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IB Maths AI HL · Paper 2

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2

Session: May 2024
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 7

Q1 Q2 Q3 Q4 Q5 Q6 Q7

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2024 TZ2 paper, which your school or the IB can provide.

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Hide our working until you've had a go.

Questions

Questions 1–7 · 110 marks

Question 1

14 marks

Three attractions in an amusement park: distances, a bearing and the fountain

Geometry and trigonometry

(a)

1 mark

$B (12, 20)$ and $C (12, 0)$ are on the same vertical line.

B C = 20 m

(b)

2 marks

Use Pythagoras with $A (0, 16)$ and $B (12, 20)$ .

A B = 1 2^{2} + 4^{2} = 160 \approx 12.6 m

(c)

3 marks

From $B$ to $A$ the walk goes 12 m west and 4 m south. The acute angle with the west direction (below the horizontal) is
West is the bearing $27 0^{\circ}$ , and the direction is rotated towards south, so subtract.

tan^{- 1} \frac{4}{12} \approx 18. 4^{\circ} 270 - 18.4 \approx 25 2^{\circ}

(d)

8 marks

(i) The gradient of $[A C]$ with $A (0, 16)$ and $C (12, 0)$ .

(ii) The midpoint of $[A C]$ .

(iii) The fountain is on the perpendicular bisector of $[A C]$ , which has gradient $\frac{3}{4}$ and passes through $(6, 8)$ .

It is also on the perpendicular bisector of $[B C]$ . Since $[B C]$ is vertical with midpoint $(12, 10)$ , that bisector is the horizontal line $y = 10$ .

The fountain is at

m_{A C} = \frac{0 - 16}{12 - 0} = - \frac{4}{3} (6, 8) y - 8 = \frac{3}{4} (x - 6) 10 - 8 = \frac{3}{4} (x - 6) \Rightarrow x = 6 + \frac{8}{3} = \frac{26}{3} (\frac{26}{3}, 10) \approx (8.67, 10)

Question 2

15 marks

A water slide: the trapezoidal rule, an exact area and the percentage error

Calculus

(a)

5 marks

Evaluate $f (x) = - \frac{1}{4} x^{2} + 2 x$ at $x = 0, 1, 2, 3, 4$ .
Use the trapezoidal rule with width 1.

0, 1.75, 3, 3.75, 4 A \approx \frac{1}{2} [0 + 4 + 2 (1.75 + 3 + 3.75)] = \frac{1}{2} (4 + 17) = 10.5 m^{2}

(b)

3 marks

Integrate term by term.

\int (- \frac{1}{4} x^{2} + 2 x) d x = - \frac{x ^{3}}{12} + x^{2} + c

(c)

4 marks

The area for $0 \leq x \leq 4$ is
For $4 \leq x \leq 5$ the height is constant at 4, a rectangle of area $1 \times 4 = 4$ .
For $5 \leq x \leq 12$ the height falls in a straight line from 4 to 0, a triangle of area $\frac{1}{2} \times 7 \times 4 = 14$ .

[- \frac{x ^{3}}{12} + x^{2}]_{0}^{4} = - \frac{64}{12} + 16 = \frac{32}{3} Area = \frac{32}{3} + 4 + 14 = \frac{86}{3} \approx 28.67 m^{2}

(d)

3 marks

The approximate total area uses 10.5 in place of $\frac{32}{3}$ .

10.5 + 4 + 14 = 28.5 \frac{28.6667 - 28.5}{28.6667} \times 100 \approx 0.581%

Question 3

22 marks

A skip as a trapezoidal prism: volume, a formula and maximum volume

Geometry and trigonometry
Calculus

(a)

9 marks

(i) The top edge is 11 m, so the sloping edge reaches 1 m across, with height $4^{2} - 1^{2} = 15$ .

(ii) The height is 3.2 m, so the sloping edge reaches $4^{2} - 3. 2^{2} = 2.4$ m across and the top is 12.4 m.

(iii) With $θ = \frac{π}{3}$ : height $= 4 sin \frac{π}{3}$ and extra top length $= 4 cos \frac{π}{3} = 2$ .

V = 3 \times \frac{1}{2} (10 + 11) 15 \approx 122 m^{3} V = 3 \times \frac{1}{2} (10 + 12.4) (3.2) \approx 108 m^{3} V = 3 \times \frac{1}{2} (10 + 12) (4 sin \frac{π}{3}) \approx 114 m^{3}

(b)

2 marks

The volume is the cross-section area times the width 3.

