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Worked solutions

IB Maths AI HL · Paper 1

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2

Session: May 2022
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 17

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2022 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–17 · 110 marks

Question 1

5 marks

A baseball's height: quadratic model and domain

Functions

(a)

1 mark

At the instant of the hit $t = 0$ .

h (0) = 1.2 m

(b)

2 marks

The ball hits the ground when $h (t) = 0$ .

- 4.8 t^{2} + 21 t + 1.2 = 0 \Rightarrow t \approx 4.43 s (reject t < 0)

(c)

2 marks

Time starts at the hit and ends when the ball lands.

0 \leq t \leq 4.43

Question 2

5 marks

Concert tickets: forming and solving three equations

Number and algebra

(a)

3 marks

The total number of tickets, the total income and the ratio are:

x + y + z = 600, 15 x + 10 y + 12 z = 7816, x = 2 y

(b)

2 marks

Substitute $x = 2 y$ into the first two equations.
Eliminate $z = 600 - 3 y$ .

3 y + z = 600, 40 y + 12 z = 7816 40 y + 12 (600 - 3 y) = 7816 \Rightarrow 4 y = 616 \Rightarrow y = 154 x = 308 adult, y = 154 child, z = 138 student

Question 3

7 marks

Is the die fair? Chi-squared goodness of fit

Statistics and probability

(a)

1 mark

The null hypothesis says the die is fair, so each face has probability $\frac{1}{6}$ . The alternative says it is not fair.

H_{0} : P (each number) = \frac{1}{6}, H_{1} : the probabilities are not all \frac{1}{6}

(b)

1 mark

Six categories and no parameters estimated from the data.

ν = 6 - 1 = 5

(c)

1 mark

A fair die gives each face one sixth of the 60 rolls.

E = 60 \times \frac{1}{6} = 10

(d)

2 marks

All six expected frequencies are 10. Add up $\frac{( O - E ) ^{2}}{E}$ .
With $ν = 5$ the upper tail gives the p-value.

χ^{2} = \frac{4 + 9 + 16 + 25 + 4 + 4}{10} = 6.2 p \approx 0.287

(e)

2 marks

Compare $p = 0.287$ with the significance level $0.05$ .
There is insufficient evidence to reject $H_{0}$ : the data are consistent with the die being fair.

0.287 > 0.05

Mentor tip

Say "insufficient evidence to reject", not "the die is proved fair".

Question 4

7 marks

Two socks from a drawer: tree diagram and conditional probability

Statistics and probability

(a)

2 marks

Seven socks: 3 brown and 4 black. After the first sock, six remain.
If the first is brown, the second is brown with $\frac{2}{6}$ and black with $\frac{4}{6}$ .
If the first is black, the second is brown with $\frac{3}{6}$ and black with $\frac{3}{6}$ .

(b)

2 marks

Same colour means brown-brown or black-black.

P (same) = \frac{3}{7} \times \frac{2}{6} + \frac{4}{7} \times \frac{3}{6} = \frac{6}{42} + \frac{12}{42} = \frac{3}{7} \approx 0.429

(c)

3 marks

Use $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ with $A$ = two brown and $B$ = same colour.

P (brown, brown ∣ same) = \frac{\frac{6}{42}}{\frac{18}{42}} = \frac{1}{3}

Mentor tip

Check that $\frac{1}{3}$ is less than 1 and that the two "same colour" pieces add to your answer to (b).

Question 5

7 marks

Bouncing ball: geometric sequence and total distance

Number and algebra

(a)

2 marks

The bounce heights form a geometric sequence with common ratio $0.85$ .

h_{6} = 1.8 \times 0.8 5^{6} = 0.6789 \dots m = 67.9 cm \approx 68 cm

(b)

2 marks

Find the largest $n$ with $1.8 \times 0.8 5^{n} > 0.1$ .
So 17 bounces reach more than 10 cm.

1.8 \times 0.8 5^{17} = 0.114 m > 0.1, 1.8 \times 0.8 5^{18} = 0.0966 m < 0.1

(c)

3 marks

Up-and-down distances for bounces 1, 2 and 3 form a geometric series with first term $1.8 \times 0.85 = 1.53$ .
Add the first drop and double the bounce distances.

