Loading…

Worked solutions

IB Maths AA HL · Paper 3

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2 TZ3

Session: May 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 May 2025 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

23 marks

Probability generating polynomials: dice, balls and coins

Statistics and probability

(a)

2 marks

Add each value times its probability.

E (M) = 1 (\frac{1}{16}) + 2 (\frac{3}{16}) + 3 (\frac{5}{16}) + 4 (\frac{7}{16}) = \frac{1 + 6 + 15 + 28}{16} = \frac{25}{8}

(b)

1 mark

The probabilities add to 1.

G (1) = \frac{1}{16} + \frac{3}{16} + \frac{5}{16} + \frac{7}{16} = 1

(c)(i)

2 marks

Differentiate $G (t) = \frac{1}{16} t + \frac{3}{16} t^{2} + \frac{5}{16} t^{3} + \frac{7}{16} t^{4}$ .

G^{'} (t) = \frac{1}{16} + \frac{6}{16} t + \frac{15}{16} t^{2} + \frac{28}{16} t^{3}

(c)(ii)

3 marks

Substitute $t = 1$ .
This is $\sum m P (M = m)$ , which is $E (M)$ from part (a).

G^{'} (1) = \frac{1 + 6 + 15 + 28}{16} = \frac{25}{8}

(d)

5 marks

There are $(2 5) = 10$ equally likely pairs of balls. The coefficient of $t^{x}$ is $P (X = x)$ .
No red balls: both balls are yellow, $(2 3) = 3$ pairs.
One red ball: one of the 2 red and one of the 3 yellow, $2 \times 3 = 6$ pairs.
Two red balls: both red, $(2 2) = 1$ pair (or $\frac{2}{5} \times \frac{1}{4}$ ).
So the polynomial is

P (X = 0) = \frac{3}{10} P (X = 1) = \frac{6}{10} = \frac{3}{5} P (X = 2) = \frac{1}{10} G_{X} (t) = \frac{3}{10} + \frac{3}{5} t + \frac{1}{10} t^{2}

(e)(i)

2 marks

Two tails need a tail on both coins: $\frac{1}{2} \times p$ .

\frac{p}{2} = \frac{1}{3} \Rightarrow p = \frac{2}{3}

(e)(ii)

4 marks

No tails: $\frac{1}{2} (1 - p) = \frac{1}{6}$ . One tail: tail on the fair coin and head on the other, or head on the fair coin and tail on the biased coin.

P (Y = 1) = \frac{1}{2} (\frac{1}{3}) + \frac{1}{2} (\frac{2}{3}) = \frac{1}{2} G_{Y} (t) = \frac{1}{6} + \frac{1}{2} t + \frac{1}{3} t^{2}

(f)

4 marks

Multiply the polynomials, $G_{Z} (t) = G_{X} (t) G_{Y} (t)$ , and differentiate. By the product rule, $G_{Z}^{'} (1) = G_{X}^{'} (1) G_{Y} (1) + G_{X} (1) G_{Y}^{'} (1) = G_{X}^{'} (1) + G_{Y}^{'} (1)$ since $G (1) = 1$ .

G_{X}^{'} (t) = \frac{3}{5} + \frac{1}{5} t \Rightarrow G_{X}^{'} (1) = \frac{4}{5} G_{Y}^{'} (t) = \frac{1}{2} + \frac{2}{3} t \Rightarrow G_{Y}^{'} (1) = \frac{7}{6} E (Z) = G_{Z}^{'} (1) = \frac{4}{5} + \frac{7}{6} = \frac{59}{30} \approx 1.97

Question 2

32 marks

The curvature of a curve: lines, parabolas, ln x, e^x and circles

Calculus

(a)

2 marks

For $g (x) = m x + c$ we have $g^{'} (x) = m$ and $g^{''} (x) = 0$ .

k (x) = \frac{∣0∣}{( 1 + m ^{2} ) ^{3/2}} = 0

(b)(i)

1 mark

With $k^{'} (x) = - 12 a (2 a x + b) \cdot (a factor that is never zero)$ , setting $k^{'} (x) = 0$ requires $2 a x + b = 0$ .

x = - \frac{b}{2 a}

(b)(ii)

2 marks

At $x = - \frac{b}{2 a}$ the gradient $2 a x + b = 0$ , so

k_{m a x} = \frac{2∣ a ∣}{( 1 + 0 ) ^{3/2}} = 2∣ a ∣

(b)(iii)

