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IB Maths AI HL · Paper 1

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2

Session: May 2024
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 15

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

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2024 TZ2 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–15 · 110 marks

Question 1

9 marks

Savings, then a car loan: compound interest and repayments

Number and algebra

(a)

3 marks

Six years is 72 months.

3000 (1 + \frac{0.0125}{12})^{72} \approx $3233.53

(b)

2 marks

Solve $3000 (1 + \frac{0.0125}{12})^{N} = 3550$ for $N$ .

N \approx 161.7 \Rightarrow 162 months

(c)

1 mark

The car costs $22 000 and the partial payment is $3550.

22000 - 3550 = $18 450

(d)

3 marks

Use the finance solver with $N = 96$ , $I = 12.6%$ , $P V = 18450$ , $F V = 0$ , 12 payments a year.

P M T \approx $306

Question 2

6 marks

Flawed chocolates in a batch: binomial probabilities and expected income

Statistics and probability

(a)

4 marks

(i) Let $X \sim B (20, 0.04)$ .

(ii) More than two flawed.

P (X = 2) = (2 20) (0.04)^{2} (0.96)^{18} \approx 0.146 P (X > 2) = 1 - P (X \leq 2) \approx 0.0439

(b)

2 marks

The expected numbers are $20 (0.96)$ perfect and $20 (0.04)$ flawed.

50 (20 \times 0.96) + 15 (20 \times 0.04) = 960 + 12 = 972 pesos

Question 3

7 marks

The pH scale: a logarithm and its inverse

Functions

(a)

2 marks

Substitute.

pH = - lo g_{10} (0.0003) \approx 3.52

(b)

2 marks

Rearrange for the concentration.

[H^{+}] = 1 0^{- 6.6} \approx 2.51 \times 1 0^{- 7} moles per litre

(c)

3 marks

The concentrations are $1 0^{- 2}$ for the lemon and $1 0^{- 4.5}$ for the tomato. Divide.
The acid in a lemon is about 316 times stronger.

\frac{1 0 ^{- 2}}{1 0 ^{- 4.5}} = 1 0^{2.5} \approx 316

Question 4

7 marks

A coin and a die: final points, a table and the expected value

Statistics and probability

(a)

2 marks

A score of 7 only happens with a head and a 6.

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

(b)

3 marks

A score of 1 needs tails and a 1. A score of 7 needs heads and a 6. Each of 2 to 6 can come from tails and that value, or heads and one less.
Check: the seven probabilities add to $\frac{1}{12} + 5 (\frac{1}{6}) + \frac{1}{12} = 1$ .

P (1) = \frac{1}{12}, P (7) = \frac{1}{12} P (2) = P (3) = P (4) = P (5) = P (6) = \frac{1}{12} + \frac{1}{12} = \frac{1}{6}

(c)

2 marks

Add each score times its probability.
Check: the die has mean 3.5 and heads adds 1 half of the time, so $3.5 + 0.5 = 4$ .

E = 1 (\frac{1}{12}) + (2 + 3 + 4 + 5 + 6) (\frac{1}{6}) + 7 (\frac{1}{12}) = \frac{8}{12} + \frac{20}{6} = 4

Question 5

8 marks

A tree's yearly growth: a geometric sequence and its limit

Number and algebra

(a)

2 marks

The growth over the past year was $42 - 37 = 5$ m, and the growth in the coming year is 80% of this.

0.8 \times 5 = 4 m

(b)

4 marks

The growths in the next 6 years are a geometric series with $a = 4$ and $r = 0.8$ .

total growth = \frac{4 ( 1 - 0. 8 ^{6} )}{1 - 0.8} \approx 14.757 height = 42 + 14.757 = 56.757 \approx 56.76 m

(c)

2 marks

The height can never exceed the current height plus the sum to infinity of the growths.

