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EST 2 Math Level 1

Provider · Academic Assessment Ltd. Format · EST II · Subject Test Chapters · 8 Level · Advanced Mathematics

Test format & strategy

🇬🇧 EST 2 Math — Domain Weightings (44 questions, 80 min)

Domain	% of Test	Key Topics
Advanced Algebra	~20%	Logarithms, exponentials, polynomial & rational equations, systems
Functions	~20%	Composite, inverse, logarithmic, transformations, piecewise
Sequences, Series & Complex Numbers	~10%	AP/GP, sigma notation, complex arithmetic
Geometry — Lines, Triangles & Polygons	~10%	Angle theorems, similarity, congruence, area
Trigonometry	~15%	Unit circle, identities, graphs, law of sines/cosines
Circles & Coordinate Geometry	~10%	Circle equations, tangents, conics, coordinate proofs
Solid Geometry & Similarity	~5%	3D surface area & volume, scale factors
Statistics & Probability	~10%	Counting, combinatorics, probability, data analysis

The EST II Mathematics subject test is a rigorous examination for students in international and American-diploma programmes who wish to demonstrate advanced mathematical competency for university admissions. It covers algebra, geometry, trigonometry, and data analysis at a depth beyond the core EST I test.

What to expect

EST II Math is taken as a standalone subject test on the EST II examination day. It contains both multiple-choice and short constructed-response questions. The content aligns closely with pre-calculus and advanced secondary mathematics.

Level

Advanced secondary / pre-calculus

Topics

Algebra · Geometry · Trig · Stats

Format

MCQ + constructed response

Depth

Beyond EST I

⏱ EST II Math Level 1 Pacing & Calculator Strategy

Pacing Baseline: 50 questions in 60 minutes = 72 seconds per question. There is zero negative marking — never leave any question blank.
Three-Pass Execution: Pass 1 (Qs 1–25, ~20 min): bank routine algebra, basic trigonometry, and straight arithmetic. Pass 2 (Qs 26–45, ~25 min): solve conic equations, solid geometry, and multi-step probability. Pass 3 (Qs 46–50 & review, ~15 min): tackle non-routine proofs, vectors, and ensure all grid-in/bubble items are complete.
Calculator Mastery: An approved scientific or graphing calculator is permitted throughout. Use it to verify complex number computations, evaluate trigonometric functions in proper degree/radian modes, and solve quadratic systems.

▲ The 4 Core Commandments for Scoring 750+ on EST II Math

1. Never Leave Unanswered Items: With no penalty for incorrect answers, eliminate obvious extremes and make an educated guess whenever stumped.
2. Verify Calculator Angle Mode: Trig questions in EST II frequently switch between radian measure (e.g., $5\pi/6$ ) and degree measure (e.g., $150°$ ). Confirm angle mode before performing numerical evaluations.
3. Conic Identification by Inspection: Check the coefficients of $x^2$ and $y^2$ : equal signs and equal coefficients indicate a circle; equal signs but different coefficients indicate an ellipse; opposite signs indicate a hyperbola.
4. Memorize 3D Solid & Circle Relations: Sphere volume $\frac{4}{3}\pi r^3$ and surface area $4\pi r^2$ ; cone volume $\frac{1}{3}\pi r^2 h$ ; inscribed circle angle $=\frac{1}{2}\text{central}$ ; cyclic quadrilateral opposite angles sum to $180°$ .

EST II Mathematics domain distribution. Advanced Algebra (20%) and Functions (20%) together form 40% of the exam, complemented by Trigonometry (15%), Sequences & Complex Numbers (10%), Geometry (10%), Circles & Coordinate Geometry (10%), Statistics (10%), and Solid Geometry (5%).

An EST II Mathematics examination consists of 50 questions with an allocated testing time of 60 minutes. If a student completes the first 30 questions in 33 minutes, what is the maximum average time, in seconds, the student can spend on each of the remaining 20 questions?

Step 1: Compute total available time in seconds: $60\text{ min} \times 60\text{ s/min} = 3600\text{ seconds}.$

Step 2: Compute time spent on first 30 questions: $33\text{ min} \times 60\text{ s/min} = 1980\text{ seconds}.$

Step 3: Calculate remaining time: $3600 - 1980 = 1620\text{ seconds}.$

Step 4: Determine average time per remaining question: $\frac{1620\text{ seconds}}{20\text{ questions}} = 81\text{ seconds per question}.$

On the EST II Math exam, Advanced Algebra and Functions together account for 40% of the questions. In a 50-question test, if a student targets an accuracy rate of 90% in these two high-yield domains, how many questions from these domains can the student afford to miss at most?

Step 1: Find total questions from Advanced Algebra and Functions: $50 \times 0.40 = 20\text{ questions}.$

Step 2: Calculate 90% accuracy target: $20 \times 0.90 = 18\text{ questions correct}.$

Step 3: Find maximum allowed missed questions: $20 - 18 = 2\text{ questions}.$

📝 Chapter Quiz

Questions Level

Advanced Algebra

🇬🇧 Advanced Algebra ~20% · Logarithms & exponentials · Polynomial long division · Rational equations · Nonlinear systems

Beyond linear equations, the EST II tests rational expressions, absolute value equations, and advanced polynomial manipulation.

Rational expression

\dfrac{P(x)}{Q(x)},\ Q\neq 0

Absolute value

|x-a|=b \Rightarrow x=a\pm b

Remainder theorem

f(k) =

remainder of

f\div(x-k)

Factor theorem

f(k)=0 \Leftrightarrow (x-k)\mid f(x)

Worked example 1.A

Simplify $\dfrac{x^2-9}{x^2-x-6}$ . $\frac{(x-3)(x+3)}{(x-3)(x+2)}=\frac{x+3}{x+2},\quad x\neq 3.$

Systems and inequalities

Graphical solutions of systems of inequalities define a feasible region. To find vertices, solve pairs of boundary lines simultaneously. EST II questions may ask for the maximum or minimum of a linear expression over the feasible region — an introduction to linear programming.