S_{3} = \frac{1.53 ( 1 - 0.8 5 ^{3} )}{1 - 0.85} = 3.935925 d = 1.8 + 2 (3.935925) = 9.67 m

Question 6

8 marks

Rope on a pole: vector from A to F, length and angle

Geometry and trigonometry

(a)

1 mark

Subtract the coordinates of A from those of F.

A F = - 3.2 - 4.5 6.1

(b)

2 marks

The length is the magnitude of the vector.

∣ A F ∣ = 3. 2^{2} + 4. 5^{2} + 6. 1^{2} = 67.7 \approx 8.23 m

(c)

5 marks

Both vectors start at A.
Use the scalar product.

A O = - 3.2 - 4.5 0.3 A F \cdot A O = 10.24 + 20.25 + 1.83 = 32.32 cos θ = \frac{32.32}{67.7 30.58} = 0.7103 θ = 0.781 rad = 44. 7^{\circ}

Mentor tip

If you used $O A$ by mistake, $cos θ$ comes out negative: the angle is $18 0^{\circ} - 44. 7^{\circ}$ .

Question 7

4 marks

Slope field: where the maximum and minimum points lie

Calculus

(a)

2 marks

Turning points have zero gradient.
This is a downward parabola through the origin with its maximum at $(0, 0)$ . Draw it across the field.

x^{2} + \frac{y}{2} = 0 \Rightarrow y = - 2 x^{2}

(b)

2 marks

Start at $(0, - 2)$ and follow the little segments. The solution drops to a local minimum on the parabola for negative $x$ , rises towards the parabola, then goes down again.
Sketch: a curve through $(0, - 2)$ with a local maximum and a local minimum, each lying on $y = - 2 x^{2}$ .

Mentor tip

A solution curve is horizontal exactly where it meets the parabola, so check that your two turning points are on it.

Question 8

6 marks

Curve with an exponential: derivatives and inflexion

Calculus

(a)

4 marks

Expand the curve: $y = 8 x - 2 x e^{x}$ .

\frac{d y}{d x} = 8 - 2 e^{x} - 2 x e^{x} \frac{d ^{2} y}{d x ^{2}} = - 2 e^{x} - 2 e^{x} - 2 x e^{x} = - 2 e^{x} (2 + x)

(b)

2 marks

An inflexion needs $\frac{d ^{2} y}{d x ^{2}} = 0$ .

- 2 e^{a} (2 + a) = 0 \Rightarrow a = - 2

Mentor tip

$e^{a}$ is never 0, so the only factor that can vanish is $2 + a$ .

Question 9

6 marks

Bags of sugar: single bag and the mean of five

Statistics and probability

(a)

2 marks

In grams $X \sim N (1000, 10 0^{2})$ .

P (X < 950) = 0.3085 \approx 0.309

(b)

4 marks

For the mean of 5 bags:
So $\overset{ˉ}{X} \sim N (1000, 2000)$ .

E (\overset{ˉ}{X}) = 1000, Var (\overset{ˉ}{X}) = \frac{10 0 ^{2}}{5} = 2000 P (\overset{ˉ}{X} > 950) = 0.8682 \approx 0.868

Mentor tip

Cross-check with the total weight: $T \sim N (5000, 50000)$ and $P (T > 4750)$ is the same 0.868.

Question 10

7 marks

Spiral on an Argand diagram: z = (θ/2)e^{iθ}

Number and algebra

(a)

3 marks

The modulus is $\frac{1}{2} θ$ and the argument is $θ$ .

θ = \frac{π}{2} : ∣ z ∣ = \frac{π}{4} \approx 0.79, ar g z = \frac{π}{2} (positive imaginary axis) θ = π : ∣ z ∣ = \frac{π}{2} \approx 1.57, ar g z = π (negative real axis) θ = \frac{3 π}{2} : ∣ z ∣ = \frac{3 π}{4} \approx 2.36, ar g z = \frac{3 π}{2} (negative imaginary axis)

(b)

4 marks

A modulus of 4 needs $\frac{1}{2} θ = 4$ .
Eight radians is $8 - 2 π \approx 1.72$ rad past the positive real axis, which is about $9 8^{\circ}$ . That is in the second quadrant, just beyond the positive imaginary axis, on the circle of radius 4.

θ = 8

Mentor tip

Subtract $2 π$ to find where an angle sits. $8 - 2 π \approx 1.72$ is a little over $\frac{π}{2}$ .

