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IB Maths AA SL · Paper 1

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2

Session: May 2021
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 2021 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 · 37 marks

Question 1

6 marks

A circle sector with a given perimeter

Geometry and trigonometry

(a)

3 marks

The shaded sector has an arc of length $r θ = r$ and two straight edges of length $r$ .

r + r + r = 3 r = 12 \Rightarrow r = 4

(b)

3 marks

The non-shaded region is the whole circle minus the sector of angle 1 radian.

π (4)^{2} - \frac{1}{2} (4)^{2} (1) = 16 π - 8 = 8 (2 π - 1)

Question 2

4 marks

Difference of consecutive squares equals their sum

Number and algebra

4 marks

Subtract the squares in the right order, expanding $(n + 1)^{2}$ .
The sum of the two integers is $n + (n + 1) = 2 n + 1$ , the same expression.

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

Mentor tip

Factorising works too: $(n + 1)^{2} - n^{2} = ((n + 1) - n) ((n + 1) + n) = 1 \cdot (2 n + 1)$ .

Question 3

6 marks

Rewriting a trig equation as a quadratic, then solving it

Geometry and trigonometry

(a)

1 mark

Replace $cos^{2} x$ by $1 - sin^{2} x$ in the equation.

2 (1 - sin^{2} x) + 5 sin x = 4 \Rightarrow 2 sin^{2} x - 5 sin x + 2 = 0

(b)

5 marks

Factorise the quadratic in $sin x$ .
$sin x = 2$ is impossible. That leaves $sin x = \frac{1}{2}$ in $0 \leq x \leq 2 π$ .

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

Question 4

5 marks

Finding a constant from one binomial coefficient

Number and algebra

5 marks

In $(x + k)^{7}$ the term in $x^{5}$ comes from choosing $x$ five times and $k$ twice.
Set the coefficient equal to 63.

(2 7) x^{5} k^{2} = 21 k^{2} x^{5} 21 k^{2} = 63 \Rightarrow k^{2} = 3 \Rightarrow k = \pm 3

Question 5

9 marks

A logarithmic curve: where it crosses and where it is steep

Calculus

(a)

3 marks

The graph meets the $x$ -axis where $f (x) = 0$ , that is, where the log's argument is 1.
Since $x > 4$ , only the positive root is allowed, so $a = 17$ .

ln (x^{2} - 16) = 0 \Rightarrow x^{2} - 16 = 1 \Rightarrow x = 17

(b)

6 marks

Differentiate with the chain rule, then set the gradient equal to $\frac{1}{3}$ .
Cross-multiply and solve the quadratic.

f^{'} (x) = \frac{2 x}{x ^{2} - 16} = \frac{1}{3} 6 x = x^{2} - 16 \Rightarrow x^{2} - 6 x - 16 = 0 \Rightarrow (x - 8) (x + 2) = 0 x = 8 (since x > 4)

Mentor tip

The domain check is the last line of the answer. Do it every time.

Question 6

7 marks

Area of a triangle with a ratio of sides and a known cosine

Number and algebra

7 marks

Apply the cosine rule at $C$ , opposite the side $A B = 10$ , with $B C = x$ and $A C = 2 x$ .
From $cos C = \frac{3}{4}$ , $sin C = 1 - \frac{9}{16} = \frac{7}{4}$ .
The area is $\frac{1}{2} \cdot B C \cdot A C \cdot sin C = \frac{1}{2} \cdot x \cdot 2 x \cdot sin C = x^{2} sin C$ .

100 = x^{2} + 4 x^{2} - 2 (x) (2 x) \cdot \frac{3}{4} = 5 x^{2} - 3 x^{2} = 2 x^{2} x^{2} = 50 Area = 50 \cdot \frac{7}{4} = \frac{25}{2} 7

Mentor tip

Check the form asked for: a rational number times a square root, $p q$ with $p = \frac{25}{2}$ and $q = 7$ .

Section B · Extended response

Questions 7–9 · 43 marks

Question 7

14 marks

Homework hours from a cumulative frequency graph

Statistics and probability

(a)

2 marks

The median is the value at half of 80, which is 40 on the cumulative frequency axis. Read across to the curve and down.

median = 14 hours

(b)

3 marks

If 10% spend more than $k$ hours, then 90% spend less. That is 72 of the 80 students. Read the curve at a cumulative frequency of 72.

k = 18

(c)

4 marks

The cumulative frequency at 15 hours is 60, which equals the first two classes added: $4 + p = 60$ .
The cumulative frequency at 21 hours is 76, so the last class has the remaining students.

p = 56 q = 80 - 76 = 4 (check: 4 + 56 + 16 + 4 = 80)

(d)

3 marks

In the sample, 20 of 80 students (a quarter) spend more than 15 hours. Scale up to 320.

