Pearson Edexcel International Advanced Level Mathematics: Functions, modulus and graph transformations (YMA01) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover topics 1.1 to 1.4 of Unit P3 in the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics specification, Issue 3 (April 2019). The specification lists P3 as compulsory for IAL Mathematics and IAL Pure Mathematics, with IAS weighting N/A. Read the study guide first, or use the revision notes. Course pages: the Edexcel A-level Mathematics hub and the printable checklist. Find weak topics with the free 10-minute diagnostics.

Questions

1. Simplify (4x² − 9)/(2x² + x − 6). [3]

2. Write (x³ + 2x² − 5x + 1)/(x² − 3) in the form Ax + B + (Cx + D)/(x² − 3), where A, B, C and D are constants. [4]

3. The function h is defined by h(x) = x² − 6x + 13 for all real x. Explain why h has no inverse, and find the smallest value of k for which h has an inverse when its domain is x ≥ k. [3]

4. Two functions are given by f(x) = 5 − 2x and g(x) = x² + 3, each for all real x.

(a) Find fg(x). [2] (b) Find gf(x), giving your answer in expanded form. [2] (c) Solve fg(x) = −19. [2] (d) State the range of fg. [1]

5. The function f is defined by f(x) = (5x − 2)/(x + 3), x ∈ ℝ, x ≠ −3.

(a) Find f⁻¹(x). [3] (b) State the range of f. [1] (c) Show that f⁻¹f(x) = x. [3]

6. The function g is defined by g(x) = x² + 8x + 3 for x ≥ −4.

(a) Write g(x) in the form (x + a)² + b, and state the range of g. [3] (b) Find g⁻¹(x) and state its domain. [3] (c) Solve g(x) = g⁻¹(x), giving your answer in exact form. [3]

7. (a) Sketch y = |7 − 3x|, giving the coordinates of the vertex and the y-intercept. [2] (b) Solve |7 − 3x| = x + 3. [3] (c) Hence solve |7 − 3x| > x + 3. [2]

8. Let f(x) = x² − 4x − 5.

(a) Sketch y = |f(x)|, giving the coordinates of the intercepts and the turning point. [3] (b) Sketch y = f(|x|), giving the coordinates of the intercepts and the turning points. [3] (c) Solve |f(x)| = 9. [3]

9. The curve y = f(x) has a maximum at A(−1, 4) and crosses the x-axis at B(−3, 0) and C(2, 0). Write down the coordinates of the images of A, B and C on each curve.

(a) y = −2f(x) [2] (b) y = f(x/2) + 3 [3] (c) y = 2 − f(x) [2]

10. (a) For y = 2 + cos 3x, state the maximum value, the minimum value and the period. [3] (b) Describe the sequence of transformations that maps y = sin x to y = −sin(x − π/3), then sketch the graph for 0 ≤ x ≤ 2π, giving the y-intercept, the x-intercepts and the coordinates of the turning points. [4]

11. The function f is defined by f(x) = 4 − |x − 2| for all real x.

(a) Sketch y = f(x), giving the coordinates of the vertex and the intercepts, and state the range of f. [3] (b) Solve f(x) = x/2 + 1. [4] (c) Hence solve f(x) > x/2 + 1. [2] (d) The domain of f is now restricted to x ≥ 2. Find f⁻¹(x) and state its domain. [3]

Answers

1. 4x² − 9 = (2x − 3)(2x + 3) [1]. 2x² + x − 6 = (2x − 3)(x + 2) [1]. Cancel (2x − 3): (2x + 3)/(x + 2) [1]. [3] Examiner insight: Write both factorisations on the page before cancelling, and cancel only brackets that match exactly. Check by substituting x = 1 into the original and the simplified forms: both give 5/3.

