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IB Maths AA HL · Paper 3

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2021
Marks: 55
Duration: 1 hour 15 min
Calculator: Allowed
Questions: 2

Q1 Q2

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2021 TZ1 paper, which your school or the IB can provide.

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Questions

Questions 1–2 · 55 marks

Question 1

27 marks

Cubics $x^{3} - 3 c x + d$ : turning points and how many times they cross the axis

Functions
Calculus

(a)(i)

3 marks

For $c = 1$ , $f (x) = x^{3} - 3 x + 2$ and $f^{'} (x) = 3 x^{2} - 3$ , which is zero at $x = \pm 1$ . The $y$ -intercept is $(0, 2)$ .
So the sketch rises from the lower left to a local maximum at $(- 1, 4)$ , falls through $(0, 2)$ to a local minimum at $(1, 0)$ that just touches the axis, then rises again.

f (- 1) = - 1 + 3 + 2 = 4, f (1) = 1 - 3 + 2 = 0

(a)(ii)

3 marks

For $c = 2$ , $f (x) = x^{3} - 6 x + 2$ and $f^{'} (x) = 3 x^{2} - 6$ , which is zero at $x = \pm 2$ .
The $y$ -intercept is $(0, 2)$ . The sketch has a local maximum at $(- 2, 2 + 42)$ and a local minimum at $(2, 2 - 42)$ below the axis, so it crosses the axis three times.

f (- 2) = 2 + 42 \approx 7.66, f (2) = 2 - 42 \approx - 3.66

(b)

1 mark

Differentiate term by term.

f^{'} (x) = 3 x^{2} - 3 c

(c)(i)

1 mark

A point of inflexion with zero gradient needs $f^{'} (x) = 0$ to have exactly one (repeated) solution, which happens when $3 x^{2} = 3 c$ has just the root $x = 0$ .

c = 0

(c)(ii)

2 marks

One local maximum and one local minimum needs two distinct solutions of $x^{2} = c$ .

c > 0

(c)(iii)

1 mark

No stationary points needs $x^{2} = c$ to have no real solution.

c < 0

(d)(i)

3 marks

For $c > 0$ the stationary points are at $x = \pm c$ . The maximum is on the left, at $x = - c$ .
Since $f^{'}$ goes from positive to negative there, this is a local maximum.

f (- c) = - c c + 3 c c + 2 = 2 c c + 2 = 2 c^{3/2} + 2

(d)(ii)

1 mark

At $x = c$ :

f (c) = c c - 3 c c + 2 = - 2 c^{3/2} + 2

(e)(i)

2 marks

The maximum value $2 + 2 c^{3/2}$ is positive for $c > 0$ . So exactly one intercept happens only when the minimum value is positive too: $2 - 2 c^{3/2} > 0$ .

0 < c < 1

(e)(ii)

2 marks

Two intercepts means the graph touches the axis at the local minimum, so that value is zero.

2 - 2 c^{3/2} = 0 \Rightarrow c = 1

(e)(iii)

2 marks

Three intercepts means the minimum is below the axis, with the maximum above it.

2 - 2 c^{3/2} < 0 \Rightarrow c > 1

(f)

6 marks

Repeat the working with a general $d$ . For $c > 0$ , the local maximum is $d + 2 c^{3/2}$ and the local minimum is $d - 2 c^{3/2}$ .
Exactly one intercept needs both turning values on the same side of the axis: either the maximum is negative ( $d < - 2 c^{3/2}$ ) or the minimum is positive ( $d > 2 c^{3/2}$ ). That is $∣ d ∣ > 2 c^{3/2}$ , or $d^{2} > 4 c^{3}$ .
For $c < 0$ there are no stationary points, so $g$ is strictly increasing and crosses the axis exactly once, for every $d$ . For $c = 0$ , $g (x) = x^{3} + d$ is also strictly increasing, so again one intercept for every $d$ .

exactly one intercept ⟺ c \leq 0 (any d), or c > 0 and ∣ d ∣ > 2 c^{3/2}

Mentor tip

The cases $c \leq 0$ are covered by "no gradient sign change", which is a different reason from the turning-value one. Give both.