● Advanced Algebra

Polynomial long division: divide $p(x)$ by $d(x)$ to get quotient $q(x)$ and remainder $r(x)$ . Rational equations: multiply through by LCD, then solve, checking for extraneous solutions. Logarithmic equations: use log laws to combine terms, then convert to exponential form.

Worked example — Rational equation with extraneous solution

Solve $\dfrac{x}{x-2} + \dfrac{1}{x+2} = \dfrac{8}{x^2-4}$ .

LCD $= (x-2)(x+2)$ . Multiply through: $x(x+2)+(x-2) = 8$ .

$x^2+2x+x-2 = 8 \Rightarrow x^2+3x-10 = 0 \Rightarrow (x+5)(x-2)=0$ .

$x=2$ is extraneous (denominator = 0). Solution: $x = -5$ .

Graphical representation of the nonlinear system

y = x^2 - 4x + 3

and

y = -2x + 6

. The parabola and line intersect at

(-1, 8)

and

(3, 0)

▲ EST II Math Traps & Pitfalls — Rational Equations & Polynomial Theorems

Extraneous Solutions in Rational Equations: When multiplying across by the LCD, you may introduce solutions that make an original denominator zero. Always substitute candidate solutions back into the original denominators to verify they are non-zero.
Remainder Theorem Sign Inversion: When dividing $P(x)$ by $(x - c)$ , the remainder is $P(c)$ . But when dividing by $(x + c)$ , the divisor is $(x - (-c))$ , so the remainder is $P(-c)$ . Misplacing this minus sign is the #1 error on EST II polynomial questions.
Absolute Value Inequalities: $|x - a| \le b$ translates to the compound inequality $-b \le x - a \le b$ (an intersection/interval). In contrast, $|x - a| \ge b$ translates to $x - a \le -b$ OR $x - a \ge b$ (a union of disjoint rays).
Nonlinear Systems Intersection Count: When solving a system of a parabola $y = ax^2 + bx + c$ and line $y = mx + k$ , substitute to obtain a single quadratic. Check the discriminant $\Delta = B^2 - 4AC$ : $\Delta > 0$ gives 2 real solutions, $\Delta = 0$ gives 1 tangent point, and $\Delta < 0$ gives no real intersections.

Solve

\log_2(x+4) - \log_2 x = 3

\log_2\dfrac{x+4}{x} = 3 \Rightarrow \dfrac{x+4}{x} = 8 \Rightarrow x+4 = 8x \Rightarrow x = \dfrac{4}{7}

Solve the nonlinear system of equations:

\begin{cases} y = x^2 - 4x + 3 \\ 2x + y = 6 \end{cases}

Find all points of intersection

(x, y)

Substitution: Express $y$ from the linear equation: $y = 6 - 2x$ . Substitute into the quadratic: $6 - 2x = x^2 - 4x + 3 \implies x^2 - 2x - 3 = 0.$

Factoring: $(x - 3)(x + 1) = 0 \implies x_1 = 3,\quad x_2 = -1.$

Find corresponding $y$ values:

For $x = 3$ : $y = 6 - 2(3) = 0$ → $(3, 0)$ .
For $x = -1$ : $y = 6 - 2(-1) = 8$ → $(-1, 8)$ .

The two intersection points are $(3, 0)$ and $(-1, 8)$ .

📝 Chapter Quiz

Questions Level

Functions: Composite, Inverse & Logarithmic

🇬🇧 Functions ~20% · Composite & inverse · Logarithmic functions · Transformations · Domain & range

EST II places heavy emphasis on function behaviour, including composite and inverse functions, domain and range analysis, and the properties of logarithms.

Composition

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

Inverse

f(f^{-1}(x))=x

Log product

\log_b(mn)=\log_b m+\log_b n

Change of base

\log_b a=\dfrac{\ln a}{\ln b}

Worked example 2.A — Inverse

Find $f^{-1}(x)$ for $f(x)=3x-2$ . $y=3x-2 \implies x=\frac{y+2}{3} \implies f^{-1}(x)=\frac{x+2}{3}.$

Worked example 2.B — Logarithm

Solve $\log_2(x+3)=5$ . $x+3=2^5=32 \implies x=29.$

⊙ Explorer · A function and its inverseinteractive

point x₀: 100

f(x₀) = 2^x₀2.000

f⁻¹(f(x₀))1.000

Rational function

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

. The vertical asymptote is at

x = 1

(where denominator is zero) and the horizontal asymptote is at

y = 3

(ratio of leading coefficients).

▲ EST II Math Traps & Pitfalls — Composite, Inverse & Logarithmic Functions

Domain of Composite Functions: For $(f \circ g)(x) = f(g(x))$ , $x$ must reside in the domain of $g$ AND $g(x)$ must fall within the domain of $f$ . Never simplify algebraically before checking initial domain constraints.
Inverse Functions and One-to-One: $f^{-1}(x)$ exists only if $f$ passes the Horizontal Line Test. When swapping $x$ and $y$ , remember that the range of $f$ becomes the domain of $f^{-1}$ . For restricted domains like $f(x) = x^2, x \ge 0$ , the inverse is $f^{-1}(x) = \sqrt{x}, x \ge 0$ .
Logarithm Laws vs. False Analogies: $\log(a + b) \neq \log a + \log b$ . Only multiplication inside yields addition outside: $\log(ab) = \log a + \log b$ . Similarly, $\frac{\log a}{\log b} \neq \log(a - b)$ — it is the change-of-base formula $\log_b a$ .
Function Transformations Order: For $y = a f(b(x - c)) + d$ , horizontal stretch/compression by $1/|b|$ and shift by $c$ occur inside, while vertical stretch by $a$ and vertical shift by $d$ occur outside. $f(x - c)$ shifts right by $c$ ; $f(x + c)$ shifts left.

Given

f(x) = e^x

and

g(x) = \ln(x+1)

, find the domain of

g(f(x))

and simplify.

g(f(x)) = \ln(e^x+1)

. Since

e^x > 0

for all

x

e^x+1 > 1 > 0

. Domain: all reals. Cannot simplify further.

