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Worked solutions

IB Maths AA HL · Paper 3

Mathematics: Analysis & Approaches, Higher Level

Session: November 2021
Marks: 55
Duration: 1 hour 15 min
Calculator: Allowed
Questions: 2

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Questions

Questions 1–2 · 55 marks

Question 1

25 marks

Hyperbolic functions as relatives of sine and cosine

Calculus

(a)

2 marks

Differentiate $f (t) = \frac{1}{2} (e^{t} + e^{- t})$ twice.
So $u = f (t)$ satisfies $\frac{d ^{2} u}{d t ^{2}} = u$ .

f^{'} (t) = \frac{1}{2} (e^{t} - e^{- t}), f^{''} (t) = \frac{1}{2} (e^{t} + e^{- t}) = f (t)

(b)

3 marks

Expand both squares.

f (t)^{2} + g (t)^{2} = \frac{( e ^{t} + e ^{- t} ) ^{2} + ( e ^{t} - e ^{- t} ) ^{2}}{4} = \frac{2 e ^{2 t} + 2 e ^{- 2 t}}{4} = \frac{e ^{2 t} + e ^{- 2 t}}{2} = f (2 t)

(c)(i)

3 marks

Use $e^{i u} = cos u + i sin u$ and $e^{- i u} = cos u - i sin u$ .

f (i u) = \frac{e ^{i u} + e ^{- i u}}{2} = cos u

(c)(ii)

2 marks

In the same way, the real parts cancel.

g (i u) = \frac{e ^{i u} - e ^{- i u}}{2} = i sin u

(d)

2 marks

Use the results of part (c), noting $i^{2} = - 1$ .

f (i u)^{2} + g (i u)^{2} = cos^{2} u - sin^{2} u = cos 2 u

(e)

4 marks

For the left side, expand the squares.
For the right side, use part (c).
Both sides equal 1, so they are equal.

f (t)^{2} - g (t)^{2} = \frac{( e ^{t} + e ^{- t} ) ^{2} - ( e ^{t} - e ^{- t} ) ^{2}}{4} = \frac{4}{4} = 1 f (i u)^{2} - g (i u)^{2} = cos^{2} u + sin^{2} u = 1

(f)

4 marks

The curve $x^{2} - y^{2} = 1$ is a hyperbola with two branches, one on the right of $x = 1$ and one on the left of $x = - 1$ . It crosses the $x$ -axis at $(1, 0)$ and $(- 1, 0)$ and doesn't meet the $y$ -axis. The branches approach the asymptotes $y = x$ and $y = - x$ .

(g)

5 marks

Rotating the hyperbola through $4 5^{\circ}$ turns its asymptotes $y = \pm x$ into the axes, so it becomes $x y = k$ . Rotate the point $(1, 0)$ by $4 5^{\circ}$ in each direction.
Then $k = x y$ at that point.

(1, 0) \to (\frac{1}{2}, \frac{1}{2}) or (\frac{1}{2}, - \frac{1}{2}) k = \frac{1}{2} or k = - \frac{1}{2}

Question 2

30 marks

Solving a system of linear differential equations step by step

Calculus

(a)(i)

3 marks

With $a = 0$ , $\frac{d y}{d t} = y$ . Separate and integrate.

\int \frac{1}{y} d y = \int d t \Rightarrow ln y = t + c \Rightarrow y = A e^{t}

(a)(ii)

1 mark

Put $y = A e^{t}$ into $\frac{d x}{d t} = x - y$ .

\frac{d x}{d t} - x = - A e^{t}

(a)(iii)

4 marks

The integrating factor is $e^{\int - 1 d t} = e^{- t}$ .

\frac{d}{d t} (x e^{- t}) = - A \Rightarrow x e^{- t} = - A t + D x = (D - A t) e^{t}

(b)(i)

3 marks

With $a = - 1$ , $\frac{d y}{d t} = - x + y$ . Differentiate with respect to $t$ and replace $\frac{d x}{d t} = x - y$ .
The bracket is $\frac{d y}{d t}$ again.

\frac{d ^{2} y}{d t ^{2}} = - \frac{d x}{d t} + \frac{d y}{d t} = - (x - y) + \frac{d y}{d t} = (- x + y) + \frac{d y}{d t} \frac{d ^{2} y}{d t ^{2}} = 2 \frac{d y}{d t}

(b)(ii)

3 marks

With $Y = \frac{d y}{d t}$ this is $\frac{d Y}{d t} = 2 Y$ . Separate and integrate.

\int \frac{1}{Y} d Y = \int 2 d t \Rightarrow ln Y = 2 t + c \Rightarrow Y = B e^{2 t}

(b)(iii)

2 marks

Integrate $\frac{d y}{d t} = B e^{2 t}$ .

y = \frac{B}{2} e^{2 t} + C

(b)(iv)

3 marks

From $\frac{d y}{d t} = - x + y$ we get $x = y - \frac{d y}{d t}$ .

x = \frac{B}{2} e^{2 t} + C - B e^{2 t} = - \frac{B}{2} e^{2 t} + C

(c)(i)

3 marks

With $a = - 4$ , $\frac{d y}{d t} = - 4 x + y$ . Differentiate and replace $\frac{d x}{d t}$ .
Now eliminate $x$ using $- 4 x = \frac{d y}{d t} - y$ .

