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IB Maths AI HL · Paper 1

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2

Session: May 2021
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 16

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

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2021 TZ2 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–16 · 110 marks

Question 1

7 marks

A perpendicular bisector and a bus stop on a road

Geometry and trigonometry

(a)

5 marks

The gradient of AB is $\frac{24 - 20}{14 - 2} = \frac{1}{3}$ , so the bisector's gradient is $- 3$ . The midpoint of AB is $(8, 22)$ .

y - 22 = - 3 (x - 8) \Rightarrow y = - 3 x + 46

(b)

2 marks

The road is $- x + y = 4$ , that is $y = x + 4$ . Set it equal to the bisector.

x + 4 = - 3 x + 46 \Rightarrow x = 10.5, y = 14.5 (10.5, 14.5)

Question 2

7 marks

Range and inverse of a reciprocal-type function with a gap

Functions

(a)

3 marks

Find the end values: $f (- 7) = 2 - \frac{12}{- 2} = 8$ and $f (7) = 2 - \frac{12}{12} = 1$ .
On the right of $x = - 5$ the function rises from $- \infty$ up to $f (7) = 1$ . On the left it falls from $+ \infty$ down to $f (- 7) = 8$ .

f (x) \leq 1 or f (x) \geq 8

(b)

3 marks

Write $y = 2 - \frac{12}{x + 5}$ , swap $x$ and $y$ , and solve for $y$ .

x = 2 - \frac{12}{y + 5} \Rightarrow y + 5 = \frac{12}{2 - x} f^{- 1} (x) = \frac{12}{2 - x} - 5

(c)

1 mark

The range of the inverse is the domain of $f$ .

- 7 \leq f^{- 1} (x) \leq 7, f^{- 1} (x) \neq = - 5

Question 3

6 marks

Are chinchilla rabbits heavier than sable rabbits? A two-sample $t$ -test

Statistics and probability

(a)

2 marks

Let $μ_{c}$ and $μ_{s}$ be the population mean weights.

H_{0} : μ_{c} = μ_{s}, H_{1} : μ_{c} > μ_{s}

(b)

2 marks

Run a two-sample $t$ -test on the calculator with pooled variances and the upper-tail alternative.

p \approx 0.0408

(c)

2 marks

Compare $p$ with the 5% level.
So reject $H_{0}$ : there is sufficient evidence that chinchilla rabbits are generally heavier than sable rabbits.

0.0408 < 0.05

Mentor tip

Use words in the conclusion, not just "reject".

Question 4

6 marks

Distances from two angles of elevation

Geometry and trigonometry

(a)

2 marks

In the right-angled triangle HAC, $tan 2 5^{\circ} = \frac{380}{AC}$ .

AC = \frac{380}{tan 2 5 ^{\circ}} \approx 815 m

(b)

3 marks

In the right-angled triangle HAB, $tan 4 0^{\circ} = \frac{380}{AB}$ , so $AB \approx 453$ m. Minta swam from C to B.

BC = AC - AB = 814.9 - 452.9 \approx 362 m

(c)

1 mark

She covered this in 15 minutes, which is 0.25 hours.

speed = \frac{362.05}{0.25} \approx 1450 m per hour

Question 5

5 marks

Two laptops depreciating at different rates

Number and algebra

(a)

2 marks

Each year the value is multiplied by 0.9.

495 (0.9)^{5} \approx £ 292

(b)

2 marks

Set Roger's value equal to Chloe's after $k$ years.

495 (0.9)^{k} = 2200 (0.85)^{k} \Rightarrow (\frac{0.9}{0.85})^{k} = \frac{2200}{495} k = \frac{ln ( 2200/495 )}{ln ( 0.9/0.85 )} \approx 26.1

(c)

1 mark

A depreciation rate is unlikely to stay constant for 26 years, and laptops are unlikely to be worth anything useful or usable by then. So the answer isn't reliable.

Question 6

7 marks

A quadratic from its vertex and an intercept

Functions

(a)

1 mark

The axis of symmetry is $x = 0.5$ , which is halfway between $- 2$ and $p$ .

