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IB Maths AI HL · Paper 1

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2

Session: May 2023
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 2023 TZ2 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

7 marks

A thrown basketball: maximum height, landing and a catch

Functions

(a)

2 marks

The maximum of a quadratic is at $t = - \frac{b}{2 a}$ .

t = \frac{8.75}{2 ( 4.75 )} = 0.921 s

(b)

2 marks

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

- 4.75 t^{2} + 8.75 t + 1.5 = 0 \Rightarrow t = 2 s

(c)

2 marks

Set $h (t) = 1.2$ .
The other root is negative, so it is not used.

- 4.75 t^{2} + 8.75 t + 0.3 = 0 \Rightarrow t = 1.88 s

(d)

1 mark

The model ignores air resistance and spin, and it predicts negative heights after the ball would reach the ground. It is only valid until the ball hits the ground.

Question 2

4 marks

Is food quality independent of the meal? A chi-squared test

Statistics and probability

(a)

2 marks

Enter the observed table as a matrix and run a $χ^{2}$ test for independence. There are $(3 - 1) (3 - 1) = 4$ degrees of freedom.

χ^{2} \approx 11.0

(b)

2 marks

Compare with the critical value.
Reject $H_{0}$ : there is significant evidence at the 5% level that the quality of the food depends on which meal it is cooked for.

11.0 > 9.488

Question 3

7 marks

Two independent events on a Venn diagram

Statistics and probability

(a)

3 marks

The region for $R$ only has probability $x$ , so $P (R) = x + 0.2$ .

P (R \cap S) = P (R) P (S) \Rightarrow 0.2 = 0.8 (x + 0.2) x + 0.2 = 0.25 \Rightarrow x = 0.05

(b)

2 marks

All the regions add to 1.

0.05 + 0.2 + 0.6 + y = 1 \Rightarrow y = 0.15

(c)

2 marks

The regions outside $S$ are $x$ and $y$ , so $P (S^{'}) = 0.2$ . Then

P (R^{'} ∣ S^{'}) = \frac{P ( R ^{'} \cap S ^{'} )}{P ( S ^{'} )} = \frac{0.15}{0.2} = \frac{3}{4}

Mentor tip

Independence gives the same: $P (R^{'} ∣ S^{'}) = P (R^{'}) = 1 - 0.25 = 0.75$ .

Question 4

6 marks

Savings with quarterly interest or a phone that loses value

Number and algebra

(a)

3 marks

Compound interest with 20 quarters.

A = 520 (1 + \frac{0.012}{4})^{20} = $552.11

(b)

3 marks

The phone's value falls from $520 to $30 in 5 years.

520 (1 - r)^{5} = 30 \Rightarrow 1 - r = (\frac{30}{520})^{1/5} r = 0.435 (43.5%)

Question 5

6 marks

Mango seed lengths: normal probabilities and inverse normal

Statistics and probability

(a)

2 marks

Use $X \sim N (4, 0.2 5^{2})$ .

P (X < 3.7) = 0.115

(b)

2 marks

30% above $k$ means 70% below.

k = 4 + 0.25 (0.5244) \approx 4.13

(c)

2 marks

The interval $4 \pm m$ holds 60%, leaving 20% in each tail, so $P (d < 4 + m) = 0.8$ .

m = 0.25 \times 0.8416 \approx 0.210

Question 6

5 marks

An ice cream scoop and the cone it sits on

Geometry and trigonometry

(a)

3 marks

Volume of the sphere of radius 3.4.
Four fifths of it is not inside the cone.

V = \frac{4}{3} π (3.4)^{3} = 164.6 cm^{3} \frac{4}{5} (164.6) \approx 132 cm^{3}

(b)

2 marks

The outside of the cone is its curved surface, with $r = 3$ and slant height $l = 11$ .

A = π r l = π (3) (11) \approx 104 cm^{2}

Question 7

6 marks

Getting a seat on the bus: binomial expectation and probabilities

Statistics and probability

(a)

1 mark

Let $X \sim B (56, 0.86)$ .

E (X) = 56 \times 0.86 = 48.16

(b)

3 marks

At least 50 is the complement of at most 49.

