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Worked solutions

IB Maths AI HL · Paper 3

Mathematics: Applications & Interpretation, Higher Level

Session: November 2023
Marks: 55
Duration: 1 hour 15 min
Calculator: Allowed
Questions: 2

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2023 Global paper, which your school or the IB can provide.

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Questions

Questions 1–2 · 55 marks

Question 1

28 marks

A skydiver in free fall: differential equations with and without air resistance

Calculus

(a)(i)

1 mark

Use the chain rule with $v = \frac{d s}{d t}$ .

\frac{d v}{d t} = \frac{d v}{d s} \frac{d s}{d t} = v \frac{d v}{d s}

(a)(ii)

4 marks

Separate the variables in $v \frac{d v}{d s} = g$ and integrate.
Use $v = 10$ when $s = 0$ .

\int v d v = \int g d s \Rightarrow \frac{v ^{2}}{2} = g s + c c = 50 \Rightarrow v = 2 g s + 100

(a)(iii)

3 marks

Find $s$ when $v = 330$ .
Since $5551 < 40000$ , the model predicts that Felix reaches the speed of sound before $s = 40000$ m.

33 0^{2} = 2 (9.8) s + 100 \Rightarrow s = 5551

(b)(i)

2 marks

With $\frac{d v}{d t} = g$ constant, $v = g t + c$ .
The graph of $v$ against $t$ should be a straight line with gradient $g$ .

(b)(ii)

1 mark

The plotted data are not a straight line (only roughly linear for small $t$ ), so the model does not appear to be valid.

(c)(i)

5 marks

Use $v \frac{d v}{d s} = g - k v^{2}$ and separate the variables.
Rearrange for $v$ .
Use $v = 10$ at $s = 0$ : $A = g - 100 k$ .

\int \frac{v}{g - k v ^{2}} d v = \int d s \Rightarrow - \frac{1}{2 k} ln (g - k v^{2}) = s + c g - k v^{2} = A e^{- 2 k s} v = \frac{g - ( g - 100 k ) e ^{- 2 k s}}{k}

(c)(ii)

2 marks

With air resistance the acceleration is $g - k v^{2}$ . At $v = 40$ the gradient of the $v$ - $t$ graph is 9.672.

9.672 = 9.8 - 1600 k \Rightarrow k = \frac{9.8 - 9.672}{1600} = 8 \times 1 0^{- 5}

(c)(iii)

2 marks

As $s \to \infty$ , $e^{- 2 k s} \to 0$ .

v \to \frac{g}{k} = \frac{9.8}{8 \times 1 0 ^{- 5}} = 350 m s^{- 1}

(c)(iv)

2 marks

The velocity increases with $s$ , so the upper bound on $0 < s \leq 40000$ is at $s = 40000$ .

v (40000) = \frac{9.8 - ( 9.8 - 0.008 ) e ^{- 6.4}}{8 \times 1 0 ^{- 5}} \approx 349.7 m s^{- 1}

(d)

4 marks

Write the new model as $\frac{d v}{d s} = \frac{1}{v} (\frac{3.98 \times 1 0 ^{14}}{( 6.41 \times 1 0 ^{6} - s ) ^{2}} - 8 \times 1 0^{- 5} v^{2})$ . Euler's method with step 4000 in $s$ :
Ten steps from $v_{0} = 10$ at $s_{0} = 0$ give

s_{n + 1} = s_{n} + 4000, v_{n + 1} = v_{n} + 4000 (\frac{3.98 \times 1 0 ^{14}}{v _{n} ( 6.41 \times 1 0 ^{6} - s _{n} ) ^{2}} - 8 \times 1 0^{- 5} v_{n}) v \approx 361 m s^{- 1}

(e)(i)

1 mark

Use a smaller step length (or a better method such as Runge-Kutta).

(e)(ii)

1 mark

The model ignores the parachute, wind and air currents, and treats Felix as a point object falling straight down. Any one of these could cause a difference from the data.