2. x³ ÷ x² = x, and x(x² − 3) = x³ − 3x, leaving 2x² − 2x + 1 [1]. 2x² ÷ x² = 2, and 2(x² − 3) = 2x² − 6, leaving −2x + 7 [1]. Quotient x + 2 [1]. So the expression is x + 2 + (7 − 2x)/(x² − 3), with A = 1, B = 2, C = −2, D = 7 [1]. [4] Examiner insight: Keep the remainder’s sign inside the numerator. Multiply back, (x + 2)(x² − 3) + 7 − 2x, to confirm you recover the original numerator.

3. h(x) = (x − 3)² + 4. For example h(2) = 5 = h(4) [1], so h is many-one and has no inverse [1]. The turning point is at x = 3, so the smallest value is k = 3 [1]. [3] Examiner insight: An explanation needs a reason with evidence, such as a pair of inputs with the same output. “It is a quadratic” alone gives no justification.

4. (a) fg(x) = 5 − 2(x² + 3) [1] = −2x² − 1 [1]. [2] (b) gf(x) = (5 − 2x)² + 3 [1] = 4x² − 20x + 28 [1]. [2] (c) −2x² − 1 = −19 gives x² = 9 [1], so x = 3 or x = −3 [1]. [2] (d) x² ≥ 0, so fg(x) ≤ −1: range fg(x) ≤ −1 [1]. [1] Examiner insight: fg means g first. Swapping the order gives the answer to part (b) in part (a). Expand brackets only when the question asks.

5. (a) y(x + 3) = 5x − 2 [1], so xy + 3y = 5x − 2 and x(y − 5) = −3y − 2 [1]. f⁻¹(x) = (3x + 2)/(5 − x), x ≠ 5 [1]. [3] (b) f(x) = 5 − 17/(x + 3), which never equals 5: range f(x) ≠ 5 [1]. [1] (c) f⁻¹(f(x)) = [3(5x − 2)/(x + 3) + 2] ÷ [5 − (5x − 2)/(x + 3)] [1]. Multiply top and bottom by (x + 3): numerator 15x − 6 + 2x + 6 = 17x, denominator 5(x + 3) − (5x − 2) = 17 [1]. So the result is 17x/17 = x [1]. [3] Examiner insight: For a “show that”, every line of working should be visible and the last line should be exactly the stated result. State the excluded value of x to complete a domain or range statement.

6. (a) g(x) = (x + 4)² − 13 [2]. Since x ≥ −4 the minimum is −13, so range g(x) ≥ −13 [1]. [3] (b) y = (x + 4)² − 13 gives x + 4 = √(y + 13) [1], taking the positive root as x ≥ −4. So g⁻¹(x) = −4 + √(x + 13) [1], with domain x ≥ −13 [1]. [3] (c) g is increasing for x ≥ −4, so g and g⁻¹ meet on y = x [1]. x² + 8x + 3 = x gives x² + 7x + 3 = 0 [1], so x = (−7 ± √37)/2. Only x = (−7 + √37)/2 is at least −4 (the other is about −6.5) [1]. [3] Examiner insight: Say why the other root is rejected, referring to the domain. For g(x) = g⁻¹(x), explain why using y = x is valid.

7. (a) A V shape above the x-axis [1] with vertex at (7/3, 0) and y-intercept (0, 7) [1]. [2] (b) 7 − 3x = x + 3 gives x = 1 [1]. 3x − 7 = x + 3 gives x = 5 [1]. Check: |7 − 3| = 4 = 1 + 3 and |7 − 15| = 8 = 5 + 3, so both are valid [1]. [3] (c) The V is above the line outside the two intersections [1]: x < 1 or x > 5 [1]. [2] Examiner insight: Use the sketch to decide between “and” and “or” in an inequality answer. Test a value such as x = 3: |−2| = 2 is less than 6, which confirms the middle region fails.

