Loading…

Worked solutions

IB Maths AA SL · Paper 1

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2

Session: November 2023
Marks: 80
Duration: 1 hour 30 min
Calculator: Not allowed
Questions: 9

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2023 TZ2 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 1–6 · 35 marks

Question 1

7 marks

Reading a cosine graph: amplitude, period and a value

Functions

(a)

1 mark

The graph reaches a maximum of 5.

a = 5

(b)

3 marks

One full wave takes $π$ .
The period is $\frac{2 π}{b}$ .

period = π \frac{2 π}{b} = π \Rightarrow b = 2

(c)

3 marks

Substitute into $f (x) = 5 cos (2 x)$ .

f (\frac{π}{6}) = 5 cos \frac{π}{3} = 5 \times \frac{1}{2} = \frac{5}{2}

Question 2

5 marks

A composite function with a constant, solved for k

Functions

(a)

2 marks

Substitute $f (x) = x - 3$ into $g$ .

(g \circ f) (x) = (x - 3)^{2} + k^{2}

(b)

3 marks

Put $x = 2$ .

(2 - 3)^{2} + k^{2} = 10 \Rightarrow k^{2} = 9 k = \pm 3

Question 3

4 marks

Union and complements of two events

Statistics and probability

(a)

2 marks

Use the addition rule.

P (A \cup B) = 0.65 + 0.75 - 0.6 = 0.8

(b)

2 marks

The event "neither A nor B" is the complement of the union.

P (A^{'} \cap B^{'}) = 1 - 0.8 = 0.2

Question 4

7 marks

An arithmetic sequence from its sum formula

Number and algebra

(a)

5 marks

Use the two given sums.
Subtract.

S_{4} = 16 p - 4 q = 40 \Rightarrow 4 p - q = 10 S_{5} = 25 p - 5 q = 65 \Rightarrow 5 p - q = 13 p = 3, q = 2

(b)

2 marks

The fifth term is the difference between consecutive sums.

u_{5} = S_{5} - S_{4} = 65 - 40 = 25

Question 5

6 marks

Area of a triangle from two sides and the cosine of the included angle

Number and algebra

6 marks

With $cos B A C = \frac{1}{5}$ and the angle acute:
Area of the triangle.

sin B A C = 1 - \frac{1}{25} = \frac{2 6}{5} Area = \frac{1}{2} (6) (10) sin B A C = 56 \cdot \frac{2 6}{5} = 2 (6) = 12 cm^{2}

Question 6

6 marks

Finding n and k from two terms of a binomial expansion

Number and algebra

6 marks

The coefficient of $x$ in $(1 + k x)^{n}$ is $nk$ , and of $x^{2}$ is $(2 n) k^{2}$ .
Cancel $k^{2}$ (since $k \neq = 0$ ) in the second equation.

nk = 12, (2 n) k^{2} = 28 k^{2} \frac{n ( n - 1 )}{2} = 28 \Rightarrow n^{2} - n - 56 = 0 \Rightarrow (n - 8) (n + 7) = 0 \Rightarrow n = 8 k = \frac{12}{8} = 1.5

Section B · Extended response

Questions 7–9 · 45 marks

Question 7

17 marks

Nutcracker ballet performances: grouped data, a cumulative frequency graph, sampling and variance

Statistics and probability

(a)

2 marks

The frequencies add to 60.
The modal class has the highest frequency, 30.

p = 60 - 3 - 18 - 30 = 9 600 < n \leq 800

(b)

4 marks

Read the median at cumulative frequency 30.
80% of 800 is 640. Read the graph at 640: 40 performances sold fewer than that.

median = 600 60 - 40 = 20 performances

(c)

4 marks

One disadvantage: the first 5% of the audience through the door may be people from the same part of the theatre or the same group, so the sample is not representative (a convenience sample).
Systematic sampling of 5%: choose every 20th person leaving the theatre, since $\frac{100}{5} = 20$ .
To represent the proportions of children and adults, use quota sampling.