Question 2

27 marks

Information theory: logarithms, expected information and the twelve-ball puzzle

Calculus
Statistics and probability

(a)(i)

3 marks

The graph of $I = - ln p$ on $0 < p \leq 1$ is decreasing and convex, with the vertical axis ( $p = 0$ ) as an asymptote and an intercept at $p = 1$ .
Sketch: a curve starting very high near $p = 0$ , falling and bending to meet the $p$ -axis at $(1, 0)$ , and stopping there.

(a)(ii)

3 marks

Differentiate.
For $0 < p \leq 1$ this is negative, so $I$ is a decreasing function of $p$ .

\frac{d I}{d p} = - \frac{1}{p}

(a)(iii)

1 mark

More information is gained from a rarer event (a smaller probability), and less from a more likely event.

(b)(i)

1 mark

Five of the integers 1 to 10 are odd.

P (odd) = \frac{5}{10} = \frac{1}{2}

(b)(ii)

2 marks

The information gained is $- ln p$ with $p = \frac{1}{2}$ .

I = - ln \frac{1}{2} = ln 2 \approx 0.693

(b)(iii)

2 marks

Being told that $y$ is not 7 has probability $\frac{9}{10}$ .

I = - ln \frac{9}{10} \approx 0.105

(c)(i)

2 marks

There are two outcomes, with probabilities $\frac{1}{10}$ and $\frac{9}{10}$ .

E (I) = - (\frac{1}{10} ln \frac{1}{10} + \frac{9}{10} ln \frac{9}{10}) \approx 0.325

(c)(ii)

2 marks

The two outcomes of asking whether $x$ is odd (odd or even) each have probability $\frac{1}{2}$ .
Since $0.693 > 0.325$ , her expected information gain is greater.

E (I) = - (\frac{1}{2} ln \frac{1}{2} + \frac{1}{2} ln \frac{1}{2}) = ln 2 \approx 0.693

(d)(i)

1 mark

Tails has probability $1 - p$ .

I = - ln (1 - p)

(d)(ii)

1 mark

Add the two outcomes weighted by their probabilities.

E (I) = - p ln p - (1 - p) ln (1 - p)

(d)(iii)

2 marks

Maximise $E (I)$ with a graph or by calculus: the maximum is where heads and tails are equally likely.

p = \frac{1}{2}

(e)(i)

2 marks

With 3 balls on each side, 6 balls are not weighed. The scales balance only if the odd ball is among those 6.

x = \frac{6}{12} = \frac{1}{2}

(e)(ii)

2 marks

The three probabilities in the row add to 1.

x + 2 y = 1 \Rightarrow y = \frac{1}{4}

(e)(iii)

2 marks

For 2 balls on each side the probabilities are $\frac{2}{3}$ , $\frac{1}{6}$ and $\frac{1}{6}$ .

z = - (\frac{2}{3} ln \frac{2}{3} + \frac{1}{6} ln \frac{1}{6} + \frac{1}{6} ln \frac{1}{6}) \approx 0.868

(e)(iv)

1 mark

The table shows the largest expected information, 1.099, when 4 balls are on each side, so the best first weighing is to put 4 balls on each side.

Our analysis

Where the marks are

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

Calculus35 marks · 64%
Q1, Q2
Statistics and probability20 marks · 36%
Q2

Mentor insights

Traps to watch for

01
Chain rule trick · Q1
$\frac{d v}{d t} = \frac{d v}{d s} \cdot \frac{d s}{d t} = v \frac{d v}{d s}$ turns a velocity-time equation into one in $v$ and $s$ .
02
Euler's method needs s as the variable · Q1
With step 4000 in $s$ , update $v_{n + 1} = v_{n} + 4000 \frac{d v}{d s}$ with $\frac{d v}{d s} = \frac{g ( s )}{v} - k v$ .
03
Rarer events carry more information · Q2
$I = - ln p$ is larger when $p$ is smaller.
04
Expected information · Q2
$E (I) = - \sum p_{r} ln p_{r}$ over all outcomes.
05
Balanced probability · Q2
In the weighing puzzle the scales balance only if the odd ball is among the balls not weighed.