Question 2

28 marks

Polygons whose area and perimeter are the same number

Geometry and trigonometry
Calculus
Number and algebra

(a)

3 marks

The area is $s^{2}$ and the perimeter is $4 s$ . Set them equal and cancel one factor of $s$ (valid since $s > 0$ ).

s^{2} = 4 s \Rightarrow s = 4

(b)

1 mark

The triangle has two sides of length $x$ with the included angle $\frac{2 π}{n}$ , so use $\frac{1}{2} ab sin C$ .

A_{T} = \frac{1}{2} x^{2} sin \frac{2 π}{n}

(c)

2 marks

Split the isosceles triangle down the middle into two right-angled triangles. Each has hypotenuse $x$ , opposite side $\frac{y}{2}$ and angle $\frac{π}{n}$ .

\frac{y}{2} = x sin \frac{π}{n} \Rightarrow y = 2 x sin \frac{π}{n}

(d)

7 marks

The polygon has $n$ triangles and $n$ sides of length $y$ . Write the area and perimeter, and use $sin \frac{2 π}{n} = 2 sin \frac{π}{n} cos \frac{π}{n}$ .
Set $A = P$ and divide by $n x sin \frac{π}{n}$ (which is not zero).
Put this back into the perimeter.

A = \frac{n}{2} x^{2} \cdot 2 sin \frac{π}{n} cos \frac{π}{n} = n x^{2} sin \frac{π}{n} cos \frac{π}{n}, P = n y = 2 n x sin \frac{π}{n} x cos \frac{π}{n} = 2 \Rightarrow x = \frac{2}{cos ( π / n )} P = 2 n \cdot \frac{2}{cos ( π / n )} sin \frac{π}{n} = 4 n tan \frac{π}{n}, A = P = 4 n tan \frac{π}{n}

(e)(i)

3 marks

Replace $tan \frac{π}{n}$ by the Maclaurin series with $x = \frac{π}{n}$ , and multiply by $4 n$ .
The terms after $4 π$ all have $n$ in the denominator, so they tend to 0.

4 n tan \frac{π}{n} = 4 n (\frac{π}{n} + \frac{1}{3} \frac{π ^{3}}{n ^{3}} + \frac{2}{15} \frac{π ^{5}}{n ^{5}} + \dots) = 4 π + \frac{4 π ^{3}}{3 n ^{2}} + \dots n \to \infty lim 4 n tan \frac{π}{n} = 4 π

(e)(ii)

1 mark

As $n \to \infty$ the polygon looks more and more like a circle. Its area and perimeter both tend to $4 π$ . A circle with $π r^{2} = 2 π r = 4 π$ has radius 2.
So the polygon becomes a circle of radius 2.

(f)

7 marks

The legs are $a$ and $b$ , the hypotenuse is $a^{2} + b^{2}$ , the area is $\frac{1}{2} ab$ and the perimeter is $a + b + a^{2} + b^{2}$ . Set $A = P$ and isolate the square root, then square.
The $a^{2}$ and $b^{2}$ cancel. Divide by $ab$ (non-zero) and multiply by 4.
Collect $a$ .

\frac{1}{2} ab - a - b = a^{2} + b^{2} \frac{1}{4} a^{2} b^{2} + a^{2} + b^{2} - a^{2} b - a b^{2} + 2 ab = a^{2} + b^{2} \frac{1}{4} ab - a - b + 2 = 0 \Rightarrow ab - 4 a - 4 b + 8 = 0 a (b - 4) = 4 b - 8 \Rightarrow a = \frac{4 b - 8}{b - 4} = 4 + \frac{8}{b - 4}

(g)(i)

3 marks

For $a$ to be a positive whole number with $a \geq b$ , $(b - 4)$ must divide 8 and be positive. So $b - 4 \in {1, 2, 4, 8}$ .
The pairs with $a \geq b$ are $(12, 5)$ and $(8, 6)$ . The hypotenuses are $169 = 13$ and $100 = 10$ .

b = 5 \Rightarrow a = 12; b = 6 \Rightarrow a = 8; b = 8 \Rightarrow a = 6; b = 12 \Rightarrow a = 5 (5, 12, 13) and (6, 8, 10)

(g)(ii)

1 mark

Check the area and perimeter of each.

