Loading…

Worked solutions

IB Maths AA HL · Paper 1

Mathematics: Analysis & Approaches, Higher Level

Session: November 2021
Marks: 110
Duration: 2 hours
Calculator: Not allowed
Questions: 12

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.

Section A · Short response

Questions 1–9 · 58 marks

Question 1

4 marks

A simple differential equation with an initial condition

Calculus

4 marks

Integrate $cos (x - \frac{π}{4})$ with respect to $x$ .
Use $y = 2$ at $x = \frac{3 π}{4}$ , where $sin \frac{π}{2} = 1$ .

y = sin (x - \frac{π}{4}) + c 2 = 1 + c \Rightarrow c = 1 y = sin (x - \frac{π}{4}) + 1

Question 2

9 marks

A rational function: asymptotes, sketch, and a self-inverse member of a family

Functions

(a)

2 marks

(i) The denominator is zero at $x = 3$ .

(ii) As $x \to \infty$ , $f (x) \to \frac{2 x}{- x}$ .

x = 3 y = - 2

(b)

2 marks

(i) The graph crosses the $x$ -axis where $2 x + 4 = 0$ .

(ii) At $x = 0$ , $f (0) = \frac{4}{3}$ .

(- 2, 0) (0, \frac{4}{3})

(c)

1 mark

The sketch has the vertical asymptote $x = 3$ and the horizontal asymptote $y = - 2$ . The left branch approaches $y = - 2$ from above as $x \to - \infty$ , passes through $(- 2, 0)$ and $(0, \frac{4}{3})$ , and rises to $+ \infty$ as $x \to 3^{-}$ . The right branch starts at $- \infty$ just right of $x = 3$ and rises towards $y = - 2$ from below as $x \to \infty$ .

(d)

4 marks

Write $y = \frac{a x + 4}{3 - x}$ , swap $x$ and $y$ , and solve for $y$ .
For $g = g^{- 1}$ , compare $\frac{a x + 4}{3 - x}$ with $\frac{3 x - 4}{x + a}$ . Writing the first as $\frac{- a x - 4}{x - 3}$ and matching with a denominator of $x - 3$ requires $a = - 3$ .

x (3 - y) = a y + 4 \Rightarrow 3 x - 4 = y (x + a) \Rightarrow g^{- 1} (x) = \frac{3 x - 4}{x + a} \frac{- 3 x + 4}{3 - x} = \frac{3 x - 4}{x - 3} = g^{- 1} (x) \Rightarrow a = - 3

Question 3

5 marks

A logarithm equation with mixed bases

Number and algebra

5 marks

Rewrite $lo g_{3} x = \frac{1}{2} lo g_{3} x$ , and use $\frac{1}{lo g _{2} 3} = lo g_{3} 2$ for the fraction. Expand $lo g_{3} (4 x^{3}) = 2 lo g_{3} 2 + 3 lo g_{3} x$ .
Collect the terms in $lo g_{3} x$ on one side.

\frac{1}{2} lo g_{3} x = \frac{1}{2} lo g_{3} 2 + 2 lo g_{3} 2 + 3 lo g_{3} x - \frac{5}{2} lo g_{3} x = \frac{5}{2} lo g_{3} 2 \Rightarrow lo g_{3} x = - lo g_{3} 2 = lo g_{3} \frac{1}{2} x = \frac{1}{2}

Question 4

5 marks

Two boxes of balls: a total probability and an independence test

Statistics and probability

(a)

3 marks

Either box is chosen with probability $\frac{1}{2}$ , then a red ball is drawn.

P (R) = \frac{1}{2} \cdot \frac{5}{7} + \frac{1}{2} \cdot \frac{4}{7} = \frac{9}{14}

(b)

2 marks

Compare $P (A) P (R)$ with $P (A \cap R)$ .
They are not equal, so $A$ and $R$ are not independent. (Box 1 has a higher proportion of red balls.)

P (A) P (R) = \frac{1}{2} \cdot \frac{9}{14} = \frac{9}{28}, P (A \cap R) = \frac{1}{2} \cdot \frac{5}{7} = \frac{5}{14} = \frac{10}{28}

Question 5

7 marks

A tangent to $f$ and the tangent to a composite

Calculus

(a)

1 mark

The tangent has gradient 6.

f^{'} (4) = 6

(b)

1 mark

The tangent passes through the graph at $x = 4$ .

f (4) = 6 (4) - 1 = 23

(c)

2 marks

First $g (4) = 16 - 12 = 4$ .

h (4) = f (g (4)) = f (4) = 23

(d)

3 marks

Use the chain rule with $g^{'} (x) = 2 x - 3$ , so $g^{'} (4) = 5$ .
The tangent at $x = 4$ passes through $(4, 23)$ with gradient 30.

h^{'} (4) = f^{'} (g (4)) g^{'} (4) = 6 \times 5 = 30 y - 23 = 30 (x - 4) \Rightarrow y = 30 x - 97

Question 6

7 marks

A rational identity, used to solve a trig equation

Number and algebra
Geometry and trigonometry

(a)

2 marks

Put the right-hand side over the common denominator.

