{pmatrix}4 & 3\\ 2 & 2\end{pmatrix}

. Find

A^{-1}

and compute the vector

\mathbf{x} = A^{-1}\begin{pmatrix}10\\ 6\end{pmatrix}

.

1. Compute

\det(A) = (4)(2) - (3)(2) = 8 - 6 = 2

.
2. Form

A^{-1}

:

A^{-1} = \frac{1}{2}\begin{pmatrix}2 & -3\\ -2 & 4\end{pmatrix} = \begin{pmatrix}1 & -1.5\\ -1 & 2\end{pmatrix}.

3. Multiply by

\begin{pmatrix}10\\ 6\end{pmatrix}

:

\mathbf{x} = \frac{1}{2}\begin{pmatrix}2(10) + (-3)(6)\\ -2(10) + 4(6)\end{pmatrix} = \frac{1}{2}\begin{pmatrix}20 - 18\\ -20 + 24\end{pmatrix} = \frac{1}{2}\begin{pmatrix}2\\ 4\end{pmatrix} = \begin{pmatrix}1\\ 2\end{pmatrix}.

📝 Chapter Quiz

Questions Level

5

Introductory Calculus

EST II Math Level 2 previews calculus in two pieces: limits (what a function approaches, not just what it equals), and the derivative — defined directly as a limit, the same object every calculus course builds everything else on. This chapter also introduces, at the lightest possible level, what a differential equation even is.

● Definition 5.1 — Limit

$\displaystyle\lim_{x\to a}f(x)=L$ means: as $x$ approaches $a$ , $f(x)$ approaches $L$ . For a polynomial or rational function with $a$ in its domain, find the limit by direct substitution: $\lim_{x\to a}f(x)=f(a)$ .

Worked example 5.A — Direct substitution

Find $\displaystyle\lim_{x\to2}\big(x^2+3x-1\big)$ .

$\lim_{x\to2}\big(x^2+3x-1\big)=4+6-1=9.$

● Definition 5.2 — Indeterminate form 0/0

If direct substitution gives $\tfrac00$ , the limit is indeterminate, not automatically zero or undefined. For rational functions, factor and cancel the offending factor, then substitute again.

Worked example 5.B — Factor and cancel

Find $\displaystyle\lim_{x\to3}\dfrac{x^2-9}{x-3}$ .

Substitution gives $\tfrac00$ . Factor: $\dfrac{x^2-9}{x-3}=\dfrac{(x-3)(x+3)}{x-3}=x+3$ ( $x\neq3$ ). So $\lim_{x\to3}\frac{x^2-9}{x-3}=\lim_{x\to3}(x+3)=6.$

● Definition 5.3 — The derivative

The derivative of $f$ at $x$ , written $f'(x)$ , is defined as the limit of the difference quotient:

$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h},$

when that limit exists. It measures the instantaneous rate of change of $f$ at $x$ — the slope of the line tangent to $f$ 's graph there.

▲ Theorem 5.1 — Derivative of

x^2

If $f(x)=x^2$ , then $f'(x)=2x$ .

Proof

By Definition 5.3, $f'(x)=\lim_{h\to0}\frac{(x+h)^2-x^2}{h}=\lim_{h\to0}\frac{x^2+2xh+h^2-x^2}{h}=\lim_{h\to0}\frac{2xh+h^2}{h}.$ For $h\neq0$ (which is all that matters for a limit as $h\to0$ ), factor and cancel: $\dfrac{2xh+h^2}{h}=\dfrac{h(2x+h)}{h}=2x+h$ . So $f'(x)=\lim_{h\to0}(2x+h)=2x+0=2x. \quad\blacksquare$

Worked example 5.C — Evaluate a derivative

Using Theorem 5.1, find the slope of the tangent line to $f(x)=x^2$ at $x=4$ .

$f'(4)=2(4)=8.$ (Numerically: the difference quotient at $x=4$ with tiny $h$ — say $h=0.001$ — gives $\approx8.001$ , confirming the limit is $8$ .)

