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IB Maths AA SL · Paper 1

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2

Session: May 2022
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 2022 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 · 34 marks

Question 1

5 marks

Reading values from a table of two functions

Functions

(a)

1 mark

Read the table at $x = 0$ .

g (0) = - 2

(b)

2 marks

Evaluate the inner function first.

(f \circ g) (0) = f (g (0)) = f (- 2) = 8

(c)

2 marks

Look along the row for $f (x)$ to find the value 0.

x = 3

Question 2

5 marks

An arithmetic sequence from its general term

Number and algebra

(a)

1 mark

Put $n = 1$ .

u_{1} = 15 - 3 = 12

(b)

2 marks

Solve $15 - 3 n = - 33$ .

3 n = 48 \Rightarrow n = 16

(c)

2 marks

The difference between consecutive terms.

d = u_{2} - u_{1} = 9 - 12 = - 3

Question 3

6 marks

Sums of consecutive integers and of their squares

Number and algebra

(a)

2 marks

Add the three integers.
This is a multiple of 3, so it is always divisible by 3.

(n - 1) + n + (n + 1) = 3 n

(b)

4 marks

Expand the squares and add.
The term $3 n^{2}$ is divisible by 3, but 2 is not, so the sum leaves remainder 2 on division by 3. It is never divisible by 3.

(n - 1)^{2} + n^{2} + (n + 1)^{2} = n^{2} - 2 n + 1 + n^{2} + n^{2} + 2 n + 1 = 3 n^{2} + 2

Question 4

6 marks

A rational function: asymptotes, sketch and inequality

Functions

(a)

2 marks

The denominator is zero at $x = - 1$ . For large $∣ x ∣$ , $\frac{2 x - 1}{x + 1} \to 2$ .

x = - 1, y = 2

(b)

3 marks

Axis intercepts: $y = - 1$ when $x = 0$ , and $x = \frac{1}{2}$ when $y = 0$ .
Sketch: two branches. The right branch rises through $(0, - 1)$ and $(\frac{1}{2}, 0)$ towards $y = 2$ from below. The left branch sits above $y = 2$ , left of $x = - 1$ .

(c)

1 mark

We need the part of the graph between $y = 0$ and $y = 2$ . For $x > - 1$ , $f (x) < 2$ always, and $f (x) > 0$ for $x > \frac{1}{2}$ .

x > \frac{1}{2}

Question 5

5 marks

The least positive solution of a cosine equation

Number and algebra

5 marks

Cosine equals $\frac{1}{2}$ when the angle is $\frac{π}{4}$ (reference angle), or $- \frac{π}{4}$ , then repeating every $2 π$ . We need $\frac{x}{2} + \frac{π}{3} \geq \frac{π}{3}$ , so $\frac{π}{4}$ would give a negative $x$ and is rejected.
The next angle with the same cosine is $2 π - \frac{π}{4} = \frac{7 π}{4}$ .

\frac{x}{2} + \frac{π}{3} = \frac{π}{4} \Rightarrow x < 0 \frac{x}{2} = \frac{7 π}{4} - \frac{π}{3} = \frac{17 π}{12} \Rightarrow x = \frac{17 π}{6}

Question 6

7 marks

A binomial expansion: a coefficient and terms in a mean

Number and algebra

(a)

2 marks

The term in $x^{5}$ in $(x + 1)^{7}$ is $(2 7) x^{5} \cdot 1^{2}$ .

b = (2 7) = \frac{7 \times 6}{2} = 21

(b)

5 marks

The coefficient $a = (1 7) = 7$ . The terms are $7 x^{6}$ , $21 x^{5}$ and $35 x^{4}$ . The third term is the mean of the second and fourth.
Move everything to one side and divide by $7 x^{4}$ (valid since $x \neq = 0$ ).