42 + \frac{4}{1 - 0.8} = 42 + 20 = 62 k = 62

Question 6

6 marks

Triangle field against sector field: which is larger

Number and algebra

6 marks

Area of the sheep's field (triangle $A B C$ ).
The goat's sector has angle $28 2^{\circ}$ and radius 8.
The goat's field is larger, by

\frac{1}{2} (15) (21) sin 7 8^{\circ} \approx 154.06 m^{2} \frac{282}{360} π (8)^{2} \approx 157.50 m^{2} 157.50 - 154.06 \approx 3.44 m^{2}

Question 7

4 marks

Euler's method for dy/dx = xy - 1

Number and algebra

4 marks

Start at $(1, 2)$ and take five steps of 0.1 to reach $x = 1.5$ .
For example, $y (1.1) = 2 + 0.1 (1 \times 2 - 1) = 2.1$ and $y (1.2) = 2.1 + 0.1 (1.1 \times 2.1 - 1) = 2.231$ .

x : 1, 1.1, 1.2, 1.3, 1.4, 1.5 y : 2, 2.1, 2.231, 2.3987, 2.6106, 2.8760 y (1.5) \approx 2.88

Question 8

10 marks

A quadrilateral in 3D: angle, area and a parallelogram check

Number and algebra

(a)

1 mark

Subtract coordinates.

A D = 2 - 1 4 - 3 7 - 5 = 112

(b)

7 marks

(i) $A B = (3, 4, 0)$ , with $∣ A B ∣ = 5$ , and $∣ A D ∣ = 6$ . Use the scalar product.

(ii) Use the vector product.

cos B A D = \frac{3 + 4 + 0}{5 6} = \frac{7}{5 6} \Rightarrow B A D \approx 55. 1^{\circ} A B \times A D = 4 \cdot 2 - 0 \cdot 1 0 \cdot 1 - 3 \cdot 2 3 \cdot 1 - 4 \cdot 1 = 8 - 6 - 1 Area = \frac{1}{2} 64 + 36 + 1 = \frac{101}{2} \approx 5.02

(c)

2 marks

Find the other two sides.
Both pairs of opposite sides are equal and parallel, so $A B C D$ is a parallelogram.

B C = 112 = A D, D C = 340 = A B

Question 9

6 marks

Traffic accidents: Poisson probabilities and the right model for 8 weeks

Statistics and probability

(a)

4 marks

(i) Let $X \sim P o (0.76)$ be the accidents in a week.

(ii) In a 4-week period the mean is $4 \times 0.76 = 3.04$ , so $F \sim P o (3.04)$ .

P (X \geq 2) = 1 - P (X \leq 1) \approx 0.177 P (F = 3) \approx 0.224

(b)

2 marks

Each of the 8 weeks is an independent trial, and the success is that fewer than 2 accidents occur that week, which has probability $P (X \leq 1) \approx 0.823$ . The number of successful weeks follows a binomial distribution.

B (8, 0.823)

Question 10

9 marks

Volume of revolution of x sqrt(3 - x^2) and after a pair of stretches

Functions

(a)

2 marks

$f (x) = x 3 - x^{2}$ on $- 3 \leq x \leq 3$ is an odd function with zeros at $x = 0$ and $x = \pm 3$ .
Sketch: a curve through the origin that rises to a maximum at about $(1.22, 1.5)$ and returns to the axis at $(3, 0)$ with a vertical tangent. The left half is the same shape turned through $18 0^{\circ}$ about the origin: a minimum at $(- 1.22, - 1.5)$ and ending at $(- 3, 0)$ .

(b)

4 marks

(i) The area between the curve and the axis is rotated, so the volume is

(ii) The integrand is even, so double the integral from 0.

V = π \int_{- 3}^{3} (x 3 - x^{2})^{2} d x = π \int_{- 3}^{3} x^{2} (3 - x^{2}) d x V = 2 π [x^{3} - \frac{x ^{5}}{5}]_{0}^{3} = 2 π (33 - \frac{9 3}{5}) = \frac{12 3 π}{5} \approx 13.1

(c)

3 marks

The new function is $g (x) = \frac{1}{2} f (\frac{x}{2})$ , on $- 23 \leq x \leq 23$ .

