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Worked solutions

IB Maths AA HL · Paper 2

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2 TZ3

Session: May 2025
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 12

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

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2025 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Section A · Short response

Questions 1–9 · 55 marks

Question 1

6 marks

A hollow cone made from a sector of a circle

Geometry and trigonometry

(a)

3 marks

The angle $21 0^{\circ}$ is $\frac{7 π}{6}$ radians.

A = \frac{1}{2} r^{2} θ = \frac{1}{2} (19.5)^{2} (\frac{7 π}{6}) \approx 697 cm^{2}

(b)

3 marks

The arc length of the sector is the circumference of the base of the cone.

arc = 19.5 \times \frac{7 π}{6} \approx 71.47 cm 2 π r = 71.47 \Rightarrow r = \frac{91}{8} = 11.375 \approx 11.4 cm

Question 2

4 marks

A tangent function through two points

Functions

(a)

1 mark

The period is $\frac{π}{2}$ .

\frac{π}{2} \approx 1.57

(b)

3 marks

The points are $(\frac{π}{12}, 5)$ and $(\frac{π}{3}, 7)$ . Substitute into $f (x) = a tan 2 x + b$ .
Subtract the second equation from the first.
Then $b = 7 + 3 a = 7 - \frac{3}{2}$ .

a tan \frac{π}{6} + b = 5, a tan \frac{2 π}{3} + b = 7 \frac{a}{3} + b = 5, - 3 a + b = 7 a (\frac{1}{3} + 3) = - 2 \Rightarrow \frac{4 a}{3} = - 2 \Rightarrow a = - \frac{3}{2} \approx - 0.866 b = \frac{11}{2}

Question 3

6 marks

A population with a decaying rate of change

Number and algebra

6 marks

The start of 2026 is $t = 4$ . The rate is negative, so the population falls.
Add the change to the starting population.

Δ P = \int_{0}^{4} - 104000 e^{- 0.0145 t} d t = [\frac{104000}{0.0145} e^{- 0.0145 t}]_{0}^{4} \approx - 404166 P (4) = 6.78 \times 1 0^{6} - 404166 \approx 6375834 P (4) \approx 6.38 \times 1 0^{6}

Question 4

8 marks

Arm span and foot length: two regression lines, the mean point and a normal spread

Statistics and probability

(a)

2 marks

Use the line of $A$ on $F$ with $F = 19.8$ .

A = 2.89 (19.8) + 99.3 \approx 157 cm

(b)

3 marks

The mean point lies on both lines, so solve them together.

F = 0.335 (2.89 F + 99.3) - 32.6 \Rightarrow F (1 - 0.96815) = 0.66655 \overset{ˉ}{F} \approx 20.9 cm, \overset{ˉ}{A} \approx 160 cm

(c)

3 marks

The interval 153 to 173 is symmetric about the mean 163, so 6% lies below 153 and 6% above 173.

P (H < 153) = 0.06 \Rightarrow \frac{153 - 163}{σ} = - 1.5548 σ \approx 6.43 cm

Question 5

6 marks

The gradient of a tangent to a cosine curve and a given angle with the x-axis

Calculus

(a)

3 marks

Differentiate $h (x) = 15 cos (\frac{π x}{50}) + 15$ with the chain rule.

h^{'} (x) = - 15 \cdot \frac{π}{50} sin (\frac{π x}{50}) = - \frac{3 π}{10} sin (\frac{π x}{50}) gradient of T_{k} = - \frac{3 π}{10} sin (\frac{π k}{50})

(b)

3 marks

The tangent slopes downwards (the curve falls from 30 to 0), so its gradient is $- tan \frac{π}{8} = - 0.4142$ .
Solve on $0 \leq k \leq 50$ , where $\frac{π k}{50}$ is between 0 and $π$ . There are two solutions, symmetric about $k = 25$ .

- \frac{3 π}{10} sin (\frac{π k}{50}) = - 0.4142 \Rightarrow sin (\frac{π k}{50}) = 0.4395 k \approx 7.24 or k \approx 42.8

Question 6

7 marks

f(x) = 4 cot x + sin x: rewriting, a sketch, an inverse value and a value from sec

Functions

(a)

3 marks

(i) Replace $cot x$ .

(ii) The graph on $0 < x < π$ falls from large positive values near $x = 0$ , passes through the $x$ -axis, and goes to large negative values near $x = π$ .

