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IB Maths AI HL · Paper 1

Mathematics: Applications & Interpretation, Higher Level

Session: November 2024
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 November 2024 Global paper, which your school or the IB can provide.

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Questions

Questions 1–17 · 110 marks

Question 1

5 marks

A triangle with sides 35, 36 and 42: an angle and the area

Geometry and trigonometry

(a)

3 marks

The angle $C A B$ is opposite the side $B C = 36$ .

cos A = \frac{4 2 ^{2} + 3 5 ^{2} - 3 6 ^{2}}{2 ( 42 ) ( 35 )} = \frac{1693}{2940} \approx 0.5758 C A B \approx 54. 8^{\circ}

(b)

2 marks

Use the two sides next to the angle.

Area = \frac{1}{2} (42) (35) sin 54.8 4^{\circ} \approx 601 cm^{2}

Question 2

6 marks

Decay of radioactive carbon in a fossil

Functions

(a)

1 mark

Put $t = 0$ .

m (0) = 120 ng

(b)

2 marks

Put $t = 20000$ .

m (20000) = 120 e^{- 0.000121 \times 20000} \approx 10.7 ng

(c)

3 marks

More than half has decayed when $m (t) < 60$ .
So the smallest whole number of years is

120 e^{- 0.000121 t} < 60 \Rightarrow t > \frac{ln 2}{0.000121} \approx 5728.5 5729 years

Question 3

7 marks

Masses of Manx cats: normal probabilities

Statistics and probability

(a)

2 marks

Use the calculator.

P (W > 3.5) \approx 0.994

(b)

3 marks

(i) $P (W < a) = 0.1$ , so use the inverse normal with area $0.1$ .

(ii) $P (W > b) = 0.1$ , so use the inverse normal with area $0.9$ to the left.

a \approx 3.99 b \approx 5.01

(c)

2 marks

First find $P (W < 3.5) = 1 - 0.99379 = 0.00621$ . The two cats are independent.

P (both < 3.5) = (0.00621)^{2} \approx 0.0000386

Question 4

5 marks

Simple interest as an arithmetic sequence

Number and algebra

(a)

2 marks

The yearly interest is 5% of the original $$25 000$ .

d = 0.05 \times 25000 = $1250

(b)

3 marks

The balance after $k$ complete years is $25000 + 1250 k$ .

25000 + 1250 k > 44000 \Rightarrow k > \frac{19 000}{1250} = 15.2 k = 16

Question 5

6 marks

A Voronoi diagram of eight hospitals: missing edges and cell areas

Geometry and trigonometry

(a)

2 marks

The edge between $A (2, 7)$ and $B (1, 6)$ is the perpendicular bisector of $A B$ . The line $A B$ has gradient 1 and the midpoint is $(1.5, 6.5)$ , so the edge has gradient $- 1$ .
It is drawn from the vertex $(2, 6)$ to the point $(0, 8)$ .
The edge between $B (1, 6)$ and $E (1, 4)$ is the horizontal line through their midpoint $(1, 5)$ , which is $y = 5$ . It is drawn from the vertex $(2, 5)$ to the point $(0, 5)$ .

y = - x + 8

(b)

3 marks

(i) The cell of $F$ is the rectangle $2 \leq x \leq 4$ , $2 \leq y \leq 5$ , which is $2 \times 3 = 6$ square units.

(ii) The cell of $C$ has corners $(2, 5)$ , $(4, 5)$ , $(4, 8)$ and $(2, 6)$ . It is a trapezium with parallel sides 1 and 3 and width 2.

6 \times 4 = 24 km^{2} \frac{1 + 3}{2} \times 2 = 4 square units 4 \times 4 = 16 km^{2}

(c)

1 mark

The number of patients depends on how many people live in each cell, not just its area. The two cells could have different population densities (for example, cell $C$ could be in a more crowded part of the city).

Question 6

4 marks

A museum's membership: a piecewise cost function

Number and algebra

(a)

2 marks

(i) Each extra visit costs $$30$ .