Let

f(x) = \dfrac{2x + 1}{x - 3}

for

x \neq 3

. (a) Determine the inverse function

f^{-1}(x)

. (b) State the domain and range of

f^{-1}(x)

(a) Find the inverse function: Set $y = \dfrac{2x + 1}{x - 3}$ . Swap $x$ and $y$ : $x = \dfrac{2y + 1}{y - 3} \implies x(y - 3) = 2y + 1 \implies xy - 3x = 2y + 1.$ Group $y$ terms: $xy - 2y = 3x + 1 \implies y(x - 2) = 3x + 1 \implies f^{-1}(x) = \dfrac{3x + 1}{x - 2}.$

(b) Domain and Range: The denominator of $f^{-1}(x)$ cannot be zero, so the domain is $(-\infty, 2) \cup (2, \infty)$ or $x \neq 2$ . The range of $f^{-1}$ equals the domain of $f$ , which is $(-\infty, 3) \cup (3, \infty)$ or $y \neq 3$ .

📝 Chapter Quiz

Questions Level

Sequences, Series & Complex Numbers

🇬🇧 Sequences, Series & Complex Numbers ~10% · Arithmetic & geometric sequences · Sigma notation · Complex arithmetic

EST II tests arithmetic and geometric sequences, sigma notation, and arithmetic with complex numbers.

Arithmetic

a_n

a_1+(n-1)d

Geometric

a_n

a_1\cdot r^{n-1}

Arithmetic sum

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

Complex product

(a+bi)(c+di)

— FOIL,

i^2=-1

Worked example 3.A

Find the 10th term of the arithmetic sequence $3, 7, 11, \ldots$ $a_{10}=3+(10-1)\cdot 4=3+36=39.$

Geometric series

The sum of a finite geometric series with first term

a_1

, ratio

r

, and

n

terms is

S_n=\dfrac{a_1(1-r^n)}{1-r}

. For

|r|<1

, the infinite geometric series converges to

S=\dfrac{a_1}{1-r}

● Sequences, Series & Complex Numbers

Arithmetic series sum: $S_n = \dfrac{n}{2}(a_1+a_n)$ . Geometric series sum: $S_n = \dfrac{a_1(1-r^n)}{1-r}$ ; sum to infinity ( $|r|<1$ ): $S_\infty = \dfrac{a_1}{1-r}$ . Complex: $(a+bi)+(c+di)=(a+c)+(b+d)i$ ; $(a+bi)(c+di)=(ac-bd)+(ad+bc)i$ .

Worked example — Sum to infinity

Find the sum to infinity of $12 + 4 + \dfrac{4}{3} + \cdots$

$r = 1/3$ . $S_\infty = \dfrac{12}{1-1/3} = \dfrac{12}{2/3} = 18$ .

Convergence interval

|r| < 1

for an infinite geometric series. The sum to infinity

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

converges strictly inside

(-1, 1)

with open endpoints; the series diverges for

|r| \ge 1

▲ EST II Math Traps & Pitfalls — Sequences, Series & Complex Numbers

Infinite Geometric Series Convergence: The sum formula $S_\infty = \frac{a_1}{1 - r}$ is valid only when $|r| < 1$ . If $|r| \ge 1$ , the series diverges and has no finite sum. If a question asks for the sum of $3 - 6 + 12 - 24 + \cdots$ , the answer is "diverges", not $\frac{3}{1 - (-2)} = 1$ .
Powers of $i$ Modulo 4: The imaginary unit follows a 4-term repeating cycle: $i^1 = i$ , $i^2 = -1$ , $i^3 = -i$ , and $i^4 = 1$ . To evaluate $i^n$ , divide $n$ by 4 and take the remainder $r$ : $i^n = i^r$ (where remainder 0 corresponds to $i^0 = 1$ ). For negative powers: $i^{-1} = -i$ .
Complex Division & Conjugate: Never divide terms individually. To evaluate $\frac{a + bi}{c + di}$ , multiply numerator and denominator by the complex conjugate $c - di$ . The denominator becomes the real number $c^2 + d^2$ .
Sigma Indexing Traps: Watch the lower limit of summation! $\sum_{k=1}^n c = nc$ , but $\sum_{k=0}^n c = (n+1)c$ . In general, the number of terms is $\text{upper} - \text{lower} + 1$ .

The sum of the first

n

terms of an arithmetic sequence is

S_n = 2n^2 + 3n

. Find the 10th term.

a_n = S_n - S_{n-1} = (2n^2+3n) - (2(n-1)^2+3(n-1)) = 4n+1

. So

a_{10} = 41

Let

z = \dfrac{5 + 2i}{2 - 3i} + i^{43}

. (a) Express

z

in standard form

a + bi

. (b) Compute the modulus

|z|

(a) Simplify the quotient and power of $i$ :

First evaluate $\dfrac{5+2i}{2-3i}$ : multiply numerator and denominator by the conjugate $2+3i$ : $\dfrac{(5+2i)(2+3i)}{(2-3i)(2+3i)} = \dfrac{10 + 15i + 4i + 6i^2}{4 - 9i^2} = \dfrac{10 - 6 + 19i}{4 + 9} = \dfrac{4 + 19i}{13}.$
Now evaluate $i^{43}$ : $43 = 4(10) + 3$ , remainder 3, so $i^{43} = i^3 = -i = -\dfrac{13i}{13}$ .
Combine terms: $z = \dfrac{4 + 19i}{13} - i = \dfrac{4 + 19i - 13i}{13} = \dfrac{4 + 6i}{13} = \dfrac{4}{13} + \dfrac{6}{13}i.$

(b) Modulus $|z|$ : $|z| = \sqrt{\left(\frac{4}{13}\right)^2 + \left(\frac{6}{13}\right)^2} = \sqrt{\frac{16 + 36}{169}} = \sqrt{\frac{52}{169}} = \dfrac{\sqrt{52}}{13} = \dfrac{2\sqrt{13}}{13}.$

📝 Chapter Quiz

Questions Level

Lines, Triangles & Polygons

🇬🇧 Lines, Triangles & Polygons ~10% · Angle theorems · Triangle congruence/similarity · Area of polygons

Advanced geometry on the EST II includes angle theorems, triangle congruence and similarity, and properties of quadrilaterals and regular polygons.