The $x$ -intercept is where $4 cos x + sin^{2} x = 0$ .

Sketch: a decreasing curve with vertical asymptotes at $x = 0$ and $x = π$ crossing the $x$ -axis at $(1.81, 0)$ .

f (x) = \frac{4 cos x}{sin x} + sin x x \approx 1.81

(b)

1 mark

$f^{- 1} (2)$ is the $x$ with $f (x) = 2$ . Solve with the calculator on $0 < x < π$ .

f^{- 1} (2) \approx 1.32

(c)

3 marks

Since $sec α = 1.5$ , $cos α = \frac{2}{3}$ , and $α$ is in the first quadrant, so $sin α = 1 - \frac{4}{9} = \frac{5}{3}$ .

cot α = \frac{2/3}{5 /3} = \frac{2}{5} f (α) = 4 \cdot \frac{2}{5} + \frac{5}{3} = \frac{8 5}{5} + \frac{5}{3} = \frac{29 5}{15} \approx 4.32

Question 7

5 marks

A helicopter: speed and the angle of descent

Number and algebra

(a)

2 marks

The velocity vector is $4 10 - 25 0 = 40 - 100 0$ .

speed = 4 0^{2} + 10 0^{2} = 11600 = 2029 \approx 108 km h^{- 1}

(b)

3 marks

During the descent the horizontal speed is still $2029$ and the vertical speed is 16 km h $^{- 1}$ downwards.

tan β = \frac{16}{20 29} = \frac{4}{5 29} β \approx 8.4 5^{\circ}

Question 8

7 marks

Adding two sine waves with complex numbers

Number and algebra

(a)

2 marks

Write each sine as the imaginary part of a complex exponential.
Add and factor out $e^{2 t i}$ , using that the imaginary part of a sum is the sum of the imaginary parts.

f (t) = Im (e^{i (2 t + 1)}), g (t) = Im (e^{i (2 t + 3)}) h (t) = Im (e^{2 t i} e^{i} + e^{2 t i} e^{3 i}) = Im (e^{2 t i} (e^{i} + e^{3 i}))

(b)

2 marks

Factor out $e^{2 i}$ from $e^{i} + e^{3 i}$ .

e^{i} + e^{3 i} = e^{2 i} (e^{- i} + e^{i}) = 2 cos (1) e^{2 i} r = 2 cos 1 \approx 1.08, θ = 2

(c)

3 marks

Substitute into the result of (a).
Here $p = 2 cos 1$ and $q = 2$ , which is in the interval $0 < q < 2 π$ .

h (t) = Im (e^{2 t i} \times 2 cos (1) e^{2 i}) = Im (2 cos (1) e^{i (2 t + 2)}) h (t) = 2 cos (1) sin (2 t + 2) \approx 1.08 sin (2 t + 2)

Question 9

6 marks

Euler's method for dy/dx = 2x/(x^2 + y): an estimate and its accuracy

Calculus

(a)

3 marks

Start at $(1, 0)$ and take four steps of 0.25.

y (1.25) \approx 0 + 0.25 (\frac{2}{1}) = 0.5 y (1.5) \approx 0.5 + 0.25 (\frac{2.5}{1.5625 + 0.5}) \approx 0.8030 y (1.75) \approx 0.8030 + 0.25 (\frac{3}{2.25 + 0.8030}) \approx 1.0487 y (2) \approx 1.0487 + 0.25 (\frac{3.5}{3.0625 + 1.0487}) \approx 1.26

(b)

3 marks

(i) The solution curve is concave down, so each tangent line segment lies above the curve and the steps get further above the curve. The estimate is an overestimate.

(ii) At $(1, 0)$ the gradient is $\frac{2}{1} = 2 > 0$ , so Euler's method moves upwards, along the upper part of the solution curve, and it cannot reach the lower part of the curve with negative $y$ .