V = 3 \times \frac{1}{2} (10 + 10 + 4 cos θ) (4 sin θ) = 6 sin θ (20 + 4 cos θ) V = 24 sin θ (5 + cos θ)

(c)

1 mark

If $θ = 0$ the sloping edge would lie flat along the ground, so the skip would have no height and zero volume, and it would hold nothing.

(d)

4 marks

(i) Sketch: starting at $(0, 0)$ , the graph of $V$ rises to a maximum of about 122 near $θ = 1.38$ and then falls slightly to $V = 120$ at $θ = \frac{π}{2}$ . The horizontal axis runs from 0 to $\frac{π}{2}$ .

(ii) Find the maximum on the calculator.

θ \approx 1.38 (79. 3^{\circ}), V_{m a x} \approx 122 m^{3}

(e)

6 marks

At the maximum $\frac{d V}{d θ} = 0$ . Differentiate with the product rule.
Replace $sin^{2} θ$ by $1 - cos^{2} θ$ .
Divide by 24.

\frac{d V}{d θ} = 24 cos θ (5 + cos θ) + 24 sin θ (- sin θ) = 120 cos θ + 24 cos^{2} θ - 24 sin^{2} θ 120 cos θ + 24 cos^{2} θ - 24 (1 - cos^{2} θ) = 0 48 cos^{2} θ + 120 cos θ - 24 = 0 2 cos^{2} θ + 5 cos θ - 1 = 0

Question 4

12 marks

A route through six towns: nearest neighbour and a lower bound

Geometry and trigonometry

(a)

4 marks

From $A$ the nearest town is $E$ (22). From $E$ the nearest unvisited is $D$ (21). From $D$ it is $C$ (19). From $C$ the nearest unvisited is $F$ (24), then $B$ (25), and finally back to $A$ (31).

A \to E \to D \to C \to F \to B \to A 22 + 21 + 19 + 24 + 25 + 31 = 142 km

(b)

5 marks

Delete $A$ and apply Prim's algorithm from $B$ . Add the shortest edge to the tree each time: $B D$ (20), then $D C$ (19), then $D E$ (21), then $D F$ (22).
Reconnect $A$ with its two cheapest edges, $A E$ (22) and $A F$ (23).

20 + 19 + 21 + 22 = 82 lower bound = 82 + 22 + 23 = 127 km

(c)

3 marks

In the tree plus the two edges at $A$ , the town $D$ has four edges ( $D B$ , $D C$ , $D E$ , $D F$ ) and the towns $B$ and $C$ only have one each. A journey that visits each town once has exactly two edges at every town, so these edges do not make a single cycle.
So the route of length 127 is not achievable, and the true shortest journey is longer than the lower bound.

Question 5

16 marks

A van parked at headquarters or at home: a Markov chain and its eigenvectors

Number and algebra

(a)

2 marks

Order the states (headquarters, home) and put the probabilities from each state in a column.

T = (0.88 0.12 0.08 0.92)

(b)

3 marks

Monday morning the van is at headquarters, so the state is $(1, 0)$ . By Friday evening it has been parked five nights.

T^{5} (10) \approx (0.5966 0.4034) P (home on Friday evening) \approx 0.403

(c)

2 marks

Find $det (T - λ I)$ .

(0.88 - λ) (0.92 - λ) - (0.08) (0.12) = λ^{2} - 1.8 λ + 0.8

(d)

4 marks

The characteristic polynomial factorises as $(λ - 1) (λ - 0.8)$ , so the eigenvalues are 1 and 0.8.
For $λ = 0.8$ , solve $0.88 x + 0.08 y = 0.8 x$ , which gives $y = - x$ .
For $λ = 1$ , solve $0.88 x + 0.08 y = x$ , which gives $0.08 y = 0.12 x$ .

v_{0.8} = (1 - 1) v_{1} = (23)

(e)

2 marks

Put the eigenvectors in the columns of $P$ and the matching eigenvalues on the diagonal of $D$ .

P = (23 1 - 1), D = (10 0 0.8)

(f)

3 marks

As $n \to \infty$ , $D^{n} \to (1000)$ , so $T^{n} \to P (1000) P^{- 1} = \frac{1}{5} (2323)$ .
Every column is $(0.4 0.6)$ , whatever the starting state.

long-term probability of home = 0.6

Question 6

14 marks

Triangle numbers and a pair of discs from a bag

Number and algebra
Statistics and probability

(a)

4 marks

(i) Add the first five integers.

