Seventeen lessons on algebra, functions, trigonometry, probability, statistics, calculus and proof for the Analysis and Approaches course.
A skills course on common problem structures across the five topic areas at standard-level depth. It does not replace the syllabus. Higher-level-only content (induction, complex numbers, vectors and Paper 3 topics) is not covered.
Algebra, coordinate geometry and basic trigonometry.
Course outline
Exponent and logarithm laws
Rewrite expressions and solve equations using the laws of exponents and logarithms.
Sequences and series
Identify arithmetic and geometric sequences and use sum formulas, including sums to infinity.
Binomial expansion
Expand (a + b)^n using the binomial theorem and find a specific term.
Functions: domain, range, composite and inverse
Find domain and range and build composite and inverse functions correctly.
Transformations: read g(x) = a f(x - h) + k as a recipe
Describe and apply translations, stretches and reflections to graphs.
Quadratics: the discriminant tells you before you solve
Use the discriminant, vertex form and factorisation to analyse roots and graphs.
Exponential and logarithmic models
Build models of growth and decay and interpret their parameters.
Radians, arcs and the unit circle
Convert between degrees and radians and use arc length, sector area and exact values.
Trig equations and identities
Solve trig equations over a given interval and use the Pythagorean and double-angle identities.
Sine rule, cosine rule and area of a triangle
Choose the right rule for a non-right triangle and handle the ambiguous case.
Probability: conditional, independent and tree diagrams
Use Venn diagrams, tree diagrams and the conditional probability formula.
Binomial and normal distributions
Recognise binomial settings and use the normal distribution with technology and standardisation.
Descriptive statistics and regression
Summarise data, interpret correlation and use a regression line responsibly.
Differentiation: rules, tangents and normals
Differentiate with the power, chain, product and quotient rules and find tangent and normal lines.
Applications of differentiation: shape and optimisation
Locate stationary points, classify them and solve optimisation problems.
Integration: antiderivatives, areas and kinematics
Integrate common functions and use definite integrals for area and motion.
Proof, reasoning and the mathematical exploration
Construct a short proof, use counterexamples and plan an exploration you can finish.
Sources and curriculum note
Reviewed October 5, 2026 against the IB subject brief for the current course (first assessments 2021). The brief gives SL Paper 1 40 percent, Paper 2 40 percent and the exploration 20 percent, with a recommended 15 hours for the exploration. The IB lists a new Mathematics: analysis and approaches course with first teaching August 2027 and first assessment May 2029; its content is not described here. Check the official IB page for changes.
Read every lesson below. The interactive reader above contains the same explanations, with visual tools and quizzes.
1. Exponent and logarithm laws
Learning goal: Rewrite expressions and solve equations using the laws of exponents and logarithms.
A logarithm is the inverse of an exponent. The statement log_a(b) = c means exactly that a^c = b, with a > 0 and a not equal to 1, and b > 0. Everything about logarithms follows from this translation. When a question gives you a logarithm equation, convert it into exponent form to isolate the unknown. When it gives an exponential equation with the unknown in the power, take logarithms of both sides to bring the power down.
The exponent laws are a^m x a^n = a^(m+n), a^m / a^n = a^(m-n) and (a^m)^n = a^(mn). The logarithm laws mirror them: log(xy) = log x + log y, log(x/y) = log x - log y and log(x^n) = n log x. Two further facts are log_a(1) = 0 and log_a(a) = 1. A common mistake is to write log(x + y) as log x + log y, which is false. No law exists for the logarithm of a sum.
To change base use log_a(x) = log_b(x) / log_b(a). This is how a calculator with only natural or base-10 logarithms evaluates log_2(7), by computing log(7) / log(2), which is about 2.807. Natural logarithms use the base e, written ln, and ln(e^x) = x. When solving 3^x = 20, take ln of both sides to get x ln 3 = ln 20, so x = ln 20 / ln 3, which is about 2.73.
When solving logarithm equations, combine the logarithms into a single one first, then convert to exponent form. Always check the answers against the domain: the argument of every logarithm must be positive. For example, log(x) + log(x - 3) = 1 gives x(x - 3) = 10, so x^2 - 3x - 10 = 0 and x = 5 or x = -2. The value x = -2 is rejected, because log(-2) is undefined. Quoting the rejection and the reason shows complete reasoning.
Worked example
Solve 2^x = 50. Give the answer to three significant figures.
Take natural logs: x ln 2 = ln 50.
x = ln 50 / ln 2.
ln 50 = 3.912 and ln 2 = 0.6931.
x = 5.64.
Practice problem and solution
Solve log_3(x) = 4. Enter the value of x.
Convert to exponent form: x = 3^4 = 81.
Mental model: Convert between log and exponent forms.
Common trap: Splitting log(x + y).
2. Sequences and series
Learning goal: Identify arithmetic and geometric sequences and use sum formulas, including sums to infinity.
A sequence is an ordered list of numbers, and a series is the sum of the terms. In an arithmetic sequence each term differs from the previous by a constant common difference d. The nth term is u_n = u_1 + (n - 1)d. The sum of the first n terms is S_n = n/2 (2u_1 + (n - 1)d), which can also be written as n/2 (u_1 + u_n). When the question gives two terms, form two equations from the nth term formula and solve for u_1 and d.