V_{g} = π \int_{- 23}^{23} \frac{1}{4} f (\frac{x}{2})^{2} d x = \frac{1}{4} π \cdot 2 \int_{- 3}^{3} f (u)^{2} d u = \frac{1}{2} V V_{g} = \frac{6 3 π}{5} \approx 6.53

Question 11

9 marks

The normal at A_k on a parabola and its second intersection

Calculus

(a)

6 marks

The curve is $y = k x^{2} - x$ with $A_{k} (\frac{1}{k}, 0)$ . Differentiate.
The normal has gradient $- 1$ and passes through $A_{k}$ .
Set the normal equal to the curve.
The solution $x = \frac{1}{k}$ is $A_{k}$ . So the other intersection $B_{k}$ has $x = - \frac{1}{k}$ .

\frac{d y}{d x} = 2 k x - 1 \Rightarrow at x = \frac{1}{k}, \frac{d y}{d x} = 1 y = - (x - \frac{1}{k}) k x^{2} - x = - x + \frac{1}{k} \Rightarrow k x^{2} = \frac{1}{k} \Rightarrow k^{2} x^{2} - 1 = 0 (k x - 1) (k x + 1) = 0 \Rightarrow x = \frac{1}{k} or x = - \frac{1}{k}

(b)

3 marks

For $k = 2$ the curve is $y = 2 x^{2} - x$ and the normal is $y = - x + \frac{1}{2}$ , with the intersections at $x = - \frac{1}{2}$ and $x = \frac{1}{2}$ . The normal is above the curve between them.

Area = \int_{- 1/2}^{1/2} (- x + \frac{1}{2} - 2 x^{2} + x) d x = \int_{- 1/2}^{1/2} (\frac{1}{2} - 2 x^{2}) d x [\frac{x}{2} - \frac{2 x ^{3}}{3}]_{- 1/2}^{1/2} = 2 (\frac{1}{4} - \frac{1}{12}) = \frac{1}{3}

Question 12

7 marks

A duck and a hawk: speed, a flight path and an interception

Number and algebra

(a)

2 marks

The speed is the length of the velocity vector $(6, 6, 3)$ .

6^{2} + 6^{2} + 3^{2} = 81 = 9 m s^{- 1}

(b)

1 mark

The hawk starts at $(- 38, 134, 315)$ with constant velocity $(15, - 20, - 60)$ .

r_{h} = - 38 134315 + t 15 - 20 - 60

(c)

4 marks

Equate the $x$ -components of the duck and the hawk.
Check the other components at $t = 5$ : duck $y = 4 + 30 = 34$ and hawk $y = 134 - 100 = 34$ ; duck $z = 15$ and hawk $z = 315 - 300 = 15$ .
The position vector of the interception is

7 + 6 t = - 38 + 15 t \Rightarrow 9 t = 45 \Rightarrow t = 5 373415

Question 13

8 marks

A particle with velocity 3 sin t (1 + cos t): rest, maximum speed and average speed

Calculus

(a)

2 marks

Set $v = 3 sin t (1 + cos t) = 0$ for $t > 0$ .

sin t = 0 or cos t = - 1 \Rightarrow a = π \approx 3.14

(b)

2 marks

On $0 \leq t \leq π$ the velocity is non-negative. Find the maximum from a sketch of $v$ or by solving $\frac{d v}{d t} = 0$ .

v_{m a x} \approx 3.90 m s^{- 1}

(c)

4 marks

The velocity is never negative between 0 and $π$ , so the distance is the integral of $v$ .
Average speed is distance over time.

d = \int_{0}^{π} 3 sin t (1 + cos t) d t = 3 [- cos t + \frac{1}{2} sin^{2} t]_{0}^{π} = 3 (1 + 1) = 6 m \frac{6}{π} \approx 1.91 m s^{- 1}

Question 14

6 marks

A cooling cup of water: a power model against an exponential model

Number and algebra

(a)

1 mark

As $t$ grows, $T$ approaches 25. So 25 is the room temperature, the temperature the water cools towards.

