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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: 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 2021 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–17 · 110 marks

Question 1

4 marks

Fish caught in a day: a Poisson probability

Number and algebra

4 marks

For 8 hours the mean number of fish is $8 \times 1.1 = 8.8$ , so $X \sim Po (8.8)$ .
"More than 9" is the complement of "at most 9".

P (X > 9) = 1 - P (X \leq 9) \approx 0.386

Mentor tip

Using $P (X \geq 9)$ by mistake gives 0.518. Write the inequality out before entering it.

Question 2

6 marks

Voronoi towers: nearest-tower regions and the edge between two towers

Geometry and trigonometry

(a)

1 mark

Every tower is identical, so the signal is strongest from the nearest one. Every point in the shaded region is closer to T4 than to any other tower.

(b)

2 marks

The edge $y = 3$ is the perpendicular bisector of T2 and T4, so T4 is the reflection of T2 in that line. T2 is 2 units above the line, so T4 is 2 units below it.

T_{4} = (- 9, 1)

(c)

3 marks

First find the gradient of the line joining T1 and T2.
The edge between them is perpendicular to this line, so its gradient is the negative reciprocal.

m = \frac{5 - 3}{- 9 - ( - 13 )} = \frac{1}{2} - 2

Question 3

8 marks

Two training plans: arithmetic against geometric growth

Number and algebra

(a)

5 marks

Charlie's distances are arithmetic, with $u_{1} = 500$ and $d = 100$ .
Daniella's are geometric with first term 500 and common ratio 1.02.

u_{20} = 500 + 19 (100) = 2400 m u_{20} = 500 (1.02)^{19} \approx 728 m

(b)

3 marks

Daniella is ahead on day $n$ when her distance exceeds Charlie's.
Solve on the calculator, by graphing both sides or tabulating values.

500 (1.02)^{n - 1} > 500 + 100 (n - 1) n > 184.2 \Rightarrow n = 185

Mentor tip

The first whole day above the crossing point is 185, not 184.

Question 4

6 marks

Forgetting modelled by an exponential decay

Calculus

(a)

2 marks

After 1 day 50% remains, so $R (1) = 50$ .

100 e^{- p} = 50 \Rightarrow e^{- p} = \frac{1}{2} \Rightarrow p = ln 2 \approx 0.693

(b)

2 marks

36 hours is 1.5 days, since $t$ is in days.

R (1.5) = 100 e^{- 1.5 l n 2} \approx 35.4%

(c)

1 mark

An exponential function is always positive, so $R (t) > 0$ for every $t$ . The graph approaches the axis $R = 0$ but never reaches it.

(d)

1 mark

One valid limit: the model gives $R (0) = 100%$ , which is not realistic immediately at the start, or large values of $t$ are unrealistic because the model keeps predicting some retention forever.

Mentor tip

The limitation has to be about the domain $t \geq 0$ , not about people in general.

Question 5

6 marks

Area of a triangular sunshade from the cross product

Geometry and trigonometry

(a)

2 marks

Subtract coordinates, end point minus start point.

C A = - 3 - 4 - 1, C B = 3 - 4 - 1

(b)

2 marks

Use the determinant rule for the cross product.

C A \times C B = (- 4) (- 1) - (- 1) (- 4) (- 1) (3) - (- 3) (- 1) (- 3) (- 4) - (- 4) (3) = 0 - 6 24

(c)

2 marks

The area of the triangle is half the length of the cross product.

Area = \frac{1}{2} 0^{2} + 6^{2} + 2 4^{2} = \frac{1}{2} 612 \approx 12.4 m^{2}

Question 6

5 marks

The largest area when every measurement is rounded

Number and algebra

5 marks

The sides can be at most 56.5 m and 82.5 m. The angle can be anywhere from $102. 5^{\circ}$ to $107. 5^{\circ}$ . Over that range $sin$ is decreasing (since the angle is past $9 0^{\circ}$ ), so the sine is largest at $102. 5^{\circ}$ .
To three significant figures that is about 2280 m².