In a geometric sequence each term is the previous one multiplied by a common ratio r. The nth term is u_n = u_1 r^(n-1) and the sum of the first n terms is S_n = u_1 (r^n - 1) / (r - 1), or equivalently u_1 (1 - r^n) / (1 - r). To identify the type, compute the differences and the ratios of consecutive terms. If the differences match, it is arithmetic. If the ratios match, it is geometric. For compound interest and population growth, the multiplier is geometric.
The sum to infinity of a geometric series exists only when the absolute value of r is less than 1. Then S_infinity = u_1 / (1 - r). For example, 8 + 4 + 2 + 1 + ... has u_1 = 8 and r = 1/2, so its sum to infinity is 8 / (1/2) = 16. Always check the condition on r before using the formula, and quote the condition in your answer. If r is 1 or greater in absolute value, the terms do not shrink, and the series does not converge.
Sigma notation writes a sum compactly: the sum from k = 1 to n of an expression in k. Identify the first term by substituting the lower limit, then work out the common difference or ratio from the second term. The number of terms in a sum from k = a to k = b is b - a + 1, a typical off-by-one trap. Problems about money often ask for the first year in which a total passes a target. Set up an inequality with the sum formula and solve with logarithms for a geometric series, or with the quadratic formula for an arithmetic one, then round up to a whole number of terms.
Worked example
Find the sum of the first 10 terms of the arithmetic sequence 3, 7, 11, ...
u_1 = 3 and d = 4.
S_10 = 10/2 (2 x 3 + 9 x 4).
= 5 x (6 + 36).
S_10 = 210.
Practice problem and solution
Find the sum to infinity of the geometric series 18 + 6 + 2 + ... Enter the value.
u_1 = 18, r = 1/3. S = 18 / (1 - 1/3) = 18 / (2/3) = 27.
Mental model: Constant difference or constant ratio decides the type.
Common trap: Using the sum to infinity when r is not less than 1.
3. Binomial expansion
Learning goal: Expand (a + b)^n using the binomial theorem and find a specific term.
The binomial theorem expands (a + b)^n as the sum of terms of the form nCr a^(n-r) b^r, for r from 0 to n. The coefficients nCr are the number of ways to choose r items from n and equal n! / (r!(n - r)!). They also appear in rows of Pascal's triangle, where each entry is the sum of the two above it. The first row is 1, the next 1 1, then 1 2 1, then 1 3 3 1, then 1 4 6 4 1. The row for n has n + 1 entries, and the powers of a decrease while the powers of b increase.
To expand (2x + 3)^4, write each term with its binomial coefficient, then evaluate the power of each part, including the constant. The term for r = 2 is 6 x (2x)^2 x 3^2 = 6 x 4x^2 x 9 = 216x^2. A common error is forgetting to raise the coefficient to the power, for example writing 2x^2 instead of 4x^2. Write the pieces in brackets before multiplying, and take extra care with negative signs when b is negative.
The general term is T_(r+1) = nCr a^(n-r) b^r. To find a specific term, such as the coefficient of x^5 in (1 + 2x)^8, set up the power of x and solve for r. Here r = 5, so the term is 8C5 x 1^3 x (2x)^5 = 56 x 32x^5 = 1792x^5. For terms with x in both parts, such as (x + 2/x)^6, write the power of x as an expression in r and solve for the value that gives the required power, then check that r is a whole number between 0 and n.
Binomial expansions are used to approximate powers and in probability. The sum of the coefficients of (a + b)^n is 2^n, which you can check by substituting a = 1 and b = 1. In calculations, use the calculator for nCr to avoid arithmetic slips, but write the setup. If a question asks for the constant term, find the value of r that makes the power of x equal to zero. State the result with the sign. Method marks are given for the correct general term, so show it even when you use technology for the arithmetic.
Worked example
Find the coefficient of x^2 in the expansion of (1 + 3x)^5.
General term: 5Cr x 1^(5-r) x (3x)^r.
For x^2, r = 2.
5C2 = 10 and 3^2 = 9.
Coefficient = 10 x 9 = 90.
Practice problem and solution
Find the coefficient of x^3 in the expansion of (2 + x)^5. Enter the number.
General term 5Cr x 2^(5-r) x^r. For r = 3: 5C3 = 10 and 2^2 = 4, so the coefficient is 40.
Mental model: Write the general term first.
Common trap: Forgetting to raise the numerical part to its power.
4. Functions: domain, range, composite and inverse
Learning goal: Find domain and range and build composite and inverse functions correctly.
A function assigns each input in its domain exactly one output. The domain is the set of allowed inputs, and the range is the set of outputs. Domain restrictions come from three sources: division by zero, even roots of negative numbers and logarithms of non-positive numbers. For f(x) = sqrt(x - 2), the domain is x at least 2 and the range is values at least 0. When a function is described by a graph, read the domain along the horizontal axis and the range along the vertical axis. Use inequality notation, and be careful with open and closed endpoints.
A composite function f(g(x)) means apply g first, then f. Substitute the whole expression g(x) into f. The order matters: f(g(x)) is usually not equal to g(f(x)). The domain of the composite is limited by the domain of g and by the requirement that g(x) lies in the domain of f. For example, if f(x) = 1/x and g(x) = x - 1, then f(g(x)) = 1 / (x - 1) is undefined at x = 1, which is a restriction that comes from the composite.
