10 free IB Mathematics: Analysis & Approaches Functions questions written in the style of IB Mathematics: Analysis & Approaches papers, each with its answer and mark scheme. 3 of them are A* challenge questions, built to catch the mistakes strong students still make.
This topic covers Functions, composites & inverses, Quadratics & the discriminant, Rational functions, Exponential & log functions, Transformations of graphs, Polynomials & roots, Modulus, odd/even & inequalities. Work through the questions below, then generate a full Functions paper with a fresh set every time.
Generate a full Functions paper →The roots of the equation x3 + 3x2 + kx − 8 = 0 form an arithmetic sequence.
Find the roots of the equation and the value of k.
| Part | Answer and marking points | Marks |
|---|---|---|
| roots −4, −1, 2; k = −6 let the roots be α − d, α, α + d M1 sum of roots: 3α = −3 ⇒ α = −1 A1 product: α(α2 − d2) = 8 M1 d = ±3 A1 roots −4, −1, 2 A1 k = sum of pair products = −6 A1 | 6 |
The line y = kx − 2 is a tangent to the curve y = x2 − x − 1.
Find the possible values of k.
| Part | Answer and marking points | Marks |
|---|---|---|
| k = −3 or k = 1 x2 − x − 1 = kx − 2 M1 x2 + (−1 − k)x + 1 = 0 A1 discriminant = 0 M1 (−1 − k)2 = 4 (A1) k = −3 or k = 1 A1 | 5 |
Let f(x) = x2 − 6x + 10, for x ≥ k.
Find the least value of k for which f has an inverse function.
For this value of k, find f−1(x), stating its domain.
Solve f(x) = f−1(x).
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | k = 3 f is one-to-one on x ≥ vertex (M1) k = 3 A1 | 2 |
| (b) | f−1(x) = 3 + √x − 1, x ≥ 1 f(x) = (x − 3)2 + 1 (A1) interchanging x and y and rearranging M1 f−1(x) = 3 + √x − 1 A1 domain x ≥ 1 A1 | 4 |
| (c) | x = 5 the graphs meet on y = x, so f(x) = x M1 x2 − 7x + 10 = 0 ⇒ x = 2 or x = 5 A1 x = 2 < 3 is not in the domain, so x = 5 R1 | 3 |
Let f(x) = 2x + 32x − 5, x ≠ 52.
Write down the equation of the vertical asymptote and the equation of the horizontal asymptote of the graph of f.
Find f−1(x).
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | x = 52, y = 1 x = 52 A1 y = 1 A1 | 2 |
| (b) | f−1(x) = 5x + 32x − 2 interchanging x and y (M1) x(2y − 5) = 2y + 3 ⇒ y(2x − 2) = 5x + 3 (A1) f−1(x) = 5x + 32x − 2 A1 | 3 |
Solve the inequality |4x − 3| > |x + 5|.
| Part | Answer and marking points | Marks |
|---|---|---|
| x < −25 or x > 83 squaring both sides, or considering cases M1 (4x − 3)2 − (x + 5)2 > 0 (A1) critical values x = −25, x = 83 A1A1 x < −25 or x > 83 A1 | 5 |
Let p(x) = x3 + ax2 + bx + 12, where a, b ∈ ℝ. Given that (x − 1) is a factor of p(x), and that the remainder when p(x) is divided by (x + 2) is 18,
find the value of a and the value of b;
hence factorise p(x) fully.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | a = −2, b = −11 p(1) = 0: a + b = −13 M1A1 p(−2) = 18: 4a − 2b = 14 A1 attempt to solve simultaneously (M1) a = −2, b = −11 A1 | 5 |
| (b) | (x − 1)(x − 4)(x + 3) attempt to divide by the known factor (M1) (x − 1)(x − 4)(x + 3) A1 | 2 |
The point P(4, −2) lies on the graph of y = f(x). The graph of y = f(x) is transformed to the graph of y = −2f(2x − 6) + 3.
Find the coordinates of the image of P.
Describe fully a sequence of transformations that maps the graph of y = f(x) onto this graph.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | (5, 7) 2x − 6 = 4 (M1) x = 5 A1 y = 7 A1 | 3 |
| (b) | horizontal stretch, scale factor 12, then horizontal translation of 3; reflection in the x-axis and vertical stretch, scale factor 2; vertical translation of 3 horizontal stretch, scale factor 12 A1 followed by a horizontal translation of 3 (or: translation of 6 first, then the stretch) A1 reflection in the x-axis and vertical stretch, scale factor 2 A1 vertical translation of 3 (after the vertical stretch) A1 | 4 |
The number of bacteria in a culture is modelled by N(t) = 500ekt, where t is in hours. After 3 hours there are 1500 bacteria.
Find the value of k.
Find the time taken for the number of bacteria to reach 6000.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 0.366 1500 = 500e3k (M1) k = 13ln1500500 (A1) k = 0.366 A1 | 3 |
| (b) | 6.79 hours 500ekt = 6000 (M1) t = ln(6000/500)k (A1) 6.79 hours A1 | 3 |
Let f(x) = 2x − 5 and g(x) = x2 − 1, for x ∈ ℝ.
Find (f∘g)(x).
Solve (g∘f)(x) = 35.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 2x2 − 7 f(x2 − 1) = 2(x2 − 1) − 5 (M1) 2x2 − 7 A1 | 2 |
| (b) | x = 112, x = −12 (2x − 5)2 − 1 = 35 M1 (2x − 5)2 = 36 (A1) 2x − 5 = ±6 (M1) x = 112, x = −12 A1 | 4 |
The graph of y = f(x) passes through the point P(4, 0).
The graph of y = f(x) is translated by −21. Find the image of P.
The graph of y = f(x) is transformed to y = 2f(x − 1). Find the image of P.
Describe fully a transformation that maps y = f(x) onto y = f(−x).
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | (2, 1) x-coordinate 2 A1 y-coordinate 1 A1 | 2 |
| (b) | (5, 0) x: 5 A1 y: 0 A1 | 2 |
| (c) | reflection in the y-axis reflection in the y-axis A1 | 1 |
In the SL & HL syllabus, Functions covers Functions, composites & inverses, Quadratics & the discriminant, Rational functions, Exponential & log functions, Transformations of graphs, Polynomials & roots, Modulus, odd/even & inequalities.
This page has 10 Functions questions with answers. For more, the free generator on mentorriyaz.in builds a new Functions paper with a full mark scheme each time you click.
Riyaz Shaikh has mentored IGCSE, IB and A Level students for 15+ years, one-on-one, online and in Mumbai. The hardest questions on this site are the ones he uses to find a student's weak points.
Meet Riyaz →Original IB-style practice questions written for mentorriyaz.in. These are not IB past paper questions, and this site is not affiliated with or endorsed by the International Baccalaureate Organization. Updated 2026-11-04. Paper generator · mentorriyaz.in