Question Simulator

AS & A Level Mathematics (9709)

Functions Banker

54 questions · about 3% of the marks we indexed · examined in 9 of 9 sessions.

Questions by year

2026

  • 2026 May/June · Paper 11 · Q33 marksTransformations of graphs
  • 2026 May/June · Paper 11 · Q9 (a, b(i))6 marksComposite function · Inverse function
  • 2026 May/June · Paper 12 · Q2 (a, b)7 marksTransformations of graphs
  • 2026 May/June · Paper 12 · Q10 (b(i), b(ii))6 marksDomain & range · Inverse function
  • 2026 May/June · Paper 13 · Q6 (b)3 marksTransformations of graphs
  • 2026 May/June · Paper 15 · Q4 (b, c)4 marksTransformations of graphs
  • 2026 May/June · Paper 15 · Q9 (b(i), b(ii))5 marksDomain & range · Inverse function

2025

  • 2025 Oct/Nov · Paper 11 · Q4 (b)3 marksTransformations of graphs
  • 2025 Oct/Nov · Paper 11 · Q6 (a, b, c)7 marksComposite function · Domain & range · Inverse function
  • 2025 Oct/Nov · Paper 12 · Q3 (a, b)5 marksTransformations of graphs
  • 2025 Oct/Nov · Paper 12 · Q9 (b, c, d)4 marksComposite function · Domain & range · Inverse function
  • 2025 Oct/Nov · Paper 13 · Q6 (a, b)8 marksTransformations of graphs
  • 2025 Oct/Nov · Paper 13 · Q74 marksInverse function
  • 2025 Oct/Nov · Paper 13 · Q10 (a(ii))1 markDomain & range
  • 2025 Oct/Nov · Paper 15 · Q10 (a, b, c)9 marksComposite function · Inverse function
  • 2025 Oct/Nov · Paper 15 · Q11 (c(i), c(ii))3 marksTransformations of graphs
  • 2025 May/June · Paper 11 · Q10 (a, b, c, d, e, f)12 marksComposite function · Domain & range · Inverse function · Transformations of graphs
  • 2025 May/June · Paper 12 · Q14 marksTransformations of graphs
  • 2025 May/June · Paper 12 · Q11 (b(i), b(ii))7 marksComposite function · Inverse function
  • 2025 May/June · Paper 13 · Q10 (c(i), c(ii))4 marksTransformations of graphs
  • 2025 May/June · Paper 13 · Q11 (a, b)10 marksComposite function · Domain & range
  • 2025 May/June · Paper 15 · Q9 (a, c, d)9 marksComposite function · Inverse function · Transformations of graphs

2024

  • 2024 Oct/Nov · Paper 11 · Q11 (b, d)6 marksInverse function · Transformations of graphs
  • 2024 Oct/Nov · Paper 12 · Q5 (a(i), a(ii), a(iii), b)10 marksComposite function · Domain & range · Inverse function
  • 2024 Oct/Nov · Paper 13 · Q5 (a, b)8 marksTransformations of graphs
  • 2024 Oct/Nov · Paper 13 · Q8 (b, c, d)7 marksComposite function · Domain & range · Inverse function
  • 2024 May/June · Paper 11 · Q2 (a, b)6 marksOfficial paper ↗Transformations of graphs
  • 2024 May/June · Paper 11 · Q6 (a, b, c, d)7 marksOfficial paper ↗Inverse function
  • 2024 May/June · Paper 12 · Q25 marksTransformations of graphs
  • 2024 May/June · Paper 12 · Q4 (a, b, c)7 marksComposite function · Domain & range · Inverse function
  • 2024 May/June · Paper 13 · Q6 (b)3 marksTransformations of graphs
  • 2024 May/June · Paper 13 · Q11 (a)3 marksDomain & range

