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IB Maths AI HL · Paper 1

Mathematics: Applications & Interpretation, Higher Level

Session: November 2023
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 November 2023 Global 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 seashell's growth modelled by a geometric sequence

Number and algebra

(a)

2 marks

The monthly increases form a geometric sequence with $u_{1} = 4$ and $r = 0.8$ .

u_{3} = 4 (0.8)^{2} = 2.56 mm

(b)

2 marks

Sum of the first 12 terms.

S_{12} = \frac{4 ( 1 - 0. 8 ^{12} )}{1 - 0.8} \approx 18.6 mm

(c)

3 marks

The total possible increase is the sum to infinity.
Add the starting width of 25 mm.

S_{\infty} = \frac{4}{1 - 0.8} = 20 mm 25 + 20 = 45 mm

Question 2

7 marks

A function from a graph, its inverse and a linear inverse

Functions

(a)

1 mark

Read the value of $f$ at $x = 2$ from the graph.

f (2) = 4

(b)

2 marks

Reflect the graph of $f$ in the line $y = x$ : each point $(a, b)$ becomes $(b, a)$ .
Sketch: the inverse passes through $(0, 0)$ and $(4, 2)$ (from $(0, 0)$ and $(2, 4)$ on $f$ ) and through $(5, 5)$ , as a reflection of the graph in $y = x$ .

(c)

2 marks

Write $y = 3 x - 1$ , swap $x$ and $y$ and solve.

x = 3 y - 1 \Rightarrow g^{- 1} (x) = \frac{x + 1}{3}

(d)

2 marks

Look for a point where the graph of $f^{- 1}$ meets the line $y = \frac{x + 1}{3}$ . At $x = 2$ , $g^{- 1} (2) = \frac{3}{3} = 1$ , and the graph of $f$ passes through $(1, 2)$ , so $f^{- 1} (2) = 1$ .

x = 2

Question 3

6 marks

The Great Pyramid: bounds on a conversion factor and a volume

Number and algebra

(a)

2 marks

A value of 0.52 correct to two decimal places lies between 0.515 and 0.525.

lower bound 0.515 m, upper bound 0.525 m

(b)

4 marks

The smallest volume uses 1 REC $= 0.515$ m.
Volume of a square-based pyramid.

base = 440 \times 0.515 = 226.6 m, height = 280 \times 0.515 = 144.2 m V = \frac{1}{3} (226.6)^{2} (144.2) \approx 2.47 \times 1 0^{6} m^{3}

Question 4

6 marks

A boat's journey with bearings: an angle and the return distance

Geometry and trigonometry

(a)

2 marks

At B the north line gives alternate angles: $18 0^{\circ} - 11 2^{\circ} = 6 8^{\circ}$ on one side and $4 1^{\circ}$ on the other.

A B C = 68 + 41 = 10 9^{\circ}

(b)

4 marks

C is due east of A, so the angle at A between AB (bearing $11 2^{\circ}$ ) and AC (bearing $09 0^{\circ}$ ) is $2 2^{\circ}$ . The angles in the triangle give
Use the sine rule with AC opposite the angle at B.

A C B = 180 - 109 - 22 = 4 9^{\circ} \frac{A C}{sin 10 9 ^{\circ}} = \frac{100}{sin 4 9 ^{\circ}} \Rightarrow A C \approx 125 km

Question 5

9 marks

A quadratic model through three points and a fence

Calculus

(a)

1 mark

The point $(3, 2.36)$ is on the curve.

2.36 = 9 a + 3 b + c

(b)

3 marks

Do the same for the other points.
Solve the three equations.

5 = 100 a + 10 b + c, 7.16 = 289 a + 17 b + c a = - 0.00490, b = 0.441, c = 1.08 f (x) = - 0.00490 x^{2} + 0.441 x + 1.08

(c)

3 marks

Evaluate the model at the fence, 80 m away.
Since $5.00 > 4$ , the ball is above the 4 m fence when it reaches it, so it goes over.

f (80) = - 0.0049 (6400) + 0.441 (80) + 1.08 = 5.00

(d)

2 marks

Solve $f (x) = 0$ and take the positive root.

- 0.0049 x^{2} + 0.441 x + 1.08 = 0 \Rightarrow x \approx 92.4 m

Question 6

8 marks

Voronoi diagrams of farmhouses: a bisector and the best place for a turbine

Geometry and trigonometry

(a)

1 mark

Read the point where the three boundaries meet on Diagram 1.

