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IB Maths AA HL · Paper 3

Mathematics: Analysis & Approaches, Higher Level

Session: November 2024
Marks: 55
Duration: 1 hour 15 min
Calculator: Allowed
Questions: 2

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Questions

Questions 1–2 · 55 marks

Question 1

27 marks

Trout in a lake: a recurrence, a closed form, and proof by induction

Number and algebra

(a)

2 marks

Use $T_{2} = 1.1 T_{1} - 500$ with $T_{1} = 6000$ .

T_{2} = 1.1 (6000) - 500 = 6600 - 500 = 6100

(b)(i)

1 mark

Use $T_{3} = 1.1 T_{2} - 500$ .

T_{3} = 1.1 (6100) - 500 = 6710 - 500 = 6210

(b)(ii)

2 marks

Use $T_{4} = 1.1 T_{3} - 500$ .

T_{4} = 1.1 (6210) - 500 = 6831 - 500 = 6331

(c)(i)

2 marks

The formula is $T_{n} = 6000 (1.1)^{n - 1} - 500 (1 + 1.1 + \dots + 1. 1^{n - 2})$ . The bracket is a geometric sum with ratio 1.1 and $n - 1$ terms.
Substitute.

1 + 1.1 + \dots + 1. 1^{n - 2} = \frac{1. 1 ^{n - 1} - 1}{0.1} = 10 (1. 1^{n - 1} - 1) T_{n} = 6000 (1.1)^{n - 1} - 5000 (1. 1^{n - 1} - 1) = 1000 (1.1)^{n - 1} + 5000

(c)(ii)

2 marks

Put $n = 6$ .

T_{6} = 1000 (1.1)^{5} + 5000 = 6610.51 \approx 6611

(d)(i)

3 marks

Use the given formula for $D_{n}$ .
Factor out $1. 1^{n - 1}$ .

D_{n + 1} - D_{n} = - 1500 (1.1)^{n} + 7500 - [- 1500 (1.1)^{n - 1} + 7500] - 1500 (1.1)^{n - 1} (1.1 - 1) = - 150 (1.1)^{n - 1}

(d)(ii)

1 mark

Since $(1.1)^{n - 1} > 0$ , we have $D_{n + 1} - D_{n} < 0$ for every $n$ . So $D_{n + 1} < D_{n}$ : the predicted number at the start of any year is greater than at the start of the next year.

(e)

4 marks

Find where $D_{n}$ stops being positive.
Check: $D_{17} \approx 607.5$ and $D_{18} \approx - 81.7$ .
At the start of year 17 there are about 607 trout. After growing by 10% there are about 668, which is fewer than the 750 to be removed. So there are no trout remaining after the removal at the end of year 17.

D_{n} = - 1500 (1.1)^{n - 1} + 7500 \leq 0 \Rightarrow 1. 1^{n - 1} \geq 5 \Rightarrow n - 1 \geq \frac{ln 5}{ln 1.1} \approx 16.89 year 17 (the formula first goes negative at the start of year 18)

(f)

3 marks

Write $C_{n}$ as a multiple of $1. 1^{n - 1}$ plus a constant.
The population is constant when the coefficient of $1. 1^{n - 1}$ is zero.

C_{n} = (6000 - 10 d) (1.1)^{n - 1} + 10 d 6000 - 10 d = 0 \Rightarrow d = 600

(g)

7 marks

Base case $n = 1$ : the right side is $r^{0} u_{1} - \frac{d ( r ^{0} - 1 )}{r - 1} = u_{1} - 0 = u_{1}$ , which equals the left side, so the formula is true for $n = 1$ .
Assume true for $n = k$ .
Use $u_{k + 1} = r u_{k} - d$ .
Put the last two terms over a common denominator.
This is the formula with $n = k + 1$ . It is true for $n = 1$ , and if it is true for $n = k$ then it is true for $n = k + 1$ , so by induction it is true for all $n \in Z^{+}$ .