21 x^{5} = \frac{7 x ^{6} + 35 x ^{4}}{2} \Rightarrow 42 x^{5} = 7 x^{6} + 35 x^{4} x^{2} - 6 x + 5 = 0 \Rightarrow (x - 1) (x - 5) = 0 \Rightarrow x = 1 or x = 5

Section B · Extended response

Questions 7–9 · 46 marks

Question 7

15 marks

A quadratic with known vertex and a function built from its integral

Functions
Calculus

(a)

1 mark

The axis of symmetry is the vertical line through the vertex.

x = 3

(b)

4 marks

Vertex form with the vertex $(3, 4)$ .
Put in $Q (5, 12)$ .

h = 3, k = 4 12 = a (5 - 3)^{2} + 4 \Rightarrow 4 a = 8 \Rightarrow a = 2

(c)

4 marks

Differentiate $f (x) = 2 (x - 3)^{2} + 4$ .
The tangent through $(5, 12)$ with gradient 8:

f^{'} (x) = 4 (x - 3) \Rightarrow f^{'} (5) = 8 y - 12 = 8 (x - 5) \Rightarrow y = 8 x - 28

(d)

3 marks

$g^{'} (x) = f (x) - d = 2 (x - 3)^{2} + 4 - d$ has its smallest value $4 - d$ at $x = 3$ .
For $g$ to be increasing, $g^{'} (x) \geq 0$ for all $x$ , so $4 - d \geq 0$ .

d \leq 4

Mentor tip

At $d = 4$ the derivative touches 0 only at $x = 3$ , so $g$ is still increasing. Some solutions use $d < 4$ , which makes $g^{'}$ positive everywhere.

(e)

3 marks

The second derivative is $g^{''} (x) = f^{'} (x) = 4 (x - 3)$ . Concave up means $g^{''} (x) > 0$ .

4 (x - 3) > 0 \Rightarrow x > 3

Question 8

15 marks

Hyperbola and a line: intersection and a region

Functions
Calculus

(a)

5 marks

Set $f (x) = g (x)$ .
At $x = 5$ , $g (5) = 2$ .

\frac{1}{x - 4} + 1 = x - 3 \Rightarrow \frac{1}{x - 4} = x - 4 \Rightarrow (x - 4)^{2} = 1 x = 3 (point A) or x = 5 B = (5, 2)

(b)

10 marks

Area under the line from $A$ to $B$ is a triangle with base 2 and height 2.
Area under the curve from $x = 5$ to $x = k$ :
Total area:
The matching choice is $k - 3 = 8$ and $p = k - 4$ .

\frac{1}{2} (2) (2) = 2 \int_{5}^{k} (\frac{1}{x - 4} + 1) d x = [ln ∣ x - 4∣ + x]_{5}^{k} = ln (k - 4) + k - 5 ln (k - 4) + k - 3 = ln p + 8 k = 11, p = 7

Question 9

16 marks

A biased die, a dice game, and two pairs of dice with the same sum distribution

Statistics and probability

(a)

5 marks

The probabilities add to 1, and the mean is 2.
Subtract the first equation from the second: $2 q = 0.4$ .

p + 0.3 + q + 0.1 = 1 \Rightarrow p + q = 0.6 E (X) = p + 0.6 + 3 q + 0.4 = 2 \Rightarrow p + 3 q = 1 q = 0.2, p = 0.4

(b)

2 marks

$X > 2$ means 3 or 4.

P (X > 2) = 0.2 + 0.1 = 0.3

(c)

5 marks

With a score of 4 after three rolls, Nicky needs at least 6 more from her last two rolls, which can total 6, 7 or 8.
Add them.

P (total 6) = 2 (0.3) (0.1) + 0. 2^{2} = 0.1, P (total 7) = 2 (0.2) (0.1) = 0.04, P (total 8) = 0. 1^{2} = 0.01 P (wins) = 0.1 + 0.04 + 0.01 = 0.15

(d)

2 marks

The largest sum for the yellow pair is 8, and for the red pair it is $3 + b$ .