2 marks

As $x \to \pm \infty$ the gradient $2 a x + b \to \pm \infty$ and the curvature tends to 0.
Far from the vertex the parabola is almost straight, so its curvature is nearly the same as that of a line.

x \to \infty lim k (x) = 0

(b)(iv)

2 marks

The maximum curvature is $2∣ a ∣$ . For $p$ , $a = - 2$ and for $q$ , $a = 2$ .
So statement C is true: the maximum curvatures are equal.

k_{m a x} (p) = 4 = k_{m a x} (q)

(c)(i)

2 marks

For $v (x) = ln x$ , $k (x) = \frac{x}{( x ^{2} + 1 ) ^{3/2}}$ . Differentiate with the product rule.

k^{'} (x) = (x^{2} + 1)^{- 5/2} [(x^{2} + 1) - 3 x^{2}] = \frac{1 - 2 x ^{2}}{( x ^{2} + 1 ) ^{5/2}} k^{'} (x) = 0 \Rightarrow x^{2} = \frac{1}{2} \Rightarrow x = \frac{1}{2}

(c)(ii)

4 marks

Substitute $x = \frac{1}{2}$ into $k (x)$ , with $x^{2} + 1 = \frac{3}{2}$ .

k_{m a x} = \frac{1/ 2}{( 3/2 ) ^{3/2}} = \frac{1}{2} \cdot \frac{2 2}{3 3} = \frac{2}{3 3} = \frac{2 3}{9}

(d)(i)

5 marks

For $w (x) = e^{x}$ , $k (x) = \frac{e ^{x}}{( 1 + e ^{2 x} ) ^{3/2}}$ . Differentiate.
Set $k^{'} (x) = 0$ .
Substitute, with $1 + e^{2 x} = \frac{3}{2}$ .

k^{'} (x) = e^{x} (1 + e^{2 x})^{- 5/2} [(1 + e^{2 x}) - 3 e^{2 x}] = \frac{e ^{x} ( 1 - 2 e ^{2 x} )}{( 1 + e ^{2 x} ) ^{5/2}} e^{2 x} = \frac{1}{2} \Rightarrow e^{x} = \frac{1}{2} k_{m a x} = \frac{1/ 2}{( 3/2 ) ^{3/2}} = \frac{2 3}{9}

(d)(ii)

1 mark

The functions $v (x) = ln x$ and $w (x) = e^{x}$ are inverses of each other, so their graphs are reflections in the line $y = x$ . Reflecting a curve does not change its shape, so the curvature is the same at corresponding points.

(e)(i)

6 marks

The curve is the upper half circle $x^{2} + y^{2} = r^{2}$ . Differentiate implicitly.
Differentiate again with the quotient rule.
Use $x^{2} + y^{2} = r^{2}$ .

2 x + 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = - \frac{x}{y} \frac{d ^{2} y}{d x ^{2}} = - \frac{y - x \frac{d y}{d x}}{y ^{2}} = - \frac{y + \frac{x ^{2}}{y}}{y ^{2}} = - \frac{y ^{2} + x ^{2}}{y ^{3}} \frac{d ^{2} y}{d x ^{2}} = - \frac{r ^{2}}{y ^{3}}

(e)(ii)

5 marks

Substitute into the curvature, with $1 + (\frac{d y}{d x})^{2} = 1 + \frac{x ^{2}}{y ^{2}} = \frac{y ^{2} + x ^{2}}{y ^{2}} = \frac{r ^{2}}{y ^{2}}$ .
Since $y > 0$ , $y^{3} = (y^{2})^{3/2}$ . The curvature is the constant $\frac{1}{r}$ for the whole family.

k = \frac{r ^{2} / y ^{3}}{( r ^{2} / y ^{2} ) ^{3/2}} = \frac{r ^{2} / y ^{3}}{r ^{3} / y ^{3}} = \frac{1}{r}

Our analysis

Where the marks are

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

Calculus32 marks · 58%
Q2
Statistics and probability23 marks · 42%
Q1

Mentor insights

Traps to watch for

01
Generating polynomial · Q1
$G (t) = \sum P (M = m) t^{m}$ . Differentiating at $t = 1$ gives the expected value.
02
Product of generating polynomials · Q1
For independent variables, the generating polynomial of a sum is the product of the polynomials.
03
Curvature · Q2
$k (x) = \frac{∣ f ^{''} ( x ) ∣}{( 1 + f ^{'} ( x ) ^{2} ) ^{3/2}}$ measures how far a curve is from being straight.
04
Maximum curvature · Q2
Solve $k^{'} (x) = 0$ using the derivative that is given.
05
Implicit differentiation · Q2
Differentiate $x^{2} + y^{2} = r^{2}$ twice to find $y^{'}$ and $y^{''}$ .