The inverse function f^(-1) reverses f, so f^(-1)(f(x)) = x. To find it, write y = f(x), swap x and y, then solve for y. A function has an inverse only if it is one-to-one, which means that it passes the horizontal line test. The graph of the inverse is the reflection of the original in the line y = x. The domain of the inverse is the range of the original, and the range of the inverse is the domain of the original. Use this to state the domain correctly, especially after restricting a quadratic to make it one-to-one.
Functions can also be classified as even or odd, but at this stage focus on three skills. First, evaluate and simplify composites without losing brackets, particularly when g(x) is substituted into a squared term. Second, find the inverse by algebra and check by composing: f(f^(-1)(x)) should equal x. Third, link points between a function and its inverse by swapping the coordinates. If (2, 5) lies on y = f(x), then (5, 2) lies on y = f^(-1)(x). A frequent mistake is to confuse the inverse with the reciprocal, since f^(-1)(x) does not mean 1 / f(x).
Worked example
f(x) = 2x + 3. Find f^(-1)(x).
y = 2x + 3.
Swap: x = 2y + 3.
Solve: y = (x - 3) / 2.
f^(-1)(x) = (x - 3) / 2.
Practice problem and solution
f(x) = x^2 and g(x) = x + 3. Find f(g(2)). Enter the value.
g(2) = 5, then f(5) = 25.
Mental model: Domain, range, and the order of composition all need checking.
Common trap: Treating the inverse as 1 / f(x).
5. Transformations: read g(x) = a f(x - h) + k as a recipe
Learning goal: Describe and apply translations, stretches and reflections to graphs.
Transformations change the graph of y = f(x) in predictable ways. A vertical translation is y = f(x) + k: positive k moves the graph up. A horizontal translation is y = f(x - h): the graph moves right by h. The sign inside the brackets is the opposite of what intuition suggests, because f(x - 3) takes the value that f had at x - 3, so features appear three units later. Always check the direction by following a single point, such as the vertex or intercept, through the transformation.
A vertical stretch by factor a is y = a f(x), which multiplies all y-coordinates by a. A horizontal stretch by factor 1/b is y = f(bx), which divides the x-coordinates by b. For example, y = f(2x) compresses the graph horizontally, so every x-coordinate is halved. Reflections are special cases: y = -f(x) reflects in the x-axis and y = f(-x) reflects in the y-axis. Transformations inside the brackets affect x and behave in the reverse way to what the sign suggests.
When several transformations are combined, the order matters. For y = a f(b(x - h)) + k, a standard order is: apply the horizontal changes first, then the vertical ones. Write the transformation of a point (x, y) as ((x / b) + h, a y + k) and check with a known point. For a quadratic in vertex form, y = a(x - h)^2 + k, the vertex is at (h, k), a determines the stretch and direction, and the axis of symmetry is x = h. These links make the vertex form a transformation recipe.
To describe a transformation in words, name the type, the direction and the factor or distance. Use words such as translation, vertical stretch with scale factor, and reflection in the x-axis. When sketching, mark the transformed key points: intercepts, turning points and asymptotes. Asymptotes also move: for y = 1/x translated by (h, k), the vertical asymptote is x = h and the horizontal asymptote is y = k. Include the new coordinates of features on the sketch, because examiners reward labelled key points.
Worked example
The point (4, 6) lies on y = f(x). Find the image on y = 3f(x - 2) + 1.
Horizontal: x = 4 + 2 = 6.
Vertical: y = 3 x 6 + 1 = 19.
The image is (6, 19).
Check using the order of operations.
Practice problem and solution
The vertex of y = f(x) is (1, -2). Find the y-coordinate of the vertex of y = 2f(x) + 5.
The vertex y-coordinate becomes 2 x (-2) + 5 = 1.
Mental model: Follow one key point through the recipe.
Common trap: Changing the x-coordinate for a vertical transformation.
6. Quadratics: the discriminant tells you before you solve
Learning goal: Use the discriminant, vertex form and factorisation to analyse roots and graphs.
A quadratic function can be written in three forms. The standard form ax^2 + bx + c shows the y-intercept c. The factored form a(x - p)(x - q) shows the roots p and q, which are the x-intercepts. The vertex form a(x - h)^2 + k shows the vertex (h, k) and the axis of symmetry x = h. Choose the form according to what the question asks: roots suggest factored form, maximum or minimum suggests vertex form, and intercepts with the axes suggest standard form.
The roots of ax^2 + bx + c = 0 come from the quadratic formula, x = (-b plus or minus the square root of (b^2 - 4ac)) / (2a). The expression b^2 - 4ac is the discriminant. If it is positive, there are two distinct real roots; if zero, there is one repeated root, and the graph touches the x-axis; if negative, there are no real roots, and the graph does not cross the axis. This information is available without solving, which is useful for questions on the number of intersections of a line and a curve.
To find where a line meets a curve, set the expressions equal, rearrange to zero and examine the discriminant. For a tangent, the line touches the curve once, so the discriminant equals zero. For questions with a parameter, such as finding k so that x^2 + kx + 9 = 0 has one repeated root, write k^2 - 36 = 0, giving k = 6 or k = -6. Present the condition first, then solve. Remember that sign matters for the number of solutions: for no real roots, the discriminant must be less than zero.