(4, 8)

(b)

3 marks

Midpoint of $A (0, 3)$ and $C (8, 13)$ is $(4, 8)$ . The gradient of AC is $\frac{13 - 3}{8 - 0} = \frac{5}{4}$ , so the bisector has gradient $- \frac{4}{5}$ .

y - 8 = - \frac{4}{5} (x - 4) \Rightarrow y = - 0.8 x + 11.2

(c)

4 marks

Find the distance from each vertex to its nearest farmhouses.
The vertex with the largest distance is $(\frac{9}{11}, \frac{116}{11})$ , so the turbine goes there, and the distance to the nearest farmhouse is

(4, 1) : to A 4^{2} + 2^{2} = 20 = 4.47 (11, 8) : to B or D 34 = 5.83 (\frac{9}{11}, \frac{116}{11}) : to A (\frac{9}{11})^{2} + (\frac{116}{11} - 3)^{2} = 7.59 7.59 km

Question 7

7 marks

A coin and a die game: a tree diagram and a conditional probability

Statistics and probability

(a)

2 marks

The missing probabilities: tails $\frac{1}{2}$ ; on tails the die gives a win with probability $\frac{2}{6} = \frac{1}{3}$ and no win with probability $\frac{2}{3}$ .

(b)

2 marks

No prize needs tails and then not a five or six.

P (no win) = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}

(c)

3 marks

Use conditional probability. Winning happens with heads ( $\frac{1}{2}$ ) or tails then five or six ( $\frac{1}{2} \times \frac{1}{3}$ ).

P (heads ∣ win) = \frac{\frac{1}{2}}{\frac{1}{2} + \frac{1}{2} \times \frac{1}{3}} = \frac{\frac{1}{2}}{\frac{2}{3}} = \frac{3}{4}

Question 8

6 marks

Polar form of sqrt 3 - i and the imaginary part of a sum

Number and algebra

(a)

2 marks

The modulus and argument of $z = 3 - i$ .

r = 3 + 1 = 2, θ = - arctan \frac{1}{3} = - \frac{π}{6} z = 2 e^{- iπ /6}

(b)

4 marks

With $z_{1} = e^{2 t i}$ and $z_{2} = 2 e^{i (2 t - π /6)}$ , factor out $e^{2 t i}$ .
Find the modulus and argument of the bracket.
Take the imaginary part.

z_{1} + z_{2} = e^{2 t i} (1 + 2 e^{- iπ /6}) 1 + 2 e^{- iπ /6} = 1 + 3 - i = 2.91 e^{- 0.351 i} Im (z_{1} + z_{2}) = 2.91 sin (2 t - 0.351)

Question 9

8 marks

Waves at a point: a cosine model, derivatives and the steepest rise

Functions
Calculus

(a)

2 marks

The maximum is 3.5 and the minimum 0.5.

p = \frac{3.5 - 0.5}{2} = 1.5, q = \frac{3.5 + 0.5}{2} = 2

(b)

3 marks

Differentiate $h (t) = 1.5 cos \frac{π t}{6} + 2$ with the chain rule.

h^{'} (t) = - \frac{π}{4} sin \frac{π t}{6} = - 0.785 sin \frac{π t}{6} h^{''} (t) = - \frac{π ^{2}}{24} cos \frac{π t}{6} = - 0.411 cos \frac{π t}{6}

(c)

3 marks

The rate of change $h^{'} (t)$ is largest when $h^{''} (t) = 0$ and the slope of $h^{'}$ is positive, so $\frac{π t}{6} = \frac{3 π}{2}$ .
The height at that time.

k = 9 h (9) = 1.5 cos \frac{3 π}{2} + 2 = 2 m

Question 10

7 marks

Decay of an isotope: power regression and linearising with logs

Functions

(a)

2 marks

Use power regression on the data.

a = 1000, b = - 0.600

(b)

2 marks

Do linear regression of $y$ on $x$ in Table 2.

y = - 0.600 x + 6.908

(c)

3 marks

Since $x = ln t$ and $y = ln M$ :
Take the exponential of both sides.
This is $M = 1000 t^{- 0.6}$ , the same as the power regression (with $a \approx 1000$ and $b = - 0.6$ ).

ln M = - 0.600 ln t + 6.908 M = e^{6.908} \times t^{- 0.600} = 1000.2 t^{- 0.6}

Question 11

7 marks

Area, a transformed graph and the volume of a solid of revolution

Calculus
Functions

(a)

2 marks

The region is below the axis on $0 < x < 1$ and above on $- 1 < x < 0$ .

