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Functions, composites & inverses: IB Mathematics: Analysis & Approaches questions with answers Free Functions, composites & inverses questions for IB Mathematics: Analysis & Approaches (SL & HL), part of Functions. Each question has the answer and the marking points an examiner would look for.
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Functions, composites & inverses practice questions Question 1 Structured · 9 marks A* challenge
Let f(x) = x2 − 6x, for x ≥ k.
(a) Find the least value of k for which f has an inverse function.
[2] (b) For this value of k, find f−1 (x), stating its domain.
[4] Show answers and mark scheme Part Answer and marking points Marks (a) k = 3f is one-to-one on x ≥ vertex (M1) k = 3 A1
2 (b) f−1 (x) = 3 + √x + 9 , x ≥ −9f(x) = (x − 3)2 − 9 (A1) interchanging x and y and rearranging M1 f−1 (x) = 3 + √x + 9 A1 domain x ≥ −9 A1
4 (c) x = 7the graphs meet on y = x, so f(x) = x M1 x2 − 7x = 0 ⇒ x = 0 or x = 7 A1 x = 0 < 3 is not in the domain, so x = 7 R1
3
Question 2 Structured · 6 marks
Let f(x) = 5 − x2 and g(x) = 2x , for −4 ≤ x ≤ 4.
(a) Find the x-coordinates of the points where the graphs of f and g intersect.
[3] (b) Hence solve the inequality f(x) > g(x).
[3] Show answers and mark scheme Part Answer and marking points Marks (a) x = −2.19, x = 1.48attempt to solve f(x) = g(x), e.g. graph or solver (M1) x = −2.19, x = 1.48 A2 award A1 if one value is missing or incorrect
3 (b) −2.19 < x < 1.48using the graph to choose the correct regions (M1) correct critical values used as endpoints (A1) −2.19 < x < 1.48 A1
3
Question 3 Structured · 6 marks
Let f(x) = −2x + 3 and g(x) = x2 − 6, for x ∈ ℝ.
Show answers and mark scheme Part Answer and marking points Marks (a) −2x2 + 15f(x2 − 6) = −2(x2 − 6) + 3 (M1) −2x2 + 15 A1
2 (b) x = −3 2 , x = 9 2 (−2x + 3)2 − 6 = 30 M1 (−2x + 3)2 = 36 (A1) −2x + 3 = ±6 (M1) x = −3 2 , x = 9 2 A1
4
Question 4 Structured · 7 marks
Let f(x) = 4x + 1 x − 4 , x ≠ 4.
(a) Write down the equation of the vertical asymptote and the equation of the horizontal asymptote of the graph of f.
[2] (c) Hence state, with a reason, a special property of f.
[2] Show answers and mark scheme Part Answer and marking points Marks (a) x = 4, y = 4x = 4 A1 y = 4 A1
2 (b) f−1 (x) = 4x + 1 x − 4 interchanging x and y (M1) x(y − 4) = 4y + 1 ⇒ y(x − 4) = 4x + 1 (A1) f−1 (x) = 4x + 1 x − 4 A1
3 (c) f is self-inversef−1 (x) = f(x) R1 f is self-inverse (its graph is symmetric in y = x) A1
2
Question 5 Structured · 9 marks A* challenge
Let f(x) = x2 − 8x + 14, for x ≥ k.
(a) Find the least value of k for which f has an inverse function.
[2] (b) For this value of k, find f−1 (x), stating its domain.
[4] Show answers and mark scheme Part Answer and marking points Marks (a) k = 4f is one-to-one on x ≥ vertex (M1) k = 4 A1
2 (b) f−1 (x) = 4 + √x + 2 , x ≥ −2f(x) = (x − 4)2 − 2 (A1) interchanging x and y and rearranging M1 f−1 (x) = 4 + √x + 2 A1 domain x ≥ −2 A1
4 (c) x = 7the graphs meet on y = x, so f(x) = x M1 x2 − 9x + 14 = 0 ⇒ x = 2 or x = 7 A1 x = 2 < 4 is not in the domain, so x = 7 R1
3
Question 6 Structured · 6 marks
Let f(x) = x3 − 4x + 1 and g(x) = e−x , for −3 ≤ x ≤ 3.
(a) Find the x-coordinates of the points where the graphs of f and g intersect.
[3] (b) Hence solve the inequality f(x) < g(x).
[3] Show answers and mark scheme Part Answer and marking points Marks (a) x = −1.37, x = 0, x = 1.88attempt to solve f(x) = g(x), e.g. graph or solver (M1) x = −1.37, x = 0, x = 1.88 A2 award A1 if one value is missing or incorrect
3 (b) −3 ≤ x < −1.37 or 0 < x < 1.88using the graph to choose the correct regions (M1) correct critical values used as endpoints (A1) −3 ≤ x < −1.37 or 0 < x < 1.88 A1
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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