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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 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

5 marks

A function, its composition with itself and its inverse

Functions

(a)

3 marks

Read the graph of $f$ : it passes through $(1, 3)$ , $(2, 2)$ , $(4, 1)$ , $(0, 5)$ and $(10, 0)$ .

f (4) = 1 (f \circ f) (4) = f (1) = 3 f^{- 1} (3) = 1

(b)

2 marks

Swap the coordinates of the points on $f$ .
Sketch: a curve from the $y$ -intercept $(0, 10)$ that falls steeply through $(1, 4)$ , $(2, 2)$ and $(3, 1)$ to the $x$ -intercept $(5, 0)$ . It is concave up, and it is the mirror image of the graph of $f$ in the line $y = x$ .

(0, 5) \to (5, 0), (1, 3) \to (3, 1), (2, 2) \to (2, 2), (4, 1) \to (1, 4), (10, 0) \to (0, 10)

Question 2

4 marks

Solving tan(2x - 5 degrees) = 1

Number and algebra

4 marks

The reference angle is $tan^{- 1} 1 = 4 5^{\circ}$ . In the interval for $2 x - 5^{\circ}$ the solutions are $4 5^{\circ}$ and $22 5^{\circ}$ .

2 x - 5 = 45 \Rightarrow x = 2 5^{\circ} 2 x - 5 = 225 \Rightarrow x = 11 5^{\circ}

Question 3

5 marks

A quadratic and an exponential equation

Number and algebra

(a)

3 marks

Factorise.

3 m^{2} + 5 m - 2 = (3 m - 1) (m + 2) = 0 \Rightarrow m = \frac{1}{3} or m = - 2

(b)

2 marks

With $m = 3^{x}$ the equation is the one in part (a).
Since $3^{x} > 0$ , only $m = \frac{1}{3}$ gives a solution.

3^{x} = \frac{1}{3} \Rightarrow x = - 1

Question 4

7 marks

A symmetrical quadrilateral OABC: midpoint, line OB and area

Geometry and trigonometry

(a)

4 marks

The midpoint of $[A C]$ with $A (6, 0)$ and $C (3, 33)$ .
The line through $O$ and $M$ has gradient

M = (\frac{9}{2}, \frac{3 3}{2}) m = \frac{3 3 /2}{9/2} = \frac{3}{3} y = \frac{3}{3} x

(b)

3 marks

$[O A]$ is perpendicular to $[A B]$ , so B lies directly above A, at $x = 6$ .
Triangle OAB has base 6 and height $23$ . The quadrilateral is two such triangles (OAB and OCB, by symmetry).

y_{B} = \frac{3}{3} \times 6 = 23 Area = 2 \times \frac{1}{2} \times 6 \times 23 = 123

Question 5

6 marks

Bird nesting in two seasons: a tree diagram and a quadratic in k

Statistics and probability

(a)

2 marks

The probability of not nesting is one minus the probability of nesting.

Spring: 1 - k, Summer (on both branches): 1 - \frac{k}{2}

(b)

4 marks

The seasons are multiplied along the branches for not nesting in both.
Expand and clear fractions.
The probability of nesting in Spring is $k$ , which cannot exceed 1. Since $\frac{8}{3} > 1$ , only $k = \frac{1}{3}$ is valid.

(1 - k) (1 - \frac{k}{2}) = \frac{5}{9} 1 - \frac{3 k}{2} + \frac{k ^{2}}{2} = \frac{5}{9} \Rightarrow 18 - 27 k + 9 k^{2} = 10 \Rightarrow 9 k^{2} - 27 k + 8 = 0

Question 6

8 marks

A rational function against a line: the number of intersections

Functions

(a)

1 mark

As $x \to \pm \infty$ , $f (x) = \frac{2 ( x + 3 )}{3 ( x + 2 )} \to \frac{2}{3}$ .

y = \frac{2}{3}

(b)

7 marks

Set $f (x) = g (x)$ and multiply by $3 (x + 2)$ .
The roots are $x = 0$ and $x = - \frac{6 m + 1}{3 m}$ .
(i) For $m > 0$ the second root is negative and not equal to $- 2$ , so there are two different solutions.
(ii) One solution needs the two roots to coincide.
This agrees with $f^{'} (0)$ : since $f^{'} (x) = \frac{2}{3} \cdot \frac{( x + 2 ) - ( x + 3 )}{( x + 2 ) ^{2}} = - \frac{2}{3 ( x + 2 ) ^{2}}$ , we get $f^{'} (0) = - \frac{1}{6}$ .
(iii) Two solutions with $x \geq 0$ need the second root to be positive.