Question 11

5 marks

Eigenvector and the long-term probability of a switch

Number and algebra

(a)

3 marks

Solve $(M - I) v = 0$ .
An eigenvector with integer entries is $(78)$ .

(- 0.8 0.8 0.7 - 0.7) (a b) = 0 \Rightarrow 8 a = 7 b

(b)

2 marks

The long-term probabilities are proportional to $(7, 8)$ and must sum to 1. The total is $7 + 8 = 15$ .

P (state A) = \frac{7}{15}

Mentor tip

Check with $M (7/15 8/15)$ : the first entry is $0.2 \cdot \frac{7}{15} + 0.7 \cdot \frac{8}{15} = \frac{7}{15}$ .

Question 12

8 marks

Earthquakes: Gutenberg-Richter equation and a Poisson wait

Number and algebra
Statistics and probability

(a)

2 marks

Substitute $N = 100$ and $M = 3$ .

lo g_{10} 100 = a - 3 \Rightarrow 2 = a - 3 \Rightarrow a = 5

(b)

2 marks

Undo the logarithm.

N = 1 0^{5 - M} = \frac{1 0 ^{5}}{1 0 ^{M}} \Rightarrow b = 100000

(c)

1 mark

Put $M = 7.2$ into the equation.

N = 1 0^{5 - 7.2} = 1 0^{- 2.2} \approx 0.00631

(d)

3 marks

$Y > 100$ means there is no earthquake of this size in the next 100 years.
Let $X$ be the number in 100 years. Then $X \sim Po (100 \times 0.0063096) = Po (0.63096)$ .

P (Y > 100) = P (X = 0) = e^{- 0.63096} \approx 0.532

Question 13

5 marks

A ship on a bearing of 045 degrees

Geometry and trigonometry

(a)

1 mark

Start position plus time times velocity.

r = (14) + t (1.2 - 0.6)

(b)

4 marks

On a bearing of $04 5^{\circ}$ the ship is as far east as it is north.
Add this to 1:00 pm.

1 + 1.2 t = 4 - 0.6 t 1.8 t = 3 \Rightarrow t = \frac{5}{3} h = 1 h 40 min time = 2 : 40 pm

Question 14

8 marks

Integrating (1/(x+2) + 1)^2 and a volume of revolution

Calculus

(a)

4 marks

Expand with $u = x + 2$ .
Integrate term by term.

(\frac{1}{u} + 1)^{2} = \frac{1}{u ^{2}} + \frac{2}{u} + 1 \int (\frac{1}{x + 2} + 1)^{2} d x = - \frac{1}{x + 2} + 2 ln (x + 2) + x + c

(b)

4 marks

Volume about the $x$ -axis is $V = π \int_{0}^{2} y^{2} d x$ .
Write it in the requested form.

V = π [- \frac{1}{x + 2} + 2 ln (x + 2) + x]_{0}^{2} V = π [(- \frac{1}{4} + 2 ln 4 + 2) - (- \frac{1}{2} + 2 ln 2)] = π (\frac{9}{4} + 2 ln 2) V = \frac{π}{4} (9 + 8 ln 2), a = 9, b = 8, c = 2

Question 15

7 marks

Cars at a junction: Poisson test and its error probabilities

Statistics and probability

(a)

1 mark

Under $H_{0}$ the rate is 5.4 per minute for 60 minutes.

X \sim Po (324)

(b)

2 marks

A Type I error rejects $H_{0}$ when it is true: $X \leq 300$ under $Po (324)$ .

P (X \leq 300) \approx 0.0947

(c)

4 marks

Now the mean is $4.5 \times 60 = 270$ . A Type II error is $X > 300$ .

P (X \geq 301) = 1 - P (X \leq 300) \approx 0.0334

Question 16

7 marks

Wind chill: rate of change using the chain rule

Calculus

(a)

2 marks

Power rule on $v^{0.16}$ .

\frac{d W}{d v} = - 7.405 \times 0.16 v^{- 0.84} = - 1.1848 v^{- 0.84}

(b)

5 marks

Use the chain rule.
The wind chill is falling at about $0.856^{\circ} C$ per minute.

\frac{d W}{d t} = \frac{d W}{d v} \times \frac{d v}{d t} \frac{d W}{d t} = - 1.1848 \times 1 0^{- 0.84} \times 5 \approx - 0.856

Mentor tip

A negative sign means it is getting colder: say "decreasing" in words as well.

