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IB Maths AA SL · Paper 1

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2

Session: May 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 May 2023 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 · 36 marks

Question 1

5 marks

Midpoint and a perpendicular line

Geometry and trigonometry

(a)

2 marks

Midpoint of $P (- 3, 2)$ and $Q (15, - 8)$ .

M = (\frac{- 3 + 15}{2}, \frac{2 - 8}{2}) = (6, - 3)

(b)

2 marks

Gradient of PQ.

m_{P Q} = \frac{- 8 - 2}{15 + 3} = - \frac{5}{9} \Rightarrow m_{L} = \frac{9}{5}

(c)

1 mark

Use the point M $(6, - 3)$ .

y + 3 = \frac{9}{5} (x - 6) \Rightarrow y = \frac{9}{5} x - \frac{69}{5}

Question 2

7 marks

A rational function: zero, asymptotes and inverse

Functions

(a)

2 marks

A zero of $f (x) = \frac{7 x + 7}{2 x - 4}$ needs the numerator to be zero.

7 x + 7 = 0 \Rightarrow x = - 1

(b)

2 marks

The vertical asymptote is where the denominator is zero. For large $∣ x ∣$ the function tends to the ratio of the leading coefficients.

x = 2, y = \frac{7}{2}

(c)

3 marks

Write $y = \frac{7 x + 7}{2 x - 4}$ and swap $x$ and $y$ .

x = \frac{7 y + 7}{2 y - 4} \Rightarrow 2 x y - 4 x = 7 y + 7 \Rightarrow y (2 x - 7) = 4 x + 7 f^{- 1} (x) = \frac{4 x + 7}{2 x - 7}, x \neq = \frac{7}{2}

Question 3

6 marks

Rides on The Dragon: probability, expected value and capacity

Statistics and probability

(a)

4 marks

Of the 40 visitors, $6$ did not ride.
The expected number of rides per visitor:

P (rides) = \frac{40 - 6}{40} = \frac{34}{40} = 0.85 E = \frac{0 ( 6 ) + 1 ( 16 ) + 2 ( 13 ) + 3 ( 2 ) + 4 ( 3 )}{40} = \frac{60}{40} = 1.5

(b)

2 marks

For 1000 visitors the expected number of rides is $1000 \times 1.5 = 1500$ . Each run carries up to 10 people.

\frac{1500}{10} = 150 runs

Question 4

6 marks

cos 2x = sin x as a quadratic in sin x

Functions

(a)

1 mark

Replace $cos 2 x$ with $1 - 2 sin^{2} x$ .

1 - 2 sin^{2} x = sin x \Rightarrow 2 sin^{2} x + sin x - 1 = 0

(b)

5 marks

Factorise.

(2 sin x - 1) (sin x + 1) = 0 \Rightarrow sin x = \frac{1}{2} or sin x = - 1 x = \frac{π}{6}, \frac{5 π}{6}, - \frac{π}{2}

Question 5

6 marks

When does e^(2x) + ln k = 3e^x have a real solution?

Number and algebra

6 marks

Put $A = e^{x}$ with $A > 0$ .
Real roots need the discriminant to be non-negative.
The roots add to 3, which is positive, so at least one root is positive whenever there are real roots. Also $k > 0$ .

A^{2} - 3 A + ln k = 0 9 - 4 ln k \geq 0 \Rightarrow ln k \leq \frac{9}{4} 0 < k \leq e^{9/4}

Question 6

6 marks

f(x) = sin qx: finding m and sketching a horizontal stretch

Functions

(a)

2 marks

One full period of $sin q x$ is $\frac{2 π}{q}$ , and this is $4 m$ .

4 m = \frac{2 π}{q} \Rightarrow m = \frac{π}{2 q}

(b)

4 marks

The new period is $\frac{2 π}{2 q /3} = \frac{3 π}{q} = 6 m$ , so the graph is stretched horizontally by a factor of $\frac{3}{2}$ , and vertically by a factor of 3.
Sketch: a sine curve through $(0, 0)$ , with its maximum 3 at $x = \frac{3}{2} m$ , an intercept at $x = 3 m$ , its minimum $- 3$ at $x = \frac{9}{2} m$ and ending at $(6 m, 0)$ .