(ii) The free visits end at 10.

a = 30 p = 10

(b)

2 marks

The 11th visit costs $200 + 30 = $230$ , so use $C (11) = 30 (11) + b = 230$ .

330 + b = 230 \Rightarrow b = - 100

Question 7

6 marks

An annuity fund: the balance and the number of withdrawals

Number and algebra

(a)

3 marks

Enter $N = 60$ (5 years of monthly withdrawals), $I = 4.5%$ , $P V = 400000$ , $P M T = - 3600$ , $P / Y = C / Y = 12$ into the finance solver, and find $F V$ .

F V \approx $258 994 \approx $259 000

(b)

3 marks

The fund runs out when $F V = 0$ . Solve for $N$ with the same inputs.
So Daniel can make 144 withdrawals.

N = 144

Question 8

7 marks

Hours of daylight in London: a sine model from its extremes

Functions

(a)

2 marks

The midline.

d = \frac{16.63 + 7.83}{2} = 12.23

(b)

1 mark

The amplitude.

a = \frac{16.63 - 7.83}{2} = 4.4

(c)

2 marks

The maximum is on day 173 and the minimum on day 356, which is half a period.

period = 2 (356 - 173) = 366 b = \frac{2 π}{366} \approx 0.0172

(d)

2 marks

The sine is at its maximum on day 173, so $b (173 - c) = \frac{π}{2}$ .

173 - c = \frac{π /2}{2 π /366} = 91.5 c = 81.5

Question 9

5 marks

A complex number on the Argand diagram: modulus, argument and multiplying by i

Number and algebra

(a)

1 mark

Plot $z = 2 - 3 i$ at the point $(2, - 3)$ : 2 to the right of the origin and 3 below the real axis.

(b)

2 marks

(i) The modulus.

(ii) The argument, in the fourth quadrant.

r = 2^{2} + 3^{2} = 13 \approx 3.61 θ = tan^{- 1} (\frac{- 3}{2}) \approx - 0.983

(c)

1 mark

Multiply out, using $i^{2} = - 1$ .

z i = (2 - 3 i) i = 2 i - 3 i^{2} = 3 + 2 i

(d)

1 mark

The point $z i = 3 + 2 i$ is $z = 2 - 3 i$ rotated anticlockwise through $\frac{π}{2}$ (90 $^{\circ}$ ) about the origin.

Question 10

8 marks

Transformations of a log graph, a composite function and an equation

Functions

(a)

3 marks

Write $f (x) = 4 ln (2 x - 3) = 4 ln [2 (x - \frac{3}{2})]$ . Starting from $y = ln x$ , apply these three transformations.
A vertical stretch with scale factor 4.
A horizontal stretch with scale factor $\frac{1}{2}$ .
A translation $(1.5 0)$ (1.5 units to the right).
(Equivalently, translate by $(30)$ first and then apply the horizontal stretch with scale factor $\frac{1}{2}$ .)

(b)

2 marks

Read from the graph: $g (2) = 5$ .

(f \circ g) (2) = f (5) = 4 ln (7) \approx 7.78

(c)

3 marks

Write the equation.
Divide by 4 and remove the logarithms.
From the graph, $g (1) = 3$ .

4 ln (2 g (x) - 3) = 2 ln 9 = 4 ln 3 2 g (x) - 3 = 3 \Rightarrow g (x) = 3 x = 1

Question 11

8 marks

Eigenvalues and diagonalisation of a 2 by 2 matrix

Number and algebra

(a)

3 marks

Write the characteristic equation for $M = (- 4 - 3 23)$ .

det (M - λ I) = (- 4 - λ) (3 - λ) + 6 = λ^{2} + λ - 6 = 0 (λ - 2) (λ + 3) = 0 \Rightarrow λ = 2, λ = - 3

(b)

5 marks

(i) The diagonal matrix of eigenvalues.