⊙ Explorer · Tangent line to f(x) = x²interactive

x₀: 3.0

slope f′(x₀) = 2x₀6.00

Drag x₀ and watch the amber tangent line's slope always equal 2x₀ — Theorem 5.1's f′(x) = 2x, visualized rather than just stated.

● Definition 5.4 — Differential equation

A differential equation is an equation involving a function and its derivative(s). Solving one means finding the function(s) $y$ that make the equation true — not a single number, but a whole function (usually a whole family of functions, differing by a constant).

Worked example 5.D — Verify a solution to a differential equation

Show that $y=x^2+5$ solves $\dfrac{dy}{dx}=2x$ .

By Theorem 5.1, the derivative of $x^2$ is $2x$ , and the derivative of the constant $5$ is $0$ (a constant never changes, so its instantaneous rate of change is $0$ ). So $\frac{dy}{dx}=2x+0=2x,$ exactly matching the equation. Any $y=x^2+C$ for a constant $C$ works the same way — this is the “whole family” Definition 5.4 mentions.

Quick reference

Difference quotient

\dfrac{f(x+h)-f(x)}{h}

, the building block of

f'(x)

Derivative of a constant

always

0

— constants don't change

Solving a differential equation

find the function(s)

y

, not a number

Parabola f(x) = x² with its tangent line y = 4x - 4 at the point of tangency (2, 4), showing instantaneous rate of change m = 4.

⚠ Introductory Calculus Pitfalls

Indeterminate Form Confusion: If direct substitution yields

\dfrac{0}{0}

, this does not mean the limit does not exist. You must algebraically factor, expand, or rationalize to eliminate the zero denominator.

Slope vs Function Value: Do not confuse the derivative

f'(a)

(the instantaneous slope) with the function output

f(a)

(the height on the graph). At a local maximum or minimum,

f'(a) = 0

, but

f(a)

is typically non-zero.

Power Rule Exponent: Remember

\dfrac{d}{dx}[x^n] = n x^{n-1}

. For negative or fractional powers, subtract 1:

\dfrac{d}{dx}[x^{1/2}] = \dfrac{1}{2}x^{-1/2} = \dfrac{1}{2\sqrt{x}}

.

Find the equation of the tangent line to the parabola

f(x) = 2x^2 - 3x + 1

at the point where

x = 2

.

1. Find the point on the curve:

f(2) = 2(2)^2 - 3(2) + 1 = 8 - 6 + 1 = 3

. The point is

(2, 3)

.
2. Compute the derivative:

f'(x) = 4x - 3

.
3. Evaluate the slope at

x = 2

:

m = f'(2) = 4(2) - 3 = 5

.
4. Write tangent line using point-slope form:

y - 3 = 5(x - 2) \implies y - 3 = 5x - 10 \implies y = 5x - 7

.

📝 Chapter Quiz

Questions Level

6

Probability

Probability starts from one simple ratio — favorable outcomes over total outcomes — and builds up to counting techniques (permutations, combinations) for when listing every outcome by hand isn't practical, plus rules for combining the probabilities of two events.

● Definition 6.1 — Probability of an event

For a sample space $S$ of equally likely outcomes and an event $E\subseteq S$ , the probability of $E$ is

$P(E)=\frac{n(E)}{n(S)},$

where $n(E)$ and $n(S)$ count outcomes. Always $0\leq P(E)\leq1$ .

Worked example 6.A — Basic probability

Two fair six-sided dice are rolled. Find the probability their sum is $7$ .

Total outcomes: $n(S)=6\times6=36$ . Favorable outcomes (sum $=7$ ): $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$ — that's $6$ . So $P(\text{sum}=7)=\frac{6}{36}=\frac16.$

● Definition 6.2 — Complement rule

The complement of event $E$ , written $E'$ , is “ $E$ does not happen.” Since $E$ and $E'$ together account for every outcome exactly once, $P(E')=1-P(E).$

● Definition 6.3 — Permutations and combinations

A permutation counts arrangements where order matters: $P(n,r)=\dfrac{n!}{(n-r)!}$ , the number of ways to arrange $r$ items chosen from $n$ . A combination counts selections where order does not matter: $\displaystyle\binom{n}{r}=\frac{n!}{r!(n-r)!}$ .

