Free IB Diploma Programme (International Baccalaureate) Exam Questions and Answers
The IB Diploma Programme, from the International Baccalaureate, is a two-year pre-university qualification for students aged 16 to 19, recognised by universities worldwide. Students take six subjects, three at Higher Level and three at Standard Level, drawn from language and literature, language acquisition, individuals and societies, sciences, mathematics and the arts, plus three core elements: Theory of Knowledge, the Extended Essay and Creativity, Activity, Service. Each subject is graded from 1 to 7 and the core adds up to 3 points, for a maximum of 45; 24 is the minimum for the diploma. Final exams are held in May and November.
If you searched for IB Diploma Programme (International Baccalaureate) dumps, an ExamTopics discussion or a free PDF, this is the question bank for it: practice questions with verified answers and explanations, a timed practice test and updates whenever General changes the exam.
Last updated: October 21, 2026
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Question #1
Prove by mathematical induction that 5²ⁿ − 2³ⁿ is divisible by 17 for all n ∈ ℤ⁺.
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Correct answer: r = √60
Explanation
After removing the sector (angle π/3), remaining angle is 5π/3. Shaded area=½(36)(5π/3)=30π. Unshaded=½r²(5π/3)−30π. Setting ratio to 2/3: (5πr²/6−30π)/30π=2/3 gives r²=60, r=√60.
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Question #2
Find ∫ x(ln x)² dx.
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Correct answer: a = 3/2, b = 9
Explanation
f ''(x) = 6ax + 2b. Inflexion at x = −2: f ''(−2) = 0 → b = 6a ...(1). f(−2) = −2c+d+24 → −8a+4b = 24 → −2a+b = 6 ...(2). Sub (1) into (2): 4a = 6 → a = 3/2, b = 9.
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Question #3
Consider the function f(x) = sin²(kx)/x², where x ≠ 0 and k ∈ ℝ⁺. Given that lim (x→0) f(x) = 16, find the value of k.
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Correct answer: μ = √3
Explanation
θ = π/6, F = 100√3: 100√3 = 300 / (√3/2 + μ/2) = 600 / (√3 + μ). √3 + μ = 600/(100√3) = 2√3 → μ = √3.
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Question #4
The functions f and g are defined by f(x) = ln(2x − 9), where x > 9/2 g(x) = 2 ln x − ln d, where x > 0, d ∈ ℝ⁺. State the equation of the vertical asymptote to the graph of y = g(x).
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Correct answer: a = 3/2, b = 9
Explanation
f″(x)=6ax+2b. Inflection at x=−2: −12a+2b=0 → b=6a (eq.1). y-coord condition: −8a+4b=24 → −2a+b=6 (eq.2). Solving: a=3/2, b=9.
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Question #5
The functions f and g are defined by f(x) = ln(2x − 9), where x > 9/2 g(x) = 2 ln x − ln d, where x > 0, d ∈ ℝ⁺. The graphs of y = f(x) and y = g(x) intersect at two distinct points. Show that, at the points of intersection, x² − 2dx + 9d = 0.
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Correct answer: c = 27/2
Explanation
f′(x)=(9/2)x²+18x+c. Horizontal tangent at x=−3: f′(−3)=81/2−54+c=0, so c=27/2.
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Question #6
The functions f and g are defined by f(x) = ln(2x − 9), where x > 9/2 g(x) = 2 ln x − ln d, where x > 0, d ∈ ℝ⁺. Hence show that d² − 9d > 0.
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Correct answer: k = −1
Explanation
f′(x)=0: (9/2)x²+18x+27/2=0 → x²+4x+3=0 → (x+1)(x+3)=0. So x=−1 or x=−3 (known). Thus k=−1.
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Question #7
The functions f and g are defined by f(x) = ln(2x − 9), where x > 9/2 g(x) = 2 ln x − ln d, where x > 0, d ∈ ℝ⁺. Find the range of possible values of d.
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Correct answer: Local minimum at x=−1
Explanation
f″(−1)=9(−1)+18=9>0, so the graph is concave up at x=−1, confirming a local minimum.
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Question #8
The following diagram shows part of the graphs of y = f(x) and y = g(x). The graphs intersect at x = p and x = q, where p < q. In the case where d = 10, find the value of q - p. Express your answer in the form a√b, where a, b ∈ ℤ⁺.
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Correct answer: Tangent passes through P
Explanation
P=(0,d) is the y-intercept. Computing f(−3)=(3/2)(−27)+81−(81/2)+d=d. The horizontal tangent y=d at x=−3 passes through (0,d)=P.
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Question #9
Consider the function f(x) = e^(cos 2x), where -π/4 ≤ x ≤ 5π/4. Find the coordinates of the points on the curve y = f(x) where the gradient is zero.
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Correct answer: cot2θ ≡ ½(cotθ−1/cotθ)
Explanation
LHS = 1/tan2θ = (1−tan²θ)/(2tanθ) = ½(1/tanθ−tanθ) = ½(cotθ−tanθ) = ½(cotθ−1/cotθ) = RHS.
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Question #10
Consider the function f(x) = e^(cos 2x), where -π/4 ≤ x ≤ 5π/4. Using the second derivative at each point found in part (a), show that the curve y = f(x) has two local maximum points and one local minimum point.
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Correct answer: θ = π/4, π/3, 2π/3, 3π/4
Explanation
Substitute (a): (t²−1)(3t²−1)=0 where t=cotθ, giving t=±1 or t=±1/√3. These yield tanθ=±1 or ±√3, so θ=π/4, π/3, 2π/3, 3π/4.
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Discussion
Explain your reasoning, not just the letterOther General certifications
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FAQ
- Q1: What is the IB Diploma Programme (International Baccalaureate) exam?
- A: IB Diploma Programme (International Baccalaureate) is a General certification exam. Judging by the questions in our bank, it concentrates on graphs, curve, points and functions.
- Q2: What topics does the IB Diploma Programme (International Baccalaureate) exam cover?
- A: Questions in our IB Diploma Programme (International Baccalaureate) bank cluster around graphs, curve, points and functions. Working through the full set is the quickest way to find which of these you are weakest on.
- Q3: How should I prepare for IB Diploma Programme (International Baccalaureate)?
- A: Work through the IB Diploma Programme (International Baccalaureate) practice questions here, checking your answer on each one, then sit the practice test to rehearse the exam under timed conditions before the real thing.
- Q4: Are these real IB Diploma Programme (International Baccalaureate) exam questions?
- A: They are drawn from officially released past questions and from community members who have sat IB Diploma Programme (International Baccalaureate). Answers are verified and updated weekly.
- Q5: Where do I register for the IB Diploma Programme (International Baccalaureate) exam?
- A: Register through General directly. Exampractice is not affiliated with General and does not administer the exam.
- Q6: Is there a free IB Diploma Programme (International Baccalaureate) sample?
- A: Yes. Every IB Diploma Programme (International Baccalaureate) page shows a free sample of real questions. Upgrading opens the full bank and the practice test.