Area = \frac{1}{2} (56.5) (82.5) sin 102. 5^{\circ} \approx 2275 m^{2}

Mentor tip

Check by trying both ends: $sin 107. 5^{\circ} \approx 0.954$ is smaller than $sin 102. 5^{\circ} \approx 0.976$ .

Question 7

8 marks

A gift box: area in terms of one variable, then minimised

Calculus

(a)

4 marks

The top is a rectangle of area $8 \times 6 = 48$ plus two right-angled triangles with areas $\frac{1}{2} \cdot 6 \cdot q = 3 q$ and $\frac{1}{2} \cdot 8 \cdot p = 4 p$ .
The two triangles are similar, so their sides are in proportion.
Replace $p$ in the area.

A = 3 q + 4 p + 48 \frac{p}{6} = \frac{8}{q} \Rightarrow pq = 48 \Rightarrow p = \frac{48}{q} A = 3 q + \frac{192}{q} + 48

(b)

2 marks

Differentiate, writing $\frac{192}{q} = 192 q^{- 1}$ .

\frac{d A}{d q} = 3 - \frac{192}{q ^{2}}

(c)

2 marks

For the minimum, set the derivative to zero.

3 - \frac{192}{q ^{2}} = 0 \Rightarrow q^{2} = 64 \Rightarrow q = 8 cm

Mentor tip

$q > 0$ is a length, so reject $- 8$ . The second derivative $\frac{384}{q ^{3}}$ is positive, so this is a minimum.

Question 8

7 marks

The larger of two dice: the distribution, a conditional and the mean

Statistics and probability

(a)

2 marks

The score is $t$ when at least one die shows $t$ and neither shows more. That happens in $2 t - 1$ of the 36 outcomes.

P (T = t) = \frac{2 t - 1}{36} for t = 1, \dots, 6 : \frac{1}{36}, \frac{3}{36}, \frac{5}{36}, \frac{7}{36}, \frac{9}{36}, \frac{11}{36}

(b)

3 marks

(i) "At least 3" has probability $\frac{5 + 7 + 9 + 11}{36}$ .

(ii) A conditional probability: divide by the probability of the condition.

P (T \geq 3) = \frac{32}{36} = \frac{8}{9} P (T = 6 ∣ T \geq 3) = \frac{11/36}{32/36} = \frac{11}{32} \approx 0.344

(c)

2 marks

Multiply each score by its probability and add.

E (T) = \frac{1 ( 1 ) + 2 ( 3 ) + 3 ( 5 ) + 4 ( 7 ) + 5 ( 9 ) + 6 ( 11 )}{36} = \frac{161}{36} \approx 4.47

Question 9

8 marks

A transformation $w = i z + 1$ on the Argand diagram

Number and algebra

(a)

3 marks

(i) For $z = 2 i$ , use $i^{2} = - 1$ .

(ii) For $z = 1 + i$ :

w = i (2 i) + 1 = - 2 + 1 = - 1 w = i (1 + i) + 1 = i - 1 + 1 = i

(b)

3 marks

Multiplying by $i$ rotates the point $9 0^{\circ}$ anticlockwise about the origin. Adding 1 then shifts it one unit to the right.
So the order is: a rotation of $9 0^{\circ}$ anticlockwise about the origin first, and then a translation by $(10)$ .

(c)

2 marks

Solve $w = 2 - i$ for $z$ by undoing the steps in reverse order.

i z = 1 - i \Rightarrow z = \frac{1 - i}{i} = - i (1 - i) = - 1 - i

Question 10

7 marks

A lower bound for a travelling-salesperson route by Prim's algorithm

Number and algebra

(a)

6 marks

Delete A and run Prim's algorithm starting at B. At each step take the shortest edge joining the chosen vertices to a new one.
The weight of the tree is the sum of those edges.
Now add the two shortest edges at A: AB with 55 and AC with 63.

