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IB Maths AI HL · Paper 2

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2

Session: May 2021
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 2021 TZ1 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

19 marks

Savings and a loan: compound interest and amortisation

Number and algebra

(a)

3 marks

Two years is 8 quarters, each at $1.6%$ interest.

37000 (1 + \frac{0.064}{4})^{8} \approx 42010 AUD

(b)

4 marks

Solve $37000 (1.016)^{n} > 50000$ for the number of quarters $n$ , on the calculator or by logs.
Interest is only added at the end of each quarter, so convert to months.

n > \frac{ln ( 50000/37000 )}{ln 1.016} \approx 18.97 \Rightarrow n = 19 quarters m = 19 \times 3 = 57

(c)

1 mark

The deposit is 25% of 200 000, so the loan is the remaining 75%.

0.75 \times 200000 = 150000 AUD

(d)

5 marks

(i) Over 10 years there are 120 payments of 1700, so the interest is what Daisy pays beyond the loan.

(ii) Use the calculator's finance solver: $N = 120$ , $PV = 150000$ , $PMT = - 1700$ , $FV = 0$ , payments and compounding 12 per year.

120 \times 1700 - 150000 = 54000 AUD r \approx 6.46%

(e)

3 marks

After 5 years she has made 60 payments. Use the same solver with $N = 60$ , $I = 6.4578%$ , $PV = 150000$ , $PMT = - 1700$ , and read off the future value, which is the balance still owed.

final payment \approx 86973 AUD

(f)

3 marks

Without the early payoff she would have paid $120 \times 1700 = 204000$ . With it, she pays 60 payments plus the final amount.

204000 - (60 \times 1700 + 86973) = 204000 - 188973 = 15027 AUD

Question 2

16 marks

A tunnel profile: maximum height, trapezoidal rule and exact area

Calculus

(a)

6 marks

(i) Differentiate each term.

(ii) At a maximum the derivative is zero.

Substitute this into the model.

\frac{d y}{d x} = - 0.3 x^{2} + 1.6 x x (- 0.3 x + 1.6) = 0 \Rightarrow x = \frac{16}{3} \approx 5.33 y = - 0.1 (\frac{16}{3})^{3} + 0.8 (\frac{16}{3})^{2} \approx 7.59 m

(b)

3 marks

Substitute into $y = - 0.1 x^{3} + 0.8 x^{2}$ .

y (4) = - 6.4 + 12.8 = 6.4 m, y (6) = - 21.6 + 28.8 = 7.2 m

(c)

3 marks

Three intervals between $x = 2$ and $x = 8$ gives $h = 2$ . The heights at $x = 2, 4, 6, 8$ are $2.4$ , $6.4$ , $7.2$ , $0$ .

A \approx \frac{1}{2} (2) [(2.4 + 0) + 2 (6.4 + 7.2)] = 29.6 m^{2}

(d)

4 marks

(i) The area under the curve is a definite integral.

(ii) Integrate and evaluate.

A = \int_{2}^{8} (- 0.1 x^{3} + 0.8 x^{2}) d x [- \frac{x ^{4}}{40} + \frac{0.8 x ^{3}}{3}]_{2}^{8} \approx 34.133 - 1.733 = 32.4 m^{2}

Mentor tip

The trapezoidal estimate, 29.6, is below the exact area, 32.4. The curve is concave down for $x > \frac{8}{3}$ , so the chords sit under it over most of the range.

Question 3

13 marks

Do bicycle stopping distances fit a normal model? A goodness-of-fit test

Statistics and probability

(a)

3 marks

With $X \sim N (6.76, 0.1 2^{2})$ , use the normal cumulative distribution.

P (X < 6.5) \approx 0.0151, P (X > 7) \approx 0.0228

(b)

3 marks

Multiply the interval probabilities by 1000.

1000 P (6.5 < X < 6.75) \approx 451.7, 1000 P (6.75 < X < 7) \approx 510.5

(c)

2 marks

State the model as the null hypothesis.

