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IB Maths AA HL · Paper 3

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ3

Session: November 2025
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 2025 TZ3 paper, which your school or the IB can provide.

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Questions

Questions 1–2 · 55 marks

Question 1

26 marks

Repeated sine functions: sketches, maxima, a derivative pattern and a limit

Functions

(a)

4 marks

The graph of $S_{1} = sin x$ on $0 \leq x \leq 2 π$ is one full sine wave. It crosses the axes at $(0, 0)$ , $(π, 0)$ and $(2 π, 0)$ , with a maximum of 1 at $x = \frac{π}{2}$ and a minimum of $- 1$ at $x = \frac{3 π}{2}$ .
The graph of $S_{2} = sin (sin x)$ has the same shape but is squashed, because $sin x$ lies in $[- 1, 1]$ . It has the same intercepts $(0, 0)$ , $(π, 0)$ and $(2 π, 0)$ , with a maximum of $sin 1 \approx 0.841$ at $x = \frac{π}{2}$ and a minimum of $- 0.841$ at $x = \frac{3 π}{2}$ .
Label each curve, and mark the intercepts $0$ , $π$ and $2 π$ on the $x$ -axis (the $y$ -intercept is 0).

(b)(i)

1 mark

The largest value of $sin x$ .

1

(b)(ii)

1 mark

The largest input to the outer sine is 1, and sine is increasing on $[0, 1]$ .

sin 1 \approx 0.841

(b)(iii)

1 mark

Apply sine again to the previous maximum.

sin (0.8415) \approx 0.746

(c)

3 marks

The maximum of $S_{n}$ is found by applying sine $n$ times to 1. Use the calculator.
The first value below 0.6 is at $n = 6$ .

S_{4} : 0.678, S_{5} : 0.628, S_{6} : 0.587 n = 6

(d)

6 marks

Differentiate $y = sin (sin x)$ using the chain rule.
Set this to zero. Either $cos x = 0$ or $cos (sin x) = 0$ .
The first gives $x = \frac{π}{2}$ and $x = \frac{3 π}{2}$ in the interval.
For $cos (sin x) = 0$ we would need $sin x = \frac{π}{2}$ (or another odd multiple of $\frac{π}{2}$ ), but $∣ sin x ∣ \leq 1 < \frac{π}{2}$ . So there are no more solutions.
So there are exactly two points of zero gradient, at $x = \frac{π}{2}$ and $x = \frac{3 π}{2}$ .

\frac{d y}{d x} = cos (sin x) cos x

(e)

1 mark

Use the given pattern with $n = 3$ .

S_{3}^{'} (x) = cos (S_{2} (x)) cos (S_{1} (x)) cos x = cos (sin (sin x)) cos (sin x) cos x

(f)

6 marks

Base case $n = 1$ : $S_{1} = sin x$ , so $S_{1}^{'} (x) = cos x$ , which is the right side (a product with only the factor $cos x$ ).
Assume that the result is true for $n = k$ .
Now $S_{k + 1} (x) = sin (S_{k} (x))$ . Use the chain rule.
Use the assumption for $S_{k}^{'} (x)$ .
This is the formula for $n = k + 1$ . 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 positive integers $n$ .

S_{k}^{'} (x) = cos (S_{k - 1} (x)) cos (S_{k - 2} (x)) \dots cos (S_{1} (x)) cos x S_{k + 1}^{'} (x) = cos (S_{k} (x)) \cdot S_{k}^{'} (x) S_{k + 1}^{'} (x) = cos (S_{k} (x)) cos (S_{k - 1} (x)) \dots cos (S_{1} (x)) cos x

(g)

3 marks

As $x \to 0$ , $S_{n} (x) \to 0$ and $x \to 0$ , so the limit has the form $\frac{0}{0}$ . Apply l'Hopital's rule.
Each factor tends to $cos 0 = 1$ .

x \to 0 lim \frac{S _{n} ( x )}{x} = x \to 0 lim \frac{S _{n}^{'} ( x )}{1} = x \to 0 lim cos (S_{n - 1} (x)) \dots cos (S_{1} (x)) cos x 1 \times 1 \times \dots \times 1 = 1

Question 2

29 marks

Maclaurin series of arctan x and approximating pi with error bounds

Calculus

(a)

2 marks

This is geometric with first term 1 and ratio $- x^{2}$ , which has size less than 1.