Triangle angle sum

180^\circ

Polygon angle sum

(n-2)\times 180^\circ

Area of triangle

\tfrac{1}{2}bh

Law of cosines

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

Worked example 4.A

In triangle $ABC$ , $a=7$ , $b=10$ , $C=60^\circ$ . Find $c$ . $c^2=49+100-2(7)(10)\cos 60^\circ=149-70=79,\quad c=\sqrt{79}.$

● Lines, Triangles & Polygons

Parallel lines cut by a transversal: alternate interior angles are equal; co-interior (same-side) angles are supplementary. Triangle congruence: SSS, SAS, ASA, AAS, RHS. Similarity: AA, SAS, SSS — corresponding sides proportional. Exterior angle = sum of two non-adjacent interior angles.

Worked example — Exterior angle theorem

In triangle $ABC$ , $\angle A = 48°$ and $\angle B = 63°$ . Find the exterior angle at $C$ .

Exterior angle $= \angle A + \angle B = 48° + 63° = 111°$ .

Oblique triangle

ABC

with base

c = AB = 8

. The Law of Cosines

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

connects the three sides and any included angle.

▲ EST II Math Traps & Pitfalls — Triangles, Polygons & Area

Linear vs. Area vs. Volume Scale Factors: If two similar polygons have a linear scale factor of $k$ , their perimeters scale by $k$ , but their areas scale by $k^2$ . On EST II, questions often provide the ratio of areas (e.g. $9:16$ ) and ask for perimeter — take the square root ( $3:4$ ) first!
Regular Polygon Angle Inversions: Interior angle is $\frac{(n-2)180°}{n}$ , while exterior angle is $\frac{360°}{n}$ . Notice that $\text{interior} + \text{exterior} = 180°$ . It is almost always faster to compute the exterior angle first and subtract from $180°$ .
Triangle Inequality Theorem: For any valid triangle with sides $a, b, c$ , the sum of any two sides must be strictly greater than the third: $a + b > c$ , $a + c > b$ , $b + c > a$ . This implies $|a - b| < c < a + b$ .
Heron's Formula vs. Trigonometric Area: If all 3 sides are known, use Heron's formula: $A = \sqrt{s(s-a)(s-b)(s-c)}$ where semi-perimeter $s = \frac{a+b+c}{2}$ . If two sides and included angle are known, use $\frac{1}{2}ab\sin C$ .

Two sides of a triangle are 7 cm and 10 cm with an included angle of 90°. Find the hypotenuse and all angles.

c = \sqrt{49+100} = \sqrt{149} \approx 12.2

cm.

\tan A = 7/10 \Rightarrow A \approx 35.0°

B \approx 55.0°

A regular hexagon has side length

s = 6

cm. (a) Calculate the exact area of the hexagon. (b) Find the length of the long diagonal connecting two opposite vertices.

(a) Exact area: A regular hexagon is composed of 6 congruent equilateral triangles of side length $s = 6$ . The area of one equilateral triangle is $\dfrac{\sqrt{3}}{4}s^2 = \dfrac{\sqrt{3}}{4}(36) = 9\sqrt{3}$ . Total area is: $A = 6 \times 9\sqrt{3} = 54\sqrt{3}\text{ cm}^2 \approx 93.53\text{ cm}^2.$

(b) Long diagonal: The long diagonal passes through the center of the circumscribed circle and equals twice the side length: $d = 2s = 2(6) = 12\text{ cm}$ .

📝 Chapter Quiz

Questions Level

Trigonometry

🇬🇧 Trigonometry ~15% · Unit circle & radians · Trig identities · Law of sines/cosines · Graphs

EST II trigonometry goes beyond right-triangle ratios to include the unit circle, radian measure, graphing, and trigonometric identities.

Pythagorean identity

\sin^2\theta+\cos^2\theta=1

Double angle sin

\sin 2\theta=2\sin\theta\cos\theta

Period of sin/cos

2\pi

Radian–degree

180^\circ=\pi

Worked example 5.A

Simplify $\dfrac{\sin^2\theta}{1-\cos\theta}$ . $\frac{1-\cos^2\theta}{1-\cos\theta}=\frac{(1-\cos\theta)(1+\cos\theta)}{1-\cos\theta}=1+\cos\theta.$

⊙ Explorer · The unit circle, in radiansinteractive

angle θ: 79

cos θ0.707

sin θ0.707

tan θ1.000

EST II works in radians: a full turn is $2\pi\approx6.28$ , not $360°$ . $P=(\cos\theta,\sin\theta)$ as above.

Graphs of trigonometric functions

The function

y=A\sin(Bx+C)+D

has amplitude

|A|

, period

\dfrac{2\pi}{B}

, phase shift

-C/B

, and vertical shift

D

. Knowing how to read these parameters from a graph or equation is a core EST II skill.

● Advanced Trigonometry

Unit circle: $(\cos\theta, \sin\theta)$ . Key identities: $\tan\theta = \sin\theta/\cos\theta$ ; $\sec\theta = 1/\cos\theta$ . Double angle: $\sin 2\theta = 2\sin\theta\cos\theta$ ; $\cos 2\theta = 1-2\sin^2\theta$ . Law of cosines: $c^2 = a^2+b^2-2ab\cos C$ . Area: $\tfrac{1}{2}ab\sin C$ .

Worked example — Non-right triangle (Law of Sines)

In triangle $ABC$ : $a=8$ , $B=35°$ , $C=75°$ . Find side $b$ .

$A = 180°-35°-75° = 70°$ . By sine rule: $\dfrac{b}{\sin B} = \dfrac{a}{\sin A} \Rightarrow b = \dfrac{8\sin 35°}{\sin 70°} \approx \dfrac{4.589}{0.9397} \approx 4.88$ .