5 - 12 - 13 : A = 30, P = 30; 6 - 8 - 10 : A = 24, P = 24

Mentor tip

That's quick to check: $\frac{1}{2} \cdot 5 \cdot 12 = 30 = 5 + 12 + 13$ and $\frac{1}{2} \cdot 6 \cdot 8 = 24 = 6 + 8 + 10$ .

Our analysis

Where the marks are

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

Functions18 marks · 33%
Q1
Geometry and trigonometry14 marks · 25%
Q2
Calculus12 marks · 22%
Q1, Q2
Number and algebra11 marks · 20%
Q2

Mentor insights

Traps to watch for

01
Zero gradient is not the same as a turning point · Q1
At $c = 0$ we have $f^{'} (0) = 0$ , but the gradient does not change sign there. It is an inflexion, not a maximum or a minimum. Check the sign on both sides.
02
Don't forget the cases without turning points · Q1
For $c \leq 0$ the cubic is always increasing, so it has exactly one intercept whatever $d$ is. Many answers to (f) only look at $c > 0$ .
03
Divide only by what can't be zero · Q2
Cancelling $n x sin \frac{π}{n}$ is fine because $n$ , $x$ and that sine are all positive. Say so, or the cancellation is unjustified.
04
Squaring can add solutions · Q2
In (f), squaring $\frac{1}{2} ab - a - b = a^{2} + b^{2}$ gives an equation that can have extra solutions. Check that the triangles you find really have area equal to perimeter.
05
Show that, so use what you are given · Q2
In (d), the area and perimeter come from parts (b) and (c). Set them equal, cancel the common factor $n x sin \frac{π}{n}$ , and the side $x$ falls out.

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 HL · Paper 3

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2021
Marks: 55
Duration: 1 hour 15 min
Calculator: Allowed
Questions: 2

Q1 Q2

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2021 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–2 · 55 marks

Question 1

27 marks

Cubics $x^{3} - 3 c x + d$ : turning points and how many times they cross the axis

Functions
Calculus

(a)(i)

3 marks

For $c = 1$ , $f (x) = x^{3} - 3 x + 2$ and $f^{'} (x) = 3 x^{2} - 3$ , which is zero at $x = \pm 1$ . The $y$ -intercept is $(0, 2)$ .
So the sketch rises from the lower left to a local maximum at $(- 1, 4)$ , falls through $(0, 2)$ to a local minimum at $(1, 0)$ that just touches the axis, then rises again.

f (- 1) = - 1 + 3 + 2 = 4, f (1) = 1 - 3 + 2 = 0

(a)(ii)

3 marks

For $c = 2$ , $f (x) = x^{3} - 6 x + 2$ and $f^{'} (x) = 3 x^{2} - 6$ , which is zero at $x = \pm 2$ .
The $y$ -intercept is $(0, 2)$ . The sketch has a local maximum at $(- 2, 2 + 42)$ and a local minimum at $(2, 2 - 42)$ below the axis, so it crosses the axis three times.

f (- 2) = 2 + 42 \approx 7.66, f (2) = 2 - 42 \approx - 3.66

(b)

1 mark

Differentiate term by term.

f^{'} (x) = 3 x^{2} - 3 c

(c)(i)

1 mark

A point of inflexion with zero gradient needs $f^{'} (x) = 0$ to have exactly one (repeated) solution, which happens when $3 x^{2} = 3 c$ has just the root $x = 0$ .

c = 0

(c)(ii)

2 marks

One local maximum and one local minimum needs two distinct solutions of $x^{2} = c$ .

c > 0

(c)(iii)

1 mark

No stationary points needs $x^{2} = c$ to have no real solution.