S_{\infty} = \frac{1}{1 - ( - x ^{2} )} = \frac{1}{1 + x ^{2}}

(b)

3 marks

Since $\frac{d}{d x} arctan x = \frac{1}{1 + x ^{2}}$ , integrate the series term by term, using $arctan 0 = 0$ .

arctan x = \int_{0}^{x} (1 - t^{2} + t^{4} - t^{6} + \dots) d t = x - \frac{x ^{3}}{3} + \frac{x ^{5}}{5} - \frac{x ^{7}}{7} + \dots

(c)

3 marks

Use $x = \frac{1}{3}$ , for which $arctan x = \frac{π}{6}$ , and the first three terms.

\frac{π}{6} \approx \frac{1}{3} (1 - \frac{1}{9} + \frac{1}{45}) = 0.52606 π \approx 6 (0.52606) \approx 3.156

(d)

3 marks

With $n$ terms the error is at most the next term, $\frac{x ^{2 n + 1}}{2 n + 1}$ . For $x = \frac{1}{3}$ :
The first one below $0.0001$ is $n = 6$ , so six non-zero terms are needed.

n = 5 : \frac{3 ^{- 5.5}}{11} \approx 2.2 \times 1 0^{- 4}, n = 6 : \frac{3 ^{- 6.5}}{13} \approx 6.1 \times 1 0^{- 5}

(e)

4 marks

Integrate by parts with $u = arctan x$ and $d v = d x$ .
The last integral is $\frac{1}{2} ln (1 + x^{2})$ .

\int_{0}^{1/ 3} arctan x d x = [x arctan x]_{0}^{1/ 3} - \int_{0}^{1/ 3} \frac{x}{1 + x ^{2}} d x \frac{1}{3} \cdot \frac{π}{6} - \frac{1}{2} ln \frac{4}{3} = \frac{π}{6 3} - \frac{1}{2} ln \frac{4}{3}

(f)

2 marks

Integrate the four terms.

\int_{0}^{1/ 3} (x - \frac{x ^{3}}{3} + \frac{x ^{5}}{5} - \frac{x ^{7}}{7}) d x = [\frac{x ^{2}}{2} - \frac{x ^{4}}{12} + \frac{x ^{6}}{30} - \frac{x ^{8}}{56}]_{0}^{1/ 3} \frac{1}{6} - \frac{1}{108} + \frac{1}{810} - \frac{1}{4536} \approx 0.158422

(g)

2 marks

Equate the results of parts (e) and (f).

\frac{π}{6 3} - \frac{1}{2} ln \frac{4}{3} \approx 0.158422 π \approx 63 (0.158422 + \frac{1}{2} ln \frac{4}{3}) \approx 3.1412

(h)

5 marks

The theorem says that the error is at most the next integral, $\int_{0}^{1/ 3} \frac{x ^{9}}{9} d x$ .
The actual error is the difference between the exact value from part (e) and the sum from part (f).
Since $3.69 \times 1 0^{- 5} < 4.57 \times 1 0^{- 5}$ , the theorem holds in this case.

[\frac{x ^{10}}{90}]_{0}^{1/ 3} = \frac{1}{90 \times 243} \approx 4.57 \times 1 0^{- 5} ∣ 0.158459 - 0.158422 ∣ \approx 3.69 \times 1 0^{- 5}

(i)

5 marks

Using $m$ non-zero terms, the error is at most the next integral.
Check values.
The first $m$ that works is 7.

\int_{0}^{1/ 3} \frac{x ^{2 m + 1}}{2 m + 1} d x = \frac{3 ^{- (m + 1)}}{( 2 m + 1 ) ( 2 m + 2 )} < 1 0^{- 6} m = 6 : \frac{3 ^{- 7}}{13 \times 14} \approx 2.5 \times 1 0^{- 6} (too big), m = 7 : \frac{3 ^{- 8}}{15 \times 16} \approx 6.4 \times 1 0^{- 7} 7 non-zero terms

Our analysis

Where the marks are

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

Calculus29 marks · 53%
Q2
Functions26 marks · 47%
Q1

Mentor insights

Traps to watch for

01
Composition · Q1
$S_{n + 1} (x) = sin (S_{n} (x))$ , so each new function is the sine of the last.
02
Chain rule · Q1
$\frac{d}{d x} sin (S_{n} (x)) = cos (S_{n} (x)) S_{n}^{'} (x)$ .
03
Geometric series · Q2
$a + a r + a r^{2} + \dots = \frac{a}{1 - r}$ for $∣ r ∣ < 1$ .
04
Alternating series · Q2
The error is no bigger than the first term that was left out.
05
By parts · Q2
Take $u = arctan x$ and $d v = d x$ .