(ii) For $λ = 2$ : $- 6 x + 2 y = 0$ , so $y = 3 x$ and an eigenvector is $(13)$ .

For $λ = - 3$ : $- x + 2 y = 0$ , so $x = 2 y$ and an eigenvector is $(21)$ .

D = (20 0 - 3) P = (1321)

Question 12

7 marks

A bouncing ball: a model, residuals and exponential regression

Functions

(a)

2 marks

Use $h (n) = 0.613 (\frac{0.514}{0.613})^{n - 1}$ .

h (3) = 0.613 (0.8385)^{2} \approx 0.431, h (4) = 0.613 (0.8385)^{3} \approx 0.361

(b)

2 marks

Add the squares of the residuals for $n = 3$ and $n = 4$ (the first two are zero).

S S_{r es} = (0.43099 - 0.439)^{2} + (0.36138 - 0.377)^{2} \approx 3.08 \times 1 0^{- 4}

(c)

3 marks

(i) Use exponential regression on the calculator.

(ii) The height the ball was dropped from is the value when $n = 0$ .

h = 0.716 \times 0.85 1^{n} h (0) = 0.716 m

Question 13

6 marks

Beavers in Scotland: finding the carrying capacity of a logistic model

Functions

6 marks

The survey in 2017 is $t = 9$ and the survey in 2020 is $t = 12$ .
Write each as $L = P (1 + C e^{- 0.28 t})$ and equate.
Solve on the calculator.

\frac{L}{1 + C e ^{- 0.28 (9)}} = 433, \frac{L}{1 + C e ^{- 0.28 (12)}} = 954 433 (1 + C e^{- 2.52}) = 954 (1 + C e^{- 3.36}) C \approx 306.2 L = 433 (1 + 306.2 e^{- 2.52}) \approx 11100

Question 14

8 marks

A mound of sand with a flagpole: the normal to a curve

Calculus

(a)

2 marks

Solve $f (a) = 0.25$ .

0.3 e^{- 2 a^{2}} = 0.25 \Rightarrow a \approx 0.302

(b)

2 marks

Differentiate with the chain rule.

f^{'} (x) = 0.3 (- 4 x) e^{- 2 x^{2}} = - 1.2 x e^{- 2 x^{2}}

(c)

4 marks

The gradient of the tangent at $P$ .
The flagpole is the normal, with gradient the negative reciprocal.
The angle with the level beach satisfies $tan θ = 3.31$ .

f^{'} (0.302) \approx - 0.302 m = \frac{1}{0.302} \approx 3.31 θ = tan^{- 1} (3.31) \approx 1.28 radians (\approx 73. 2^{\circ})

Question 15

8 marks

A metal ring: normal distributions of radii and the width of the ring

Statistics and probability

(a)

3 marks

The radius is half the diameter. The mean is halved and the standard deviation is halved.

R \sim N (15, 0. 4^{2})

(b)

5 marks

The width is $W = R - r$ , where $r \sim N (12, 0.2 5^{2})$ is the radius of the hole.
The variances add, even for a difference of independent normal variables.
So $W \sim N (3, 0.2225)$ .

E (W) = 15 - 12 = 3 Var (W) = 0. 4^{2} + 0.2 5^{2} = 0.2225 P (W < 2.5) \approx 0.145

Question 16

7 marks

The volume of a wine glass from a rotated curve

Calculus

(a)

1 mark

The $y$ -intercept is at $x = 0$ .

f (0) = 4 - 2 ln 5 \approx 0.781

(b)

6 marks

Make $x$ the subject of $y = 4 - 2 ln (5 - x)$ .
The region runs from $y = 0.781$ up to $y = 11.8$ . Rotate about the $y$ -axis.
Evaluate on the calculator.