2023

  • 2023 Oct/Nov · Paper 11 · Q4 (a, b, c)6 marksTransformations of graphs
  • 2023 Oct/Nov · Paper 11 · Q9 (b, c, d)7 marksComposite function · Inverse function
  • 2023 Oct/Nov · Paper 12 · Q6 (b, c)6 marksTransformations of graphs
  • 2023 Oct/Nov · Paper 12 · Q8 (a, b(i), b(ii), c)8 marksComposite function · Domain & range · Inverse function
  • 2023 Oct/Nov · Paper 13 · Q7 (a, b, c)7 marksComposite function · Domain & range · Inverse function
  • 2023 Oct/Nov · Paper 13 · Q8 (b)3 marksTransformations of graphs
  • 2023 May/June · Paper 11 · Q34 marksTransformations of graphs
  • 2023 May/June · Paper 11 · Q8 (a, b, c)8 marksComposite function · Domain & range · Inverse function
  • 2023 May/June · Paper 12 · Q8 (a, b, c, d)12 marksDomain & range · Inverse function · Transformations of graphs
  • 2023 May/June · Paper 13 · Q14 marksTransformations of graphs
  • 2023 May/June · Paper 13 · Q7 (a, b, c)8 marksComposite function · Domain & range · Inverse function

2022

  • 2022 Oct/Nov · Paper 11 · Q8 (b, c)5 marksInverse function
  • 2022 Oct/Nov · Paper 11 · Q9 (d)4 marksTransformations of graphs
  • 2022 Oct/Nov · Paper 12 · Q5 (a, b)5 marksTransformations of graphs
  • 2022 Oct/Nov · Paper 12 · Q9 (a, b, c, d, e)8 marksComposite function · Inverse function
  • 2022 Oct/Nov · Paper 13 · Q2 (b, c)4 marksDomain & range · Inverse function
  • 2022 Oct/Nov · Paper 13 · Q5 (a, b, c, d)6 marksTransformations of graphs
  • 2022 May/June · Paper 11 · Q6 (a, b, c)8 marksComposite function · Domain & range · Inverse function
  • 2022 May/June · Paper 11 · Q8 (a)5 marksTransformations of graphs
  • 2022 May/June · Paper 12 · Q10 (a, b, c, d)7 marksComposite function · Domain & range · Inverse function
  • 2022 May/June · Paper 13 · Q4 (a, b)5 marksTransformations of graphs
  • 2022 May/June · Paper 13 · Q6 (b, c, d)6 marksComposite function · Domain & range · Inverse function

References only, not question text. Open Official paper ↗ where Cambridge publishes the paper, or see the official Cambridge past-papers page; for other papers, ask your school. “Paper 42” means Paper 4, variant 2.

Practice questions

Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.