Question 17

8 marks

Finding p, q and r in f(t) = p e^{q cos(rt)}

Calculus

8 marks

At A, $cos 0 = 1$ .
B is a minimum, so $cos (5.2 r) = - 1$ .
At B the value is
Divide the two equations to eliminate $p$ .
Multiply the equations to eliminate $q$ .

p e^{q} = 6.5 5.2 r = π \Rightarrow r = \frac{π}{5.2} \approx 0.604 p e^{- q} = 0.2 e^{2 q} = \frac{6.5}{0.2} = 32.5 \Rightarrow q = \frac{1}{2} ln 32.5 \approx 1.74 p^{2} = 6.5 \times 0.2 = 1.3 \Rightarrow p = 1.14

Mentor tip

Check: $1.14 e^{1.74} \approx 6.5$ .

Our analysis

Where the marks are

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

Calculus33 marks · 30%
Q7, Q8, Q14, Q16, Q17
Statistics and probability30 marks · 27%
Q3, Q4, Q9, Q12, Q15
Number and algebra29 marks · 26%
Q2, Q5, Q10, Q11, Q12
Geometry and trigonometry13 marks · 12%
Q6, Q13
Functions5 marks · 5%
Q1

Mentor insights

Traps to watch for

01
Expected frequencies come from the null · Q3
If the die is fair every face is expected $60 \div 6 = 10$ times. Do not use the observed counts as expected ones.
02
Angle at A needs vectors that both start at A · Q6
The angle FAO is between $A F$ and $A O$ . Using $O A$ gives the supplementary angle.
03
Waiting more than 100 years means none in 100 years · Q12
Scale the Poisson mean to the time period: 100 years gives mean $100 \times 0.0063096$ .
04
Square before you integrate · Q14
$(\frac{1}{u} + 1)^{2}$ is not $\frac{1}{u ^{2}} + 1$ . Expand to $\frac{1}{u ^{2}} + \frac{2}{u} + 1$ .
05
Use the shape, not just the points · Q17
A maximum at A and the next minimum at B means $cos (r t)$ goes from 1 to $- 1$ at $t = 5.2$ , so $5.2 r = π$ .

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 1

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2

Session: May 2022
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 17

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2022 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–17 · 110 marks

Question 1

5 marks

A baseball's height: quadratic model and domain

Functions

(a)

1 mark

At the instant of the hit $t = 0$ .

h (0) = 1.2 m

(b)

2 marks

The ball hits the ground when $h (t) = 0$ .

- 4.8 t^{2} + 21 t + 1.2 = 0 \Rightarrow t \approx 4.43 s (reject t < 0)

(c)

2 marks

Time starts at the hit and ends when the ball lands.

0 \leq t \leq 4.43

Question 2

5 marks

Concert tickets: forming and solving three equations

Number and algebra

(a)

3 marks

The total number of tickets, the total income and the ratio are:

x + y + z = 600, 15 x + 10 y + 12 z = 7816, x = 2 y

(b)

2 marks

Substitute $x = 2 y$ into the first two equations.
Eliminate $z = 600 - 3 y$ .

3 y + z = 600, 40 y + 12 z = 7816 40 y + 12 (600 - 3 y) = 7816 \Rightarrow 4 y = 616 \Rightarrow y = 154 x = 308 adult, y = 154 child, z = 138 student

Question 3

7 marks

Is the die fair? Chi-squared goodness of fit

Statistics and probability

(a)

1 mark

The null hypothesis says the die is fair, so each face has probability $\frac{1}{6}$ . The alternative says it is not fair.

H_{0} : P (each number) = \frac{1}{6}, H_{1} : the probabilities are not all \frac{1}{6}

(b)

1 mark

Six categories and no parameters estimated from the data.

ν = 6 - 1 = 5

(c)

1 mark

A fair die gives each face one sixth of the 60 rolls.

E = 60 \times \frac{1}{6} = 10

(d)

2 marks

All six expected frequencies are 10. Add up $\frac{( O - E ) ^{2}}{E}$ .
With $ν = 5$ the upper tail gives the p-value.

χ^{2} = \frac{4 + 9 + 16 + 25 + 4 + 4}{10} = 6.2 p \approx 0.287

(e)

2 marks

Compare $p = 0.287$ with the significance level $0.05$ .
There is insufficient evidence to reject $H_{0}$ : the data are consistent with the die being fair.