Sinusoidal wave

y = 2\sin(2x) + 1

with amplitude

A = 2

, period

T = \frac{2\pi}{B} = \pi \approx 3.14

, and midline

y = 1

▲ EST II Math Traps & Pitfalls — Trigonometry & Angle Measures

Radian vs. Degree Calculator Mode: EST II problems frequently present angles in radians (e.g., $5\pi/6$ ) alongside degree options or vice versa. Always check your calculator's active mode before computing numerical values.
Double-Angle Identity Selection: $\cos(2\theta)$ has three identities: $\cos^2\theta - \sin^2\theta$ , $2\cos^2\theta - 1$ , and $1 - 2\sin^2\theta$ . In equations containing $\sin\theta$ , choose $1 - 2\sin^2\theta$ to produce a single-variable quadratic in sine; in equations containing $\cos\theta$ , choose $2\cos^2\theta - 1$ .
Quadrant Signs (ASTC Rule): Remember: All positive in Q I; Sine positive in Q II; Tangent positive in Q III; Cosine positive in Q IV. For $\sin\theta = -1/2$ , the solutions lie strictly in Quadrants III and IV.
Range of Inverse Trigonometric Functions: Standard principal value ranges: $\arcsin x \in [-\pi/2, \pi/2]$ , $\arccos x \in [0, \pi]$ , and $\arctan x \in (-\pi/2, \pi/2)$ . Calculator evaluations will return only the principal value.

Find all solutions of

\cos 2\theta = \cos\theta

for

0° \leq \theta \leq 360°

2\cos^2\theta-1 = \cos\theta \Rightarrow 2\cos^2\theta-\cos\theta-1=0 \Rightarrow (2\cos\theta+1)(\cos\theta-1)=0

\cos\theta=1 \Rightarrow \theta=0°

;

\cos\theta=-1/2 \Rightarrow \theta=120°, 240°

Solve the trigonometric equation

2\cos^2\theta + 3\sin\theta - 3 = 0

for all solutions in the interval

0 \le \theta < 2\pi

Use Pythagorean identity: Substitute $\cos^2\theta = 1 - \sin^2\theta$ : $2(1 - \sin^2\theta) + 3\sin\theta - 3 = 0 \implies 2 - 2\sin^2\theta + 3\sin\theta - 3 = 0.$ Multiply by $-1$ : $2\sin^2\theta - 3\sin\theta + 1 = 0 \implies (2\sin\theta - 1)(\sin\theta - 1) = 0.$

Case 1: $2\sin\theta - 1 = 0 \implies \sin\theta = \dfrac{1}{2}$ . In $[0, 2\pi)$ , sine is positive in Quadrants I and II with reference angle $\pi/6$ : $\theta = \dfrac{\pi}{6},\quad \theta = \pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}.$

Case 2: $\sin\theta - 1 = 0 \implies \sin\theta = 1$ . In $[0, 2\pi)$ : $\theta = \dfrac{\pi}{2}.$

The complete solution set is $\left\{ \dfrac{\pi}{6},\, \dfrac{\pi}{2},\, \dfrac{5\pi}{6} \right\}$ .

📝 Chapter Quiz

Questions Level

Circles & Coordinate Geometry

🇬🇧 Circles & Coordinate Geometry ~10% · Circle equation & properties · Tangent lines · Coordinate proofs

Circle theorems and coordinate geometry of conic sections are significant topics on the EST II Mathematics subject test.

Circle equation

(x-h)^2+(y-k)^2=r^2

Arc length

r\theta

(θ in radians)

Sector area

\tfrac{1}{2}r^2\theta

Inscribed angle

=\tfrac{1}{2}\times \text{central angle}

Worked example 6.A

Write the equation of a circle with centre $(3,-2)$ and radius 5. $(x-3)^2+(y+2)^2=25.$

⊙ Explorer · Circle equation, centre-radiusinteractive

centre h: 3

centre k: -2

radius r: 5

equation(x-3)² + (y+2)² = 25

Ellipses and hyperbolas

An ellipse:

\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1

. A hyperbola:

\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1

. EST II may ask you to identify the type of conic from a standard-form equation and state key features like vertices, foci, or asymptotes.

● Circles & Coordinate Geometry

Circle equation: $(x-h)^2+(y-k)^2=r^2$ . A tangent to a circle at point $P$ is perpendicular to the radius at $P$ . Circle theorems: angle at centre = twice angle at circumference; angles in same segment are equal; tangent-radius angle = 90°. Chord perpendicular bisector passes through the centre.

Worked example — Tangent from external point

A circle has centre $O(3,2)$ and radius 5. Point $P(10,2)$ lies outside the circle. Find the length of the tangent from $P$ to the circle.

$OP = \sqrt{(10-3)^2+(2-2)^2} = 7$ . Tangent length $= \sqrt{OP^2 - r^2} = \sqrt{49-25} = \sqrt{24} = 2\sqrt{6}$ .

Ellipse centered at

(1, -2)

with horizontal major axis length

2a = 8

and vertical minor axis length

2b = 6

. The foci lie at

(1 \pm \sqrt{7}, -2)

▲ EST II Math Traps & Pitfalls — Circles & Conic Sections

Completing the Square for Circles: When converting general form $x^2 + y^2 + Dx + Ey + F = 0$ to standard form $(x - h)^2 + (y - k)^2 = r^2$ , remember to add $(D/2)^2$ and $(E/2)^2$ to both sides of the equation. Forgetting to add them to the right-hand side produces an incorrect radius.
Tangent Perpendicularity: A line tangent to a circle at point $P(x_1, y_1)$ is always perpendicular to the radius $OP$ . If the radius has slope $m$ , the tangent line has slope $-1/m$ .
Conic Identification from General Quadratic: In $Ax^2 + Cy^2 + Dx + Ey + F = 0$ : if $A = C$ , it is a circle; if $A \neq C$ but both have the same sign ( $AC > 0$ ), it is an ellipse; if $A$ and $C$ have opposite signs ( $AC < 0$ ), it is a hyperbola; if either $A = 0$ or $C = 0$ (but not both), it is a parabola.
Focal Distance Formulas: For an ellipse, $c^2 = a^2 - b^2$ (foci are inside the ellipse, so $c < a$ ). For a hyperbola, $c^2 = a^2 + b^2$ (foci are outside the vertices, so $c > a$ ). Do not confuse these two relations!