Q1Transformations of graphs · Core · Paper 1 style · 4 marks
  1. (a) The point P(4, −3) lies on the curve y = f(x). State the coordinates of the image of P on the curve y = f(x + 2). [1]
  2. (b) State the coordinates of the image of P on the curve y = 3f(x). [1]
  3. (c) State the coordinates of the image of P on the curve y = f(−x) − 1. [2]
Show answer
  1. (a) (2, −3)
    B1: translation by 2 units in the negative x-direction.
  2. (b) (4, −9)
    B1: stretch parallel to the y-axis, y-coordinate multiplied by 3.
  3. (c) (−4, −4)
    B1 x-coordinate −4 (reflection in the y-axis); B1 y-coordinate −4 (translation 1 unit down).
Q2Transformations of graphs · Core · Paper 1 style · 3 marks
  1. (a) Describe fully the single transformation that maps the curve y = sin x onto the curve y = sin(x/3). [2]
  2. (b) Describe fully the single transformation that maps the curve y = sin x onto the curve y = −sin x. [1]
Show answer
  1. (a) A stretch parallel to the x-axis with scale factor 3 (y-axis invariant).
    B1 stretch parallel to the x-axis; B1 scale factor 3.
  2. (b) A reflection in the x-axis.
    B1 reflection in the x-axis (or y = 0).
Q3Transformations of graphs · Core · Paper 1 style · 4 marks
  1. (a) The curve y = x² − 4x + 7 is translated by the vector (−1, 3), written as a column vector with x-component −1 and y-component 3. Find the equation of the new curve in the form y = x² + bx + c. [3]
  2. (b) State the coordinates of the minimum point of the new curve. [1]
Show answer
  1. (a) y = x² − 2x + 7
    M1 replace x by (x + 1): y = (x + 1)² − 4(x + 1) + 7; M1 add 3; A1 y = x² − 2x + 7.
  2. (b) (1, 6)
    B1: original minimum (2, 3) moves to (1, 6); or complete the square (x − 1)² + 6.
Q4Transformations of graphs · Core · Paper 1 style · 4 marks
  1. (a) The curve y = x² − 5x is reflected in the x-axis and then translated by the vector (0, 6). Find the equation of the new curve. [2]
  2. (b) Find the x-coordinates of the points where the new curve meets the x-axis. [2]
Show answer
  1. (a) y = −x² + 5x + 6
    B1 reflection: y = −x² + 5x; B1 add 6.
  2. (b) x = −1 and x = 6
    M1 −(x² − 5x − 6) = −(x − 6)(x + 1) = 0; A1 x = 6, x = −1.
Q5Transformations of graphs · Standard · Paper 1 style · 6 marks
  1. (a) The curve y = f(x) is transformed to the curve y = 2f(x − 1) + 3. Describe fully a sequence of transformations that maps y = f(x) onto y = 2f(x − 1) + 3, making clear the order in which they are applied. [4]
  2. (b) The point (3, 5) lies on y = f(x). Find the coordinates of its image on y = 2f(x − 1) + 3. [2]
Show answer
  1. (a) Translation by (1, 0); then stretch parallel to the y-axis, scale factor 2; then translation by (0, 3). (Equivalent: stretch in y, factor 2, then translation (1, 3).)
    B1 translation 1 unit in positive x-direction; B1 stretch parallel to the y-axis, factor 2; B1 translation 3 units in positive y-direction; B1 correct order (the y-translation must come after the stretch).
  2. (b) (4, 13)
    B1 x = 3 + 1 = 4; B1 y = 2 × 5 + 3 = 13.
Q6Transformations of graphs · Standard · Paper 1 style · 6 marks
  1. (a) Express 2x² − 12x + 13 in the form a(x − b)² + c, where a, b and c are constants. [3]
  2. (b) Hence describe fully a sequence of two transformations that maps the curve y = x² onto the curve y = 2x² − 12x + 13. [3]
Show answer
  1. (a) 2(x − 3)² − 5
    B1 a = 2; B1 b = 3; B1 c = −5.
  2. (b) Stretch parallel to the y-axis with scale factor 2, followed by a translation by the vector (3, −5).
    B1 stretch, parallel to y-axis, factor 2; B1 translation (3, −5); B1 order correct (stretch first). [Translation (3, 0), stretch factor 2, translation (0, −5) also acceptable.]