\frac{- 2 + p}{2} = 0.5 \Rightarrow p = 3

(b)

5 marks

Write the function in factorised form $f (x) = a (x + 2) (x - 3)$ and use the vertex $(0.5, - 12.5)$ .
Expand to read off $b$ and $c$ .

f (0.5) = a (2.5) (- 2.5) = - 6.25 a = - 12.5 \Rightarrow a = 2 f (x) = 2 (x^{2} - x - 6) = 2 x^{2} - 2 x - 12 a = 2, b = - 2, c = - 12

(c)

1 mark

The axis of symmetry passes through the vertex.

x = 0.5

Question 7

6 marks

A hot air balloon that slows down: a geometric series

Number and algebra

(a)

3 marks

The distances 450, 450(0.82), 450(0.82)², … form a geometric sequence with $r = 0.82$ . The height after 10 minutes is the sum of the first 10 terms.

S_{10} = \frac{450 ( 1 - 0.8 2 ^{10} )}{1 - 0.82} \approx 2160 m

(b)

2 marks

The most the balloon could ever reach is the sum to infinity.
Since $2500 < 2520$ , the balloon will not reach the required height.

S_{\infty} = \frac{450}{1 - 0.82} = 2500

(c)

1 mark

The model ignores wind, and horizontal movement, and it treats the slowing down as a fixed percentage each minute. In practice it will not be that regular.

Question 8

7 marks

Perpendicular lines, and whether they meet

Geometry and trigonometry

(a)

3 marks

The direction vectors are $p 2 p 4$ and $p + 4 4 - 7$ . Perpendicular lines have a dot product of zero.

p (p + 4) + 8 p - 28 = 0 \Rightarrow p^{2} + 12 p - 28 = 0 (p + 14) (p - 2) = 0 \Rightarrow p = - 14 or p = 2

(b)

4 marks

For $p = - 14$ the lines are $r = 2 - 5 - 3 + λ - 14 - 28 4$ and $r = 147 - 2 + μ - 10 4 - 7$ . Equate the first two components.
Solving these two equations gives $λ = - \frac{1}{2}$ and $μ = \frac{1}{2}$ . Now test the third component.
The third component doesn't match, so the lines do not intersect.

2 - 14 λ = 14 - 10 μ, - 5 - 28 λ = 7 + 4 μ - 3 + 4 (- \frac{1}{2}) = - 5, - 2 - 7 (\frac{1}{2}) = - 5.5

Mentor tip

They are not parallel either, so they are skew lines.

Question 9

8 marks

A newspaper vendor's model tested with a goodness-of-fit test

Statistics and probability

(a)

1 mark

The total sold is $74 + 97 + 91 + 86 + 112 = 460$ . Under the model each day is equally likely.

\frac{460}{5} = 92

(b)

7 marks

(i) State the hypotheses about the model.

(ii) There are 5 categories and no parameters were estimated from the data, so $df = 5 - 1 = 4$ .

(iii) On the calculator, $χ^{2} = \sum \frac{( O - 92 ) ^{2}}{92} \approx 8.54$ with $p \approx 0.0736$ .

So there is insufficient evidence to reject $H_{0}$ : the data is consistent with the vendor's model.

H_{0} : the data fits the model (same number each day), H_{1} : the data does not fit the model 8.54 < 9.49 (or p = 0.0736 > 0.05)

Question 10

6 marks

A confidence interval for the mean number of chocolates

Statistics and probability

(a)

1 mark

The unbiased estimate of the mean is the sample mean.

\overset{x}{ˉ} = \frac{2506}{30} \approx 83.5

(b)

2 marks

Use the formula given.

s_{n - 1}^{2} = \frac{209738 - \frac{250 6 ^{2}}{30}}{29} \approx 13.9

(c)

2 marks

Enter the summary statistics into a $t$ -interval on the calculator, with $n = 30$ and $s_{n - 1} = 13.9126 \approx 3.73$ .