\frac{d ^{2} y}{d t ^{2}} = - 4 \frac{d x}{d t} + \frac{d y}{d t} = - 4 (x - y) + \frac{d y}{d t} \frac{d ^{2} y}{d t ^{2}} = (\frac{d y}{d t} - y) + 4 y + \frac{d y}{d t} = 2 \frac{d y}{d t} + 3 y \frac{d ^{2} y}{d t ^{2}} - 2 \frac{d y}{d t} - 3 y = 0

(c)(ii)

4 marks

Put $y = F e^{λ t}$ , so $y^{'} = λ F e^{λ t}$ and $y^{''} = λ^{2} F e^{λ t}$ .

F e^{λ t} (λ^{2} - 2 λ - 3) = 0 \Rightarrow λ^{2} - 2 λ - 3 = 0 (λ - 3) (λ + 1) = 0 \Rightarrow λ = 3 or λ = - 1

(c)(iii)

4 marks

Let $λ_{1} = 3$ and $λ_{2} = - 1$ , and $y = F e^{3 t} + G e^{- t}$ . Then
Substitute into $y^{''} - 2 y^{'} - 3 y$ .
So $y = F e^{λ_{1} t} + G e^{λ_{2} t}$ is a solution.

y^{'} = 3 F e^{3 t} - G e^{- t}, y^{''} = 9 F e^{3 t} + G e^{- t} (9 - 6 - 3) F e^{3 t} + (1 + 2 - 3) G e^{- t} = 0

Our analysis

Where the marks are

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

Calculus55 marks · 100%
Q1, Q2

Mentor insights

Traps to watch for

01
Express in $e$ first · Q1
Every part is an identity about $e^{t}$ and $e^{- t}$ . Substitute the definitions, expand, and let the cross terms cancel. For the complex arguments use $e^{i u} = cos u + i sin u$ .
02
Which sign gives which curve · Q1
$x^{2} - y^{2} = 1$ has two separate branches and asymptotes $y = \pm x$ . A sign error here gives a circle or an ellipse instead.
03
Use the equation you are given · Q2
Each part prepares the next. In (b) you differentiate $\frac{d y}{d t} = - x + y$ and then replace $\frac{d x}{d t}$ using the first equation.
04
Name the first constant · Q2
In (a)(iii) and (b)(iii) the constants must be written with the right letter. The first constant of integration is the one in the formula you were asked to find, and the second is a new one.
05
Try $e^{λ t}$ · Q2
In (c) the pattern suggests $y = F e^{λ t}$ . Substitute it into the second order equation to get the quadratic in $λ$ .

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 2021
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 2021 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

25 marks

Hyperbolic functions as relatives of sine and cosine

Calculus

(a)

2 marks

Differentiate $f (t) = \frac{1}{2} (e^{t} + e^{- t})$ twice.
So $u = f (t)$ satisfies $\frac{d ^{2} u}{d t ^{2}} = u$ .

f^{'} (t) = \frac{1}{2} (e^{t} - e^{- t}), f^{''} (t) = \frac{1}{2} (e^{t} + e^{- t}) = f (t)

(b)

3 marks

Expand both squares.

f (t)^{2} + g (t)^{2} = \frac{( e ^{t} + e ^{- t} ) ^{2} + ( e ^{t} - e ^{- t} ) ^{2}}{4} = \frac{2 e ^{2 t} + 2 e ^{- 2 t}}{4} = \frac{e ^{2 t} + e ^{- 2 t}}{2} = f (2 t)

(c)(i)

3 marks

Use $e^{i u} = cos u + i sin u$ and $e^{- i u} = cos u - i sin u$ .

f (i u) = \frac{e ^{i u} + e ^{- i u}}{2} = cos u

(c)(ii)

2 marks

In the same way, the real parts cancel.

g (i u) = \frac{e ^{i u} - e ^{- i u}}{2} = i sin u

(d)

2 marks

Use the results of part (c), noting $i^{2} = - 1$ .

f (i u)^{2} + g (i u)^{2} = cos^{2} u - sin^{2} u = cos 2 u

(e)

4 marks

For the left side, expand the squares.
For the right side, use part (c).
Both sides equal 1, so they are equal.

f (t)^{2} - g (t)^{2} = \frac{( e ^{t} + e ^{- t} ) ^{2} - ( e ^{t} - e ^{- t} ) ^{2}}{4} = \frac{4}{4} = 1 f (i u)^{2} - g (i u)^{2} = cos^{2} u + sin^{2} u = 1

(f)

4 marks

(g)

5 marks

Rotating the hyperbola through $4 5^{\circ}$ turns its asymptotes $y = \pm x$ into the axes, so it becomes $x y = k$ . Rotate the point $(1, 0)$ by $4 5^{\circ}$ in each direction.
Then $k = x y$ at that point.