0.287 > 0.05

Mentor tip

Say "insufficient evidence to reject", not "the die is proved fair".

Question 4

7 marks

Two socks from a drawer: tree diagram and conditional probability

Statistics and probability

(a)

2 marks

Seven socks: 3 brown and 4 black. After the first sock, six remain.
If the first is brown, the second is brown with $\frac{2}{6}$ and black with $\frac{4}{6}$ .
If the first is black, the second is brown with $\frac{3}{6}$ and black with $\frac{3}{6}$ .

(b)

2 marks

Same colour means brown-brown or black-black.

P (same) = \frac{3}{7} \times \frac{2}{6} + \frac{4}{7} \times \frac{3}{6} = \frac{6}{42} + \frac{12}{42} = \frac{3}{7} \approx 0.429

(c)

3 marks

Use $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ with $A$ = two brown and $B$ = same colour.

P (brown, brown ∣ same) = \frac{\frac{6}{42}}{\frac{18}{42}} = \frac{1}{3}

Mentor tip

Check that $\frac{1}{3}$ is less than 1 and that the two "same colour" pieces add to your answer to (b).

Question 5

7 marks

Bouncing ball: geometric sequence and total distance

Number and algebra

(a)

2 marks

The bounce heights form a geometric sequence with common ratio $0.85$ .

h_{6} = 1.8 \times 0.8 5^{6} = 0.6789 \dots m = 67.9 cm \approx 68 cm

(b)

2 marks

Find the largest $n$ with $1.8 \times 0.8 5^{n} > 0.1$ .
So 17 bounces reach more than 10 cm.

1.8 \times 0.8 5^{17} = 0.114 m > 0.1, 1.8 \times 0.8 5^{18} = 0.0966 m < 0.1

(c)

3 marks

Up-and-down distances for bounces 1, 2 and 3 form a geometric series with first term $1.8 \times 0.85 = 1.53$ .
Add the first drop and double the bounce distances.

S_{3} = \frac{1.53 ( 1 - 0.8 5 ^{3} )}{1 - 0.85} = 3.935925 d = 1.8 + 2 (3.935925) = 9.67 m

Question 6

8 marks

Rope on a pole: vector from A to F, length and angle

Geometry and trigonometry

(a)

1 mark

Subtract the coordinates of A from those of F.

A F = - 3.2 - 4.5 6.1

(b)

2 marks

The length is the magnitude of the vector.

∣ A F ∣ = 3. 2^{2} + 4. 5^{2} + 6. 1^{2} = 67.7 \approx 8.23 m

(c)

5 marks

Both vectors start at A.
Use the scalar product.

A O = - 3.2 - 4.5 0.3 A F \cdot A O = 10.24 + 20.25 + 1.83 = 32.32 cos θ = \frac{32.32}{67.7 30.58} = 0.7103 θ = 0.781 rad = 44. 7^{\circ}

Mentor tip

If you used $O A$ by mistake, $cos θ$ comes out negative: the angle is $18 0^{\circ} - 44. 7^{\circ}$ .

Question 7

4 marks

Slope field: where the maximum and minimum points lie

Calculus

(a)

2 marks

Turning points have zero gradient.
This is a downward parabola through the origin with its maximum at $(0, 0)$ . Draw it across the field.

x^{2} + \frac{y}{2} = 0 \Rightarrow y = - 2 x^{2}

(b)

2 marks

Start at $(0, - 2)$ and follow the little segments. The solution drops to a local minimum on the parabola for negative $x$ , rises towards the parabola, then goes down again.
Sketch: a curve through $(0, - 2)$ with a local maximum and a local minimum, each lying on $y = - 2 x^{2}$ .

Mentor tip

A solution curve is horizontal exactly where it meets the parabola, so check that your two turning points are on it.

Question 8

6 marks

Curve with an exponential: derivatives and inflexion

Calculus

(a)

4 marks

Expand the curve: $y = 8 x - 2 x e^{x}$ .

\frac{d y}{d x} = 8 - 2 e^{x} - 2 x e^{x} \frac{d ^{2} y}{d x ^{2}} = - 2 e^{x} - 2 e^{x} - 2 x e^{x} = - 2 e^{x} (2 + x)

(b)

2 marks

An inflexion needs $\frac{d ^{2} y}{d x ^{2}} = 0$ .