(d)

3 marks

From the box plot, 75% of the audience spent between $3 and $25.
Half the audience spent less than the median, read from the box plot.

0.75 \times 36000 = 27000 a = 7

(e)

4 marks

The old mean is 600 tickets. Each performance sells 17 more.
Adding the same constant to every value does not change the spread, so there is no effect on the variance.

600 + 17 = 617 tickets

Question 8

15 marks

Two logarithmic graphs that meet twice

Functions

(a)

1 mark

$g (x) = 2 ln x - ln d$ is undefined at $x = 0$ .

x = 0

(b)

9 marks

At an intersection $ln (2 x - 9) = 2 ln x - ln d = ln \frac{x ^{2}}{d}$ .
Two distinct intersection points mean the discriminant is positive.
Factorise: $d (d - 9) > 0$ . Since $d > 0$ this means

2 x - 9 = \frac{x ^{2}}{d} \Rightarrow x^{2} - 2 d x + 9 d = 0 (- 2 d)^{2} - 4 (1) (9 d) > 0 \Rightarrow 4 d^{2} - 36 d > 0 \Rightarrow d^{2} - 9 d > 0 d > 9

(c)

5 marks

With $d = 10$ : $x^{2} - 20 x + 90 = 0$ .
The difference between the roots:

x = \frac{20 \pm 400 - 360}{2} = 10 \pm 10 q - p = 210

Question 9

13 marks

A rational function: derivative, integral and the area between f and g (area 27/2)

Calculus

(a)

4 marks

Write $f (x) = 8 x (x^{2} + 1)^{- 3}$ and use the product rule with the chain rule.

f^{'} (x) = 8 (x^{2} + 1)^{- 3} - 48 x^{2} (x^{2} + 1)^{- 4} = \frac{8 [ ( x ^{2} + 1 ) - 6 x ^{2} ]}{( x ^{2} + 1 ) ^{4}} f^{'} (x) = \frac{8 ( 1 - 5 x ^{2} )}{( x ^{2} + 1 ) ^{4}}

(b)

4 marks

Let $u = x^{2} + 1$ , so $d u = 2 x d x$ and $8 x d x = 4 d u$ .

\int \frac{8 x}{( x ^{2} + 1 ) ^{3}} d x = 4 \int u^{- 3} d u = - 2 u^{- 2} + c = - \frac{2}{( x ^{2} + 1 ) ^{2}} + c

(c)

5 marks

Since $g^{'} (x) = f^{'} (x)$ , we have $g (x) = f (x) + C$ . The vertical distance between the graphs is $∣ C ∣$ , so

Area = 3∣ C ∣ = \frac{27}{2} \Rightarrow ∣ C ∣ = \frac{9}{2} g (x) = \frac{8 x}{( x ^{2} + 1 ) ^{3}} + \frac{9}{2} or g (x) = \frac{8 x}{( x ^{2} + 1 ) ^{3}} - \frac{9}{2}

Our analysis

Where the marks are

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

Functions27 marks · 34%
Q1, Q2, Q8
Statistics and probability21 marks · 26%
Q3, Q7
Number and algebra19 marks · 24%
Q4, Q5, Q6
Calculus13 marks · 16%
Q9

Mentor insights

Traps to watch for

01
Amplitude is the maximum height · Q1
Read it off the graph.
02
Two equations for two unknowns · Q4
Substitute $n = 4$ and $n = 5$ into $S_{n} = p n^{2} - q n$ .
03
Sampling methods · Q7
Systematic sampling takes every $k$ th person. Quota sampling keeps the sample in proportion to subgroups such as children and adults.
04
Combine the logs · Q8
$2 ln x - ln d = ln \frac{x ^{2}}{d}$ , so equal logs give equal arguments.
05
Same derivative means differ by a constant · Q9
If $g^{'} = f^{'}$ then $g (x) = f (x) + C$ .

