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IB Maths AA HL · Paper 3

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2024
Marks: 55
Duration: 1 hour 15 min
Calculator: Allowed
Questions: 2

Q1 Q2

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Questions

Questions 1–2 · 55 marks

Question 1

27 marks

Boosted actions in a game: geometric and rising-probability models

Statistics and probability

(a)(i)

2 marks

The first two actions are not boosted and the third is.

(0.9) (0.9) (0.1) = 0.081

(a)(ii)

3 marks

Use the complement: no boosts in six actions has probability $0. 9^{6}$ .

1 - 0. 9^{6} \approx 0.469

(b)(i)

1 mark

The first boost is on action $x$ exactly when the first $x - 1$ actions are not boosted, each with probability $1 - p$ , and then action $x$ is boosted, with probability $p$ . The actions are independent, so the probability is $(1 - p)^{x - 1} p$ .

(b)(ii)

1 mark

The expected value is the sum of each value times its probability.

E (X) = x = 1 \sum \infty x p (1 - p)^{x - 1}

(c)(i)

4 marks

Differentiate the left side of $\sum_{n = 0}^{\infty} a r^{n} = \frac{a}{1 - r}$ term by term, with the constant term vanishing.
Differentiate the right side, $a (1 - r)^{- 1}$ .

\frac{d}{d r} (a + a r + a r^{2} + \dots) = a + 2 a r + 3 a r^{2} + \dots = n = 1 \sum \infty na r^{n - 1} \frac{d}{d r} (a (1 - r)^{- 1}) = a (1 - r)^{- 2} n = 1 \sum \infty na r^{n - 1} = \frac{a}{( 1 - r ) ^{2}}

(c)(ii)

2 marks

In $E (X) = \sum x p (1 - p)^{x - 1}$ put $a = p$ , $r = 1 - p$ and $n = x$ .

E (X) = \frac{p}{( 1 - ( 1 - p ) ) ^{2}} = \frac{p}{p ^{2}} = \frac{1}{p}

(d)

2 marks

With $p = 0.1$ .

E (X) = \frac{1}{0.1} = 10, Var (X) = \frac{1 - 0.1}{0. 1 ^{2}} = 90

(e)

2 marks

The chance of a boost is 0.2 on the first action, 0.4 on the second and 0.6 on the third. The first boost on the third action needs two actions not boosted, then a boost.

(1 - 0.2) (1 - 0.4) (0.6) = 0.8 \times 0.6 \times 0.6 = 0.288

(f)

1 mark

After four actions that were not boosted, the chance of a boost on the fifth action is $0.2 + 4 (0.2) = 1$ , so the fifth action is certain to be boosted. The first boost cannot come later than the fifth action.

(g)(i)

2 marks

For $Y = 2$ : not boosted on the first, then boosted with chance 0.4.
For $Y = 4$ : not boosted three times, then boosted with chance 0.8.

m = 0.8 \times 0.4 = 0.32 n = 0.8 \times 0.6 \times 0.4 \times 0.8 = 0.1536

(g)(ii)

2 marks

Sum $y P (Y = y)$ .

E (Y) = 1 (0.2) + 2 (0.32) + 3 (0.288) + 4 (0.1536) + 5 (0.0384) = 0.2 + 0.64 + 0.864 + 0.6144 + 0.192 = 2.5104

(g)(iii)

2 marks

Use $Var (Y) = E (Y^{2}) - [E (Y)]^{2}$ .

E (Y^{2}) = 0.2 + 4 (0.32) + 9 (0.288) + 16 (0.1536) + 25 (0.0384) = 7.4896 Var (Y) = 7.4896 - 2.510 4^{2} \approx 1.19

(h)(i)

1 mark

Set $E (X) = \frac{1}{p} = 2.5104$ .

p = \frac{1}{2.5104} \approx 0.398

(h)(ii)

1 mark

Use the variance formula.

Var (X) = \frac{1 - p}{p ^{2}} \approx 3.79

(h)(iii)

1 mark

The two models have the same mean, but $Var (Y) \approx 1.19 < 3.79 \approx Var (X)$ . The second model has the smaller spread, so it gives a more consistent experience.

Question 2

28 marks

The line through the points of inflexion of a quartic and a ratio of lengths

Calculus

(a)

3 marks

Differentiate twice.