- 2 e^{a} (2 + a) = 0 \Rightarrow a = - 2

Mentor tip

$e^{a}$ is never 0, so the only factor that can vanish is $2 + a$ .

Question 9

6 marks

Bags of sugar: single bag and the mean of five

Statistics and probability

(a)

2 marks

In grams $X \sim N (1000, 10 0^{2})$ .

P (X < 950) = 0.3085 \approx 0.309

(b)

4 marks

For the mean of 5 bags:
So $\overset{ˉ}{X} \sim N (1000, 2000)$ .

E (\overset{ˉ}{X}) = 1000, Var (\overset{ˉ}{X}) = \frac{10 0 ^{2}}{5} = 2000 P (\overset{ˉ}{X} > 950) = 0.8682 \approx 0.868

Mentor tip

Cross-check with the total weight: $T \sim N (5000, 50000)$ and $P (T > 4750)$ is the same 0.868.

Question 10

7 marks

Spiral on an Argand diagram: z = (θ/2)e^{iθ}

Number and algebra

(a)

3 marks

The modulus is $\frac{1}{2} θ$ and the argument is $θ$ .

θ = \frac{π}{2} : ∣ z ∣ = \frac{π}{4} \approx 0.79, ar g z = \frac{π}{2} (positive imaginary axis) θ = π : ∣ z ∣ = \frac{π}{2} \approx 1.57, ar g z = π (negative real axis) θ = \frac{3 π}{2} : ∣ z ∣ = \frac{3 π}{4} \approx 2.36, ar g z = \frac{3 π}{2} (negative imaginary axis)

(b)

4 marks

A modulus of 4 needs $\frac{1}{2} θ = 4$ .
Eight radians is $8 - 2 π \approx 1.72$ rad past the positive real axis, which is about $9 8^{\circ}$ . That is in the second quadrant, just beyond the positive imaginary axis, on the circle of radius 4.

θ = 8

Mentor tip

Subtract $2 π$ to find where an angle sits. $8 - 2 π \approx 1.72$ is a little over $\frac{π}{2}$ .

Question 11

5 marks

Eigenvector and the long-term probability of a switch

Number and algebra

(a)

3 marks

Solve $(M - I) v = 0$ .
An eigenvector with integer entries is $(78)$ .

(- 0.8 0.8 0.7 - 0.7) (a b) = 0 \Rightarrow 8 a = 7 b

(b)

2 marks

The long-term probabilities are proportional to $(7, 8)$ and must sum to 1. The total is $7 + 8 = 15$ .

P (state A) = \frac{7}{15}

Mentor tip

Check with $M (7/15 8/15)$ : the first entry is $0.2 \cdot \frac{7}{15} + 0.7 \cdot \frac{8}{15} = \frac{7}{15}$ .

Question 12

8 marks

Earthquakes: Gutenberg-Richter equation and a Poisson wait

Number and algebra
Statistics and probability

(a)

2 marks

Substitute $N = 100$ and $M = 3$ .

lo g_{10} 100 = a - 3 \Rightarrow 2 = a - 3 \Rightarrow a = 5

(b)

2 marks

Undo the logarithm.

N = 1 0^{5 - M} = \frac{1 0 ^{5}}{1 0 ^{M}} \Rightarrow b = 100000

(c)

1 mark

Put $M = 7.2$ into the equation.

N = 1 0^{5 - 7.2} = 1 0^{- 2.2} \approx 0.00631

(d)

3 marks

$Y > 100$ means there is no earthquake of this size in the next 100 years.
Let $X$ be the number in 100 years. Then $X \sim Po (100 \times 0.0063096) = Po (0.63096)$ .

P (Y > 100) = P (X = 0) = e^{- 0.63096} \approx 0.532

Question 13

5 marks

A ship on a bearing of 045 degrees

Geometry and trigonometry

(a)

1 mark

Start position plus time times velocity.

r = (14) + t (1.2 - 0.6)

(b)

4 marks

On a bearing of $04 5^{\circ}$ the ship is as far east as it is north.
Add this to 1:00 pm.

1 + 1.2 t = 4 - 0.6 t 1.8 t = 3 \Rightarrow t = \frac{5}{3} h = 1 h 40 min time = 2 : 40 pm

Question 14

8 marks

Integrating (1/(x+2) + 1)^2 and a volume of revolution

Calculus

(a)

4 marks

Expand with $u = x + 2$ .
Integrate term by term.