8. (a) f(x) = (x − 5)(x + 1). The graph is a W-shape with x-intercepts (−1, 0), (5, 0) [1], y-intercept (0, 5) [1] and a local maximum (2, 9) [1]. [3] (b) For x ≥ 0, keep f(x) and mirror it in the y-axis. x-intercepts (−5, 0), (5, 0) [1], y-intercept (0, −5) [1], minimum points (−2, −9) and (2, −9) [1]. [3] (c) f(x) = 9 gives x² − 4x − 14 = 0, so x = 2 ± 3√2 [1]. f(x) = −9 gives (x − 2)² = 0, so x = 2 [1]. There are three solutions in total [1]. [3] Examiner insight: A modulus equation has two cases, and each needs dealing with explicitly. Here f(x) = −9 touches the turning point, giving one repeated solution. The minimum points of y = f(|x|) are (±2, −9), and its y-intercept (0, −5) is where the two halves meet.

9. (a) A → (−1, −8) [1]. B and C stay at (−3, 0) and (2, 0) [1]. [2] (b) x-coordinates double, then y-coordinates increase by 3. A → (−2, 7) [1], B → (−6, 3) [1], C → (4, 3) [1]. [3] (c) y = 2 − f(x) reflects in the x-axis, then adds 2. A → (−1, −2) [1]. B → (−3, 2) and C → (2, 2) [1]. [2] Examiner insight: Doubling the x-coordinates for f(x/2) is the common point to check, because the scale factor is 2 and not 1/2. Show the order of your steps when two transformations are combined.

10. (a) Maximum 3 [1], minimum 1 [1], period 2π/3 [1]. [3] (b) Translate y = sin x right by π/3 [1], then reflect in the x-axis [1]. y-intercept (0, √3/2) [1]. x-intercepts (π/3, 0) and (4π/3, 0); minimum (5π/6, −1) and maximum (11π/6, 1) [1]. [4] Examiner insight: Give exact values in terms of π, and mark the end-point values of the interval on the sketch. Inside the bracket, x − π/3 means a shift to the right, which you can confirm by checking that the first zero is at x = π/3.

11. (a) An inverted V with vertex (2, 4) [1], y-intercept (0, 2) and x-intercepts (−2, 0), (6, 0) [1]. Range f(x) ≤ 4 [1]. [3] (b) For x ≥ 2: 6 − x = x/2 + 1 [1], so x = 10/3 [1]. For x < 2: x + 2 = x/2 + 1, so x = −2 [1]. Both lie in their cases, and both check: f(−2) = 0 = −1 + 1 and f(10/3) = 8/3 = 5/3 + 1 [1]. [4] (c) The inverted V lies above the line between the two intersections [1]: −2 < x < 10/3 [1]. [2] (d) For x ≥ 2, f(x) = 6 − x [1]. Let y = 6 − x, so f⁻¹(x) = 6 − x [1], with domain x ≤ 4 (the range of f) [1]. [3] Examiner insight: Write each branch of the modulus separately with its range of x, and state that the answers satisfy it. In part (d), the domain of f⁻¹ comes from the range of the restricted f, not from the formula 6 − x.

Where marks are usually lost

• Cancelling a term rather than a factor, or forgetting an excluded value.
• Putting the remainder’s sign in the wrong place after algebraic division.
• Reading fg as “f first”.
• Giving a domain for f⁻¹ that comes from f’s domain.
• Choosing the wrong square root sign when inverting a quadratic with a restricted domain.
• Not testing modulus solutions or giving “and” for an “or” inequality.
• Sketching y = f(|x|) with a turning point at x = 0 that is not there, or leaving the reflected part out.
• Doubling instead of halving, or the reverse, in f(ax), and moving f(x + a) the wrong way.
• Leaving coordinates and equations of asymptotes off a sketch.

Next steps

• Revisit the revision notes for any question you lost marks on.
• Go back to the relevant section of the study guide.
• Tick off topics on the printable checklist and see the course hub.
• Take the free 10-minute diagnostics to compare topics.
• Want a teacher to go through your working? Book a free trial class.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics, Specification Issue 3, April 2019, Unit P3 content section 1 (Algebra and functions), topics 1.1 to 1.4. Pearson Education Limited.