(ii) This is an arithmetic series.

T_{5} = 1 + 2 + 3 + 4 + 5 = 15 T_{k} = \frac{k}{2} (1 + k) = \frac{1}{2} k^{2} + \frac{1}{2} k

(b)

3 marks

(i) $T_{4} = 10$ .

(ii) Add the formulas.

T_{5} + T_{4} = 15 + 10 = 25 T_{k} + T_{k - 1} = \frac{k ( k + 1 )}{2} + \frac{( k - 1 ) k}{2} = \frac{k [ ( k + 1 ) + ( k - 1 ) ]}{2} = k^{2}

(c)

3 marks

There are 25 discs: 15 red and 10 blue. Add the two orders.

P = \frac{15}{25} \times \frac{10}{24} + \frac{10}{25} \times \frac{15}{24} = \frac{300}{600} = \frac{1}{2}

(d)

4 marks

There are $T_{k} + T_{k - 1} = k^{2}$ discs, of which $T_{k}$ are red and $T_{k - 1}$ are blue.
Now $T_{k} T_{k - 1} = \frac{k ( k + 1 )}{2} \times \frac{( k - 1 ) k}{2} = \frac{k ^{2} ( k ^{2} - 1 )}{4}$ .
The answer is $\frac{1}{2}$ for every $k$ , so it does not depend on $k$ .

P = \frac{T _{k}}{k ^{2}} \times \frac{T _{k - 1}}{k ^{2} - 1} + \frac{T _{k - 1}}{k ^{2}} \times \frac{T _{k}}{k ^{2} - 1} = \frac{2 T _{k} T _{k - 1}}{k ^{2} ( k ^{2} - 1 )} P = \frac{2 \cdot \frac{k ^{2} ( k ^{2} - 1 )}{4}}{k ^{2} ( k ^{2} - 1 )} = \frac{1}{2}

Question 7

17 marks

Containment walls: normal slabs and a lead layer, with sums of independent variables

Statistics and probability

(a)

2 marks

Use the normal distribution with mean 350 and standard deviation 10.

P (X < 340) \approx 0.159

(b)

3 marks

The symmetric 95% interval has 2.5% in each tail.

P (X < 330.4) = 0.025, P (X < 369.6) = 0.975 330 mm to 370 mm

(c)

7 marks

(i) Let $W = C_{1} + C_{2} + L$ with independent parts.

(ii) The sum of independent normal variables is normal, so $W \sim N (800, 1 5^{2})$ .

E (W) = 350 + 350 + 100 = 800 Var (W) = 1 0^{2} + 1 0^{2} + 5^{2} = 225, SD (W) = 15 P (780 < W < 810) \approx 0.656

(d)

4 marks

The mean of the total is 810, with the two slabs at 350.
The variance of the total is $1 6^{2} = 256$ .

810 = 350 + 350 + E (L) \Rightarrow E (L) = 110 256 = 100 + 100 + Var (L) \Rightarrow Var (L) = 56 SD (L) = 56 \approx 7.48

(e)

1 mark

Use the inverse normal for $L \sim N (110, 56)$ with area 0.8.

k \approx 116

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Geometry and trigonometry carries the most marks on this paper.

Geometry and trigonometry35 marks · 32%
Q1, Q3, Q4
Calculus28 marks · 25%
Q2, Q3
Statistics and probability24 marks · 22%
Q6, Q7
Number and algebra23 marks · 21%
Q5, Q6

Mentor insights

Traps to watch for

01
Perpendicular bisectors · Q1
A point equally far from two points lies on the perpendicular bisector of the segment joining them.
02
Trapezoidal rule · Q2
With strips of width 1, area $\approx \frac{1}{2} [y_{0} + y_{n} + 2 (middle values)]$ .
03
Cross-section · Q3
The cross-section is a trapezoid with parallel sides 10 and $10 + 4 cos θ$ and height $4 sin θ$ .
04
Transition matrix · Q5
Columns give where the van goes from each state, so each column adds to 1.
05
Sum of independent normals · Q7
Means add, and so do the variances, but standard deviations do not.

Solutions and tips are GeorgeTheTutor's own; the questions are not reproduced here. Not affiliated with or endorsed by the International Baccalaureate Organization.