BC 46, BD 58, DE 23, EF 47 46 + 58 + 23 + 47 = 174 lower bound = 174 + 55 + 63 = 292 minutes

(b)

1 mark

Repeat the process after deleting a different vertex, and use the largest lower bound found.

Mentor tip

Each vertex gives a valid lower bound. The best one is the largest, because the true tour is at least as long as all of them.

Question 11

7 marks

Is 1% of the toys faulty? A one-tailed hypothesis test

Statistics and probability

(a)

1 mark

The manager used one box that was easy to get, so this is convenience sampling.

(b)

2 marks

The factory claims the faulty proportion is $p = 0.01$ and the manager suspects it is higher.

H_{0} : p = 0.01, H_{1} : p > 0.01

(c)

2 marks

Under $H_{0}$ the number of faulty toys is $X \sim B (200, 0.01)$ . The $p$ -value is the probability of four or more.

P (X \geq 4) \approx 0.142

(d)

2 marks

Compare the $p$ -value with the significance level.
So there is insufficient evidence to reject $H_{0}$ . The factory's claim of 1% is reasonable.

0.142 > 0.10

Question 12

8 marks

A leaking tank: solving $\frac{d V}{d t} = - k V$

Calculus

(a)

6 marks

Separate the variables and integrate.
At $t = 0$ , $V = 400$ , so $2 \times 20 = c$ and $c = 40$ . At $t = 10$ , $V = 324$ , so $2 \times 18 = - 10 k + 40$ and $k = 0.4$ .
Square both sides, writing $0.2 = \frac{1}{5}$ .

\int V^{- 1/2} d V = - \int k d t \Rightarrow 2 V = - k t + c 2 V = 40 - 0.4 t \Rightarrow V = 20 - 0.2 t V = (20 - \frac{t}{5})^{2}

(b)

2 marks

The tank is empty when $V = 0$ .

20 - \frac{t}{5} = 0 \Rightarrow t = 100 minutes

Question 13

7 marks

A submarine's path to the surface

Geometry and trigonometry

(a)

2 marks

A line is a starting point plus a multiple of the direction vector.

r = 0.8 1.3 - 0.3 + λ - 2 - 3 1

(b)

5 marks

(i) At the surface $z = 0$ , so $- 0.3 + λ = 0$ and $λ = 0.3$ . Substitute into $x$ and $y$ .

(ii) The distance from the ship at the origin is the length of $O P$ .

x = 0.8 - 0.6 = 0.2, y = 1.3 - 0.9 = 0.4 P (0.2, 0.4, 0) OP = 0. 2^{2} + 0. 4^{2} + 0^{2} = 0.2 \approx 0.447 km

Question 14

6 marks

Bags of six apples: the normal distribution of a sum

Statistics and probability

(a)

2 marks

The mean of the total of six apples is six times the mean of one.

6 \times 158 = 948 g

(b)

2 marks

Variances of independent weights add, so the variance of a bag is $6 \times 1 3^{2}$ .

σ = 6 \times 1 3^{2} = 136 \approx 31.8 g

(c)

2 marks

Let $B \sim N (948, 31.8 4^{2})$ be the bag's weight. A kilogram is 1000 g.

P (B > 1000) \approx 0.0512

Question 15

5 marks

Reading a slope field: where the solutions turn

Calculus

(a)

3 marks

A turning point needs $\frac{d y}{d x} = sin (x + y) = 0$ . The minima are where the slope changes from negative to positive; in the picture they lie where $x + y = 0$ .

x + y = 0 \Rightarrow y = - x

(b)

2 marks

The maxima are where $sin (x + y)$ changes from positive to negative, the next zero, at $x + y = π$ .

y = - x + π

Mentor tip

The two lines are parallel, since both have gradient $- 1$ and they differ by $π$ in the intercept.

Question 16

5 marks

An ant walking on a triangular prism frame: adjacency matrix and walks

Number and algebra

(a)

3 marks

Write 1 when two vertices share an edge and 0 otherwise. The edges are AB, AC, AD, BC, BE, CF, DE, DF and EF.

