IB DP Mathematics: Analysis and Approaches – Limits, further differentiation and integration, differential equations and Maclaurin series (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 calculus 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 5.12–5.19, 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 unless the question says otherwise.

Learn the methods first in the further calculus study guide and the revision notes. The IB DP Maths AA course hub and the printable syllabus checklist show where this unit sits.

Questions

1. (calculator-free) Let f(x) = x³ − 2x. Find f′(x) from first principles. [4]

2. (calculator-free)

(a) Use l’Hôpital’s rule to find lim (x→0) (x − sin x)/x³. [3] (b) Use Maclaurin series to find lim (x→0) (e²ˣ − 1 − 2x)/(1 − cos x). [3]

3. (calculator-free) The curve C has equation x²y + y³ = 10. Find the equation of the normal to C at the point (1, 2), in the form ax + by + c = 0 with a, b, c ∈ ℤ. [5]

4. (calculator allowed) A container is an inverted right circular cone with radius 30 cm and height 60 cm. Water flows in at 200 cm³ s⁻¹. Find the rate at which the depth of water is increasing when the depth is 20 cm. [5]

5. (calculator-free)

(a) Find dy/dx when y = log₂ x + sec 3x. [2] (b) Find ∫ 1/(x² − 4x + 8) dx. [3]

6. (calculator-free) Show that ∫₂⁴ (5x + 1)/((x − 1)(x + 2)) dx = ln(243/8). [6]

7. (calculator-free)

(a) Find ∫ x² e⁻ˣ dx. [4] (b) Use the substitution u = √x to find ∫₁⁴ 1/(x + √x) dx, giving your answer in the form ln k. [4]

8. (calculator-free) Use the substitution y = vx to solve dy/dx = (y² + 2xy)/x², for 0 < x < 2, given that y = 1 when x = 1. Give y in terms of x. [6]

9. (calculator allowed) Region R is enclosed by the curve y = ln x, the y-axis, the x-axis and the line y = 2.

(a) Find the area of R. [2] (b) R is rotated through 2π about the y-axis. Find the volume generated. [3] (c) Region S is enclosed by y = ln x, the x-axis and the line x = e. S is rotated through 2π about the x-axis. Find the volume generated. [3]

10. (calculator allowed) A curve satisfies dy/dx = y/x + x for x ≥ 1, and y = 2 when x = 1.

(a) Use Euler’s method with h = 0.25 to estimate y when x = 1.5. [3] (b) Use an integrating factor to find y in terms of x. [5] (c) Find the percentage error in your estimate from (a). [2]

11. (calculator allowed) A population N, in thousands, satisfies dN/dt = 0.002N(300 − N), where t is in years, and N = 60 when t = 0.

(a) Show that N = 300/(1 + 4e^(−0.6t)). [6] (b) Find the time when N = 240. [2] (c) Find the value of N at which the population grows fastest, and that greatest rate. [2]

12. (calculator-free) The function y = f(x) satisfies dy/dx = x² + y², with y = 1 when x = 0.

(a) Find the Maclaurin series for y up to and including the term in x³. [5] (b) Use your series to estimate y when x = 0.1, to 4 significant figures. [1]

Answers

1. f′(x) = lim (h→0) [(x + h)³ − 2(x + h) − x³ + 2x]/h [1] = lim (3x²h + 3xh² + h³ − 2h)/h [1] = lim (3x² + 3xh + h² − 2) [1] = 3x² − 2 [1] Examiner insight: The limit notation must appear until h is removed; writing “= 3x² + 3xh + h² − 2 = 3x² − 2” without “lim” can lose the final mark.

2. (a) Form 0/0; l’Hôpital gives (1 − cos x)/(3x²), still 0/0 [1]. Again: (sin x)/(6x), still 0/0 [1]. Again: (cos x)/6 → 1/6 [1] (b) e²ˣ − 1 − 2x = 2x² + (4/3)x³ + … [1]; 1 − cos x = x²/2 − … [1]. Divide by x²: limit = 2 ÷ (1/2) = 4 [1] Examiner insight: When a part says “use Maclaurin series”, answering by l’Hôpital earns no marks even with the right value.

3. 2xy + x² dy/dx [1] + 3y² dy/dx = 0 [1]. At (1, 2): 4 + 13 dy/dx = 0, dy/dx = −4/13 [1]. Normal gradient 13/4 [1]; y − 2 = (13/4)(x − 1), so 13x − 4y − 5 = 0 [1] Examiner insight: The product rule on x²y is a separate method mark; differentiating it as 2x dy/dx loses it and every later accuracy mark.

4. By similar triangles r = h/2 [1], so V = (1/3)π(h/2)²h = πh³/12 [1]. dV/dt = (πh²/4) dh/dt [1]. At h = 20: 200 = 100π dh/dt [1], dh/dt = 2/π ≈ 0.637 cm s⁻¹ [1] Examiner insight: Substituting h = 20 before differentiating makes V a constant and scores only the first two marks.

5. (a) dy/dx = 1/(x ln 2) [1] + 3 sec 3x tan 3x [1] (b) x² − 4x + 8 = (x − 2)² + 4 [1]; use ∫ 1/(u² + a²) du = (1/a) arctan(u/a) [1]; (1/2) arctan((x − 2)/2) + C [1] Examiner insight: Omitting “+ C” on an indefinite integral usually costs the final accuracy mark.

