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Worked solutions

IB Maths AA HL · Paper 3

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2023
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 2023 TZ1 paper, which your school or the IB can provide.

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Questions

Questions 1–2 · 55 marks

Question 1

25 marks

The family x^n e^(-x): areas, a limit and an induction proof for n factorial

Calculus

(a)

4 marks

Differentiate $f_{1} (x) = x e^{- x}$ .
The maximum value is $f_{1} (1) = e^{- 1} \approx 0.368$ .
Sketch: a curve starting at the origin, rising to the maximum at $(1, 0.368)$ and then decreasing towards the $x$ -axis without touching it.

f_{1}^{'} (x) = e^{- x} (1 - x) = 0 \Rightarrow x = 1 local maximum (1, e^{- 1})

(b)

6 marks

Integrate by parts with $u = x$ and $d v = e^{- x} d x$ .

\int_{0}^{b} x e^{- x} d x = [- x e^{- x}]_{0}^{b} + \int_{0}^{b} e^{- x} d x = - b e^{- b} + [- e^{- x}]_{0}^{b} = - b e^{- b} - e^{- b} + 1 Area = 1 - \frac{b + 1}{e ^{b}} = \frac{e ^{b} - b - 1}{e ^{b}}

(c)(i)

2 marks

As $b \to \infty$ the top and bottom both tend to infinity. Apply l'Hopital's rule twice.

b \to \infty lim \frac{e ^{b} - b - 1}{e ^{b}} = b \to \infty lim \frac{e ^{b} - 1}{e ^{b}} = b \to \infty lim \frac{e ^{b}}{e ^{b}} = 1

(c)(ii)

1 mark

The total area is this limit.

A_{1} = 1

(d)(i)

2 marks

Use the calculator with a large upper limit, such as 30, for $\int_{0}^{b} x^{4} e^{- x} d x$ .

A_{4} = 24

(d)(ii)

1 mark

Use the same method with $x^{5}$ in place of $x^{4}$ .

A_{5} = 120

(e)

1 mark

The values $1, 2, 6, 24, 120$ are the factorials.

A_{n} = n!

(f)

8 marks

Claim: $A_{n} = n!$ for all $n \in Z^{+}$ .
Base case $n = 1$ : $A_{1} = 1 = 1!$ from part (c).
Assume $A_{k} = k!$ , that is $\int_{0}^{\infty} x^{k} e^{- x} d x = k!$ .
Then for $n = k + 1$ , integrate by parts with $u = x^{k + 1}$ and $d v = e^{- x} d x$ .
The boundary term tends to 0 because $lim_{x \to \infty} x^{k + 1} e^{- x} = 0$ , and it is also 0 at $x = 0$ .
The statement is true for $n = 1$ and, if true for $n = k$ , it is true for $n = k + 1$ . By mathematical induction $A_{n} = n!$ for all $n \in Z^{+}$ .

A_{k + 1} = b \to \infty lim ([- x^{k + 1} e^{- x}]_{0}^{b} + (k + 1) \int_{0}^{b} x^{k} e^{- x} d x) A_{k + 1} = (k + 1) A_{k} = (k + 1) k! = (k + 1)!

Question 2

30 marks

The maximum product of positive numbers with a fixed sum

Calculus
Number and algebra

(a)

2 marks

Since $x_{2} = 12 - x_{1}$ :

f (x_{1}) = x_{1} (12 - x_{1})

(b)(i)

1 mark

Differentiate and set to zero.

f^{'} (x_{1}) = 12 - 2 x_{1} = 0 \Rightarrow x_{1} = 6

(b)(ii)

1 mark

Substitute $x_{1} = 6$ .

f (6) = 6 \times 6 = 36

(c)

1 mark

Put $S = 12$ into $(\frac{S}{2})^{2}$ .

(\frac{12}{2})^{2} = 36

(d)(i)

2 marks

Suppose every number equals $x$ . Then the geometric mean and the arithmetic mean are
They are equal.

(x^{n})^{1/ n} = x, \frac{n x}{n} = x

(d)(ii)

4 marks

The numbers have sum $S$ , so the arithmetic mean is $\frac{S}{n}$ . The geometric mean is at most this.
Equality holds when all $x_{i} = \frac{S}{n}$ (by part (i)), and these numbers do have sum $S$ . So the bound is reached.

(x_{1} x_{2} \dots x_{n})^{1/ n} \leq \frac{S}{n} \Rightarrow x_{1} x_{2} \dots x_{n} \leq (\frac{S}{n})^{n} M_{n} (S) = (\frac{S}{n})^{n}

(e)(i)

1 mark

Use $M_{n} (S) = (\frac{S}{n})^{n}$ with $S = 12$ and $n = 3$ .

M_{3} (12) = 4^{3} = 64

(e)(ii)

1 mark

Now $n = 4$ .

M_{4} (12) = 3^{4} = 81

(e)(iii)

1 mark

Now $n = 5$ .