P (X \geq 50) = 1 - P (X \leq 49) \approx 0.316

(c)

2 marks

Check the cumulative probabilities.
The smallest $n$ is

P (X \leq 45) = 0.1526, P (X \leq 46) = 0.2525 n = 46

Question 8

7 marks

Walks on a road network between villages

Number and algebra

(a)

3 marks

Read each directed road from the diagram: A to C and E, B to C and D, C to B and E, D to A and C, E to D.

M = 0001000100110100100110100

(b)

4 marks

Visiting seven villages after E means a walk of seven roads, so use $M^{7}$ . The entry in the row for E and the column for A is 2.
One route is found by following the roads: E, D, C, B, C, B, D, A.

2 routes E \to D \to C \to B \to C \to B \to D \to A

Mentor tip

Check each step is a road: E to D, D to C, C to B, B to C, C to B, B to D, D to A.

Question 9

9 marks

Age and 100 m time: Spearman's rank correlation and the coefficient of determination

Statistics and probability

(a)

2 marks

Rank the ages with the oldest as 1, and the times with the slowest as 1.

Age rank: 7, 6, 3, 5, 4, 2, 8, 1 Time rank: 3.5, 2, 3.5, 6, 7, 8, 1, 5

(b)

2 marks

Spearman's coefficient is the correlation of the ranks.

r_{s} = - 0.671

(c)

1 mark

The coefficient is negative: older athletes in this sample tend to have faster (shorter) times, although the evidence is not strong with only 8 athletes.

(d)

1 mark

The relationship between age and time is unlikely to be linear (it is also affected by outliers such as the slowest time, 17.0 s, for the youngest athlete), and the data may not come from a bivariate normal distribution.

(e)

3 marks

The coefficient of determination is $r^{2}$ for the regression of time on age.
About 26% of the variation in the times is explained by the athletes' ages.

r^{2} = 0.264

Question 10

6 marks

A chi-squared goodness of fit test for flavour preference

Statistics and probability

(a)

2 marks

The null hypothesis says that each flavour is chosen with probability $\frac{1}{4}$ .

H_{0} : all four flavours are equally preferred, H_{1} : they are not all equally preferred

(b)

4 marks

Each expected frequency is $\frac{246}{4} = 61.5$ .
Compare with the critical value.
Do not reject $H_{0}$ : there is insufficient evidence at the 5% level to say that people prefer some flavours more than others.

χ^{2} = \frac{( 75 - 61.5 ) ^{2} + ( 59 - 61.5 ) ^{2} + ( 46 - 61.5 ) ^{2} + ( 66 - 61.5 ) ^{2}}{61.5} = \frac{449}{61.5} = 7.30 7.30 < 7.82

Question 11

6 marks

Adding two AC voltages with different phases

Functions

(a)

2 marks

Add the two waves as vectors $30∠6 0^{\circ}$ and $60∠1 0^{\circ}$ : the combined amplitude is the length of the sum.

(30 cos 6 0^{\circ} + 60 cos 1 0^{\circ}, 30 sin 6 0^{\circ} + 60 sin 1 0^{\circ}) = (74.09, 36.40) V_{m a x} = 74.0 9^{2} + 36.4 0^{2} \approx 82.5

(b)

4 marks

The angle of the vector gives the phase.

b = arctan \frac{36.40}{74.09} \approx 26. 2^{\circ} V = 82.5 sin (t + 26. 2^{\circ})

Question 12

5 marks

A balloon being inflated: volume and rate of change of the radius

Calculus

(a)

2 marks

Set the volume equal to $288 π$ .

\frac{4}{3} π r^{3} = 288 π \Rightarrow r^{3} = 216 \Rightarrow r = 6 cm

(b)

3 marks

Use the chain rule.

15 = 4 π (6)^{2} \frac{d r}{d t} \Rightarrow \frac{d r}{d t} = \frac{15}{144 π} \approx 0.0332 cm s^{- 1}

Question 13

7 marks

Two transformations undone, and the area of the original triangle

Geometry and trigonometry

(a)

4 marks

The combined transformation is $QP$ .
Its determinant is $- 84 + 6 = - 78$ . The inverse undoes the transformation.

QP = (- 4 1 13) (3012) = (- 12 3 - 2 7) (QP)^{- 1} = \frac{1}{- 78} (7 - 3 2 - 12) = (- \frac{7}{78} \frac{1}{26} - \frac{1}{39} \frac{2}{13})

(b)

3 marks

The area of $T^{'}$ is $∣ det (QP) ∣ = 78$ times the area of $T$ .