Q7Transformations of graphs · Standard · Paper 1 style · 5 marks
  1. (a) The function f is defined by f(x) = x² + 2x for x ∈ ℝ. The curve y = f(x) is reflected in the y-axis and then translated by the vector (3, 0). The resulting curve is y = g(x). Find g(x) in the form x² + px + q. [3]
  2. (b) Find the x-coordinates of the points where y = g(x) crosses the x-axis. [2]
Show answer
  1. (a) g(x) = x² − 8x + 15
    M1 reflection gives y = f(−x) = x² − 2x; M1 replace x by (x − 3): (x − 3)² − 2(x − 3); A1 x² − 8x + 15.
  2. (b) x = 3 and x = 5
    M1 factorise (x − 3)(x − 5) = 0; A1 both values.
Q8Transformations of graphs · Standard · Paper 1 style · 4 marks
  1. (a) Show that √(4x − 8) can be written in the form k√(x − 2), stating the value of k. [1]
  2. (b) Hence describe fully a sequence of two transformations that maps the curve y = √x onto the curve y = √(4x − 8). [3]
Show answer
  1. (a) k = 2
    B1 √(4(x − 2)) = 2√(x − 2).
  2. (b) Translation by the vector (2, 0) and a stretch parallel to the y-axis with scale factor 2 (in either order).
    B1 translation (2, 0); B1 stretch parallel to the y-axis; B1 scale factor 2. [Alternative: translation (8, 0) followed by stretch parallel to x-axis, factor ¼.]
Q9Transformations of graphs · Challenge · Paper 1 style · 6 marks
  1. (a) The curve y = x³ is transformed by the following sequence: a translation by the vector (2, 0); then a stretch parallel to the y-axis with scale factor ½; then a reflection in the x-axis. Find the equation of the resulting curve. [3]
  2. (b) Find the coordinates of the point where the resulting curve crosses the y-axis. [1]
  3. (c) Find the image of the point (1, 1) under the same sequence of transformations. [2]
Show answer
  1. (a) y = −½(x − 2)³
    B1 y = (x − 2)³; B1 y = ½(x − 2)³; B1 y = −½(x − 2)³.
  2. (b) (0, 4)
    B1 y = −½(−2)³ = 4.
  3. (c) (3, −½)
    B1 (3, 1) after translation, (3, ½) after stretch; B1 (3, −½).
Q10Transformations of graphs · Challenge · Paper 1 style · 8 marks
  1. (a) The curve y = f(x) has a maximum point at (6, 10) and crosses the x-axis at (2, 0). The function g is defined by g(x) = 5 − 2f(½x). Find the coordinates of the stationary point on y = g(x) and state whether it is a maximum or a minimum. [3]
  2. (b) Find the coordinates of the image of (2, 0) on y = g(x). [1]
  3. (c) Describe fully a sequence of transformations that maps y = f(x) onto y = g(x). [4]
Show answer
  1. (a) (12, −15), a minimum point
    B1 x = 6 × 2 = 12; B1 y = 5 − 2 × 10 = −15; B1 minimum (the factor −2 reflects the curve in the x-axis).
  2. (b) (4, 5)
    B1 x = 4, y = 5 − 0 = 5.
  3. (c) Stretch parallel to the x-axis, factor 2; stretch parallel to the y-axis, factor 2; reflection in the x-axis; translation by (0, 5) — the translation applied last.
    B1 x-stretch factor 2; B1 y-stretch factor 2; B1 reflection in x-axis; B1 translation (0, 5) after the stretch and reflection (the stretch/reflection may be combined as a y-stretch with factor −2).
Q11Transformations of graphs · Challenge · Paper 1 style · 6 marks
  1. (a) The function f is defined by f(x) = x² − 6x + 5 for x ∈ ℝ. The curve y = f(x) is mapped onto the curve y = f(x + k) + m by a translation, and the vertex of the new curve is at (−1, 2). Find the values of k and m. [3]
  2. (b) Find the y-intercept of the new curve. [1]
  3. (c) Determine whether the new curve meets the line y = x + 4, giving a reason. [2]
Show answer
  1. (a) k = 4, m = 6
    M1 complete the square: f(x) = (x − 3)² − 4, vertex (3, −4); A1 k = 4 (3 − k = −1); A1 m = 6 (−4 + m = 2).