(b)

3 marks

Subtract 25 from each temperature (34, 27, 21, 15, 10, 4) and fit a power regression $a t^{b}$ .

a \approx 244, b \approx - 1.03

(c)

1 mark

Fit an exponential regression $k c^{t}$ to the same adjusted data.

k \approx 61.1, c \approx 0.923

(d)

1 mark

The exponential model has a larger coefficient of determination ( $r^{2} \approx 0.923$ compared with $0.766$ for the power model, on the data with 25 taken off), so it fits the data better. (Also, the power model gives an impossibly large temperature as $t \to 0$ .)

Question 15

8 marks

z = -1 + i: modulus-argument form, a spiral similarity and powers

Number and algebra

(a)

2 marks

The modulus is $2$ and, since $z$ is in the second quadrant, the argument is $π - \frac{π}{4}$ .

z = 2 e^{i 3 π /4}

(b)

5 marks

(i) $z^{2} = 2 e^{i 3 π /2}$ , so going from $z$ to $z^{2}$ multiplies the modulus by $2$ and adds $\frac{3 π}{4}$ to the argument. The mapping is an enlargement with scale factor $2$ and a rotation through $\frac{3 π}{4}$ anticlockwise, both about the origin.

(ii) Multiply the matrices of the two transformations.

(2002) (cos \frac{3 π}{4} sin \frac{3 π}{4} - sin \frac{3 π}{4} cos \frac{3 π}{4}) = 2 (- \frac{2}{2} \frac{2}{2} - \frac{2}{2} - \frac{2}{2}) (- 1 1 - 1 - 1)

(c)

1 mark

$z^{n}$ has argument $\frac{3 nπ}{4}$ , which must be a multiple of $2 π$ for a positive real number.

\frac{3 n}{4} = 2 k \Rightarrow n = 8

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 algebra58 marks · 53%
Q1, Q5, Q6, Q7, Q8, Q12, Q14, Q15
Statistics and probability19 marks · 17%
Q2, Q4, Q9
Calculus17 marks · 15%
Q11, Q13
Functions16 marks · 15%
Q3, Q10

Mentor insights

Traps to watch for

01
Nominal rate · Q1
A nominal annual rate of 1.25% compounded monthly is $\frac{1.25}{12} %$ per month.
02
Growth is geometric · Q5
Each year's growth is 0.8 times the last. The heights are the starting height plus a sum of growths.
03
Vectors from A · Q8
Use $A B$ and $A D$ for the angle and the area.
04
Volume about the x-axis · Q10
$V = π \int f (x)^{2} d x$ , so the square root disappears.
05
Gradient of the normal · Q11
At $A_{k}$ the curve has gradient 1, so the normal has gradient $- 1$ .

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 2024
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 15

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

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2024 TZ2 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–15 · 110 marks

Question 1

9 marks

Savings, then a car loan: compound interest and repayments

Number and algebra

(a)

3 marks

Six years is 72 months.

3000 (1 + \frac{0.0125}{12})^{72} \approx $3233.53

(b)

2 marks

Solve $3000 (1 + \frac{0.0125}{12})^{N} = 3550$ for $N$ .

N \approx 161.7 \Rightarrow 162 months

(c)

1 mark

The car costs $22 000 and the partial payment is $3550.

22000 - 3550 = $18 450

(d)

3 marks

Use the finance solver with $N = 96$ , $I = 12.6%$ , $P V = 18450$ , $F V = 0$ , 12 payments a year.

P M T \approx $306

Question 2

6 marks

Flawed chocolates in a batch: binomial probabilities and expected income

Statistics and probability

(a)

4 marks

(i) Let $X \sim B (20, 0.04)$ .

(ii) More than two flawed.

P (X = 2) = (2 20) (0.04)^{2} (0.96)^{18} \approx 0.146 P (X > 2) = 1 - P (X \leq 2) \approx 0.0439

(b)

2 marks

The expected numbers are $20 (0.96)$ perfect and $20 (0.04)$ flawed.

50 (20 \times 0.96) + 15 (20 \times 0.04) = 960 + 12 = 972 pesos

Question 3

7 marks

The pH scale: a logarithm and its inverse

Functions

(a)

2 marks

Substitute.

pH = - lo g_{10} (0.0003) \approx 3.52

(b)

2 marks

Rearrange for the concentration.