Area = \int_{- 1}^{1} x^{3} - x d x = 2 \int_{0}^{1} (x - x^{3}) d x 2 [\frac{x ^{2}}{2} - \frac{x ^{4}}{4}]_{0}^{1} = 2 (\frac{1}{4}) = 0.5

(b)

2 marks

Shift the graph 1 unit to the right (the zeros move to $x = 0, 1, 2$ ) and double all the heights.
Sketch: a cubic shape on $0 \leq x \leq 2$ through $(0, 0)$ , $(1, 0)$ and $(2, 0)$ , rising to a maximum near $(0.42, 0.77)$ and falling to a minimum near $(1.58, - 0.77)$ .

(c)

3 marks

Volume of revolution about the $x$ -axis.

V = π \int_{- 1}^{1} (x^{3} - x)^{2} d x = \frac{16 π}{105} \approx 0.479

Question 12

4 marks

A differential equation with a tangent and a solution curve

Calculus

(a)

2 marks

At $(- 1, - 1)$ the gradient is
The tangent line:

\frac{d y}{d x} = 3 (- 1) - (- 1) + 1 = - 1 y + 1 = - (x + 1) \Rightarrow x + y + 2 = 0

(b)

2 marks

Start at $(- 1, - 1)$ with gradient $- 1$ and follow the slope marks.
The exact solution through the point is $y = 3 x - 2 + \frac{4}{e} e^{- x}$ . Sketch: the curve falls through $(- 1, - 1)$ with gradient $- 1$ , reaches a minimum near $(- 0.7, - 1.1)$ , and then rises with gradient close to 3, following the line $y = 3 x - 2$ . To the left of $x = - 1$ it rises steeply.

Question 13

8 marks

A particle with a piecewise velocity

Calculus

(a)

2 marks

Set the two pieces equal.
Since $T > 0$ :

2 T^{2} - T^{3} = 8 - 4 T \Rightarrow T^{3} - 2 T^{2} - 4 T + 8 = 0 \Rightarrow (T - 2)^{2} (T + 2) = 0 T = 2

(b)

2 marks

Differentiate each piece.
The derivatives are equal.

u_{1}^{'} (t) = 4 t - 3 t^{2} \Rightarrow u_{1}^{'} (2) = 8 - 12 = - 4 u_{2}^{'} (t) = - 4 \Rightarrow u_{2}^{'} (2) = - 4

(c)

4 marks

The displacement at time $k > 2$ is the integral of the velocity. Set it equal to zero.

\int_{0}^{2} (2 t^{2} - t^{3}) d t + \int_{2}^{k} (8 - 4 t) d t = 0 \frac{4}{3} + (8 k - 2 k^{2} - 8) = 0 \Rightarrow 3 k^{2} - 12 k + 10 = 0 k = 2 + \frac{6}{3} \approx 2.82 s

Question 14

7 marks

The point on a line nearest to a point, and a vector perpendicular to two directions

Geometry and trigonometry

(a)

4 marks

A general point of $L$ is $(1 + λ, 3 + λ, 2 λ)$ . The vector from $Q (11, - 1, 3)$ to it is
This is perpendicular to the direction $(1, 1, 2)$ at the closest point.

QX = λ - 10 λ + 4 2 λ - 3 (λ - 10) + (λ + 4) + 2 (2 λ - 3) = 6 λ - 12 = 0 \Rightarrow λ = 2 P = (3, 5, 4)

(b)

3 marks

Find $P Q$ and take the vector product with the direction of $L$ .
Any multiple of this vector is perpendicular to both $L$ and the line $P Q$ .

P Q = 8 - 6 - 1, 8 - 6 - 1 \times 112 = - 11 - 17 14

Question 15

7 marks

Vegetarian or not: a Markov chain with its eigenvalues and eigenvectors

Statistics and probability

(a)

1 mark

The matrix is $T = (0.8 0.2 0.1 0.9)$ with the first row and column for vegetarians. The entry 0.9 is the probability that a non-vegetarian remains a non-vegetarian the following month.