(82.1, 84.9)

(d)

1 mark

The claimed mean of 85 is outside the 95% confidence interval. So there is evidence that the true mean is lower than the manufacturer's claim.

Mentor tip

A confidence interval and a significance test at 5% agree here, so you can use either to justify the conclusion.

Question 11

7 marks

Delivering leaflets along every road: the route inspection problem

Number and algebra

(a)

5 marks

The total of all the road lengths is 582 m. The odd-degree junctions are A, B, D and H, so these must be paired and the paths between them repeated. List all three pairings with the shortest path for each pair.
The cheapest pairing is AD and BH, adding 65 m.

AB + DH : (19 + 16 + 19) + (16 + 27) = 54 + 43 = 97 AD + BH : 19 + (19 + 27) = 19 + 46 = 65 AH + BD : (19 + 16 + 27) + (19 + 16) = 62 + 35 = 97 582 + 65 = 647 m

(b)

2 marks

If he starts and finishes at different junctions, those two should be odd vertices, so that the path between them need not be repeated. The cheapest is to start and finish at B and H, leaving only A and D to pair, with a repeat of 19 m.

total = 582 + 19 = 601 m

Mentor tip

Choosing B and H avoids the expensive 46 m repeat between B and H.

Question 12

8 marks

Complex numbers in polar form: powers, quotients and a real product

Number and algebra

(a)

5 marks

(i) Cube the modulus and triple the argument.

(ii) For $n = 2$ we have $z_{2} = 2 cis \frac{π}{8}$ . Divide the numbers, then take the fourth power.

z_{1}^{3} = 27 cis \frac{9 π}{4} = 27 e^{i π /4} \frac{z _{1}}{z _{2}} = \frac{3}{2} cis (\frac{3 π}{4} - \frac{π}{8}) = \frac{3}{2} cis \frac{5 π}{8} (\frac{z _{1}}{z _{2}})^{4} = \frac{81}{16} cis \frac{5 π}{2} = \frac{81}{16} e^{i π /2}

(b)

3 marks

The product has argument $\frac{3 π}{4} + \frac{nπ}{16} = \frac{( 12 + n ) π}{16}$ . For a positive real number the argument must be a multiple of $2 π$ .
The smallest positive $n$ comes from $k = 1$ .

\frac{12 + n}{16} = 2 k \Rightarrow 12 + n = 32 k n = 20

Question 13

8 marks

A mouse in a maze: transition matrix and long-run time at F

Number and algebra

(a)

3 marks

From each junction, list the paths and their weights, and divide by the total. A has three normal paths. B has a normal path and a sugar path (weights 1 and 2). C and E each have two normal paths. D has two normal paths. F has three normal paths and one sugar path (weights 1, 1, 1, 2).
As a matrix, with the columns showing where the mouse goes from each junction (A, B, C, D, E, F):

A: B, E, F \frac{1}{3} each; B: A \frac{1}{3}, C \frac{2}{3}; C: A, D \frac{1}{2} each D: C, F \frac{1}{2} each; E: C, D \frac{1}{2} each; F: B, C, E \frac{1}{5} each, D \frac{2}{5} 0 \frac{1}{3} 00 \frac{1}{3} \frac{1}{3} \frac{1}{3} 0 \frac{2}{3} 000 \frac{1}{2} 00 \frac{1}{2} 00 00 \frac{1}{2} 00 \frac{1}{2} 00 \frac{1}{2} \frac{1}{2} 00 0 \frac{1}{5} \frac{1}{5} \frac{2}{5} \frac{1}{5} 0

(b)

3 marks

Raise the matrix to a large power on the calculator and read off the steady state. The fraction of time at F is the last entry.
So the mouse spends about 17.3% of the time at F.

(0.157, 0.0868, 0.256, 0.241, 0.0868, 0.173)

(c)

2 marks

The model treats each rest as the same length and assumes that travel between junctions is instant, so the real time at F would differ. The mouse may also eat the sugar, which would change the probabilities over time.

Mentor tip

Give one limitation and say how it would change the answer.