The sum of the roots is -b / a and the product is c / a. These let you form an equation from its roots or check solutions. For the vertex, the x-coordinate is -b / (2a), and substituting gives the y-coordinate. Completing the square rewrites standard form as vertex form: x^2 + 6x + 5 = (x + 3)^2 - 4, with vertex (-3, -4). For applications, such as the height of a thrown ball, the vertex gives the maximum height, and the roots give the times when the ball is at ground level. Interpret each value in context, with units.
Worked example
Find the values of k for which x^2 + kx + 16 = 0 has exactly one solution.
One solution means the discriminant is zero.
k^2 - 4 x 1 x 16 = 0.
k^2 = 64.
k = 8 or k = -8.
Practice problem and solution
Find the x-coordinate of the vertex of y = 2x^2 - 12x + 7.
x = -b / (2a) = 12 / 4 = 3.
Mental model: Pick the form that shows what you need.
Common trap: Using the wrong sign for b.
7. Exponential and logarithmic models
Learning goal: Build models of growth and decay and interpret their parameters.
An exponential model has the form y = A x b^t or y = A e^(kt). The constant A is the value at t = 0, and b or k controls the rate. If b is greater than 1 or k is positive, the model shows growth; if b is between 0 and 1 or k is negative, it shows decay. The key feature is that equal time steps multiply the quantity by the same factor, so the percentage change per period is constant. Linear models, by contrast, add a constant amount each period.
To interpret a model, translate each number. For a population P = 500 x 1.04^t, with t in years, the initial population is 500 and it grows by 4 percent each year. For P = 500 e^(0.04t), the continuous growth rate is 0.04 per year, but the equivalent annual factor is e^0.04, about 1.0408. Do not call these the same percentage. When asked to compare two models, calculate the same quantity, such as the value after 10 years, and also state the units.
To find the time at which a quantity reaches a target, set up an equation and take logarithms. For 500 x 1.04^t = 1000, divide by 500 to get 1.04^t = 2, then t = ln 2 / ln 1.04, which is about 17.7 years. Half-life, the time for a decaying quantity to halve, follows the same method: for A e^(-kt), solve e^(-kt) = 1/2 to get t = ln 2 / k. Notice that half-life does not depend on the starting amount, which is a signature of exponential decay.
The graph of an exponential function has a horizontal asymptote, y = 0 for the basic form, since the quantity approaches zero but never reaches it. In a model with an added constant, such as y = A e^(-kt) + c, the asymptote is y = c, which represents a limiting value like room temperature in a cooling model. When fitting a model to data, use technology to find the parameters, then check the fit with a residual or by comparing predictions with known points. Comment on the limitations: real populations cannot grow without bound, so the model applies only within a limited range.
Worked example
A quantity decays as Q = 80 e^(-0.2t). Find the half-life to three significant figures.
Set Q = 40: 40 = 80 e^(-0.2t).
e^(-0.2t) = 0.5.
-0.2t = ln 0.5 = -0.6931.
t = 3.47.
Practice problem and solution
A population is P = 100 x 2^t, t in years. Find the time in years for P to reach 800.
2^t = 8, so t = 3.
Mental model: Equal time steps give equal multipliers.
Common trap: Confusing continuous rates with annual percentages.
8. Radians, arcs and the unit circle
Learning goal: Convert between degrees and radians and use arc length, sector area and exact values.
A radian is the angle at the centre of a circle subtended by an arc equal in length to the radius. A full turn is 2 pi radians, which equals 360 degrees, so pi radians equal 180 degrees. To convert from degrees to radians multiply by pi / 180, and for the reverse multiply by 180 / pi. Many problems in calculus use radians, since derivatives of trigonometric functions take their simple forms only when the angle is in radians. Check the calculator mode before computing, as this is a frequent source of wrong answers.
With the angle theta in radians, the arc length is l = r theta and the area of a sector is A = (1/2) r^2 theta. These formulas are simpler than the degree versions, which include a factor theta / 360 multiplied by the circumference or the area. For a circle of radius 6 cm and a sector angle of 2 radians, the arc length is 12 cm and the sector area is 36 cm^2. The perimeter of a sector includes the two radii as well as the arc, which is often forgotten.
The unit circle has radius 1, and a point at angle theta has coordinates (cos theta, sin theta). Tangent is sin theta / cos theta, which is the gradient of the line from the origin. Signs follow the quadrants: sine is positive in the first and second quadrants, cosine is positive in the first and fourth, and tangent is positive in the first and third. Use the reference angle, the acute angle made with the horizontal axis, to find values in other quadrants.
Exact values for 30, 45 and 60 degrees (pi/6, pi/4 and pi/3) should be known. Sine of pi/6 is 1/2, sine of pi/4 is the square root of 2 over 2 and sine of pi/3 is the square root of 3 over 2. Cosine takes the same values in reverse order. Tangent of pi/4 is 1. Using the triangle diagrams for 45-45-90 and 30-60-90 triangles reconstructs these values if memory fails. In a no-technology paper, leave answers in exact form unless asked otherwise, and write fractions of pi in simplest form.
Worked example
A sector has radius 5 cm and angle 1.2 radians. Find the arc length and area.
Arc length = r theta = 5 x 1.2 = 6 cm.
Area = (1/2) r^2 theta.