  2. (b) 3
    B1 (0 + 1)² + 2 = 3.
  3. (c) Yes, at two points: x² + x − 1 = 0 has discriminant 5 > 0.
    M1 (x + 1)² + 2 = x + 4 gives x² + x − 1 = 0; A1 discriminant 1 + 4 = 5 > 0, so two points of intersection.
Q12Inverse function · Core · Paper 1 style · 3 marks
  1. (a) The function f is defined by f(x) = 5x − 3 for x ∈ ℝ. Find an expression for f⁻¹(x). [2]
  2. (b) Find the value of f⁻¹(12). [1]
Show answer
  1. (a) f⁻¹(x) = (x + 3)/5
    M1 y = 5x − 3 ⇒ x = (y + 3)/5; A1.
  2. (b) 3
    B1 (12 + 3)/5 = 3.
Q13Inverse function · Core · Paper 1 style · 5 marks
  1. (a) The function f is defined by f(x) = (2x + 1)/(x − 3) for x ≠ 3. Find an expression for f⁻¹(x). [3]
  2. (b) Verify your answer by finding f(5) and then f⁻¹ of your result. [2]
Show answer
  1. (a) f⁻¹(x) = (3x + 1)/(x − 2), x ≠ 2
    M1 y(x − 3) = 2x + 1; M1 collect x terms: x(y − 2) = 3y + 1; A1 f⁻¹(x) = (3x + 1)/(x − 2).
  2. (b) f(5) = 11/2 and f⁻¹(11/2) = 5
    B1 f(5) = 11/2; B1 f⁻¹(11/2) = (33/2 + 1)/(11/2 − 2) = (35/2)/(7/2) = 5.
Q14Inverse function · Core · Paper 1 style · 5 marks
  1. (a) The function g is defined by g(x) = 7 − 3x for −2 ≤ x ≤ 4. Find the range of g. [2]
  2. (b) Find an expression for g⁻¹(x) and state the domain of g⁻¹. [3]
Show answer
  1. (a) −5 ≤ g(x) ≤ 13
    M1 g(4) = −5, g(−2) = 13; A1 correct inequality.
  2. (b) g⁻¹(x) = (7 − x)/3, −5 ≤ x ≤ 13
    M1 x = (7 − y)/3; A1; B1 domain equals the range of g.
Q15Inverse function · Standard · Paper 1 style · 6 marks
  1. (a) The function f is defined by f(x) = x² − 8x + 21 for x ≥ 4. Express f(x) in the form (x − a)² + b. [2]
  2. (b) State the range of f. [1]
  3. (c) Find an expression for f⁻¹(x) and state the domain of f⁻¹. [3]
Show answer
  1. (a) (x − 4)² + 5
    B1 a = 4; B1 b = 5.
  2. (b) f(x) ≥ 5
    B1 (minimum value 5 at x = 4).
  3. (c) f⁻¹(x) = 4 + √(x − 5), domain x ≥ 5
    M1 y − 5 = (x − 4)²; A1 x = 4 + √(y − 5) (positive root as x ≥ 4); B1 domain x ≥ 5.
Q16Inverse function · Standard · Paper 1 style · 6 marks
  1. (a) The function f is defined by f(x) = 3 − 2x² for x ≤ 0. Explain why f has an inverse, but the function g(x) = 3 − 2x² for x ∈ ℝ does not. [2]
  2. (b) State the range of f. [1]
  3. (c) Find an expression for f⁻¹(x). [3]
Show answer
  1. (a) f is one-one on x ≤ 0 (strictly increasing there); g is not one-one, e.g. g(1) = g(−1) = 1.
    B1 f is one-one / increasing for x ≤ 0; B1 g is many-one, with a correct example or reference to symmetry about x = 0.
  2. (b) f(x) ≤ 3
    B1.
  3. (c) f⁻¹(x) = −√((3 − x)/2), for x ≤ 3
    M1 x² = (3 − y)/2; M1 square root; A1 negative root chosen because the domain of f is x ≤ 0.
Q17Inverse function · Standard · Paper 1 style · 7 marks
  1. (a) The function f is defined by f(x) = 2x² + 12x + 11 for x ≥ k. Express 2x² + 12x + 11 in the form a(x + b)² + c. [3]
  2. (b) State the smallest value of k for which f has an inverse. [1]
  3. (c) For this value of k, find an expression for f⁻¹(x) and state its domain. [3]
Show answer
  1. (a) 2(x + 3)² − 7
    B1 a = 2; B1 b = 3; B1 c = −7.
  2. (b) k = −3
    B1 (x-coordinate of the vertex).
  3. (c) f⁻¹(x) = −3 + √((x + 7)/2), x ≥ −7
    M1 (x + 3)² = (y + 7)/2; A1 x = −3 + √((y + 7)/2); B1 domain x ≥ −7.
Q18Inverse function · Standard · Paper 1 style · 6 marks
  1. (a) The function f is defined by f(x) = 2 + 4/(x − 1) for x > 1. State the range of f. [1]
  2. (b) Find an expression for f⁻¹(x). [3]
  3. (c) State the domain and the range of f⁻¹. [2]
Show answer
  1. (a) f(x) > 2
    B1.
  2. (b) f⁻¹(x) = 1 + 4/(x − 2)
    M1 y − 2 = 4/(x − 1); M1 x − 1 = 4/(y − 2); A1 f⁻¹(x) = 1 + 4/(x − 2).
