IGCSE Additional Mathematics: Functions Practice Questions
A function maps each value in its domain to exactly one value in its range. A composite fg(x) applies g first, an inverse reverses the original mapping, and the modulus function returns the size of a value without its sign.
Topic 1 of Cambridge IGCSE Additional Mathematics 0606 sets up the notation used across the whole syllabus. Range questions are where marks are lost, because a range must be justified rather than guessed. The questions below cover notation, composites, inverses and the modulus.
What you need to know for Functions
- Domain and rangeThe domain is the set of allowed inputs. The range is the set of outputs produced. A restricted domain is often given so that an inverse exists.
- Composite functionfg(x) means apply g first and then f. Substitute the whole of g(x) wherever x appears in f.
- Inverse functionWrite y = f(x), rearrange to make x the subject, then swap the letters. An inverse exists only if the function is one to one.
- Finding a rangeComplete the square for a quadratic, or consider the behaviour at the ends of the domain. State the range as an inequality in f(x).
- Modulus functionThe modulus of a value is its size without regard to sign. To solve an equation containing a modulus, set the inside equal to both the positive and the negative value.
IGCSE Additional Mathematics Functions questions and answers
4 exam-style questions written to the 0606 syllabus. Try each one on paper first, then open the worked answer to check your method against the marks.
The function f is defined by f(x) = 2x minus 5. Find f-1(x).
Show the worked answer
- Write y = 2x minus 5.
- Rearrange to make x the subject: y + 5 = 2x, so x = (y + 5) divided by 2.
- Swap the letters to express the inverse as a function of x: f-1(x) = (x + 5) divided by 2. Check: f(4) = 3 and f-1(3) = 4.
f(x) = x2 + 1 for x greater than or equal to 0, and g(x) = 3x minus 2. Find fg(2) and gf(2).
Show the worked answer
- For fg(2), apply g first: g(2) = 3 x 2 minus 2 = 4.
- Then apply f: f(4) = 42 + 1 = 17.
- For gf(2), apply f first: f(2) = 22 + 1 = 5.
- Then apply g: g(5) = 3 x 5 minus 2 = 13. The two differ, so composition is not commutative.
Find the range of f(x) = x2 minus 4x + 7, stating your method.
Show the worked answer
- Complete the square. Half the coefficient of x is minus 2, so start with (x minus 2)2.
- (x minus 2)2 = x2 minus 4x + 4, which is 3 short of the constant, so f(x) = (x minus 2)2 + 3.
- A squared term is never negative, so its least value is 0, occurring at x = 2.
- The least value of f(x) is therefore 3, and the range is f(x) greater than or equal to 3.
Solve the equation the modulus of (2x minus 3) equals 7.
Show the worked answer
- The expression inside the modulus can be either 7 or minus 7, since both have size 7.
- Case one: 2x minus 3 = 7, so 2x = 10 and x = 5.
- Case two: 2x minus 3 = minus 7, so 2x = minus 4 and x = minus 2. Both satisfy the original equation.
Common mistakes in this topic
- Stating a range in terms of x rather than f(x).
- Applying composite functions in the wrong order.
- Treating the inverse notation as a reciprocal.
- Solving only one case of a modulus equation.
- Forgetting that an inverse requires a one to one function, which is why domains are restricted.
Exam tips
- For a range, complete the square first. The turning point gives the boundary directly.
- Write the intermediate value in a composite. It is usually worth a method mark.
- Check an inverse by feeding a number through f and then through f inverse.
- For modulus equations, always write both cases before solving.
- State ranges and domains as inequalities, using the correct variable each time.
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Functions FAQs
How do I find the range of a quadratic function?
Complete the square to find the turning point. Since a squared term has a minimum of zero, the constant added afterwards gives the least value of the function when the coefficient is positive. State the range as an inequality in f(x), not in x.
What does fg(x) mean?
Apply g first and then f. Substitute the whole expression for g(x) wherever x appears in f. Because the function nearest the bracket acts first, fg(x) and gf(x) usually give different results.
How do I find an inverse function?
Set y equal to the function, rearrange to make x the subject, then swap the letters so the result is written in terms of x. An inverse exists only when the function is one to one, which is why domains are often restricted.
How do I solve an equation containing a modulus?
The expression inside the modulus can equal either the positive or the negative of the value on the right. Write both cases, solve each separately, then check both solutions in the original equation.
Continue through the IGCSE Additional Mathematics syllabus
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Written to the published Cambridge IGCSE Additional Mathematics (0606) syllabus. Check your school entry code and syllabus year, because Core and Extended candidates are assessed on different content. Last reviewed 2026-08-12.