= 0.5 x 25 x 1.2.
Area = 15 cm^2.
Practice problem and solution
Convert 150 degrees to radians and give the answer as a multiple of pi. Enter the multiplier as a decimal to three places.
150 x pi / 180 = 5 pi / 6, and 5/6 = 0.833.
Mental model: Radians make arc and sector formulas simple.
Common trap: Using degree mode for radian problems.
9. Trig equations and identities
Learning goal: Solve trig equations over a given interval and use the Pythagorean and double-angle identities.
To solve a trig equation such as sin x = 0.5 over a given interval, find the principal value from the calculator, then find all other solutions in the interval using symmetry. For sine, if x is a solution then pi - x is also a solution, plus multiples of 2 pi. For cosine, if x is a solution then 2 pi - x (or -x) is also a solution. For tangent, solutions repeat every pi. Draw a quick sketch of the graph over the interval to see how many solutions to expect before you calculate them.
When the equation contains a multiple angle, such as sin(2x) = 0.5, solve for the whole angle first and then divide. If x lies between 0 and 2 pi, then 2x lies between 0 and 4 pi, so there are more solutions for 2x than for x. List all values of 2x in the extended interval, then divide each by 2. A frequent error is to divide the interval boundaries but not extend them, which loses half of the solutions.
Two identities are essential. The Pythagorean identity says sin^2 x + cos^2 x = 1, which lets you replace sin^2 x by 1 - cos^2 x. The tangent identity says tan x = sin x / cos x. The double angle identities are sin 2x = 2 sin x cos x and cos 2x = cos^2 x - sin^2 x, which can also be written 2cos^2 x - 1 or 1 - 2 sin^2 x. Choose the form of the cosine identity that matches the other terms in the equation, so only one trigonometric function remains.
Equations with a squared trig function behave like quadratics. For 2 sin^2 x - sin x - 1 = 0, substitute s = sin x to get 2s^2 - s - 1 = 0, which factors as (2s + 1)(s - 1) = 0. So sin x = -1/2 or sin x = 1. Reject any value outside the range of the function, since sin x cannot exceed 1 or fall below -1. For sin x = -1/2 on the interval 0 to 2 pi, the solutions are 7 pi / 6 and 11 pi / 6, and for sin x = 1 the solution is pi / 2. Check each value by substitution.
Worked example
Solve 2 cos x = 1 for x between 0 and 2 pi.
cos x = 1/2.
Principal value x = pi / 3.
Second solution = 2 pi - pi / 3 = 5 pi / 3.
Solutions: pi / 3 and 5 pi / 3.
Practice problem and solution
How many solutions does sin x = 0.3 have for x between 0 and 4 pi (in radians)? Enter the number.
Two solutions in each interval of length 2 pi, so four in 0 to 4 pi.
Mental model: Sketch first to count solutions.
Common trap: Dividing the interval for a multiple angle but not extending it.
10. Sine rule, cosine rule and area of a triangle
Learning goal: Choose the right rule for a non-right triangle and handle the ambiguous case.
For a right-angled triangle, basic trigonometry is enough. For other triangles, two rules apply. The sine rule says that a / sin A = b / sin B = c / sin C, where each side is opposite the angle with the same letter. Use it when you know two angles and a side, or two sides and a non-included angle. The cosine rule says c^2 = a^2 + b^2 - 2ab cos C. Use it when you know two sides and the included angle, to find the third side, or three sides, to find an angle.
To find an angle with the cosine rule, rearrange to cos C = (a^2 + b^2 - c^2) / (2ab). If the result is negative, the angle is obtuse. Keep the sides labelled consistently, with the angle you are finding opposite the side on the left of the equation. Mixing up the sides is the main source of error. In sine rule problems, put the unknown in the numerator by turning the equation upside down, so you only need to multiply, which reduces slips.
The area of a triangle is (1/2) a b sin C, where C is the included angle between sides a and b. This works for any triangle, and it replaces the base times height formula when the height is not known. For problems with a sector or segment, subtract the area of the triangle from the sector area. For bearings and navigation problems, draw a clear diagram, mark all the given angles and lengths and add the angles from the compass lines before choosing a rule.
The ambiguous case can arise with the sine rule when you know two sides and a non-included angle. The sine of an angle also equals the sine of its supplement, so two triangles may fit: one with an acute angle and one with an obtuse angle. If the sum of the obtuse angle and the given angle is below 180 degrees, both triangles exist. Check by calculating the angle B = sin^-1(...) and also 180 minus B, then test whether the angle sum is valid. In exam questions this is often signalled by a phrase such as 'find the two possible values'.
Worked example
In triangle ABC, a = 7 cm, b = 9 cm and angle C = 60 degrees. Find side c.
Cosine rule: c^2 = a^2 + b^2 - 2ab cos C.
c^2 = 49 + 81 - 2 x 7 x 9 x 0.5.
c^2 = 130 - 63 = 67.
c = 8.19 cm.
Practice problem and solution
Two sides of a triangle are 8 cm and 10 cm with an included angle of 30 degrees. Find the area in cm^2.
Area = (1/2) x 8 x 10 x sin 30 = 40 x 0.5 = 20.
Mental model: The data decide the rule.
Common trap: Using a non-included angle in the area formula.