3 + b = 8 \Rightarrow b = 5

(e)

2 marks

Compare the probability of a sum of 5. For the yellow pair $P (S = 5) = \frac{4}{16}$ . For the red pair a sum of 5 comes from the 2 on the first die (probability $\frac{2}{4}$ ) with the $a$ on the second die (probability $\frac{2}{4}$ ) when $a + 2 = 5$ , which gives $\frac{1}{2} \cdot \frac{1}{2} = \frac{4}{16}$ .
Check the other sums: with faces ${1, 2, 2, 3}$ and ${1, 3, 3, 5}$ the sums 2 to 8 have probabilities $\frac{1}{16}, \frac{2}{16}, \frac{3}{16}, \frac{4}{16}, \frac{3}{16}, \frac{2}{16}, \frac{1}{16}$ , as required.

a = 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 algebra23 marks · 29%
Q2, Q3, Q5, Q6
Functions21 marks · 26%
Q1, Q4, Q7, Q8
Calculus20 marks · 25%
Q7, Q8
Statistics and probability16 marks · 20%
Q9

Mentor insights

Traps to watch for

01
Expand in terms of $n$ · Q3
Write the three integers as $n - 1$ , $n$ , $n + 1$ and expand. For (b), show the result is $3 n^{2} + 2$ , which leaves remainder 2 on division by 3.
02
Use the term in the question · Q6
"The third term is the mean of the second and the fourth" gives a quadratic once you cancel $x^{4}$ . Discard $x = 0$ , which the question excludes.
03
Tangent uses the derivative · Q7
Differentiate $f$ and evaluate at $x = 5$ for the gradient.
04
Split the region · Q8
Under the line from $A$ to $B$ is a triangle. After $B$ the curve is the upper boundary.
05
Add the ways to reach each total · Q9
In (c) Nicky needs the last two rolls to total at least 6. List each pair of rolls giving 6, 7 and 8, using the probabilities from (a).

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: May 2022
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 2022 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 · 34 marks

Question 1

5 marks

Reading values from a table of two functions

Functions

(a)

1 mark

Read the table at $x = 0$ .

g (0) = - 2

(b)

2 marks

Evaluate the inner function first.

(f \circ g) (0) = f (g (0)) = f (- 2) = 8

(c)

2 marks

Look along the row for $f (x)$ to find the value 0.

x = 3

Question 2

5 marks

An arithmetic sequence from its general term

Number and algebra

(a)

1 mark

Put $n = 1$ .

u_{1} = 15 - 3 = 12

(b)

2 marks

Solve $15 - 3 n = - 33$ .

3 n = 48 \Rightarrow n = 16

(c)

2 marks

The difference between consecutive terms.

d = u_{2} - u_{1} = 9 - 12 = - 3

Question 3

6 marks

Sums of consecutive integers and of their squares

Number and algebra

(a)

2 marks

Add the three integers.
This is a multiple of 3, so it is always divisible by 3.

(n - 1) + n + (n + 1) = 3 n

(b)

4 marks

Expand the squares and add.
The term $3 n^{2}$ is divisible by 3, but 2 is not, so the sum leaves remainder 2 on division by 3. It is never divisible by 3.

(n - 1)^{2} + n^{2} + (n + 1)^{2} = n^{2} - 2 n + 1 + n^{2} + n^{2} + 2 n + 1 = 3 n^{2} + 2

Question 4

6 marks

A rational function: asymptotes, sketch and inequality

Functions

(a)

2 marks

The denominator is zero at $x = - 1$ . For large $∣ x ∣$ , $\frac{2 x - 1}{x + 1} \to 2$ .

x = - 1, y = 2

(b)

3 marks

Axis intercepts: $y = - 1$ when $x = 0$ , and $x = \frac{1}{2}$ when $y = 0$ .
Sketch: two branches. The right branch rises through $(0, - 1)$ and $(\frac{1}{2}, 0)$ towards $y = 2$ from below. The left branch sits above $y = 2$ , left of $x = - 1$ .

