IB DP Mathematics: Analysis and Approaches – Polynomials, rational functions, odd/even functions and modulus graphs (HL) 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 – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

This practice set covers the HL further functions unit of IB Diploma Programme Mathematics: Analysis and Approaches. It is aligned to the IB Mathematics: analysis and approaches guide, first assessment 2021, syllabus sections 2.12–2.16, and every question is HL only (AHL). It follows the IB guide for first assessment 2021, which remains the examined syllabus until the new course is first assessed in May 2029, so it applies to the May and November 2026, 2027 and 2028 HL sessions.

HL Paper 1 allows no technology; Papers 2 and 3 require it. Questions are labelled “(calculator-free)” or “(calculator allowed)” to match. Give calculator-allowed answers exactly or to 3 significant figures.

Learn the methods first in the further functions study guide and the revision notes. The IB DP Maths AA course hub and the printable syllabus checklist show where this unit sits. For SL function skills, use the functions practice questions.

Questions

1. (calculator-free) When p(x) = x³ + kx² − 4x + 3 is divided by (x − 2), the remainder is 11. Find k. [2]

2. (calculator-free) The equation 3x³ + kx² + 5x − 12 = 0 has roots α, β and γ, where α + β + γ = 2. Find k and the value of αβγ. [3]

3. (calculator-free) Determine whether each function is odd, even or neither. Justify your answers.

(a) f(x) = x⁴ − 3x² [1] (b) g(x) = x cos x [2] (c) h(x) = x² + x [2]

4. (calculator-free) Let f(x) = (3x + 4)/(2x − 3), x ≠ 3/2.

(a) Show that f(f(x)) = x. [3] (b) Hence write down f⁻¹(x). [1]

5. (calculator-free) Let f(x) = 2x² + 8x + 3, x ≥ k.

(a) Write down the smallest value of k for which f⁻¹ exists. [1] (b) For this value of k, find f⁻¹(x) and state its domain. [4]

6. (calculator-free) Let g(x) = (3x − 6)/(x² − 2x − 8). Find the equations of all the asymptotes of y = g(x) and the coordinates of its axis intercepts. [5]

7. (calculator-free) Solve the inequality x³ − 4x ≥ 3x² − 12. [5]

8. (calculator-free)

(a) Solve |x − 4| = 2x + 1. [4] (b) Solve |x + 3| > 2|x − 1|. [4]

9. (calculator allowed) Solve ln(x + 3) > |x − 1|. [5]

10. (calculator-free) Let f(x) = x² − 2x − 3.

(a) Find the coordinates of the minimum point on y = f(2x + 1). [2] (b) Find the x-intercepts of y = f(|x|). [2] (c) Find the coordinates of the local maximum and the two minimum points on y = [f(x)]². [2] (d) For y = 1/f(x), write down the equations of the asymptotes and the coordinates of the local maximum point. [3]

11. (calculator allowed) Let f(x) = (x² + 3)/(x − 1), x ≠ 1.

(a) Write f(x) in the form x + p + q/(x − 1). [2] (b) Write down the equations of the asymptotes of y = f(x). [2] (c) Find the coordinates of the local maximum and local minimum points, and hence the range of f. [4] (d) Solve f(x) > 2x − 1. [3]

12. (calculator-free) Let p(x) = x³ − x² − 8x + 12.

(a) Show that (x − 2) is a factor of p(x). [2] (b) Hence factorise p(x) fully. [2] (c) Verify that the roots of p(x) = 0 agree with the formulas for the sum and product of the roots. [2] (d) For y = 1/p(x), write down the equations of the asymptotes and the y-intercept. [3] (e) Solve p(x) ≤ 12. [3]

Answers

1. p(2) = 8 + 4k − 8 + 3 = 4k + 3 [1]. 4k + 3 = 11, so k = 2 [1]. [2] Examiner insight: substituting x = −2 instead of x = 2 is a method error, so both marks are lost, not just the accuracy mark.

2. Sum of roots = −k/3 [1]. −k/3 = 2, so k = −6 [1]. αβγ = (−1)³(−12)/3 = 4 [1]. [3] Examiner insight: the product mark needs the correct sign; −4 from forgetting (−1)³ scores 0 for that line.

3. (a) f(−x) = (−x)⁴ − 3(−x)² = x⁴ − 3x² = f(x), so even [1]. (b) g(−x) = (−x)cos(−x) [1] = −x cos x = −g(x), so odd [1]. (c) h(1) = 2 and h(−1) = 0 [1]. Since h(−1) ≠ h(1) and h(−1) ≠ −h(1), h is neither [1]. Examiner insight: in (b) you must use cos(−x) = cos x explicitly in general x; a numerical check earns nothing for “odd”, whereas in (c) one counterexample is the full proof.

4. (a) Numerator: 3(3x + 4)/(2x − 3) + 4 = (9x + 12 + 8x − 12)/(2x − 3) = 17x/(2x − 3) [1]. Denominator: 2(3x + 4)/(2x − 3) − 3 = (6x + 8 − 6x + 9)/(2x − 3) = 17/(2x − 3) [1]. f(f(x)) = 17x/17 = x [1]. (b) f is self-inverse, so f⁻¹(x) = (3x + 4)/(2x − 3), x ≠ 3/2 [1]. Examiner insight: this is a “show that”, so each simplified fraction must appear; jumping from the substitution to “= x” loses the middle marks.