[H^{+}] = 1 0^{- 6.6} \approx 2.51 \times 1 0^{- 7} moles per litre

(c)

3 marks

The concentrations are $1 0^{- 2}$ for the lemon and $1 0^{- 4.5}$ for the tomato. Divide.
The acid in a lemon is about 316 times stronger.

\frac{1 0 ^{- 2}}{1 0 ^{- 4.5}} = 1 0^{2.5} \approx 316

Question 4

7 marks

A coin and a die: final points, a table and the expected value

Statistics and probability

(a)

2 marks

A score of 7 only happens with a head and a 6.

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

(b)

3 marks

A score of 1 needs tails and a 1. A score of 7 needs heads and a 6. Each of 2 to 6 can come from tails and that value, or heads and one less.
Check: the seven probabilities add to $\frac{1}{12} + 5 (\frac{1}{6}) + \frac{1}{12} = 1$ .

P (1) = \frac{1}{12}, P (7) = \frac{1}{12} P (2) = P (3) = P (4) = P (5) = P (6) = \frac{1}{12} + \frac{1}{12} = \frac{1}{6}

(c)

2 marks

Add each score times its probability.
Check: the die has mean 3.5 and heads adds 1 half of the time, so $3.5 + 0.5 = 4$ .

E = 1 (\frac{1}{12}) + (2 + 3 + 4 + 5 + 6) (\frac{1}{6}) + 7 (\frac{1}{12}) = \frac{8}{12} + \frac{20}{6} = 4

Question 5

8 marks

A tree's yearly growth: a geometric sequence and its limit

Number and algebra

(a)

2 marks

The growth over the past year was $42 - 37 = 5$ m, and the growth in the coming year is 80% of this.

0.8 \times 5 = 4 m

(b)

4 marks

The growths in the next 6 years are a geometric series with $a = 4$ and $r = 0.8$ .

total growth = \frac{4 ( 1 - 0. 8 ^{6} )}{1 - 0.8} \approx 14.757 height = 42 + 14.757 = 56.757 \approx 56.76 m

(c)

2 marks

The height can never exceed the current height plus the sum to infinity of the growths.

42 + \frac{4}{1 - 0.8} = 42 + 20 = 62 k = 62

Question 6

6 marks

Triangle field against sector field: which is larger

Number and algebra

6 marks

Area of the sheep's field (triangle $A B C$ ).
The goat's sector has angle $28 2^{\circ}$ and radius 8.
The goat's field is larger, by

\frac{1}{2} (15) (21) sin 7 8^{\circ} \approx 154.06 m^{2} \frac{282}{360} π (8)^{2} \approx 157.50 m^{2} 157.50 - 154.06 \approx 3.44 m^{2}

Question 7

4 marks

Euler's method for dy/dx = xy - 1

Number and algebra

4 marks

Start at $(1, 2)$ and take five steps of 0.1 to reach $x = 1.5$ .
For example, $y (1.1) = 2 + 0.1 (1 \times 2 - 1) = 2.1$ and $y (1.2) = 2.1 + 0.1 (1.1 \times 2.1 - 1) = 2.231$ .

x : 1, 1.1, 1.2, 1.3, 1.4, 1.5 y : 2, 2.1, 2.231, 2.3987, 2.6106, 2.8760 y (1.5) \approx 2.88

Question 8

10 marks

A quadrilateral in 3D: angle, area and a parallelogram check

Number and algebra

(a)

1 mark

Subtract coordinates.

A D = 2 - 1 4 - 3 7 - 5 = 112

(b)

7 marks

(i) $A B = (3, 4, 0)$ , with $∣ A B ∣ = 5$ , and $∣ A D ∣ = 6$ . Use the scalar product.

