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Worked solutions

IB Maths AA SL · Paper 1

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2

Session: May 2024
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 May 2024 TZ1 paper, which your school or the IB can provide.

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Hide our working until you've had a go.

Section A · Short response

Questions 1–6 · 35 marks

Question 1

4 marks

An arithmetic sequence from its second and fourth terms

Number and algebra

(a)

2 marks

The fourth term is two steps on from the second term.

2 d = 22 - 10 = 12 \Rightarrow d = 6

(b)

2 marks

The first term is $u_{1} = u_{2} - d = 10 - 6 = 4$ . Use $u_{n} = u_{1} + (n - 1) d$ .

u_{n} = 4 + 6 (n - 1) = 6 n - 2

Question 2

6 marks

A die rolled sixteen times: finding two frequencies from the mean

Statistics and probability

(a)

5 marks

The frequencies add to 16.
The mean is 3, so the total of the scores is $3 \times 16 = 48$ .
Subtract the two equations.

p + q + 4 + 2 + 0 + 3 = 16 \Rightarrow p + q = 7 p + 2 q + 12 + 8 + 0 + 18 = 48 \Rightarrow p + 2 q = 10 q = 3, p = 4

(b)

1 mark

Multiplying each score by 10 multiplies the mean by 10.

3 \times 10 = 30

Question 3

5 marks

Logarithms of 1/a and with base 1000 given log of a

Functions

(a)

2 marks

Use $lo g_{10} \frac{1}{a} = - lo g_{10} a$ with $lo g_{10} a = \frac{1}{3}$ .

lo g_{10} \frac{1}{a} = - \frac{1}{3}

(b)

3 marks

Change the base: $lo g_{1000} a = \frac{lo g _{10} a}{lo g _{10} 1000}$ .

lo g_{1000} a = \frac{1/3}{3} = \frac{1}{9}

Question 4

8 marks

A circular sector with perimeter 10 and area 6.25

Geometry and trigonometry

(a)

4 marks

Write the two conditions.
From the first, $r θ = 10 - 2 r$ . Substitute into the area.

2 r + r θ = 10, \frac{1}{2} r^{2} θ = 6.25 \frac{1}{2} r (10 - 2 r) = 6.25 \Rightarrow 5 r - r^{2} = 6.25 \Rightarrow 4 r^{2} - 20 r + 25 = 0

(b)

4 marks

Factorise: $(2 r - 5)^{2} = 0$ .
Then $r θ = 10 - 5 = 5$ .

r = \frac{5}{2} θ = \frac{5}{2.5} = 2

Question 5

7 marks

Intersections of cos x and sin 2x and the area between them

Functions
Calculus

(a)

3 marks

Set $cos x = sin 2 x = 2 sin x cos x$ .
The point B at $x = \frac{π}{2}$ comes from $cos x = 0$ . The other solutions in $0 \leq x \leq π$ are

cos x (1 - 2 sin x) = 0 \Rightarrow cos x = 0 or sin x = \frac{1}{2} x_{A} = \frac{π}{6}, x_{C} = \frac{5 π}{6}

(b)

4 marks

Between B and C the curve of $f$ lies above $g$ .

Area = \int_{π /2}^{5 π /6} (cos x - sin 2 x) d x = [sin x + \frac{1}{2} cos 2 x]_{π /2}^{5 π /6} (\frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2}) - (1 - \frac{1}{2}) = \frac{3}{4} - \frac{1}{2} = \frac{1}{4}

Question 6

5 marks

Sums of the geometric sequence with ratio 10 and a sum of sums

Number and algebra

(a)

1 mark

Use $S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1}$ with $a = 1$ and $r = 10$ .

S_{n} = \frac{1 0 ^{n} - 1}{9}

(b)

4 marks

Add the first $n$ sums.
The geometric series has first term 10, ratio 10 and $n$ terms.

S_{1} + S_{2} + \dots + S_{n} = \frac{10 - 1}{9} + \frac{1 0 ^{2} - 1}{9} + \dots + \frac{1 0 ^{n} - 1}{9} = \frac{1}{9} [(10 + 1 0^{2} + \dots + 1 0^{n}) - n] 10 + 1 0^{2} + \dots + 1 0^{n} = \frac{10 ( 1 0 ^{n} - 1 )}{9} S_{1} + \dots + S_{n} = \frac{1}{9} [\frac{10 ( 1 0 ^{n} - 1 )}{9} - n] = \frac{1 0 ^{n + 1} - 10 - 9 n}{81}

Section B · Extended response

Questions 7–9 · 45 marks

Question 7

14 marks

A cubic curve: local maximum, inflexion and normal

Calculus

(a)

3 marks

Differentiate $y = x^{3} - x^{2} - x + 1$ .

