HL · 110 marks · 2h · Section A short + Section B long. Each generated paper follows the SL & HL Paper 1 · HL (no calculator) · exams to Nov 2028 format, with the same number and style of questions as the real exam, and comes with an IB-style mark scheme.
Generate Paper 1 · HL (no calculator) · exams to Nov 2028 →The graph of y = 1x is transformed to the graph of y = 1x − 2 + 1.
Describe the transformation.
State the domain and range of the new function.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | translation by 21 translation A1 by 21 A1 | 2 |
| (b) | x ≠ 2, y ≠ 1 domain x ≠ 2 A1 range y ≠ 1 A1 | 2 |
Using the Maclaurin series for standard functions, find the Maclaurin series for exln(1 + x), up to and including the term in x3.
| Part | Answer and marking points | Marks |
|---|---|---|
| x + 12x2 + 13x3 writing down the standard series (to x3) A1 attempt to multiply the series M1 x + 12x2 + 13x3 A3 award A1 for each correct non-zero term | 5 |
Show that 12r − 1 − 12r + 1 = 2(2r − 1)(2r + 1).
Hence find the exact value of Σr=1491(2r − 1)(2r + 1).
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | shown (2r + 1) − (2r − 1)(2r − 1)(2r + 1) M1 = 2(2r − 1)(2r + 1) AG A1 | 2 |
| (b) | 4999 12Σr=1n(12r − 1 − 12r + 1) M1 telescoping: 12(1 − 199) (A1) 4999 A1 | 3 |
Let sin θ = 2029, where π2 < θ < π.
Find the exact value of cos θ.
Find the exact value of sin 2θ.
Find the exact value of tan 2θ.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | −2129 use of sin2θ + cos2θ = 1 (M1) cos θ = −2129 (negative in the second quadrant) A1 | 2 |
| (b) | −840841 2(2029)(−2129) (M1) −840841 A1 | 2 |
| (c) | −84041 cos 2θ = 41841 (M1)(A1) tan 2θ = −84041 A1 | 3 |
Solve the equation 2cos2x + 5sin x + 1 = 0 for 0 ≤ x ≤ 2π.
| Part | Answer and marking points | Marks |
|---|---|---|
| x = 7π6, 11π6 use of cos2x = 1 − sin2x M1 2sin2x − 5sin x − 3 = 0 A1 (2sin x + 1)(sin x − 3) = 0 (M1) sin x = −12 A1 sin x = 3 has no solutions since |sin x| ≤ 1 R1 x = 7π6, 11π6 A2 | 7 |
Find ∫0π2sin2x cos x dx.
| Part | Answer and marking points | Marks |
|---|---|---|
| 13 substitution u = sin x, du = cos x dx (M1) ∫01u2du A1 [u33]01 (A1) 13 A1 | 4 |
All IB Mathematics: Analysis & Approaches papers and topics
Riyaz Shaikh has mentored IGCSE, IB and A Level students for 15+ years, one-on-one, online and in Mumbai. The hardest questions on this site are the ones he uses to find a student's weak points.
Meet Riyaz →Original IB-style practice questions written for mentorriyaz.in. These are not IB past paper questions, and this site is not affiliated with or endorsed by the International Baccalaureate Organization. Updated 2026-11-04. Paper generator · mentorriyaz.in