Question 14

7 marks

A geometric transformation: image, a pair of fixed numbers, and area

Number and algebra

(a)

2 marks

Apply the matrix and then add the translation.

(72 - 10 - 3) (6 - 2) + (- 5 4) = (6218) + (- 5 4) = (5722)

(b)

3 marks

Setting the image equal to twice the point gives two equations.
Substitute $p = 1 + 2 q$ into the second.

7 p - 10 q - 5 = 2 p \Rightarrow p - 2 q = 1 2 p - 3 q + 4 = 2 q \Rightarrow 2 p - 5 q = - 4 2 + 4 q - 5 q = - 4 \Rightarrow q = 6, p = 13

(c)

2 marks

The area scale factor of the matrix is the absolute value of its determinant.
The area factor is 1, and the translation doesn't change area, so L and its image have the same area.

det (72 - 10 - 3) = - 21 + 20 = - 1, ∣ - 1 ∣ = 1

Question 15

7 marks

A coffee shop discount: a Poisson hypothesis test

Statistics and probability

(a)

1 mark

With a mean of 22 an hour, the mean for 5 hours is 110.

H_{0} : μ = 110, H_{1} : μ > 110

(b)

4 marks

Under $H_{0}$ , $X \sim Po (110)$ . Find the smallest value $c$ with $P (X \geq c) \leq 0.05$ .
So the critical region is $X \geq 129$ , and the probability of a type I error is the size of this region.

P (X \geq 128) \approx 0.0502 > 0.05, P (X \geq 129) \approx 0.0415 < 0.05 P (type I error) \approx 0.0415

(c)

2 marks

Compare the observed 126 with the critical region, or compute its tail probability.
So there is insufficient evidence to reject $H_{0}$ : no evidence that sales have increased.

P (X \geq 126) \approx 0.072 > 0.05

Question 16

8 marks

Euler's method, then an exact solution and the percentage error

Calculus

(a)

3 marks

Euler's method with $h = 0.1$ : $x_{n + 1} = x_{n} + h x_{n} cos t_{n} e^{- s i n t_{n}}$ , starting from $t_{0} = 0$ and $x_{0} = \frac{1}{e}$ .
So $x (0.3) \approx 0.477$ .

x_{1} \approx 0.404667, x_{2} \approx 0.441106, x_{3} \approx 0.476548

(b)

5 marks

Separate the variables and integrate. The right side is the derivative of $- e^{- s i n t}$ .
At $t = 0$ , $x = \frac{1}{e}$ , so $- 1 = - 1 + c$ and $c = 0$ .
The percentage error is the difference between the approximation and the exact value, divided by the exact value.

\int \frac{1}{x} d x = \int cos t e^{- s i n t} d t \Rightarrow ln x = - e^{- s i n t} + c x = e^{- e^{- s i n t}} \Rightarrow x (0.3) \approx 0.475141 \frac{0.476548 - 0.475141}{0.475141} \times 100% \approx 0.296%

Our analysis

Where the marks are

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

Number and algebra41 marks · 37%
Q5, Q7, Q11, Q12, Q13, Q14
Statistics and probability27 marks · 25%
Q3, Q9, Q10, Q15
Geometry and trigonometry20 marks · 18%
Q1, Q4, Q8
Functions14 marks · 13%
Q2, Q6
Calculus8 marks · 7%
Q16

Mentor insights

Traps to watch for

01
Equidistant means on the bisector · Q1
The point equally far from two schools lies on the perpendicular bisector of the segment joining them. The bus stop must also lie on the road, so it is where the two lines meet.
02
Expected counts come from the model · Q9
If the same number sells each day, the expected value for each day is the total divided by 5. The degrees of freedom are the number of categories minus 1, since nothing else was estimated.
03
Add arguments, then reduce the angle · Q12
When you raise to a power or multiply, the angle can leave $(- π, π]$ . Subtract multiples of $2 π$ to bring it back before you give the answer.
04
Sugar doubles the weight · Q13
A path with sugar is twice as likely as a normal path, so give it weight 2 and the others weight 1, then divide by the total weight at that junction.
05
Separate and integrate once · Q16
The equation separates as $\frac{d x}{x} = cos t e^{- s i n t} d t$ , whose right side is the derivative of $- e^{- s i n t}$ . Then the initial condition fixes the constant.