\frac{d y}{d x} = 4 x^{3} - 9 x^{2} + 3, \frac{d ^{2} y}{d x ^{2}} = 12 x^{2} - 18 x

(b)

4 marks

Set $y^{''} = 0$ .
Find the $y$ -values.

6 x (2 x - 3) = 0 \Rightarrow x = 0 or x = \frac{3}{2} y (0) = 0, y (\frac{3}{2}) = \frac{81}{16} - \frac{81}{8} + \frac{9}{2} = - \frac{9}{16} B (0, 0), C (\frac{3}{2}, - \frac{9}{16})

(c)

2 marks

The line passes through the origin, so $y = m x$ with the gradient of $B C$ .

m = \frac{- 9/16}{3/2} = - \frac{3}{8} = - 0.375 y = - 0.375 x

(d)

2 marks

Set $x^{4} - 3 x^{3} + 3 x = - 0.375 x$ and divide by $x$ .
$x = 1.5$ is a root, so factorise: $(x - 1.5) (x^{2} - 1.5 x - 2.25) = 0$ . The quadratic gives $x = \frac{1.5 \pm 11.25}{2}$ .

x^{3} - 3 x^{2} + 3.375 = 0 x_{D} = \frac{3 + 3 5}{4} \approx 2.427

(e)

3 marks

For $y = x^{4} - m x^{3} + n x$ :

y^{'} = 4 x^{3} - 3 m x^{2} + n, y^{''} = 12 x^{2} - 6 m x = 6 x (2 x - m) x = 0 and x = \frac{m}{2}

(f)(i)

1 mark

At $x = 0$ , $y = 0$ .

B = (0, 0)

(f)(ii)

2 marks

Substitute $x = \frac{m}{2}$ .

y = \frac{m ^{4}}{16} - \frac{m ^{4}}{8} + \frac{nm}{2} = \frac{nm}{2} - \frac{m ^{4}}{16} C = (\frac{m}{2}, \frac{nm}{2} - \frac{m ^{4}}{16})

(g)

2 marks

The line passes through the origin B, so its gradient is the $y$ -coordinate of C divided by its $x$ -coordinate.

gradient = \frac{\frac{nm}{2} - \frac{m ^{4}}{16}}{\frac{m}{2}} = n - \frac{m ^{3}}{8} y = (n - \frac{m ^{3}}{8}) x

(h)

7 marks

Set the curve equal to the line and bring everything to one side.
Factor out $x$ (the root at B).
The point C is on both curves, so $x = \frac{m}{2}$ is a root of the cubic. Divide it out.
The remaining roots come from the quadratic formula.
A is on the left of B, so it takes the smaller root.

x^{4} - m x^{3} + n x = (n - \frac{m ^{3}}{8}) x \Rightarrow x^{4} - m x^{3} + \frac{m ^{3}}{8} x = 0 x (x^{3} - m x^{2} + \frac{m ^{3}}{8}) = 0 x^{3} - m x^{2} + \frac{m ^{3}}{8} = (x - \frac{m}{2}) (x^{2} - \frac{m}{2} x - \frac{m ^{2}}{4}) x = \frac{\frac{m}{2} \pm \frac{m ^{2}}{4} + m ^{2}}{2} = \frac{m ( 1 \pm 5 )}{4} x_{A} = \frac{m ( 1 - 5 )}{4}

(i)

2 marks

Use $x_{B} = 0$ , $x_{C} = \frac{m}{2}$ and $x_{A} = \frac{m ( 1 - 5 )}{4}$ .

\frac{x _{B} - x _{A}}{x _{C} - x _{B}} = \frac{0 - \frac{m ( 1 - 5 )}{4}}{\frac{m}{2}} = \frac{5 - 1}{2}

Our analysis

Where the marks are

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

Calculus28 marks · 51%
Q2
Statistics and probability27 marks · 49%
Q1

Mentor insights

Traps to watch for

01
First success on trial x · Q1
The first boost on action $x$ means $x - 1$ failures and then a success: $p (1 - p)^{x - 1}$ .
02
Differentiate a series · Q1
Differentiate $\sum a r^{n}$ term by term to get $\sum na r^{n - 1}$ , and differentiate the closed form too.
03
Probability that changes · Q1
In the second model the chance of a boost goes up 0.2 after each action that is not boosted.
04
Inflexion · Q2
Solve $y^{''} = 0$ and check that the sign of $y^{''}$ changes.
05
Intersections of a line and the curve · Q2
Set the line equal to the curve. The inflexion point C is a repeated root, so factor it out.