Section B · Extended response

Questions 7–9 · 44 marks

Question 7

13 marks

h(x) = 2xe^x + 3: intercept, derivative, minimum and concavity

Calculus

(a)

2 marks

Put $x = 0$ .

h (0) = 2 (0) e^{0} + 3 = 3

(b)

2 marks

Use the product rule on $2 x e^{x}$ .

h^{'} (x) = 2 e^{x} + 2 x e^{x} = 2 (x + 1) e^{x}

(c)

5 marks

At the minimum $h^{'} (x) = 0$ . Since $e^{x} > 0$ , $x + 1 = 0$ .

x = - 1 h (- 1) = - 2 e^{- 1} + 3 \approx 2.26 A = (- 1, 3 - \frac{2}{e})

(d)

4 marks

Differentiate $h^{'} (x) = 2 e^{x} + 2 x e^{x}$ again.
The graph is concave up when $h^{''} (x) > 0$ . Since $e^{x} > 0$ this needs $2 x + 4 > 0$ .

h^{''} (x) = 2 e^{x} + 2 e^{x} + 2 x e^{x} = (2 x + 4) e^{x} x > - 2

Question 8

14 marks

A sequence from its sum formula, then a geometric sequence

Number and algebra

(a)

4 marks

Put $n = 5$ and $n = 6$ .

S_{5} = 25 + 20 = 45 u_{6} = S_{6} - S_{5} = 60 - 45 = 15

(b)

2 marks

The first term is the first sum.

u_{1} = S_{1} = 1 + 4 = 5

(c)

3 marks

$S_{2} = 4 + 8 = 12$ , so $u_{2} = 12 - 5 = 7$ and the common difference is $d = 2$ .

u_{n} = 5 + 2 (n - 1) = 2 n + 3

(d)

3 marks

$v_{2} = 5$ and $v_{4} = 15$ , and $v_{4} = v_{2} r^{2}$ .

r^{2} = 3 \Rightarrow r = \pm 3

(e)

2 marks

$v_{99} = v_{2} r^{97}$ with $v_{2} = 5 > 0$ , so $v_{99} < 0$ needs $r < 0$ .

r = - 3, v_{5} = v_{4} r = 15 (- 3) = - 153

Question 9

17 marks

An object's velocity: displacement, acceleration, speed and distance

Calculus

(a)

5 marks

Integrate $v (t) = - t^{3} + \frac{7}{2} t^{2} - 2 t + 6$ with $s (0) = 0$ .

s (t) = - \frac{t ^{4}}{4} + \frac{7 t ^{3}}{6} - t^{2} + 6 t s (1) = - \frac{1}{4} + \frac{7}{6} - 1 + 6 = \frac{71}{12} \approx 5.92 m

(b)

2 marks

The acceleration is the derivative of the velocity.

a (t) = - 3 t^{2} + 7 t - 2

(c)

5 marks

The speed has a possible maximum where $a (t) = 0$ .
Compare $v (\frac{1}{3}) \approx 5.7$ and $v (2) = - 8 + 14 - 4 + 6 = 8$ . The object comes to rest later than $t = 2$ , so the greatest speed before then is

- 3 t^{2} + 7 t - 2 = 0 \Rightarrow (3 t - 1) (t - 2) = 0 \Rightarrow t = \frac{1}{3} or t = 2 8 m s^{- 1}

(d)

2 marks

After coming to rest the velocity is negative. Check the end value $t = 4$ .
The greatest speed on $0 \leq t \leq 4$ is

v (4) = - 64 + 56 - 8 + 6 = - 10 \Rightarrow speed 10 10 m s^{- 1}

(e)

3 marks

The acceleration $a (t) = - (3 t - 1) (t - 2)$ is positive for $\frac{1}{3} < t < 2$ and negative for $t > 2$ . The speed is increasing when $v$ and $a$ have the same sign: for $\frac{1}{3} < t < 2$ ( $v > 0$ , $a > 0$ ) and for $k < t < 4$ ( $v < 0$ , $a < 0$ ).

distance = \int_{1/3}^{2} v (t) d t - \int_{k}^{4} v (t) d t

Our analysis

Where the marks are

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

Calculus30 marks · 38%
Q7, Q9
Number and algebra20 marks · 25%
Q5, Q8
Functions19 marks · 24%
Q2, Q4, Q6
Statistics and probability6 marks · 8%
Q3
Geometry and trigonometry5 marks · 6%
Q1

Mentor insights

Traps to watch for

01
Zero is where the numerator is zero · Q2
Set $7 x + 7 = 0$ , as long as the denominator is not also zero there.
02
Use the sample as a model · Q3
The table is assumed representative of next day's visitors, so use the relative frequencies as probabilities.
03
Product rule · Q7
Differentiate $2 x e^{x}$ as $2 e^{x} + 2 x e^{x}$ .
04
Term from sums · Q8
$u_{n} = S_{n} - S_{n - 1}$ , and $u_{1} = S_{1}$ .
05
Speed is the size of the velocity · Q9
The greatest speed may occur at a turning point of the velocity or at the end of the interval.