M_{5} (12) = 2. 4^{5} = 79.6

(f)

2 marks

Compare these values: 36 for $n = 2$ , 64, 81 and 79.6. The values fall after $n = 4$ .

P (12) = 81, n = 4

(g)

3 marks

Evaluate $(\frac{20}{n})^{n}$ for several $n$ .

n = 6 : 1371.7, n = 7 : 1554.3, n = 8 : 1525.9 P (20) \approx 1554, n = 7

(h)

6 marks

Maximising $g (x) = (\frac{S}{x})^{x}$ is the same as maximising $ln g (x) = x ln \frac{S}{x}$ . Differentiate.
Set it to zero.
This is a maximum because the derivative $ln \frac{S}{x} - 1$ is positive for $x < \frac{S}{e}$ and negative for $x > \frac{S}{e}$ .

\frac{d}{d x} (x ln \frac{S}{x}) = ln \frac{S}{x} + x \cdot (- \frac{1}{x}) = ln \frac{S}{x} - 1 ln \frac{S}{x} = 1 \Rightarrow \frac{S}{x} = e \Rightarrow x = \frac{S}{e}

(i)

2 marks

For a whole number $x$ , part (d)(ii) gives $M_{x} (S) = (\frac{S}{x})^{x}$ , which is $g (x)$ .

(j)

3 marks

For $S = 100$ the best real $x$ is $\frac{100}{e} = 36.8$ , so use the nearest whole number, 37 numbers each equal to $\frac{100}{37}$ .

M_{37} (100) = (\frac{100}{37})^{37} \approx 9.47 \times 1 0^{15}

Mentor tip

The real-number maximum is $g (36.8) = e^{36.8} \approx 9.48 \times 1 0^{15}$ , only a little bigger.

Our analysis

Where the marks are

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

Calculus38 marks · 69%
Q1, Q2
Number and algebra17 marks · 31%
Q2

Mentor insights

Traps to watch for

01
Integration by parts for the area · Q1
$\int x e^{- x} d x = - x e^{- x} - e^{- x}$ .
02
A limit that needs l'Hopital twice · Q1
The first derivative still gives an infinity over infinity form.
03
Express the product in one variable · Q2
With $x_{1} + x_{2} = 12$ , write $x_{2} = 12 - x_{1}$ .
04
Use AM-GM · Q2
The geometric mean is at most the arithmetic mean, with equality when all the numbers are equal. That gives the maximum product.
05
Integer versus real n · Q2
$M_{n} (S) = (S / n)^{n}$ is defined for whole $n$ , but $g (x) = (S / x)^{x}$ lets you use calculus to find the best real $x$ .

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 2023
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 2023 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

25 marks

The family x^n e^(-x): areas, a limit and an induction proof for n factorial

Calculus

(a)

4 marks

Differentiate $f_{1} (x) = x e^{- x}$ .
The maximum value is $f_{1} (1) = e^{- 1} \approx 0.368$ .
Sketch: a curve starting at the origin, rising to the maximum at $(1, 0.368)$ and then decreasing towards the $x$ -axis without touching it.

f_{1}^{'} (x) = e^{- x} (1 - x) = 0 \Rightarrow x = 1 local maximum (1, e^{- 1})

(b)

6 marks

Integrate by parts with $u = x$ and $d v = e^{- x} d x$ .

\int_{0}^{b} x e^{- x} d x = [- x e^{- x}]_{0}^{b} + \int_{0}^{b} e^{- x} d x = - b e^{- b} + [- e^{- x}]_{0}^{b} = - b e^{- b} - e^{- b} + 1 Area = 1 - \frac{b + 1}{e ^{b}} = \frac{e ^{b} - b - 1}{e ^{b}}

(c)(i)

2 marks

As $b \to \infty$ the top and bottom both tend to infinity. Apply l'Hopital's rule twice.

b \to \infty lim \frac{e ^{b} - b - 1}{e ^{b}} = b \to \infty lim \frac{e ^{b} - 1}{e ^{b}} = b \to \infty lim \frac{e ^{b}}{e ^{b}} = 1

(c)(ii)

1 mark

The total area is this limit.

A_{1} = 1

(d)(i)

2 marks

Use the calculator with a large upper limit, such as 30, for $\int_{0}^{b} x^{4} e^{- x} d x$ .

A_{4} = 24

(d)(ii)

1 mark

Use the same method with $x^{5}$ in place of $x^{4}$ .

A_{5} = 120

(e)

1 mark

The values $1, 2, 6, 24, 120$ are the factorials.

A_{n} = n!

(f)

8 marks

Claim: $A_{n} = n!$ for all $n \in Z^{+}$ .
Base case $n = 1$ : $A_{1} = 1 = 1!$ from part (c).
Assume $A_{k} = k!$ , that is $\int_{0}^{\infty} x^{k} e^{- x} d x = k!$ .
Then for $n = k + 1$ , integrate by parts with $u = x^{k + 1}$ and $d v = e^{- x} d x$ .
The boundary term tends to 0 because $lim_{x \to \infty} x^{k + 1} e^{- x} = 0$ , and it is also 0 at $x = 0$ .
The statement is true for $n = 1$ and, if true for $n = k$ , it is true for $n = k + 1$ . By mathematical induction $A_{n} = n!$ for all $n \in Z^{+}$ .