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 AI HL · Paper 3

Mathematics: Applications & Interpretation, Higher Level

Session: November 2023
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 November 2023 Global 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

28 marks

A skydiver in free fall: differential equations with and without air resistance

Calculus

(a)(i)

1 mark

Use the chain rule with $v = \frac{d s}{d t}$ .

\frac{d v}{d t} = \frac{d v}{d s} \frac{d s}{d t} = v \frac{d v}{d s}

(a)(ii)

4 marks

Separate the variables in $v \frac{d v}{d s} = g$ and integrate.
Use $v = 10$ when $s = 0$ .

\int v d v = \int g d s \Rightarrow \frac{v ^{2}}{2} = g s + c c = 50 \Rightarrow v = 2 g s + 100

(a)(iii)

3 marks

Find $s$ when $v = 330$ .
Since $5551 < 40000$ , the model predicts that Felix reaches the speed of sound before $s = 40000$ m.

33 0^{2} = 2 (9.8) s + 100 \Rightarrow s = 5551

(b)(i)

2 marks

With $\frac{d v}{d t} = g$ constant, $v = g t + c$ .
The graph of $v$ against $t$ should be a straight line with gradient $g$ .

(b)(ii)

1 mark

The plotted data are not a straight line (only roughly linear for small $t$ ), so the model does not appear to be valid.

(c)(i)

5 marks

Use $v \frac{d v}{d s} = g - k v^{2}$ and separate the variables.
Rearrange for $v$ .
Use $v = 10$ at $s = 0$ : $A = g - 100 k$ .

\int \frac{v}{g - k v ^{2}} d v = \int d s \Rightarrow - \frac{1}{2 k} ln (g - k v^{2}) = s + c g - k v^{2} = A e^{- 2 k s} v = \frac{g - ( g - 100 k ) e ^{- 2 k s}}{k}

(c)(ii)

2 marks

With air resistance the acceleration is $g - k v^{2}$ . At $v = 40$ the gradient of the $v$ - $t$ graph is 9.672.

9.672 = 9.8 - 1600 k \Rightarrow k = \frac{9.8 - 9.672}{1600} = 8 \times 1 0^{- 5}

(c)(iii)

2 marks

As $s \to \infty$ , $e^{- 2 k s} \to 0$ .

v \to \frac{g}{k} = \frac{9.8}{8 \times 1 0 ^{- 5}} = 350 m s^{- 1}

(c)(iv)

2 marks

The velocity increases with $s$ , so the upper bound on $0 < s \leq 40000$ is at $s = 40000$ .

v (40000) = \frac{9.8 - ( 9.8 - 0.008 ) e ^{- 6.4}}{8 \times 1 0 ^{- 5}} \approx 349.7 m s^{- 1}

(d)

4 marks

Write the new model as $\frac{d v}{d s} = \frac{1}{v} (\frac{3.98 \times 1 0 ^{14}}{( 6.41 \times 1 0 ^{6} - s ) ^{2}} - 8 \times 1 0^{- 5} v^{2})$ . Euler's method with step 4000 in $s$ :
Ten steps from $v_{0} = 10$ at $s_{0} = 0$ give

s_{n + 1} = s_{n} + 4000, v_{n + 1} = v_{n} + 4000 (\frac{3.98 \times 1 0 ^{14}}{v _{n} ( 6.41 \times 1 0 ^{6} - s _{n} ) ^{2}} - 8 \times 1 0^{- 5} v_{n}) v \approx 361 m s^{- 1}

(e)(i)

1 mark

Use a smaller step length (or a better method such as Runge-Kutta).

(e)(ii)

1 mark

The model ignores the parachute, wind and air currents, and treats Felix as a point object falling straight down. Any one of these could cause a difference from the data.