▲ Theorem 6.1 — Combinations from permutations

$\binom{n}{r}=\frac{P(n,r)}{r!}.$

Proof

Every combination (an unordered group of $r$ items) corresponds to exactly $r!$ different permutations (every way of ordering that same group). So the permutations of $n$ items taken $r$ at a time can be built in two steps: first choose which $r$ items (a combination, $\binom{n}{r}$ ways), then arrange those $r$ items in order ( $r!$ ways). By the multiplication principle, $P(n,r)=\binom{n}{r}\cdot r!$ , so dividing both sides by $r!$ gives $\binom{n}{r}=\dfrac{P(n,r)}{r!}=\dfrac{n!}{(n-r)!\,r!}$ . $\blacksquare$

Worked example 6.B — Combinations

How many ways can a committee of $3$ be chosen from $8$ people?

Order doesn't matter (a committee is a group, not a sequence), so use combinations: $\binom{8}{3}=\frac{8!}{3!\,5!}=\frac{8\cdot7\cdot6}{3\cdot2\cdot1}=56.$

● Definition 6.4 — Independent events

Events $A$ and $B$ are independent if one occurring doesn't change the probability of the other. For independent events, $P(A\text{ and }B)=P(A)\cdot P(B).$

● Definition 6.5 — Addition rule

For any two events $A,B$ (whether or not they can happen together),

$P(A\text{ or }B)=P(A)+P(B)-P(A\text{ and }B).$

The subtracted term avoids double-counting outcomes where both happen. If $A$ and $B$ are mutually exclusive (can't both happen), $P(A\text{ and }B)=0$ and the rule simplifies to $P(A\text{ or }B)=P(A)+P(B)$ .

Worked example 6.C — Addition rule with a standard deck

A card is drawn from a standard $52$ -card deck. Find $P(\text{King or Heart})$ .

$P(\text{King})=\tfrac{4}{52}$ , $P(\text{Heart})=\tfrac{13}{52}$ , and $P(\text{King and Heart})=\tfrac{1}{52}$ (the King of Hearts, counted in both). By Definition 6.5:

$P(\text{King or Heart})=\frac{4}{52}+\frac{13}{52}-\frac{1}{52}=\frac{16}{52}=\frac{4}{13}.$

Quick reference

Order matters?

Yes → permutation

P(n,r)

; No → combination

\binom{n}{r}

Independent events

P(A\text{ and }B)=P(A)P(B)

Mutually exclusive

P(A\text{ and }B)=0,\ \ P(A\text{ or }B)=P(A)+P(B)

Venn diagram of overlapping probability events A and B illustrating P(A ∪ B) = P(A) + P(B) - P(A ∩ B).

⚠ Probability & Counting Traps

Independent vs Mutually Exclusive: Mutually exclusive events (which cannot happen together,

P(A \cap B) = 0

) are never independent if both have non-zero probability, because the occurrence of one prevents the other.

Double Counting in Addition Rule: For non-exclusive events, you must subtract the intersection:

P(A \cup B) = P(A) + P(B) - P(A \cap B)

. Simply adding

P(A) + P(B)

double-counts shared outcomes.

Permutations vs Combinations: Ask 'Does order matter?' Order matters for titles, podium finishes, passwords, and assigned positions (permutations

nPr

); order does not matter for committees, hands of cards, and chosen groups (combinations

nCr

).

A committee of 4 people is to be selected from a group of 6 men and 5 women. What is the probability that the committee consists of exactly 2 men and 2 women?

1. Find the total number of ways to choose 4 people from 11:

\binom{11}{4} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330.

2. Find the number of ways to choose 2 men from 6 and 2 women from 5:

\binom{6}{2} \times \binom{5}{2} = 15 \times 10 = 150.

3. Compute the probability:

P = \frac{150}{330} = \frac{15}{33} = \frac{5}{11} \approx 0.4545.

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