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

Worked solutions

IB Maths AA HL · Paper 3

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2 TZ3

Session: May 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 May 2025 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

23 marks

Probability generating polynomials: dice, balls and coins

Statistics and probability

(a)

2 marks

Add each value times its probability.

E (M) = 1 (\frac{1}{16}) + 2 (\frac{3}{16}) + 3 (\frac{5}{16}) + 4 (\frac{7}{16}) = \frac{1 + 6 + 15 + 28}{16} = \frac{25}{8}

(b)

1 mark

The probabilities add to 1.

G (1) = \frac{1}{16} + \frac{3}{16} + \frac{5}{16} + \frac{7}{16} = 1

(c)(i)

2 marks

Differentiate $G (t) = \frac{1}{16} t + \frac{3}{16} t^{2} + \frac{5}{16} t^{3} + \frac{7}{16} t^{4}$ .

G^{'} (t) = \frac{1}{16} + \frac{6}{16} t + \frac{15}{16} t^{2} + \frac{28}{16} t^{3}

(c)(ii)

3 marks

Substitute $t = 1$ .
This is $\sum m P (M = m)$ , which is $E (M)$ from part (a).

G^{'} (1) = \frac{1 + 6 + 15 + 28}{16} = \frac{25}{8}

(d)

5 marks

There are $(2 5) = 10$ equally likely pairs of balls. The coefficient of $t^{x}$ is $P (X = x)$ .
No red balls: both balls are yellow, $(2 3) = 3$ pairs.
One red ball: one of the 2 red and one of the 3 yellow, $2 \times 3 = 6$ pairs.
Two red balls: both red, $(2 2) = 1$ pair (or $\frac{2}{5} \times \frac{1}{4}$ ).
So the polynomial is

P (X = 0) = \frac{3}{10} P (X = 1) = \frac{6}{10} = \frac{3}{5} P (X = 2) = \frac{1}{10} G_{X} (t) = \frac{3}{10} + \frac{3}{5} t + \frac{1}{10} t^{2}

(e)(i)

2 marks

Two tails need a tail on both coins: $\frac{1}{2} \times p$ .

\frac{p}{2} = \frac{1}{3} \Rightarrow p = \frac{2}{3}

(e)(ii)

4 marks

No tails: $\frac{1}{2} (1 - p) = \frac{1}{6}$ . One tail: tail on the fair coin and head on the other, or head on the fair coin and tail on the biased coin.

P (Y = 1) = \frac{1}{2} (\frac{1}{3}) + \frac{1}{2} (\frac{2}{3}) = \frac{1}{2} G_{Y} (t) = \frac{1}{6} + \frac{1}{2} t + \frac{1}{3} t^{2}

(f)

4 marks

G_{X}^{'} (t) = \frac{3}{5} + \frac{1}{5} t \Rightarrow G_{X}^{'} (1) = \frac{4}{5} G_{Y}^{'} (t) = \frac{1}{2} + \frac{2}{3} t \Rightarrow G_{Y}^{'} (1) = \frac{7}{6} E (Z) = G_{Z}^{'} (1) = \frac{4}{5} + \frac{7}{6} = \frac{59}{30} \approx 1.97

Question 2

32 marks

The curvature of a curve: lines, parabolas, ln x, e^x and circles

Calculus

(a)

2 marks

For $g (x) = m x + c$ we have $g^{'} (x) = m$ and $g^{''} (x) = 0$ .

k (x) = \frac{∣0∣}{( 1 + m ^{2} ) ^{3/2}} = 0

(b)(i)

1 mark

With $k^{'} (x) = - 12 a (2 a x + b) \cdot (a factor that is never zero)$ , setting $k^{'} (x) = 0$ requires $2 a x + b = 0$ .

x = - \frac{b}{2 a}

(b)(ii)

2 marks

At $x = - \frac{b}{2 a}$ the gradient $2 a x + b = 0$ , so

k_{m a x} = \frac{2∣ a ∣}{( 1 + 0 ) ^{3/2}} = 2∣ a ∣

(b)(iii)