  3. (c) Domain x > 2; range f⁻¹(x) > 1
    B1 domain; B1 range.
Q19Inverse function · Challenge · Paper 1 style · 7 marks
  1. (a) The function f is defined by f(x) = x² − 4x + 6 for x ≥ 2. Find an expression for f⁻¹(x) and state its domain. [3]
  2. (b) Explain why the solutions of f(x) = f⁻¹(x) are the same as the solutions of f(x) = x for this function. [1]
  3. (c) Hence solve the equation f(x) = f⁻¹(x). [3]
Show answer
  1. (a) f⁻¹(x) = 2 + √(x − 2), x ≥ 2
    M1 complete the square: (x − 2)² + 2; A1 f⁻¹(x) = 2 + √(x − 2); B1 domain x ≥ 2.
  2. (b) f is increasing, so the graphs of y = f(x) and y = f⁻¹(x) (reflections of each other in y = x) can only meet on the line y = x.
    B1 reference to reflection in y = x with f increasing.
  3. (c) x = 2 and x = 3
    M1 x² − 4x + 6 = x ⇒ x² − 5x + 6 = 0; A1 (x − 2)(x − 3) = 0; A1 x = 2, 3 (both in the domain).
Q20Inverse function · Challenge · Paper 1 style · 5 marks
  1. (a) The function f is defined by f(x) = 3 + 2 sin x for −½π ≤ x ≤ ½π. State the range of f. [1]
  2. (b) Find an expression for f⁻¹(x). [2]
  3. (c) Find the exact value of f⁻¹(4). [1]
  4. (d) Explain why f would not have an inverse if its domain were changed to 0 ≤ x ≤ π. [1]
Show answer
  1. (a) 1 ≤ f(x) ≤ 5
    B1.
  2. (b) f⁻¹(x) = sin⁻¹((x − 3)/2), 1 ≤ x ≤ 5
    M1 sin x = (y − 3)/2; A1.
  3. (c) π/6
    B1 sin⁻¹(½) = π/6.
  4. (d) On 0 ≤ x ≤ π the function is not one-one, e.g. f(π/6) = f(5π/6) = 4.
    B1 not one-one with a reason.
Q21Composite function · Core · Paper 1 style · 4 marks
  1. (a) The functions f and g are defined by f(x) = 2x − 5 for x ∈ ℝ and g(x) = 1/(x + 1) for x ≠ −1. Find the value of gf(3). [2]
  2. (b) Find an expression for fg(x) as a single fraction. [2]
Show answer
  1. (a) ½
    M1 f(3) = 1; A1 g(1) = ½.
  2. (b) fg(x) = (−5x − 3)/(x + 1)
    M1 2/(x + 1) − 5; A1 (2 − 5x − 5)/(x + 1) = (−5x − 3)/(x + 1).
Q22Composite function · Standard · Paper 1 style · 6 marks
  1. (a) The functions f and g are defined for x ∈ ℝ by f(x) = 3x + 2 and g(x) = x² − 1. Find expressions for fg(x) and gf(x). [3]
  2. (b) Solve the equation fg(x) = gf(x), giving your answers in exact form. [3]
Show answer
  1. (a) fg(x) = 3x² − 1; gf(x) = 9x² + 12x + 3
    B1 fg(x) = 3(x² − 1) + 2 = 3x² − 1; M1 (3x + 2)² − 1; A1 9x² + 12x + 3.
  2. (b) x = −1 ± √3/3
    M1 6x² + 12x + 4 = 0 ⇒ 3x² + 6x + 2 = 0; M1 formula; A1 x = (−6 ± √12)/6 = −1 ± √3/3 (≈ −0.423, −1.58).
Q23Composite function · Standard · Paper 1 style · 5 marks
  1. (a) The functions f and g are defined for x ∈ ℝ by f(x) = x² + 4x and g(x) = 2x − 3. Find and simplify an expression for gf(x). [2]
  2. (b) Solve the equation fg(x) = 12. [3]
Show answer
  1. (a) gf(x) = 2x² + 8x − 3
    M1 2(x² + 4x) − 3; A1.
  2. (b) x = 5/2 or x = −3/2
    M1 (2x − 3)² + 4(2x − 3) = 12; M1 put u = 2x − 3: u² + 4u − 12 = 0, u = 2 or −6; A1 x = 5/2, −3/2.
Q24Composite function · Standard · Paper 1 style · 4 marks
  1. (a) The function f is defined by f(x) = (x + 3)/(x − 1) for x ≠ 1. Show that ff(x) = x. [3]
  2. (b) Hence state an expression for f⁻¹(x). [1]
Show answer
  1. (a) ff(x) = x (shown)
    M1 substitute: ((x + 3)/(x − 1) + 3) / ((x + 3)/(x − 1) − 1); M1 multiply through by (x − 1): (4x)/(4); A1 = x.
  2. (b) f⁻¹(x) = (x + 3)/(x − 1)
    B1 f is self-inverse.
Q25Composite function · Challenge · Paper 1 style · 7 marks
  1. (a) The function f is defined by f(x) = x² − 6x + 10 for x ≥ 3, and g is defined by g(x) = 2x + k for x ≥ 0, where k is a constant. State the range of f and the range of g in terms of k. [2]