\frac{20}{80} \times 320 = 80 students

(e)

2 marks

(i) Only Year 12 students were surveyed, and the homework load probably differs between Year 6 and Year 12, so the sample doesn't represent the whole school.

(ii) A stratified sample, with students chosen from each year group in proportion to its size.

Question 8

15 marks

Area under a cosine curve, then a cone with the same surface area

Calculus
Geometry and trigonometry

(a)

3 marks

The graph touches the $x$ -axis where $f (x) = 0$ , that is, where $cos x = - 1$ in $0 \leq x \leq 4 π$ .

6 + 6 cos x = 0 \Rightarrow cos x = - 1 \Rightarrow x = π, 3 π

(b)

5 marks

The area is the integral between those two points.

\int_{π}^{3 π} (6 + 6 cos x) d x = [6 x + 6 sin x]_{π}^{3 π} (18 π + 0) - (6 π + 0) = 12 π

Mentor tip

Both sine terms vanish at multiples of $π$ , so write that rather than skipping it.

(c)

3 marks

The total surface area is the base plus the curved surface, $π r^{2} + π r l$ , with $r = 2$ . Set it equal to $12 π$ .

4 π + 2 π l = 12 π \Rightarrow l = 4

(d)

4 marks

Find the height from $r^{2} + h^{2} = l^{2}$ , then use $V = \frac{1}{3} π r^{2} h$ .

h = 16 - 4 = 23 V = \frac{1}{3} π (2^{2}) (23) = \frac{8 3 π}{3}

Question 9

14 marks

Two particles on a line: distance and a matching time

Calculus

(a)

2 marks

The particle is at the origin when $s = 0$ .

8 t - t^{2} = 0 \Rightarrow t (8 - t) = 0 \Rightarrow p = 8

(b)

4 marks

(i) A particle changes direction when $v = \frac{d s}{d t} = 0$ .

(ii) Substitute into the displacement.

v = 8 - 2 t = 0 \Rightarrow q = 4 s (4) = 32 - 16 = 16 m

(c)

2 marks

At $t = 10$ , $s = 80 - 100 = - 20$ , so the particle is 20 m from the origin (on the other side).

distance = 20 m

(d)

2 marks

A goes forward 16 m to $s = 16$ , then back to $s = - 20$ , a distance of $16 + 20 = 36$ m.

d = 16 + 36 = 52 m

(e)

4 marks

Particle B has $v = 14 - 2 t$ , which is zero at $t = 7$ . The distance it travels forward is the area of a triangle.
After $t = 7$ it comes back. In the next $x$ seconds the speed is $2 (t - 7)$ , so the extra distance is the area of a triangle with base $x$ and height $2 x$ , which is $x^{2}$ .

\int_{0}^{7} (14 - 2 t) d t = 49 49 + x^{2} = 52 \Rightarrow x = 3 k = 7 + 3 \approx 8.73

Our analysis

Where the marks are

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

Calculus31 marks · 39%
Q5, Q8, Q9
Geometry and trigonometry19 marks · 24%
Q1, Q3, Q8
Number and algebra16 marks · 20%
Q2, Q4, Q6
Statistics and probability14 marks · 18%
Q7

Mentor insights

Traps to watch for

01
Reject the root outside the domain · Q5
In part (b) the quadratic has $x = 8$ and $x = - 2$ . The function is only defined for $x > 4$ , so only 8 is valid.
02
Find $x$ first · Q6
The cosine rule on the 10-unit side gives $x^{2}$ directly, and the area $\frac{1}{2} ab sin C$ needs only $x^{2}$ as well. You never need $x$ itself.
03
Read the graph at the right height · Q7
The median is at half the total, 40 of 80 students. "The top 10%" is the point where 72 of 80 students are below.
04
Total surface area includes the base · Q8
The cone's surface area is the curved part $π r l$ plus the circular base $π r^{2}$ . Using only $π r l$ gives the wrong slant height.
05
Direction changes mean split the journey · Q9
Particle A turns round when its velocity is zero. The distance travelled is the out-and-back total, not the final distance from the origin.