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

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Worked solutions

IB Maths AA HL · Paper 3

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2024
Marks: 55
Duration: 1 hour 15 min
Calculator: Allowed
Questions: 2

Q1 Q2

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2024 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–2 · 55 marks

Question 1

27 marks

Boosted actions in a game: geometric and rising-probability models

Statistics and probability

(a)(i)

2 marks

The first two actions are not boosted and the third is.

(0.9) (0.9) (0.1) = 0.081

(a)(ii)

3 marks

Use the complement: no boosts in six actions has probability $0. 9^{6}$ .

1 - 0. 9^{6} \approx 0.469

(b)(i)

1 mark

(b)(ii)

1 mark

The expected value is the sum of each value times its probability.

E (X) = x = 1 \sum \infty x p (1 - p)^{x - 1}

(c)(i)

4 marks

Differentiate the left side of $\sum_{n = 0}^{\infty} a r^{n} = \frac{a}{1 - r}$ term by term, with the constant term vanishing.
Differentiate the right side, $a (1 - r)^{- 1}$ .

\frac{d}{d r} (a + a r + a r^{2} + \dots) = a + 2 a r + 3 a r^{2} + \dots = n = 1 \sum \infty na r^{n - 1} \frac{d}{d r} (a (1 - r)^{- 1}) = a (1 - r)^{- 2} n = 1 \sum \infty na r^{n - 1} = \frac{a}{( 1 - r ) ^{2}}

(c)(ii)

2 marks

In $E (X) = \sum x p (1 - p)^{x - 1}$ put $a = p$ , $r = 1 - p$ and $n = x$ .

E (X) = \frac{p}{( 1 - ( 1 - p ) ) ^{2}} = \frac{p}{p ^{2}} = \frac{1}{p}

(d)

2 marks

With $p = 0.1$ .

E (X) = \frac{1}{0.1} = 10, Var (X) = \frac{1 - 0.1}{0. 1 ^{2}} = 90

(e)

2 marks

The chance of a boost is 0.2 on the first action, 0.4 on the second and 0.6 on the third. The first boost on the third action needs two actions not boosted, then a boost.

(1 - 0.2) (1 - 0.4) (0.6) = 0.8 \times 0.6 \times 0.6 = 0.288

(f)

1 mark

(g)(i)

2 marks

For $Y = 2$ : not boosted on the first, then boosted with chance 0.4.
For $Y = 4$ : not boosted three times, then boosted with chance 0.8.

m = 0.8 \times 0.4 = 0.32 n = 0.8 \times 0.6 \times 0.4 \times 0.8 = 0.1536

(g)(ii)

2 marks

Sum $y P (Y = y)$ .

E (Y) = 1 (0.2) + 2 (0.32) + 3 (0.288) + 4 (0.1536) + 5 (0.0384) = 0.2 + 0.64 + 0.864 + 0.6144 + 0.192 = 2.5104

(g)(iii)

2 marks

Use $Var (Y) = E (Y^{2}) - [E (Y)]^{2}$ .

E (Y^{2}) = 0.2 + 4 (0.32) + 9 (0.288) + 16 (0.1536) + 25 (0.0384) = 7.4896 Var (Y) = 7.4896 - 2.510 4^{2} \approx 1.19

(h)(i)

1 mark

Set $E (X) = \frac{1}{p} = 2.5104$ .

p = \frac{1}{2.5104} \approx 0.398

(h)(ii)

1 mark

Use the variance formula.

Var (X) = \frac{1 - p}{p ^{2}} \approx 3.79

(h)(iii)

1 mark

The two models have the same mean, but $Var (Y) \approx 1.19 < 3.79 \approx Var (X)$ . The second model has the smaller spread, so it gives a more consistent experience.

Question 2

28 marks

The line through the points of inflexion of a quartic and a ratio of lengths

Calculus

(a)

3 marks

Differentiate twice.

\frac{d y}{d x} = 4 x^{3} - 9 x^{2} + 3, \frac{d ^{2} y}{d x ^{2}} = 12 x^{2} - 18 x

(b)

4 marks

Set $y^{''} = 0$ .
Find the $y$ -values.