Section B · Extended response

Questions 10–12 · 55 marks

Question 10

16 marks

A pyramid of playing cards: a quadratic number sequence

Number and algebra

(a)

1 mark

From the diagram of the stack with 3 rows, count the cards.

t_{3} = 2 + 5 + 8 = 15

(b)

2 marks

A fourth row adds 11 cards (the differences between rows are 3).

t_{4} = 15 + 11 = 26

(c)

3 marks

The numbers of cards in the rows form the arithmetic sequence $2, 5, 8, 11, \dots$ with first term 2 and common difference 3. Use the sum of $n$ terms.

t_{n} = \frac{n}{2} [2 (2) + (n - 1) 3] = \frac{n}{2} (3 n + 1)

(d)

3 marks

The cards available are $14 \times 52 = 728$ . Solve $\frac{n ( 3 n + 1 )}{2} \leq 728$ .
The most rows is 21.

t_{21} = \frac{21 ( 64 )}{2} = 672 \leq 728, t_{22} = \frac{22 ( 67 )}{2} = 737 > 728

(e)

2 marks

The number of cards $t_{n}$ must be a multiple of 52. List $t_{n}$ : 2, 7, 15, 26, 40, 57, 77, 100, 126, 155, 187, 222, 260.
The first multiple of 52 is $t_{13} = 260 = 5 \times 52$ (so 5 packs).

n = 13

(f)

5 marks

Each row is an equilateral triangle with side 88 mm, so its height is
For the stack to be taller than 2000 mm we need $76.21 n > 2000$ .
The number of cards for 27 rows.

88 sin 6 0^{\circ} = 443 \approx 76.21 mm n > 26.2 \Rightarrow n = 27 t_{27} = \frac{27 ( 82 )}{2} = 1107

Question 11

19 marks

Marathon times: a cosine-based probability density function

Statistics and probability

(a)

4 marks

(i) Integrate the first branch of the density.

(ii) The maximum of the graph is at $t = 4.5$ .

(iii) The probability of finishing before the mode is $\frac{3}{7} < 0.5$ , so more than half the area is to the right of 4.5. The median is greater than the mode (the median is about 4.69).

\int_{2.25}^{4.5} f (t) d t = \frac{4}{21} [t - \frac{9}{4 π} sin (\frac{4 π}{9} (t - 2.25))]_{2.25}^{4.5} = \frac{4}{21} (2.25) = \frac{3}{7} \approx 0.429 mode = 4.5 hours

(b)

2 marks

A fast runner finishes in 3.5 hours or less.

P (T \leq 3.5) = \int_{2.25}^{3.5} f (t) d t \approx 0.104

(c)

3 marks

Use conditional probability.

P (T \leq 3 ∣ T \leq 3.5) = \frac{P ( T \leq 3 )}{P ( T \leq 3.5 )} = \frac{0.02472}{0.10375} \approx 0.238

(d)

3 marks

The lower quartile $q$ satisfies $\int_{2.25}^{q} f (t) d t = 0.25$ .
Solve with the calculator (the lower quartile is below the mode, so use the first branch).

\frac{4}{21} [t - \frac{9}{4 π} sin (\frac{4 π}{9} (t - 2.25))]_{2.25}^{q} = 0.25 q \approx 4.01 hours

(e)

5 marks

The expected score is $E (P) = a - b E (T)$ .
The maximum score of 150 goes to the fastest possible time, 2.25 hours.
Subtract the first equation from the second: $2.473 b = 50$ .

a - 4.723 b = 100 a - 2.25 b = 150 b \approx 20.2, a \approx 195.5 a = 195, b = 20

(f)

2 marks

Use $Var (a - b T) = b^{2} Var (T)$ with $b = 20.22$ .

Var (P) = (20.22)^{2} (0.906) \approx 370

Question 12

20 marks

A family of polynomials, an infinite series, an inverse function and an enclosed area

Number and algebra

(a)

3 marks

Replace $x$ by $- x$ in $f_{n} (x) = \sum_{r = 0}^{n} (- 2 x^{2})^{r}$ .
Each term is unchanged, so $f_{n} (- x) = f_{n} (x)$ and $f_{n}$ is even for every $n$ .

(- 2 (- x)^{2})^{r} = (- 2 x^{2})^{r}

(b)

2 marks

(i) Put $n = 3$ .

(ii) Add the next term, $(- 2 x^{2})^{4} = 16 x^{8}$ .

f_{3} (x) = 1 + (- 2 x^{2}) + (- 2 x^{2})^{2} + (- 2 x^{2})^{3} = 1 - 2 x^{2} + 4 x^{4} - 8 x^{6} f_{4} (x) = 1 - 2 x^{2} + 4 x^{4} - 8 x^{6} + 16 x^{8}

(c)

5 marks

(i) As $n \to \infty$ , $f_{n} (x)$ is a geometric series with first term 1 and common ratio $- 2 x^{2}$ . It converges when $- 2 x^{2} < 1$ .

(ii) The sum to infinity is $\frac{1}{1 - r}$ .