(b)

3 marks

Solve $det (T - λ I) = 0$ .

(0.8 - λ) (0.9 - λ) - 0.1 (0.2) = λ^{2} - 1.7 λ + 0.7 = 0 λ = 1 or λ = 0.7

(c)

3 marks

The given eigenvector $(1, - 1)$ belongs to $λ = 0.7$ . For $λ = 1$ solve $(T - I) v = 0$ .
In the long term the ratio of vegetarians to non-vegetarians is $1 : 2$ : one third of the students will be vegetarian and two thirds will not.

- 0.2 a + 0.1 b = 0 \Rightarrow b = 2 a \Rightarrow v = (12)

Question 16

6 marks

Basketball shots: Poisson probabilities with a binomial success rate

Statistics and probability

(a)

2 marks

In 12 minutes $X \sim Po (5)$ .

P (X < 7) = P (X \leq 6) \approx 0.762

(b)

4 marks

For 6 minutes the number of shots $N \sim Po (2.5)$ . Given $N = n$ , the chance of at least one success is $1 - 0. 6^{n}$ .

P = n = 1 \sum 3 P (N = n) (1 - 0. 6^{n}) P = P (1) (0.4) + P (2) (0.64) + P (3) (0.784) \approx 0.414

Our analysis

Where the marks are

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

Calculus32 marks · 29%
Q5, Q9, Q11, Q12, Q13
Geometry and trigonometry21 marks · 19%
Q4, Q6, Q14
Statistics and probability20 marks · 18%
Q7, Q15, Q16
Number and algebra19 marks · 17%
Q1, Q3, Q8
Functions18 marks · 16%
Q2, Q9, Q10, Q11

Mentor insights

Traps to watch for

01
The sum to infinity gives the limit · Q1
The total increase over all months is $\frac{u _{1}}{1 - r}$ . Add the starting width.
02
Three points give three equations · Q5
Substitute each point into $f (x) = a x^{2} + b x + c$ and solve the system.
03
Bisector of AC · Q6
Use the midpoint and the negative reciprocal gradient.
04
Amplitude and midline · Q9
Half the difference between the maximum 3.5 and minimum 0.5, and their mean.
05
Join smoothly · Q13
At $t = T$ both the values and the derivatives of the two pieces agree.

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

Session: November 2023
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 November 2023 Global 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 seashell's growth modelled by a geometric sequence

Number and algebra

(a)

2 marks

The monthly increases form a geometric sequence with $u_{1} = 4$ and $r = 0.8$ .

u_{3} = 4 (0.8)^{2} = 2.56 mm

(b)

2 marks

Sum of the first 12 terms.

S_{12} = \frac{4 ( 1 - 0. 8 ^{12} )}{1 - 0.8} \approx 18.6 mm

(c)

3 marks

The total possible increase is the sum to infinity.
Add the starting width of 25 mm.

S_{\infty} = \frac{4}{1 - 0.8} = 20 mm 25 + 20 = 45 mm

Question 2

7 marks

A function from a graph, its inverse and a linear inverse

Functions

(a)

1 mark

Read the value of $f$ at $x = 2$ from the graph.

f (2) = 4

(b)

2 marks

Reflect the graph of $f$ in the line $y = x$ : each point $(a, b)$ becomes $(b, a)$ .
Sketch: the inverse passes through $(0, 0)$ and $(4, 2)$ (from $(0, 0)$ and $(2, 4)$ on $f$ ) and through $(5, 5)$ , as a reflection of the graph in $y = x$ .

(c)

2 marks

Write $y = 3 x - 1$ , swap $x$ and $y$ and solve.

x = 3 y - 1 \Rightarrow g^{- 1} (x) = \frac{x + 1}{3}

(d)

2 marks

x = 2

Question 3

6 marks

The Great Pyramid: bounds on a conversion factor and a volume

Number and algebra

(a)

2 marks

A value of 0.52 correct to two decimal places lies between 0.515 and 0.525.

lower bound 0.515 m, upper bound 0.525 m

(b)

4 marks

The smallest volume uses 1 REC $= 0.515$ m.
Volume of a square-based pyramid.

base = 440 \times 0.515 = 226.6 m, height = 280 \times 0.515 = 144.2 m V = \frac{1}{3} (226.6)^{2} (144.2) \approx 2.47 \times 1 0^{6} m^{3}

Question 4

6 marks

A boat's journey with bearings: an angle and the return distance

Geometry and trigonometry

(a)

2 marks

At B the north line gives alternate angles: $18 0^{\circ} - 11 2^{\circ} = 6 8^{\circ}$ on one side and $4 1^{\circ}$ on the other.