2 ln (5 - x) = 4 - y \Rightarrow 5 - x = e^{(4 - y) /2} \Rightarrow x = 5 - e^{(4 - y) /2} V = π \int_{0.781}^{11.8} (5 - e^{(4 - y) /2})^{2} d y V \approx 631 cm^{3}

Question 17

7 marks

Fish and insects: Euler's method for coupled differential equations

Calculus

(a)

5 marks

With $h = 4$ weeks, use $F_{n + 1} = F_{n} + 4 (0.000004 F_{n} L_{n} - 0.2 F_{n})$ and $L_{n + 1} = L_{n} + 4 (0.06 L_{n} - 0.00003 F_{n} L_{n})$ , starting from $F_{0} = 5000$ and $L_{0} = 80000$ . Rounded values are:
Carry on to $t = 52$ : the fish keep falling towards 0 while the insects grow again after $t = 16$ .
(i) The greatest value of $F$ is at $t = 8$ .
(ii) The least value of $L$ is at $t = 16$ .

t = 0 : (5000, 80000), t = 4 : (7400, 51200), t = 8 : (7542, 18022) t = 12 : (3683, 6037), t = 16 : (1092, 4817), t = 20 : (303, 5342) F_{ma x} \approx 7542 L_{min} \approx 4817

(b)

2 marks

(i) The model predicts that the fish population falls towards zero (the fish die out).

(ii) One reason: Euler's method is only an approximation, so the true behaviour may differ. Another: other factors, such as new food sources, restocking or changes in the lake, are not in the model.

Our analysis

Where the marks are

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

Functions34 marks · 31%
Q2, Q8, Q10, Q12, Q13
Number and algebra28 marks · 25%
Q4, Q6, Q7, Q9, Q11
Calculus22 marks · 20%
Q14, Q16, Q17
Statistics and probability15 marks · 14%
Q3, Q15
Geometry and trigonometry11 marks · 10%
Q1, Q5

Mentor insights

Traps to watch for

01
Normal · Q3
$W \sim N (4.5, 0. 4^{2})$ .
02
Composite · Q10
$(f \circ g) (x) = f (g (x))$ .
03
Characteristic equation · Q11
Solve $det (M - λ I) = 0$ .
04
Chain rule · Q14
$\frac{d}{d x} e^{u} = u^{'} e^{u}$ .
05
Halving · Q15
Radius is half the diameter, so the mean and standard deviation are halved.

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 2024
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 November 2024 Global 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

5 marks

A triangle with sides 35, 36 and 42: an angle and the area

Geometry and trigonometry

(a)

3 marks

The angle $C A B$ is opposite the side $B C = 36$ .

cos A = \frac{4 2 ^{2} + 3 5 ^{2} - 3 6 ^{2}}{2 ( 42 ) ( 35 )} = \frac{1693}{2940} \approx 0.5758 C A B \approx 54. 8^{\circ}

(b)

2 marks

Use the two sides next to the angle.

Area = \frac{1}{2} (42) (35) sin 54.8 4^{\circ} \approx 601 cm^{2}

Question 2

6 marks

Decay of radioactive carbon in a fossil

Functions

(a)

1 mark

Put $t = 0$ .

m (0) = 120 ng

(b)

2 marks

Put $t = 20000$ .

m (20000) = 120 e^{- 0.000121 \times 20000} \approx 10.7 ng

(c)

3 marks

More than half has decayed when $m (t) < 60$ .
So the smallest whole number of years is

120 e^{- 0.000121 t} < 60 \Rightarrow t > \frac{ln 2}{0.000121} \approx 5728.5 5729 years

Question 3

7 marks

Masses of Manx cats: normal probabilities

Statistics and probability

(a)

2 marks

Use the calculator.

P (W > 3.5) \approx 0.994

(b)

3 marks

(i) $P (W < a) = 0.1$ , so use the inverse normal with area $0.1$ .

(ii) $P (W > b) = 0.1$ , so use the inverse normal with area $0.9$ to the left.

a \approx 3.99 b \approx 5.01

(c)

2 marks

First find $P (W < 3.5) = 1 - 0.99379 = 0.00621$ . The two cats are independent.

