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IB Maths AA HL · Paper 3

Mathematics: Analysis & Approaches, Higher Level

Session: November 2022
Marks: 55
Duration: 1 hour 15 min
Calculator: Allowed
Questions: 2

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Questions

Questions 1–2 · 55 marks

Question 1

28 marks

Sums of powers: patterns, induction on derivatives and l'Hopital

Number and algebra
Calculus

(a)

1 mark

This is an arithmetic series with first term 1, last term $n$ and $n$ terms.

i = 1 \sum n i = \frac{n}{2} (1 + n)

(b)(i)

1 mark

The entry $p$ is the sum of the first three squares.

p = 1 + 4 + 9 = 14

(b)(ii)

3 marks

Put $n = 1, 2, 3$ into $a_{1} n + a_{2} n^{2} + a_{3} n^{3}$ and use the sums 1, 5, 14.

a_{1} + a_{2} + a_{3} = 1 2 a_{1} + 4 a_{2} + 8 a_{3} = 5 3 a_{1} + 9 a_{2} + 27 a_{3} = 14

(b)(iii)

2 marks

Solve the system.

a_{1} = \frac{1}{6}, a_{2} = \frac{1}{2}, a_{3} = \frac{1}{3}

(c)

1 mark

Differentiate $f (x) = 1 + x + x^{2} + \dots + x^{n}$ term by term and multiply by $x$ .

f^{'} (x) = 1 + 2 x + 3 x^{2} + \dots + n x^{n - 1} x f^{'} (x) = x + 2 x^{2} + 3 x^{3} + \dots + n x^{n}

(d)(i)

2 marks

From (c), $f_{1} (x) = \sum_{i = 1}^{n} i x^{i}$ , so $f_{1}^{'} (x) = \sum_{i = 1}^{n} i^{2} x^{i - 1}$ .

f_{2} (x) = x f_{1}^{'} (x) = i = 1 \sum n i^{2} x^{i}

(d)(ii)

6 marks

Claim: $f_{q} (x) = \sum_{i = 1}^{n} i^{q} x^{i}$ for all $q \in Z^{+}$ .
Base case $q = 1$ : $f_{1} (x) = x f^{'} (x) = \sum i x^{i}$ from (c), so it holds.
Assume it holds for $q = k$ : $f_{k} (x) = \sum_{i = 1}^{n} i^{k} x^{i}$ .
Then, by the definition of the family,
So it holds for $q = k + 1$ .
It is true for $q = 1$ and, if true for $q = k$ , it is true for $q = k + 1$ . By mathematical induction it is true for all $q \in Z^{+}$ .

f_{k + 1} (x) = x f_{k}^{'} (x) = x i = 1 \sum n i^{k + 1} x^{i - 1} = i = 1 \sum n i^{k + 1} x^{i}

(d)(iii)

1 mark

Put $x = 1$ in the result of (d)(ii).

f_{q} (1) = i = 1 \sum n i^{q}

(e)

2 marks

$f$ is a geometric series with first term 1, ratio $x$ and $n + 1$ terms.

f (x) = \frac{x ^{n + 1} - 1}{x - 1}

(f)

3 marks

Differentiate with the quotient rule.
Multiply by $x$ .

f^{'} (x) = \frac{( n + 1 ) x ^{n} ( x - 1 ) - ( x ^{n + 1} - 1 )}{( x - 1 ) ^{2}} = \frac{n x ^{n + 1} - ( n + 1 ) x ^{n} + 1}{( x - 1 ) ^{2}} f_{1} (x) = \frac{n x ^{n + 2} - ( n + 1 ) x ^{n + 1} + x}{( x - 1 ) ^{2}}

(g)(i)

1 mark

At $x = 1$ the numerator is $n - (n + 1) + 1 = 0$ and the denominator is $0$ , so the limit has the form $\frac{0}{0}$ , which is indeterminate.

(g)(ii)

5 marks

Apply l'Hopital's rule: differentiate the numerator and the denominator.
At $x = 1$ the numerator is $n (n + 2) - (n + 1)^{2} + 1 = 0$ and the denominator is 0 again. Apply the rule a second time.
At $x = 1$ :

x \to 1 lim f_{1} (x) = x \to 1 lim \frac{n ( n + 2 ) x ^{n + 1} - ( n + 1 ) ^{2} x ^{n} + 1}{2 ( x - 1 )} x \to 1 lim \frac{n ( n + 1 ) ( n + 2 ) x ^{n} - n ( n + 1 ) ^{2} x ^{n - 1}}{2} \frac{n ( n + 1 ) [ ( n + 2 ) - ( n + 1 ) ]}{2} = \frac{n ( n + 1 )}{2}

Question 2

27 marks

Surface areas of revolution: cone, sphere and ellipsoid

Calculus
Geometry and trigonometry
Functions

(a)

2 marks

With $m = 2$ and $h = 3$ : $A = 2 π \int_{0}^{3} 2 x 1 + 4 d x$ .