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 2021
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 16

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

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2021 TZ2 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–16 · 110 marks

Question 1

7 marks

A perpendicular bisector and a bus stop on a road

Geometry and trigonometry

(a)

5 marks

The gradient of AB is $\frac{24 - 20}{14 - 2} = \frac{1}{3}$ , so the bisector's gradient is $- 3$ . The midpoint of AB is $(8, 22)$ .

y - 22 = - 3 (x - 8) \Rightarrow y = - 3 x + 46

(b)

2 marks

The road is $- x + y = 4$ , that is $y = x + 4$ . Set it equal to the bisector.

x + 4 = - 3 x + 46 \Rightarrow x = 10.5, y = 14.5 (10.5, 14.5)

Question 2

7 marks

Range and inverse of a reciprocal-type function with a gap

Functions

(a)

3 marks

Find the end values: $f (- 7) = 2 - \frac{12}{- 2} = 8$ and $f (7) = 2 - \frac{12}{12} = 1$ .
On the right of $x = - 5$ the function rises from $- \infty$ up to $f (7) = 1$ . On the left it falls from $+ \infty$ down to $f (- 7) = 8$ .

f (x) \leq 1 or f (x) \geq 8

(b)

3 marks

Write $y = 2 - \frac{12}{x + 5}$ , swap $x$ and $y$ , and solve for $y$ .

x = 2 - \frac{12}{y + 5} \Rightarrow y + 5 = \frac{12}{2 - x} f^{- 1} (x) = \frac{12}{2 - x} - 5

(c)

1 mark

The range of the inverse is the domain of $f$ .

- 7 \leq f^{- 1} (x) \leq 7, f^{- 1} (x) \neq = - 5

Question 3

6 marks

Are chinchilla rabbits heavier than sable rabbits? A two-sample $t$ -test

Statistics and probability

(a)

2 marks

Let $μ_{c}$ and $μ_{s}$ be the population mean weights.

H_{0} : μ_{c} = μ_{s}, H_{1} : μ_{c} > μ_{s}

(b)

2 marks

Run a two-sample $t$ -test on the calculator with pooled variances and the upper-tail alternative.

p \approx 0.0408

(c)

2 marks

Compare $p$ with the 5% level.
So reject $H_{0}$ : there is sufficient evidence that chinchilla rabbits are generally heavier than sable rabbits.

0.0408 < 0.05

Mentor tip

Use words in the conclusion, not just "reject".

Question 4

6 marks

Distances from two angles of elevation

Geometry and trigonometry

(a)

2 marks

In the right-angled triangle HAC, $tan 2 5^{\circ} = \frac{380}{AC}$ .

AC = \frac{380}{tan 2 5 ^{\circ}} \approx 815 m

(b)

3 marks

In the right-angled triangle HAB, $tan 4 0^{\circ} = \frac{380}{AB}$ , so $AB \approx 453$ m. Minta swam from C to B.

BC = AC - AB = 814.9 - 452.9 \approx 362 m

(c)

1 mark

She covered this in 15 minutes, which is 0.25 hours.

speed = \frac{362.05}{0.25} \approx 1450 m per hour

Question 5

5 marks

Two laptops depreciating at different rates

Number and algebra

(a)

2 marks

Each year the value is multiplied by 0.9.

495 (0.9)^{5} \approx £ 292

(b)

2 marks

Set Roger's value equal to Chloe's after $k$ years.

495 (0.9)^{k} = 2200 (0.85)^{k} \Rightarrow (\frac{0.9}{0.85})^{k} = \frac{2200}{495} k = \frac{ln ( 2200/495 )}{ln ( 0.9/0.85 )} \approx 26.1

(c)

1 mark

A depreciation rate is unlikely to stay constant for 26 years, and laptops are unlikely to be worth anything useful or usable by then. So the answer isn't reliable.