2 x^{2} < 1 \Rightarrow ∣ x ∣ < \frac{1}{2} K = \frac{1}{2} = \frac{2}{2} f (x) = \frac{1}{1 - ( - 2 x ^{2} )} = \frac{1}{1 + 2 x ^{2}}, a = 1, b = 2

(d)

6 marks

(i) For $0 \leq x < K$ , $g (x) = \frac{1}{1 + 2 x ^{2}}$ is strictly decreasing and has no turning points, so it is one-to-one, and the inverse exists.

(ii) Swap $x$ and $y$ and solve for $y$ .

The domain of $g$ gives $y \geq 0$ , so take the positive root.

The domain of $g^{- 1}$ is the range of $g$ . Since $g (0) = 1$ and $g (K) = \frac{1}{2}$ ,

x = \frac{1}{1 + 2 y ^{2}} \Rightarrow 1 + 2 y^{2} = \frac{1}{x} \Rightarrow y^{2} = \frac{1 - x}{2 x} g^{- 1} (x) = \frac{1 - x}{2 x} \frac{1}{2} < x \leq 1

(e)

4 marks

The curves $y = g (x)$ and $y = g^{- 1} (x)$ are reflections in $y = x$ and meet where $g (x) = x$ .
By symmetry, the area is twice the area between $y = g (x)$ and $y = x$ from 0 to 0.5898.

x \approx 0.5898 Area = 2 \int_{0}^{0.5898} (\frac{1}{1 + 2 x ^{2}} - x) d x \approx 0.635

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 algebra54 marks · 49%
Q3, Q7, Q8, Q10, Q12
Statistics and probability27 marks · 25%
Q4, Q11
Calculus12 marks · 11%
Q5, Q9
Functions11 marks · 10%
Q2, Q6
Geometry and trigonometry6 marks · 5%
Q1

Mentor insights

Traps to watch for

01
Which line to use · Q4
To estimate $A$ from $F$ use the regression line of $A$ on $F$ .
02
Cotangent · Q6
$cot x = \frac{cos x}{sin x}$ .
03
Counting · Q10
Look at how many cards are added to make each new row, and find the pattern.
04
Probability from a pdf · Q11
Integrate the density over the interval.
05
Even function · Q12
$f (- x) = f (x)$ , which holds when $x$ only appears as $x^{2}$ .

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 AA HL · Paper 2

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2 TZ3

Session: May 2025
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 12

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

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2025 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Section A · Short response

Questions 1–9 · 55 marks

Question 1

6 marks

A hollow cone made from a sector of a circle

Geometry and trigonometry

(a)

3 marks

The angle $21 0^{\circ}$ is $\frac{7 π}{6}$ radians.

A = \frac{1}{2} r^{2} θ = \frac{1}{2} (19.5)^{2} (\frac{7 π}{6}) \approx 697 cm^{2}

(b)

3 marks

The arc length of the sector is the circumference of the base of the cone.

arc = 19.5 \times \frac{7 π}{6} \approx 71.47 cm 2 π r = 71.47 \Rightarrow r = \frac{91}{8} = 11.375 \approx 11.4 cm

Question 2

4 marks

A tangent function through two points

Functions

(a)

1 mark

The period is $\frac{π}{2}$ .

\frac{π}{2} \approx 1.57

(b)

3 marks

The points are $(\frac{π}{12}, 5)$ and $(\frac{π}{3}, 7)$ . Substitute into $f (x) = a tan 2 x + b$ .
Subtract the second equation from the first.
Then $b = 7 + 3 a = 7 - \frac{3}{2}$ .

a tan \frac{π}{6} + b = 5, a tan \frac{2 π}{3} + b = 7 \frac{a}{3} + b = 5, - 3 a + b = 7 a (\frac{1}{3} + 3) = - 2 \Rightarrow \frac{4 a}{3} = - 2 \Rightarrow a = - \frac{3}{2} \approx - 0.866 b = \frac{11}{2}

Question 3

6 marks

A population with a decaying rate of change

Number and algebra

6 marks

The start of 2026 is $t = 4$ . The rate is negative, so the population falls.
Add the change to the starting population.

Δ P = \int_{0}^{4} - 104000 e^{- 0.0145 t} d t = [\frac{104000}{0.0145} e^{- 0.0145 t}]_{0}^{4} \approx - 404166 P (4) = 6.78 \times 1 0^{6} - 404166 \approx 6375834 P (4) \approx 6.38 \times 1 0^{6}

Question 4

8 marks

Arm span and foot length: two regression lines, the mean point and a normal spread

Statistics and probability

(a)

2 marks

Use the line of $A$ on $F$ with $F = 19.8$ .

