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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

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Questions

Questions 1–2 · 55 marks

Question 1

31 marks

The family $f_{n} (x) = x^{n} (a - x)^{n}$ : shape, stationary points, and four solutions

Functions
Calculus

(a)

3 marks

$f_{1} (x) = x (2 - x)$ is a downward parabola crossing the axis at $x = 0$ and $x = 2$ . Its vertex is halfway, at $x = 1$ .

f_{1} (1) = 1, intercepts (0, 0), (2, 0), local maximum (1, 1)

(b)

6 marks

Exploring on the calculator, the graph is symmetric about $x = 1$ with its single peak at $x = 1$ , where $f_{n} (1) = 1$ .
For odd $n$ (3 and 5), the graph flattens and keeps going the same way through $x = 0$ and $x = 2$ , so those are points of inflexion with zero gradient. For even $n$ (2 and 4), the graph touches the axis at 0 and 2 and bounces back, so those are minima.

n = 3, 5 n = 2, 4 local max 11 local min 02 inflexion, zero gradient 20

(c)

5 marks

Use the product rule with the chain rule on $(a - x)^{n}$ , which has derivative $- n (a - x)^{n - 1}$ .
Factor out the common $n x^{n - 1} (a - x)^{n - 1}$ .

f_{n}^{'} (x) = n x^{n - 1} (a - x)^{n} - n x^{n} (a - x)^{n - 1} f_{n}^{'} (x) = n x^{n - 1} (a - x)^{n - 1} [(a - x) - x] = n x^{n - 1} (a - 2 x) (a - x)^{n - 1}

(d)

2 marks

Each factor can be zero.

x = 0, x = \frac{a}{2}, x = a

(e)

3 marks

At $x = \frac{a}{2}$ both factors $x$ and $a - x$ equal $\frac{a}{2}$ , which is positive.
A positive $y$ -value means the point is above the horizontal axis.

f_{n} (\frac{a}{2}) = (\frac{a}{2})^{n} (\frac{a}{2})^{n} = (\frac{a ^{2}}{4})^{n} > 0

(f)

2 marks

At $x = \frac{a}{4}$ each factor in $f_{n}^{'}$ is positive: $(\frac{a}{4})^{n - 1}$ , $a - \frac{a}{2} = \frac{a}{2}$ and $(\frac{3 a}{4})^{n - 1}$ , with $n > 0$ .

f_{n}^{'} (\frac{a}{4}) = n (\frac{a}{4})^{n - 1} \cdot \frac{a}{2} \cdot (\frac{3 a}{4})^{n - 1} > 0

(g)(i)

3 marks

The gradient is positive just to the right of $0$ (by part (f), since there are no other stationary points between 0 and $\frac{a}{2}$ ). For $x = - 1$ , $f_{n}^{'} (- 1) = n (- 1)^{n - 1} (a + 2) (a + 1)^{n - 1}$ and its sign is that of $(- 1)^{n - 1}$ .
For even $n$ , $n - 1$ is odd, so $f_{n}^{'} (- 1) < 0$ . The gradient changes from negative to positive at $x = 0$ , so $(0, 0)$ is a local minimum.

(g)(ii)

2 marks

For odd $n$ , $n - 1$ is even, so $f_{n}^{'} (- 1) > 0$ . The gradient is positive on both sides of $x = 0$ and is zero there (since $n > 1$ ), so $(0, 0)$ is a point of inflexion with zero gradient.

(h)

5 marks

Four solutions needs the line $y = k$ to cut the graph four times. The graph has two minima at height 0 and a maximum of $(\frac{a ^{2}}{4})^{n}$ in the middle, and only when $n$ is even.
For odd $n$ the graph has only one turning peak and falls away on both sides, so a horizontal line meets it in at most two points.

n even, 0 < k < (\frac{a}{2})^{2 n}

Mentor tip

For even $n$ the graph is W-shaped. $k$ at the peak gives 3 solutions, $k = 0$ gives 2, and $k$ above the peak gives 2. Only a $k$ strictly between 0 and the peak gives 4.

Question 2

24 marks

Roots of $z^{n} = 1$ and the product of chord lengths

Number and algebra
Geometry and trigonometry

(a)(i)

2 marks

Expand the product.