Area (T) = \frac{273}{78} = 3.5 cm^{2}

Question 14

8 marks

Two ships: bearing and when one is south or south-east of the other

Geometry and trigonometry

(a)

3 marks

The velocity of A is $2 i - 3 j$ . The angle from south is $arctan \frac{2}{3} = 33. 7^{\circ}$ , towards the east.

bearing = 18 0^{\circ} - 33. 7^{\circ} = 14 6^{\circ}

(b)

2 marks

Equal east coordinates.

1 + 2 t = - 2 + 4 t \Rightarrow t = 1.5

(c)

3 marks

The position of B relative to A is
South-east means the east and south parts are equal: $2 t - 3 = - (4 t - 7)$ .
Check: the east part is $\frac{1}{3} > 0$ and the north part is $- \frac{1}{3} < 0$ .

r_{B} - r_{A} = (2 t - 3) i + (4 t - 7) j 2 t - 3 = 7 - 4 t \Rightarrow t = \frac{5}{3}

Question 15

6 marks

Weights of coffee granules: a confidence interval

Statistics and probability

(a)

3 marks

The sample mean and the unbiased standard deviation of the eight weights.
95% $t$ -interval for the mean with $ν = 7$ .

\overset{x}{ˉ} = 224.25, s_{n - 1} = 2.605 (222.1, 226.4)

(b)

1 mark

The weights are assumed to be normally distributed (and the sample random).

(c)

2 marks

The label claims 226 g. This value is inside the 95% confidence interval $(222.1, 226.4)$ , so the data are consistent with the claim on the label.

Question 16

9 marks

Volume of revolution of a region between cos x and sqrt(x)/2 about the y-axis

Functions
Calculus

(a)

2 marks

Solve $cos x = \frac{x}{2}$ on a graph.

P = (1.04, 0.509)

(b)

5 marks

From $y = \frac{x}{2}$ we get $x = 4 y^{2}$ . From $y = cos x$ we get $x = arccos y$ .
For $0 \leq y \leq 0.509$ the region extends from the $y$ -axis to $x = 4 y^{2}$ . For $0.509 \leq y \leq 1$ it extends to $x = arccos y$ .

V = π \int_{0}^{0.509} (4 y^{2})^{2} d y + π \int_{0.509}^{1} (arccos y)^{2} d y

(c)

2 marks

Evaluate each integral with the calculator.

V = 0.344 + 0.803 \approx 1.15

Question 17

6 marks

A separable equation with a starting singularity

Calculus

(a)

1 mark

At $x = 0$ , $y = 1$ the slope is $\frac{d y}{d x} = \frac{0}{2 ( 1 ) - 2} = \frac{0}{0}$ , which is undefined, so Euler's method cannot start.

(b)

4 marks

Separate the variables.
At $x = 0$ , $y = 1$ : $1 - 2 = c$ , so $c = - 1$ . Complete the square.
Since $y \geq 1$ take the positive root.

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

(c)

1 mark

Substitute $x = 0.1$ into the solution.

y (0.1) = 1 + \frac{ln 1.01}{2} \approx 1.0705

Our analysis

Where the marks are

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

Statistics and probability44 marks · 40%
Q2, Q3, Q5, Q7, Q9, Q10, Q15
Geometry and trigonometry20 marks · 18%
Q6, Q13, Q14
Calculus18 marks · 16%
Q12, Q16, Q17
Functions15 marks · 14%
Q1, Q11, Q16
Number and algebra13 marks · 12%
Q4, Q8

Mentor insights

Traps to watch for

01
Use the quadratic formula for the time · Q1
Reject the negative root, because time starts at 0.
02
Independent events · Q3
$P (R \cap S) = P (R) P (S)$ . The overlap is 0.2 and $P (S) = 0.2 + 0.6 = 0.8$ .
03
Ties share the average rank · Q9
Two athletes with a time of 13.4 s share ranks 3 and 4, so each gets 3.5.
04
Bearing of a velocity · Q14
The velocity of ship A is $(2, - 3)$ : east 2 and south 3, so the bearing is more than 90 degrees.
05
Rotate about the y-axis means integrating in y · Q16
Write each boundary as $x$ in terms of $y$ , and split at the intersection point.

