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IB Maths AA SL · Paper 1

Mathematics: Analysis & Approaches, Standard Level

Session: November 2021
Marks: 80
Duration: 1 hour 30 min
Calculator: Not allowed
Questions: 9

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2021 Global 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

7 marks

A quadratic in factorised form: intercepts, vertex and completed square

Functions

(a)

5 marks

(i) $f (x) = 0$ when

(ii) The vertex is halfway between $- 3$ and 1, so at $x = - 1$ .

- 2 (x - 1) (x + 3) = 0 \Rightarrow x = 1 or x = - 3 f (- 1) = - 2 (- 2) (2) = 8, vertex (- 1, 8)

(b)

2 marks

The vertex form has the vertex at $(h, k)$ .

h = - 1, k = 8

Question 2

4 marks

A simple differential equation with an initial condition

Calculus

4 marks

Integrate $cos (x - \frac{π}{4})$ with respect to $x$ .
Use $y = 2$ at $x = \frac{3 π}{4}$ , where $sin \frac{π}{2} = 1$ .

y = sin (x - \frac{π}{4}) + c 2 = 1 + c \Rightarrow c = 1 y = sin (x - \frac{π}{4}) + 1

Question 3

5 marks

A rational function: asymptotes, intercepts and a sketch

Functions

(a)

2 marks

The function is $f (x) = \frac{2 x + 4}{3 - x}$ . (i) The denominator is zero at
(ii) As $x \to \infty$ , $f (x) \to \frac{2 x}{- x}$ .

x = 3 y = - 2

(b)

2 marks

(i) The graph crosses the $x$ -axis where $2 x + 4 = 0$ .

(ii) At $x = 0$ , $f (0) = \frac{4}{3}$ .

(- 2, 0) (0, \frac{4}{3})

(c)

1 mark

Draw the asymptotes $x = 3$ and $y = - 2$ . The left branch approaches $y = - 2$ from above as $x \to - \infty$ , passes through $(- 2, 0)$ and $(0, \frac{4}{3})$ and rises to $+ \infty$ as $x \to 3^{-}$ . The right branch starts at $- \infty$ just right of $x = 3$ and rises towards $y = - 2$ from below.

Question 4

5 marks

Two boxes of balls: a total probability and an independence test

Statistics and probability

(a)

3 marks

Either box is chosen with probability $\frac{1}{2}$ , then a red ball is drawn.

P (R) = \frac{1}{2} \cdot \frac{5}{7} + \frac{1}{2} \cdot \frac{4}{7} = \frac{9}{14}

(b)

2 marks

Compare $P (A) P (R)$ with $P (A \cap R)$ .
They are not equal, so $A$ and $R$ are not independent. (Box 1 has a higher proportion of red balls.)

P (A) P (R) = \frac{1}{2} \cdot \frac{9}{14} = \frac{9}{28}, P (A \cap R) = \frac{1}{2} \cdot \frac{5}{7} = \frac{5}{14} = \frac{10}{28}

Question 5

7 marks

A tangent to $f$ and the tangent to a composite

Calculus

(a)

1 mark

The tangent has gradient 6.

f^{'} (4) = 6

(b)

1 mark

The tangent passes through the graph at $x = 4$ .

f (4) = 6 (4) - 1 = 23

(c)

2 marks

First $g (4) = 16 - 12 = 4$ .

h (4) = f (g (4)) = f (4) = 23

(d)

3 marks

Use the chain rule with $g^{'} (x) = 2 x - 3$ , so $g^{'} (4) = 5$ .
The tangent at $x = 4$ passes through $(4, 23)$ with gradient 30.

h^{'} (4) = f^{'} (g (4)) g^{'} (4) = 6 \times 5 = 30 y - 23 = 30 (x - 4) \Rightarrow y = 30 x - 97

Question 6

7 marks

A rational identity, used to solve a trig equation

Number and algebra
Geometry and trigonometry

(a)

2 marks

Put the right-hand side over the common denominator.

