Loading…

Worked solutions

IB Maths AA HL · Paper 2

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2022
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 2022 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

4 marks

Exercise hours: a missing frequency and a standard deviation

Statistics and probability

(a)

2 marks

The frequencies for 2, 3, 4 hours add to $5 + 1 + 4 = 10$ . For the median to be 4.5, half the students (10) must have exercised 4 hours or less, so the total is 20.

5 + 1 + 4 + 3 + x = 20 \Rightarrow x = 7

(b)

2 marks

The mean is $\frac{2 ( 5 ) + 3 ( 1 ) + 4 ( 4 ) + 5 ( 3 ) + 6 ( 7 )}{20} = 4.3$ . Enter the frequency table in the calculator for the standard deviation.

σ \approx 1.58

Question 2

6 marks

A logo made by removing two circular segments

Geometry and trigonometry

(a)

3 marks

Take the sector and subtract the triangle OAB.

segment = \frac{1}{2} (2)^{2} θ - \frac{1}{2} (2)^{2} sin θ = 2 θ - 2 sin θ

(b)

3 marks

The logo is the $5 \times 4 = 20$ rectangle minus two equal segments.
Solve on the calculator in radians, for $0 < θ < π$ .

20 - 2 (2 θ - 2 sin θ) = 13.4 \Rightarrow θ - sin θ = 1.65 θ \approx 2.36

Question 3

6 marks

A probability distribution with a parameter

Statistics and probability

(a)

1 mark

The probabilities add to 1.

0.41 + (k - 0.28) + 0.46 + (0.29 - 2 k^{2}) = 1 \Rightarrow 0.88 + k - 2 k^{2} = 1 \Rightarrow 2 k^{2} - k + 0.12 = 0

(b)

3 marks

Solve the quadratic.
If $k = 0.2$ then $k - 0.28 = - 0.08 < 0$ , which is not a probability. So

k = \frac{1 \pm 1 - 0.96}{4} = \frac{1 \pm 0.2}{4} \Rightarrow k = 0.3 or k = 0.2 k = 0.3

(c)

2 marks

The probabilities are $0.41$ , $0.02$ , $0.46$ , $0.11$ for $x = 0, 1, 2, 3$ .

E (X) = 0 (0.41) + 1 (0.02) + 2 (0.46) + 3 (0.11) = 1.27

Question 4

7 marks

A particle's velocity: rest, acceleration and total distance

Calculus

(a)

2 marks

The particle is at rest when $v = 0$ . Solve $e^{s i n t} + 4 sin t = 0$ on the calculator for $0 \leq t \leq 6$ , taking the first value.

t \approx 3.35 s

(b)

3 marks

The particle changes direction when $v = 0$ , so at $t = 3.347$ . The acceleration is $v^{'} (t) = cos t e^{s i n t} + 4 cos t$ .

a (3.347) \approx - 4.71 m s^{- 2}

(c)

2 marks

The total distance is the integral of the speed.

\int_{0}^{6} ∣ v ∣ d t = \int_{0}^{3.347} v d t - \int_{3.347}^{6} v d t \approx 20.8 m

Question 5

6 marks

Independent events and a conditional probability

Statistics and probability

(a)

4 marks

Let $P (A \cap B) = x$ . Then $P (A) = x + 0.16$ and $P (B) = x + 0.36$ .

x = (x + 0.16) (x + 0.36) \Rightarrow x^{2} - 0.48 x + 0.0576 = 0 \Rightarrow (x - 0.24)^{2} = 0 x = 0.24

(b)

2 marks

Independent events have independent complements, so $P (A^{'} ∣ B^{'}) = P (A^{'})$ . Here $P (A) = 0.24 + 0.16 = 0.4$ .