H_{0} : the stopping distances follow N (6.76, 0.1 2^{2}), H_{1} : they do not

(d)

3 marks

The expected counts are about 15.1, 451.7, 510.5 and 22.8. Run the $χ^{2}$ goodness-of-fit test on the calculator with these and the observed 12, 428, 527, 33. There are 4 classes, so $df = 3$ .

p \approx 0.073

Mentor tip

With the expected counts rounded as above, $p = 0.0727$ . With unrounded ones it is 0.0706. Both are above 0.05, so the conclusion is the same.

(e)

2 marks

Compare with the 5% level.
So there is insufficient evidence to reject $H_{0}$ : the normal model is a reasonable fit.

p > 0.05

Question 4

14 marks

Modelling a cupcake and its volume of revolution

Functions
Calculus

(a)

2 marks

The gradient between $(0, 3.5)$ and $(4, 6)$ is $\frac{6 - 3.5}{4} = 0.625$ and the $y$ -intercept is 3.5.

y = 0.625 x + 3.5

(b)

3 marks

(i) Enter the four points into the calculator's quadratic regression.

(ii) The gradient at $x = 4$ is $- 1.95 (4) + 9.56 \approx 1.76$ , which is positive. So the curve is still rising after the peak at $(4, 6)$ , meaning it goes above 6, which does not match the cupcake.

y \approx - 0.975 x^{2} + 9.56 x - 16.7

(c)

4 marks

Write the quadratic with its maximum at $(4, 6)$ as $y = a (x - 4)^{2} + 6$ . It passes through $(7.5, 0)$ .

0 = a (3.5)^{2} + 6 \Rightarrow a = - \frac{24}{49} y = - \frac{24}{49} (x - 4)^{2} + 6

(d)

5 marks

(i) Rotate each piece about the $x$ -axis. The line is used for $0 \leq x \leq 4$ and the new quadratic for $4 \leq x \leq 7.5$ .

(ii) Evaluate on the calculator.

V = π \int_{0}^{4} (0.625 x + 3.5)^{2} d x + π \int_{4}^{7.5} (- \frac{24}{49} (x - 4)^{2} + 6)^{2} d x V \approx 501 cm^{3}

Question 5

13 marks

A sunny/not-sunny Markov chain: eigenvalues and the long-run share

Number and algebra

(a)

2 marks

Today is sunny, so the state is $(10)$ . Apply $T$ three times.
The probability of sun in three days is 0.65.

T^{3} (10) = (0.65 0.35)

(b)

5 marks

Solve $det (T - λ I) = 0$ .
For $λ = 1$ : $- 0.2 x + 0.3 y = 0$ , so an eigenvector is $(32)$ . For $λ = 0.5$ : $0.3 x + 0.3 y = 0$ , so an eigenvector is $(1 - 1)$ .

(0.8 - λ) (0.7 - λ) - 0.06 = λ^{2} - 1.5 λ + 0.5 = 0 \Rightarrow λ = 1, 0.5

(c)

2 marks

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

P = (32 1 - 1), D = (10 0 0.5)

(d)

4 marks

As $n \to \infty$ , $D^{n} \to (1000)$ because $0. 5^{n} \to 0$ . Then $T^{n} = P D^{n} P^{- 1}$ tends to a fixed matrix.
Each column is the long-run split, whatever the start.

T^{n} \to P (1000) P^{- 1} = (0.6 0.4 0.6 0.4) 60% of days are sunny

Mentor tip

This agrees with the eigenvector $(3, 2)$ , scaled so the entries add to 1: $(\frac{3}{5}, \frac{2}{5})$ .

Question 6

18 marks

An ice-skater on a spiral path

Calculus

(a)

3 marks

Differentiate each component with the product rule.

v = (ab e^{b t} cos t - a e^{b t} sin t ab e^{b t} sin t + a e^{b t} cos t)

(b)

4 marks

Square and add the components, factoring out $a^{2} e^{2 b t}$ .
Expand the brackets: the cross terms $\mp 2 b sin t cos t$ cancel.
Taking the square root gives the speed.