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

(a)(ii)

2 marks

$ω$ is a root of $z^{3} = 1$ , so $ω^{3} - 1 = 0$ . Then $(ω - 1) (ω^{2} + ω + 1) = 0$ , and $ω \neq = 1$ .

ω^{2} + ω + 1 = 0

(b)

3 marks

The lengths are $P_{0} P_{1} = ∣1 - ω ∣$ and $P_{0} P_{2} = ∣1 - ω^{2} ∣$ . Their product is the modulus of the product of these two numbers.
By part (a), $ω + ω^{2} = - 1$ .

(1 - ω) (1 - ω^{2}) = 1 - ω - ω^{2} + ω^{3} = 2 - (ω + ω^{2}) P_{0} P_{1} \times P_{0} P_{2} = ∣2 + 1∣ = 3

(c)

2 marks

$z^{4} - 1 = (z - 1) (z^{3} + z^{2} + z + 1)$ . Since $ω$ is a root of $z^{4} - 1$ and $ω \neq = 1$ , the second factor is zero at $ω$ .

ω^{3} + ω^{2} + ω + 1 = 0

(d)

4 marks

For $n = 4$ the points are $1$ , $i$ , $- 1$ , $- i$ on the unit circle. Find each distance from $P_{0} = 1$ .

P_{0} P_{1} = ∣1 - i ∣ = 2, P_{0} P_{2} = ∣1 - (- 1) ∣ = 2, P_{0} P_{3} = ∣1 + i ∣ = 2 2 \times 2 \times 2 = 4

(e)

1 mark

The pattern from $n = 3, 4, 5$ is 3, 4, 5.

n

(f)(i)

2 marks

Each distance is the modulus of $1 - ω^{k}$ .

P_{0} P_{2} = ∣1 - ω^{2} ∣, P_{0} P_{3} = ∣1 - ω^{3} ∣

(f)(ii)

1 mark

Following the same pattern, the last one has index $n - 1$ .

P_{0} P_{n - 1} = ∣1 - ω^{n - 1} ∣

(g)(i)

3 marks

Since $z^{n} - 1 = (z - 1) (z^{n - 1} + \dots + z + 1)$ , the second factor is zero exactly at the other roots $ω, ω^{2}, \dots, ω^{n - 1}$ . It has degree $n - 1$ and leading coefficient 1, so it factorises with those $n - 1$ linear factors.

z^{n - 1} + z^{n - 2} + \dots + z + 1 = (z - ω) (z - ω^{2}) \dots (z - ω^{n - 1})

(g)(ii)

4 marks

Put $z = 1$ in the identity from part (g)(i). The left side is a sum of $n$ ones.
Take the modulus of both sides, using that the modulus of a product is the product of the moduli.

n = (1 - ω) (1 - ω^{2}) \dots (1 - ω^{n - 1}) n = ∣1 - ω ∣ ∣1 - ω^{2} ∣ \dots ∣1 - ω^{n - 1} ∣ = P_{0} P_{1} \times P_{0} P_{2} \times \dots \times P_{0} P_{n - 1}

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 algebra20 marks · 36%
Q2
Calculus17 marks · 31%
Q1
Functions14 marks · 25%
Q1
Geometry and trigonometry4 marks · 7%
Q2

Mentor insights

Traps to watch for

01
Which sign decides odd versus even · Q1
The gradient's sign just left of 0 is that of $(- 1)^{n - 1}$ . An even $n$ gives a negative gradient before 0, so a minimum. An odd $n$ gives a positive one, so an inflexion.
02
Strict inequalities in (h) · Q1
$k = 0$ and $k$ at the maximum both give fewer than four solutions. The condition on $k$ must be strict at both ends.
03
Odd and even $n$ behave differently · Q1
At $x = 0$ the factor $x^{n - 1}$ has a sign that depends on whether $n$ is odd or even. That one fact decides between a local minimum and an inflexion.
04
Say why $ω$ is not 1 · Q2
When you cancel $ω - 1$ from $(ω - 1) (ω^{2} + ω + 1) = 0$ , state that $ω \neq = 1$ . That is the whole justification.
05
Modulus of a product · Q2
In (g)(ii) the product of lengths appears only because you take the modulus of both sides, and $n$ is already real and positive.