2 (x + 3) = 3 (x + 2) (m x + 1) \Rightarrow 3 m x^{2} + (6 m + 1) x = 0 \Rightarrow x (3 m x + 6 m + 1) = 0 2 solutions - \frac{6 m + 1}{3 m} = 0 \Rightarrow m = - \frac{1}{6} - \frac{6 m + 1}{3 m} > 0 \Rightarrow - \frac{1}{6} < m < 0

Section B · Extended response

Questions 7–9 · 45 marks

Question 7

12 marks

Sea temperatures modelled by sine curves

Functions

(a)

1 mark

The temperature repeats every 12 months, so the period is 12.

\frac{2 π}{b} = 12 \Rightarrow b = \frac{π}{6}

(b)

2 marks

The model $f$ has amplitude 5 and midline 15 (from the graph of the temperatures, with the highest at 20 and the lowest at 10).

a = 5, c = 15

(c)

6 marks

(i) 1st October is $x = 5$ .

(ii) The highest value of $sin$ is 1, so the highest temperature is

Since $14.5 < 15$ , the water is never warm enough for Alex to swim.

g (5) = 3.5 sin \frac{5 π}{6} + 11 = 3.5 (\frac{1}{2}) + 11 = 12.75^{\circ} C g_{m a x} = 3.5 + 11 = 14.5^{\circ} C

(d)

3 marks

Compare $0.7 f (x) + q = 3.5 sin \frac{π x}{6} + 10.5 + q$ with $g (x) = 3.5 sin \frac{π x}{6} + 11$ .

10.5 + q = 11 \Rightarrow q = 0.5

Question 8

17 marks

A function given by its derivative: sign, concavity, an antiderivative and a normal

Calculus

(a)

3 marks

(i) Complete the square.

(ii) $f^{'} (x) = \frac{2 x + 2}{x ^{2} + 2 x + 2}$ with a positive denominator, so $f^{'} > 0$ when $2 x + 2 > 0$ .

x^{2} + 2 x + 2 = (x + 1)^{2} + 1 \geq 1 > 0 x > - 1

(b)

7 marks

(i) The numerator of $f^{'}$ is zero.

(ii) Use the quotient rule.

(iii) At $x = - 1$ :

Since $f^{'} (- 1) = 0$ and $f^{''} (- 1) > 0$ , $x = - 1$ is a local minimum.

x = - 1 f^{''} (x) = \frac{2 ( x ^{2} + 2 x + 2 ) - ( 2 x + 2 ) ( 2 x + 2 )}{( x ^{2} + 2 x + 2 ) ^{2}} f^{''} (x) = \frac{2 x ^{2} + 4 x + 4 - 4 x ^{2} - 8 x - 4}{( x ^{2} + 2 x + 2 ) ^{2}} = \frac{- 2 x ^{2} - 4 x}{( x ^{2} + 2 x + 2 ) ^{2}} f^{''} (- 1) = \frac{- 2 + 4}{( 1 - 2 + 2 ) ^{2}} = 2 > 0

(c)

4 marks

The numerator $2 x + 2$ is the derivative of the denominator, so integrate.
Use $f (2) = 3 + ln 10$ , where $x^{2} + 2 x + 2 = 10$ .

f (x) = \int \frac{2 x + 2}{x ^{2} + 2 x + 2} d x = ln (x^{2} + 2 x + 2) + c ln 10 + c = 3 + ln 10 \Rightarrow c = 3 f (x) = ln (x^{2} + 2 x + 2) + 3

(d)

3 marks

At $x = 2$ the gradient is $f^{'} (2) = \frac{6}{10} = \frac{3}{5}$ , so the normal has gradient $- \frac{5}{3}$ .

y - (3 + ln 10) = - \frac{5}{3} (x - 2)

Question 9

16 marks

An arithmetic and a geometric sequence combined into a new arithmetic sequence

Number and algebra

(a)

2 marks

The common difference gives $p - a = q - p$ .

2 p - q = a

(b)

2 marks

The common ratio gives $\frac{s}{a} = \frac{t}{s}$ .

s^{2} = a t

(c)

2 marks

With $q = t = 1$ , part (a) gives $a = 2 p - 1$ and part (b) gives $s^{2} = a$ .
Since $s \neq = 0$ , $s^{2} > 0$ .