\frac{d y}{d x} = 3 x^{2} - 2 x - 1 \frac{d ^{2} y}{d x ^{2}} = 6 x - 2

(b)

6 marks

Set $\frac{d y}{d x} = 0$ and factorise.
At $x = - \frac{1}{3}$ the second derivative is $6 (- \frac{1}{3}) - 2 = - 4 < 0$ , so this is a local maximum. (At $x = 1$ it is $4 > 0$ , a local minimum.)
Find the $y$ -coordinate of the local maximum.

3 x^{2} - 2 x - 1 = (3 x + 1) (x - 1) = 0 \Rightarrow x = - \frac{1}{3} or x = 1 y = - \frac{1}{27} - \frac{1}{9} + \frac{1}{3} + 1 = \frac{32}{27} A = (- \frac{1}{3}, \frac{32}{27})

(c)

2 marks

A point of inflexion has $\frac{d ^{2} y}{d x ^{2}} = 0$ .

6 x - 2 = 0 \Rightarrow x = \frac{1}{3}

(d)

3 marks

The gradient of the tangent at $(0, 1)$ is $\frac{d y}{d x} = - 1$ , so the normal has gradient $1$ .

y - 1 = 1 (x - 0) \Rightarrow y = x + 1

Question 8

14 marks

A rational function, a quadratic and the number of intersections

Functions

(a)

5 marks

$f (x) = \frac{4 x + 2}{x - 2}$ . The $y$ -intercept is $f (0) = - 1$ and the $x$ -intercept is at $4 x + 2 = 0$ .
The vertical asymptote is $x = 2$ and the horizontal asymptote is $y = 4$ (the ratio of the leading coefficients).
Sketch: a branch to the left of $x = 2$ through $(- \frac{1}{2}, 0)$ and $(0, - 1)$ , falling to $- \infty$ as $x \to 2^{-}$ and approaching $y = 4$ from below as $x \to - \infty$ . A second branch to the right of $x = 2$ coming down from $+ \infty$ towards $y = 4$ .

(- \frac{1}{2}, 0), (0, - 1)

(b)

1 mark

The function takes every value except the horizontal asymptote.

f (x) \in R, f (x) \neq = 4

(c)

1 mark

The axis of symmetry $x = 2$ is the midpoint of the roots.

\frac{- \frac{1}{2} + p}{2} = 2 \Rightarrow p = \frac{9}{2}

(d)

3 marks

Write $g (x) = (x + \frac{1}{2}) (x - \frac{9}{2})$ and expand.

g (x) = x^{2} - 4 x - \frac{9}{4} \Rightarrow b = - 4, c = - \frac{9}{4}

(e)

2 marks

The vertex is on the axis of symmetry, so substitute $x = 2$ .

g (2) = 4 - 8 - \frac{9}{4} = - \frac{25}{4}

(f)

2 marks

Sketch the graphs of $f$ and $g$ on the same axes and count the crossings. Clearing the fraction in $f (x) = g (x)$ gives a cubic equation, and the number of real roots of the cubic is the number of solutions.
The cubic changes sign between $- 1$ and 0, between 0 and 1, and between 5 and 6, so it has three roots, none equal to 2.

4 x + 2 = (x - 2) (x^{2} - 4 x - \frac{9}{4}) \Rightarrow x^{3} - 6 x^{2} + \frac{7}{4} x + \frac{5}{2} = 0 3 solutions

Question 9

17 marks

Buttons added to the bag: a tree, conditional probability and a first blue

Statistics and probability

(a)

2 marks

After a first blue button is chosen, it is replaced and one more blue is added, so the bag then holds 3 red and 2 blue buttons.

p = \frac{3}{5}, q = \frac{2}{5}

(b)

2 marks

Two of the same colour means red then red, or blue then blue.

P (same) = \frac{3}{4} \times \frac{4}{5} + \frac{1}{4} \times \frac{2}{5} = \frac{12}{20} + \frac{2}{20} = \frac{14}{20} = \frac{7}{10}

(c)

3 marks

Use conditional probability.