(1, 0) \to (\frac{1}{2}, \frac{1}{2}) or (\frac{1}{2}, - \frac{1}{2}) k = \frac{1}{2} or k = - \frac{1}{2}

Question 2

30 marks

Solving a system of linear differential equations step by step

Calculus

(a)(i)

3 marks

With $a = 0$ , $\frac{d y}{d t} = y$ . Separate and integrate.

\int \frac{1}{y} d y = \int d t \Rightarrow ln y = t + c \Rightarrow y = A e^{t}

(a)(ii)

1 mark

Put $y = A e^{t}$ into $\frac{d x}{d t} = x - y$ .

\frac{d x}{d t} - x = - A e^{t}

(a)(iii)

4 marks

The integrating factor is $e^{\int - 1 d t} = e^{- t}$ .

\frac{d}{d t} (x e^{- t}) = - A \Rightarrow x e^{- t} = - A t + D x = (D - A t) e^{t}

(b)(i)

3 marks

With $a = - 1$ , $\frac{d y}{d t} = - x + y$ . Differentiate with respect to $t$ and replace $\frac{d x}{d t} = x - y$ .
The bracket is $\frac{d y}{d t}$ again.

\frac{d ^{2} y}{d t ^{2}} = - \frac{d x}{d t} + \frac{d y}{d t} = - (x - y) + \frac{d y}{d t} = (- x + y) + \frac{d y}{d t} \frac{d ^{2} y}{d t ^{2}} = 2 \frac{d y}{d t}

(b)(ii)

3 marks

With $Y = \frac{d y}{d t}$ this is $\frac{d Y}{d t} = 2 Y$ . Separate and integrate.

\int \frac{1}{Y} d Y = \int 2 d t \Rightarrow ln Y = 2 t + c \Rightarrow Y = B e^{2 t}

(b)(iii)

2 marks

Integrate $\frac{d y}{d t} = B e^{2 t}$ .

y = \frac{B}{2} e^{2 t} + C

(b)(iv)

3 marks

From $\frac{d y}{d t} = - x + y$ we get $x = y - \frac{d y}{d t}$ .

x = \frac{B}{2} e^{2 t} + C - B e^{2 t} = - \frac{B}{2} e^{2 t} + C

(c)(i)

3 marks

With $a = - 4$ , $\frac{d y}{d t} = - 4 x + y$ . Differentiate and replace $\frac{d x}{d t}$ .
Now eliminate $x$ using $- 4 x = \frac{d y}{d t} - y$ .

\frac{d ^{2} y}{d t ^{2}} = - 4 \frac{d x}{d t} + \frac{d y}{d t} = - 4 (x - y) + \frac{d y}{d t} \frac{d ^{2} y}{d t ^{2}} = (\frac{d y}{d t} - y) + 4 y + \frac{d y}{d t} = 2 \frac{d y}{d t} + 3 y \frac{d ^{2} y}{d t ^{2}} - 2 \frac{d y}{d t} - 3 y = 0

(c)(ii)

4 marks

Put $y = F e^{λ t}$ , so $y^{'} = λ F e^{λ t}$ and $y^{''} = λ^{2} F e^{λ t}$ .

F e^{λ t} (λ^{2} - 2 λ - 3) = 0 \Rightarrow λ^{2} - 2 λ - 3 = 0 (λ - 3) (λ + 1) = 0 \Rightarrow λ = 3 or λ = - 1

(c)(iii)

4 marks

Let $λ_{1} = 3$ and $λ_{2} = - 1$ , and $y = F e^{3 t} + G e^{- t}$ . Then
Substitute into $y^{''} - 2 y^{'} - 3 y$ .
So $y = F e^{λ_{1} t} + G e^{λ_{2} t}$ is a solution.

y^{'} = 3 F e^{3 t} - G e^{- t}, y^{''} = 9 F e^{3 t} + G e^{- t} (9 - 6 - 3) F e^{3 t} + (1 + 2 - 3) G e^{- t} = 0

Our analysis

Where the marks are

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

Calculus55 marks · 100%
Q1, Q2

Mentor insights

Traps to watch for

01
Express in $e$ first · Q1
Every part is an identity about $e^{t}$ and $e^{- t}$ . Substitute the definitions, expand, and let the cross terms cancel. For the complex arguments use $e^{i u} = cos u + i sin u$ .
02
Which sign gives which curve · Q1
$x^{2} - y^{2} = 1$ has two separate branches and asymptotes $y = \pm x$ . A sign error here gives a circle or an ellipse instead.
03
Use the equation you are given · Q2
Each part prepares the next. In (b) you differentiate $\frac{d y}{d t} = - x + y$ and then replace $\frac{d x}{d t}$ using the first equation.
04
Name the first constant · Q2
In (a)(iii) and (b)(iii) the constants must be written with the right letter. The first constant of integration is the one in the formula you were asked to find, and the second is a new one.
05
Try $e^{λ t}$ · Q2
In (c) the pattern suggests $y = F e^{λ t}$ . Substitute it into the second order equation to get the quadratic in $λ$ .

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