2 p - 1 = s^{2} > 0 p > \frac{1}{2}

(d)

4 marks

With $a = 9$ and $q = 1$ , $p = \frac{a + q}{2} = 5$ , and the common difference is $- 4$ .
With $s^{2} = 9 \times 1$ and $s > 0$ , $s = 3$ and the ratio is $\frac{1}{3}$ .

9, 5, 1, - 3 9, 3, 1, \frac{1}{3}

(e)

6 marks

The common difference of $u_{n}$ .
Check against the next pair: $u_{3} - u_{2} = (1 + 0) - (5 + ln 3) = - 4 - ln 3$ .
Sum the first ten terms with $u_{1} = 9 + 2 ln 3$ .

d = u_{2} - u_{1} = (5 + ln 3) - (9 + ln 9) = - 4 + ln 3 - 2 ln 3 = - 4 - ln 3 S_{10} = \frac{10}{2} [2 (9 + 2 ln 3) + 9 (- 4 - ln 3)] = 5 (18 + 4 ln 3 - 36 - 9 ln 3) i = 1 \sum 10 u_{i} = 5 (- 18 - 5 ln 3) = - 90 - 25 ln 3

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 algebra25 marks · 31%
Q2, Q3, Q9
Functions25 marks · 31%
Q1, Q6, Q7
Calculus17 marks · 21%
Q8
Geometry and trigonometry7 marks · 9%
Q4
Statistics and probability6 marks · 8%
Q5

Mentor insights

Traps to watch for

01
Symmetry about a diagonal · Q4
The diagonal OB is the perpendicular bisector of $[A C]$ , so it passes through the midpoint of $[A C]$ .
02
Intersections as a quadratic · Q6
Setting $f (x) = g (x)$ and clearing the fraction gives $3 m x^{2} + (6 m + 1) x = 0$ , which has the root $x = 0$ .
03
Sine model · Q7
$a sin b x + c$ has amplitude $a$ , period $\frac{2 π}{b}$ and midline $y = c$ .
04
Complete the square · Q8
$x^{2} + 2 x + 2 = (x + 1)^{2} + 1$ is always at least 1.
05
Equal differences · Q9
In $a, p, q$ we have $p - a = q - p$ . In $a, s, t$ the ratios are equal: $\frac{s}{a} = \frac{t}{s}$ .

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 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

5 marks

A function, its composition with itself and its inverse

Functions

(a)

3 marks

Read the graph of $f$ : it passes through $(1, 3)$ , $(2, 2)$ , $(4, 1)$ , $(0, 5)$ and $(10, 0)$ .

f (4) = 1 (f \circ f) (4) = f (1) = 3 f^{- 1} (3) = 1

(b)

2 marks

Swap the coordinates of the points on $f$ .
Sketch: a curve from the $y$ -intercept $(0, 10)$ that falls steeply through $(1, 4)$ , $(2, 2)$ and $(3, 1)$ to the $x$ -intercept $(5, 0)$ . It is concave up, and it is the mirror image of the graph of $f$ in the line $y = x$ .

(0, 5) \to (5, 0), (1, 3) \to (3, 1), (2, 2) \to (2, 2), (4, 1) \to (1, 4), (10, 0) \to (0, 10)

Question 2

4 marks

Solving tan(2x - 5 degrees) = 1

Number and algebra

4 marks

The reference angle is $tan^{- 1} 1 = 4 5^{\circ}$ . In the interval for $2 x - 5^{\circ}$ the solutions are $4 5^{\circ}$ and $22 5^{\circ}$ .

2 x - 5 = 45 \Rightarrow x = 2 5^{\circ} 2 x - 5 = 225 \Rightarrow x = 11 5^{\circ}

Question 3

5 marks

A quadratic and an exponential equation

Number and algebra

(a)

3 marks

Factorise.

3 m^{2} + 5 m - 2 = (3 m - 1) (m + 2) = 0 \Rightarrow m = \frac{1}{3} or m = - 2

(b)

2 marks

With $m = 3^{x}$ the equation is the one in part (a).
Since $3^{x} > 0$ , only $m = \frac{1}{3}$ gives a solution.

