IB · Mathematics: Analysis & Approaches SL & HL · N1S

IB Mathematics: Analysis & Approaches Paper 1 · SL (no calculator) · 2029 syllabus practice papers

2029 syllabus · SL · 75 marks · 1h 30m · Section A short + Section B long. Each generated paper follows the SL & HL Paper 1 · SL (no calculator) · 2029 syllabus format, with the same number and style of questions as the real exam, and comes with an IB-style mark scheme.

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Sample Paper 1 · SL (no calculator) · 2029 syllabus questions

Question 1Structured · 6 marksA* challenge

Solve the equation 2log4(x + 4) − log2(x − 5) = 1.

[6]
Show answers and mark scheme
PartAnswer and marking pointsMarks
x = 14
change of base: log4(x + 4) = log2(x + 4)log24 M1
= 12log2(x + 4), so the equation becomes log2(x + 4) − log2(x − 5) = 1 A1
log2x + 4x − 5 = 1 (M1)
x + 4 = 2(x − 5) A1
x = 14 A1
check: x − 5 = 9 > 0, so the solution is valid R1
6
Question 2Structured · 5 marksA* challenge

The line y = kx − 4 is a tangent to the curve y = x2 − x + 5.

Find the possible values of k.

[5]
Show answers and mark scheme
PartAnswer and marking pointsMarks
k = −7 or k = 5
x2 − x + 5 = kx − 4 M1
x2 + (−1 − k)x + 9 = 0 A1
discriminant = 0 M1
(−1 − k)2 = 36 (A1)
k = −7 or k = 5 A1
5
Question 3Structured · 7 marks

The points A, B and C have coordinates A(2, 1, 2), B(3, −1, 1) and C(4, 3, −1).

(a)

Find AB × AC.

[3]
(b)

Hence find the Cartesian equation of the plane containing A, B and C.

[2]
(c)

Find the exact area of triangle ABC.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)816
AB = 1−2−1, AC = 22−3 A1
attempt at the cross product (M1)
816 A1
3
(b)8x + y + 6z = 29
using the normal and a point, r·n = a·n (M1)
8x + y + 6z = 29 A1
2
(c)√1012
12|AB × AC| (M1)
12√101 = √1012 A1
2
Question 4Structured · 6 marks

Events A and B are such that P(A) = 0.6, P(B) = 0.5 and P(A ∪ B) = 0.72.

(a)

Find P(A ∩ B).

[2]
(b)

Find P(A | B).

[2]
(c)

Determine whether A and B are independent. Justify your answer.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)0.38
0.6 + 0.5 − P(A ∩ B) = 0.72 (M1)
0.38 A1
2
(b)0.76
0.380.5 (M1)
0.76 A1
2
(c)not independent
P(A)P(B) = 0.3 ≠ P(A ∩ B) R1
not independent A1
2
Question 5Structured · 5 marks

Solve dydx = xy, given that y = 4 when x = 1. Give y in terms of x.

[5]
Show answers and mark scheme
PartAnswer and marking pointsMarks
y = √x2 + 15
∫y dy = ∫x dx M1
y22 = x22 + c A1A1
substituting (1, 4) M1
y = √x2 + 15 (positive root since y > 0) A1
5
Question 6Structured · 5 marks

Solve the inequality |x + 2| > |3x − 5|.

[5]
Show answers and mark scheme
PartAnswer and marking pointsMarks
34 < x < 72
squaring both sides, or considering cases M1
(x + 2)2 − (3x − 5)2 > 0 (A1)
critical values x = 34, x = 72 A1A1
34 < x < 72 A1
5

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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