Find the equation of the circle that passes through

(0,0)

(6,0)

and

(0,8)

Let

(x-h)^2+(y-k)^2=r^2

. Using

(0,0)

h^2+k^2=r^2

. Using

(6,0)

(6-h)^2+k^2=r^2

gives

h=3

. Using

(0,8)

h^2+(8-k)^2=r^2

gives

k=4

. So

r^2=25

. Equation:

(x-3)^2+(y-4)^2=25

Find the equation of the tangent line to the circle

x^2 + y^2 = 25

at the point

P(3, 4)

Radius slope: The center is the origin $O(0, 0)$ . The slope of the radius connecting $(0, 0)$ and $(3, 4)$ is $m_{\text{radius}} = \dfrac{4 - 0}{3 - 0} = \dfrac{4}{3}.$

Tangent line slope: Since the tangent is perpendicular to the radius, its slope is the negative reciprocal: $m_{\text{tangent}} = -\dfrac{3}{4}.$

Point-slope equation: Using point $P(3, 4)$ : $y - 4 = -\dfrac{3}{4}(x - 3) \implies 4(y - 4) = -3(x - 3) \implies 4y - 16 = -3x + 9 \implies 3x + 4y = 25.$

📝 Chapter Quiz

Questions Level

Solid Geometry & Similarity

🇬🇧 Solid Geometry & Similarity ~5% · Surface area & volume of 3D solids · Scale factors for area/volume

Three-dimensional geometry and similarity ratios appear at the end of the EST II test, often combined with algebraic reasoning.

Sphere volume

\tfrac{4}{3}\pi r^3

Cone volume

\tfrac{1}{3}\pi r^2 h

Cylinder volume

\pi r^2 h

Similarity ratio

lengths

k

: areas

k^2

: volumes

k^3

Worked example 7.A

Two similar spheres have radii 3 and 6. Find the ratio of their volumes. $\text{Ratio of radii}=1:2,\quad\text{Volume ratio}=1^3:2^3=1:8.$

● Solid Geometry & Similarity

Sphere: SA $=4\pi r^2$ , V $=\tfrac{4}{3}\pi r^3$ . Cylinder: SA $=2\pi r^2+2\pi rh$ , V $=\pi r^2 h$ . Cone: SA $=\pi r^2+\pi rl$ , V $=\tfrac{1}{3}\pi r^2 h$ . For similar solids with scale factor $k$ : area scales as $k^2$ , volume scales as $k^3$ .

Worked example — Volume of composite solid

A solid consists of a cylinder (radius 4 cm, height 6 cm) with a hemisphere on top. Find the total volume.

Cylinder: $\pi(16)(6) = 96\pi$ . Hemisphere: $\tfrac{1}{2}\cdot\tfrac{4}{3}\pi(64) = \tfrac{128\pi}{3}$ .

Total $= 96\pi + \tfrac{128\pi}{3} = \dfrac{288\pi+128\pi}{3} = \dfrac{416\pi}{3} \approx 435.6$ cm $^3$ .

Right circular cylinder of radius

r = 4

and height

h = 10

. Volume

V = \pi r^2 h = 160\pi

, lateral area

L = 2\pi rh = 80\pi

, total surface area

A = 2\pi r^2 + 2\pi rh = 112\pi

▲ EST II Math Traps & Pitfalls — 3D Solids & Similarity Ratios

Similar Solids Power Laws: When two solids are similar with linear scale factor $k = L_1/L_2$ : corresponding surface areas scale as $k^2$ , and volumes scale as $k^3$ . If a question states that volume increases by a factor of 8, the linear dimensions double ( $\sqrt[3]{8} = 2$ ), and the surface area quadruples ( $2^2 = 4$ ).
Cone Slant Height vs. Vertical Height: The volume formula $V = \frac{1}{3}\pi r^2 h$ requires the vertical altitude $h$ , NOT the slant height $l$ . Use the Pythagorean theorem $r^2 + h^2 = l^2$ to solve for $h$ before calculating volume.
Total Surface Area vs. Lateral Surface Area: Questions often ask for "lateral surface area" (excluding the circular base(s)) or "open cylinder" (one base). Always read carefully whether one, two, or zero circular bases are included.
Composite Solids Volume vs. Surface Area: While volumes of combined parts add directly ( $V_{\text{total}} = V_1 + V_2$ ), surface areas do NOT add directly because the joining interface/contact face is interior and must be subtracted from both solids.

Two similar cones have surface areas in the ratio

4:9

. If the smaller cone has volume

16\pi

^3

, find the volume of the larger.

Area ratio

= k^2 = 4/9

, so

k = 2/3

. Volume ratio

= k^3 = 8/27

. Larger volume

= 16\pi \times 27/8 = 54\pi

^3

A right circular cone has base radius

r = 6

cm and slant height

l = 10

cm. (a) Determine the vertical height

h

and volume

V

of the cone in terms of

\pi

. (b) Calculate the total surface area of the cone.

(a) Height and Volume: The radius, vertical height, and slant height form a right triangle: $h = \sqrt{l^2 - r^2} = \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8\text{ cm}.$ The volume is: $V = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi (36)(8) = 96\pi\text{ cm}^3.$

(b) Total Surface Area: Sum of base area and lateral area: $A = \pi r^2 + \pi r l = \pi(6^2) + \pi(6)(10) = 36\pi + 60\pi = 96\pi\text{ cm}^2.$

📝 Chapter Quiz

Questions Level

Statistics & Probability

🇬🇧 Statistics & Probability ~10% · Permutations & combinations · Probability rules · Data distributions

Data analysis and probability on the EST II go beyond basic means and medians to include distributions, graphical representations, and counting methods.

Permutations

_nP_r=\dfrac{n!}{(n-r)!}

Combinations

\binom{n}{r}=\dfrac{n!}{r!(n-r)!}

Conditional probability

P(A|B)=\dfrac{P(A\cap B)}{P(B)}

Standard deviation

measures spread from mean

Worked example 8.A

In how many ways can 3 students be chosen from a group of 8? $\binom{8}{3}=\frac{8!}{3!\cdot 5!}=\frac{8\times7\times6}{6}=56.$

● Statistics, Counting & Probability

Permutations (order matters): $^nP_r = \dfrac{n!}{(n-r)!}$ . Combinations (order irrelevant): $^nC_r = \dfrac{n!}{r!(n-r)!}$ . Conditional probability: $P(A|B) = \dfrac{P(A \cap B)}{P(B)}$ . Independent events: $P(A \cap B) = P(A) \cdot P(B)$ . Normal distribution: ~68% within $1\sigma$ , ~95% within $2\sigma$ .