P (R R ∣ same) = \frac{12/20}{14/20} = \frac{6}{7}

(d)

2 marks

The probability of $X = 0$ is blue then blue, $\frac{1}{10}$ . The probability of $X = 2$ is red then red, and $X = 1$ is the rest.

b = \frac{12}{20} = \frac{3}{5}, a = 1 - \frac{1}{10} - \frac{3}{5} = \frac{3}{10}

(e)

2 marks

Add each value times its probability.

E (X) = 0 (\frac{1}{10}) + 1 (\frac{3}{10}) + 2 (\frac{3}{5}) = \frac{3}{10} + \frac{12}{10} = \frac{3}{2}

(f)

1 mark

After two reds the bag holds 5 red and 1 blue button.

\frac{1}{6}

(g)

5 marks

Suppose the first $n - 1$ selections are red. After $k$ red selections the bag has $3 + k$ red and 1 blue, so the next red has probability $\frac{3 + k}{4 + k}$ .
After $n - 1$ red selections the bag has $n + 2$ red and 1 blue, a total of $n + 3$ , so the next one is blue with probability $\frac{1}{n + 3}$ .

P (n - 1 reds) = \frac{3}{4} \times \frac{4}{5} \times \dots \times \frac{n + 1}{n + 2} = \frac{3}{n + 2} P (first blue on selection n) = \frac{3}{n + 2} \times \frac{1}{n + 3} = \frac{3}{56} (n + 2) (n + 3) = 56 \Rightarrow n = 5

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 probability23 marks · 29%
Q2, Q9
Functions22 marks · 28%
Q3, Q5, Q8
Calculus18 marks · 23%
Q5, Q7
Number and algebra9 marks · 11%
Q1, Q6
Geometry and trigonometry8 marks · 10%
Q4

Mentor insights

Traps to watch for

01
Sector formulas · Q4
Perimeter $2 r + r θ$ and area $\frac{1}{2} r^{2} θ$ , with $θ$ in radians.
02
Double angle · Q5
Replace $sin 2 x$ with $2 sin x cos x$ and factorise rather than dividing by $cos x$ .
03
Second derivative test · Q7
At a stationary point, $f^{''} < 0$ means a local maximum and $f^{''} > 0$ a local minimum.
04
Asymptotes and intercepts · Q8
Set $y = 0$ and $x = 0$ for the intercepts. The vertical asymptote is where the denominator is zero.
05
Pólya-style urn · Q9
After each pick the chosen colour is replaced and one more of that colour is added, so the bag grows by one button every time.

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

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Worked solutions

IB Maths AA SL · Paper 1

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2

Session: May 2024
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 May 2024 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 1–6 · 35 marks

Question 1

4 marks

An arithmetic sequence from its second and fourth terms

Number and algebra

(a)

2 marks

The fourth term is two steps on from the second term.

2 d = 22 - 10 = 12 \Rightarrow d = 6

(b)

2 marks

The first term is $u_{1} = u_{2} - d = 10 - 6 = 4$ . Use $u_{n} = u_{1} + (n - 1) d$ .

u_{n} = 4 + 6 (n - 1) = 6 n - 2

Question 2

6 marks

A die rolled sixteen times: finding two frequencies from the mean

Statistics and probability

(a)

5 marks

The frequencies add to 16.
The mean is 3, so the total of the scores is $3 \times 16 = 48$ .
Subtract the two equations.

p + q + 4 + 2 + 0 + 3 = 16 \Rightarrow p + q = 7 p + 2 q + 12 + 8 + 0 + 18 = 48 \Rightarrow p + 2 q = 10 q = 3, p = 4

(b)

1 mark

Multiplying each score by 10 multiplies the mean by 10.

3 \times 10 = 30

Question 3

5 marks

Logarithms of 1/a and with base 1000 given log of a

Functions

(a)

2 marks

Use $lo g_{10} \frac{1}{a} = - lo g_{10} a$ with $lo g_{10} a = \frac{1}{3}$ .

lo g_{10} \frac{1}{a} = - \frac{1}{3}

(b)

3 marks

Change the base: $lo g_{1000} a = \frac{lo g _{10} a}{lo g _{10} 1000}$ .

lo g_{1000} a = \frac{1/3}{3} = \frac{1}{9}

Question 4

8 marks

A circular sector with perimeter 10 and area 6.25

Geometry and trigonometry

(a)

4 marks

Write the two conditions.
From the first, $r θ = 10 - 2 r$ . Substitute into the area.