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Worked solutions

IB Maths AI HL · Paper 1

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2

Session: May 2023
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 2023 TZ2 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

7 marks

A thrown basketball: maximum height, landing and a catch

Functions

(a)

2 marks

The maximum of a quadratic is at $t = - \frac{b}{2 a}$ .

t = \frac{8.75}{2 ( 4.75 )} = 0.921 s

(b)

2 marks

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

- 4.75 t^{2} + 8.75 t + 1.5 = 0 \Rightarrow t = 2 s

(c)

2 marks

Set $h (t) = 1.2$ .
The other root is negative, so it is not used.

- 4.75 t^{2} + 8.75 t + 0.3 = 0 \Rightarrow t = 1.88 s

(d)

1 mark

The model ignores air resistance and spin, and it predicts negative heights after the ball would reach the ground. It is only valid until the ball hits the ground.

Question 2

4 marks

Is food quality independent of the meal? A chi-squared test

Statistics and probability

(a)

2 marks

Enter the observed table as a matrix and run a $χ^{2}$ test for independence. There are $(3 - 1) (3 - 1) = 4$ degrees of freedom.

χ^{2} \approx 11.0

(b)

2 marks

Compare with the critical value.
Reject $H_{0}$ : there is significant evidence at the 5% level that the quality of the food depends on which meal it is cooked for.

11.0 > 9.488

Question 3

7 marks

Two independent events on a Venn diagram

Statistics and probability

(a)

3 marks

The region for $R$ only has probability $x$ , so $P (R) = x + 0.2$ .

P (R \cap S) = P (R) P (S) \Rightarrow 0.2 = 0.8 (x + 0.2) x + 0.2 = 0.25 \Rightarrow x = 0.05

(b)

2 marks

All the regions add to 1.

0.05 + 0.2 + 0.6 + y = 1 \Rightarrow y = 0.15

(c)

2 marks

The regions outside $S$ are $x$ and $y$ , so $P (S^{'}) = 0.2$ . Then

P (R^{'} ∣ S^{'}) = \frac{P ( R ^{'} \cap S ^{'} )}{P ( S ^{'} )} = \frac{0.15}{0.2} = \frac{3}{4}

Mentor tip

Independence gives the same: $P (R^{'} ∣ S^{'}) = P (R^{'}) = 1 - 0.25 = 0.75$ .

Question 4

6 marks

Savings with quarterly interest or a phone that loses value

Number and algebra

(a)

3 marks

Compound interest with 20 quarters.

A = 520 (1 + \frac{0.012}{4})^{20} = $552.11

(b)

3 marks

The phone's value falls from $520 to $30 in 5 years.

520 (1 - r)^{5} = 30 \Rightarrow 1 - r = (\frac{30}{520})^{1/5} r = 0.435 (43.5%)

Question 5

6 marks

Mango seed lengths: normal probabilities and inverse normal

Statistics and probability

(a)

2 marks

Use $X \sim N (4, 0.2 5^{2})$ .

P (X < 3.7) = 0.115

(b)

2 marks

30% above $k$ means 70% below.

k = 4 + 0.25 (0.5244) \approx 4.13

(c)

2 marks

The interval $4 \pm m$ holds 60%, leaving 20% in each tail, so $P (d < 4 + m) = 0.8$ .

m = 0.25 \times 0.8416 \approx 0.210

Question 6

5 marks

An ice cream scoop and the cone it sits on

Geometry and trigonometry

(a)

3 marks

Volume of the sphere of radius 3.4.
Four fifths of it is not inside the cone.

V = \frac{4}{3} π (3.4)^{3} = 164.6 cm^{3} \frac{4}{5} (164.6) \approx 132 cm^{3}

(b)

2 marks

The outside of the cone is its curved surface, with $r = 3$ and slant height $l = 11$ .

A = π r l = π (3) (11) \approx 104 cm^{2}

Question 7

6 marks

Getting a seat on the bus: binomial expectation and probabilities

Statistics and probability

(a)

1 mark

Let $X \sim B (56, 0.86)$ .

E (X) = 56 \times 0.86 = 48.16

(b)

3 marks

At least 50 is the complement of at most 49.