2 x - 3 - \frac{6}{x - 1} = \frac{( 2 x - 3 ) ( x - 1 ) - 6}{x - 1} = \frac{2 x ^{2} - 5 x + 3 - 6}{x - 1} = \frac{2 x ^{2} - 5 x - 3}{x - 1}

(b)

5 marks

Put $x = sin 2 θ$ in the equation $2 sin 2 θ - 3 - \frac{6}{sin 2 θ - 1} = 0$ . By part (a) it becomes a quadratic.
$sin 2 θ = 3$ is impossible, so $sin 2 θ = - \frac{1}{2}$ for $0 \leq 2 θ \leq 2 π$ .

\frac{2 sin ^{2} 2 θ - 5 sin 2 θ - 3}{sin 2 θ - 1} = 0 \Rightarrow (2 sin 2 θ + 1) (sin 2 θ - 3) = 0 2 θ = \frac{7 π}{6}, \frac{11 π}{6} \Rightarrow θ = \frac{7 π}{12}, \frac{11 π}{12}

Mentor tip

Neither answer is $\frac{π}{4}$ , the excluded value, where $sin 2 θ = 1$ would make the denominator zero.

Section B · Extended response

Questions 7–9 · 45 marks

Question 7

16 marks

A particle's velocity: maximum speed, distance, graph and total distance

Calculus

(a)

7 marks

(i) The velocity is largest where $v^{'} = 0$ .

(ii) Integrate $v$ from 0 (since $P$ starts at the origin) up to $t = \frac{2}{3}$ .

v^{'} = 4 - 6 t = 0 \Rightarrow t = \frac{2}{3} x = \int_{0}^{2/3} (4 + 4 t - 3 t^{2}) d t = [4 t + 2 t^{2} - t^{3}]_{0}^{2/3} \frac{8}{3} + \frac{8}{9} - \frac{8}{27} = \frac{72 + 24 - 8}{27} = \frac{88}{27} m

(b)

4 marks

The graph is a downward-opening parabola for $0 \leq t \leq 3$ . It starts at $(0, 4)$ and has a maximum at $t = \frac{2}{3}$ with $v = \frac{16}{3}$ . It crosses the $t$ -axis where $3 t^{2} - 4 t - 4 = 0$ .
It ends at $t = 3$ with $v = 4 + 12 - 27 = - 11$ .

(3 t + 2) (t - 2) = 0 \Rightarrow t = 2

(c)

5 marks

$v > 0$ for $0 < t < 2$ and $v < 0$ for $2 < t \leq 3$ . Integrate each piece.

\int_{0}^{2} v d t = 8, \int_{2}^{3} v d t = - 5 total distance = 8 + 5 = 13 m

Question 8

15 marks

An exponential function, its inverse and a geometric sequence hidden in logarithms

Functions
Number and algebra

(a)

2 marks

Substitute the point $(\frac{2}{3}, 4)$ into $f (x) = a^{x}$ .

a^{2/3} = 4 \Rightarrow a^{2} = 4^{3} = 64 \Rightarrow a = 8

(b)

1 mark

The inverse of $8^{x}$ is a logarithm.

f^{- 1} (x) = lo g_{8} x

(c)

3 marks

Use $32 = 3 2^{1/2} = 2^{5/2}$ and $8 = 2^{3}$ .

f^{- 1} (32) = lo g_{8} 2^{5/2} = \frac{5/2}{3} = \frac{5}{6}

(d)

9 marks

(i) The logarithms form an arithmetic sequence, so consecutive differences are equal.

The ratios between consecutive terms are all equal, so $27, p, q, 125$ are consecutive terms of a geometric sequence.

(ii) With first term 27 and fourth term 125, the common ratio satisfies $27 r^{3} = 125$ .

lo g_{8} p - lo g_{8} 27 = lo g_{8} q - lo g_{8} p = lo g_{8} 125 - lo g_{8} q lo g_{8} \frac{p}{27} = lo g_{8} \frac{q}{p} = lo g_{8} \frac{125}{q} \Rightarrow \frac{p}{27} = \frac{q}{p} = \frac{125}{q} r^{3} = \frac{125}{27} \Rightarrow r = \frac{5}{3} p = 27 \cdot \frac{5}{3} = 45, q = 45 \cdot \frac{5}{3} = 75

Question 9

14 marks

Reading the graph of a derivative: increasing, turning points and a value of $f$

Calculus

(a)

2 marks

$f$ is increasing where $f^{'} (x) > 0$ , that is, where the graph of $f^{'}$ is above the $x$ -axis. That is between its intercepts at $p$ and 0.

p < x < 0

(b)

1 mark

$f^{'}$ changes from positive to negative at $x = 0$ , so $f$ has a local maximum there.

x = 0

(c)

5 marks

(i) At $x = p$ we have $f^{'} (p) = 0$ , and $f^{'}$ changes from negative to positive, so $f$ has a local minimum.