A = 45 π [\frac{x ^{2}}{2}]_{0}^{3} = 45 π \cdot \frac{9}{2} = 185 π

(b)(i)

1 mark

The radius of the cone is the height of the line at $x = h$ .

r = mh

(b)(ii)

2 marks

The slant height is the length of the line from $(0, 0)$ to $(h, mh)$ .

l = h^{2} + m^{2} h^{2} = h 1 + m^{2}

(b)(iii)

3 marks

Put $y = m x$ into the integral.

A = 2 π \int_{0}^{h} m x 1 + m^{2} d x = 2 π m 1 + m^{2} \cdot \frac{h ^{2}}{2} = π (mh) (h 1 + m^{2}) A = π r l

(c)

2 marks

Differentiate $y = (r^{2} - x^{2})^{1/2}$ with the chain rule.

\frac{d y}{d x} = \frac{- x}{r ^{2} - x ^{2}}

(d)

4 marks

Substitute $y$ and $\frac{d y}{d x}$ into the area formula.

A = 2 π \int_{- r}^{r} r^{2} - x^{2} 1 + \frac{x ^{2}}{r ^{2} - x ^{2}} d x = 2 π \int_{- r}^{r} r^{2} - x^{2} + x^{2} d x A = 2 π r \int_{- r}^{r} d x = 2 π r (2 r) = 4 π r^{2}

(e)(i)

2 marks

Replacing $x$ by $k x$ is a horizontal stretch (parallel to the $x$ -axis) with scale factor $\frac{1}{k}$ .

(e)(ii)

1 mark

The graph meets the axis where $f (k x) = 0$ , so $k x = \pm r$ .

x = \pm \frac{r}{k}

(e)(iii)

2 marks

Now $y = r^{2} - k^{2} x^{2}$ .

\frac{d y}{d x} = \frac{- k ^{2} x}{r ^{2} - k ^{2} x ^{2}}

(e)(iv)

4 marks

Substitute into the area formula.
Combine under one square root.

A = 2 π \int_{- r / k}^{r / k} r^{2} - k^{2} x^{2} 1 + \frac{k ^{4} x ^{2}}{r ^{2} - k ^{2} x ^{2}} d x A = 2 π \int_{- r / k}^{r / k} r^{2} + k^{2} (k^{2} - 1) x^{2} d x, p (x) = r^{2} + k^{2} (k^{2} - 1) x^{2}

(e)(v)

4 marks

The equator has radius $r = \frac{12756}{2} = 6378$ . The poles are on the $x$ -axis at $x = \pm \frac{r}{k} = \pm \frac{12714}{2} = \pm 6357$ , so $k = \frac{6378}{6357} = 1.00330$ .
Use the formula from (iv) with these values.

A = 2 π \int_{- 6357}^{6357} 637 8^{2} + k^{2} (k^{2} - 1) x^{2} d x \approx 5.101 \times 1 0^{8} km^{2}

Our analysis

Where the marks are

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

Calculus40 marks · 73%
Q1, Q2
Number and algebra9 marks · 16%
Q1
Functions3 marks · 5%
Q2
Geometry and trigonometry3 marks · 5%
Q2

Mentor insights

Traps to watch for

01
Link the sum to a derivative · Q1
Differentiating $1 + x + \dots + x^{n}$ and multiplying by $x$ turns each $x^{i}$ into $i x^{i}$ . Repeating raises the power each time.
02
Induction layout · Q1
State the assumption for $q = k$ , derive $q = k + 1$ with the operation $f_{k + 1} = x f_{k}^{'}$ , and write a conclusion.
03
0/0 can need l'Hopital twice · Q1
If the numerator and denominator are both 0 after the first derivative, differentiate again.
04
The formula needs the derivative · Q2
Surface area is $2 π \int y 1 + (d y / d x)^{2} d x$ . For the sphere the square root simplifies to a constant.
05
Radius and semi-axis · Q2
For the Earth, the equator gives $r = 6378$ and the pole-to-pole distance gives the semi-axis $\frac{r}{k} = 6357$ .