P (X \geq 50) = 1 - P (X \leq 49) \approx 0.316

(c)

2 marks

Check the cumulative probabilities.
The smallest $n$ is

P (X \leq 45) = 0.1526, P (X \leq 46) = 0.2525 n = 46

Question 8

7 marks

Walks on a road network between villages

Number and algebra

(a)

3 marks

Read each directed road from the diagram: A to C and E, B to C and D, C to B and E, D to A and C, E to D.

M = 0001000100110100100110100

(b)

4 marks

Visiting seven villages after E means a walk of seven roads, so use $M^{7}$ . The entry in the row for E and the column for A is 2.
One route is found by following the roads: E, D, C, B, C, B, D, A.

2 routes E \to D \to C \to B \to C \to B \to D \to A

Mentor tip

Check each step is a road: E to D, D to C, C to B, B to C, C to B, B to D, D to A.

Question 9

9 marks

Age and 100 m time: Spearman's rank correlation and the coefficient of determination

Statistics and probability

(a)

2 marks

Rank the ages with the oldest as 1, and the times with the slowest as 1.

Age rank: 7, 6, 3, 5, 4, 2, 8, 1 Time rank: 3.5, 2, 3.5, 6, 7, 8, 1, 5

(b)

2 marks

Spearman's coefficient is the correlation of the ranks.

r_{s} = - 0.671

(c)

1 mark

The coefficient is negative: older athletes in this sample tend to have faster (shorter) times, although the evidence is not strong with only 8 athletes.

(d)

1 mark

(e)

3 marks

The coefficient of determination is $r^{2}$ for the regression of time on age.
About 26% of the variation in the times is explained by the athletes' ages.

r^{2} = 0.264

Question 10

6 marks

A chi-squared goodness of fit test for flavour preference

Statistics and probability

(a)

2 marks

The null hypothesis says that each flavour is chosen with probability $\frac{1}{4}$ .

H_{0} : all four flavours are equally preferred, H_{1} : they are not all equally preferred

(b)

4 marks

Each expected frequency is $\frac{246}{4} = 61.5$ .
Compare with the critical value.
Do not reject $H_{0}$ : there is insufficient evidence at the 5% level to say that people prefer some flavours more than others.

χ^{2} = \frac{( 75 - 61.5 ) ^{2} + ( 59 - 61.5 ) ^{2} + ( 46 - 61.5 ) ^{2} + ( 66 - 61.5 ) ^{2}}{61.5} = \frac{449}{61.5} = 7.30 7.30 < 7.82

Question 11

6 marks

Adding two AC voltages with different phases

Functions

(a)

2 marks

Add the two waves as vectors $30∠6 0^{\circ}$ and $60∠1 0^{\circ}$ : the combined amplitude is the length of the sum.

(30 cos 6 0^{\circ} + 60 cos 1 0^{\circ}, 30 sin 6 0^{\circ} + 60 sin 1 0^{\circ}) = (74.09, 36.40) V_{m a x} = 74.0 9^{2} + 36.4 0^{2} \approx 82.5

(b)

4 marks

The angle of the vector gives the phase.

b = arctan \frac{36.40}{74.09} \approx 26. 2^{\circ} V = 82.5 sin (t + 26. 2^{\circ})

Question 12

5 marks

A balloon being inflated: volume and rate of change of the radius

Calculus

(a)

2 marks

Set the volume equal to $288 π$ .

\frac{4}{3} π r^{3} = 288 π \Rightarrow r^{3} = 216 \Rightarrow r = 6 cm

(b)

3 marks

Use the chain rule.

15 = 4 π (6)^{2} \frac{d r}{d t} \Rightarrow \frac{d r}{d t} = \frac{15}{144 π} \approx 0.0332 cm s^{- 1}

Question 13

7 marks

Two transformations undone, and the area of the original triangle

Geometry and trigonometry

(a)

4 marks

The combined transformation is $QP$ .
Its determinant is $- 84 + 6 = - 78$ . The inverse undoes the transformation.

QP = (- 4 1 13) (3012) = (- 12 3 - 2 7) (QP)^{- 1} = \frac{1}{- 78} (7 - 3 2 - 12) = (- \frac{7}{78} \frac{1}{26} - \frac{1}{39} \frac{2}{13})

(b)

3 marks

The area of $T^{'}$ is $∣ det (QP) ∣ = 78$ times the area of $T$ .