u_{k} = r^{k - 1} u_{1} - \frac{d ( r ^{k - 1} - 1 )}{r - 1} u_{k + 1} = r^{k} u_{1} - \frac{d r ( r ^{k - 1} - 1 )}{r - 1} - d u_{k + 1} = r^{k} u_{1} - \frac{d r ^{k} - d r + d ( r - 1 )}{r - 1} = r^{k} u_{1} - \frac{d r ^{k} - d}{r - 1} u_{k + 1} = r^{k} u_{1} - \frac{d ( r ^{k} - 1 )}{r - 1}

Question 2

28 marks

Palindromic polynomials: reciprocal roots and factorisation

Number and algebra

(a)(i)

3 marks

Use the quadratic formula on $x^{2} + 4 x + 1 = 0$ .

x = \frac{- 4 \pm 16 - 4}{2} = - 2 \pm 3

(a)(ii)

2 marks

Multiply the two roots.
Their product is 1, so each root is the reciprocal of the other.

(- 2 + 3) (- 2 - 3) = 4 - 3 = 1

(b)

2 marks

The roots of $x^{2} + 1 = 0$ are $x = i$ and $x = - i$ . The reciprocal of $i$ is
So the two roots are reciprocals of each other.

\frac{1}{i} = \frac{- i}{- i \cdot i} = - i

(c)

2 marks

Work out $x^{2} p (\frac{1}{x})$ .

x^{2} p (\frac{1}{x}) = x^{2} (\frac{a}{x ^{2}} + \frac{b}{x} + a) = a + b x + a x^{2} = p (x)

(d)

2 marks

Suppose $p (α) = 0$ with $α \neq = 0$ . Use $p (x) = x^{n} p (\frac{1}{x})$ at $x = α$ .
Since $α^{n} \neq = 0$ , we get $p (\frac{1}{α}) = 0$ , so $\frac{1}{α}$ is also a root.

0 = p (α) = α^{n} p (\frac{1}{α})

(e)

4 marks

The polynomial $f = pq$ has degree $n + m$ . Use the given property for $p$ and $q$ .
So $f (x) = x^{n + m} f (\frac{1}{x})$ , and $f$ is palindromic.

x^{n + m} f (\frac{1}{x}) = x^{n} p (\frac{1}{x}) \cdot x^{m} q (\frac{1}{x}) = p (x) q (x) = f (x)

(f)

6 marks

Expand the product.
Compare with $x^{4} + 2 x^{3} - x^{2} + 2 x + 1$ .
So $u$ and $v$ are the roots of $t^{2} - 2 t - 3 = 0$ , which factorises as $(t - 3) (t + 1) = 0$ . Since $u < v$ :

(x^{2} + ux + 1) (x^{2} + v x + 1) = x^{4} + (u + v) x^{3} + (2 + uv) x^{2} + (u + v) x + 1 u + v = 2, 2 + uv = - 1 \Rightarrow uv = - 3 u = - 1, v = 3

(g)

3 marks

The polynomial is $(x^{2} - x + 1) (x^{2} + 3 x + 1)$ .
From $x^{2} - x + 1 = 0$ the complex roots are
From $x^{2} + 3 x + 1 = 0$ the real roots are

x = \frac{1 \pm i 3}{2} x = \frac{- 3 \pm 5}{2}

(h)

4 marks

Let $p (x) = x^{n} + a_{1} x^{n - 1} + \dots + a_{1} x + 1$ , which is palindromic, so $p (x) = x^{n} p (\frac{1}{x})$ .
Put $x = - 1$ .
Since $n$ is odd, $(- 1)^{n} = - 1$ .
So $- 1$ is always a root.

p (- 1) = (- 1)^{n} p (- 1) p (- 1) = - p (- 1) \Rightarrow 2 p (- 1) = 0 \Rightarrow p (- 1) = 0

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 algebra55 marks · 100%
Q1, Q2

Mentor insights

Traps to watch for

01
Recurrence · Q1
Each year the population grows by 10% and then a fixed number are removed.
02
Constant population · Q1
The population stays the same when the coefficient of $1. 1^{n - 1}$ is zero.
03
Induction · Q1
Prove the first case, assume the case $n = k$ , then show $n = k + 1$ .
04
Palindromic · Q2
The coefficients read the same forwards and backwards, so $p (x) = x^{n} p (\frac{1}{x})$ .
05
Reciprocal pair · Q2
Two numbers whose product is 1.