(ii) Points of inflexion of $f$ are where $f^{'}$ has a turning point, since $f^{''}$ changes sign there. The turning points of $f^{'}$ are at

x = p x = q, x = r, x = t

(d)

6 marks

The area between $f^{'}$ and the axis is the sum of the sizes of the integrals from $p$ to 0 (above the axis) and from 0 to $t$ (below).
Evaluate each integral as a change in $f$ .
Put in $f (p) + f (t) = 4$ .

\int_{p}^{0} f^{'} (x) d x - \int_{0}^{t} f^{'} (x) d x = 20 [f (0) - f (p)] - [f (t) - f (0)] = 2 f (0) - (f (p) + f (t)) = 20 2 f (0) - 4 = 20 \Rightarrow f (0) = 12

Our analysis

Where the marks are

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

Calculus41 marks · 51%
Q2, Q5, Q7, Q9
Functions18 marks · 23%
Q1, Q3, Q8
Number and algebra11 marks · 14%
Q6, Q8
Geometry and trigonometry5 marks · 6%
Q6
Statistics and probability5 marks · 6%
Q4

Mentor insights

Traps to watch for

01
Vertex is halfway between the roots · Q1
For a quadratic the vertex lies on the axis of symmetry, which is the mean of the two intercepts.
02
Chain rule for the composite · Q5
$h = f \circ g$ , so $h^{'} (x) = f^{'} (g (x)) g^{'} (x)$ . At $x = 4$ we need $g (4)$ and $g^{'} (4)$ .
03
Total distance needs the sign change · Q7
The velocity changes sign at $t = 2$ . Integrate each side separately and add the sizes.
04
Equal differences of logs mean a constant ratio · Q8
In an arithmetic sequence of logarithms the differences are equal. Equal differences of logs are equal log of ratios, so the ratios are equal.
05
Justify with the derivative's graph · Q9
Say " $f^{'} > 0$ so $f$ is increasing". Always refer to the graph of $f^{'}$ , not the graph of $f$ , to avoid mixing up the two gradients.

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

Session: November 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 November 2021 Global 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

7 marks

A quadratic in factorised form: intercepts, vertex and completed square

Functions

(a)

5 marks

(i) $f (x) = 0$ when

(ii) The vertex is halfway between $- 3$ and 1, so at $x = - 1$ .

- 2 (x - 1) (x + 3) = 0 \Rightarrow x = 1 or x = - 3 f (- 1) = - 2 (- 2) (2) = 8, vertex (- 1, 8)

(b)

2 marks

The vertex form has the vertex at $(h, k)$ .

h = - 1, k = 8

Question 2

4 marks

A simple differential equation with an initial condition

Calculus

4 marks

Integrate $cos (x - \frac{π}{4})$ with respect to $x$ .
Use $y = 2$ at $x = \frac{3 π}{4}$ , where $sin \frac{π}{2} = 1$ .

y = sin (x - \frac{π}{4}) + c 2 = 1 + c \Rightarrow c = 1 y = sin (x - \frac{π}{4}) + 1

Question 3

5 marks

A rational function: asymptotes, intercepts and a sketch

Functions

(a)

2 marks

The function is $f (x) = \frac{2 x + 4}{3 - x}$ . (i) The denominator is zero at
(ii) As $x \to \infty$ , $f (x) \to \frac{2 x}{- x}$ .

x = 3 y = - 2

(b)

2 marks

(i) The graph crosses the $x$ -axis where $2 x + 4 = 0$ .

(ii) At $x = 0$ , $f (0) = \frac{4}{3}$ .

(- 2, 0) (0, \frac{4}{3})

(c)

1 mark

Question 4

5 marks

Two boxes of balls: a total probability and an independence test

Statistics and probability

(a)

3 marks

Either box is chosen with probability $\frac{1}{2}$ , then a red ball is drawn.