(\frac{1}{u} + 1)^{2} = \frac{1}{u ^{2}} + \frac{2}{u} + 1 \int (\frac{1}{x + 2} + 1)^{2} d x = - \frac{1}{x + 2} + 2 ln (x + 2) + x + c

(b)

4 marks

Volume about the $x$ -axis is $V = π \int_{0}^{2} y^{2} d x$ .
Write it in the requested form.

V = π [- \frac{1}{x + 2} + 2 ln (x + 2) + x]_{0}^{2} V = π [(- \frac{1}{4} + 2 ln 4 + 2) - (- \frac{1}{2} + 2 ln 2)] = π (\frac{9}{4} + 2 ln 2) V = \frac{π}{4} (9 + 8 ln 2), a = 9, b = 8, c = 2

Question 15

7 marks

Cars at a junction: Poisson test and its error probabilities

Statistics and probability

(a)

1 mark

Under $H_{0}$ the rate is 5.4 per minute for 60 minutes.

X \sim Po (324)

(b)

2 marks

A Type I error rejects $H_{0}$ when it is true: $X \leq 300$ under $Po (324)$ .

P (X \leq 300) \approx 0.0947

(c)

4 marks

Now the mean is $4.5 \times 60 = 270$ . A Type II error is $X > 300$ .

P (X \geq 301) = 1 - P (X \leq 300) \approx 0.0334

Question 16

7 marks

Wind chill: rate of change using the chain rule

Calculus

(a)

2 marks

Power rule on $v^{0.16}$ .

\frac{d W}{d v} = - 7.405 \times 0.16 v^{- 0.84} = - 1.1848 v^{- 0.84}

(b)

5 marks

Use the chain rule.
The wind chill is falling at about $0.856^{\circ} C$ per minute.

\frac{d W}{d t} = \frac{d W}{d v} \times \frac{d v}{d t} \frac{d W}{d t} = - 1.1848 \times 1 0^{- 0.84} \times 5 \approx - 0.856

Mentor tip

A negative sign means it is getting colder: say "decreasing" in words as well.

Question 17

8 marks

Finding p, q and r in f(t) = p e^{q cos(rt)}

Calculus

8 marks

At A, $cos 0 = 1$ .
B is a minimum, so $cos (5.2 r) = - 1$ .
At B the value is
Divide the two equations to eliminate $p$ .
Multiply the equations to eliminate $q$ .

p e^{q} = 6.5 5.2 r = π \Rightarrow r = \frac{π}{5.2} \approx 0.604 p e^{- q} = 0.2 e^{2 q} = \frac{6.5}{0.2} = 32.5 \Rightarrow q = \frac{1}{2} ln 32.5 \approx 1.74 p^{2} = 6.5 \times 0.2 = 1.3 \Rightarrow p = 1.14

Mentor tip

Check: $1.14 e^{1.74} \approx 6.5$ .

Our analysis

Where the marks are

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

Calculus33 marks · 30%
Q7, Q8, Q14, Q16, Q17
Statistics and probability30 marks · 27%
Q3, Q4, Q9, Q12, Q15
Number and algebra29 marks · 26%
Q2, Q5, Q10, Q11, Q12
Geometry and trigonometry13 marks · 12%
Q6, Q13
Functions5 marks · 5%
Q1

Mentor insights

Traps to watch for

01
Expected frequencies come from the null · Q3
If the die is fair every face is expected $60 \div 6 = 10$ times. Do not use the observed counts as expected ones.
02
Angle at A needs vectors that both start at A · Q6
The angle FAO is between $A F$ and $A O$ . Using $O A$ gives the supplementary angle.
03
Waiting more than 100 years means none in 100 years · Q12
Scale the Poisson mean to the time period: 100 years gives mean $100 \times 0.0063096$ .
04
Square before you integrate · Q14
$(\frac{1}{u} + 1)^{2}$ is not $\frac{1}{u ^{2}} + 1$ . Expand to $\frac{1}{u ^{2}} + \frac{2}{u} + 1$ .
05
Use the shape, not just the points · Q17
A maximum at A and the next minimum at B means $cos (r t)$ goes from 1 to $- 1$ at $t = 5.2$ , so $5.2 r = π$ .

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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