P (both < 3.5) = (0.00621)^{2} \approx 0.0000386

Question 4

5 marks

Simple interest as an arithmetic sequence

Number and algebra

(a)

2 marks

The yearly interest is 5% of the original $$25 000$ .

d = 0.05 \times 25000 = $1250

(b)

3 marks

The balance after $k$ complete years is $25000 + 1250 k$ .

25000 + 1250 k > 44000 \Rightarrow k > \frac{19 000}{1250} = 15.2 k = 16

Question 5

6 marks

A Voronoi diagram of eight hospitals: missing edges and cell areas

Geometry and trigonometry

(a)

2 marks

The edge between $A (2, 7)$ and $B (1, 6)$ is the perpendicular bisector of $A B$ . The line $A B$ has gradient 1 and the midpoint is $(1.5, 6.5)$ , so the edge has gradient $- 1$ .
It is drawn from the vertex $(2, 6)$ to the point $(0, 8)$ .
The edge between $B (1, 6)$ and $E (1, 4)$ is the horizontal line through their midpoint $(1, 5)$ , which is $y = 5$ . It is drawn from the vertex $(2, 5)$ to the point $(0, 5)$ .

y = - x + 8

(b)

3 marks

(i) The cell of $F$ is the rectangle $2 \leq x \leq 4$ , $2 \leq y \leq 5$ , which is $2 \times 3 = 6$ square units.

(ii) The cell of $C$ has corners $(2, 5)$ , $(4, 5)$ , $(4, 8)$ and $(2, 6)$ . It is a trapezium with parallel sides 1 and 3 and width 2.

6 \times 4 = 24 km^{2} \frac{1 + 3}{2} \times 2 = 4 square units 4 \times 4 = 16 km^{2}

(c)

1 mark

Question 6

4 marks

A museum's membership: a piecewise cost function

Number and algebra

(a)

2 marks

(i) Each extra visit costs $$30$ .

(ii) The free visits end at 10.

a = 30 p = 10

(b)

2 marks

The 11th visit costs $200 + 30 = $230$ , so use $C (11) = 30 (11) + b = 230$ .

330 + b = 230 \Rightarrow b = - 100

Question 7

6 marks

An annuity fund: the balance and the number of withdrawals

Number and algebra

(a)

3 marks

Enter $N = 60$ (5 years of monthly withdrawals), $I = 4.5%$ , $P V = 400000$ , $P M T = - 3600$ , $P / Y = C / Y = 12$ into the finance solver, and find $F V$ .

F V \approx $258 994 \approx $259 000

(b)

3 marks

The fund runs out when $F V = 0$ . Solve for $N$ with the same inputs.
So Daniel can make 144 withdrawals.

N = 144

Question 8

7 marks

Hours of daylight in London: a sine model from its extremes

Functions

(a)

2 marks

The midline.

d = \frac{16.63 + 7.83}{2} = 12.23

(b)

1 mark

The amplitude.

a = \frac{16.63 - 7.83}{2} = 4.4

(c)

2 marks

The maximum is on day 173 and the minimum on day 356, which is half a period.

period = 2 (356 - 173) = 366 b = \frac{2 π}{366} \approx 0.0172

(d)

2 marks

The sine is at its maximum on day 173, so $b (173 - c) = \frac{π}{2}$ .

173 - c = \frac{π /2}{2 π /366} = 91.5 c = 81.5

Question 9

5 marks

A complex number on the Argand diagram: modulus, argument and multiplying by i

Number and algebra

(a)

1 mark

Plot $z = 2 - 3 i$ at the point $(2, - 3)$ : 2 to the right of the origin and 3 below the real axis.

(b)

2 marks

(i) The modulus.

(ii) The argument, in the fourth quadrant.

r = 2^{2} + 3^{2} = 13 \approx 3.61 θ = tan^{- 1} (\frac{- 3}{2}) \approx - 0.983

(c)

1 mark

Multiply out, using $i^{2} = - 1$ .

z i = (2 - 3 i) i = 2 i - 3 i^{2} = 3 + 2 i

(d)

1 mark

The point $z i = 3 + 2 i$ is $z = 2 - 3 i$ rotated anticlockwise through $\frac{π}{2}$ (90 $^{\circ}$ ) about the origin.

