2029 syllabus · SL · 75 marks · 1h 30m · Section A short + Section B long. Each generated paper follows the SL & HL Paper 2 · SL (calculator) · 2029 syllabus format, with the same number and style of questions as the real exam, and comes with an IB-style mark scheme.
Generate Paper 2 · SL (calculator) · 2029 syllabus →A random variable X is normally distributed with mean μ and standard deviation σ. P(X < 49.6) = 0.15 and P(X > 66.7) = 0.25.
Find the value of μ and the value of σ.
| Part | Answer and marking points | Marks |
|---|---|---|
| μ = 60.0, σ = 9.99 attempt to standardise (M1) 49.6 − μσ = −1.036 A1 66.7 − μσ = 0.6745 A1 attempt to solve simultaneously (M1) μ = 60.0 A1 σ = 9.99 A1 | 6 |
Consider the infinite geometric series 1 + (2x + 25) + (2x + 25)2 + (2x + 25)3 + …
Find the set of values of x for which the series has a finite sum.
The sum to infinity of the series is 52. Find the value of x.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | −72 < x < 32 |2x + 25| < 1 (M1) −5 < 2x + 2 < 5 (A1) −72 < x < 32 A1 | 3 |
| (b) | x = 12 11 − r = 52 (M1) r = 35 (A1) 2x + 2 = 3 ⇒ x = 12 A1 | 3 |
Offer A: a starting salary of $32000 with an increase of $1200 each year.
Offer B: a starting salary of $32000 with an increase of 4% each year.
Find the salary in year 5 for each offer.
Find the total amount earned over 15 years under each offer.
Find the first year in which the salary under Offer B is greater than the salary under Offer A.
Find the minimum number of complete years after which the total earned under Offer B exceeds the total earned under Offer A.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | A: $36800; B: $37435 A: 32000 + 4(1200) (M1) $36800 A1 B: 32000(1.04)4 (M1) $37435 A1 | 4 |
| (b) | A: $606000; B: $640755 S15 = 152(2(32000) + 14(1200)) (M1) $606000 A1 S15 = 32000(1.0415 − 1)1.04 − 1 (M1) $640755 A1 | 4 |
| (c) | year 2 32000(1.04)n−1 > 32000 + (n − 1)(1200) (M1) attempt to solve, e.g. table or graph (M1) year 2 A1 | 3 |
| (d) | 2 years equating or comparing the sums (M1) attempt to solve (M1) 2 A1 | 3 |
Consider the system of equations
x + 2z = 3
y − z = 5
x + y + kz = c
where k, c ∈ ℝ.
Find the value of k for which the system does not have a unique solution.
For this value of k, find the value of c for which the system has infinitely many solutions.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | k = 1 attempt to eliminate a variable, or set the determinant equal to 0 M1 row 3 = 1×row 1 + 1×row 2 for x and y (A1) k = 1 A1 | 3 |
| (b) | c = 8 c = (3) + (5) (M1) c = 8 A1 | 2 |
The graph of y = f(x) passes through the point P(0, 0).
The graph of y = f(x) is translated by 21. Find the image of P.
The graph of y = f(x) is transformed to y = 3f(x − 1). Find the image of P.
Describe fully a transformation that maps y = f(x) onto y = f(−x).
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | (2, 1) x-coordinate 2 A1 y-coordinate 1 A1 | 2 |
| (b) | (1, 0) x: 1 A1 y: 0 A1 | 2 |
| (c) | reflection in the y-axis reflection in the y-axis A1 | 1 |
A ladder of length 5 m leans against a vertical wall. The bottom of the ladder slides away from the wall at a constant 0.5 m s−1.
Find the rate at which the top of the ladder is sliding down the wall when the bottom is 3 m from the wall.
| Part | Answer and marking points | Marks |
|---|---|---|
| 0.375 m s−1 x2 + y2 = 25 (A1) 2xdxdt + 2ydydt = 0 M1A1 y = 4 when x = 3 (A1) dydt = −0.375, so sliding down at 0.375 m s−1 A1 | 5 |
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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