A = 2.89 (19.8) + 99.3 \approx 157 cm

(b)

3 marks

The mean point lies on both lines, so solve them together.

F = 0.335 (2.89 F + 99.3) - 32.6 \Rightarrow F (1 - 0.96815) = 0.66655 \overset{ˉ}{F} \approx 20.9 cm, \overset{ˉ}{A} \approx 160 cm

(c)

3 marks

The interval 153 to 173 is symmetric about the mean 163, so 6% lies below 153 and 6% above 173.

P (H < 153) = 0.06 \Rightarrow \frac{153 - 163}{σ} = - 1.5548 σ \approx 6.43 cm

Question 5

6 marks

The gradient of a tangent to a cosine curve and a given angle with the x-axis

Calculus

(a)

3 marks

Differentiate $h (x) = 15 cos (\frac{π x}{50}) + 15$ with the chain rule.

h^{'} (x) = - 15 \cdot \frac{π}{50} sin (\frac{π x}{50}) = - \frac{3 π}{10} sin (\frac{π x}{50}) gradient of T_{k} = - \frac{3 π}{10} sin (\frac{π k}{50})

(b)

3 marks

The tangent slopes downwards (the curve falls from 30 to 0), so its gradient is $- tan \frac{π}{8} = - 0.4142$ .
Solve on $0 \leq k \leq 50$ , where $\frac{π k}{50}$ is between 0 and $π$ . There are two solutions, symmetric about $k = 25$ .

- \frac{3 π}{10} sin (\frac{π k}{50}) = - 0.4142 \Rightarrow sin (\frac{π k}{50}) = 0.4395 k \approx 7.24 or k \approx 42.8

Question 6

7 marks

f(x) = 4 cot x + sin x: rewriting, a sketch, an inverse value and a value from sec

Functions

(a)

3 marks

(i) Replace $cot x$ .

(ii) The graph on $0 < x < π$ falls from large positive values near $x = 0$ , passes through the $x$ -axis, and goes to large negative values near $x = π$ .

The $x$ -intercept is where $4 cos x + sin^{2} x = 0$ .

Sketch: a decreasing curve with vertical asymptotes at $x = 0$ and $x = π$ crossing the $x$ -axis at $(1.81, 0)$ .

f (x) = \frac{4 cos x}{sin x} + sin x x \approx 1.81

(b)

1 mark

$f^{- 1} (2)$ is the $x$ with $f (x) = 2$ . Solve with the calculator on $0 < x < π$ .

f^{- 1} (2) \approx 1.32

(c)

3 marks

Since $sec α = 1.5$ , $cos α = \frac{2}{3}$ , and $α$ is in the first quadrant, so $sin α = 1 - \frac{4}{9} = \frac{5}{3}$ .

cot α = \frac{2/3}{5 /3} = \frac{2}{5} f (α) = 4 \cdot \frac{2}{5} + \frac{5}{3} = \frac{8 5}{5} + \frac{5}{3} = \frac{29 5}{15} \approx 4.32

Question 7

5 marks

A helicopter: speed and the angle of descent

Number and algebra

(a)

2 marks

The velocity vector is $4 10 - 25 0 = 40 - 100 0$ .

speed = 4 0^{2} + 10 0^{2} = 11600 = 2029 \approx 108 km h^{- 1}

(b)

3 marks

During the descent the horizontal speed is still $2029$ and the vertical speed is 16 km h $^{- 1}$ downwards.

tan β = \frac{16}{20 29} = \frac{4}{5 29} β \approx 8.4 5^{\circ}

Question 8

7 marks

Adding two sine waves with complex numbers

Number and algebra

(a)

2 marks

Write each sine as the imaginary part of a complex exponential.
Add and factor out $e^{2 t i}$ , using that the imaginary part of a sum is the sum of the imaginary parts.

f (t) = Im (e^{i (2 t + 1)}), g (t) = Im (e^{i (2 t + 3)}) h (t) = Im (e^{2 t i} e^{i} + e^{2 t i} e^{3 i}) = Im (e^{2 t i} (e^{i} + e^{3 i}))

(b)

2 marks

Factor out $e^{2 i}$ from $e^{i} + e^{3 i}$ .