P (R) = \frac{1}{2} \cdot \frac{5}{7} + \frac{1}{2} \cdot \frac{4}{7} = \frac{9}{14}

(b)

2 marks

Compare $P (A) P (R)$ with $P (A \cap R)$ .
They are not equal, so $A$ and $R$ are not independent. (Box 1 has a higher proportion of red balls.)

P (A) P (R) = \frac{1}{2} \cdot \frac{9}{14} = \frac{9}{28}, P (A \cap R) = \frac{1}{2} \cdot \frac{5}{7} = \frac{5}{14} = \frac{10}{28}

Question 5

7 marks

A tangent to $f$ and the tangent to a composite

Calculus

(a)

1 mark

The tangent has gradient 6.

f^{'} (4) = 6

(b)

1 mark

The tangent passes through the graph at $x = 4$ .

f (4) = 6 (4) - 1 = 23

(c)

2 marks

First $g (4) = 16 - 12 = 4$ .

h (4) = f (g (4)) = f (4) = 23

(d)

3 marks

Use the chain rule with $g^{'} (x) = 2 x - 3$ , so $g^{'} (4) = 5$ .
The tangent at $x = 4$ passes through $(4, 23)$ with gradient 30.

h^{'} (4) = f^{'} (g (4)) g^{'} (4) = 6 \times 5 = 30 y - 23 = 30 (x - 4) \Rightarrow y = 30 x - 97

Question 6

7 marks

A rational identity, used to solve a trig equation

Number and algebra
Geometry and trigonometry

(a)

2 marks

Put the right-hand side over the common denominator.

2 x - 3 - \frac{6}{x - 1} = \frac{( 2 x - 3 ) ( x - 1 ) - 6}{x - 1} = \frac{2 x ^{2} - 5 x + 3 - 6}{x - 1} = \frac{2 x ^{2} - 5 x - 3}{x - 1}

(b)

5 marks

Put $x = sin 2 θ$ in the equation $2 sin 2 θ - 3 - \frac{6}{sin 2 θ - 1} = 0$ . By part (a) it becomes a quadratic.
$sin 2 θ = 3$ is impossible, so $sin 2 θ = - \frac{1}{2}$ for $0 \leq 2 θ \leq 2 π$ .

\frac{2 sin ^{2} 2 θ - 5 sin 2 θ - 3}{sin 2 θ - 1} = 0 \Rightarrow (2 sin 2 θ + 1) (sin 2 θ - 3) = 0 2 θ = \frac{7 π}{6}, \frac{11 π}{6} \Rightarrow θ = \frac{7 π}{12}, \frac{11 π}{12}

Mentor tip

Neither answer is $\frac{π}{4}$ , the excluded value, where $sin 2 θ = 1$ would make the denominator zero.

Section B · Extended response

Questions 7–9 · 45 marks

Question 7

16 marks

A particle's velocity: maximum speed, distance, graph and total distance

Calculus

(a)

7 marks

(i) The velocity is largest where $v^{'} = 0$ .

(ii) Integrate $v$ from 0 (since $P$ starts at the origin) up to $t = \frac{2}{3}$ .

v^{'} = 4 - 6 t = 0 \Rightarrow t = \frac{2}{3} x = \int_{0}^{2/3} (4 + 4 t - 3 t^{2}) d t = [4 t + 2 t^{2} - t^{3}]_{0}^{2/3} \frac{8}{3} + \frac{8}{9} - \frac{8}{27} = \frac{72 + 24 - 8}{27} = \frac{88}{27} m

(b)

4 marks

The graph is a downward-opening parabola for $0 \leq t \leq 3$ . It starts at $(0, 4)$ and has a maximum at $t = \frac{2}{3}$ with $v = \frac{16}{3}$ . It crosses the $t$ -axis where $3 t^{2} - 4 t - 4 = 0$ .
It ends at $t = 3$ with $v = 4 + 12 - 27 = - 11$ .

(3 t + 2) (t - 2) = 0 \Rightarrow t = 2

(c)

5 marks

$v > 0$ for $0 < t < 2$ and $v < 0$ for $2 < t \leq 3$ . Integrate each piece.