P (A^{'} ∣ B^{'}) = 1 - 0.4 = 0.6

Question 6

6 marks

An odd function and an inequality between two functions

Functions

(a)

2 marks

Replace $x$ by $- x$ .
So $f$ is an odd function.

f (- x) = 2^{- x} - \frac{1}{2 ^{- x}} = \frac{1}{2 ^{x}} - 2^{x} = - f (x)

(b)

4 marks

Find where $f (x) = g (x)$ on the calculator.
Test a value in each interval around these and the asymptotes $x = - 1$ and $x = 3$ : $f (x) - g (x)$ is positive on $(- 1.27, - 1)$ , negative on $(- 1, 0.178)$ , positive on $(0.178, 3)$ , and negative on $(3, 3.06)$ and positive after 3.06.

x \approx - 1.27, x \approx 0.178, x \approx 3.06 - 1.27 \leq x < - 1, 0.178 \leq x < 3, x \geq 3.06

Question 7

9 marks

Vectors: the range of a sum, its minimum, and a perpendicular vector

Geometry and trigonometry

(a)

2 marks

The length of $a$ is $1 2^{2} + 5^{2} = 13$ . When $b$ points the opposite way the length is $15 - 13$ , and the same way it is $15 + 13$ .

2 \leq ∣ a + b ∣ \leq 28

(b)

2 marks

The minimum happens when $b$ is opposite to $a$ , so $b = - \frac{15}{13} a$ .

p = a + b = (1 - \frac{15}{13}) a = - \frac{2}{13} (12 - 5) = (- \frac{24}{13} \frac{10}{13})

(c)

5 marks

$q$ is perpendicular to $a$ , so $q \cdot a = 12 x - 5 y = 0$ and $y = \frac{12}{5} x$ . Its length is 15.

x^{2} + (\frac{12}{5} x)^{2} = 1 5^{2} \Rightarrow \frac{169}{25} x^{2} = 225 \Rightarrow x = \frac{75}{13} q = (\frac{75}{13} \frac{180}{13}) \approx (5.77 13.8)

Question 8

4 marks

A quadratic with one positive and one negative root

Number and algebra

(a)

1 mark

For $k x^{2} - (k + 3) x + (2 k + 9) = 0$ , the product of roots is $\frac{c}{a}$ .

\frac{2 k + 9}{k}

(b)

3 marks

One root positive and one negative means the product is negative.
The critical values are $k = - \frac{9}{2}$ and $k = 0$ , and the fraction is negative between them.

\frac{2 k + 9}{k} < 0 - \frac{9}{2} < k < 0

Question 9

7 marks

Seating five people in a row of ten seats

Statistics and probability

(a)

3 marks

Treat the five people as a block of 5 adjacent seats: there are 6 positions for the block in a row of 10, and $5!$ orders inside.

6 \times 5! = 720

(b)

4 marks

Peter at an end (2 choices of seat): the one seat next to him is out, leaving 8 seats for the four girls, in order.
Peter in one of the 8 other seats: both neighbouring seats are out, leaving 7 seats for the four girls.

2 \times^{8} P_{4} = 2 \times 1680 = 3360 8 \times^{7} P_{4} = 8 \times 840 = 6720 3360 + 6720 = 10080

Section B · Extended response

Questions 10–12 · 55 marks

Question 10

20 marks

Rotating $y = x^{2} - 1$ to model a water container

Functions
Calculus

(a)

2 marks

The curve starts at $(1, 0)$ and rises with decreasing gradient to $(2, 3)$ , where $3 \approx 1.73$ .

(b)

5 marks

(i) Interchange $x$ and $y$ in $y = x^{2} - 1$ and solve for $y$ .

(ii) The domain of $f^{- 1}$ is the range of $f$ , and its range is the domain of $f$ .

x = y^{2} - 1 \Rightarrow x^{2} = y^{2} - 1 \Rightarrow f^{- 1} (x) = x^{2} + 1 domain: 0 \leq x \leq 3, range: 1 \leq y \leq 2

(c)

5 marks

(i) Rotating about the $y$ -axis, the cross-sectional radius at height $y$ is $x = y^{2} + 1$ .

(ii) The maximum height is $h = 3$ .