(ii) Use the vector product.

cos B A D = \frac{3 + 4 + 0}{5 6} = \frac{7}{5 6} \Rightarrow B A D \approx 55. 1^{\circ} A B \times A D = 4 \cdot 2 - 0 \cdot 1 0 \cdot 1 - 3 \cdot 2 3 \cdot 1 - 4 \cdot 1 = 8 - 6 - 1 Area = \frac{1}{2} 64 + 36 + 1 = \frac{101}{2} \approx 5.02

(c)

2 marks

Find the other two sides.
Both pairs of opposite sides are equal and parallel, so $A B C D$ is a parallelogram.

B C = 112 = A D, D C = 340 = A B

Question 9

6 marks

Traffic accidents: Poisson probabilities and the right model for 8 weeks

Statistics and probability

(a)

4 marks

(i) Let $X \sim P o (0.76)$ be the accidents in a week.

(ii) In a 4-week period the mean is $4 \times 0.76 = 3.04$ , so $F \sim P o (3.04)$ .

P (X \geq 2) = 1 - P (X \leq 1) \approx 0.177 P (F = 3) \approx 0.224

(b)

2 marks

B (8, 0.823)

Question 10

9 marks

Volume of revolution of x sqrt(3 - x^2) and after a pair of stretches

Functions

(a)

2 marks

$f (x) = x 3 - x^{2}$ on $- 3 \leq x \leq 3$ is an odd function with zeros at $x = 0$ and $x = \pm 3$ .
Sketch: a curve through the origin that rises to a maximum at about $(1.22, 1.5)$ and returns to the axis at $(3, 0)$ with a vertical tangent. The left half is the same shape turned through $18 0^{\circ}$ about the origin: a minimum at $(- 1.22, - 1.5)$ and ending at $(- 3, 0)$ .

(b)

4 marks

(i) The area between the curve and the axis is rotated, so the volume is

(ii) The integrand is even, so double the integral from 0.

V = π \int_{- 3}^{3} (x 3 - x^{2})^{2} d x = π \int_{- 3}^{3} x^{2} (3 - x^{2}) d x V = 2 π [x^{3} - \frac{x ^{5}}{5}]_{0}^{3} = 2 π (33 - \frac{9 3}{5}) = \frac{12 3 π}{5} \approx 13.1

(c)

3 marks

The new function is $g (x) = \frac{1}{2} f (\frac{x}{2})$ , on $- 23 \leq x \leq 23$ .

V_{g} = π \int_{- 23}^{23} \frac{1}{4} f (\frac{x}{2})^{2} d x = \frac{1}{4} π \cdot 2 \int_{- 3}^{3} f (u)^{2} d u = \frac{1}{2} V V_{g} = \frac{6 3 π}{5} \approx 6.53

Question 11

9 marks

The normal at A_k on a parabola and its second intersection

Calculus

(a)

6 marks

The curve is $y = k x^{2} - x$ with $A_{k} (\frac{1}{k}, 0)$ . Differentiate.
The normal has gradient $- 1$ and passes through $A_{k}$ .
Set the normal equal to the curve.
The solution $x = \frac{1}{k}$ is $A_{k}$ . So the other intersection $B_{k}$ has $x = - \frac{1}{k}$ .

\frac{d y}{d x} = 2 k x - 1 \Rightarrow at x = \frac{1}{k}, \frac{d y}{d x} = 1 y = - (x - \frac{1}{k}) k x^{2} - x = - x + \frac{1}{k} \Rightarrow k x^{2} = \frac{1}{k} \Rightarrow k^{2} x^{2} - 1 = 0 (k x - 1) (k x + 1) = 0 \Rightarrow x = \frac{1}{k} or x = - \frac{1}{k}

(b)

3 marks

Area = \int_{- 1/2}^{1/2} (- x + \frac{1}{2} - 2 x^{2} + x) d x = \int_{- 1/2}^{1/2} (\frac{1}{2} - 2 x^{2}) d x [\frac{x}{2} - \frac{2 x ^{3}}{3}]_{- 1/2}^{1/2} = 2 (\frac{1}{4} - \frac{1}{12}) = \frac{1}{3}

Question 12

7 marks

A duck and a hawk: speed, a flight path and an interception

Number and algebra

(a)

2 marks

The speed is the length of the velocity vector $(6, 6, 3)$ .