Worked example — Combinations with restrictions

From 4 boys and 5 girls, choose a team of 4 with at least 2 girls. How many ways?

Exactly 2G: $\binom{5}{2}\binom{4}{2}=10 \times 6=60$ . Exactly 3G: $\binom{5}{3}\binom{4}{1}=10 \times 4=40$ . Exactly 4G: $\binom{5}{4}\binom{4}{0}=5$ . Total: $105$ .

Normal distribution with mean

\mu = 100

and standard deviation

\sigma = 15

. Approximately 68.2% of data falls within

[85, 115]

(shaded), 95.4% within

[70, 130]

, and 99.7% within

[55, 145]

▲ EST II Math Traps & Pitfalls — Statistics & Counting Methods

Permutation vs. Combination: If order or role assignment matters (e.g. President/VP/Treasurer, race positions, arrangements in a line), use permutations $_nP_r = \frac{n!}{(n-r)!}$ . If selection order does not distinguish groups (e.g. committees, hand of cards, team rosters), use combinations $\binom{n}{r} = \frac{n!}{r!(n-r)!}$ .
Mutually Exclusive vs. Independent Events: Two events cannot be both mutually exclusive and independent (unless one has probability 0)! Mutually exclusive means $P(A \cap B) = 0$ , so $P(A \cup B) = P(A) + P(B)$ . Independent means $P(A \cap B) = P(A) \cdot P(B)$ .
Conditional Probability Denominator: In $P(A|B) = \frac{P(A \cap B)}{P(B)}$ , the conditioning event $B$ becomes the new, restricted sample space in the denominator. Never divide by $P(A)$ when conditioning on $B$ .
Standard Deviation vs. Variance: Variance is $\sigma^2$ (squared units), while standard deviation is $\sigma = \sqrt{\text{variance}}$ (original units). Standard deviation measures dispersion: adding a constant to all data values leaves $\sigma$ unchanged, while multiplying by $c$ multiplies $\sigma$ by $|c|$ .

Events

A

and

B

are such that

P(A)=0.6

P(B)=0.5

P(A|B)=0.4

. Find

P(A \cup B)

P(A \cap B) = P(A|B)\cdot P(B) = 0.4 \times 0.5 = 0.2

P(A \cup B) = 0.6+0.5-0.2 = 0.9

A committee of 5 is to be formed from 6 men and 8 women. (a) In how many ways can the committee be formed if it must contain exactly 3 women? (b) What is the probability that the committee contains at least 4 women?

(a) Exactly 3 women: We must choose 3 women from 8 and 2 men from 6: $\binom{8}{3} \times \binom{6}{2} = 56 \times 15 = 840\text{ ways}.$

(b) Probability of at least 4 women:

Total ways to form any 5-member committee from 14 people: $\binom{14}{5} = \dfrac{14 \times 13 \times 12 \times 11 \times 10}{5 \times 4 \times 3 \times 2 \times 1} = 2002.$
Exactly 4 women (4W, 1M): $\binom{8}{4}\binom{6}{1} = 70 \times 6 = 420$ .
Exactly 5 women (5W, 0M): $\binom{8}{5}\binom{6}{0} = 56 \times 1 = 56$ .
Favorable outcomes: $420 + 56 = 476$ .
Probability: $P(\text{at least 4 women}) = \dfrac{476}{2002} = \dfrac{34}{143} \approx 0.2377\, (23.8\%).$

📝 Chapter Quiz

Questions Level

Practice set

Fifty EST II Mathematics problems spanning advanced algebra, functions, sequences, trigonometry, geometry, circles, solids, and statistics — graded by difficulty.

Factor completely: $x^3 - 4x$ .

Intermediate

x(x^2-4)=x(x-2)(x+2).

Simplify $\dfrac{x^2-9}{x^2-x-6}$ .

Intermediate

\dfrac{(x-3)(x+3)}{(x-3)(x+2)}=\dfrac{x+3}{x+2},\quad x\neq 3.

When $f(x)=x^3-2x^2+x-5$ is divided by $(x-2)$ , find the remainder.

Intermediate

f(2)=8-8+2-5=-3.

Solve $|3x-1|=8$ .

Intermediate

3x-1=8\Rightarrow x=3;\quad 3x-1=-8\Rightarrow x=-\tfrac{7}{3}.

Is $(x-1)$ a factor of $x^3 - 3x + 2$ ?

Advanced

$f(1)=1-3+2=0$ , so yes.

\text{Yes — }(x-1)\text{ is a factor.}

Solve $2x^2 - 3x - 5 = 0$ .

Intermediate

(2x-5)(x+1)=0 \Rightarrow x=\tfrac{5}{2}\text{ or }x=-1.

Solve the system $x^2+y=5$ , $x+y=3$ .

Advanced

$y=3-x$ . Substitute: $x^2+3-x=5\Rightarrow x^2-x-2=0\Rightarrow x=2\text{ or }x=-1$ .

(2,1)\text{ and }(-1,4).

Solve $\dfrac{2x}{x-1}=3$ .

Advanced

2x=3(x-1)=3x-3 \Rightarrow x=3.

Solve $\sqrt{2x+3}=5$ .

Intermediate

2x+3=25 \Rightarrow x=11.

Find the range of $f(x)=x^2+3$ .

Intermediate

[3,\infty).

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

Intermediate

f(g(x))=2(x+3)-1=2x+5.

Find $f^{-1}(x)$ for $f(x)=\dfrac{x+2}{3}$ .

Intermediate

y=\dfrac{x+2}{3}\Rightarrow x=3y-2 \Rightarrow f^{-1}(x)=3x-2.

Solve $\log_2(x-1)=3$ .

Intermediate

x-1=2^3=8 \Rightarrow x=9.

Simplify $\log_5 125 - \log_5 25$ .

Intermediate

\log_5\!\dfrac{125}{25}=\log_5 5=1.