V = π \int_{0}^{h} x^{2} d y = π \int_{0}^{h} (y^{2} + 1) d y = π (\frac{h ^{3}}{3} + h) V_{m a x} = π (\frac{3 3}{3} + 3) = 23 π \approx 10.9 m^{3}

(d)

2 marks

Divide the volume by the filling rate.

t = \frac{2 3 π}{0.4} = 53 π \approx 27.2 s

(e)

6 marks

Half the maximum volume is $3 π$ . Find the height.
Use the chain rule: $\frac{d V}{d t} = \frac{d V}{d h} \frac{d h}{d t}$ with $\frac{d V}{d h} = π (h^{2} + 1)$ .

π (\frac{h ^{3}}{3} + h) = 3 π \Rightarrow h \approx 1.182 0.4 = π (h^{2} + 1) \frac{d h}{d t} \Rightarrow \frac{d h}{d t} = \frac{0.4}{π ( 1.18 2 ^{2} + 1 )} \approx 0.0531 m s^{- 1}

Question 11

16 marks

Chocolate and banana muffins: normal probabilities and an unknown standard deviation

Statistics and probability

(a)

2 marks

With $C \sim N (62, 2. 9^{2})$ :

P (C < 61) \approx 0.365

(b)

2 marks

The number of light muffins among 12 is $B (12, 0.365)$ .

P (X = 5) \approx 0.214

(c)

7 marks

(i) For banana muffins, $P (B < 61) = 0.0198$ with $B \sim N (68, 3. 4^{2})$ . Chocolate muffins are 60% of the total.

(ii) A conditional probability: the chocolate share of the muffins under 61 g.

P (muffin < 61) = 0.6 (0.3651) + 0.4 (0.01976) \approx 0.227 P (chocolate ∣< 61) = \frac{0.6 ( 0.3651 )}{0.227} \approx 0.965

(d)

5 marks

The overall probability is now 0.157, so $0.6 P (C < 61) + 0.4 (0.01976) = 0.157$ .
Use the inverse normal: $z = - 0.6792$ , so $\frac{61 - 62}{σ} = - 0.6792$ .

P (C < 61) = \frac{0.157 - 0.00790}{0.6} \approx 0.2485 σ = \frac{1}{0.6792} \approx 1.47 g

Question 12

19 marks

A differential equation $\frac{d y}{d x} = \frac{y ^{2}}{x ^{2}} - 2$ : Euler, substitution and exact solution

Calculus

(a)

4 marks

Euler's method with $h = 0.1$ : $y_{n + 1} = y_{n} + 0.1 (\frac{y _{n}^{2}}{x _{n}^{2}} - 2)$ from $(1, 3)$ .

y (1.1) = 3.7, y (1.2) \approx 4.631, y (1.3) \approx 5.921, y (1.4) \approx 7.795 y (1.5) \approx 10.7

(b)

3 marks

Let $y = v x$ , so $\frac{d y}{d x} = v + x \frac{d v}{d x}$ . Substitute into the equation, with $\frac{y ^{2}}{x ^{2}} = v^{2}$ .

v + x \frac{d v}{d x} = v^{2} - 2 \Rightarrow x \frac{d v}{d x} = v^{2} - v - 2

(c)

12 marks

(i) Separate the variables and use partial fractions.

At $x = 1$ , $y = 3$ , so $v = 3$ : $c = \frac{1}{3} (ln 1 - ln 4) = - \frac{1}{3} ln 4$ .

Replace $v = \frac{y}{x}$ and solve for $y$ .

(ii) At $x = 1.5$ :

(iii) The graph has a vertical asymptote at $x = 34 \approx 1.587$ , close to 1.5. The gradient grows very rapidly there, so Euler's method with steps of 0.1 falls well short of the true value.