Question 6

7 marks

A quadratic from its vertex and an intercept

Functions

(a)

1 mark

The axis of symmetry is $x = 0.5$ , which is halfway between $- 2$ and $p$ .

\frac{- 2 + p}{2} = 0.5 \Rightarrow p = 3

(b)

5 marks

Write the function in factorised form $f (x) = a (x + 2) (x - 3)$ and use the vertex $(0.5, - 12.5)$ .
Expand to read off $b$ and $c$ .

f (0.5) = a (2.5) (- 2.5) = - 6.25 a = - 12.5 \Rightarrow a = 2 f (x) = 2 (x^{2} - x - 6) = 2 x^{2} - 2 x - 12 a = 2, b = - 2, c = - 12

(c)

1 mark

The axis of symmetry passes through the vertex.

x = 0.5

Question 7

6 marks

A hot air balloon that slows down: a geometric series

Number and algebra

(a)

3 marks

The distances 450, 450(0.82), 450(0.82)², … form a geometric sequence with $r = 0.82$ . The height after 10 minutes is the sum of the first 10 terms.

S_{10} = \frac{450 ( 1 - 0.8 2 ^{10} )}{1 - 0.82} \approx 2160 m

(b)

2 marks

The most the balloon could ever reach is the sum to infinity.
Since $2500 < 2520$ , the balloon will not reach the required height.

S_{\infty} = \frac{450}{1 - 0.82} = 2500

(c)

1 mark

The model ignores wind, and horizontal movement, and it treats the slowing down as a fixed percentage each minute. In practice it will not be that regular.

Question 8

7 marks

Perpendicular lines, and whether they meet

Geometry and trigonometry

(a)

3 marks

The direction vectors are $p 2 p 4$ and $p + 4 4 - 7$ . Perpendicular lines have a dot product of zero.

p (p + 4) + 8 p - 28 = 0 \Rightarrow p^{2} + 12 p - 28 = 0 (p + 14) (p - 2) = 0 \Rightarrow p = - 14 or p = 2

(b)

4 marks

For $p = - 14$ the lines are $r = 2 - 5 - 3 + λ - 14 - 28 4$ and $r = 147 - 2 + μ - 10 4 - 7$ . Equate the first two components.
Solving these two equations gives $λ = - \frac{1}{2}$ and $μ = \frac{1}{2}$ . Now test the third component.
The third component doesn't match, so the lines do not intersect.

2 - 14 λ = 14 - 10 μ, - 5 - 28 λ = 7 + 4 μ - 3 + 4 (- \frac{1}{2}) = - 5, - 2 - 7 (\frac{1}{2}) = - 5.5

Mentor tip

They are not parallel either, so they are skew lines.

Question 9

8 marks

A newspaper vendor's model tested with a goodness-of-fit test

Statistics and probability

(a)

1 mark

The total sold is $74 + 97 + 91 + 86 + 112 = 460$ . Under the model each day is equally likely.

\frac{460}{5} = 92

(b)

7 marks

(i) State the hypotheses about the model.

(ii) There are 5 categories and no parameters were estimated from the data, so $df = 5 - 1 = 4$ .

(iii) On the calculator, $χ^{2} = \sum \frac{( O - 92 ) ^{2}}{92} \approx 8.54$ with $p \approx 0.0736$ .

So there is insufficient evidence to reject $H_{0}$ : the data is consistent with the vendor's model.

H_{0} : the data fits the model (same number each day), H_{1} : the data does not fit the model 8.54 < 9.49 (or p = 0.0736 > 0.05)

Question 10

6 marks

A confidence interval for the mean number of chocolates

Statistics and probability

(a)

1 mark

The unbiased estimate of the mean is the sample mean.

\overset{x}{ˉ} = \frac{2506}{30} \approx 83.5

(b)

2 marks

Use the formula given.

s_{n - 1}^{2} = \frac{209738 - \frac{250 6 ^{2}}{30}}{29} \approx 13.9

(c)

2 marks

Enter the summary statistics into a $t$ -interval on the calculator, with $n = 30$ and $s_{n - 1} = 13.9126 \approx 3.73$ .