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.

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Worked solutions

IB Maths AA SL · Paper 1

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2

Session: May 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 May 2023 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 · 36 marks

Question 1

5 marks

Midpoint and a perpendicular line

Geometry and trigonometry

(a)

2 marks

Midpoint of $P (- 3, 2)$ and $Q (15, - 8)$ .

M = (\frac{- 3 + 15}{2}, \frac{2 - 8}{2}) = (6, - 3)

(b)

2 marks

Gradient of PQ.

m_{P Q} = \frac{- 8 - 2}{15 + 3} = - \frac{5}{9} \Rightarrow m_{L} = \frac{9}{5}

(c)

1 mark

Use the point M $(6, - 3)$ .

y + 3 = \frac{9}{5} (x - 6) \Rightarrow y = \frac{9}{5} x - \frac{69}{5}

Question 2

7 marks

A rational function: zero, asymptotes and inverse

Functions

(a)

2 marks

A zero of $f (x) = \frac{7 x + 7}{2 x - 4}$ needs the numerator to be zero.

7 x + 7 = 0 \Rightarrow x = - 1

(b)

2 marks

The vertical asymptote is where the denominator is zero. For large $∣ x ∣$ the function tends to the ratio of the leading coefficients.

x = 2, y = \frac{7}{2}

(c)

3 marks

Write $y = \frac{7 x + 7}{2 x - 4}$ and swap $x$ and $y$ .

x = \frac{7 y + 7}{2 y - 4} \Rightarrow 2 x y - 4 x = 7 y + 7 \Rightarrow y (2 x - 7) = 4 x + 7 f^{- 1} (x) = \frac{4 x + 7}{2 x - 7}, x \neq = \frac{7}{2}

Question 3

6 marks

Rides on The Dragon: probability, expected value and capacity

Statistics and probability

(a)

4 marks

Of the 40 visitors, $6$ did not ride.
The expected number of rides per visitor:

P (rides) = \frac{40 - 6}{40} = \frac{34}{40} = 0.85 E = \frac{0 ( 6 ) + 1 ( 16 ) + 2 ( 13 ) + 3 ( 2 ) + 4 ( 3 )}{40} = \frac{60}{40} = 1.5

(b)

2 marks

For 1000 visitors the expected number of rides is $1000 \times 1.5 = 1500$ . Each run carries up to 10 people.

\frac{1500}{10} = 150 runs

Question 4

6 marks

cos 2x = sin x as a quadratic in sin x

Functions

(a)

1 mark

Replace $cos 2 x$ with $1 - 2 sin^{2} x$ .

1 - 2 sin^{2} x = sin x \Rightarrow 2 sin^{2} x + sin x - 1 = 0

(b)

5 marks

Factorise.

(2 sin x - 1) (sin x + 1) = 0 \Rightarrow sin x = \frac{1}{2} or sin x = - 1 x = \frac{π}{6}, \frac{5 π}{6}, - \frac{π}{2}

Question 5

6 marks

When does e^(2x) + ln k = 3e^x have a real solution?

Number and algebra

6 marks

Put $A = e^{x}$ with $A > 0$ .
Real roots need the discriminant to be non-negative.
The roots add to 3, which is positive, so at least one root is positive whenever there are real roots. Also $k > 0$ .

A^{2} - 3 A + ln k = 0 9 - 4 ln k \geq 0 \Rightarrow ln k \leq \frac{9}{4} 0 < k \leq e^{9/4}

Question 6

6 marks

f(x) = sin qx: finding m and sketching a horizontal stretch

Functions

(a)

2 marks

One full period of $sin q x$ is $\frac{2 π}{q}$ , and this is $4 m$ .

4 m = \frac{2 π}{q} \Rightarrow m = \frac{π}{2 q}

(b)

4 marks

The new period is $\frac{2 π}{2 q /3} = \frac{3 π}{q} = 6 m$ , so the graph is stretched horizontally by a factor of $\frac{3}{2}$ , and vertically by a factor of 3.
Sketch: a sine curve through $(0, 0)$ , with its maximum 3 at $x = \frac{3}{2} m$ , an intercept at $x = 3 m$ , its minimum $- 3$ at $x = \frac{9}{2} m$ and ending at $(6 m, 0)$ .