\int \frac{d v}{( v - 2 ) ( v + 1 )} = \int \frac{d x}{x}, \frac{1}{( v - 2 ) ( v + 1 )} = \frac{1}{3} (\frac{1}{v - 2} - \frac{1}{v + 1}) \frac{1}{3} (ln (v - 2) - ln (v + 1)) = ln x + c ln \frac{v - 2}{v + 1} = 3 ln x - ln 4 \Rightarrow \frac{v - 2}{v + 1} = \frac{x ^{3}}{4} 4 (y - 2 x) = x^{3} (y + x) \Rightarrow y (4 - x^{3}) = x^{4} + 8 x \Rightarrow y = \frac{x ^{4} + 8 x}{4 - x ^{3}} y = \frac{5.0625 + 12}{4 - 3.375} = 27.3

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 probability39 marks · 35%
Q1, Q3, Q5, Q9, Q11
Calculus39 marks · 35%
Q4, Q10, Q12
Geometry and trigonometry15 marks · 14%
Q2, Q7
Functions13 marks · 12%
Q6, Q10
Number and algebra4 marks · 4%
Q8

Mentor insights

Traps to watch for

01
Radian mode · Q4
The velocity involves $sin t$ with $t$ in seconds, so the calculator must be in radians. The acceleration is the derivative of $v$ .
02
Triangle inequality · Q7
The length of $a + b$ is largest when the vectors point the same way and smallest when they point opposite ways.
03
Rotate about the $y$ -axis · Q10
The volume uses $π \int x^{2} d y$ , so write $x^{2}$ in terms of $y$ first. Then the filling rate relates $\frac{d V}{d t}$ and $\frac{d h}{d t}$ by the chain rule.
04
Total probability, then Bayes · Q11
In (c)(i) combine the two muffin types using their shares of 60% and 40%. In (c)(ii) divide the chocolate part of that sum by the total.
05
Substitute $y = v x$ · Q12
With $y = v x$ the right-hand side depends only on $v$ , so the equation becomes separable. Use the product rule on $y = v x$ .

Solutions and tips are GeorgeTheTutor's own; the questions are not reproduced here. Not affiliated with or endorsed by the International Baccalaureate Organization.

Want someone to go through your paper with you?

Our maths mentors took these exams recently. One-to-one, they'll find where your marks are going and fix it before May.

Find your maths mentor More past papers

Worked solutions

IB Maths AA HL · Paper 2

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2022
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 2022 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

4 marks

Exercise hours: a missing frequency and a standard deviation

Statistics and probability

(a)

2 marks

The frequencies for 2, 3, 4 hours add to $5 + 1 + 4 = 10$ . For the median to be 4.5, half the students (10) must have exercised 4 hours or less, so the total is 20.

5 + 1 + 4 + 3 + x = 20 \Rightarrow x = 7

(b)

2 marks

The mean is $\frac{2 ( 5 ) + 3 ( 1 ) + 4 ( 4 ) + 5 ( 3 ) + 6 ( 7 )}{20} = 4.3$ . Enter the frequency table in the calculator for the standard deviation.

σ \approx 1.58

Question 2

6 marks

A logo made by removing two circular segments

Geometry and trigonometry

(a)

3 marks

Take the sector and subtract the triangle OAB.

segment = \frac{1}{2} (2)^{2} θ - \frac{1}{2} (2)^{2} sin θ = 2 θ - 2 sin θ

(b)

3 marks

The logo is the $5 \times 4 = 20$ rectangle minus two equal segments.
Solve on the calculator in radians, for $0 < θ < π$ .

20 - 2 (2 θ - 2 sin θ) = 13.4 \Rightarrow θ - sin θ = 1.65 θ \approx 2.36

Question 3

6 marks

A probability distribution with a parameter

Statistics and probability

(a)

1 mark

The probabilities add to 1.

0.41 + (k - 0.28) + 0.46 + (0.29 - 2 k^{2}) = 1 \Rightarrow 0.88 + k - 2 k^{2} = 1 \Rightarrow 2 k^{2} - k + 0.12 = 0

(b)

3 marks

Solve the quadratic.
If $k = 0.2$ then $k - 0.28 = - 0.08 < 0$ , which is not a probability. So

k = \frac{1 \pm 1 - 0.96}{4} = \frac{1 \pm 0.2}{4} \Rightarrow k = 0.3 or k = 0.2 k = 0.3

(c)

2 marks

The probabilities are $0.41$ , $0.02$ , $0.46$ , $0.11$ for $x = 0, 1, 2, 3$ .