2 r + r θ = 10, \frac{1}{2} r^{2} θ = 6.25 \frac{1}{2} r (10 - 2 r) = 6.25 \Rightarrow 5 r - r^{2} = 6.25 \Rightarrow 4 r^{2} - 20 r + 25 = 0

(b)

4 marks

Factorise: $(2 r - 5)^{2} = 0$ .
Then $r θ = 10 - 5 = 5$ .

r = \frac{5}{2} θ = \frac{5}{2.5} = 2

Question 5

7 marks

Intersections of cos x and sin 2x and the area between them

Functions
Calculus

(a)

3 marks

Set $cos x = sin 2 x = 2 sin x cos x$ .
The point B at $x = \frac{π}{2}$ comes from $cos x = 0$ . The other solutions in $0 \leq x \leq π$ are

cos x (1 - 2 sin x) = 0 \Rightarrow cos x = 0 or sin x = \frac{1}{2} x_{A} = \frac{π}{6}, x_{C} = \frac{5 π}{6}

(b)

4 marks

Between B and C the curve of $f$ lies above $g$ .

Area = \int_{π /2}^{5 π /6} (cos x - sin 2 x) d x = [sin x + \frac{1}{2} cos 2 x]_{π /2}^{5 π /6} (\frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2}) - (1 - \frac{1}{2}) = \frac{3}{4} - \frac{1}{2} = \frac{1}{4}

Question 6

5 marks

Sums of the geometric sequence with ratio 10 and a sum of sums

Number and algebra

(a)

1 mark

Use $S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1}$ with $a = 1$ and $r = 10$ .

S_{n} = \frac{1 0 ^{n} - 1}{9}

(b)

4 marks

Add the first $n$ sums.
The geometric series has first term 10, ratio 10 and $n$ terms.

S_{1} + S_{2} + \dots + S_{n} = \frac{10 - 1}{9} + \frac{1 0 ^{2} - 1}{9} + \dots + \frac{1 0 ^{n} - 1}{9} = \frac{1}{9} [(10 + 1 0^{2} + \dots + 1 0^{n}) - n] 10 + 1 0^{2} + \dots + 1 0^{n} = \frac{10 ( 1 0 ^{n} - 1 )}{9} S_{1} + \dots + S_{n} = \frac{1}{9} [\frac{10 ( 1 0 ^{n} - 1 )}{9} - n] = \frac{1 0 ^{n + 1} - 10 - 9 n}{81}

Section B · Extended response

Questions 7–9 · 45 marks

Question 7

14 marks

A cubic curve: local maximum, inflexion and normal

Calculus

(a)

3 marks

Differentiate $y = x^{3} - x^{2} - x + 1$ .

\frac{d y}{d x} = 3 x^{2} - 2 x - 1 \frac{d ^{2} y}{d x ^{2}} = 6 x - 2

(b)

6 marks

Set $\frac{d y}{d x} = 0$ and factorise.
At $x = - \frac{1}{3}$ the second derivative is $6 (- \frac{1}{3}) - 2 = - 4 < 0$ , so this is a local maximum. (At $x = 1$ it is $4 > 0$ , a local minimum.)
Find the $y$ -coordinate of the local maximum.

3 x^{2} - 2 x - 1 = (3 x + 1) (x - 1) = 0 \Rightarrow x = - \frac{1}{3} or x = 1 y = - \frac{1}{27} - \frac{1}{9} + \frac{1}{3} + 1 = \frac{32}{27} A = (- \frac{1}{3}, \frac{32}{27})

(c)

2 marks

A point of inflexion has $\frac{d ^{2} y}{d x ^{2}} = 0$ .

6 x - 2 = 0 \Rightarrow x = \frac{1}{3}

(d)

3 marks

The gradient of the tangent at $(0, 1)$ is $\frac{d y}{d x} = - 1$ , so the normal has gradient $1$ .

y - 1 = 1 (x - 0) \Rightarrow y = x + 1

Question 8

14 marks

A rational function, a quadratic and the number of intersections

Functions

(a)

5 marks

$f (x) = \frac{4 x + 2}{x - 2}$ . The $y$ -intercept is $f (0) = - 1$ and the $x$ -intercept is at $4 x + 2 = 0$ .
The vertical asymptote is $x = 2$ and the horizontal asymptote is $y = 4$ (the ratio of the leading coefficients).
Sketch: a branch to the left of $x = 2$ through $(- \frac{1}{2}, 0)$ and $(0, - 1)$ , falling to $- \infty$ as $x \to 2^{-}$ and approaching $y = 4$ from below as $x \to - \infty$ . A second branch to the right of $x = 2$ coming down from $+ \infty$ towards $y = 4$ .