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 SL · Paper 1

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2

Session: November 2023
Marks: 80
Duration: 1 hour 30 min
Calculator: Not allowed
Questions: 9

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2023 TZ2 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 1–6 · 35 marks

Question 1

7 marks

Reading a cosine graph: amplitude, period and a value

Functions

(a)

1 mark

The graph reaches a maximum of 5.

a = 5

(b)

3 marks

One full wave takes $π$ .
The period is $\frac{2 π}{b}$ .

period = π \frac{2 π}{b} = π \Rightarrow b = 2

(c)

3 marks

Substitute into $f (x) = 5 cos (2 x)$ .

f (\frac{π}{6}) = 5 cos \frac{π}{3} = 5 \times \frac{1}{2} = \frac{5}{2}

Question 2

5 marks

A composite function with a constant, solved for k

Functions

(a)

2 marks

Substitute $f (x) = x - 3$ into $g$ .

(g \circ f) (x) = (x - 3)^{2} + k^{2}

(b)

3 marks

Put $x = 2$ .

(2 - 3)^{2} + k^{2} = 10 \Rightarrow k^{2} = 9 k = \pm 3

Question 3

4 marks

Union and complements of two events

Statistics and probability

(a)

2 marks

Use the addition rule.

P (A \cup B) = 0.65 + 0.75 - 0.6 = 0.8

(b)

2 marks

The event "neither A nor B" is the complement of the union.

P (A^{'} \cap B^{'}) = 1 - 0.8 = 0.2

Question 4

7 marks

An arithmetic sequence from its sum formula

Number and algebra

(a)

5 marks

Use the two given sums.
Subtract.

S_{4} = 16 p - 4 q = 40 \Rightarrow 4 p - q = 10 S_{5} = 25 p - 5 q = 65 \Rightarrow 5 p - q = 13 p = 3, q = 2

(b)

2 marks

The fifth term is the difference between consecutive sums.

u_{5} = S_{5} - S_{4} = 65 - 40 = 25

Question 5

6 marks

Area of a triangle from two sides and the cosine of the included angle

Number and algebra

6 marks

With $cos B A C = \frac{1}{5}$ and the angle acute:
Area of the triangle.

sin B A C = 1 - \frac{1}{25} = \frac{2 6}{5} Area = \frac{1}{2} (6) (10) sin B A C = 56 \cdot \frac{2 6}{5} = 2 (6) = 12 cm^{2}

Question 6

6 marks

Finding n and k from two terms of a binomial expansion

Number and algebra

6 marks

The coefficient of $x$ in $(1 + k x)^{n}$ is $nk$ , and of $x^{2}$ is $(2 n) k^{2}$ .
Cancel $k^{2}$ (since $k \neq = 0$ ) in the second equation.

nk = 12, (2 n) k^{2} = 28 k^{2} \frac{n ( n - 1 )}{2} = 28 \Rightarrow n^{2} - n - 56 = 0 \Rightarrow (n - 8) (n + 7) = 0 \Rightarrow n = 8 k = \frac{12}{8} = 1.5

Section B · Extended response

Questions 7–9 · 45 marks

Question 7

17 marks

Nutcracker ballet performances: grouped data, a cumulative frequency graph, sampling and variance

Statistics and probability

(a)

2 marks

The frequencies add to 60.
The modal class has the highest frequency, 30.

p = 60 - 3 - 18 - 30 = 9 600 < n \leq 800

(b)

4 marks

Read the median at cumulative frequency 30.
80% of 800 is 640. Read the graph at 640: 40 performances sold fewer than that.

median = 600 60 - 40 = 20 performances

(c)

4 marks

One disadvantage: the first 5% of the audience through the door may be people from the same part of the theatre or the same group, so the sample is not representative (a convenience sample).
Systematic sampling of 5%: choose every 20th person leaving the theatre, since $\frac{100}{5} = 20$ .
To represent the proportions of children and adults, use quota sampling.