M = 011100101010110001100011010101001110

(b)

2 marks

The number of walks of length 6 from A back to A is the A–A entry of $M^{6}$ , found on the calculator.

(M^{6})_{AA} = 143

Question 17

7 marks

A translated logarithm through two points

Number and algebra

7 marks

The translated function is $g (x) = ln (x - a) + b$ . Using the two points gives two equations.
Subtract the first from the second so that $b$ drops out.
Put $- a = e^{3}$ back into the first equation.

ln (- a) + b = 1, ln (e^{3} - a) + b = 1 + ln 2 ln (e^{3} - a) - ln (- a) = ln 2 \Rightarrow \frac{e ^{3} - a}{- a} = 2 e^{3} - a = - 2 a \Rightarrow a = - e^{3} \approx - 20.1 ln e^{3} + b = 1 \Rightarrow b = 1 - 3 = - 2

Mentor tip

Check the second point: $ln (2 e^{3}) - 2 = ln 2 + 3 - 2 = 1 + ln 2$ .

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 algebra44 marks · 40%
Q1, Q3, Q6, Q9, Q10, Q16, Q17
Calculus27 marks · 25%
Q4, Q7, Q12, Q15
Statistics and probability20 marks · 18%
Q8, Q11, Q14
Geometry and trigonometry19 marks · 17%
Q2, Q5, Q13

Mentor insights

Traps to watch for

01
Day 20 is 19 steps from day 1 · Q3
Day 20 means 19 increases from day 1, not 20. The $n^{th}$ term uses the exponent $n - 1$ .
02
Link $p$ and $q$ first · Q7
The area has two variables. A similar-triangles equation, $pq = 48$ , lets you write it in one variable before you differentiate.
03
Count the outcomes · Q8
For a score of $t$ , count how many of the 36 outcomes have a larger number equal to $t$ . That count is $2 t - 1$ .
04
Multiply first, then add · Q9
The rule applies $\times i$ (a rotation) first and then $+ 1$ (a translation). To reverse it, undo the translation first.
05
Use both conditions · Q12
The two given volumes serve different purposes: $V = 400$ at $t = 0$ finds the constant of integration, and $V = 324$ at $t = 10$ finds $k$ .

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: 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 2021 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–17 · 110 marks

Question 1

4 marks

Fish caught in a day: a Poisson probability

Number and algebra

4 marks

For 8 hours the mean number of fish is $8 \times 1.1 = 8.8$ , so $X \sim Po (8.8)$ .
"More than 9" is the complement of "at most 9".

P (X > 9) = 1 - P (X \leq 9) \approx 0.386

Mentor tip

Using $P (X \geq 9)$ by mistake gives 0.518. Write the inequality out before entering it.

Question 2

6 marks

Voronoi towers: nearest-tower regions and the edge between two towers

Geometry and trigonometry

(a)

1 mark

Every tower is identical, so the signal is strongest from the nearest one. Every point in the shaded region is closer to T4 than to any other tower.

(b)

2 marks

The edge $y = 3$ is the perpendicular bisector of T2 and T4, so T4 is the reflection of T2 in that line. T2 is 2 units above the line, so T4 is 2 units below it.

T_{4} = (- 9, 1)

(c)

3 marks

First find the gradient of the line joining T1 and T2.
The edge between them is perpendicular to this line, so its gradient is the negative reciprocal.

m = \frac{5 - 3}{- 9 - ( - 13 )} = \frac{1}{2} - 2

Question 3

8 marks

Two training plans: arithmetic against geometric growth

Number and algebra

(a)

5 marks

Charlie's distances are arithmetic, with $u_{1} = 500$ and $d = 100$ .
Daniella's are geometric with first term 500 and common ratio 1.02.

u_{20} = 500 + 19 (100) = 2400 m u_{20} = 500 (1.02)^{19} \approx 728 m

(b)

3 marks

Daniella is ahead on day $n$ when her distance exceeds Charlie's.
Solve on the calculator, by graphing both sides or tabulating values.