Question 10

8 marks

Transformations of a log graph, a composite function and an equation

Functions

(a)

3 marks

Write $f (x) = 4 ln (2 x - 3) = 4 ln [2 (x - \frac{3}{2})]$ . Starting from $y = ln x$ , apply these three transformations.
A vertical stretch with scale factor 4.
A horizontal stretch with scale factor $\frac{1}{2}$ .
A translation $(1.5 0)$ (1.5 units to the right).
(Equivalently, translate by $(30)$ first and then apply the horizontal stretch with scale factor $\frac{1}{2}$ .)

(b)

2 marks

Read from the graph: $g (2) = 5$ .

(f \circ g) (2) = f (5) = 4 ln (7) \approx 7.78

(c)

3 marks

Write the equation.
Divide by 4 and remove the logarithms.
From the graph, $g (1) = 3$ .

4 ln (2 g (x) - 3) = 2 ln 9 = 4 ln 3 2 g (x) - 3 = 3 \Rightarrow g (x) = 3 x = 1

Question 11

8 marks

Eigenvalues and diagonalisation of a 2 by 2 matrix

Number and algebra

(a)

3 marks

Write the characteristic equation for $M = (- 4 - 3 23)$ .

det (M - λ I) = (- 4 - λ) (3 - λ) + 6 = λ^{2} + λ - 6 = 0 (λ - 2) (λ + 3) = 0 \Rightarrow λ = 2, λ = - 3

(b)

5 marks

(i) The diagonal matrix of eigenvalues.

(ii) For $λ = 2$ : $- 6 x + 2 y = 0$ , so $y = 3 x$ and an eigenvector is $(13)$ .

For $λ = - 3$ : $- x + 2 y = 0$ , so $x = 2 y$ and an eigenvector is $(21)$ .

D = (20 0 - 3) P = (1321)

Question 12

7 marks

A bouncing ball: a model, residuals and exponential regression

Functions

(a)

2 marks

Use $h (n) = 0.613 (\frac{0.514}{0.613})^{n - 1}$ .

h (3) = 0.613 (0.8385)^{2} \approx 0.431, h (4) = 0.613 (0.8385)^{3} \approx 0.361

(b)

2 marks

Add the squares of the residuals for $n = 3$ and $n = 4$ (the first two are zero).

S S_{r es} = (0.43099 - 0.439)^{2} + (0.36138 - 0.377)^{2} \approx 3.08 \times 1 0^{- 4}

(c)

3 marks

(i) Use exponential regression on the calculator.

(ii) The height the ball was dropped from is the value when $n = 0$ .

h = 0.716 \times 0.85 1^{n} h (0) = 0.716 m

Question 13

6 marks

Beavers in Scotland: finding the carrying capacity of a logistic model

Functions

6 marks

The survey in 2017 is $t = 9$ and the survey in 2020 is $t = 12$ .
Write each as $L = P (1 + C e^{- 0.28 t})$ and equate.
Solve on the calculator.

\frac{L}{1 + C e ^{- 0.28 (9)}} = 433, \frac{L}{1 + C e ^{- 0.28 (12)}} = 954 433 (1 + C e^{- 2.52}) = 954 (1 + C e^{- 3.36}) C \approx 306.2 L = 433 (1 + 306.2 e^{- 2.52}) \approx 11100

Question 14

8 marks

A mound of sand with a flagpole: the normal to a curve

Calculus

(a)

2 marks

Solve $f (a) = 0.25$ .