Section B · Extended response

Questions 7–9 · 44 marks

Question 7

13 marks

h(x) = 2xe^x + 3: intercept, derivative, minimum and concavity

Calculus

(a)

2 marks

Put $x = 0$ .

h (0) = 2 (0) e^{0} + 3 = 3

(b)

2 marks

Use the product rule on $2 x e^{x}$ .

h^{'} (x) = 2 e^{x} + 2 x e^{x} = 2 (x + 1) e^{x}

(c)

5 marks

At the minimum $h^{'} (x) = 0$ . Since $e^{x} > 0$ , $x + 1 = 0$ .

x = - 1 h (- 1) = - 2 e^{- 1} + 3 \approx 2.26 A = (- 1, 3 - \frac{2}{e})

(d)

4 marks

Differentiate $h^{'} (x) = 2 e^{x} + 2 x e^{x}$ again.
The graph is concave up when $h^{''} (x) > 0$ . Since $e^{x} > 0$ this needs $2 x + 4 > 0$ .

h^{''} (x) = 2 e^{x} + 2 e^{x} + 2 x e^{x} = (2 x + 4) e^{x} x > - 2

Question 8

14 marks

A sequence from its sum formula, then a geometric sequence

Number and algebra

(a)

4 marks

Put $n = 5$ and $n = 6$ .

S_{5} = 25 + 20 = 45 u_{6} = S_{6} - S_{5} = 60 - 45 = 15

(b)

2 marks

The first term is the first sum.

u_{1} = S_{1} = 1 + 4 = 5

(c)

3 marks

$S_{2} = 4 + 8 = 12$ , so $u_{2} = 12 - 5 = 7$ and the common difference is $d = 2$ .

u_{n} = 5 + 2 (n - 1) = 2 n + 3

(d)

3 marks

$v_{2} = 5$ and $v_{4} = 15$ , and $v_{4} = v_{2} r^{2}$ .

r^{2} = 3 \Rightarrow r = \pm 3

(e)

2 marks

$v_{99} = v_{2} r^{97}$ with $v_{2} = 5 > 0$ , so $v_{99} < 0$ needs $r < 0$ .

r = - 3, v_{5} = v_{4} r = 15 (- 3) = - 153

Question 9

17 marks

An object's velocity: displacement, acceleration, speed and distance

Calculus

(a)

5 marks

Integrate $v (t) = - t^{3} + \frac{7}{2} t^{2} - 2 t + 6$ with $s (0) = 0$ .

s (t) = - \frac{t ^{4}}{4} + \frac{7 t ^{3}}{6} - t^{2} + 6 t s (1) = - \frac{1}{4} + \frac{7}{6} - 1 + 6 = \frac{71}{12} \approx 5.92 m

(b)

2 marks

The acceleration is the derivative of the velocity.

a (t) = - 3 t^{2} + 7 t - 2

(c)

5 marks

The speed has a possible maximum where $a (t) = 0$ .
Compare $v (\frac{1}{3}) \approx 5.7$ and $v (2) = - 8 + 14 - 4 + 6 = 8$ . The object comes to rest later than $t = 2$ , so the greatest speed before then is

- 3 t^{2} + 7 t - 2 = 0 \Rightarrow (3 t - 1) (t - 2) = 0 \Rightarrow t = \frac{1}{3} or t = 2 8 m s^{- 1}

(d)

2 marks

After coming to rest the velocity is negative. Check the end value $t = 4$ .
The greatest speed on $0 \leq t \leq 4$ is

v (4) = - 64 + 56 - 8 + 6 = - 10 \Rightarrow speed 10 10 m s^{- 1}

(e)

3 marks

distance = \int_{1/3}^{2} v (t) d t - \int_{k}^{4} v (t) d t

Our analysis

Where the marks are

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

Calculus30 marks · 38%
Q7, Q9
Number and algebra20 marks · 25%
Q5, Q8
Functions19 marks · 24%
Q2, Q4, Q6
Statistics and probability6 marks · 8%
Q3
Geometry and trigonometry5 marks · 6%
Q1

Mentor insights

Traps to watch for

01
Zero is where the numerator is zero · Q2
Set $7 x + 7 = 0$ , as long as the denominator is not also zero there.
02
Use the sample as a model · Q3
The table is assumed representative of next day's visitors, so use the relative frequencies as probabilities.
03
Product rule · Q7
Differentiate $2 x e^{x}$ as $2 e^{x} + 2 x e^{x}$ .
04
Term from sums · Q8
$u_{n} = S_{n} - S_{n - 1}$ , and $u_{1} = S_{1}$ .
05
Speed is the size of the velocity · Q9
The greatest speed may occur at a turning point of the velocity or at the end of the interval.

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