6. (5x + 1)/((x − 1)(x + 2)) = A/(x − 1) + B/(x + 2): A = 6/3 = 2 [1], B = (−9)/(−3) = 3 [1]. Integral = 2 ln|x − 1| [1] + 3 ln|x + 2| [1]. Limits: (2 ln 3 + 3 ln 6) − (0 + 3 ln 4) [1] = ln 9 + ln(27/8) = ln(243/8) [1] Examiner insight: In a “show that”, the line combining logarithms must be written; jumping from the substituted limits to ln(243/8) loses the last mark.

7. (a) u = x², v = −e⁻ˣ: −x²e⁻ˣ + ∫ 2x e⁻ˣ dx [1][1]. Parts again: ∫ 2x e⁻ˣ dx = −2x e⁻ˣ − 2e⁻ˣ [1]. Result −e⁻ˣ(x² + 2x + 2) + C [1] (b) x = u², dx = 2u du [1]; limits u = 1 to u = 2 [1]; ∫₁² 2u/(u² + u) du = ∫₁² 2/(u + 1) du [1] = 2 ln(3/2) = ln(9/4) [1] Examiner insight: Keeping the x-limits 1 and 4 after substituting is a method error; the final accuracy mark is then lost.

8. v + x dv/dx = v² + 2v [1], so ∫ 1/(v(v + 1)) dv = ∫ 1/x dx [1]. Partial fractions: 1/v − 1/(v + 1) [1]. ln|v/(v + 1)| = ln|x| + c [1]. x = 1, v = 1: c = ln(1/2), so v/(v + 1) = x/2 [1]. v = x/(2 − x), y = x²/(2 − x) [1] Examiner insight: Leaving the answer in terms of v loses the final mark; the question asks for y in terms of x.

9. (a) x = e^y, area = ∫₀² e^y dy [1] = e² − 1 ≈ 6.39 [1] (b) V = π ∫₀² x² dy [1] = π ∫₀² e^(2y) dy [1] = (π/2)(e⁴ − 1) ≈ 84.2 [1] (c) V = π ∫ (ln x)² dx [1] with limits 1 and e [1] ≈ 2.26 [1] Examiner insight: A GDC value with no integral written earns no method mark if it is wrong; always write the integral you enter.

10. (a) f(x, y) = y/x + x. y₁ = 2 + 0.25(3) = 2.75 [1]. y₂ = 2.75 + 0.25(2.75/1.25 + 1.25) [1] = 3.6125 ≈ 3.61 [1] (b) y′ − (1/x)y = x [1]; I = e^(−ln x) = 1/x [1]; d/dx(y/x) = 1 [1]; y/x = x + C [1]; y(1) = 2 gives C = 1, y = x² + x [1] (c) Exact y(1.5) = 3.75 [1]; error = (3.75 − 3.6125)/3.75 × 100 ≈ 3.67% [1] Examiner insight: In (c) the percentage error divides by the exact value 3.75; dividing by the Euler estimate gives 3.81% and loses the accuracy mark.

11. (a) ∫ 1/(N(300 − N)) dN = ∫ 0.002 dt [1]. 1/(N(300 − N)) = (1/300)(1/N + 1/(300 − N)) [1]. (1/300)[ln N − ln(300 − N)] [1] = 0.002t + c [1]. So ln(N/(300 − N)) = 0.6t + C; t = 0 gives C = ln(60/240) = −ln 4 [1]. N/(300 − N) = e^(0.6t)/4, so N = 300e^(0.6t)/(4 + e^(0.6t)) = 300/(1 + 4e^(−0.6t)) [1] (b) 1 + 4e^(−0.6t) = 1.25, e^(−0.6t) = 1/16 [1]; t = ln 16/0.6 ≈ 4.62 years [1] (c) d/dN[0.002N(300 − N)] = 0.002(300 − 2N) = 0, so N = 150 [1]; rate = 0.002 × 150 × 150 = 45 thousand per year [1] Examiner insight: The partial-fraction split earns its own method mark; integrating 1/(N(300 − N)) as a single logarithm loses it, and the “show that” cannot then be completed.

12. (a) y′(0) = 0 + 1 = 1 [1]. y″ = 2x + 2y y′ [1], so y″(0) = 2 [1]. y‴ = 2 + 2(y′)² + 2y y″, so y‴(0) = 8 [1]. y ≈ 1 + x + (2/2!)x² + (8/3!)x³ = 1 + x + x² + (4/3)x³ [1] (b) 1 + 0.1 + 0.01 + 0.001333… ≈ 1.111 [1] Examiner insight: Forgetting to divide by 2! and 3! is a common way to lose the final accuracy mark in (a).

Where marks are usually lost

• First principles with the “lim” dropped, or h set to 0 before dividing (question 1).
• l’Hôpital applied without checking the form is still 0/0 each time (question 2).
• Product rule missed on x²y in implicit differentiation (question 3).
• Related rates with values substituted too early (question 4).
• Arctan integrals without the 1/a factor or the completed square (question 5).
• Logarithms combined in one step in a “show that” (question 6).
• Old limits kept after a substitution (question 7).
• DE answers left in v, or the constant found before rearranging correctly (questions 8 and 11).
• Volumes about the y-axis using x-limits (question 9).
• Euler values rounded mid-table (question 10).

Next steps

• Recap the key facts in the further calculus revision notes.
• Re-learn any weak topic in the further calculus study guide.
• See the rest of the course on the IB DP Maths AA course hub and tick off the printable syllabus checklist.
• Plan your final weeks with the AA exam preparation guide.
• 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).