(c)

1 mark

We need the part of the graph between $y = 0$ and $y = 2$ . For $x > - 1$ , $f (x) < 2$ always, and $f (x) > 0$ for $x > \frac{1}{2}$ .

x > \frac{1}{2}

Question 5

5 marks

The least positive solution of a cosine equation

Number and algebra

5 marks

Cosine equals $\frac{1}{2}$ when the angle is $\frac{π}{4}$ (reference angle), or $- \frac{π}{4}$ , then repeating every $2 π$ . We need $\frac{x}{2} + \frac{π}{3} \geq \frac{π}{3}$ , so $\frac{π}{4}$ would give a negative $x$ and is rejected.
The next angle with the same cosine is $2 π - \frac{π}{4} = \frac{7 π}{4}$ .

\frac{x}{2} + \frac{π}{3} = \frac{π}{4} \Rightarrow x < 0 \frac{x}{2} = \frac{7 π}{4} - \frac{π}{3} = \frac{17 π}{12} \Rightarrow x = \frac{17 π}{6}

Question 6

7 marks

A binomial expansion: a coefficient and terms in a mean

Number and algebra

(a)

2 marks

The term in $x^{5}$ in $(x + 1)^{7}$ is $(2 7) x^{5} \cdot 1^{2}$ .

b = (2 7) = \frac{7 \times 6}{2} = 21

(b)

5 marks

The coefficient $a = (1 7) = 7$ . The terms are $7 x^{6}$ , $21 x^{5}$ and $35 x^{4}$ . The third term is the mean of the second and fourth.
Move everything to one side and divide by $7 x^{4}$ (valid since $x \neq = 0$ ).

21 x^{5} = \frac{7 x ^{6} + 35 x ^{4}}{2} \Rightarrow 42 x^{5} = 7 x^{6} + 35 x^{4} x^{2} - 6 x + 5 = 0 \Rightarrow (x - 1) (x - 5) = 0 \Rightarrow x = 1 or x = 5

Section B · Extended response

Questions 7–9 · 46 marks

Question 7

15 marks

A quadratic with known vertex and a function built from its integral

Functions
Calculus

(a)

1 mark

The axis of symmetry is the vertical line through the vertex.

x = 3

(b)

4 marks

Vertex form with the vertex $(3, 4)$ .
Put in $Q (5, 12)$ .

h = 3, k = 4 12 = a (5 - 3)^{2} + 4 \Rightarrow 4 a = 8 \Rightarrow a = 2

(c)

4 marks

Differentiate $f (x) = 2 (x - 3)^{2} + 4$ .
The tangent through $(5, 12)$ with gradient 8:

f^{'} (x) = 4 (x - 3) \Rightarrow f^{'} (5) = 8 y - 12 = 8 (x - 5) \Rightarrow y = 8 x - 28

(d)

3 marks

$g^{'} (x) = f (x) - d = 2 (x - 3)^{2} + 4 - d$ has its smallest value $4 - d$ at $x = 3$ .
For $g$ to be increasing, $g^{'} (x) \geq 0$ for all $x$ , so $4 - d \geq 0$ .

d \leq 4

Mentor tip

At $d = 4$ the derivative touches 0 only at $x = 3$ , so $g$ is still increasing. Some solutions use $d < 4$ , which makes $g^{'}$ positive everywhere.

(e)

3 marks

The second derivative is $g^{''} (x) = f^{'} (x) = 4 (x - 3)$ . Concave up means $g^{''} (x) > 0$ .

4 (x - 3) > 0 \Rightarrow x > 3

Question 8

15 marks

Hyperbola and a line: intersection and a region

Functions
Calculus

(a)

5 marks

Set $f (x) = g (x)$ .
At $x = 5$ , $g (5) = 2$ .