11. Probability: conditional, independent and tree diagrams
Learning goal: Use Venn diagrams, tree diagrams and the conditional probability formula.
Probability measures how likely an event is, from 0 to 1. For two events A and B, the union A or B has probability P(A) + P(B) - P(A and B). The subtraction avoids counting the overlap twice. Mutually exclusive events cannot occur together, so P(A and B) = 0 and the addition rule simplifies to P(A) + P(B). The complement has probability 1 - P(A), which is often the fastest route to an answer that includes the phrase 'at least one'.
Conditional probability is the probability of A given that B has occurred: P(A | B) = P(A and B) / P(B). It restricts the sample space to B. In a Venn diagram, it is the overlap divided by the whole of B. In a two-way table, it is the cell count divided by the row or column total. To decide which total to use, find the phrase after the word 'given', and use that group as the denominator. Reversing the condition gives a different answer in general.
Events A and B are independent if the occurrence of one does not change the probability of the other. This means P(A and B) = P(A) x P(B), and equivalently P(A | B) = P(A). Independence is not the same as being mutually exclusive: mutually exclusive events with nonzero probabilities are dependent, since knowing one occurred means the other cannot. To test for independence, calculate P(A) x P(B) and compare with P(A and B). Do not assume independence unless the problem states it or you have verified it.
Tree diagrams show sequences of events. Multiply along the branches to find the probability of a path and add the probabilities of the paths that give the required outcome. For sampling without replacement, the probabilities on the second branches change because the sample space has changed. A tree also supports Bayes-type questions: P(A | B) is the probability of the path through A and B divided by the sum of all paths that end in B. Label each branch with its probability, and check that the branches at each node sum to one.
Worked example
P(A) = 0.5, P(B) = 0.4 and P(A and B) = 0.2. Find P(A | B).
P(A | B) = P(A and B) / P(B).
= 0.2 / 0.4.
= 0.5.
This equals P(A), so A and B are independent.
Practice problem and solution
A bag has 3 red and 2 blue balls. Two are drawn without replacement. Find the probability both are red. Enter as a decimal.
(3/5) x (2/4) = 6/20 = 0.3.
Mental model: Multiply along branches, add across paths.
Common trap: Using the same denominators without replacement.
12. Binomial and normal distributions
Learning goal: Recognise binomial settings and use the normal distribution with technology and standardisation.
A binomial distribution applies when there is a fixed number n of independent trials, each with two outcomes, and the probability of success p is the same each time. The random variable X counts the successes, written X ~ B(n, p). The probability of exactly k successes is nCk p^k (1 - p)^(n - k). The mean is np and the variance is np(1 - p). Before using the model, check each condition: if the probability changes between trials, such as when drawing without replacement from a small group, the model does not fit.
On a calculator, use the binomial probability density function for P(X = k) and the cumulative function for P(X at most k). A frequent error is treating 'at least' and 'more than' as the same. For 'at least 3', calculate 1 - P(X at most 2). For 'more than 3', calculate 1 - P(X at most 3). Write the probability statement in symbols first, so you can check that you used the correct boundary. Whole numbers only: X can take values 0, 1, 2 and so on.
The normal distribution is continuous and symmetric about the mean mu, with spread controlled by the standard deviation sigma. It is written X ~ N(mu, sigma^2), where the second parameter is the variance. Probabilities are areas under the curve, so P(X = a) is zero and only intervals matter. Approximately 68 percent of values lie within one standard deviation of the mean, 95 percent within two and 99.7 percent within three. Always sketch the curve and shade the required area, so you can check that your answer is sensible.
To use standard tables or to compare values from different distributions, standardise: z = (x - mu) / sigma, which counts the number of standard deviations from the mean. For example, a score of 70 in a test with mean 60 and standard deviation 5 has z = 2. Inverse normal problems give a probability and ask for the value of x; use the inverse normal function with the correct area, which is the area to the left of x. For a statement such as 'the top 10 percent', the area to the left is 0.9. Take care to match this area with the tail you want.
Worked example
X ~ B(5, 0.4). Find P(X = 2).
P = 5C2 x 0.4^2 x 0.6^3.
5C2 = 10.
0.4^2 = 0.16 and 0.6^3 = 0.216.
P = 10 x 0.16 x 0.216 = 0.3456.
Practice problem and solution
X ~ B(20, 0.3). Find the mean of X.
Mean = np = 20 x 0.3 = 6.
Mental model: Check the conditions, then define X.
Common trap: Using the wrong boundary for 'at least'.
13. Descriptive statistics and regression
Learning goal: Summarise data, interpret correlation and use a regression line responsibly.
To summarise data, report a measure of centre and a measure of spread. The mean is the sum divided by the number of values and is affected by extreme values. The median is the middle value of the ordered data and is more resistant to outliers. The mode is the most frequent value. For spread, the range is the largest minus the smallest value, the interquartile range is the difference between the upper and lower quartiles, and the standard deviation measures typical distance from the mean. State which measure you use and why.
An outlier can be defined by a rule. A common one is that a value is an outlier if it is more than 1.5 times the interquartile range above the upper quartile or below the lower quartile. Box plots show the median, quartiles and the outliers. When comparing two data sets, compare both centre and spread, using context words. For example, say the median of class A is higher but its spread is larger, so the results are less consistent.