∣ v ∣^{2} = a^{2} e^{2 b t} [(b cos t - sin t)^{2} + (b sin t + cos t)^{2}] a^{2} e^{2 b t} [b^{2} (cos^{2} t + sin^{2} t) + (sin^{2} t + cos^{2} t)] = a^{2} e^{2 b t} (b^{2} + 1) ∣ v ∣ = a e^{b t} 1 + b^{2}

(c)

3 marks

At $t = 0$ the position is $(a, 0) = (5, 0)$ , so $a = 5$ . The velocity at $t = 0$ is $(ab, a) = (5 b, 5)$ and equals $(- 3.5, 5)$ .

5 b = - 3.5 \Rightarrow b = - 0.7

(d)

2 marks

Use the speed formula with $a = 5$ , $b = - 0.7$ and $t = 2$ .

∣ v ∣ = 5 e^{- 1.4} 1 + 0.49 \approx 1.51 m s^{- 1}

(e)

6 marks

Parallel to the $y$ -axis means the $x$ -component of velocity is zero.
The first positive $t$ with this tangent is $t = π - 0.6107 \approx 2.531$ . Find the position there.

ab e^{b t} cos t - a e^{b t} sin t = 0 \Rightarrow tan t = b = - 0.7 x = 5 e^{- 0.7 t} cos t \approx - 0.697, y = 5 e^{- 0.7 t} sin t \approx 0.488 OP = 0.69 7^{2} + 0.48 8^{2} \approx 0.850 m

Question 7

17 marks

Rabbits and foxes: exponential growth, Euler's method and equilibrium

Calculus

(a)

5 marks

Separate and integrate $\frac{d x}{d t} = 2 x$ , so $x = A e^{2 t}$ . The starting population of 100 fixes $A$ .

x (t) = 100 e^{2 t}, x (1) = 100 e^{2} \approx 739

(b)

6 marks

With $h = 0.25$ , start from $x = 1000$ , $y = 100$ and update both using the previous values.
After four steps (one year):
So after one year there are about 2840 rabbits and 58 foxes.

x_{n + 1} = x_{n} + 0.25 x_{n} (2 - 0.01 y_{n}), y_{n + 1} = y_{n} + 0.25 y_{n} (0.0002 x_{n} - 0.8) x_{4} \approx 2837 (about 2840), y_{4} \approx 58

(c)

3 marks

The graph is a closed loop in the $(x, y)$ plane. At A, both populations are increasing. At B the rabbit population is decreasing while the fox population is still increasing, since there are plenty of rabbits for the foxes to eat but the foxes are eating them faster than they breed.

(d)

3 marks

At equilibrium both rates are zero. For non-zero populations that means the brackets are zero.

2 - 0.01 y = 0 \Rightarrow y = 200, 0.0002 x - 0.8 = 0 \Rightarrow x = 4000 (x, y) = (4000, 200)

Our analysis

Where the marks are

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

Calculus56 marks · 51%
Q2, Q4, Q6, Q7
Number and algebra32 marks · 29%
Q1, Q5
Statistics and probability13 marks · 12%
Q3
Functions9 marks · 8%
Q4

Mentor insights

Traps to watch for

01
Quarterly means four periods a year · Q1
With interest compounded quarterly the rate per period is $\frac{6.4}{4} %$ and the number of periods is four times the years. The answer in months must then be a multiple of 3.
02
Use the unrounded stationary point · Q2
The maximum is at $x = \frac{16}{3}$ . If you round to 5.33 first, the height changes in the third figure.
03
A good model joins smoothly · Q4
At the join with the straight line the new curve should be flat, not still rising. A positive gradient at $x = 4$ means the curve overshoots the peak.
04
Simplify using $sin^{2} + cos^{2} = 1$ · Q6
The speed squared has four terms that collapse once you factorise. Do the algebra on the squared speed and take the root last.
05
Update both populations together · Q7
In Euler's method for a coupled system, each step must use the old values of both $x$ and $y$ , then update $x$ and $y$ at the same time.