Solutions and tips are GeorgeTheTutor's own; the questions are not reproduced here. Not affiliated with or endorsed by the International Baccalaureate Organization.

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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 TZ2 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

31 marks

The family $f_{n} (x) = x^{n} (a - x)^{n}$ : shape, stationary points, and four solutions

Functions
Calculus

(a)

3 marks

$f_{1} (x) = x (2 - x)$ is a downward parabola crossing the axis at $x = 0$ and $x = 2$ . Its vertex is halfway, at $x = 1$ .

f_{1} (1) = 1, intercepts (0, 0), (2, 0), local maximum (1, 1)

(b)

6 marks

Exploring on the calculator, the graph is symmetric about $x = 1$ with its single peak at $x = 1$ , where $f_{n} (1) = 1$ .
For odd $n$ (3 and 5), the graph flattens and keeps going the same way through $x = 0$ and $x = 2$ , so those are points of inflexion with zero gradient. For even $n$ (2 and 4), the graph touches the axis at 0 and 2 and bounces back, so those are minima.

n = 3, 5 n = 2, 4 local max 11 local min 02 inflexion, zero gradient 20

(c)

5 marks

Use the product rule with the chain rule on $(a - x)^{n}$ , which has derivative $- n (a - x)^{n - 1}$ .
Factor out the common $n x^{n - 1} (a - x)^{n - 1}$ .

f_{n}^{'} (x) = n x^{n - 1} (a - x)^{n} - n x^{n} (a - x)^{n - 1} f_{n}^{'} (x) = n x^{n - 1} (a - x)^{n - 1} [(a - x) - x] = n x^{n - 1} (a - 2 x) (a - x)^{n - 1}

(d)

2 marks

Each factor can be zero.

x = 0, x = \frac{a}{2}, x = a

(e)

3 marks

At $x = \frac{a}{2}$ both factors $x$ and $a - x$ equal $\frac{a}{2}$ , which is positive.
A positive $y$ -value means the point is above the horizontal axis.

f_{n} (\frac{a}{2}) = (\frac{a}{2})^{n} (\frac{a}{2})^{n} = (\frac{a ^{2}}{4})^{n} > 0

(f)

2 marks

At $x = \frac{a}{4}$ each factor in $f_{n}^{'}$ is positive: $(\frac{a}{4})^{n - 1}$ , $a - \frac{a}{2} = \frac{a}{2}$ and $(\frac{3 a}{4})^{n - 1}$ , with $n > 0$ .

f_{n}^{'} (\frac{a}{4}) = n (\frac{a}{4})^{n - 1} \cdot \frac{a}{2} \cdot (\frac{3 a}{4})^{n - 1} > 0

(g)(i)

3 marks

The gradient is positive just to the right of $0$ (by part (f), since there are no other stationary points between 0 and $\frac{a}{2}$ ). For $x = - 1$ , $f_{n}^{'} (- 1) = n (- 1)^{n - 1} (a + 2) (a + 1)^{n - 1}$ and its sign is that of $(- 1)^{n - 1}$ .
For even $n$ , $n - 1$ is odd, so $f_{n}^{'} (- 1) < 0$ . The gradient changes from negative to positive at $x = 0$ , so $(0, 0)$ is a local minimum.

(g)(ii)

2 marks

For odd $n$ , $n - 1$ is even, so $f_{n}^{'} (- 1) > 0$ . The gradient is positive on both sides of $x = 0$ and is zero there (since $n > 1$ ), so $(0, 0)$ is a point of inflexion with zero gradient.

(h)

5 marks

Four solutions needs the line $y = k$ to cut the graph four times. The graph has two minima at height 0 and a maximum of $(\frac{a ^{2}}{4})^{n}$ in the middle, and only when $n$ is even.
For odd $n$ the graph has only one turning peak and falls away on both sides, so a horizontal line meets it in at most two points.

n even, 0 < k < (\frac{a}{2})^{2 n}

Mentor tip

For even $n$ the graph is W-shaped. $k$ at the peak gives 3 solutions, $k = 0$ gives 2, and $k$ above the peak gives 2. Only a $k$ strictly between 0 and the peak gives 4.