e^{i} + e^{3 i} = e^{2 i} (e^{- i} + e^{i}) = 2 cos (1) e^{2 i} r = 2 cos 1 \approx 1.08, θ = 2

(c)

3 marks

Substitute into the result of (a).
Here $p = 2 cos 1$ and $q = 2$ , which is in the interval $0 < q < 2 π$ .

h (t) = Im (e^{2 t i} \times 2 cos (1) e^{2 i}) = Im (2 cos (1) e^{i (2 t + 2)}) h (t) = 2 cos (1) sin (2 t + 2) \approx 1.08 sin (2 t + 2)

Question 9

6 marks

Euler's method for dy/dx = 2x/(x^2 + y): an estimate and its accuracy

Calculus

(a)

3 marks

Start at $(1, 0)$ and take four steps of 0.25.

y (1.25) \approx 0 + 0.25 (\frac{2}{1}) = 0.5 y (1.5) \approx 0.5 + 0.25 (\frac{2.5}{1.5625 + 0.5}) \approx 0.8030 y (1.75) \approx 0.8030 + 0.25 (\frac{3}{2.25 + 0.8030}) \approx 1.0487 y (2) \approx 1.0487 + 0.25 (\frac{3.5}{3.0625 + 1.0487}) \approx 1.26

(b)

3 marks

(i) The solution curve is concave down, so each tangent line segment lies above the curve and the steps get further above the curve. The estimate is an overestimate.

Section B · Extended response

Questions 10–12 · 55 marks

Question 10

16 marks

A pyramid of playing cards: a quadratic number sequence

Number and algebra

(a)

1 mark

From the diagram of the stack with 3 rows, count the cards.

t_{3} = 2 + 5 + 8 = 15

(b)

2 marks

A fourth row adds 11 cards (the differences between rows are 3).

t_{4} = 15 + 11 = 26

(c)

3 marks

The numbers of cards in the rows form the arithmetic sequence $2, 5, 8, 11, \dots$ with first term 2 and common difference 3. Use the sum of $n$ terms.

t_{n} = \frac{n}{2} [2 (2) + (n - 1) 3] = \frac{n}{2} (3 n + 1)

(d)

3 marks

The cards available are $14 \times 52 = 728$ . Solve $\frac{n ( 3 n + 1 )}{2} \leq 728$ .
The most rows is 21.

t_{21} = \frac{21 ( 64 )}{2} = 672 \leq 728, t_{22} = \frac{22 ( 67 )}{2} = 737 > 728

(e)

2 marks

The number of cards $t_{n}$ must be a multiple of 52. List $t_{n}$ : 2, 7, 15, 26, 40, 57, 77, 100, 126, 155, 187, 222, 260.
The first multiple of 52 is $t_{13} = 260 = 5 \times 52$ (so 5 packs).

n = 13

(f)

5 marks

Each row is an equilateral triangle with side 88 mm, so its height is
For the stack to be taller than 2000 mm we need $76.21 n > 2000$ .
The number of cards for 27 rows.

88 sin 6 0^{\circ} = 443 \approx 76.21 mm n > 26.2 \Rightarrow n = 27 t_{27} = \frac{27 ( 82 )}{2} = 1107

Question 11

19 marks

Marathon times: a cosine-based probability density function

Statistics and probability

(a)

4 marks

(i) Integrate the first branch of the density.

(ii) The maximum of the graph is at $t = 4.5$ .

(iii) The probability of finishing before the mode is $\frac{3}{7} < 0.5$ , so more than half the area is to the right of 4.5. The median is greater than the mode (the median is about 4.69).

\int_{2.25}^{4.5} f (t) d t = \frac{4}{21} [t - \frac{9}{4 π} sin (\frac{4 π}{9} (t - 2.25))]_{2.25}^{4.5} = \frac{4}{21} (2.25) = \frac{3}{7} \approx 0.429 mode = 4.5 hours

(b)

2 marks

A fast runner finishes in 3.5 hours or less.

P (T \leq 3.5) = \int_{2.25}^{3.5} f (t) d t \approx 0.104

(c)

3 marks

Use conditional probability.

P (T \leq 3 ∣ T \leq 3.5) = \frac{P ( T \leq 3 )}{P ( T \leq 3.5 )} = \frac{0.02472}{0.10375} \approx 0.238

(d)

3 marks

The lower quartile $q$ satisfies $\int_{2.25}^{q} f (t) d t = 0.25$ .
Solve with the calculator (the lower quartile is below the mode, so use the first branch).