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 2021
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 2021 TZ1 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

19 marks

Savings and a loan: compound interest and amortisation

Number and algebra

(a)

3 marks

Two years is 8 quarters, each at $1.6%$ interest.

37000 (1 + \frac{0.064}{4})^{8} \approx 42010 AUD

(b)

4 marks

Solve $37000 (1.016)^{n} > 50000$ for the number of quarters $n$ , on the calculator or by logs.
Interest is only added at the end of each quarter, so convert to months.

n > \frac{ln ( 50000/37000 )}{ln 1.016} \approx 18.97 \Rightarrow n = 19 quarters m = 19 \times 3 = 57

(c)

1 mark

The deposit is 25% of 200 000, so the loan is the remaining 75%.

0.75 \times 200000 = 150000 AUD

(d)

5 marks

(i) Over 10 years there are 120 payments of 1700, so the interest is what Daisy pays beyond the loan.

(ii) Use the calculator's finance solver: $N = 120$ , $PV = 150000$ , $PMT = - 1700$ , $FV = 0$ , payments and compounding 12 per year.

120 \times 1700 - 150000 = 54000 AUD r \approx 6.46%

(e)

3 marks

After 5 years she has made 60 payments. Use the same solver with $N = 60$ , $I = 6.4578%$ , $PV = 150000$ , $PMT = - 1700$ , and read off the future value, which is the balance still owed.

final payment \approx 86973 AUD

(f)

3 marks

Without the early payoff she would have paid $120 \times 1700 = 204000$ . With it, she pays 60 payments plus the final amount.

204000 - (60 \times 1700 + 86973) = 204000 - 188973 = 15027 AUD

Question 2

16 marks

A tunnel profile: maximum height, trapezoidal rule and exact area

Calculus

(a)

6 marks

(i) Differentiate each term.

(ii) At a maximum the derivative is zero.

Substitute this into the model.

\frac{d y}{d x} = - 0.3 x^{2} + 1.6 x x (- 0.3 x + 1.6) = 0 \Rightarrow x = \frac{16}{3} \approx 5.33 y = - 0.1 (\frac{16}{3})^{3} + 0.8 (\frac{16}{3})^{2} \approx 7.59 m

(b)

3 marks

Substitute into $y = - 0.1 x^{3} + 0.8 x^{2}$ .

y (4) = - 6.4 + 12.8 = 6.4 m, y (6) = - 21.6 + 28.8 = 7.2 m

(c)

3 marks

Three intervals between $x = 2$ and $x = 8$ gives $h = 2$ . The heights at $x = 2, 4, 6, 8$ are $2.4$ , $6.4$ , $7.2$ , $0$ .

A \approx \frac{1}{2} (2) [(2.4 + 0) + 2 (6.4 + 7.2)] = 29.6 m^{2}

(d)

4 marks

(i) The area under the curve is a definite integral.

(ii) Integrate and evaluate.

A = \int_{2}^{8} (- 0.1 x^{3} + 0.8 x^{2}) d x [- \frac{x ^{4}}{40} + \frac{0.8 x ^{3}}{3}]_{2}^{8} \approx 34.133 - 1.733 = 32.4 m^{2}

Mentor tip

The trapezoidal estimate, 29.6, is below the exact area, 32.4. The curve is concave down for $x > \frac{8}{3}$ , so the chords sit under it over most of the range.

Question 3

13 marks

Do bicycle stopping distances fit a normal model? A goodness-of-fit test

Statistics and probability

(a)

3 marks

With $X \sim N (6.76, 0.1 2^{2})$ , use the normal cumulative distribution.

P (X < 6.5) \approx 0.0151, P (X > 7) \approx 0.0228

(b)

3 marks

Multiply the interval probabilities by 1000.