(d)

3 marks

From the box plot, 75% of the audience spent between $3 and $25.
Half the audience spent less than the median, read from the box plot.

0.75 \times 36000 = 27000 a = 7

(e)

4 marks

The old mean is 600 tickets. Each performance sells 17 more.
Adding the same constant to every value does not change the spread, so there is no effect on the variance.

600 + 17 = 617 tickets

Question 8

15 marks

Two logarithmic graphs that meet twice

Functions

(a)

1 mark

$g (x) = 2 ln x - ln d$ is undefined at $x = 0$ .

x = 0

(b)

9 marks

At an intersection $ln (2 x - 9) = 2 ln x - ln d = ln \frac{x ^{2}}{d}$ .
Two distinct intersection points mean the discriminant is positive.
Factorise: $d (d - 9) > 0$ . Since $d > 0$ this means

2 x - 9 = \frac{x ^{2}}{d} \Rightarrow x^{2} - 2 d x + 9 d = 0 (- 2 d)^{2} - 4 (1) (9 d) > 0 \Rightarrow 4 d^{2} - 36 d > 0 \Rightarrow d^{2} - 9 d > 0 d > 9

(c)

5 marks

With $d = 10$ : $x^{2} - 20 x + 90 = 0$ .
The difference between the roots:

x = \frac{20 \pm 400 - 360}{2} = 10 \pm 10 q - p = 210

Question 9

13 marks

A rational function: derivative, integral and the area between f and g (area 27/2)

Calculus

(a)

4 marks

Write $f (x) = 8 x (x^{2} + 1)^{- 3}$ and use the product rule with the chain rule.

f^{'} (x) = 8 (x^{2} + 1)^{- 3} - 48 x^{2} (x^{2} + 1)^{- 4} = \frac{8 [ ( x ^{2} + 1 ) - 6 x ^{2} ]}{( x ^{2} + 1 ) ^{4}} f^{'} (x) = \frac{8 ( 1 - 5 x ^{2} )}{( x ^{2} + 1 ) ^{4}}

(b)

4 marks

Let $u = x^{2} + 1$ , so $d u = 2 x d x$ and $8 x d x = 4 d u$ .

\int \frac{8 x}{( x ^{2} + 1 ) ^{3}} d x = 4 \int u^{- 3} d u = - 2 u^{- 2} + c = - \frac{2}{( x ^{2} + 1 ) ^{2}} + c

(c)

5 marks

Since $g^{'} (x) = f^{'} (x)$ , we have $g (x) = f (x) + C$ . The vertical distance between the graphs is $∣ C ∣$ , so

Area = 3∣ C ∣ = \frac{27}{2} \Rightarrow ∣ C ∣ = \frac{9}{2} g (x) = \frac{8 x}{( x ^{2} + 1 ) ^{3}} + \frac{9}{2} or g (x) = \frac{8 x}{( x ^{2} + 1 ) ^{3}} - \frac{9}{2}

Our analysis

Where the marks are

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

Functions27 marks · 34%
Q1, Q2, Q8
Statistics and probability21 marks · 26%
Q3, Q7
Number and algebra19 marks · 24%
Q4, Q5, Q6
Calculus13 marks · 16%
Q9

Mentor insights

Traps to watch for

01
Amplitude is the maximum height · Q1
Read it off the graph.
02
Two equations for two unknowns · Q4
Substitute $n = 4$ and $n = 5$ into $S_{n} = p n^{2} - q n$ .
03
Sampling methods · Q7
Systematic sampling takes every $k$ th person. Quota sampling keeps the sample in proportion to subgroups such as children and adults.
04
Combine the logs · Q8
$2 ln x - ln d = ln \frac{x ^{2}}{d}$ , so equal logs give equal arguments.
05
Same derivative means differ by a constant · Q9
If $g^{'} = f^{'}$ then $g (x) = f (x) + C$ .

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