Solve $e^x = 10$ . Give exact form.

Advanced

x=\ln 10.

Find the 15th term of the arithmetic sequence $4,\,7,\,10,\ldots$

Intermediate

a_{15}=4+(14)(3)=46.

Find the sum of the first 20 terms of $2+5+8+\cdots$

Intermediate

$a_{20}=2+19\times3=59$ .

S_{20}=\dfrac{20}{2}(2+59)=610.

Find the 6th term of the geometric sequence $3,\,6,\,12,\ldots$

Intermediate

a_6=3\cdot2^5=96.

Find the sum to infinity of $8 + 4 + 2 + \cdots$

Advanced

S_\infty=\dfrac{8}{1-\tfrac{1}{2}}=16.

An arithmetic sequence has $a_1=5$ and common difference 4. Find the sum of the first 10 terms.

Advanced

S_{10}=\dfrac{10}{2}(2\times5+9\times4)=5\times46=230.

Find the exact value of $\sin 30^\circ + \cos 60^\circ$ .

Basic

\tfrac{1}{2}+\tfrac{1}{2}=1.

Simplify $\sin^2\theta + \cos^2\theta$ .

Basic

1.

Prove: $\dfrac{\sin^2\theta}{1-\cos\theta}=1+\cos\theta$ .

Advanced

\dfrac{1-\cos^2\theta}{1-\cos\theta}=\dfrac{(1-\cos\theta)(1+\cos\theta)}{1-\cos\theta}=1+\cos\theta.\ \checkmark

In a right triangle, $\sin\theta=\dfrac{5}{13}$ . Find $\cos\theta$ .

Intermediate

\cos\theta=\sqrt{1-\tfrac{25}{169}}=\dfrac{12}{13}.

Find the exact value of $\tan 45^\circ + \sin 90^\circ$ .

Intermediate

1+1=2.

In triangle $ABC$ , $AB=6$ , $BC=8$ , and angle $B=90^\circ$ . Find $AC$ .

Intermediate

AC=\sqrt{36+64}=10.

Find the area of a triangle with sides 5, 7, and included angle $60^\circ$ .

Advanced

A=\tfrac{1}{2}\times5\times7\times\sin60^\circ=\dfrac{35\sqrt{3}}{4}.

Use the law of cosines to find $c$ when $a=5$ , $b=7$ , $C=90^\circ$ .

Advanced

c^2=25+49-0=74 \Rightarrow c=\sqrt{74}.

Find the equation of a circle centred at $(2,-3)$ with radius 5.

Intermediate

(x-2)^2+(y+3)^2=25.

Find the centre and radius of $x^2+y^2-4x+6y-12=0$ .

Advanced

Complete the square: $(x-2)^2+(y+3)^2=25$ .

\text{Centre }(2,-3),\quad r=5.

An arc of a circle with radius 6 subtends an angle of $2$ radians. Find the arc length.

Intermediate

s=r\theta=6\times2=12.

A sector has radius 4 and angle $\pi/3$ . Find its area.

Advanced

A=\tfrac{1}{2}r^2\theta=\tfrac{1}{2}(16)\tfrac{\pi}{3}=\dfrac{8\pi}{3}.

An inscribed angle intercepts an arc of $80^\circ$ . Find the angle.

Intermediate

\text{Inscribed angle}=\tfrac{1}{2}\times80^\circ=40^\circ.

Find the volume of a sphere with radius 3.

Intermediate

V=\tfrac{4}{3}\pi(3)^3=36\pi.

Find the surface area of a cone with radius 4 and slant height 5.

Advanced

SA=\pi r(r+l)=\pi(4)(9)=36\pi.

Two similar cones have radii $3$ and $6$ . Find the ratio of their volumes.

Advanced

\left(\dfrac{3}{6}\right)^3=\dfrac{1}{8}.

Find the volume of a cylinder with radius 5 and height 9.

Intermediate

V=\pi r^2 h=225\pi.

Five test scores are 70, 75, 80, 85, 90. Find the mean and range.

Basic

\bar{x}=80;\quad \text{range}=90-70=20.

How many ways can 4 students be chosen from 10?

Intermediate

\binom{10}{4}=\dfrac{10!}{4!\cdot6!}=210.

A bag contains 5 red and 3 blue balls. Two are drawn without replacement. Find $P(\text{both red})$ .

Advanced

\dfrac{5}{8}\times\dfrac{4}{7}=\dfrac{20}{56}=\dfrac{5}{14}.

Evaluate $5! \div 3!$ .

Basic

\dfrac{120}{6}=20.

Two events $A$ and $B$ are independent. $P(A)=0.4$ , $P(B)=0.5$ . Find $P(A\cap B)$ .

Intermediate

P(A\cap B)=0.4\times0.5=0.2.

Expand $(x+2)^4$ using the binomial theorem. Find the coefficient of $x^3$ .

Advanced

\binom{4}{1}2^1=8.

A complex number is $z=3+4i$ . Find $|z|$ .

Intermediate

|z|=\sqrt{9+16}=5.

Multiply $(2+3i)(1-i)$ .

Advanced

2-2i+3i-3i^2=2+i+3=5+i.

Express $0.3\overline{6}$ as a fraction. ( $0.3666\ldots$ )

Advanced

Let $x=0.3\overline{6}$ . Then $10x=3.\overline{6}$ , $9x=3.3$ .

x=\dfrac{3.3}{9}=\dfrac{11}{30}.

How many 3-digit numbers can be formed from $\{1,2,3,4,5\}$ with no repetition?

Intermediate

_5P_3=5\times4\times3=60.

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

Advanced

\dfrac{(x-2)(x+2)}{x-2}=x+2 \xrightarrow{x\to2} 4.

A geometric sequence has first term 6 and common ratio $-\tfrac{1}{2}$ . Find the 5th term.

Intermediate

a_5=6\times\left(-\tfrac{1}{2}\right)^4=6\times\tfrac{1}{16}=\tfrac{3}{8}.

Solve $2^{3x}=16^{x-1}$ .

Advanced

$2^{3x}=2^{4(x-1)} \Rightarrow 3x=4x-4$ .

x=4.

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