\frac{4}{21} [t - \frac{9}{4 π} sin (\frac{4 π}{9} (t - 2.25))]_{2.25}^{q} = 0.25 q \approx 4.01 hours

(e)

5 marks

The expected score is $E (P) = a - b E (T)$ .
The maximum score of 150 goes to the fastest possible time, 2.25 hours.
Subtract the first equation from the second: $2.473 b = 50$ .

a - 4.723 b = 100 a - 2.25 b = 150 b \approx 20.2, a \approx 195.5 a = 195, b = 20

(f)

2 marks

Use $Var (a - b T) = b^{2} Var (T)$ with $b = 20.22$ .

Var (P) = (20.22)^{2} (0.906) \approx 370

Question 12

20 marks

A family of polynomials, an infinite series, an inverse function and an enclosed area

Number and algebra

(a)

3 marks

Replace $x$ by $- x$ in $f_{n} (x) = \sum_{r = 0}^{n} (- 2 x^{2})^{r}$ .
Each term is unchanged, so $f_{n} (- x) = f_{n} (x)$ and $f_{n}$ is even for every $n$ .

(- 2 (- x)^{2})^{r} = (- 2 x^{2})^{r}

(b)

2 marks

(i) Put $n = 3$ .

(ii) Add the next term, $(- 2 x^{2})^{4} = 16 x^{8}$ .

f_{3} (x) = 1 + (- 2 x^{2}) + (- 2 x^{2})^{2} + (- 2 x^{2})^{3} = 1 - 2 x^{2} + 4 x^{4} - 8 x^{6} f_{4} (x) = 1 - 2 x^{2} + 4 x^{4} - 8 x^{6} + 16 x^{8}

(c)

5 marks

(i) As $n \to \infty$ , $f_{n} (x)$ is a geometric series with first term 1 and common ratio $- 2 x^{2}$ . It converges when $- 2 x^{2} < 1$ .

(ii) The sum to infinity is $\frac{1}{1 - r}$ .

2 x^{2} < 1 \Rightarrow ∣ x ∣ < \frac{1}{2} K = \frac{1}{2} = \frac{2}{2} f (x) = \frac{1}{1 - ( - 2 x ^{2} )} = \frac{1}{1 + 2 x ^{2}}, a = 1, b = 2

(d)

6 marks

(i) For $0 \leq x < K$ , $g (x) = \frac{1}{1 + 2 x ^{2}}$ is strictly decreasing and has no turning points, so it is one-to-one, and the inverse exists.

(ii) Swap $x$ and $y$ and solve for $y$ .

The domain of $g$ gives $y \geq 0$ , so take the positive root.

The domain of $g^{- 1}$ is the range of $g$ . Since $g (0) = 1$ and $g (K) = \frac{1}{2}$ ,

x = \frac{1}{1 + 2 y ^{2}} \Rightarrow 1 + 2 y^{2} = \frac{1}{x} \Rightarrow y^{2} = \frac{1 - x}{2 x} g^{- 1} (x) = \frac{1 - x}{2 x} \frac{1}{2} < x \leq 1

(e)

4 marks

The curves $y = g (x)$ and $y = g^{- 1} (x)$ are reflections in $y = x$ and meet where $g (x) = x$ .
By symmetry, the area is twice the area between $y = g (x)$ and $y = x$ from 0 to 0.5898.

x \approx 0.5898 Area = 2 \int_{0}^{0.5898} (\frac{1}{1 + 2 x ^{2}} - x) d x \approx 0.635

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 algebra54 marks · 49%
Q3, Q7, Q8, Q10, Q12
Statistics and probability27 marks · 25%
Q4, Q11
Calculus12 marks · 11%
Q5, Q9
Functions11 marks · 10%
Q2, Q6
Geometry and trigonometry6 marks · 5%
Q1

Mentor insights

Traps to watch for

01
Which line to use · Q4
To estimate $A$ from $F$ use the regression line of $A$ on $F$ .
02
Cotangent · Q6
$cot x = \frac{cos x}{sin x}$ .
03
Counting · Q10
Look at how many cards are added to make each new row, and find the pattern.
04
Probability from a pdf · Q11
Integrate the density over the interval.
05
Even function · Q12
$f (- x) = f (x)$ , which holds when $x$ only appears as $x^{2}$ .

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