E (X) = 0 (0.41) + 1 (0.02) + 2 (0.46) + 3 (0.11) = 1.27

Question 4

7 marks

A particle's velocity: rest, acceleration and total distance

Calculus

(a)

2 marks

The particle is at rest when $v = 0$ . Solve $e^{s i n t} + 4 sin t = 0$ on the calculator for $0 \leq t \leq 6$ , taking the first value.

t \approx 3.35 s

(b)

3 marks

The particle changes direction when $v = 0$ , so at $t = 3.347$ . The acceleration is $v^{'} (t) = cos t e^{s i n t} + 4 cos t$ .

a (3.347) \approx - 4.71 m s^{- 2}

(c)

2 marks

The total distance is the integral of the speed.

\int_{0}^{6} ∣ v ∣ d t = \int_{0}^{3.347} v d t - \int_{3.347}^{6} v d t \approx 20.8 m

Question 5

6 marks

Independent events and a conditional probability

Statistics and probability

(a)

4 marks

Let $P (A \cap B) = x$ . Then $P (A) = x + 0.16$ and $P (B) = x + 0.36$ .

x = (x + 0.16) (x + 0.36) \Rightarrow x^{2} - 0.48 x + 0.0576 = 0 \Rightarrow (x - 0.24)^{2} = 0 x = 0.24

(b)

2 marks

Independent events have independent complements, so $P (A^{'} ∣ B^{'}) = P (A^{'})$ . Here $P (A) = 0.24 + 0.16 = 0.4$ .

P (A^{'} ∣ B^{'}) = 1 - 0.4 = 0.6

Question 6

6 marks

An odd function and an inequality between two functions

Functions

(a)

2 marks

Replace $x$ by $- x$ .
So $f$ is an odd function.

f (- x) = 2^{- x} - \frac{1}{2 ^{- x}} = \frac{1}{2 ^{x}} - 2^{x} = - f (x)

(b)

4 marks

Find where $f (x) = g (x)$ on the calculator.
Test a value in each interval around these and the asymptotes $x = - 1$ and $x = 3$ : $f (x) - g (x)$ is positive on $(- 1.27, - 1)$ , negative on $(- 1, 0.178)$ , positive on $(0.178, 3)$ , and negative on $(3, 3.06)$ and positive after 3.06.

x \approx - 1.27, x \approx 0.178, x \approx 3.06 - 1.27 \leq x < - 1, 0.178 \leq x < 3, x \geq 3.06

Question 7

9 marks

Vectors: the range of a sum, its minimum, and a perpendicular vector

Geometry and trigonometry

(a)

2 marks

The length of $a$ is $1 2^{2} + 5^{2} = 13$ . When $b$ points the opposite way the length is $15 - 13$ , and the same way it is $15 + 13$ .

2 \leq ∣ a + b ∣ \leq 28

(b)

2 marks

The minimum happens when $b$ is opposite to $a$ , so $b = - \frac{15}{13} a$ .

p = a + b = (1 - \frac{15}{13}) a = - \frac{2}{13} (12 - 5) = (- \frac{24}{13} \frac{10}{13})

(c)

5 marks

$q$ is perpendicular to $a$ , so $q \cdot a = 12 x - 5 y = 0$ and $y = \frac{12}{5} x$ . Its length is 15.

x^{2} + (\frac{12}{5} x)^{2} = 1 5^{2} \Rightarrow \frac{169}{25} x^{2} = 225 \Rightarrow x = \frac{75}{13} q = (\frac{75}{13} \frac{180}{13}) \approx (5.77 13.8)

Question 8

4 marks

A quadratic with one positive and one negative root

Number and algebra

(a)

1 mark

For $k x^{2} - (k + 3) x + (2 k + 9) = 0$ , the product of roots is $\frac{c}{a}$ .