6 x (2 x - 3) = 0 \Rightarrow x = 0 or x = \frac{3}{2} y (0) = 0, y (\frac{3}{2}) = \frac{81}{16} - \frac{81}{8} + \frac{9}{2} = - \frac{9}{16} B (0, 0), C (\frac{3}{2}, - \frac{9}{16})

(c)

2 marks

The line passes through the origin, so $y = m x$ with the gradient of $B C$ .

m = \frac{- 9/16}{3/2} = - \frac{3}{8} = - 0.375 y = - 0.375 x

(d)

2 marks

Set $x^{4} - 3 x^{3} + 3 x = - 0.375 x$ and divide by $x$ .
$x = 1.5$ is a root, so factorise: $(x - 1.5) (x^{2} - 1.5 x - 2.25) = 0$ . The quadratic gives $x = \frac{1.5 \pm 11.25}{2}$ .

x^{3} - 3 x^{2} + 3.375 = 0 x_{D} = \frac{3 + 3 5}{4} \approx 2.427

(e)

3 marks

For $y = x^{4} - m x^{3} + n x$ :

y^{'} = 4 x^{3} - 3 m x^{2} + n, y^{''} = 12 x^{2} - 6 m x = 6 x (2 x - m) x = 0 and x = \frac{m}{2}

(f)(i)

1 mark

At $x = 0$ , $y = 0$ .

B = (0, 0)

(f)(ii)

2 marks

Substitute $x = \frac{m}{2}$ .

y = \frac{m ^{4}}{16} - \frac{m ^{4}}{8} + \frac{nm}{2} = \frac{nm}{2} - \frac{m ^{4}}{16} C = (\frac{m}{2}, \frac{nm}{2} - \frac{m ^{4}}{16})

(g)

2 marks

The line passes through the origin B, so its gradient is the $y$ -coordinate of C divided by its $x$ -coordinate.

gradient = \frac{\frac{nm}{2} - \frac{m ^{4}}{16}}{\frac{m}{2}} = n - \frac{m ^{3}}{8} y = (n - \frac{m ^{3}}{8}) x

(h)

7 marks

Set the curve equal to the line and bring everything to one side.
Factor out $x$ (the root at B).
The point C is on both curves, so $x = \frac{m}{2}$ is a root of the cubic. Divide it out.
The remaining roots come from the quadratic formula.
A is on the left of B, so it takes the smaller root.

x^{4} - m x^{3} + n x = (n - \frac{m ^{3}}{8}) x \Rightarrow x^{4} - m x^{3} + \frac{m ^{3}}{8} x = 0 x (x^{3} - m x^{2} + \frac{m ^{3}}{8}) = 0 x^{3} - m x^{2} + \frac{m ^{3}}{8} = (x - \frac{m}{2}) (x^{2} - \frac{m}{2} x - \frac{m ^{2}}{4}) x = \frac{\frac{m}{2} \pm \frac{m ^{2}}{4} + m ^{2}}{2} = \frac{m ( 1 \pm 5 )}{4} x_{A} = \frac{m ( 1 - 5 )}{4}

(i)

2 marks

Use $x_{B} = 0$ , $x_{C} = \frac{m}{2}$ and $x_{A} = \frac{m ( 1 - 5 )}{4}$ .

\frac{x _{B} - x _{A}}{x _{C} - x _{B}} = \frac{0 - \frac{m ( 1 - 5 )}{4}}{\frac{m}{2}} = \frac{5 - 1}{2}

Our analysis

Where the marks are

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

Calculus28 marks · 51%
Q2
Statistics and probability27 marks · 49%
Q1

Mentor insights

Traps to watch for

01
First success on trial x · Q1
The first boost on action $x$ means $x - 1$ failures and then a success: $p (1 - p)^{x - 1}$ .
02
Differentiate a series · Q1
Differentiate $\sum a r^{n}$ term by term to get $\sum na r^{n - 1}$ , and differentiate the closed form too.
03
Probability that changes · Q1
In the second model the chance of a boost goes up 0.2 after each action that is not boosted.
04
Inflexion · Q2
Solve $y^{''} = 0$ and check that the sign of $y^{''}$ changes.
05
Intersections of a line and the curve · Q2
Set the line equal to the curve. The inflexion point C is a repeated root, so factor it out.

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