3^{x} = \frac{1}{3} \Rightarrow x = - 1

Question 4

7 marks

A symmetrical quadrilateral OABC: midpoint, line OB and area

Geometry and trigonometry

(a)

4 marks

The midpoint of $[A C]$ with $A (6, 0)$ and $C (3, 33)$ .
The line through $O$ and $M$ has gradient

M = (\frac{9}{2}, \frac{3 3}{2}) m = \frac{3 3 /2}{9/2} = \frac{3}{3} y = \frac{3}{3} x

(b)

3 marks

$[O A]$ is perpendicular to $[A B]$ , so B lies directly above A, at $x = 6$ .
Triangle OAB has base 6 and height $23$ . The quadrilateral is two such triangles (OAB and OCB, by symmetry).

y_{B} = \frac{3}{3} \times 6 = 23 Area = 2 \times \frac{1}{2} \times 6 \times 23 = 123

Question 5

6 marks

Bird nesting in two seasons: a tree diagram and a quadratic in k

Statistics and probability

(a)

2 marks

The probability of not nesting is one minus the probability of nesting.

Spring: 1 - k, Summer (on both branches): 1 - \frac{k}{2}

(b)

4 marks

The seasons are multiplied along the branches for not nesting in both.
Expand and clear fractions.
The probability of nesting in Spring is $k$ , which cannot exceed 1. Since $\frac{8}{3} > 1$ , only $k = \frac{1}{3}$ is valid.

(1 - k) (1 - \frac{k}{2}) = \frac{5}{9} 1 - \frac{3 k}{2} + \frac{k ^{2}}{2} = \frac{5}{9} \Rightarrow 18 - 27 k + 9 k^{2} = 10 \Rightarrow 9 k^{2} - 27 k + 8 = 0

Question 6

8 marks

A rational function against a line: the number of intersections

Functions

(a)

1 mark

As $x \to \pm \infty$ , $f (x) = \frac{2 ( x + 3 )}{3 ( x + 2 )} \to \frac{2}{3}$ .

y = \frac{2}{3}

(b)

7 marks

Set $f (x) = g (x)$ and multiply by $3 (x + 2)$ .
The roots are $x = 0$ and $x = - \frac{6 m + 1}{3 m}$ .
(i) For $m > 0$ the second root is negative and not equal to $- 2$ , so there are two different solutions.
(ii) One solution needs the two roots to coincide.
This agrees with $f^{'} (0)$ : since $f^{'} (x) = \frac{2}{3} \cdot \frac{( x + 2 ) - ( x + 3 )}{( x + 2 ) ^{2}} = - \frac{2}{3 ( x + 2 ) ^{2}}$ , we get $f^{'} (0) = - \frac{1}{6}$ .
(iii) Two solutions with $x \geq 0$ need the second root to be positive.

2 (x + 3) = 3 (x + 2) (m x + 1) \Rightarrow 3 m x^{2} + (6 m + 1) x = 0 \Rightarrow x (3 m x + 6 m + 1) = 0 2 solutions - \frac{6 m + 1}{3 m} = 0 \Rightarrow m = - \frac{1}{6} - \frac{6 m + 1}{3 m} > 0 \Rightarrow - \frac{1}{6} < m < 0

Section B · Extended response

Questions 7–9 · 45 marks

Question 7

12 marks

Sea temperatures modelled by sine curves

Functions

(a)

1 mark

The temperature repeats every 12 months, so the period is 12.

\frac{2 π}{b} = 12 \Rightarrow b = \frac{π}{6}

(b)

2 marks

The model $f$ has amplitude 5 and midline 15 (from the graph of the temperatures, with the highest at 20 and the lowest at 10).

a = 5, c = 15

(c)

6 marks

(i) 1st October is $x = 5$ .

(ii) The highest value of $sin$ is 1, so the highest temperature is

Since $14.5 < 15$ , the water is never warm enough for Alex to swim.

g (5) = 3.5 sin \frac{5 π}{6} + 11 = 3.5 (\frac{1}{2}) + 11 = 12.75^{\circ} C g_{m a x} = 3.5 + 11 = 14.5^{\circ} C

(d)

3 marks

Compare $0.7 f (x) + q = 3.5 sin \frac{π x}{6} + 10.5 + q$ with $g (x) = 3.5 sin \frac{π x}{6} + 11$ .

10.5 + q = 11 \Rightarrow q = 0.5

Question 8

17 marks

A function given by its derivative: sign, concavity, an antiderivative and a normal

Calculus

(a)

3 marks

(i) Complete the square.

(ii) $f^{'} (x) = \frac{2 x + 2}{x ^{2} + 2 x + 2}$ with a positive denominator, so $f^{'} > 0$ when $2 x + 2 > 0$ .

x^{2} + 2 x + 2 = (x + 1)^{2} + 1 \geq 1 > 0 x > - 1

(b)

7 marks

(i) The numerator of $f^{'}$ is zero.