c < 0

(d)(i)

3 marks

For $c > 0$ the stationary points are at $x = \pm c$ . The maximum is on the left, at $x = - c$ .
Since $f^{'}$ goes from positive to negative there, this is a local maximum.

f (- c) = - c c + 3 c c + 2 = 2 c c + 2 = 2 c^{3/2} + 2

(d)(ii)

1 mark

At $x = c$ :

f (c) = c c - 3 c c + 2 = - 2 c^{3/2} + 2

(e)(i)

2 marks

The maximum value $2 + 2 c^{3/2}$ is positive for $c > 0$ . So exactly one intercept happens only when the minimum value is positive too: $2 - 2 c^{3/2} > 0$ .

0 < c < 1

(e)(ii)

2 marks

Two intercepts means the graph touches the axis at the local minimum, so that value is zero.

2 - 2 c^{3/2} = 0 \Rightarrow c = 1

(e)(iii)

2 marks

Three intercepts means the minimum is below the axis, with the maximum above it.

2 - 2 c^{3/2} < 0 \Rightarrow c > 1

(f)

6 marks

Repeat the working with a general $d$ . For $c > 0$ , the local maximum is $d + 2 c^{3/2}$ and the local minimum is $d - 2 c^{3/2}$ .
Exactly one intercept needs both turning values on the same side of the axis: either the maximum is negative ( $d < - 2 c^{3/2}$ ) or the minimum is positive ( $d > 2 c^{3/2}$ ). That is $∣ d ∣ > 2 c^{3/2}$ , or $d^{2} > 4 c^{3}$ .
For $c < 0$ there are no stationary points, so $g$ is strictly increasing and crosses the axis exactly once, for every $d$ . For $c = 0$ , $g (x) = x^{3} + d$ is also strictly increasing, so again one intercept for every $d$ .

exactly one intercept ⟺ c \leq 0 (any d), or c > 0 and ∣ d ∣ > 2 c^{3/2}

Mentor tip

The cases $c \leq 0$ are covered by "no gradient sign change", which is a different reason from the turning-value one. Give both.

Question 2

28 marks

Polygons whose area and perimeter are the same number

Geometry and trigonometry
Calculus
Number and algebra

(a)

3 marks

The area is $s^{2}$ and the perimeter is $4 s$ . Set them equal and cancel one factor of $s$ (valid since $s > 0$ ).

s^{2} = 4 s \Rightarrow s = 4

(b)

1 mark

The triangle has two sides of length $x$ with the included angle $\frac{2 π}{n}$ , so use $\frac{1}{2} ab sin C$ .

A_{T} = \frac{1}{2} x^{2} sin \frac{2 π}{n}

(c)

2 marks

Split the isosceles triangle down the middle into two right-angled triangles. Each has hypotenuse $x$ , opposite side $\frac{y}{2}$ and angle $\frac{π}{n}$ .

\frac{y}{2} = x sin \frac{π}{n} \Rightarrow y = 2 x sin \frac{π}{n}

(d)

7 marks

The polygon has $n$ triangles and $n$ sides of length $y$ . Write the area and perimeter, and use $sin \frac{2 π}{n} = 2 sin \frac{π}{n} cos \frac{π}{n}$ .
Set $A = P$ and divide by $n x sin \frac{π}{n}$ (which is not zero).
Put this back into the perimeter.

A = \frac{n}{2} x^{2} \cdot 2 sin \frac{π}{n} cos \frac{π}{n} = n x^{2} sin \frac{π}{n} cos \frac{π}{n}, P = n y = 2 n x sin \frac{π}{n} x cos \frac{π}{n} = 2 \Rightarrow x = \frac{2}{cos ( π / n )} P = 2 n \cdot \frac{2}{cos ( π / n )} sin \frac{π}{n} = 4 n tan \frac{π}{n}, A = P = 4 n tan \frac{π}{n}

(e)(i)

3 marks

Replace $tan \frac{π}{n}$ by the Maclaurin series with $x = \frac{π}{n}$ , and multiply by $4 n$ .
The terms after $4 π$ all have $n$ in the denominator, so they tend to 0.