Question 2

27 marks

Information theory: logarithms, expected information and the twelve-ball puzzle

Calculus
Statistics and probability

(a)(i)

3 marks

The graph of $I = - ln p$ on $0 < p \leq 1$ is decreasing and convex, with the vertical axis ( $p = 0$ ) as an asymptote and an intercept at $p = 1$ .
Sketch: a curve starting very high near $p = 0$ , falling and bending to meet the $p$ -axis at $(1, 0)$ , and stopping there.

(a)(ii)

3 marks

Differentiate.
For $0 < p \leq 1$ this is negative, so $I$ is a decreasing function of $p$ .

\frac{d I}{d p} = - \frac{1}{p}

(a)(iii)

1 mark

More information is gained from a rarer event (a smaller probability), and less from a more likely event.

(b)(i)

1 mark

Five of the integers 1 to 10 are odd.

P (odd) = \frac{5}{10} = \frac{1}{2}

(b)(ii)

2 marks

The information gained is $- ln p$ with $p = \frac{1}{2}$ .

I = - ln \frac{1}{2} = ln 2 \approx 0.693

(b)(iii)

2 marks

Being told that $y$ is not 7 has probability $\frac{9}{10}$ .

I = - ln \frac{9}{10} \approx 0.105

(c)(i)

2 marks

There are two outcomes, with probabilities $\frac{1}{10}$ and $\frac{9}{10}$ .

E (I) = - (\frac{1}{10} ln \frac{1}{10} + \frac{9}{10} ln \frac{9}{10}) \approx 0.325

(c)(ii)

2 marks

The two outcomes of asking whether $x$ is odd (odd or even) each have probability $\frac{1}{2}$ .
Since $0.693 > 0.325$ , her expected information gain is greater.

E (I) = - (\frac{1}{2} ln \frac{1}{2} + \frac{1}{2} ln \frac{1}{2}) = ln 2 \approx 0.693

(d)(i)

1 mark

Tails has probability $1 - p$ .

I = - ln (1 - p)

(d)(ii)

1 mark

Add the two outcomes weighted by their probabilities.

E (I) = - p ln p - (1 - p) ln (1 - p)

(d)(iii)

2 marks

Maximise $E (I)$ with a graph or by calculus: the maximum is where heads and tails are equally likely.

p = \frac{1}{2}

(e)(i)

2 marks

With 3 balls on each side, 6 balls are not weighed. The scales balance only if the odd ball is among those 6.

x = \frac{6}{12} = \frac{1}{2}

(e)(ii)

2 marks

The three probabilities in the row add to 1.

x + 2 y = 1 \Rightarrow y = \frac{1}{4}

(e)(iii)

2 marks

For 2 balls on each side the probabilities are $\frac{2}{3}$ , $\frac{1}{6}$ and $\frac{1}{6}$ .

z = - (\frac{2}{3} ln \frac{2}{3} + \frac{1}{6} ln \frac{1}{6} + \frac{1}{6} ln \frac{1}{6}) \approx 0.868

(e)(iv)

1 mark

The table shows the largest expected information, 1.099, when 4 balls are on each side, so the best first weighing is to put 4 balls on each side.

Our analysis

Where the marks are

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

Calculus35 marks · 64%
Q1, Q2
Statistics and probability20 marks · 36%
Q2

Mentor insights

Traps to watch for

01
Chain rule trick · Q1
$\frac{d v}{d t} = \frac{d v}{d s} \cdot \frac{d s}{d t} = v \frac{d v}{d s}$ turns a velocity-time equation into one in $v$ and $s$ .
02
Euler's method needs s as the variable · Q1
With step 4000 in $s$ , update $v_{n + 1} = v_{n} + 4000 \frac{d v}{d s}$ with $\frac{d v}{d s} = \frac{g ( s )}{v} - k v$ .
03
Rarer events carry more information · Q2
$I = - ln p$ is larger when $p$ is smaller.
04
Expected information · Q2
$E (I) = - \sum p_{r} ln p_{r}$ over all outcomes.
05
Balanced probability · Q2
In the weighing puzzle the scales balance only if the odd ball is among the balls not weighed.

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