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

TZ1TZ3

Session: November 2025
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 2025 TZ3 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

26 marks

Repeated sine functions: sketches, maxima, a derivative pattern and a limit

Functions

(a)

4 marks

The graph of $S_{1} = sin x$ on $0 \leq x \leq 2 π$ is one full sine wave. It crosses the axes at $(0, 0)$ , $(π, 0)$ and $(2 π, 0)$ , with a maximum of 1 at $x = \frac{π}{2}$ and a minimum of $- 1$ at $x = \frac{3 π}{2}$ .
The graph of $S_{2} = sin (sin x)$ has the same shape but is squashed, because $sin x$ lies in $[- 1, 1]$ . It has the same intercepts $(0, 0)$ , $(π, 0)$ and $(2 π, 0)$ , with a maximum of $sin 1 \approx 0.841$ at $x = \frac{π}{2}$ and a minimum of $- 0.841$ at $x = \frac{3 π}{2}$ .
Label each curve, and mark the intercepts $0$ , $π$ and $2 π$ on the $x$ -axis (the $y$ -intercept is 0).

(b)(i)

1 mark

The largest value of $sin x$ .

1

(b)(ii)

1 mark

The largest input to the outer sine is 1, and sine is increasing on $[0, 1]$ .

sin 1 \approx 0.841

(b)(iii)

1 mark

Apply sine again to the previous maximum.

sin (0.8415) \approx 0.746

(c)

3 marks

The maximum of $S_{n}$ is found by applying sine $n$ times to 1. Use the calculator.
The first value below 0.6 is at $n = 6$ .

S_{4} : 0.678, S_{5} : 0.628, S_{6} : 0.587 n = 6

(d)

6 marks

Differentiate $y = sin (sin x)$ using the chain rule.
Set this to zero. Either $cos x = 0$ or $cos (sin x) = 0$ .
The first gives $x = \frac{π}{2}$ and $x = \frac{3 π}{2}$ in the interval.
For $cos (sin x) = 0$ we would need $sin x = \frac{π}{2}$ (or another odd multiple of $\frac{π}{2}$ ), but $∣ sin x ∣ \leq 1 < \frac{π}{2}$ . So there are no more solutions.
So there are exactly two points of zero gradient, at $x = \frac{π}{2}$ and $x = \frac{3 π}{2}$ .

\frac{d y}{d x} = cos (sin x) cos x

(e)

1 mark

Use the given pattern with $n = 3$ .

S_{3}^{'} (x) = cos (S_{2} (x)) cos (S_{1} (x)) cos x = cos (sin (sin x)) cos (sin x) cos x

(f)

6 marks

Base case $n = 1$ : $S_{1} = sin x$ , so $S_{1}^{'} (x) = cos x$ , which is the right side (a product with only the factor $cos x$ ).
Assume that the result is true for $n = k$ .
Now $S_{k + 1} (x) = sin (S_{k} (x))$ . Use the chain rule.
Use the assumption for $S_{k}^{'} (x)$ .
This is the formula for $n = k + 1$ . 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 positive integers $n$ .

S_{k}^{'} (x) = cos (S_{k - 1} (x)) cos (S_{k - 2} (x)) \dots cos (S_{1} (x)) cos x S_{k + 1}^{'} (x) = cos (S_{k} (x)) \cdot S_{k}^{'} (x) S_{k + 1}^{'} (x) = cos (S_{k} (x)) cos (S_{k - 1} (x)) \dots cos (S_{1} (x)) cos x

(g)

3 marks

As $x \to 0$ , $S_{n} (x) \to 0$ and $x \to 0$ , so the limit has the form $\frac{0}{0}$ . Apply l'Hopital's rule.
Each factor tends to $cos 0 = 1$ .

x \to 0 lim \frac{S _{n} ( x )}{x} = x \to 0 lim \frac{S _{n}^{'} ( x )}{1} = x \to 0 lim cos (S_{n - 1} (x)) \dots cos (S_{1} (x)) cos x 1 \times 1 \times \dots \times 1 = 1

Question 2

29 marks

Maclaurin series of arctan x and approximating pi with error bounds

Calculus

(a)

2 marks

This is geometric with first term 1 and ratio $- x^{2}$ , which has size less than 1.