Question 2

24 marks

Roots of $z^{n} = 1$ and the product of chord lengths

Number and algebra
Geometry and trigonometry

(a)(i)

2 marks

Expand the product.

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

(a)(ii)

2 marks

$ω$ is a root of $z^{3} = 1$ , so $ω^{3} - 1 = 0$ . Then $(ω - 1) (ω^{2} + ω + 1) = 0$ , and $ω \neq = 1$ .

ω^{2} + ω + 1 = 0

(b)

3 marks

The lengths are $P_{0} P_{1} = ∣1 - ω ∣$ and $P_{0} P_{2} = ∣1 - ω^{2} ∣$ . Their product is the modulus of the product of these two numbers.
By part (a), $ω + ω^{2} = - 1$ .

(1 - ω) (1 - ω^{2}) = 1 - ω - ω^{2} + ω^{3} = 2 - (ω + ω^{2}) P_{0} P_{1} \times P_{0} P_{2} = ∣2 + 1∣ = 3

(c)

2 marks

$z^{4} - 1 = (z - 1) (z^{3} + z^{2} + z + 1)$ . Since $ω$ is a root of $z^{4} - 1$ and $ω \neq = 1$ , the second factor is zero at $ω$ .

ω^{3} + ω^{2} + ω + 1 = 0

(d)

4 marks

For $n = 4$ the points are $1$ , $i$ , $- 1$ , $- i$ on the unit circle. Find each distance from $P_{0} = 1$ .

P_{0} P_{1} = ∣1 - i ∣ = 2, P_{0} P_{2} = ∣1 - (- 1) ∣ = 2, P_{0} P_{3} = ∣1 + i ∣ = 2 2 \times 2 \times 2 = 4

(e)

1 mark

The pattern from $n = 3, 4, 5$ is 3, 4, 5.

n

(f)(i)

2 marks

Each distance is the modulus of $1 - ω^{k}$ .

P_{0} P_{2} = ∣1 - ω^{2} ∣, P_{0} P_{3} = ∣1 - ω^{3} ∣

(f)(ii)

1 mark

Following the same pattern, the last one has index $n - 1$ .

P_{0} P_{n - 1} = ∣1 - ω^{n - 1} ∣

(g)(i)

3 marks

z^{n - 1} + z^{n - 2} + \dots + z + 1 = (z - ω) (z - ω^{2}) \dots (z - ω^{n - 1})

(g)(ii)

4 marks

Put $z = 1$ in the identity from part (g)(i). The left side is a sum of $n$ ones.
Take the modulus of both sides, using that the modulus of a product is the product of the moduli.

n = (1 - ω) (1 - ω^{2}) \dots (1 - ω^{n - 1}) n = ∣1 - ω ∣ ∣1 - ω^{2} ∣ \dots ∣1 - ω^{n - 1} ∣ = P_{0} P_{1} \times P_{0} P_{2} \times \dots \times P_{0} P_{n - 1}

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 algebra20 marks · 36%
Q2
Calculus17 marks · 31%
Q1
Functions14 marks · 25%
Q1
Geometry and trigonometry4 marks · 7%
Q2

Mentor insights

Traps to watch for

01
Which sign decides odd versus even · Q1
The gradient's sign just left of 0 is that of $(- 1)^{n - 1}$ . An even $n$ gives a negative gradient before 0, so a minimum. An odd $n$ gives a positive one, so an inflexion.
02
Strict inequalities in (h) · Q1
$k = 0$ and $k$ at the maximum both give fewer than four solutions. The condition on $k$ must be strict at both ends.
03
Odd and even $n$ behave differently · Q1
At $x = 0$ the factor $x^{n - 1}$ has a sign that depends on whether $n$ is odd or even. That one fact decides between a local minimum and an inflexion.
04
Say why $ω$ is not 1 · Q2
When you cancel $ω - 1$ from $(ω - 1) (ω^{2} + ω + 1) = 0$ , state that $ω \neq = 1$ . That is the whole justification.
05
Modulus of a product · Q2
In (g)(ii) the product of lengths appears only because you take the modulus of both sides, and $n$ is already real and positive.

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