500 (1.02)^{n - 1} > 500 + 100 (n - 1) n > 184.2 \Rightarrow n = 185

Mentor tip

The first whole day above the crossing point is 185, not 184.

Question 4

6 marks

Forgetting modelled by an exponential decay

Calculus

(a)

2 marks

After 1 day 50% remains, so $R (1) = 50$ .

100 e^{- p} = 50 \Rightarrow e^{- p} = \frac{1}{2} \Rightarrow p = ln 2 \approx 0.693

(b)

2 marks

36 hours is 1.5 days, since $t$ is in days.

R (1.5) = 100 e^{- 1.5 l n 2} \approx 35.4%

(c)

1 mark

An exponential function is always positive, so $R (t) > 0$ for every $t$ . The graph approaches the axis $R = 0$ but never reaches it.

(d)

1 mark

One valid limit: the model gives $R (0) = 100%$ , which is not realistic immediately at the start, or large values of $t$ are unrealistic because the model keeps predicting some retention forever.

Mentor tip

The limitation has to be about the domain $t \geq 0$ , not about people in general.

Question 5

6 marks

Area of a triangular sunshade from the cross product

Geometry and trigonometry

(a)

2 marks

Subtract coordinates, end point minus start point.

C A = - 3 - 4 - 1, C B = 3 - 4 - 1

(b)

2 marks

Use the determinant rule for the cross product.

C A \times C B = (- 4) (- 1) - (- 1) (- 4) (- 1) (3) - (- 3) (- 1) (- 3) (- 4) - (- 4) (3) = 0 - 6 24

(c)

2 marks

The area of the triangle is half the length of the cross product.

Area = \frac{1}{2} 0^{2} + 6^{2} + 2 4^{2} = \frac{1}{2} 612 \approx 12.4 m^{2}

Question 6

5 marks

The largest area when every measurement is rounded

Number and algebra

5 marks

The sides can be at most 56.5 m and 82.5 m. The angle can be anywhere from $102. 5^{\circ}$ to $107. 5^{\circ}$ . Over that range $sin$ is decreasing (since the angle is past $9 0^{\circ}$ ), so the sine is largest at $102. 5^{\circ}$ .
To three significant figures that is about 2280 m².

Area = \frac{1}{2} (56.5) (82.5) sin 102. 5^{\circ} \approx 2275 m^{2}

Mentor tip

Check by trying both ends: $sin 107. 5^{\circ} \approx 0.954$ is smaller than $sin 102. 5^{\circ} \approx 0.976$ .

Question 7

8 marks

A gift box: area in terms of one variable, then minimised

Calculus

(a)

4 marks

The top is a rectangle of area $8 \times 6 = 48$ plus two right-angled triangles with areas $\frac{1}{2} \cdot 6 \cdot q = 3 q$ and $\frac{1}{2} \cdot 8 \cdot p = 4 p$ .
The two triangles are similar, so their sides are in proportion.
Replace $p$ in the area.

A = 3 q + 4 p + 48 \frac{p}{6} = \frac{8}{q} \Rightarrow pq = 48 \Rightarrow p = \frac{48}{q} A = 3 q + \frac{192}{q} + 48

(b)

2 marks

Differentiate, writing $\frac{192}{q} = 192 q^{- 1}$ .

\frac{d A}{d q} = 3 - \frac{192}{q ^{2}}

(c)

2 marks

For the minimum, set the derivative to zero.

3 - \frac{192}{q ^{2}} = 0 \Rightarrow q^{2} = 64 \Rightarrow q = 8 cm

Mentor tip

$q > 0$ is a length, so reject $- 8$ . The second derivative $\frac{384}{q ^{3}}$ is positive, so this is a minimum.

Question 8

7 marks

The larger of two dice: the distribution, a conditional and the mean

Statistics and probability

(a)

2 marks

The score is $t$ when at least one die shows $t$ and neither shows more. That happens in $2 t - 1$ of the 36 outcomes.