6^{2} + 6^{2} + 3^{2} = 81 = 9 m s^{- 1}

(b)

1 mark

The hawk starts at $(- 38, 134, 315)$ with constant velocity $(15, - 20, - 60)$ .

r_{h} = - 38 134315 + t 15 - 20 - 60

(c)

4 marks

Equate the $x$ -components of the duck and the hawk.
Check the other components at $t = 5$ : duck $y = 4 + 30 = 34$ and hawk $y = 134 - 100 = 34$ ; duck $z = 15$ and hawk $z = 315 - 300 = 15$ .
The position vector of the interception is

7 + 6 t = - 38 + 15 t \Rightarrow 9 t = 45 \Rightarrow t = 5 373415

Question 13

8 marks

A particle with velocity 3 sin t (1 + cos t): rest, maximum speed and average speed

Calculus

(a)

2 marks

Set $v = 3 sin t (1 + cos t) = 0$ for $t > 0$ .

sin t = 0 or cos t = - 1 \Rightarrow a = π \approx 3.14

(b)

2 marks

On $0 \leq t \leq π$ the velocity is non-negative. Find the maximum from a sketch of $v$ or by solving $\frac{d v}{d t} = 0$ .

v_{m a x} \approx 3.90 m s^{- 1}

(c)

4 marks

The velocity is never negative between 0 and $π$ , so the distance is the integral of $v$ .
Average speed is distance over time.

d = \int_{0}^{π} 3 sin t (1 + cos t) d t = 3 [- cos t + \frac{1}{2} sin^{2} t]_{0}^{π} = 3 (1 + 1) = 6 m \frac{6}{π} \approx 1.91 m s^{- 1}

Question 14

6 marks

A cooling cup of water: a power model against an exponential model

Number and algebra

(a)

1 mark

As $t$ grows, $T$ approaches 25. So 25 is the room temperature, the temperature the water cools towards.

(b)

3 marks

Subtract 25 from each temperature (34, 27, 21, 15, 10, 4) and fit a power regression $a t^{b}$ .

a \approx 244, b \approx - 1.03

(c)

1 mark

Fit an exponential regression $k c^{t}$ to the same adjusted data.

k \approx 61.1, c \approx 0.923

(d)

1 mark

Question 15

8 marks

z = -1 + i: modulus-argument form, a spiral similarity and powers

Number and algebra

(a)

2 marks

The modulus is $2$ and, since $z$ is in the second quadrant, the argument is $π - \frac{π}{4}$ .

z = 2 e^{i 3 π /4}

(b)

5 marks

(ii) Multiply the matrices of the two transformations.

(2002) (cos \frac{3 π}{4} sin \frac{3 π}{4} - sin \frac{3 π}{4} cos \frac{3 π}{4}) = 2 (- \frac{2}{2} \frac{2}{2} - \frac{2}{2} - \frac{2}{2}) (- 1 1 - 1 - 1)

(c)

1 mark

$z^{n}$ has argument $\frac{3 nπ}{4}$ , which must be a multiple of $2 π$ for a positive real number.

\frac{3 n}{4} = 2 k \Rightarrow n = 8

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 algebra58 marks · 53%
Q1, Q5, Q6, Q7, Q8, Q12, Q14, Q15
Statistics and probability19 marks · 17%
Q2, Q4, Q9
Calculus17 marks · 15%
Q11, Q13
Functions16 marks · 15%
Q3, Q10

Mentor insights

Traps to watch for

01
Nominal rate · Q1
A nominal annual rate of 1.25% compounded monthly is $\frac{1.25}{12} %$ per month.
02
Growth is geometric · Q5
Each year's growth is 0.8 times the last. The heights are the starting height plus a sum of growths.
03
Vectors from A · Q8
Use $A B$ and $A D$ for the angle and the area.
04
Volume about the x-axis · Q10
$V = π \int f (x)^{2} d x$ , so the square root disappears.
05
Gradient of the normal · Q11
At $A_{k}$ the curve has gradient 1, so the normal has gradient $- 1$ .

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