Want someone to go through your paper with you?

Our maths mentors took these exams recently. One-to-one, they'll find where your marks are going and fix it before May.

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Worked solutions

IB Maths AI HL · Paper 2

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2

Session: May 2024
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 7

Q1 Q2 Q3 Q4 Q5 Q6 Q7

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2024 TZ2 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–7 · 110 marks

Question 1

14 marks

Three attractions in an amusement park: distances, a bearing and the fountain

Geometry and trigonometry

(a)

1 mark

$B (12, 20)$ and $C (12, 0)$ are on the same vertical line.

B C = 20 m

(b)

2 marks

Use Pythagoras with $A (0, 16)$ and $B (12, 20)$ .

A B = 1 2^{2} + 4^{2} = 160 \approx 12.6 m

(c)

3 marks

From $B$ to $A$ the walk goes 12 m west and 4 m south. The acute angle with the west direction (below the horizontal) is
West is the bearing $27 0^{\circ}$ , and the direction is rotated towards south, so subtract.

tan^{- 1} \frac{4}{12} \approx 18. 4^{\circ} 270 - 18.4 \approx 25 2^{\circ}

(d)

8 marks

(i) The gradient of $[A C]$ with $A (0, 16)$ and $C (12, 0)$ .

(ii) The midpoint of $[A C]$ .

(iii) The fountain is on the perpendicular bisector of $[A C]$ , which has gradient $\frac{3}{4}$ and passes through $(6, 8)$ .

It is also on the perpendicular bisector of $[B C]$ . Since $[B C]$ is vertical with midpoint $(12, 10)$ , that bisector is the horizontal line $y = 10$ .

The fountain is at

m_{A C} = \frac{0 - 16}{12 - 0} = - \frac{4}{3} (6, 8) y - 8 = \frac{3}{4} (x - 6) 10 - 8 = \frac{3}{4} (x - 6) \Rightarrow x = 6 + \frac{8}{3} = \frac{26}{3} (\frac{26}{3}, 10) \approx (8.67, 10)

Question 2

15 marks

A water slide: the trapezoidal rule, an exact area and the percentage error

Calculus

(a)

5 marks

Evaluate $f (x) = - \frac{1}{4} x^{2} + 2 x$ at $x = 0, 1, 2, 3, 4$ .
Use the trapezoidal rule with width 1.

0, 1.75, 3, 3.75, 4 A \approx \frac{1}{2} [0 + 4 + 2 (1.75 + 3 + 3.75)] = \frac{1}{2} (4 + 17) = 10.5 m^{2}

(b)

3 marks

Integrate term by term.

\int (- \frac{1}{4} x^{2} + 2 x) d x = - \frac{x ^{3}}{12} + x^{2} + c

(c)

4 marks

The area for $0 \leq x \leq 4$ is
For $4 \leq x \leq 5$ the height is constant at 4, a rectangle of area $1 \times 4 = 4$ .
For $5 \leq x \leq 12$ the height falls in a straight line from 4 to 0, a triangle of area $\frac{1}{2} \times 7 \times 4 = 14$ .

[- \frac{x ^{3}}{12} + x^{2}]_{0}^{4} = - \frac{64}{12} + 16 = \frac{32}{3} Area = \frac{32}{3} + 4 + 14 = \frac{86}{3} \approx 28.67 m^{2}

(d)

3 marks

The approximate total area uses 10.5 in place of $\frac{32}{3}$ .

10.5 + 4 + 14 = 28.5 \frac{28.6667 - 28.5}{28.6667} \times 100 \approx 0.581%

Question 3

22 marks

A skip as a trapezoidal prism: volume, a formula and maximum volume

Geometry and trigonometry
Calculus

(a)

9 marks

(i) The top edge is 11 m, so the sloping edge reaches 1 m across, with height $4^{2} - 1^{2} = 15$ .

(ii) The height is 3.2 m, so the sloping edge reaches $4^{2} - 3. 2^{2} = 2.4$ m across and the top is 12.4 m.

(iii) With $θ = \frac{π}{3}$ : height $= 4 sin \frac{π}{3}$ and extra top length $= 4 cos \frac{π}{3} = 2$ .