P (T = t) = \frac{2 t - 1}{36} for t = 1, \dots, 6 : \frac{1}{36}, \frac{3}{36}, \frac{5}{36}, \frac{7}{36}, \frac{9}{36}, \frac{11}{36}

(b)

3 marks

(i) "At least 3" has probability $\frac{5 + 7 + 9 + 11}{36}$ .

(ii) A conditional probability: divide by the probability of the condition.

P (T \geq 3) = \frac{32}{36} = \frac{8}{9} P (T = 6 ∣ T \geq 3) = \frac{11/36}{32/36} = \frac{11}{32} \approx 0.344

(c)

2 marks

Multiply each score by its probability and add.

E (T) = \frac{1 ( 1 ) + 2 ( 3 ) + 3 ( 5 ) + 4 ( 7 ) + 5 ( 9 ) + 6 ( 11 )}{36} = \frac{161}{36} \approx 4.47

Question 9

8 marks

A transformation $w = i z + 1$ on the Argand diagram

Number and algebra

(a)

3 marks

(i) For $z = 2 i$ , use $i^{2} = - 1$ .

(ii) For $z = 1 + i$ :

w = i (2 i) + 1 = - 2 + 1 = - 1 w = i (1 + i) + 1 = i - 1 + 1 = i

(b)

3 marks

Multiplying by $i$ rotates the point $9 0^{\circ}$ anticlockwise about the origin. Adding 1 then shifts it one unit to the right.
So the order is: a rotation of $9 0^{\circ}$ anticlockwise about the origin first, and then a translation by $(10)$ .

(c)

2 marks

Solve $w = 2 - i$ for $z$ by undoing the steps in reverse order.

i z = 1 - i \Rightarrow z = \frac{1 - i}{i} = - i (1 - i) = - 1 - i

Question 10

7 marks

A lower bound for a travelling-salesperson route by Prim's algorithm

Number and algebra

(a)

6 marks

Delete A and run Prim's algorithm starting at B. At each step take the shortest edge joining the chosen vertices to a new one.
The weight of the tree is the sum of those edges.
Now add the two shortest edges at A: AB with 55 and AC with 63.

BC 46, BD 58, DE 23, EF 47 46 + 58 + 23 + 47 = 174 lower bound = 174 + 55 + 63 = 292 minutes

(b)

1 mark

Repeat the process after deleting a different vertex, and use the largest lower bound found.

Mentor tip

Each vertex gives a valid lower bound. The best one is the largest, because the true tour is at least as long as all of them.

Question 11

7 marks

Is 1% of the toys faulty? A one-tailed hypothesis test

Statistics and probability

(a)

1 mark

The manager used one box that was easy to get, so this is convenience sampling.

(b)

2 marks

The factory claims the faulty proportion is $p = 0.01$ and the manager suspects it is higher.

H_{0} : p = 0.01, H_{1} : p > 0.01

(c)

2 marks

Under $H_{0}$ the number of faulty toys is $X \sim B (200, 0.01)$ . The $p$ -value is the probability of four or more.

P (X \geq 4) \approx 0.142

(d)

2 marks

Compare the $p$ -value with the significance level.
So there is insufficient evidence to reject $H_{0}$ . The factory's claim of 1% is reasonable.

0.142 > 0.10

Question 12

8 marks

A leaking tank: solving $\frac{d V}{d t} = - k V$

Calculus

(a)

6 marks

Separate the variables and integrate.
At $t = 0$ , $V = 400$ , so $2 \times 20 = c$ and $c = 40$ . At $t = 10$ , $V = 324$ , so $2 \times 18 = - 10 k + 40$ and $k = 0.4$ .
Square both sides, writing $0.2 = \frac{1}{5}$ .

\int V^{- 1/2} d V = - \int k d t \Rightarrow 2 V = - k t + c 2 V = 40 - 0.4 t \Rightarrow V = 20 - 0.2 t V = (20 - \frac{t}{5})^{2}

(b)

2 marks

The tank is empty when $V = 0$ .