A B C = 68 + 41 = 10 9^{\circ}

(b)

4 marks

C is due east of A, so the angle at A between AB (bearing $11 2^{\circ}$ ) and AC (bearing $09 0^{\circ}$ ) is $2 2^{\circ}$ . The angles in the triangle give
Use the sine rule with AC opposite the angle at B.

A C B = 180 - 109 - 22 = 4 9^{\circ} \frac{A C}{sin 10 9 ^{\circ}} = \frac{100}{sin 4 9 ^{\circ}} \Rightarrow A C \approx 125 km

Question 5

9 marks

A quadratic model through three points and a fence

Calculus

(a)

1 mark

The point $(3, 2.36)$ is on the curve.

2.36 = 9 a + 3 b + c

(b)

3 marks

Do the same for the other points.
Solve the three equations.

5 = 100 a + 10 b + c, 7.16 = 289 a + 17 b + c a = - 0.00490, b = 0.441, c = 1.08 f (x) = - 0.00490 x^{2} + 0.441 x + 1.08

(c)

3 marks

Evaluate the model at the fence, 80 m away.
Since $5.00 > 4$ , the ball is above the 4 m fence when it reaches it, so it goes over.

f (80) = - 0.0049 (6400) + 0.441 (80) + 1.08 = 5.00

(d)

2 marks

Solve $f (x) = 0$ and take the positive root.

- 0.0049 x^{2} + 0.441 x + 1.08 = 0 \Rightarrow x \approx 92.4 m

Question 6

8 marks

Voronoi diagrams of farmhouses: a bisector and the best place for a turbine

Geometry and trigonometry

(a)

1 mark

Read the point where the three boundaries meet on Diagram 1.

(4, 8)

(b)

3 marks

Midpoint of $A (0, 3)$ and $C (8, 13)$ is $(4, 8)$ . The gradient of AC is $\frac{13 - 3}{8 - 0} = \frac{5}{4}$ , so the bisector has gradient $- \frac{4}{5}$ .

y - 8 = - \frac{4}{5} (x - 4) \Rightarrow y = - 0.8 x + 11.2

(c)

4 marks

Find the distance from each vertex to its nearest farmhouses.
The vertex with the largest distance is $(\frac{9}{11}, \frac{116}{11})$ , so the turbine goes there, and the distance to the nearest farmhouse is

(4, 1) : to A 4^{2} + 2^{2} = 20 = 4.47 (11, 8) : to B or D 34 = 5.83 (\frac{9}{11}, \frac{116}{11}) : to A (\frac{9}{11})^{2} + (\frac{116}{11} - 3)^{2} = 7.59 7.59 km

Question 7

7 marks

A coin and a die game: a tree diagram and a conditional probability

Statistics and probability

(a)

2 marks

The missing probabilities: tails $\frac{1}{2}$ ; on tails the die gives a win with probability $\frac{2}{6} = \frac{1}{3}$ and no win with probability $\frac{2}{3}$ .

(b)

2 marks

No prize needs tails and then not a five or six.

P (no win) = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}

(c)

3 marks

Use conditional probability. Winning happens with heads ( $\frac{1}{2}$ ) or tails then five or six ( $\frac{1}{2} \times \frac{1}{3}$ ).

P (heads ∣ win) = \frac{\frac{1}{2}}{\frac{1}{2} + \frac{1}{2} \times \frac{1}{3}} = \frac{\frac{1}{2}}{\frac{2}{3}} = \frac{3}{4}

Question 8

6 marks

Polar form of sqrt 3 - i and the imaginary part of a sum

Number and algebra

(a)

2 marks

The modulus and argument of $z = 3 - i$ .

r = 3 + 1 = 2, θ = - arctan \frac{1}{3} = - \frac{π}{6} z = 2 e^{- iπ /6}

(b)