\frac{1}{x - 4} + 1 = x - 3 \Rightarrow \frac{1}{x - 4} = x - 4 \Rightarrow (x - 4)^{2} = 1 x = 3 (point A) or x = 5 B = (5, 2)

(b)

10 marks

Area under the line from $A$ to $B$ is a triangle with base 2 and height 2.
Area under the curve from $x = 5$ to $x = k$ :
Total area:
The matching choice is $k - 3 = 8$ and $p = k - 4$ .

\frac{1}{2} (2) (2) = 2 \int_{5}^{k} (\frac{1}{x - 4} + 1) d x = [ln ∣ x - 4∣ + x]_{5}^{k} = ln (k - 4) + k - 5 ln (k - 4) + k - 3 = ln p + 8 k = 11, p = 7

Question 9

16 marks

A biased die, a dice game, and two pairs of dice with the same sum distribution

Statistics and probability

(a)

5 marks

The probabilities add to 1, and the mean is 2.
Subtract the first equation from the second: $2 q = 0.4$ .

p + 0.3 + q + 0.1 = 1 \Rightarrow p + q = 0.6 E (X) = p + 0.6 + 3 q + 0.4 = 2 \Rightarrow p + 3 q = 1 q = 0.2, p = 0.4

(b)

2 marks

$X > 2$ means 3 or 4.

P (X > 2) = 0.2 + 0.1 = 0.3

(c)

5 marks

With a score of 4 after three rolls, Nicky needs at least 6 more from her last two rolls, which can total 6, 7 or 8.
Add them.

P (total 6) = 2 (0.3) (0.1) + 0. 2^{2} = 0.1, P (total 7) = 2 (0.2) (0.1) = 0.04, P (total 8) = 0. 1^{2} = 0.01 P (wins) = 0.1 + 0.04 + 0.01 = 0.15

(d)

2 marks

The largest sum for the yellow pair is 8, and for the red pair it is $3 + b$ .

3 + b = 8 \Rightarrow b = 5

(e)

2 marks

Compare the probability of a sum of 5. For the yellow pair $P (S = 5) = \frac{4}{16}$ . For the red pair a sum of 5 comes from the 2 on the first die (probability $\frac{2}{4}$ ) with the $a$ on the second die (probability $\frac{2}{4}$ ) when $a + 2 = 5$ , which gives $\frac{1}{2} \cdot \frac{1}{2} = \frac{4}{16}$ .
Check the other sums: with faces ${1, 2, 2, 3}$ and ${1, 3, 3, 5}$ the sums 2 to 8 have probabilities $\frac{1}{16}, \frac{2}{16}, \frac{3}{16}, \frac{4}{16}, \frac{3}{16}, \frac{2}{16}, \frac{1}{16}$ , as required.

a = 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 algebra23 marks · 29%
Q2, Q3, Q5, Q6
Functions21 marks · 26%
Q1, Q4, Q7, Q8
Calculus20 marks · 25%
Q7, Q8
Statistics and probability16 marks · 20%
Q9

Mentor insights

Traps to watch for

01
Expand in terms of $n$ · Q3
Write the three integers as $n - 1$ , $n$ , $n + 1$ and expand. For (b), show the result is $3 n^{2} + 2$ , which leaves remainder 2 on division by 3.
02
Use the term in the question · Q6
"The third term is the mean of the second and the fourth" gives a quadratic once you cancel $x^{4}$ . Discard $x = 0$ , which the question excludes.
03
Tangent uses the derivative · Q7
Differentiate $f$ and evaluate at $x = 5$ for the gradient.
04
Split the region · Q8
Under the line from $A$ to $B$ is a triangle. After $B$ the curve is the upper boundary.
05
Add the ways to reach each total · Q9
In (c) Nicky needs the last two rolls to total at least 6. List each pair of rolls giving 6, 7 and 8, using the probabilities from (a).

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