(ii) Use the quotient rule.

(iii) At $x = - 1$ :

Since $f^{'} (- 1) = 0$ and $f^{''} (- 1) > 0$ , $x = - 1$ is a local minimum.

x = - 1 f^{''} (x) = \frac{2 ( x ^{2} + 2 x + 2 ) - ( 2 x + 2 ) ( 2 x + 2 )}{( x ^{2} + 2 x + 2 ) ^{2}} f^{''} (x) = \frac{2 x ^{2} + 4 x + 4 - 4 x ^{2} - 8 x - 4}{( x ^{2} + 2 x + 2 ) ^{2}} = \frac{- 2 x ^{2} - 4 x}{( x ^{2} + 2 x + 2 ) ^{2}} f^{''} (- 1) = \frac{- 2 + 4}{( 1 - 2 + 2 ) ^{2}} = 2 > 0

(c)

4 marks

The numerator $2 x + 2$ is the derivative of the denominator, so integrate.
Use $f (2) = 3 + ln 10$ , where $x^{2} + 2 x + 2 = 10$ .

f (x) = \int \frac{2 x + 2}{x ^{2} + 2 x + 2} d x = ln (x^{2} + 2 x + 2) + c ln 10 + c = 3 + ln 10 \Rightarrow c = 3 f (x) = ln (x^{2} + 2 x + 2) + 3

(d)

3 marks

At $x = 2$ the gradient is $f^{'} (2) = \frac{6}{10} = \frac{3}{5}$ , so the normal has gradient $- \frac{5}{3}$ .

y - (3 + ln 10) = - \frac{5}{3} (x - 2)

Question 9

16 marks

An arithmetic and a geometric sequence combined into a new arithmetic sequence

Number and algebra

(a)

2 marks

The common difference gives $p - a = q - p$ .

2 p - q = a

(b)

2 marks

The common ratio gives $\frac{s}{a} = \frac{t}{s}$ .

s^{2} = a t

(c)

2 marks

With $q = t = 1$ , part (a) gives $a = 2 p - 1$ and part (b) gives $s^{2} = a$ .
Since $s \neq = 0$ , $s^{2} > 0$ .

2 p - 1 = s^{2} > 0 p > \frac{1}{2}

(d)

4 marks

With $a = 9$ and $q = 1$ , $p = \frac{a + q}{2} = 5$ , and the common difference is $- 4$ .
With $s^{2} = 9 \times 1$ and $s > 0$ , $s = 3$ and the ratio is $\frac{1}{3}$ .

9, 5, 1, - 3 9, 3, 1, \frac{1}{3}

(e)

6 marks

The common difference of $u_{n}$ .
Check against the next pair: $u_{3} - u_{2} = (1 + 0) - (5 + ln 3) = - 4 - ln 3$ .
Sum the first ten terms with $u_{1} = 9 + 2 ln 3$ .

d = u_{2} - u_{1} = (5 + ln 3) - (9 + ln 9) = - 4 + ln 3 - 2 ln 3 = - 4 - ln 3 S_{10} = \frac{10}{2} [2 (9 + 2 ln 3) + 9 (- 4 - ln 3)] = 5 (18 + 4 ln 3 - 36 - 9 ln 3) i = 1 \sum 10 u_{i} = 5 (- 18 - 5 ln 3) = - 90 - 25 ln 3

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 algebra25 marks · 31%
Q2, Q3, Q9
Functions25 marks · 31%
Q1, Q6, Q7
Calculus17 marks · 21%
Q8
Geometry and trigonometry7 marks · 9%
Q4
Statistics and probability6 marks · 8%
Q5

Mentor insights

Traps to watch for

01
Symmetry about a diagonal · Q4
The diagonal OB is the perpendicular bisector of $[A C]$ , so it passes through the midpoint of $[A C]$ .
02
Intersections as a quadratic · Q6
Setting $f (x) = g (x)$ and clearing the fraction gives $3 m x^{2} + (6 m + 1) x = 0$ , which has the root $x = 0$ .
03
Sine model · Q7
$a sin b x + c$ has amplitude $a$ , period $\frac{2 π}{b}$ and midline $y = c$ .
04
Complete the square · Q8
$x^{2} + 2 x + 2 = (x + 1)^{2} + 1$ is always at least 1.
05
Equal differences · Q9
In $a, p, q$ we have $p - a = q - p$ . In $a, s, t$ the ratios are equal: $\frac{s}{a} = \frac{t}{s}$ .

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