V = 3 \times \frac{1}{2} (10 + 11) 15 \approx 122 m^{3} V = 3 \times \frac{1}{2} (10 + 12.4) (3.2) \approx 108 m^{3} V = 3 \times \frac{1}{2} (10 + 12) (4 sin \frac{π}{3}) \approx 114 m^{3}

(b)

2 marks

The volume is the cross-section area times the width 3.

V = 3 \times \frac{1}{2} (10 + 10 + 4 cos θ) (4 sin θ) = 6 sin θ (20 + 4 cos θ) V = 24 sin θ (5 + cos θ)

(c)

1 mark

If $θ = 0$ the sloping edge would lie flat along the ground, so the skip would have no height and zero volume, and it would hold nothing.

(d)

4 marks

(ii) Find the maximum on the calculator.

θ \approx 1.38 (79. 3^{\circ}), V_{m a x} \approx 122 m^{3}

(e)

6 marks

At the maximum $\frac{d V}{d θ} = 0$ . Differentiate with the product rule.
Replace $sin^{2} θ$ by $1 - cos^{2} θ$ .
Divide by 24.

\frac{d V}{d θ} = 24 cos θ (5 + cos θ) + 24 sin θ (- sin θ) = 120 cos θ + 24 cos^{2} θ - 24 sin^{2} θ 120 cos θ + 24 cos^{2} θ - 24 (1 - cos^{2} θ) = 0 48 cos^{2} θ + 120 cos θ - 24 = 0 2 cos^{2} θ + 5 cos θ - 1 = 0

Question 4

12 marks

A route through six towns: nearest neighbour and a lower bound

Geometry and trigonometry

(a)

4 marks

From $A$ the nearest town is $E$ (22). From $E$ the nearest unvisited is $D$ (21). From $D$ it is $C$ (19). From $C$ the nearest unvisited is $F$ (24), then $B$ (25), and finally back to $A$ (31).

A \to E \to D \to C \to F \to B \to A 22 + 21 + 19 + 24 + 25 + 31 = 142 km

(b)

5 marks

Delete $A$ and apply Prim's algorithm from $B$ . Add the shortest edge to the tree each time: $B D$ (20), then $D C$ (19), then $D E$ (21), then $D F$ (22).
Reconnect $A$ with its two cheapest edges, $A E$ (22) and $A F$ (23).

20 + 19 + 21 + 22 = 82 lower bound = 82 + 22 + 23 = 127 km

(c)

3 marks

In the tree plus the two edges at $A$ , the town $D$ has four edges ( $D B$ , $D C$ , $D E$ , $D F$ ) and the towns $B$ and $C$ only have one each. A journey that visits each town once has exactly two edges at every town, so these edges do not make a single cycle.
So the route of length 127 is not achievable, and the true shortest journey is longer than the lower bound.

Question 5

16 marks

A van parked at headquarters or at home: a Markov chain and its eigenvectors

Number and algebra

(a)

2 marks

Order the states (headquarters, home) and put the probabilities from each state in a column.

T = (0.88 0.12 0.08 0.92)

(b)

3 marks

Monday morning the van is at headquarters, so the state is $(1, 0)$ . By Friday evening it has been parked five nights.

T^{5} (10) \approx (0.5966 0.4034) P (home on Friday evening) \approx 0.403

(c)

2 marks

Find $det (T - λ I)$ .

(0.88 - λ) (0.92 - λ) - (0.08) (0.12) = λ^{2} - 1.8 λ + 0.8

(d)

4 marks

The characteristic polynomial factorises as $(λ - 1) (λ - 0.8)$ , so the eigenvalues are 1 and 0.8.
For $λ = 0.8$ , solve $0.88 x + 0.08 y = 0.8 x$ , which gives $y = - x$ .
For $λ = 1$ , solve $0.88 x + 0.08 y = x$ , which gives $0.08 y = 0.12 x$ .

v_{0.8} = (1 - 1) v_{1} = (23)

(e)

2 marks

Put the eigenvectors in the columns of $P$ and the matching eigenvalues on the diagonal of $D$ .

P = (23 1 - 1), D = (10 0 0.8)

(f)

3 marks

As $n \to \infty$ , $D^{n} \to (1000)$ , so $T^{n} \to P (1000) P^{- 1} = \frac{1}{5} (2323)$ .
Every column is $(0.4 0.6)$ , whatever the starting state.

long-term probability of home = 0.6

Question 6

14 marks

Triangle numbers and a pair of discs from a bag

Number and algebra
Statistics and probability

(a)

4 marks

(i) Add the first five integers.