20 - \frac{t}{5} = 0 \Rightarrow t = 100 minutes

Question 13

7 marks

A submarine's path to the surface

Geometry and trigonometry

(a)

2 marks

A line is a starting point plus a multiple of the direction vector.

r = 0.8 1.3 - 0.3 + λ - 2 - 3 1

(b)

5 marks

(i) At the surface $z = 0$ , so $- 0.3 + λ = 0$ and $λ = 0.3$ . Substitute into $x$ and $y$ .

(ii) The distance from the ship at the origin is the length of $O P$ .

x = 0.8 - 0.6 = 0.2, y = 1.3 - 0.9 = 0.4 P (0.2, 0.4, 0) OP = 0. 2^{2} + 0. 4^{2} + 0^{2} = 0.2 \approx 0.447 km

Question 14

6 marks

Bags of six apples: the normal distribution of a sum

Statistics and probability

(a)

2 marks

The mean of the total of six apples is six times the mean of one.

6 \times 158 = 948 g

(b)

2 marks

Variances of independent weights add, so the variance of a bag is $6 \times 1 3^{2}$ .

σ = 6 \times 1 3^{2} = 136 \approx 31.8 g

(c)

2 marks

Let $B \sim N (948, 31.8 4^{2})$ be the bag's weight. A kilogram is 1000 g.

P (B > 1000) \approx 0.0512

Question 15

5 marks

Reading a slope field: where the solutions turn

Calculus

(a)

3 marks

A turning point needs $\frac{d y}{d x} = sin (x + y) = 0$ . The minima are where the slope changes from negative to positive; in the picture they lie where $x + y = 0$ .

x + y = 0 \Rightarrow y = - x

(b)

2 marks

The maxima are where $sin (x + y)$ changes from positive to negative, the next zero, at $x + y = π$ .

y = - x + π

Mentor tip

The two lines are parallel, since both have gradient $- 1$ and they differ by $π$ in the intercept.

Question 16

5 marks

An ant walking on a triangular prism frame: adjacency matrix and walks

Number and algebra

(a)

3 marks

Write 1 when two vertices share an edge and 0 otherwise. The edges are AB, AC, AD, BC, BE, CF, DE, DF and EF.

M = 011100101010110001100011010101001110

(b)

2 marks

The number of walks of length 6 from A back to A is the A–A entry of $M^{6}$ , found on the calculator.

(M^{6})_{AA} = 143

Question 17

7 marks

A translated logarithm through two points

Number and algebra

7 marks

The translated function is $g (x) = ln (x - a) + b$ . Using the two points gives two equations.
Subtract the first from the second so that $b$ drops out.
Put $- a = e^{3}$ back into the first equation.

ln (- a) + b = 1, ln (e^{3} - a) + b = 1 + ln 2 ln (e^{3} - a) - ln (- a) = ln 2 \Rightarrow \frac{e ^{3} - a}{- a} = 2 e^{3} - a = - 2 a \Rightarrow a = - e^{3} \approx - 20.1 ln e^{3} + b = 1 \Rightarrow b = 1 - 3 = - 2

Mentor tip

Check the second point: $ln (2 e^{3}) - 2 = ln 2 + 3 - 2 = 1 + ln 2$ .

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 algebra44 marks · 40%
Q1, Q3, Q6, Q9, Q10, Q16, Q17
Calculus27 marks · 25%
Q4, Q7, Q12, Q15
Statistics and probability20 marks · 18%
Q8, Q11, Q14
Geometry and trigonometry19 marks · 17%
Q2, Q5, Q13

Mentor insights

Traps to watch for

01
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02
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03
Count the outcomes · Q8
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04
Multiply first, then add · Q9
The rule applies $\times i$ (a rotation) first and then $+ 1$ (a translation). To reverse it, undo the translation first.
05
Use both conditions · Q12
The two given volumes serve different purposes: $V = 400$ at $t = 0$ finds the constant of integration, and $V = 324$ at $t = 10$ finds $k$ .

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