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 2022
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 2022 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

28 marks

Sums of powers: patterns, induction on derivatives and l'Hopital

Number and algebra
Calculus

(a)

1 mark

This is an arithmetic series with first term 1, last term $n$ and $n$ terms.

i = 1 \sum n i = \frac{n}{2} (1 + n)

(b)(i)

1 mark

The entry $p$ is the sum of the first three squares.

p = 1 + 4 + 9 = 14

(b)(ii)

3 marks

Put $n = 1, 2, 3$ into $a_{1} n + a_{2} n^{2} + a_{3} n^{3}$ and use the sums 1, 5, 14.

a_{1} + a_{2} + a_{3} = 1 2 a_{1} + 4 a_{2} + 8 a_{3} = 5 3 a_{1} + 9 a_{2} + 27 a_{3} = 14

(b)(iii)

2 marks

Solve the system.

a_{1} = \frac{1}{6}, a_{2} = \frac{1}{2}, a_{3} = \frac{1}{3}

(c)

1 mark

Differentiate $f (x) = 1 + x + x^{2} + \dots + x^{n}$ term by term and multiply by $x$ .

f^{'} (x) = 1 + 2 x + 3 x^{2} + \dots + n x^{n - 1} x f^{'} (x) = x + 2 x^{2} + 3 x^{3} + \dots + n x^{n}

(d)(i)

2 marks

From (c), $f_{1} (x) = \sum_{i = 1}^{n} i x^{i}$ , so $f_{1}^{'} (x) = \sum_{i = 1}^{n} i^{2} x^{i - 1}$ .

f_{2} (x) = x f_{1}^{'} (x) = i = 1 \sum n i^{2} x^{i}

(d)(ii)

6 marks

Claim: $f_{q} (x) = \sum_{i = 1}^{n} i^{q} x^{i}$ for all $q \in Z^{+}$ .
Base case $q = 1$ : $f_{1} (x) = x f^{'} (x) = \sum i x^{i}$ from (c), so it holds.
Assume it holds for $q = k$ : $f_{k} (x) = \sum_{i = 1}^{n} i^{k} x^{i}$ .
Then, by the definition of the family,
So it holds for $q = k + 1$ .
It is true for $q = 1$ and, if true for $q = k$ , it is true for $q = k + 1$ . By mathematical induction it is true for all $q \in Z^{+}$ .

f_{k + 1} (x) = x f_{k}^{'} (x) = x i = 1 \sum n i^{k + 1} x^{i - 1} = i = 1 \sum n i^{k + 1} x^{i}

(d)(iii)

1 mark

Put $x = 1$ in the result of (d)(ii).

f_{q} (1) = i = 1 \sum n i^{q}

(e)

2 marks

$f$ is a geometric series with first term 1, ratio $x$ and $n + 1$ terms.

f (x) = \frac{x ^{n + 1} - 1}{x - 1}

(f)

3 marks

Differentiate with the quotient rule.
Multiply by $x$ .

f^{'} (x) = \frac{( n + 1 ) x ^{n} ( x - 1 ) - ( x ^{n + 1} - 1 )}{( x - 1 ) ^{2}} = \frac{n x ^{n + 1} - ( n + 1 ) x ^{n} + 1}{( x - 1 ) ^{2}} f_{1} (x) = \frac{n x ^{n + 2} - ( n + 1 ) x ^{n + 1} + x}{( x - 1 ) ^{2}}

(g)(i)

1 mark

At $x = 1$ the numerator is $n - (n + 1) + 1 = 0$ and the denominator is $0$ , so the limit has the form $\frac{0}{0}$ , which is indeterminate.

(g)(ii)

5 marks

Apply l'Hopital's rule: differentiate the numerator and the denominator.
At $x = 1$ the numerator is $n (n + 2) - (n + 1)^{2} + 1 = 0$ and the denominator is 0 again. Apply the rule a second time.
At $x = 1$ :

x \to 1 lim f_{1} (x) = x \to 1 lim \frac{n ( n + 2 ) x ^{n + 1} - ( n + 1 ) ^{2} x ^{n} + 1}{2 ( x - 1 )} x \to 1 lim \frac{n ( n + 1 ) ( n + 2 ) x ^{n} - n ( n + 1 ) ^{2} x ^{n - 1}}{2} \frac{n ( n + 1 ) [ ( n + 2 ) - ( n + 1 ) ]}{2} = \frac{n ( n + 1 )}{2}

Question 2

27 marks

Surface areas of revolution: cone, sphere and ellipsoid

Calculus
Geometry and trigonometry
Functions

(a)

2 marks

With $m = 2$ and $h = 3$ : $A = 2 π \int_{0}^{3} 2 x 1 + 4 d x$ .