4 marks

With $z_{1} = e^{2 t i}$ and $z_{2} = 2 e^{i (2 t - π /6)}$ , factor out $e^{2 t i}$ .
Find the modulus and argument of the bracket.
Take the imaginary part.

z_{1} + z_{2} = e^{2 t i} (1 + 2 e^{- iπ /6}) 1 + 2 e^{- iπ /6} = 1 + 3 - i = 2.91 e^{- 0.351 i} Im (z_{1} + z_{2}) = 2.91 sin (2 t - 0.351)

Question 9

8 marks

Waves at a point: a cosine model, derivatives and the steepest rise

Functions
Calculus

(a)

2 marks

The maximum is 3.5 and the minimum 0.5.

p = \frac{3.5 - 0.5}{2} = 1.5, q = \frac{3.5 + 0.5}{2} = 2

(b)

3 marks

Differentiate $h (t) = 1.5 cos \frac{π t}{6} + 2$ with the chain rule.

h^{'} (t) = - \frac{π}{4} sin \frac{π t}{6} = - 0.785 sin \frac{π t}{6} h^{''} (t) = - \frac{π ^{2}}{24} cos \frac{π t}{6} = - 0.411 cos \frac{π t}{6}

(c)

3 marks

The rate of change $h^{'} (t)$ is largest when $h^{''} (t) = 0$ and the slope of $h^{'}$ is positive, so $\frac{π t}{6} = \frac{3 π}{2}$ .
The height at that time.

k = 9 h (9) = 1.5 cos \frac{3 π}{2} + 2 = 2 m

Question 10

7 marks

Decay of an isotope: power regression and linearising with logs

Functions

(a)

2 marks

Use power regression on the data.

a = 1000, b = - 0.600

(b)

2 marks

Do linear regression of $y$ on $x$ in Table 2.

y = - 0.600 x + 6.908

(c)

3 marks

Since $x = ln t$ and $y = ln M$ :
Take the exponential of both sides.
This is $M = 1000 t^{- 0.6}$ , the same as the power regression (with $a \approx 1000$ and $b = - 0.6$ ).

ln M = - 0.600 ln t + 6.908 M = e^{6.908} \times t^{- 0.600} = 1000.2 t^{- 0.6}

Question 11

7 marks

Area, a transformed graph and the volume of a solid of revolution

Calculus
Functions

(a)

2 marks

The region is below the axis on $0 < x < 1$ and above on $- 1 < x < 0$ .

Area = \int_{- 1}^{1} x^{3} - x d x = 2 \int_{0}^{1} (x - x^{3}) d x 2 [\frac{x ^{2}}{2} - \frac{x ^{4}}{4}]_{0}^{1} = 2 (\frac{1}{4}) = 0.5

(b)

2 marks

Shift the graph 1 unit to the right (the zeros move to $x = 0, 1, 2$ ) and double all the heights.
Sketch: a cubic shape on $0 \leq x \leq 2$ through $(0, 0)$ , $(1, 0)$ and $(2, 0)$ , rising to a maximum near $(0.42, 0.77)$ and falling to a minimum near $(1.58, - 0.77)$ .

(c)

3 marks

Volume of revolution about the $x$ -axis.

V = π \int_{- 1}^{1} (x^{3} - x)^{2} d x = \frac{16 π}{105} \approx 0.479

Question 12

4 marks

A differential equation with a tangent and a solution curve

Calculus

(a)

2 marks

At $(- 1, - 1)$ the gradient is
The tangent line:

\frac{d y}{d x} = 3 (- 1) - (- 1) + 1 = - 1 y + 1 = - (x + 1) \Rightarrow x + y + 2 = 0

(b)

2 marks

Start at $(- 1, - 1)$ with gradient $- 1$ and follow the slope marks.
The exact solution through the point is $y = 3 x - 2 + \frac{4}{e} e^{- x}$ . Sketch: the curve falls through $(- 1, - 1)$ with gradient $- 1$ , reaches a minimum near $(- 0.7, - 1.1)$ , and then rises with gradient close to 3, following the line $y = 3 x - 2$ . To the left of $x = - 1$ it rises steeply.