\int_{0}^{2} v d t = 8, \int_{2}^{3} v d t = - 5 total distance = 8 + 5 = 13 m

Question 8

15 marks

An exponential function, its inverse and a geometric sequence hidden in logarithms

Functions
Number and algebra

(a)

2 marks

Substitute the point $(\frac{2}{3}, 4)$ into $f (x) = a^{x}$ .

a^{2/3} = 4 \Rightarrow a^{2} = 4^{3} = 64 \Rightarrow a = 8

(b)

1 mark

The inverse of $8^{x}$ is a logarithm.

f^{- 1} (x) = lo g_{8} x

(c)

3 marks

Use $32 = 3 2^{1/2} = 2^{5/2}$ and $8 = 2^{3}$ .

f^{- 1} (32) = lo g_{8} 2^{5/2} = \frac{5/2}{3} = \frac{5}{6}

(d)

9 marks

(i) The logarithms form an arithmetic sequence, so consecutive differences are equal.

The ratios between consecutive terms are all equal, so $27, p, q, 125$ are consecutive terms of a geometric sequence.

(ii) With first term 27 and fourth term 125, the common ratio satisfies $27 r^{3} = 125$ .

lo g_{8} p - lo g_{8} 27 = lo g_{8} q - lo g_{8} p = lo g_{8} 125 - lo g_{8} q lo g_{8} \frac{p}{27} = lo g_{8} \frac{q}{p} = lo g_{8} \frac{125}{q} \Rightarrow \frac{p}{27} = \frac{q}{p} = \frac{125}{q} r^{3} = \frac{125}{27} \Rightarrow r = \frac{5}{3} p = 27 \cdot \frac{5}{3} = 45, q = 45 \cdot \frac{5}{3} = 75

Question 9

14 marks

Reading the graph of a derivative: increasing, turning points and a value of $f$

Calculus

(a)

2 marks

$f$ is increasing where $f^{'} (x) > 0$ , that is, where the graph of $f^{'}$ is above the $x$ -axis. That is between its intercepts at $p$ and 0.

p < x < 0

(b)

1 mark

$f^{'}$ changes from positive to negative at $x = 0$ , so $f$ has a local maximum there.

x = 0

(c)

5 marks

(i) At $x = p$ we have $f^{'} (p) = 0$ , and $f^{'}$ changes from negative to positive, so $f$ has a local minimum.

(ii) Points of inflexion of $f$ are where $f^{'}$ has a turning point, since $f^{''}$ changes sign there. The turning points of $f^{'}$ are at

x = p x = q, x = r, x = t

(d)

6 marks

The area between $f^{'}$ and the axis is the sum of the sizes of the integrals from $p$ to 0 (above the axis) and from 0 to $t$ (below).
Evaluate each integral as a change in $f$ .
Put in $f (p) + f (t) = 4$ .

\int_{p}^{0} f^{'} (x) d x - \int_{0}^{t} f^{'} (x) d x = 20 [f (0) - f (p)] - [f (t) - f (0)] = 2 f (0) - (f (p) + f (t)) = 20 2 f (0) - 4 = 20 \Rightarrow f (0) = 12

Our analysis

Where the marks are

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

Calculus41 marks · 51%
Q2, Q5, Q7, Q9
Functions18 marks · 23%
Q1, Q3, Q8
Number and algebra11 marks · 14%
Q6, Q8
Geometry and trigonometry5 marks · 6%
Q6
Statistics and probability5 marks · 6%
Q4

Mentor insights

Traps to watch for

01
Vertex is halfway between the roots · Q1
For a quadratic the vertex lies on the axis of symmetry, which is the mean of the two intercepts.
02
Chain rule for the composite · Q5
$h = f \circ g$ , so $h^{'} (x) = f^{'} (g (x)) g^{'} (x)$ . At $x = 4$ we need $g (4)$ and $g^{'} (4)$ .
03
Total distance needs the sign change · Q7
The velocity changes sign at $t = 2$ . Integrate each side separately and add the sizes.
04
Equal differences of logs mean a constant ratio · Q8
In an arithmetic sequence of logarithms the differences are equal. Equal differences of logs are equal log of ratios, so the ratios are equal.
05
Justify with the derivative's graph · Q9
Say " $f^{'} > 0$ so $f$ is increasing". Always refer to the graph of $f^{'}$ , not the graph of $f$ , to avoid mixing up the two gradients.

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