\frac{2 k + 9}{k}

(b)

3 marks

One root positive and one negative means the product is negative.
The critical values are $k = - \frac{9}{2}$ and $k = 0$ , and the fraction is negative between them.

\frac{2 k + 9}{k} < 0 - \frac{9}{2} < k < 0

Question 9

7 marks

Seating five people in a row of ten seats

Statistics and probability

(a)

3 marks

Treat the five people as a block of 5 adjacent seats: there are 6 positions for the block in a row of 10, and $5!$ orders inside.

6 \times 5! = 720

(b)

4 marks

Peter at an end (2 choices of seat): the one seat next to him is out, leaving 8 seats for the four girls, in order.
Peter in one of the 8 other seats: both neighbouring seats are out, leaving 7 seats for the four girls.

2 \times^{8} P_{4} = 2 \times 1680 = 3360 8 \times^{7} P_{4} = 8 \times 840 = 6720 3360 + 6720 = 10080

Section B · Extended response

Questions 10–12 · 55 marks

Question 10

20 marks

Rotating $y = x^{2} - 1$ to model a water container

Functions
Calculus

(a)

2 marks

The curve starts at $(1, 0)$ and rises with decreasing gradient to $(2, 3)$ , where $3 \approx 1.73$ .

(b)

5 marks

(i) Interchange $x$ and $y$ in $y = x^{2} - 1$ and solve for $y$ .

(ii) The domain of $f^{- 1}$ is the range of $f$ , and its range is the domain of $f$ .

x = y^{2} - 1 \Rightarrow x^{2} = y^{2} - 1 \Rightarrow f^{- 1} (x) = x^{2} + 1 domain: 0 \leq x \leq 3, range: 1 \leq y \leq 2

(c)

5 marks

(i) Rotating about the $y$ -axis, the cross-sectional radius at height $y$ is $x = y^{2} + 1$ .

(ii) The maximum height is $h = 3$ .

V = π \int_{0}^{h} x^{2} d y = π \int_{0}^{h} (y^{2} + 1) d y = π (\frac{h ^{3}}{3} + h) V_{m a x} = π (\frac{3 3}{3} + 3) = 23 π \approx 10.9 m^{3}

(d)

2 marks

Divide the volume by the filling rate.

t = \frac{2 3 π}{0.4} = 53 π \approx 27.2 s

(e)

6 marks

Half the maximum volume is $3 π$ . Find the height.
Use the chain rule: $\frac{d V}{d t} = \frac{d V}{d h} \frac{d h}{d t}$ with $\frac{d V}{d h} = π (h^{2} + 1)$ .

π (\frac{h ^{3}}{3} + h) = 3 π \Rightarrow h \approx 1.182 0.4 = π (h^{2} + 1) \frac{d h}{d t} \Rightarrow \frac{d h}{d t} = \frac{0.4}{π ( 1.18 2 ^{2} + 1 )} \approx 0.0531 m s^{- 1}

Question 11

16 marks

Chocolate and banana muffins: normal probabilities and an unknown standard deviation

Statistics and probability

(a)

2 marks

With $C \sim N (62, 2. 9^{2})$ :

P (C < 61) \approx 0.365

(b)

2 marks

The number of light muffins among 12 is $B (12, 0.365)$ .

P (X = 5) \approx 0.214

(c)

7 marks

(i) For banana muffins, $P (B < 61) = 0.0198$ with $B \sim N (68, 3. 4^{2})$ . Chocolate muffins are 60% of the total.

(ii) A conditional probability: the chocolate share of the muffins under 61 g.

P (muffin < 61) = 0.6 (0.3651) + 0.4 (0.01976) \approx 0.227 P (chocolate ∣< 61) = \frac{0.6 ( 0.3651 )}{0.227} \approx 0.965

(d)

5 marks

The overall probability is now 0.157, so $0.6 P (C < 61) + 0.4 (0.01976) = 0.157$ .
Use the inverse normal: $z = - 0.6792$ , so $\frac{61 - 62}{σ} = - 0.6792$ .

