IB Maths AA HL · Unit 1: Number and Algebra
IB Maths AA HL Partial Fractions Questions
Exam-style IB Maths AA HL partial fractions questions with worked solutions. Start with the 3 fully worked examples below — each is solved step by step the way an IB examiner expects — then try the practice questions and check your working against the mark scheme.
- 20 questions
- Paper 1: 19
- Paper 2: 1
- 5 easy
- 7 medium
- 7 hard
- 1 starter
- 3 worked examples
Practise Partial Fractions questions →
AA HL formula booklet
What's examined in AA HL partial fractions
The question bank covers these partial fractions question types (number of questions in brackets):
- Improper & Applications (11)
- Linear Denominators (7)
- Complex Denominators (2)
Partial Fractions worked examples
Worked example 1: Decomposing into two linear factors · easy
Express $\frac{5x+4}{(x-1)(x+2)}$ in partial fractions.
1. Set up the basic algebraic identity with unknown constants: $\frac{5x+4}{(x-1)(x+2)} \equiv \frac{A}{x-1} + \frac{B}{x+2}$.
2. Multiply through by the common denominator to clear the fractions: $5x+4 \equiv A(x+2) + B(x-1)$.
3. Substitute a strategic value $x = 1$ to completely eliminate the $B$ term: $5(1) + 4 = A(1 + 2) \implies 9 = 3A$.
4. Solve for the first constant: $A = 3$.
5. Substitute the other root $x = -2$ to eliminate the $A$ term: $5(-2) + 4 = B(-2 - 1) \implies -6 = -3B$.
6. Solve for the second constant and write the final decomposition: $B = 2$, yielding $\mathbf{\frac{3}{x-1} + \frac{2}{x+2}}$.
Examiner tip: The substitution method is generally much faster than expanding and comparing coefficients for distinct linear factors, but remind students to carefully track negative signs during substitution.
Worked example 2: Solving identities with unknown parameters · medium
Given the identity $\frac{7x-4}{(2x-1)(x-2)} \equiv \frac{A}{2x-1} + \frac{B}{x-2}$, find the exact values of the constants $A$ and $B$.
1. Multiply both sides of the identity by the denominator $(2x-1)(x-2)$ to get the linear format $7x - 4 \equiv A(x - 2) + B(2x - 1)$.
2. Choose $x = 2$ to eliminate the $A$ term: $7(2) - 4 = A(0) + B(2(2) - 1)$.
3. Simplify the arithmetic to find $B$: $14 - 4 = 3B \implies 10 = 3B$, yielding $B = \frac{10}{3}$.
4. Choose $x = 0.5$ to eliminate the $B$ term: $7(0.5) - 4 = A(0.5 - 2) + B(0)$.
5. Simplify the resulting equation: $3.5 - 4 = -1.5A \implies -0.5 = -1.5A$.
6. Solve for the final constant: $A = \frac{-0.5}{-1.5} = \frac{1}{3}$. Answer: $\mathbf{A = \frac{1}{3}, B = \frac{10}{3}}$.
Examiner tip: When linear factors involve a leading coefficient, such as $2x-1$, substituting fractions can lead to arithmetic errors under exam pressure; an alternative safety-check is to compare the coefficients of $x$ on both sides.
Worked example 3: Handling improper rational functions · hard
Express $\frac{x^2+1}{x^2-1}$ in the form $C + \frac{A}{x-1} + \frac{B}{x+1}$, where $A, B,$ and $C$ are integers to be determined.
1. Recognize the fraction is improper (degree of numerator $\ge$ degree of denominator), so perform polynomial division or algebraic manipulation first.
2. Rewrite the numerator to match the denominator structure: $\frac{x^2 - 1 + 2}{x^2 - 1} = 1 + \frac{2}{x^2 - 1}$. Thus, the constant $C = 1$.
3. Factorise the new denominator using the difference of two squares: $x^2 - 1 = (x-1)(x+1)$.
4. Set up the partial fraction identity for the remainder: $\frac{2}{(x-1)(x+1)} \equiv \frac{A}{x-1} + \frac{B}{x+1}$, which rearranges to $2 \equiv A(x+1) + B(x-1)$.
5. Substitute $x = 1$ to find $A$: $2 = A(1+1) \implies 2A = 2 \implies A = 1$.
6. Substitute $x = -1$ to find $B$: $2 = B(-1-1) \implies -2B = 2 \implies B = -1$. The final form is $\mathbf{1 + \frac{1}{x-1} - \frac{1}{x+1}}$.
Examiner tip: Students almost universally forget to divide improper algebraic fractions first, directly setting up an identity equal to $\frac{A}{x-1} + \frac{B}{x+1}$ which makes finding the standalone constant $C$ impossible.
Try these IB Maths AA HL partial fractions questions
Three questions from the bank, easiest first. Mark schemes and AI marking of your written working are in the practice area.
Question 1 · easy · 3 marks · Paper 1
Decompose the rational function \(\frac{3x+5}{(x+1)(x+3)}\) into partial fractions.
Question 2 · medium · 4 marks · Paper 1
The rational function \(\frac{x^2+1}{x^2-1}\) is an improper fraction because the degree of the numerator is equal to the degree of the denominator.
By using polynomial long division or algebraic manipulation, express the function in the form \(C + \frac{Dx + E}{x^2-1}\). [1 mark]
Hence, express \(\frac{x^2+1}{x^2-1}\) fully in partial fractions. [3 marks]
Question 3 · hard · 6 marks · Paper 1
A curve \(y = f(x)\) passes through the point \((e, 2)\) and satisfies the separable differential equation:
\[\frac{dy}{dx} = \frac{y^2-1}{2x} \quad \text{for } x > 0 \text{ and } y > 1\]
Separate the variables and use partial fractions to integrate both sides. [4 marks]
Find the particular solution, expressing \(y\) explicitly in terms of \(x\). [2 marks]
All 20 partial fractions questions with mark schemes →
FAQ
How many IB Maths AA HL partial fractions questions are there?
There are 20 exam-style partial fractions questions in the AA HL question bank (Paper 1: 19 · Paper 2: 1), graded 5 easy, 7 medium, 7 hard, 1 starter. Every question has a full IB-style mark scheme (M, A and R marks).
Is partial fractions on Paper 1 or Paper 2?
Both. In the bank, Paper 1: 19 · Paper 2: 1. Paper 1 is non-calculator, so practise the exact-value algebra as well as the GDC methods.
Where can I get the mark schemes?
Open the AA HL Unit 1 practice page: every question has a step-by-step IB-style mark scheme, and you can photograph your working for instant AI marking. The worked examples on this page are free.
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