(- \frac{1}{2}, 0), (0, - 1)

(b)

1 mark

The function takes every value except the horizontal asymptote.

f (x) \in R, f (x) \neq = 4

(c)

1 mark

The axis of symmetry $x = 2$ is the midpoint of the roots.

\frac{- \frac{1}{2} + p}{2} = 2 \Rightarrow p = \frac{9}{2}

(d)

3 marks

Write $g (x) = (x + \frac{1}{2}) (x - \frac{9}{2})$ and expand.

g (x) = x^{2} - 4 x - \frac{9}{4} \Rightarrow b = - 4, c = - \frac{9}{4}

(e)

2 marks

The vertex is on the axis of symmetry, so substitute $x = 2$ .

g (2) = 4 - 8 - \frac{9}{4} = - \frac{25}{4}

(f)

2 marks

Sketch the graphs of $f$ and $g$ on the same axes and count the crossings. Clearing the fraction in $f (x) = g (x)$ gives a cubic equation, and the number of real roots of the cubic is the number of solutions.
The cubic changes sign between $- 1$ and 0, between 0 and 1, and between 5 and 6, so it has three roots, none equal to 2.

4 x + 2 = (x - 2) (x^{2} - 4 x - \frac{9}{4}) \Rightarrow x^{3} - 6 x^{2} + \frac{7}{4} x + \frac{5}{2} = 0 3 solutions

Question 9

17 marks

Buttons added to the bag: a tree, conditional probability and a first blue

Statistics and probability

(a)

2 marks

After a first blue button is chosen, it is replaced and one more blue is added, so the bag then holds 3 red and 2 blue buttons.

p = \frac{3}{5}, q = \frac{2}{5}

(b)

2 marks

Two of the same colour means red then red, or blue then blue.

P (same) = \frac{3}{4} \times \frac{4}{5} + \frac{1}{4} \times \frac{2}{5} = \frac{12}{20} + \frac{2}{20} = \frac{14}{20} = \frac{7}{10}

(c)

3 marks

Use conditional probability.

P (R R ∣ same) = \frac{12/20}{14/20} = \frac{6}{7}

(d)

2 marks

The probability of $X = 0$ is blue then blue, $\frac{1}{10}$ . The probability of $X = 2$ is red then red, and $X = 1$ is the rest.

b = \frac{12}{20} = \frac{3}{5}, a = 1 - \frac{1}{10} - \frac{3}{5} = \frac{3}{10}

(e)

2 marks

Add each value times its probability.

E (X) = 0 (\frac{1}{10}) + 1 (\frac{3}{10}) + 2 (\frac{3}{5}) = \frac{3}{10} + \frac{12}{10} = \frac{3}{2}

(f)

1 mark

After two reds the bag holds 5 red and 1 blue button.

\frac{1}{6}

(g)

5 marks

Suppose the first $n - 1$ selections are red. After $k$ red selections the bag has $3 + k$ red and 1 blue, so the next red has probability $\frac{3 + k}{4 + k}$ .
After $n - 1$ red selections the bag has $n + 2$ red and 1 blue, a total of $n + 3$ , so the next one is blue with probability $\frac{1}{n + 3}$ .

P (n - 1 reds) = \frac{3}{4} \times \frac{4}{5} \times \dots \times \frac{n + 1}{n + 2} = \frac{3}{n + 2} P (first blue on selection n) = \frac{3}{n + 2} \times \frac{1}{n + 3} = \frac{3}{56} (n + 2) (n + 3) = 56 \Rightarrow n = 5

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 probability23 marks · 29%
Q2, Q9
Functions22 marks · 28%
Q3, Q5, Q8
Calculus18 marks · 23%
Q5, Q7
Number and algebra9 marks · 11%
Q1, Q6
Geometry and trigonometry8 marks · 10%
Q4

Mentor insights

Traps to watch for

01
Sector formulas · Q4
Perimeter $2 r + r θ$ and area $\frac{1}{2} r^{2} θ$ , with $θ$ in radians.
02
Double angle · Q5
Replace $sin 2 x$ with $2 sin x cos x$ and factorise rather than dividing by $cos x$ .
03
Second derivative test · Q7
At a stationary point, $f^{''} < 0$ means a local maximum and $f^{''} > 0$ a local minimum.
04
Asymptotes and intercepts · Q8
Set $y = 0$ and $x = 0$ for the intercepts. The vertical asymptote is where the denominator is zero.
05
Pólya-style urn · Q9
After each pick the chosen colour is replaced and one more of that colour is added, so the bag grows by one button every time.

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