1000 P (6.5 < X < 6.75) \approx 451.7, 1000 P (6.75 < X < 7) \approx 510.5

(c)

2 marks

State the model as the null hypothesis.

H_{0} : the stopping distances follow N (6.76, 0.1 2^{2}), H_{1} : they do not

(d)

3 marks

The expected counts are about 15.1, 451.7, 510.5 and 22.8. Run the $χ^{2}$ goodness-of-fit test on the calculator with these and the observed 12, 428, 527, 33. There are 4 classes, so $df = 3$ .

p \approx 0.073

Mentor tip

With the expected counts rounded as above, $p = 0.0727$ . With unrounded ones it is 0.0706. Both are above 0.05, so the conclusion is the same.

(e)

2 marks

Compare with the 5% level.
So there is insufficient evidence to reject $H_{0}$ : the normal model is a reasonable fit.

p > 0.05

Question 4

14 marks

Modelling a cupcake and its volume of revolution

Functions
Calculus

(a)

2 marks

The gradient between $(0, 3.5)$ and $(4, 6)$ is $\frac{6 - 3.5}{4} = 0.625$ and the $y$ -intercept is 3.5.

y = 0.625 x + 3.5

(b)

3 marks

(i) Enter the four points into the calculator's quadratic regression.

y \approx - 0.975 x^{2} + 9.56 x - 16.7

(c)

4 marks

Write the quadratic with its maximum at $(4, 6)$ as $y = a (x - 4)^{2} + 6$ . It passes through $(7.5, 0)$ .

0 = a (3.5)^{2} + 6 \Rightarrow a = - \frac{24}{49} y = - \frac{24}{49} (x - 4)^{2} + 6

(d)

5 marks

(i) Rotate each piece about the $x$ -axis. The line is used for $0 \leq x \leq 4$ and the new quadratic for $4 \leq x \leq 7.5$ .

(ii) Evaluate on the calculator.

V = π \int_{0}^{4} (0.625 x + 3.5)^{2} d x + π \int_{4}^{7.5} (- \frac{24}{49} (x - 4)^{2} + 6)^{2} d x V \approx 501 cm^{3}

Question 5

13 marks

A sunny/not-sunny Markov chain: eigenvalues and the long-run share

Number and algebra

(a)

2 marks

Today is sunny, so the state is $(10)$ . Apply $T$ three times.
The probability of sun in three days is 0.65.

T^{3} (10) = (0.65 0.35)

(b)

5 marks

Solve $det (T - λ I) = 0$ .
For $λ = 1$ : $- 0.2 x + 0.3 y = 0$ , so an eigenvector is $(32)$ . For $λ = 0.5$ : $0.3 x + 0.3 y = 0$ , so an eigenvector is $(1 - 1)$ .

(0.8 - λ) (0.7 - λ) - 0.06 = λ^{2} - 1.5 λ + 0.5 = 0 \Rightarrow λ = 1, 0.5

(c)

2 marks

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

P = (32 1 - 1), D = (10 0 0.5)

(d)

4 marks

As $n \to \infty$ , $D^{n} \to (1000)$ because $0. 5^{n} \to 0$ . Then $T^{n} = P D^{n} P^{- 1}$ tends to a fixed matrix.
Each column is the long-run split, whatever the start.

T^{n} \to P (1000) P^{- 1} = (0.6 0.4 0.6 0.4) 60% of days are sunny

Mentor tip

This agrees with the eigenvector $(3, 2)$ , scaled so the entries add to 1: $(\frac{3}{5}, \frac{2}{5})$ .

Question 6

18 marks

An ice-skater on a spiral path

Calculus

(a)

3 marks

Differentiate each component with the product rule.

v = (ab e^{b t} cos t - a e^{b t} sin t ab e^{b t} sin t + a e^{b t} cos t)

(b)

4 marks

Square and add the components, factoring out $a^{2} e^{2 b t}$ .
Expand the brackets: the cross terms $\mp 2 b sin t cos t$ cancel.
Taking the square root gives the speed.