(82.1, 84.9)

(d)

1 mark

The claimed mean of 85 is outside the 95% confidence interval. So there is evidence that the true mean is lower than the manufacturer's claim.

Mentor tip

A confidence interval and a significance test at 5% agree here, so you can use either to justify the conclusion.

Question 11

7 marks

Delivering leaflets along every road: the route inspection problem

Number and algebra

(a)

5 marks

The total of all the road lengths is 582 m. The odd-degree junctions are A, B, D and H, so these must be paired and the paths between them repeated. List all three pairings with the shortest path for each pair.
The cheapest pairing is AD and BH, adding 65 m.

AB + DH : (19 + 16 + 19) + (16 + 27) = 54 + 43 = 97 AD + BH : 19 + (19 + 27) = 19 + 46 = 65 AH + BD : (19 + 16 + 27) + (19 + 16) = 62 + 35 = 97 582 + 65 = 647 m

(b)

2 marks

total = 582 + 19 = 601 m

Mentor tip

Choosing B and H avoids the expensive 46 m repeat between B and H.

Question 12

8 marks

Complex numbers in polar form: powers, quotients and a real product

Number and algebra

(a)

5 marks

(i) Cube the modulus and triple the argument.

(ii) For $n = 2$ we have $z_{2} = 2 cis \frac{π}{8}$ . Divide the numbers, then take the fourth power.

z_{1}^{3} = 27 cis \frac{9 π}{4} = 27 e^{i π /4} \frac{z _{1}}{z _{2}} = \frac{3}{2} cis (\frac{3 π}{4} - \frac{π}{8}) = \frac{3}{2} cis \frac{5 π}{8} (\frac{z _{1}}{z _{2}})^{4} = \frac{81}{16} cis \frac{5 π}{2} = \frac{81}{16} e^{i π /2}

(b)

3 marks

The product has argument $\frac{3 π}{4} + \frac{nπ}{16} = \frac{( 12 + n ) π}{16}$ . For a positive real number the argument must be a multiple of $2 π$ .
The smallest positive $n$ comes from $k = 1$ .

\frac{12 + n}{16} = 2 k \Rightarrow 12 + n = 32 k n = 20

Question 13

8 marks

A mouse in a maze: transition matrix and long-run time at F

Number and algebra

(a)

3 marks

From each junction, list the paths and their weights, and divide by the total. A has three normal paths. B has a normal path and a sugar path (weights 1 and 2). C and E each have two normal paths. D has two normal paths. F has three normal paths and one sugar path (weights 1, 1, 1, 2).
As a matrix, with the columns showing where the mouse goes from each junction (A, B, C, D, E, F):

A: B, E, F \frac{1}{3} each; B: A \frac{1}{3}, C \frac{2}{3}; C: A, D \frac{1}{2} each D: C, F \frac{1}{2} each; E: C, D \frac{1}{2} each; F: B, C, E \frac{1}{5} each, D \frac{2}{5} 0 \frac{1}{3} 00 \frac{1}{3} \frac{1}{3} \frac{1}{3} 0 \frac{2}{3} 000 \frac{1}{2} 00 \frac{1}{2} 00 00 \frac{1}{2} 00 \frac{1}{2} 00 \frac{1}{2} \frac{1}{2} 00 0 \frac{1}{5} \frac{1}{5} \frac{2}{5} \frac{1}{5} 0

(b)

3 marks

Raise the matrix to a large power on the calculator and read off the steady state. The fraction of time at F is the last entry.
So the mouse spends about 17.3% of the time at F.

(0.157, 0.0868, 0.256, 0.241, 0.0868, 0.173)

(c)

2 marks

Mentor tip

Give one limitation and say how it would change the answer.

Question 14

7 marks

A geometric transformation: image, a pair of fixed numbers, and area

Number and algebra

(a)

2 marks

Apply the matrix and then add the translation.