A_{k + 1} = b \to \infty lim ([- x^{k + 1} e^{- x}]_{0}^{b} + (k + 1) \int_{0}^{b} x^{k} e^{- x} d x) A_{k + 1} = (k + 1) A_{k} = (k + 1) k! = (k + 1)!

Question 2

30 marks

The maximum product of positive numbers with a fixed sum

Calculus
Number and algebra

(a)

2 marks

Since $x_{2} = 12 - x_{1}$ :

f (x_{1}) = x_{1} (12 - x_{1})

(b)(i)

1 mark

Differentiate and set to zero.

f^{'} (x_{1}) = 12 - 2 x_{1} = 0 \Rightarrow x_{1} = 6

(b)(ii)

1 mark

Substitute $x_{1} = 6$ .

f (6) = 6 \times 6 = 36

(c)

1 mark

Put $S = 12$ into $(\frac{S}{2})^{2}$ .

(\frac{12}{2})^{2} = 36

(d)(i)

2 marks

Suppose every number equals $x$ . Then the geometric mean and the arithmetic mean are
They are equal.

(x^{n})^{1/ n} = x, \frac{n x}{n} = x

(d)(ii)

4 marks

The numbers have sum $S$ , so the arithmetic mean is $\frac{S}{n}$ . The geometric mean is at most this.
Equality holds when all $x_{i} = \frac{S}{n}$ (by part (i)), and these numbers do have sum $S$ . So the bound is reached.

(x_{1} x_{2} \dots x_{n})^{1/ n} \leq \frac{S}{n} \Rightarrow x_{1} x_{2} \dots x_{n} \leq (\frac{S}{n})^{n} M_{n} (S) = (\frac{S}{n})^{n}

(e)(i)

1 mark

Use $M_{n} (S) = (\frac{S}{n})^{n}$ with $S = 12$ and $n = 3$ .

M_{3} (12) = 4^{3} = 64

(e)(ii)

1 mark

Now $n = 4$ .

M_{4} (12) = 3^{4} = 81

(e)(iii)

1 mark

Now $n = 5$ .

M_{5} (12) = 2. 4^{5} = 79.6

(f)

2 marks

Compare these values: 36 for $n = 2$ , 64, 81 and 79.6. The values fall after $n = 4$ .

P (12) = 81, n = 4

(g)

3 marks

Evaluate $(\frac{20}{n})^{n}$ for several $n$ .

n = 6 : 1371.7, n = 7 : 1554.3, n = 8 : 1525.9 P (20) \approx 1554, n = 7

(h)

6 marks

Maximising $g (x) = (\frac{S}{x})^{x}$ is the same as maximising $ln g (x) = x ln \frac{S}{x}$ . Differentiate.
Set it to zero.
This is a maximum because the derivative $ln \frac{S}{x} - 1$ is positive for $x < \frac{S}{e}$ and negative for $x > \frac{S}{e}$ .

\frac{d}{d x} (x ln \frac{S}{x}) = ln \frac{S}{x} + x \cdot (- \frac{1}{x}) = ln \frac{S}{x} - 1 ln \frac{S}{x} = 1 \Rightarrow \frac{S}{x} = e \Rightarrow x = \frac{S}{e}

(i)

2 marks

For a whole number $x$ , part (d)(ii) gives $M_{x} (S) = (\frac{S}{x})^{x}$ , which is $g (x)$ .

(j)

3 marks

For $S = 100$ the best real $x$ is $\frac{100}{e} = 36.8$ , so use the nearest whole number, 37 numbers each equal to $\frac{100}{37}$ .

M_{37} (100) = (\frac{100}{37})^{37} \approx 9.47 \times 1 0^{15}

Mentor tip

The real-number maximum is $g (36.8) = e^{36.8} \approx 9.48 \times 1 0^{15}$ , only a little bigger.

Our analysis

Where the marks are

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

Calculus38 marks · 69%
Q1, Q2
Number and algebra17 marks · 31%
Q2

Mentor insights

Traps to watch for

01
Integration by parts for the area · Q1
$\int x e^{- x} d x = - x e^{- x} - e^{- x}$ .
02
A limit that needs l'Hopital twice · Q1
The first derivative still gives an infinity over infinity form.
03
Express the product in one variable · Q2
With $x_{1} + x_{2} = 12$ , write $x_{2} = 12 - x_{1}$ .
04
Use AM-GM · Q2
The geometric mean is at most the arithmetic mean, with equality when all the numbers are equal. That gives the maximum product.
05
Integer versus real n · Q2
$M_{n} (S) = (S / n)^{n}$ is defined for whole $n$ , but $g (x) = (S / x)^{x}$ lets you use calculus to find the best real $x$ .

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