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

Session: November 2024
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 November 2024 Global 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

Trout in a lake: a recurrence, a closed form, and proof by induction

Number and algebra

(a)

2 marks

Use $T_{2} = 1.1 T_{1} - 500$ with $T_{1} = 6000$ .

T_{2} = 1.1 (6000) - 500 = 6600 - 500 = 6100

(b)(i)

1 mark

Use $T_{3} = 1.1 T_{2} - 500$ .

T_{3} = 1.1 (6100) - 500 = 6710 - 500 = 6210

(b)(ii)

2 marks

Use $T_{4} = 1.1 T_{3} - 500$ .

T_{4} = 1.1 (6210) - 500 = 6831 - 500 = 6331

(c)(i)

2 marks

The formula is $T_{n} = 6000 (1.1)^{n - 1} - 500 (1 + 1.1 + \dots + 1. 1^{n - 2})$ . The bracket is a geometric sum with ratio 1.1 and $n - 1$ terms.
Substitute.

1 + 1.1 + \dots + 1. 1^{n - 2} = \frac{1. 1 ^{n - 1} - 1}{0.1} = 10 (1. 1^{n - 1} - 1) T_{n} = 6000 (1.1)^{n - 1} - 5000 (1. 1^{n - 1} - 1) = 1000 (1.1)^{n - 1} + 5000

(c)(ii)

2 marks

Put $n = 6$ .

T_{6} = 1000 (1.1)^{5} + 5000 = 6610.51 \approx 6611

(d)(i)

3 marks

Use the given formula for $D_{n}$ .
Factor out $1. 1^{n - 1}$ .

D_{n + 1} - D_{n} = - 1500 (1.1)^{n} + 7500 - [- 1500 (1.1)^{n - 1} + 7500] - 1500 (1.1)^{n - 1} (1.1 - 1) = - 150 (1.1)^{n - 1}

(d)(ii)

1 mark

Since $(1.1)^{n - 1} > 0$ , we have $D_{n + 1} - D_{n} < 0$ for every $n$ . So $D_{n + 1} < D_{n}$ : the predicted number at the start of any year is greater than at the start of the next year.

(e)

4 marks

Find where $D_{n}$ stops being positive.
Check: $D_{17} \approx 607.5$ and $D_{18} \approx - 81.7$ .
At the start of year 17 there are about 607 trout. After growing by 10% there are about 668, which is fewer than the 750 to be removed. So there are no trout remaining after the removal at the end of year 17.

D_{n} = - 1500 (1.1)^{n - 1} + 7500 \leq 0 \Rightarrow 1. 1^{n - 1} \geq 5 \Rightarrow n - 1 \geq \frac{ln 5}{ln 1.1} \approx 16.89 year 17 (the formula first goes negative at the start of year 18)

(f)

3 marks

Write $C_{n}$ as a multiple of $1. 1^{n - 1}$ plus a constant.
The population is constant when the coefficient of $1. 1^{n - 1}$ is zero.

C_{n} = (6000 - 10 d) (1.1)^{n - 1} + 10 d 6000 - 10 d = 0 \Rightarrow d = 600

(g)

7 marks

Base case $n = 1$ : the right side is $r^{0} u_{1} - \frac{d ( r ^{0} - 1 )}{r - 1} = u_{1} - 0 = u_{1}$ , which equals the left side, so the formula is true for $n = 1$ .
Assume true for $n = k$ .
Use $u_{k + 1} = r u_{k} - d$ .
Put the last two terms over a common denominator.
This is the formula with $n = k + 1$ . It is true for $n = 1$ , and if it is true for $n = k$ then it is true for $n = k + 1$ , so by induction it is true for all $n \in Z^{+}$ .

u_{k} = r^{k - 1} u_{1} - \frac{d ( r ^{k - 1} - 1 )}{r - 1} u_{k + 1} = r^{k} u_{1} - \frac{d r ( r ^{k - 1} - 1 )}{r - 1} - d u_{k + 1} = r^{k} u_{1} - \frac{d r ^{k} - d r + d ( r - 1 )}{r - 1} = r^{k} u_{1} - \frac{d r ^{k} - d}{r - 1} u_{k + 1} = r^{k} u_{1} - \frac{d ( r ^{k} - 1 )}{r - 1}