A = 45 π [\frac{x ^{2}}{2}]_{0}^{3} = 45 π \cdot \frac{9}{2} = 185 π

(b)(i)

1 mark

The radius of the cone is the height of the line at $x = h$ .

r = mh

(b)(ii)

2 marks

The slant height is the length of the line from $(0, 0)$ to $(h, mh)$ .

l = h^{2} + m^{2} h^{2} = h 1 + m^{2}

(b)(iii)

3 marks

Put $y = m x$ into the integral.

A = 2 π \int_{0}^{h} m x 1 + m^{2} d x = 2 π m 1 + m^{2} \cdot \frac{h ^{2}}{2} = π (mh) (h 1 + m^{2}) A = π r l

(c)

2 marks

Differentiate $y = (r^{2} - x^{2})^{1/2}$ with the chain rule.

\frac{d y}{d x} = \frac{- x}{r ^{2} - x ^{2}}

(d)

4 marks

Substitute $y$ and $\frac{d y}{d x}$ into the area formula.

A = 2 π \int_{- r}^{r} r^{2} - x^{2} 1 + \frac{x ^{2}}{r ^{2} - x ^{2}} d x = 2 π \int_{- r}^{r} r^{2} - x^{2} + x^{2} d x A = 2 π r \int_{- r}^{r} d x = 2 π r (2 r) = 4 π r^{2}

(e)(i)

2 marks

Replacing $x$ by $k x$ is a horizontal stretch (parallel to the $x$ -axis) with scale factor $\frac{1}{k}$ .

(e)(ii)

1 mark

The graph meets the axis where $f (k x) = 0$ , so $k x = \pm r$ .

x = \pm \frac{r}{k}

(e)(iii)

2 marks

Now $y = r^{2} - k^{2} x^{2}$ .

\frac{d y}{d x} = \frac{- k ^{2} x}{r ^{2} - k ^{2} x ^{2}}

(e)(iv)

4 marks

Substitute into the area formula.
Combine under one square root.

A = 2 π \int_{- r / k}^{r / k} r^{2} - k^{2} x^{2} 1 + \frac{k ^{4} x ^{2}}{r ^{2} - k ^{2} x ^{2}} d x A = 2 π \int_{- r / k}^{r / k} r^{2} + k^{2} (k^{2} - 1) x^{2} d x, p (x) = r^{2} + k^{2} (k^{2} - 1) x^{2}

(e)(v)

4 marks

The equator has radius $r = \frac{12756}{2} = 6378$ . The poles are on the $x$ -axis at $x = \pm \frac{r}{k} = \pm \frac{12714}{2} = \pm 6357$ , so $k = \frac{6378}{6357} = 1.00330$ .
Use the formula from (iv) with these values.

A = 2 π \int_{- 6357}^{6357} 637 8^{2} + k^{2} (k^{2} - 1) x^{2} d x \approx 5.101 \times 1 0^{8} km^{2}

Our analysis

Where the marks are

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

Calculus40 marks · 73%
Q1, Q2
Number and algebra9 marks · 16%
Q1
Functions3 marks · 5%
Q2
Geometry and trigonometry3 marks · 5%
Q2

Mentor insights

Traps to watch for

01
Link the sum to a derivative · Q1
Differentiating $1 + x + \dots + x^{n}$ and multiplying by $x$ turns each $x^{i}$ into $i x^{i}$ . Repeating raises the power each time.
02
Induction layout · Q1
State the assumption for $q = k$ , derive $q = k + 1$ with the operation $f_{k + 1} = x f_{k}^{'}$ , and write a conclusion.
03
0/0 can need l'Hopital twice · Q1
If the numerator and denominator are both 0 after the first derivative, differentiate again.
04
The formula needs the derivative · Q2
Surface area is $2 π \int y 1 + (d y / d x)^{2} d x$ . For the sphere the square root simplifies to a constant.
05
Radius and semi-axis · Q2
For the Earth, the equator gives $r = 6378$ and the pole-to-pole distance gives the semi-axis $\frac{r}{k} = 6357$ .

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