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 2021
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 2021 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 · 37 marks

Question 1

6 marks

A circle sector with a given perimeter

Geometry and trigonometry

(a)

3 marks

The shaded sector has an arc of length $r θ = r$ and two straight edges of length $r$ .

r + r + r = 3 r = 12 \Rightarrow r = 4

(b)

3 marks

The non-shaded region is the whole circle minus the sector of angle 1 radian.

π (4)^{2} - \frac{1}{2} (4)^{2} (1) = 16 π - 8 = 8 (2 π - 1)

Question 2

4 marks

Difference of consecutive squares equals their sum

Number and algebra

4 marks

Subtract the squares in the right order, expanding $(n + 1)^{2}$ .
The sum of the two integers is $n + (n + 1) = 2 n + 1$ , the same expression.

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

Mentor tip

Factorising works too: $(n + 1)^{2} - n^{2} = ((n + 1) - n) ((n + 1) + n) = 1 \cdot (2 n + 1)$ .

Question 3

6 marks

Rewriting a trig equation as a quadratic, then solving it

Geometry and trigonometry

(a)

1 mark

Replace $cos^{2} x$ by $1 - sin^{2} x$ in the equation.

2 (1 - sin^{2} x) + 5 sin x = 4 \Rightarrow 2 sin^{2} x - 5 sin x + 2 = 0

(b)

5 marks

Factorise the quadratic in $sin x$ .
$sin x = 2$ is impossible. That leaves $sin x = \frac{1}{2}$ in $0 \leq x \leq 2 π$ .

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

Question 4

5 marks

Finding a constant from one binomial coefficient

Number and algebra

5 marks

In $(x + k)^{7}$ the term in $x^{5}$ comes from choosing $x$ five times and $k$ twice.
Set the coefficient equal to 63.

(2 7) x^{5} k^{2} = 21 k^{2} x^{5} 21 k^{2} = 63 \Rightarrow k^{2} = 3 \Rightarrow k = \pm 3

Question 5

9 marks

A logarithmic curve: where it crosses and where it is steep

Calculus

(a)

3 marks

The graph meets the $x$ -axis where $f (x) = 0$ , that is, where the log's argument is 1.
Since $x > 4$ , only the positive root is allowed, so $a = 17$ .

ln (x^{2} - 16) = 0 \Rightarrow x^{2} - 16 = 1 \Rightarrow x = 17

(b)

6 marks

Differentiate with the chain rule, then set the gradient equal to $\frac{1}{3}$ .
Cross-multiply and solve the quadratic.

f^{'} (x) = \frac{2 x}{x ^{2} - 16} = \frac{1}{3} 6 x = x^{2} - 16 \Rightarrow x^{2} - 6 x - 16 = 0 \Rightarrow (x - 8) (x + 2) = 0 x = 8 (since x > 4)

Mentor tip

The domain check is the last line of the answer. Do it every time.

Question 6

7 marks

Area of a triangle with a ratio of sides and a known cosine

Number and algebra

7 marks

Apply the cosine rule at $C$ , opposite the side $A B = 10$ , with $B C = x$ and $A C = 2 x$ .
From $cos C = \frac{3}{4}$ , $sin C = 1 - \frac{9}{16} = \frac{7}{4}$ .
The area is $\frac{1}{2} \cdot B C \cdot A C \cdot sin C = \frac{1}{2} \cdot x \cdot 2 x \cdot sin C = x^{2} sin C$ .

100 = x^{2} + 4 x^{2} - 2 (x) (2 x) \cdot \frac{3}{4} = 5 x^{2} - 3 x^{2} = 2 x^{2} x^{2} = 50 Area = 50 \cdot \frac{7}{4} = \frac{25}{2} 7

Mentor tip

Check the form asked for: a rational number times a square root, $p q$ with $p = \frac{25}{2}$ and $q = 7$ .

Section B · Extended response

Questions 7–9 · 43 marks

Question 7

14 marks

Homework hours from a cumulative frequency graph

Statistics and probability

(a)

2 marks

The median is the value at half of 80, which is 40 on the cumulative frequency axis. Read across to the curve and down.

median = 14 hours

(b)

3 marks

If 10% spend more than $k$ hours, then 90% spend less. That is 72 of the 80 students. Read the curve at a cumulative frequency of 72.

k = 18

(c)

4 marks

The cumulative frequency at 15 hours is 60, which equals the first two classes added: $4 + p = 60$ .
The cumulative frequency at 21 hours is 76, so the last class has the remaining students.

p = 56 q = 80 - 76 = 4 (check: 4 + 56 + 16 + 4 = 80)

(d)

3 marks

In the sample, 20 of 80 students (a quarter) spend more than 15 hours. Scale up to 320.