  2. (b) Find the least value of k for which the composite function fg exists. [2]
  3. (c) Using k = 3, solve the equation fg(x) = 17. [3]
Show answer
  1. (a) f(x) ≥ 1; g(x) ≥ k
    B1 f(x) = (x − 3)² + 1 ≥ 1; B1 g(x) ≥ k.
  2. (b) k = 3
    M1 range of g must lie within the domain of f (x ≥ 3); A1 k ≥ 3, least value 3.
  3. (c) x = 2
    M1 fg(x) = (2x + 3 − 3)² + 1 = 4x² + 1; M1 4x² = 16; A1 x = 2 only (x = −2 is outside the domain of g).
Q26Composite function · Challenge · Paper 1 style · 5 marks
  1. (a) The functions f and g are defined for x ∈ ℝ by f(x) = 2x + 1 and g(x) = x² − 2. Find f⁻¹(x) and gf(x). [2]
  2. (b) Solve the equation gf(x) = f⁻¹(x). [3]
Show answer
  1. (a) f⁻¹(x) = (x − 1)/2; gf(x) = 4x² + 4x − 1
    B1 f⁻¹; B1 gf.
  2. (b) x = 1/8 or x = −1
    M1 4x² + 4x − 1 = (x − 1)/2 ⇒ 8x² + 7x − 1 = 0; M1 (8x − 1)(x + 1) = 0; A1 both values.
Q27Domain & range · Core · Paper 1 style · 4 marks
  1. (a) The function f is defined by f(x) = 5 − (x − 2)² for 0 ≤ x ≤ 5. Find the range of f. [3]
  2. (b) Explain why f does not have an inverse. [1]
Show answer
  1. (a) −4 ≤ f(x) ≤ 5
    B1 maximum 5 at x = 2; M1 evaluate ends f(0) = 1, f(5) = −4; A1 −4 ≤ f(x) ≤ 5.
  2. (b) f is not one-one on 0 ≤ x ≤ 5, e.g. f(1) = f(3) = 4.
    B1.
Q28Domain & range · Standard · Paper 1 style · 5 marks
  1. (a) The function f is defined by f(x) = 3/(x + 2) for x ≥ 1. Find the range of f. [2]
  2. (b) The function g is defined by g(x) = x² − 1 for x ∈ ℝ. Find an expression for gf(x) and the range of gf. [3]
Show answer
  1. (a) 0 < f(x) ≤ 1
    B1 f(1) = 1 and f decreasing; B1 f(x) > 0 (tends to 0).
  2. (b) gf(x) = 9/(x + 2)² − 1; range −1 < gf(x) ≤ 0
    B1 gf(x); M1 use the range of f as the domain of g; A1 −1 < gf(x) ≤ 0.
Q29Domain & range · Standard · Paper 1 style · 4 marks
  1. (a) The function f is defined by f(x) = 4 sin x − 1 for 0 ≤ x ≤ 7π/6. Find the range of f. [3]
  2. (b) Find the largest value of a such that f, with domain 0 ≤ x ≤ a, has an inverse. [1]
Show answer
  1. (a) −3 ≤ f(x) ≤ 3
    B1 maximum of sin x is 1 at x = π/2, giving 3; M1 least value of sin x on the domain is sin(7π/6) = −½; A1 −3 ≤ f(x) ≤ 3.
  2. (b) a = π/2
    B1 f is increasing up to x = π/2 only.
Q30Domain & range · Challenge · Paper 1 style · 5 marks
  1. (a) The function f is defined by f(x) = x² − 6x + c for 1 ≤ x ≤ 7, where c is a constant. The range of f is −4 ≤ f(x) ≤ 12. Find the value of c and verify that the greatest value is 12. [3]
  2. (b) Explain why f does not have an inverse, and state the largest domain of the form 3 ≤ x ≤ b, with b ≤ 7, on which it would have one. [2]
Show answer
  1. (a) c = 5; f(7) = 49 − 42 + 5 = 12
    M1 minimum at x = 3: c − 9 = −4; A1 c = 5; B1 f(7) = 12 and f(1) = 0 < 12.
  2. (b) f(1) = f(5) = 0, so f is not one-one; on 3 ≤ x ≤ 7 it is one-one (b = 7).
    B1 not one-one with example; B1 b = 7.

Similar questions

Questions that test the same subtopic in different years, grouped so you can work on one skill at a time.

References only: find each question in its paper (official link where available, otherwise through your school).

Transformations of graphs Banker28 questions · 9/9 sessions
Inverse function Banker27 questions · 9/9 sessions
Composite function Banker20 questions · 9/9 sessions
Domain & range Banker20 questions · 9/9 sessions

Subtopics

Our analysis of how often each subtopic was examined and how many marks it carried, 2022 to May/June 2026.

SubtopicMarks indexedSessions
Transformations of graphs Banker1319/9
Inverse function Banker979/9
Composite function Banker619/9
Domain & range Banker339/9

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