Bivariate data pair two variables. A scatter diagram shows the form, direction and strength of the relationship. Pearson's correlation coefficient r measures the strength of a linear relationship, between -1 and 1. Values near 1 or -1 show a strong linear relationship; values near 0 show little linear relationship. Correlation does not imply causation, since a third variable may drive both. Always describe the relationship in context, and avoid saying that one variable causes another unless the data come from a controlled experiment.
A regression line y = ax + b is found by technology. The gradient a is the average change in y for a one-unit increase in x, and b is the predicted value of y when x is zero, which may not have a meaningful interpretation. Use the line to predict y from x only within the range of the data, which is interpolation. Predicting outside that range, extrapolation, is less reliable. The regression line of y on x is used to predict y, not to predict x. Give the units in every interpretation, and comment on the reliability using r and the range of the data.
Worked example
Quartiles are Q1 = 12 and Q3 = 20. Find the upper outlier boundary using the 1.5 x IQR rule.
IQR = 20 - 12 = 8.
1.5 x IQR = 12.
Upper boundary = Q3 + 12.
Upper boundary = 32.
Practice problem and solution
A regression line is y = 3x + 4. Predict y when x = 5.
y = 3 x 5 + 4 = 19.
Mental model: Use the line only within the range of the data.
Common trap: Treating correlation as proof of cause.
14. Differentiation: rules, tangents and normals
Learning goal: Differentiate with the power, chain, product and quotient rules and find tangent and normal lines.
The derivative gives the rate of change of a function, and graphically the gradient of the tangent. For a term ax^n, the derivative is a n x^(n-1). So the derivative of 4x^3 is 12x^2, a constant has derivative zero and the derivative of a sum is the sum of the derivatives. Rewrite roots and reciprocals as powers before differentiating, for example square root of x is x^(1/2) and 1 / x^2 is x^(-2). A common slip is failing to rewrite, which leads to wrong powers.
The chain rule handles composite functions: if y = f(g(x)), then dy/dx = f'(g(x)) x g'(x). In words, differentiate the outer function, keep the inside unchanged, then multiply by the derivative of the inside. For y = (3x + 1)^5, the result is 5(3x + 1)^4 x 3 = 15(3x + 1)^4. The standard derivatives include d/dx of e^x = e^x, of ln x = 1 / x, of sin x = cos x and of cos x = -sin x, with x in radians.
The product rule is d/dx (uv) = u'v + uv'. The quotient rule is d/dx (u / v) = (u'v - uv') / v^2. Choose the rule by the structure: two functions multiplied suggests the product rule, and one divided by another suggests the quotient rule, but a quotient with a constant denominator is simply a multiple. Label u, v, u' and v' before substituting, and keep brackets around each, since the minus sign in the quotient rule easily causes sign errors.
The gradient of the tangent to y = f(x) at x = a is f'(a). The equation of the tangent through (a, f(a)) is y - f(a) = f'(a)(x - a). The normal is perpendicular to the tangent, so its gradient is -1 / f'(a), and its equation is y - f(a) = (-1 / f'(a))(x - a). Compute the point first by substituting a into f, then the gradient by substituting into f'. Check your equation by confirming that the point lies on the line. If the tangent gradient is zero, the normal is vertical, with equation x = a.
Worked example
Find the gradient of y = x^3 - 2x at x = 2.
dy/dx = 3x^2 - 2.
Substitute x = 2.
3 x 4 - 2 = 10.
Gradient = 10.
Practice problem and solution
Find the derivative of y = (2x + 1)^3 at x = 1. Enter the value.
dy/dx = 3(2x + 1)^2 x 2 = 6(2x + 1)^2. At x = 1: 6 x 9 = 54.
Mental model: Name the structure, then pick the rule.
Common trap: Forgetting to multiply by the derivative of the inside.
15. Applications of differentiation: shape and optimisation
Learning goal: Locate stationary points, classify them and solve optimisation problems.
Stationary points occur where the gradient is zero, f'(x) = 0. They may be local maxima, local minima or points of inflection. To classify a stationary point, use the first derivative test: check the sign of f'(x) just before and after the point. A change from positive to negative gives a maximum, a change from negative to positive gives a minimum, and no change of sign suggests a horizontal point of inflection. This test always works, even when the second derivative is zero.
The second derivative test is quicker: if f''(x) is positive at a stationary point, the point is a local minimum, and if f''(x) is negative, it is a local maximum. If f''(x) is zero, the test is inconclusive. A function is increasing where f'(x) is positive and decreasing where f'(x) is negative. Points of inflection occur where the concavity changes, with f''(x) = 0 and a sign change in f''(x). Solve f'(x) > 0 as an inequality, and write the answer as intervals.
Optimisation turns a word problem into a one-variable function. Define the quantity to maximise or minimise, such as an area or a cost. Write it in terms of two variables, use the constraint to eliminate one and obtain a function of one variable, with a sensible domain. Differentiate, set the derivative to zero, solve, then confirm that the point is a maximum or minimum, and check the endpoints if the domain is limited. Finally, answer the question in context with units, giving the value of the quantity, not just the value of the variable.
For example, a farmer has 100 m of fencing for a rectangle against a wall, using three sides. If the width is x, the length is 100 - 2x, so the area is A = x(100 - 2x) = 100x - 2x^2. Then dA/dx = 100 - 4x = 0 gives x = 25, and d^2A/dx^2 = -4 shows a maximum. The maximum area is 25 x 50 = 1250 m^2. The domain is 0 to 50, so the maximum lies in the interior. Sketching the function helps check that the answer makes sense.