5. (a) f(x) = 2(x + 2)² − 5, vertex at x = −2, so k = −2 [1]. (b) f(x) = 2(x + 2)² − 5 [1]. Swap: x = 2(y + 2)² − 5, so (y + 2)² = (x + 5)/2 [1]. With y ≥ −2, f⁻¹(x) = −2 + √((x + 5)/2) [1], domain x ≥ −5 [1]. Examiner insight: leaving ± in the final answer, or omitting the domain, each cost a mark even when the algebra is right.

6. x² − 2x − 8 = (x − 4)(x + 2) [1]. Vertical asymptotes x = 4 and x = −2 [1]. Horizontal asymptote y = 0 [1]. x-intercept (2, 0) [1]. y-intercept −6/−8 gives (0, 3/4) [1]. [5] Examiner insight: asymptotes must be written as equations; “4 and −2” is not accepted as an answer for asymptotes.

7. x³ − 3x² − 4x + 12 ≥ 0 [1]. x²(x − 3) − 4(x − 3) ≥ 0 [1]. (x − 3)(x − 2)(x + 2) ≥ 0 [1]. −2 ≤ x ≤ 2 [1] or x ≥ 3 [1]. [5] Examiner insight: dividing by (x − 3) or (x² − 4) at the start is invalid for an inequality and earns no further method marks.

8. (a) x − 4 = 2x + 1 gives x = −5 [1]. Then 2x + 1 = −9 < 0, so reject [1]. x − 4 = −(2x + 1) gives x = 1 [1]. 2(1) + 1 = 3 > 0, so x = 1 [1]. (b) Square: x² + 6x + 9 > 4(x² − 2x + 1) [1]. 3x² − 14x − 5 < 0 [1]. (3x + 1)(x − 5) < 0 [1]. −1/3 < x < 5 [1]. Examiner insight: in (a), giving x = −5 as well as x = 1 loses the final accuracy mark, because the candidate solution was not checked against the sign of the right-hand side.

9. Graph y = ln(x + 3) and y = |x − 1| for x > −3 [1]. Intersections at x = −0.0737 [1] and x = 2.75 [1]. The logarithm is above the modulus graph between these values [1]. −0.0737 < x < 2.75 [1]. [5] Examiner insight: endpoints must be given to 3 s.f.; writing −0.07 is an accuracy error, and a bare interval with no sketch or intersection values risks losing the method marks.

10. (a) The minimum of f is at x = 1, so 2x + 1 = 1 gives x = 0 [1]. (0, −4) [1]. (b) f(|x|) = 0 when |x| = 3 or |x| = −1 (impossible) [1]. x = 3 and x = −3 [1]. (c) Minimum points (−1, 0) and (3, 0) [1]; local maximum (1, 16) [1]. (d) Vertical asymptotes x = −1 and x = 3 [1]; horizontal asymptote y = 0 [1]; local maximum (1, −1/4) [1]. Examiner insight: in (b), listing x = −1 as well as ±3 shows f(|x|) was confused with |f(x)| and loses the accuracy mark.

11. (a) x² + 3 = (x − 1)(x + 1) + 4 [1], so f(x) = x + 1 + 4/(x − 1): p = 1, q = 4 [1]. (b) x = 1 [1] and y = x + 1 [1]. (c) From the GDC: local maximum (−1, −2) [1], local minimum (3, 6) [1]. Range f(x) ≤ −2 [1] or f(x) ≥ 6 [1]. (d) f(x) − (2x − 1) = (−x² + 3x + 2)/(x − 1) > 0, or graph both sides [1]. Critical values x = −0.562, x = 1 and x = 3.56 [1]. x < −0.562 or 1 < x < 3.56 [1]. Examiner insight: in (d), multiplying through by (x − 1) without considering its sign loses the interval x < −0.562 and the final mark.

12. (a) p(2) = 8 − 4 − 16 + 12 [1] = 0, so (x − 2) is a factor [1]. (b) p(x) = (x − 2)(x² + x − 6) [1] = (x − 2)²(x + 3) [1]. (c) Roots 2, 2, −3. Sum 2 + 2 − 3 = 1 = −(−1)/1 [1]. Product 2 × 2 × (−3) = −12 = (−1)³(12)/1 [1]. (d) Vertical asymptotes x = 2 and x = −3 [1]; horizontal asymptote y = 0 [1]; y-intercept (0, 1/12) [1]. (e) x³ − x² − 8x ≤ 0, so x(x² − x − 8) ≤ 0 [1]. Zeros 0 and (1 ± √33)/2 [1]. x ≤ (1 − √33)/2 or 0 ≤ x ≤ (1 + √33)/2 [1]. Examiner insight: in (c) the repeated root 2 must be counted twice; using roots 2 and −3 only gives a sum of −1 and scores 0.

Where marks are usually lost

• Substituting the wrong sign into the remainder theorem, for example p(1) for the divisor (x + 1).
• Missing the (−1)ⁿ factor in the product of roots.
• Counting a repeated root once when using the sum or product formulas.
• Writing asymptotes as numbers instead of equations, or missing the oblique asymptote.
• Claiming “odd” or “even” from one numerical check.
• Omitting the domain of f⁻¹, or leaving ± in the answer.
• Dividing or multiplying an inequality by an expression of unknown sign.
• Not rejecting modulus solutions that make the right-hand side negative.
• Rounding intersection values to fewer than 3 significant figures.

Next steps

• Review the further functions revision notes.
• Re-read any section of the further functions study guide where you lost marks.
• See the IB DP Maths AA course hub and the printable syllabus checklist.
• Read the exam preparation guide for paper strategy.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

International Baccalaureate Organization, Diploma Programme, Mathematics: analysis and approaches guide, first assessment 2021 (published February 2019, updated November 2020), syllabus sections AHL 2.12–2.16.