\frac{20}{80} \times 320 = 80 students

(e)

2 marks

(i) Only Year 12 students were surveyed, and the homework load probably differs between Year 6 and Year 12, so the sample doesn't represent the whole school.

(ii) A stratified sample, with students chosen from each year group in proportion to its size.

Question 8

15 marks

Area under a cosine curve, then a cone with the same surface area

Calculus
Geometry and trigonometry

(a)

3 marks

The graph touches the $x$ -axis where $f (x) = 0$ , that is, where $cos x = - 1$ in $0 \leq x \leq 4 π$ .

6 + 6 cos x = 0 \Rightarrow cos x = - 1 \Rightarrow x = π, 3 π

(b)

5 marks

The area is the integral between those two points.

\int_{π}^{3 π} (6 + 6 cos x) d x = [6 x + 6 sin x]_{π}^{3 π} (18 π + 0) - (6 π + 0) = 12 π

Mentor tip

Both sine terms vanish at multiples of $π$ , so write that rather than skipping it.

(c)

3 marks

The total surface area is the base plus the curved surface, $π r^{2} + π r l$ , with $r = 2$ . Set it equal to $12 π$ .

4 π + 2 π l = 12 π \Rightarrow l = 4

(d)

4 marks

Find the height from $r^{2} + h^{2} = l^{2}$ , then use $V = \frac{1}{3} π r^{2} h$ .

h = 16 - 4 = 23 V = \frac{1}{3} π (2^{2}) (23) = \frac{8 3 π}{3}

Question 9

14 marks

Two particles on a line: distance and a matching time

Calculus

(a)

2 marks

The particle is at the origin when $s = 0$ .

8 t - t^{2} = 0 \Rightarrow t (8 - t) = 0 \Rightarrow p = 8

(b)

4 marks

(i) A particle changes direction when $v = \frac{d s}{d t} = 0$ .

(ii) Substitute into the displacement.

v = 8 - 2 t = 0 \Rightarrow q = 4 s (4) = 32 - 16 = 16 m

(c)

2 marks

At $t = 10$ , $s = 80 - 100 = - 20$ , so the particle is 20 m from the origin (on the other side).

distance = 20 m

(d)

2 marks

A goes forward 16 m to $s = 16$ , then back to $s = - 20$ , a distance of $16 + 20 = 36$ m.

d = 16 + 36 = 52 m

(e)

4 marks

Particle B has $v = 14 - 2 t$ , which is zero at $t = 7$ . The distance it travels forward is the area of a triangle.
After $t = 7$ it comes back. In the next $x$ seconds the speed is $2 (t - 7)$ , so the extra distance is the area of a triangle with base $x$ and height $2 x$ , which is $x^{2}$ .

\int_{0}^{7} (14 - 2 t) d t = 49 49 + x^{2} = 52 \Rightarrow x = 3 k = 7 + 3 \approx 8.73

Our analysis

Where the marks are

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

Calculus31 marks · 39%
Q5, Q8, Q9
Geometry and trigonometry19 marks · 24%
Q1, Q3, Q8
Number and algebra16 marks · 20%
Q2, Q4, Q6
Statistics and probability14 marks · 18%
Q7

Mentor insights

Traps to watch for

01
Reject the root outside the domain · Q5
In part (b) the quadratic has $x = 8$ and $x = - 2$ . The function is only defined for $x > 4$ , so only 8 is valid.
02
Find $x$ first · Q6
The cosine rule on the 10-unit side gives $x^{2}$ directly, and the area $\frac{1}{2} ab sin C$ needs only $x^{2}$ as well. You never need $x$ itself.
03
Read the graph at the right height · Q7
The median is at half the total, 40 of 80 students. "The top 10%" is the point where 72 of 80 students are below.
04
Total surface area includes the base · Q8
The cone's surface area is the curved part $π r l$ plus the circular base $π r^{2}$ . Using only $π r l$ gives the wrong slant height.
05
Direction changes mean split the journey · Q9
Particle A turns round when its velocity is zero. The distance travelled is the out-and-back total, not the final distance from the origin.

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