Worked example
Find the stationary point of y = x^2 - 6x + 5 and state its type.
dy/dx = 2x - 6 = 0, so x = 3.
d^2y/dx^2 = 2, which is positive.
y at x = 3: 9 - 18 + 5 = -4.
Minimum at (3, -4).
Practice problem and solution
A rectangle has perimeter 40 m. Find the maximum area in m^2.
Sides x and 20 - x; A = x(20 - x), dA/dx = 20 - 2x = 0 so x = 10. Area = 10 x 10 = 100.
Common trap: Answering with the variable instead of the quantity asked.
16. Integration: antiderivatives, areas and kinematics
Learning goal: Integrate common functions and use definite integrals for area and motion.
Integration reverses differentiation. The indefinite integral of x^n is x^(n+1) / (n + 1) + C, for n not equal to -1, where C is the constant of integration. The integral of 1 / x is ln |x| + C, of e^x is e^x + C, of cos x is sin x + C and of sin x is -cos x + C. Always include the constant of integration for an indefinite integral. To find the constant, use a given condition, such as a point on the curve, and substitute.
For a linear inner function, the reverse chain rule applies: the integral of f(ax + b) is (1 / a) F(ax + b) + C, where F is an antiderivative of f. For example, the integral of (2x + 3)^4 is (2x + 3)^5 / 10 + C, since the inner derivative is 2 and the power rises to 5. More general substitution is used when an expression and its derivative appear together, such as 2x(x^2 + 1)^3, where u = x^2 + 1. Check any integral by differentiating the answer.
A definite integral from a to b gives the signed area between the curve and the x-axis. Areas above the axis are positive and areas below are negative. To find the total area when the curve crosses the axis, split the integral at the roots and take absolute values, or use technology for the absolute value of the function. To find the area between two curves, integrate the upper function minus the lower function between the intersection points. A sketch avoids sign errors.
In kinematics, velocity is the derivative of displacement and acceleration is the derivative of velocity. So displacement is the integral of velocity, and the change in velocity is the integral of acceleration. The definite integral of velocity from t = a to t = b is the displacement, which can be negative. The total distance travelled is the integral of the absolute value of velocity. Use the initial condition to find the constant when integrating a velocity function to get displacement. State the units of the answer, such as metres for displacement.
Worked example
Find the area under y = x^2 from x = 0 to x = 3.
Integral of x^2 is x^3 / 3.
Evaluate at 3: 27 / 3 = 9.
Evaluate at 0: 0.
Area = 9.
Practice problem and solution
Find the definite integral of 2x from x = 1 to x = 4.
The integral of 2x is x^2. Evaluate: 16 - 1 = 15.
Mental model: Check by differentiating.
Common trap: Forgetting the constant of integration.
17. Proof, reasoning and the mathematical exploration
Learning goal: Construct a short proof, use counterexamples and plan an exploration you can finish.
The IB subject brief for Mathematics: analysis and approaches stresses the ability to construct, communicate and justify correct mathematical arguments. Proof is the formal version of this. A direct proof starts from known facts or definitions and moves by valid steps to the statement. A common pattern for statements about integers is to represent an even number as 2k and an odd number as 2k + 1, for any integer k, then manipulate algebraically. For example, the sum of two odd numbers 2a + 1 and 2b + 1 is 2(a + b + 1), which is even.
To show that a statement is false, one counterexample is enough. For the claim that the sum of two prime numbers is always even, the example 2 + 3 = 5 shows it to be false. By contrast, to show a statement is true, examples are never enough. Checking five cases proves nothing about all cases. Write each step with a reason, use the equals sign only between equal quantities and finish with a clear concluding statement, such as 'Therefore the sum is even'.
Communication is assessed in every paper. Use correct notation, define symbols that you introduce, and write answers to the accuracy asked, normally three significant figures unless otherwise stated. Write intermediate values with more digits to avoid rounding errors. When technology is allowed, state what you entered, such as the function and the window, so that the method can be followed. In a no-technology paper, show all algebraic steps, since marks are given for method even if the answer is wrong.
The mathematical exploration is an internal assessment worth 20 percent of the final grade at SL according to the subject brief, with a recommended 15 hours for it. Choose a topic you are curious about and can finish, with a clear aim, a personal connection and mathematics at the right level. A good exploration states a focused research question, uses the mathematics accurately, interprets the results and reflects on limitations. Avoid topics that are too large, such as a whole branch of mathematics, and avoid copying a source without your own analysis. A shorter, clear and well-argued exploration beats a long unfocused one.
Worked example
Prove that the sum of three consecutive integers is divisible by 3.
Let the integers be n, n + 1 and n + 2.
Sum = 3n + 3.
3n + 3 = 3(n + 1).
Therefore the sum is a multiple of 3.
Practice problem and solution
What fraction of the final grade at SL is the exploration worth, as a percentage according to the IB subject brief? Enter the percentage.
The IB subject brief gives the exploration a weighting of 20 percent at both SL and HL.
Mental model: Reasoning needs general steps; counterexamples need one case.
Common trap: Treating a few checked examples as a proof.