2 marks

As $x \to \pm \infty$ the gradient $2 a x + b \to \pm \infty$ and the curvature tends to 0.
Far from the vertex the parabola is almost straight, so its curvature is nearly the same as that of a line.

x \to \infty lim k (x) = 0

(b)(iv)

2 marks

The maximum curvature is $2∣ a ∣$ . For $p$ , $a = - 2$ and for $q$ , $a = 2$ .
So statement C is true: the maximum curvatures are equal.

k_{m a x} (p) = 4 = k_{m a x} (q)

(c)(i)

2 marks

For $v (x) = ln x$ , $k (x) = \frac{x}{( x ^{2} + 1 ) ^{3/2}}$ . Differentiate with the product rule.

k^{'} (x) = (x^{2} + 1)^{- 5/2} [(x^{2} + 1) - 3 x^{2}] = \frac{1 - 2 x ^{2}}{( x ^{2} + 1 ) ^{5/2}} k^{'} (x) = 0 \Rightarrow x^{2} = \frac{1}{2} \Rightarrow x = \frac{1}{2}

(c)(ii)

4 marks

Substitute $x = \frac{1}{2}$ into $k (x)$ , with $x^{2} + 1 = \frac{3}{2}$ .

k_{m a x} = \frac{1/ 2}{( 3/2 ) ^{3/2}} = \frac{1}{2} \cdot \frac{2 2}{3 3} = \frac{2}{3 3} = \frac{2 3}{9}

(d)(i)

5 marks

For $w (x) = e^{x}$ , $k (x) = \frac{e ^{x}}{( 1 + e ^{2 x} ) ^{3/2}}$ . Differentiate.
Set $k^{'} (x) = 0$ .
Substitute, with $1 + e^{2 x} = \frac{3}{2}$ .

k^{'} (x) = e^{x} (1 + e^{2 x})^{- 5/2} [(1 + e^{2 x}) - 3 e^{2 x}] = \frac{e ^{x} ( 1 - 2 e ^{2 x} )}{( 1 + e ^{2 x} ) ^{5/2}} e^{2 x} = \frac{1}{2} \Rightarrow e^{x} = \frac{1}{2} k_{m a x} = \frac{1/ 2}{( 3/2 ) ^{3/2}} = \frac{2 3}{9}

(d)(ii)

1 mark

(e)(i)

6 marks

The curve is the upper half circle $x^{2} + y^{2} = r^{2}$ . Differentiate implicitly.
Differentiate again with the quotient rule.
Use $x^{2} + y^{2} = r^{2}$ .

2 x + 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = - \frac{x}{y} \frac{d ^{2} y}{d x ^{2}} = - \frac{y - x \frac{d y}{d x}}{y ^{2}} = - \frac{y + \frac{x ^{2}}{y}}{y ^{2}} = - \frac{y ^{2} + x ^{2}}{y ^{3}} \frac{d ^{2} y}{d x ^{2}} = - \frac{r ^{2}}{y ^{3}}

(e)(ii)

5 marks

Substitute into the curvature, with $1 + (\frac{d y}{d x})^{2} = 1 + \frac{x ^{2}}{y ^{2}} = \frac{y ^{2} + x ^{2}}{y ^{2}} = \frac{r ^{2}}{y ^{2}}$ .
Since $y > 0$ , $y^{3} = (y^{2})^{3/2}$ . The curvature is the constant $\frac{1}{r}$ for the whole family.

k = \frac{r ^{2} / y ^{3}}{( r ^{2} / y ^{2} ) ^{3/2}} = \frac{r ^{2} / y ^{3}}{r ^{3} / y ^{3}} = \frac{1}{r}

Our analysis

Where the marks are

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

Calculus32 marks · 58%
Q2
Statistics and probability23 marks · 42%
Q1

Mentor insights

Traps to watch for

01
Generating polynomial · Q1
$G (t) = \sum P (M = m) t^{m}$ . Differentiating at $t = 1$ gives the expected value.
02
Product of generating polynomials · Q1
For independent variables, the generating polynomial of a sum is the product of the polynomials.
03
Curvature · Q2
$k (x) = \frac{∣ f ^{''} ( x ) ∣}{( 1 + f ^{'} ( x ) ^{2} ) ^{3/2}}$ measures how far a curve is from being straight.
04
Maximum curvature · Q2
Solve $k^{'} (x) = 0$ using the derivative that is given.
05
Implicit differentiation · Q2
Differentiate $x^{2} + y^{2} = r^{2}$ twice to find $y^{'}$ and $y^{''}$ .

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