P (C < 61) = \frac{0.157 - 0.00790}{0.6} \approx 0.2485 σ = \frac{1}{0.6792} \approx 1.47 g

Question 12

19 marks

A differential equation $\frac{d y}{d x} = \frac{y ^{2}}{x ^{2}} - 2$ : Euler, substitution and exact solution

Calculus

(a)

4 marks

Euler's method with $h = 0.1$ : $y_{n + 1} = y_{n} + 0.1 (\frac{y _{n}^{2}}{x _{n}^{2}} - 2)$ from $(1, 3)$ .

y (1.1) = 3.7, y (1.2) \approx 4.631, y (1.3) \approx 5.921, y (1.4) \approx 7.795 y (1.5) \approx 10.7

(b)

3 marks

Let $y = v x$ , so $\frac{d y}{d x} = v + x \frac{d v}{d x}$ . Substitute into the equation, with $\frac{y ^{2}}{x ^{2}} = v^{2}$ .

v + x \frac{d v}{d x} = v^{2} - 2 \Rightarrow x \frac{d v}{d x} = v^{2} - v - 2

(c)

12 marks

(i) Separate the variables and use partial fractions.

At $x = 1$ , $y = 3$ , so $v = 3$ : $c = \frac{1}{3} (ln 1 - ln 4) = - \frac{1}{3} ln 4$ .

Replace $v = \frac{y}{x}$ and solve for $y$ .

(ii) At $x = 1.5$ :

(iii) The graph has a vertical asymptote at $x = 34 \approx 1.587$ , close to 1.5. The gradient grows very rapidly there, so Euler's method with steps of 0.1 falls well short of the true value.

\int \frac{d v}{( v - 2 ) ( v + 1 )} = \int \frac{d x}{x}, \frac{1}{( v - 2 ) ( v + 1 )} = \frac{1}{3} (\frac{1}{v - 2} - \frac{1}{v + 1}) \frac{1}{3} (ln (v - 2) - ln (v + 1)) = ln x + c ln \frac{v - 2}{v + 1} = 3 ln x - ln 4 \Rightarrow \frac{v - 2}{v + 1} = \frac{x ^{3}}{4} 4 (y - 2 x) = x^{3} (y + x) \Rightarrow y (4 - x^{3}) = x^{4} + 8 x \Rightarrow y = \frac{x ^{4} + 8 x}{4 - x ^{3}} y = \frac{5.0625 + 12}{4 - 3.375} = 27.3

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 probability39 marks · 35%
Q1, Q3, Q5, Q9, Q11
Calculus39 marks · 35%
Q4, Q10, Q12
Geometry and trigonometry15 marks · 14%
Q2, Q7
Functions13 marks · 12%
Q6, Q10
Number and algebra4 marks · 4%
Q8

Mentor insights

Traps to watch for

01
Radian mode · Q4
The velocity involves $sin t$ with $t$ in seconds, so the calculator must be in radians. The acceleration is the derivative of $v$ .
02
Triangle inequality · Q7
The length of $a + b$ is largest when the vectors point the same way and smallest when they point opposite ways.
03
Rotate about the $y$ -axis · Q10
The volume uses $π \int x^{2} d y$ , so write $x^{2}$ in terms of $y$ first. Then the filling rate relates $\frac{d V}{d t}$ and $\frac{d h}{d t}$ by the chain rule.
04
Total probability, then Bayes · Q11
In (c)(i) combine the two muffin types using their shares of 60% and 40%. In (c)(ii) divide the chocolate part of that sum by the total.
05
Substitute $y = v x$ · Q12
With $y = v x$ the right-hand side depends only on $v$ , so the equation becomes separable. Use the product rule on $y = v x$ .

Solutions and tips are GeorgeTheTutor's own; the questions are not reproduced here. Not affiliated with or endorsed by the International Baccalaureate Organization.

Want someone to go through your paper with you?

Our maths mentors took these exams recently. One-to-one, they'll find where your marks are going and fix it before May.

Find your maths mentor More past papers