4 n tan \frac{π}{n} = 4 n (\frac{π}{n} + \frac{1}{3} \frac{π ^{3}}{n ^{3}} + \frac{2}{15} \frac{π ^{5}}{n ^{5}} + \dots) = 4 π + \frac{4 π ^{3}}{3 n ^{2}} + \dots n \to \infty lim 4 n tan \frac{π}{n} = 4 π

(e)(ii)

1 mark

As $n \to \infty$ the polygon looks more and more like a circle. Its area and perimeter both tend to $4 π$ . A circle with $π r^{2} = 2 π r = 4 π$ has radius 2.
So the polygon becomes a circle of radius 2.

(f)

7 marks

The legs are $a$ and $b$ , the hypotenuse is $a^{2} + b^{2}$ , the area is $\frac{1}{2} ab$ and the perimeter is $a + b + a^{2} + b^{2}$ . Set $A = P$ and isolate the square root, then square.
The $a^{2}$ and $b^{2}$ cancel. Divide by $ab$ (non-zero) and multiply by 4.
Collect $a$ .

\frac{1}{2} ab - a - b = a^{2} + b^{2} \frac{1}{4} a^{2} b^{2} + a^{2} + b^{2} - a^{2} b - a b^{2} + 2 ab = a^{2} + b^{2} \frac{1}{4} ab - a - b + 2 = 0 \Rightarrow ab - 4 a - 4 b + 8 = 0 a (b - 4) = 4 b - 8 \Rightarrow a = \frac{4 b - 8}{b - 4} = 4 + \frac{8}{b - 4}

(g)(i)

3 marks

For $a$ to be a positive whole number with $a \geq b$ , $(b - 4)$ must divide 8 and be positive. So $b - 4 \in {1, 2, 4, 8}$ .
The pairs with $a \geq b$ are $(12, 5)$ and $(8, 6)$ . The hypotenuses are $169 = 13$ and $100 = 10$ .

b = 5 \Rightarrow a = 12; b = 6 \Rightarrow a = 8; b = 8 \Rightarrow a = 6; b = 12 \Rightarrow a = 5 (5, 12, 13) and (6, 8, 10)

(g)(ii)

1 mark

Check the area and perimeter of each.

5 - 12 - 13 : A = 30, P = 30; 6 - 8 - 10 : A = 24, P = 24

Mentor tip

That's quick to check: $\frac{1}{2} \cdot 5 \cdot 12 = 30 = 5 + 12 + 13$ and $\frac{1}{2} \cdot 6 \cdot 8 = 24 = 6 + 8 + 10$ .

Our analysis

Where the marks are

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

Functions18 marks · 33%
Q1
Geometry and trigonometry14 marks · 25%
Q2
Calculus12 marks · 22%
Q1, Q2
Number and algebra11 marks · 20%
Q2

Mentor insights

Traps to watch for

01
Zero gradient is not the same as a turning point · Q1
At $c = 0$ we have $f^{'} (0) = 0$ , but the gradient does not change sign there. It is an inflexion, not a maximum or a minimum. Check the sign on both sides.
02
Don't forget the cases without turning points · Q1
For $c \leq 0$ the cubic is always increasing, so it has exactly one intercept whatever $d$ is. Many answers to (f) only look at $c > 0$ .
03
Divide only by what can't be zero · Q2
Cancelling $n x sin \frac{π}{n}$ is fine because $n$ , $x$ and that sine are all positive. Say so, or the cancellation is unjustified.
04
Squaring can add solutions · Q2
In (f), squaring $\frac{1}{2} ab - a - b = a^{2} + b^{2}$ gives an equation that can have extra solutions. Check that the triangles you find really have area equal to perimeter.
05
Show that, so use what you are given · Q2
In (d), the area and perimeter come from parts (b) and (c). Set them equal, cancel the common factor $n x sin \frac{π}{n}$ , and the side $x$ falls out.

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