(72 - 10 - 3) (6 - 2) + (- 5 4) = (6218) + (- 5 4) = (5722)

(b)

3 marks

Setting the image equal to twice the point gives two equations.
Substitute $p = 1 + 2 q$ into the second.

7 p - 10 q - 5 = 2 p \Rightarrow p - 2 q = 1 2 p - 3 q + 4 = 2 q \Rightarrow 2 p - 5 q = - 4 2 + 4 q - 5 q = - 4 \Rightarrow q = 6, p = 13

(c)

2 marks

The area scale factor of the matrix is the absolute value of its determinant.
The area factor is 1, and the translation doesn't change area, so L and its image have the same area.

det (72 - 10 - 3) = - 21 + 20 = - 1, ∣ - 1 ∣ = 1

Question 15

7 marks

A coffee shop discount: a Poisson hypothesis test

Statistics and probability

(a)

1 mark

With a mean of 22 an hour, the mean for 5 hours is 110.

H_{0} : μ = 110, H_{1} : μ > 110

(b)

4 marks

Under $H_{0}$ , $X \sim Po (110)$ . Find the smallest value $c$ with $P (X \geq c) \leq 0.05$ .
So the critical region is $X \geq 129$ , and the probability of a type I error is the size of this region.

P (X \geq 128) \approx 0.0502 > 0.05, P (X \geq 129) \approx 0.0415 < 0.05 P (type I error) \approx 0.0415

(c)

2 marks

Compare the observed 126 with the critical region, or compute its tail probability.
So there is insufficient evidence to reject $H_{0}$ : no evidence that sales have increased.

P (X \geq 126) \approx 0.072 > 0.05

Question 16

8 marks

Euler's method, then an exact solution and the percentage error

Calculus

(a)

3 marks

Euler's method with $h = 0.1$ : $x_{n + 1} = x_{n} + h x_{n} cos t_{n} e^{- s i n t_{n}}$ , starting from $t_{0} = 0$ and $x_{0} = \frac{1}{e}$ .
So $x (0.3) \approx 0.477$ .

x_{1} \approx 0.404667, x_{2} \approx 0.441106, x_{3} \approx 0.476548

(b)

5 marks

Separate the variables and integrate. The right side is the derivative of $- e^{- s i n t}$ .
At $t = 0$ , $x = \frac{1}{e}$ , so $- 1 = - 1 + c$ and $c = 0$ .
The percentage error is the difference between the approximation and the exact value, divided by the exact value.

\int \frac{1}{x} d x = \int cos t e^{- s i n t} d t \Rightarrow ln x = - e^{- s i n t} + c x = e^{- e^{- s i n t}} \Rightarrow x (0.3) \approx 0.475141 \frac{0.476548 - 0.475141}{0.475141} \times 100% \approx 0.296%

Our analysis

Where the marks are

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

Number and algebra41 marks · 37%
Q5, Q7, Q11, Q12, Q13, Q14
Statistics and probability27 marks · 25%
Q3, Q9, Q10, Q15
Geometry and trigonometry20 marks · 18%
Q1, Q4, Q8
Functions14 marks · 13%
Q2, Q6
Calculus8 marks · 7%
Q16

Mentor insights

Traps to watch for

01
Equidistant means on the bisector · Q1
The point equally far from two schools lies on the perpendicular bisector of the segment joining them. The bus stop must also lie on the road, so it is where the two lines meet.
02
Expected counts come from the model · Q9
If the same number sells each day, the expected value for each day is the total divided by 5. The degrees of freedom are the number of categories minus 1, since nothing else was estimated.
03
Add arguments, then reduce the angle · Q12
When you raise to a power or multiply, the angle can leave $(- π, π]$ . Subtract multiples of $2 π$ to bring it back before you give the answer.
04
Sugar doubles the weight · Q13
A path with sugar is twice as likely as a normal path, so give it weight 2 and the others weight 1, then divide by the total weight at that junction.
05
Separate and integrate once · Q16
The equation separates as $\frac{d x}{x} = cos t e^{- s i n t} d t$ , whose right side is the derivative of $- e^{- s i n t}$ . Then the initial condition fixes the constant.

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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