Area (T) = \frac{273}{78} = 3.5 cm^{2}

Question 14

8 marks

Two ships: bearing and when one is south or south-east of the other

Geometry and trigonometry

(a)

3 marks

The velocity of A is $2 i - 3 j$ . The angle from south is $arctan \frac{2}{3} = 33. 7^{\circ}$ , towards the east.

bearing = 18 0^{\circ} - 33. 7^{\circ} = 14 6^{\circ}

(b)

2 marks

Equal east coordinates.

1 + 2 t = - 2 + 4 t \Rightarrow t = 1.5

(c)

3 marks

The position of B relative to A is
South-east means the east and south parts are equal: $2 t - 3 = - (4 t - 7)$ .
Check: the east part is $\frac{1}{3} > 0$ and the north part is $- \frac{1}{3} < 0$ .

r_{B} - r_{A} = (2 t - 3) i + (4 t - 7) j 2 t - 3 = 7 - 4 t \Rightarrow t = \frac{5}{3}

Question 15

6 marks

Weights of coffee granules: a confidence interval

Statistics and probability

(a)

3 marks

The sample mean and the unbiased standard deviation of the eight weights.
95% $t$ -interval for the mean with $ν = 7$ .

\overset{x}{ˉ} = 224.25, s_{n - 1} = 2.605 (222.1, 226.4)

(b)

1 mark

The weights are assumed to be normally distributed (and the sample random).

(c)

2 marks

The label claims 226 g. This value is inside the 95% confidence interval $(222.1, 226.4)$ , so the data are consistent with the claim on the label.

Question 16

9 marks

Volume of revolution of a region between cos x and sqrt(x)/2 about the y-axis

Functions
Calculus

(a)

2 marks

Solve $cos x = \frac{x}{2}$ on a graph.

P = (1.04, 0.509)

(b)

5 marks

From $y = \frac{x}{2}$ we get $x = 4 y^{2}$ . From $y = cos x$ we get $x = arccos y$ .
For $0 \leq y \leq 0.509$ the region extends from the $y$ -axis to $x = 4 y^{2}$ . For $0.509 \leq y \leq 1$ it extends to $x = arccos y$ .

V = π \int_{0}^{0.509} (4 y^{2})^{2} d y + π \int_{0.509}^{1} (arccos y)^{2} d y

(c)

2 marks

Evaluate each integral with the calculator.

V = 0.344 + 0.803 \approx 1.15

Question 17

6 marks

A separable equation with a starting singularity

Calculus

(a)

1 mark

At $x = 0$ , $y = 1$ the slope is $\frac{d y}{d x} = \frac{0}{2 ( 1 ) - 2} = \frac{0}{0}$ , which is undefined, so Euler's method cannot start.

(b)

4 marks

Separate the variables.
At $x = 0$ , $y = 1$ : $1 - 2 = c$ , so $c = - 1$ . Complete the square.
Since $y \geq 1$ take the positive root.

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

(c)

1 mark

Substitute $x = 0.1$ into the solution.

y (0.1) = 1 + \frac{ln 1.01}{2} \approx 1.0705

Our analysis

Where the marks are

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

Statistics and probability44 marks · 40%
Q2, Q3, Q5, Q7, Q9, Q10, Q15
Geometry and trigonometry20 marks · 18%
Q6, Q13, Q14
Calculus18 marks · 16%
Q12, Q16, Q17
Functions15 marks · 14%
Q1, Q11, Q16
Number and algebra13 marks · 12%
Q4, Q8

Mentor insights

Traps to watch for

01
Use the quadratic formula for the time · Q1
Reject the negative root, because time starts at 0.
02
Independent events · Q3
$P (R \cap S) = P (R) P (S)$ . The overlap is 0.2 and $P (S) = 0.2 + 0.6 = 0.8$ .
03
Ties share the average rank · Q9
Two athletes with a time of 13.4 s share ranks 3 and 4, so each gets 3.5.
04
Bearing of a velocity · Q14
The velocity of ship A is $(2, - 3)$ : east 2 and south 3, so the bearing is more than 90 degrees.
05
Rotate about the y-axis means integrating in y · Q16
Write each boundary as $x$ in terms of $y$ , and split at the intersection point.

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