(ii) This is an arithmetic series.

T_{5} = 1 + 2 + 3 + 4 + 5 = 15 T_{k} = \frac{k}{2} (1 + k) = \frac{1}{2} k^{2} + \frac{1}{2} k

(b)

3 marks

(i) $T_{4} = 10$ .

(ii) Add the formulas.

T_{5} + T_{4} = 15 + 10 = 25 T_{k} + T_{k - 1} = \frac{k ( k + 1 )}{2} + \frac{( k - 1 ) k}{2} = \frac{k [ ( k + 1 ) + ( k - 1 ) ]}{2} = k^{2}

(c)

3 marks

There are 25 discs: 15 red and 10 blue. Add the two orders.

P = \frac{15}{25} \times \frac{10}{24} + \frac{10}{25} \times \frac{15}{24} = \frac{300}{600} = \frac{1}{2}

(d)

4 marks

There are $T_{k} + T_{k - 1} = k^{2}$ discs, of which $T_{k}$ are red and $T_{k - 1}$ are blue.
Now $T_{k} T_{k - 1} = \frac{k ( k + 1 )}{2} \times \frac{( k - 1 ) k}{2} = \frac{k ^{2} ( k ^{2} - 1 )}{4}$ .
The answer is $\frac{1}{2}$ for every $k$ , so it does not depend on $k$ .

P = \frac{T _{k}}{k ^{2}} \times \frac{T _{k - 1}}{k ^{2} - 1} + \frac{T _{k - 1}}{k ^{2}} \times \frac{T _{k}}{k ^{2} - 1} = \frac{2 T _{k} T _{k - 1}}{k ^{2} ( k ^{2} - 1 )} P = \frac{2 \cdot \frac{k ^{2} ( k ^{2} - 1 )}{4}}{k ^{2} ( k ^{2} - 1 )} = \frac{1}{2}

Question 7

17 marks

Containment walls: normal slabs and a lead layer, with sums of independent variables

Statistics and probability

(a)

2 marks

Use the normal distribution with mean 350 and standard deviation 10.

P (X < 340) \approx 0.159

(b)

3 marks

The symmetric 95% interval has 2.5% in each tail.

P (X < 330.4) = 0.025, P (X < 369.6) = 0.975 330 mm to 370 mm

(c)

7 marks

(i) Let $W = C_{1} + C_{2} + L$ with independent parts.

(ii) The sum of independent normal variables is normal, so $W \sim N (800, 1 5^{2})$ .

E (W) = 350 + 350 + 100 = 800 Var (W) = 1 0^{2} + 1 0^{2} + 5^{2} = 225, SD (W) = 15 P (780 < W < 810) \approx 0.656

(d)

4 marks

The mean of the total is 810, with the two slabs at 350.
The variance of the total is $1 6^{2} = 256$ .

810 = 350 + 350 + E (L) \Rightarrow E (L) = 110 256 = 100 + 100 + Var (L) \Rightarrow Var (L) = 56 SD (L) = 56 \approx 7.48

(e)

1 mark

Use the inverse normal for $L \sim N (110, 56)$ with area 0.8.

k \approx 116

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Geometry and trigonometry carries the most marks on this paper.

Geometry and trigonometry35 marks · 32%
Q1, Q3, Q4
Calculus28 marks · 25%
Q2, Q3
Statistics and probability24 marks · 22%
Q6, Q7
Number and algebra23 marks · 21%
Q5, Q6

Mentor insights

Traps to watch for

01
Perpendicular bisectors · Q1
A point equally far from two points lies on the perpendicular bisector of the segment joining them.
02
Trapezoidal rule · Q2
With strips of width 1, area $\approx \frac{1}{2} [y_{0} + y_{n} + 2 (middle values)]$ .
03
Cross-section · Q3
The cross-section is a trapezoid with parallel sides 10 and $10 + 4 cos θ$ and height $4 sin θ$ .
04
Transition matrix · Q5
Columns give where the van goes from each state, so each column adds to 1.
05
Sum of independent normals · Q7
Means add, and so do the variances, but standard deviations do not.

Solutions and tips are GeorgeTheTutor's own; the questions are not reproduced here. Not affiliated with or endorsed by the International Baccalaureate Organization.

Want someone to go through your paper with you?

Our maths mentors took these exams recently. One-to-one, they'll find where your marks are going and fix it before May.

Find your maths mentor More past papers