0.3 e^{- 2 a^{2}} = 0.25 \Rightarrow a \approx 0.302

(b)

2 marks

Differentiate with the chain rule.

f^{'} (x) = 0.3 (- 4 x) e^{- 2 x^{2}} = - 1.2 x e^{- 2 x^{2}}

(c)

4 marks

The gradient of the tangent at $P$ .
The flagpole is the normal, with gradient the negative reciprocal.
The angle with the level beach satisfies $tan θ = 3.31$ .

f^{'} (0.302) \approx - 0.302 m = \frac{1}{0.302} \approx 3.31 θ = tan^{- 1} (3.31) \approx 1.28 radians (\approx 73. 2^{\circ})

Question 15

8 marks

A metal ring: normal distributions of radii and the width of the ring

Statistics and probability

(a)

3 marks

The radius is half the diameter. The mean is halved and the standard deviation is halved.

R \sim N (15, 0. 4^{2})

(b)

5 marks

The width is $W = R - r$ , where $r \sim N (12, 0.2 5^{2})$ is the radius of the hole.
The variances add, even for a difference of independent normal variables.
So $W \sim N (3, 0.2225)$ .

E (W) = 15 - 12 = 3 Var (W) = 0. 4^{2} + 0.2 5^{2} = 0.2225 P (W < 2.5) \approx 0.145

Question 16

7 marks

The volume of a wine glass from a rotated curve

Calculus

(a)

1 mark

The $y$ -intercept is at $x = 0$ .

f (0) = 4 - 2 ln 5 \approx 0.781

(b)

6 marks

Make $x$ the subject of $y = 4 - 2 ln (5 - x)$ .
The region runs from $y = 0.781$ up to $y = 11.8$ . Rotate about the $y$ -axis.
Evaluate on the calculator.

2 ln (5 - x) = 4 - y \Rightarrow 5 - x = e^{(4 - y) /2} \Rightarrow x = 5 - e^{(4 - y) /2} V = π \int_{0.781}^{11.8} (5 - e^{(4 - y) /2})^{2} d y V \approx 631 cm^{3}

Question 17

7 marks

Fish and insects: Euler's method for coupled differential equations

Calculus

(a)

5 marks

With $h = 4$ weeks, use $F_{n + 1} = F_{n} + 4 (0.000004 F_{n} L_{n} - 0.2 F_{n})$ and $L_{n + 1} = L_{n} + 4 (0.06 L_{n} - 0.00003 F_{n} L_{n})$ , starting from $F_{0} = 5000$ and $L_{0} = 80000$ . Rounded values are:
Carry on to $t = 52$ : the fish keep falling towards 0 while the insects grow again after $t = 16$ .
(i) The greatest value of $F$ is at $t = 8$ .
(ii) The least value of $L$ is at $t = 16$ .

t = 0 : (5000, 80000), t = 4 : (7400, 51200), t = 8 : (7542, 18022) t = 12 : (3683, 6037), t = 16 : (1092, 4817), t = 20 : (303, 5342) F_{ma x} \approx 7542 L_{min} \approx 4817

(b)

2 marks

(i) The model predicts that the fish population falls towards zero (the fish die out).

(ii) One reason: Euler's method is only an approximation, so the true behaviour may differ. Another: other factors, such as new food sources, restocking or changes in the lake, are not in the model.

Our analysis

Where the marks are

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

Functions34 marks · 31%
Q2, Q8, Q10, Q12, Q13
Number and algebra28 marks · 25%
Q4, Q6, Q7, Q9, Q11
Calculus22 marks · 20%
Q14, Q16, Q17
Statistics and probability15 marks · 14%
Q3, Q15
Geometry and trigonometry11 marks · 10%
Q1, Q5

Mentor insights

Traps to watch for

01
Normal · Q3
$W \sim N (4.5, 0. 4^{2})$ .
02
Composite · Q10
$(f \circ g) (x) = f (g (x))$ .
03
Characteristic equation · Q11
Solve $det (M - λ I) = 0$ .
04
Chain rule · Q14
$\frac{d}{d x} e^{u} = u^{'} e^{u}$ .
05
Halving · Q15
Radius is half the diameter, so the mean and standard deviation are halved.

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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