∣ v ∣^{2} = a^{2} e^{2 b t} [(b cos t - sin t)^{2} + (b sin t + cos t)^{2}] a^{2} e^{2 b t} [b^{2} (cos^{2} t + sin^{2} t) + (sin^{2} t + cos^{2} t)] = a^{2} e^{2 b t} (b^{2} + 1) ∣ v ∣ = a e^{b t} 1 + b^{2}

(c)

3 marks

At $t = 0$ the position is $(a, 0) = (5, 0)$ , so $a = 5$ . The velocity at $t = 0$ is $(ab, a) = (5 b, 5)$ and equals $(- 3.5, 5)$ .

5 b = - 3.5 \Rightarrow b = - 0.7

(d)

2 marks

Use the speed formula with $a = 5$ , $b = - 0.7$ and $t = 2$ .

∣ v ∣ = 5 e^{- 1.4} 1 + 0.49 \approx 1.51 m s^{- 1}

(e)

6 marks

Parallel to the $y$ -axis means the $x$ -component of velocity is zero.
The first positive $t$ with this tangent is $t = π - 0.6107 \approx 2.531$ . Find the position there.

ab e^{b t} cos t - a e^{b t} sin t = 0 \Rightarrow tan t = b = - 0.7 x = 5 e^{- 0.7 t} cos t \approx - 0.697, y = 5 e^{- 0.7 t} sin t \approx 0.488 OP = 0.69 7^{2} + 0.48 8^{2} \approx 0.850 m

Question 7

17 marks

Rabbits and foxes: exponential growth, Euler's method and equilibrium

Calculus

(a)

5 marks

Separate and integrate $\frac{d x}{d t} = 2 x$ , so $x = A e^{2 t}$ . The starting population of 100 fixes $A$ .

x (t) = 100 e^{2 t}, x (1) = 100 e^{2} \approx 739

(b)

6 marks

With $h = 0.25$ , start from $x = 1000$ , $y = 100$ and update both using the previous values.
After four steps (one year):
So after one year there are about 2840 rabbits and 58 foxes.

x_{n + 1} = x_{n} + 0.25 x_{n} (2 - 0.01 y_{n}), y_{n + 1} = y_{n} + 0.25 y_{n} (0.0002 x_{n} - 0.8) x_{4} \approx 2837 (about 2840), y_{4} \approx 58

(c)

3 marks

(d)

3 marks

At equilibrium both rates are zero. For non-zero populations that means the brackets are zero.

2 - 0.01 y = 0 \Rightarrow y = 200, 0.0002 x - 0.8 = 0 \Rightarrow x = 4000 (x, y) = (4000, 200)

Our analysis

Where the marks are

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

Calculus56 marks · 51%
Q2, Q4, Q6, Q7
Number and algebra32 marks · 29%
Q1, Q5
Statistics and probability13 marks · 12%
Q3
Functions9 marks · 8%
Q4

Mentor insights

Traps to watch for

01
Quarterly means four periods a year · Q1
With interest compounded quarterly the rate per period is $\frac{6.4}{4} %$ and the number of periods is four times the years. The answer in months must then be a multiple of 3.
02
Use the unrounded stationary point · Q2
The maximum is at $x = \frac{16}{3}$ . If you round to 5.33 first, the height changes in the third figure.
03
A good model joins smoothly · Q4
At the join with the straight line the new curve should be flat, not still rising. A positive gradient at $x = 4$ means the curve overshoots the peak.
04
Simplify using $sin^{2} + cos^{2} = 1$ · Q6
The speed squared has four terms that collapse once you factorise. Do the algebra on the squared speed and take the root last.
05
Update both populations together · Q7
In Euler's method for a coupled system, each step must use the old values of both $x$ and $y$ , then update $x$ and $y$ at the same time.

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