Question 13

8 marks

A particle with a piecewise velocity

Calculus

(a)

2 marks

Set the two pieces equal.
Since $T > 0$ :

2 T^{2} - T^{3} = 8 - 4 T \Rightarrow T^{3} - 2 T^{2} - 4 T + 8 = 0 \Rightarrow (T - 2)^{2} (T + 2) = 0 T = 2

(b)

2 marks

Differentiate each piece.
The derivatives are equal.

u_{1}^{'} (t) = 4 t - 3 t^{2} \Rightarrow u_{1}^{'} (2) = 8 - 12 = - 4 u_{2}^{'} (t) = - 4 \Rightarrow u_{2}^{'} (2) = - 4

(c)

4 marks

The displacement at time $k > 2$ is the integral of the velocity. Set it equal to zero.

\int_{0}^{2} (2 t^{2} - t^{3}) d t + \int_{2}^{k} (8 - 4 t) d t = 0 \frac{4}{3} + (8 k - 2 k^{2} - 8) = 0 \Rightarrow 3 k^{2} - 12 k + 10 = 0 k = 2 + \frac{6}{3} \approx 2.82 s

Question 14

7 marks

The point on a line nearest to a point, and a vector perpendicular to two directions

Geometry and trigonometry

(a)

4 marks

A general point of $L$ is $(1 + λ, 3 + λ, 2 λ)$ . The vector from $Q (11, - 1, 3)$ to it is
This is perpendicular to the direction $(1, 1, 2)$ at the closest point.

QX = λ - 10 λ + 4 2 λ - 3 (λ - 10) + (λ + 4) + 2 (2 λ - 3) = 6 λ - 12 = 0 \Rightarrow λ = 2 P = (3, 5, 4)

(b)

3 marks

Find $P Q$ and take the vector product with the direction of $L$ .
Any multiple of this vector is perpendicular to both $L$ and the line $P Q$ .

P Q = 8 - 6 - 1, 8 - 6 - 1 \times 112 = - 11 - 17 14

Question 15

7 marks

Vegetarian or not: a Markov chain with its eigenvalues and eigenvectors

Statistics and probability

(a)

1 mark

The matrix is $T = (0.8 0.2 0.1 0.9)$ with the first row and column for vegetarians. The entry 0.9 is the probability that a non-vegetarian remains a non-vegetarian the following month.

(b)

3 marks

Solve $det (T - λ I) = 0$ .

(0.8 - λ) (0.9 - λ) - 0.1 (0.2) = λ^{2} - 1.7 λ + 0.7 = 0 λ = 1 or λ = 0.7

(c)

3 marks

The given eigenvector $(1, - 1)$ belongs to $λ = 0.7$ . For $λ = 1$ solve $(T - I) v = 0$ .
In the long term the ratio of vegetarians to non-vegetarians is $1 : 2$ : one third of the students will be vegetarian and two thirds will not.

- 0.2 a + 0.1 b = 0 \Rightarrow b = 2 a \Rightarrow v = (12)

Question 16

6 marks

Basketball shots: Poisson probabilities with a binomial success rate

Statistics and probability

(a)

2 marks

In 12 minutes $X \sim Po (5)$ .

P (X < 7) = P (X \leq 6) \approx 0.762

(b)

4 marks

For 6 minutes the number of shots $N \sim Po (2.5)$ . Given $N = n$ , the chance of at least one success is $1 - 0. 6^{n}$ .

P = n = 1 \sum 3 P (N = n) (1 - 0. 6^{n}) P = P (1) (0.4) + P (2) (0.64) + P (3) (0.784) \approx 0.414

Our analysis

Where the marks are

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

Calculus32 marks · 29%
Q5, Q9, Q11, Q12, Q13
Geometry and trigonometry21 marks · 19%
Q4, Q6, Q14
Statistics and probability20 marks · 18%
Q7, Q15, Q16
Number and algebra19 marks · 17%
Q1, Q3, Q8
Functions18 marks · 16%
Q2, Q9, Q10, Q11

Mentor insights

Traps to watch for

01
The sum to infinity gives the limit · Q1
The total increase over all months is $\frac{u _{1}}{1 - r}$ . Add the starting width.
02
Three points give three equations · Q5
Substitute each point into $f (x) = a x^{2} + b x + c$ and solve the system.
03
Bisector of AC · Q6
Use the midpoint and the negative reciprocal gradient.
04
Amplitude and midline · Q9
Half the difference between the maximum 3.5 and minimum 0.5, and their mean.
05
Join smoothly · Q13
At $t = T$ both the values and the derivatives of the two pieces agree.

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