S_{\infty} = \frac{1}{1 - ( - x ^{2} )} = \frac{1}{1 + x ^{2}}

(b)

3 marks

Since $\frac{d}{d x} arctan x = \frac{1}{1 + x ^{2}}$ , integrate the series term by term, using $arctan 0 = 0$ .

arctan x = \int_{0}^{x} (1 - t^{2} + t^{4} - t^{6} + \dots) d t = x - \frac{x ^{3}}{3} + \frac{x ^{5}}{5} - \frac{x ^{7}}{7} + \dots

(c)

3 marks

Use $x = \frac{1}{3}$ , for which $arctan x = \frac{π}{6}$ , and the first three terms.

\frac{π}{6} \approx \frac{1}{3} (1 - \frac{1}{9} + \frac{1}{45}) = 0.52606 π \approx 6 (0.52606) \approx 3.156

(d)

3 marks

With $n$ terms the error is at most the next term, $\frac{x ^{2 n + 1}}{2 n + 1}$ . For $x = \frac{1}{3}$ :
The first one below $0.0001$ is $n = 6$ , so six non-zero terms are needed.

n = 5 : \frac{3 ^{- 5.5}}{11} \approx 2.2 \times 1 0^{- 4}, n = 6 : \frac{3 ^{- 6.5}}{13} \approx 6.1 \times 1 0^{- 5}

(e)

4 marks

Integrate by parts with $u = arctan x$ and $d v = d x$ .
The last integral is $\frac{1}{2} ln (1 + x^{2})$ .

\int_{0}^{1/ 3} arctan x d x = [x arctan x]_{0}^{1/ 3} - \int_{0}^{1/ 3} \frac{x}{1 + x ^{2}} d x \frac{1}{3} \cdot \frac{π}{6} - \frac{1}{2} ln \frac{4}{3} = \frac{π}{6 3} - \frac{1}{2} ln \frac{4}{3}

(f)

2 marks

Integrate the four terms.

\int_{0}^{1/ 3} (x - \frac{x ^{3}}{3} + \frac{x ^{5}}{5} - \frac{x ^{7}}{7}) d x = [\frac{x ^{2}}{2} - \frac{x ^{4}}{12} + \frac{x ^{6}}{30} - \frac{x ^{8}}{56}]_{0}^{1/ 3} \frac{1}{6} - \frac{1}{108} + \frac{1}{810} - \frac{1}{4536} \approx 0.158422

(g)

2 marks

Equate the results of parts (e) and (f).

\frac{π}{6 3} - \frac{1}{2} ln \frac{4}{3} \approx 0.158422 π \approx 63 (0.158422 + \frac{1}{2} ln \frac{4}{3}) \approx 3.1412

(h)

5 marks

The theorem says that the error is at most the next integral, $\int_{0}^{1/ 3} \frac{x ^{9}}{9} d x$ .
The actual error is the difference between the exact value from part (e) and the sum from part (f).
Since $3.69 \times 1 0^{- 5} < 4.57 \times 1 0^{- 5}$ , the theorem holds in this case.

[\frac{x ^{10}}{90}]_{0}^{1/ 3} = \frac{1}{90 \times 243} \approx 4.57 \times 1 0^{- 5} ∣ 0.158459 - 0.158422 ∣ \approx 3.69 \times 1 0^{- 5}

(i)

5 marks

Using $m$ non-zero terms, the error is at most the next integral.
Check values.
The first $m$ that works is 7.

\int_{0}^{1/ 3} \frac{x ^{2 m + 1}}{2 m + 1} d x = \frac{3 ^{- (m + 1)}}{( 2 m + 1 ) ( 2 m + 2 )} < 1 0^{- 6} m = 6 : \frac{3 ^{- 7}}{13 \times 14} \approx 2.5 \times 1 0^{- 6} (too big), m = 7 : \frac{3 ^{- 8}}{15 \times 16} \approx 6.4 \times 1 0^{- 7} 7 non-zero terms

Our analysis

Where the marks are

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

Calculus29 marks · 53%
Q2
Functions26 marks · 47%
Q1

Mentor insights

Traps to watch for

01
Composition · Q1
$S_{n + 1} (x) = sin (S_{n} (x))$ , so each new function is the sine of the last.
02
Chain rule · Q1
$\frac{d}{d x} sin (S_{n} (x)) = cos (S_{n} (x)) S_{n}^{'} (x)$ .
03
Geometric series · Q2
$a + a r + a r^{2} + \dots = \frac{a}{1 - r}$ for $∣ r ∣ < 1$ .
04
Alternating series · Q2
The error is no bigger than the first term that was left out.
05
By parts · Q2
Take $u = arctan x$ and $d v = d x$ .

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