Question 2

28 marks

Palindromic polynomials: reciprocal roots and factorisation

Number and algebra

(a)(i)

3 marks

Use the quadratic formula on $x^{2} + 4 x + 1 = 0$ .

x = \frac{- 4 \pm 16 - 4}{2} = - 2 \pm 3

(a)(ii)

2 marks

Multiply the two roots.
Their product is 1, so each root is the reciprocal of the other.

(- 2 + 3) (- 2 - 3) = 4 - 3 = 1

(b)

2 marks

The roots of $x^{2} + 1 = 0$ are $x = i$ and $x = - i$ . The reciprocal of $i$ is
So the two roots are reciprocals of each other.

\frac{1}{i} = \frac{- i}{- i \cdot i} = - i

(c)

2 marks

Work out $x^{2} p (\frac{1}{x})$ .

x^{2} p (\frac{1}{x}) = x^{2} (\frac{a}{x ^{2}} + \frac{b}{x} + a) = a + b x + a x^{2} = p (x)

(d)

2 marks

Suppose $p (α) = 0$ with $α \neq = 0$ . Use $p (x) = x^{n} p (\frac{1}{x})$ at $x = α$ .
Since $α^{n} \neq = 0$ , we get $p (\frac{1}{α}) = 0$ , so $\frac{1}{α}$ is also a root.

0 = p (α) = α^{n} p (\frac{1}{α})

(e)

4 marks

The polynomial $f = pq$ has degree $n + m$ . Use the given property for $p$ and $q$ .
So $f (x) = x^{n + m} f (\frac{1}{x})$ , and $f$ is palindromic.

x^{n + m} f (\frac{1}{x}) = x^{n} p (\frac{1}{x}) \cdot x^{m} q (\frac{1}{x}) = p (x) q (x) = f (x)

(f)

6 marks

Expand the product.
Compare with $x^{4} + 2 x^{3} - x^{2} + 2 x + 1$ .
So $u$ and $v$ are the roots of $t^{2} - 2 t - 3 = 0$ , which factorises as $(t - 3) (t + 1) = 0$ . Since $u < v$ :

(x^{2} + ux + 1) (x^{2} + v x + 1) = x^{4} + (u + v) x^{3} + (2 + uv) x^{2} + (u + v) x + 1 u + v = 2, 2 + uv = - 1 \Rightarrow uv = - 3 u = - 1, v = 3

(g)

3 marks

The polynomial is $(x^{2} - x + 1) (x^{2} + 3 x + 1)$ .
From $x^{2} - x + 1 = 0$ the complex roots are
From $x^{2} + 3 x + 1 = 0$ the real roots are

x = \frac{1 \pm i 3}{2} x = \frac{- 3 \pm 5}{2}

(h)

4 marks

Let $p (x) = x^{n} + a_{1} x^{n - 1} + \dots + a_{1} x + 1$ , which is palindromic, so $p (x) = x^{n} p (\frac{1}{x})$ .
Put $x = - 1$ .
Since $n$ is odd, $(- 1)^{n} = - 1$ .
So $- 1$ is always a root.

p (- 1) = (- 1)^{n} p (- 1) p (- 1) = - p (- 1) \Rightarrow 2 p (- 1) = 0 \Rightarrow p (- 1) = 0

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 algebra55 marks · 100%
Q1, Q2

Mentor insights

Traps to watch for

01
Recurrence · Q1
Each year the population grows by 10% and then a fixed number are removed.
02
Constant population · Q1
The population stays the same when the coefficient of $1. 1^{n - 1}$ is zero.
03
Induction · Q1
Prove the first case, assume the case $n = k$ , then show $n = k + 1$ .
04
Palindromic · Q2
The coefficients read the same forwards and backwards, so $p (x) = x^{n} p (\frac{1}{x})$ .
05
Reciprocal pair · Q2
Two numbers whose product is 1.

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