2 x - 3 - \frac{6}{x - 1} = \frac{( 2 x - 3 ) ( x - 1 ) - 6}{x - 1} = \frac{2 x ^{2} - 5 x + 3 - 6}{x - 1} = \frac{2 x ^{2} - 5 x - 3}{x - 1}

(b)

5 marks

Put $x = sin 2 θ$ in the equation $2 sin 2 θ - 3 - \frac{6}{sin 2 θ - 1} = 0$ . By part (a) it becomes a quadratic.
$sin 2 θ = 3$ is impossible, so $sin 2 θ = - \frac{1}{2}$ for $0 \leq 2 θ \leq 2 π$ .

\frac{2 sin ^{2} 2 θ - 5 sin 2 θ - 3}{sin 2 θ - 1} = 0 \Rightarrow (2 sin 2 θ + 1) (sin 2 θ - 3) = 0 2 θ = \frac{7 π}{6}, \frac{11 π}{6} \Rightarrow θ = \frac{7 π}{12}, \frac{11 π}{12}

Mentor tip

Neither answer is $\frac{π}{4}$ , the excluded value, where $sin 2 θ = 1$ would make the denominator zero.

Question 7

7 marks

A quadratic with two distinct roots

Number and algebra

(a)

5 marks

Rewrite as $3 p x^{2} + 2 p x + (1 - p) = 0$ and take the discriminant.
Factorise and find where it is positive.

Δ = (2 p)^{2} - 4 (3 p) (1 - p) = 16 p^{2} - 12 p > 0 4 p (4 p - 3) > 0 \Rightarrow p < 0 or p > \frac{3}{4}

(b)

2 marks

For $p = 4$ the equation is $12 x^{2} + 8 x - 3 = 0$ .

x = \frac{- 8 \pm 64 + 144}{24} = \frac{- 8 \pm 4 13}{24} = \frac{- 2 \pm 13}{6} a = - 2

Question 8

7 marks

A first-order linear differential equation with an integrating factor

Number and algebra

7 marks

Rewrite as $\frac{d y}{d x} + \frac{2}{x} y = \frac{ln 2 x}{x ^{2}}$ . The integrating factor is $e^{\int 2/ x d x} = e^{2 l n x} = x^{2}$ .
Integrate the right side, with $\int ln 2 x d x = x ln 2 x - x$ by parts.
Use $y = 4$ at $x = \frac{1}{2}$ : the log term is $ln 1 = 0$ .

\frac{d}{d x} (x^{2} y) = ln 2 x x^{2} y = x ln 2 x - x + c \frac{1}{4} (4) = 0 - \frac{1}{2} + c \Rightarrow c = \frac{3}{2} y = \frac{ln 2 x - 1}{x} + \frac{3}{2 x ^{2}}

Question 9

7 marks

A binomial expansion with a parameter, and when it is valid

Number and algebra

(a)

6 marks

Expand each part to the $x^{2}$ term.
Subtract the series.
Compare with $4 b x + b x^{2}$ . From the $x^{2}$ coefficient $b = \frac{3 a ^{2} + 1}{8}$ , and from the $x$ coefficient $4 b = \frac{1 - a}{2}$ .

(1 + a x)^{- 1/2} = 1 - \frac{a}{2} x + \frac{3}{8} a^{2} x^{2} + \dots, 1 - x = 1 - \frac{1}{2} x - \frac{1}{8} x^{2} - \dots \frac{1}{1 + a x} - 1 - x = \frac{1 - a}{2} x + \frac{3 a ^{2} + 1}{8} x^{2} + \dots \frac{3 a ^{2} + 1}{2} = \frac{1 - a}{2} \Rightarrow 3 a^{2} + a = 0 \Rightarrow a = - \frac{1}{3} (a \neq = 0) b = \frac{3 \cdot \frac{1}{9} + 1}{8} = \frac{1}{6}

(b)

1 mark

The series for $(1 + a x)^{- 1/2}$ needs $∣ a x ∣ < 1$ , which is $∣ x ∣ < 3$ . The series for $1 - x$ needs $∣ x ∣ < 1$ . Both hold only if

∣ x ∣ < 1

Section B · Extended response

Questions 10–12 · 52 marks

Question 10

16 marks

A particle's velocity: maximum speed, distance, graph and total distance

Calculus

(a)

7 marks

(i) The velocity is largest where $v^{'} = 0$ .

(ii) Integrate $v$ from 0 (since $P$ starts at the origin) up to $t = \frac{2}{3}$ .

v^{'} = 4 - 6 t = 0 \Rightarrow t = \frac{2}{3} x = \int_{0}^{2/3} (4 + 4 t - 3 t^{2}) d t = [4 t + 2 t^{2} - t^{3}]_{0}^{2/3} \frac{8}{3} + \frac{8}{9} - \frac{8}{27} = \frac{72 + 24 - 8}{27} = \frac{88}{27} m

(b)

4 marks

The graph is a downward-opening parabola for $0 \leq t \leq 3$ . It starts at $(0, 4)$ and has a maximum at $t = \frac{2}{3}$ with $v = \frac{16}{3}$ . It crosses the $t$ -axis where $3 t^{2} - 4 t - 4 = 0$ .
It ends at $t = 3$ with $v = 4 + 12 - 27 = - 11$ .

(3 t + 2) (t - 2) = 0 \Rightarrow t = 2

(c)

5 marks

$v > 0$ for $0 < t < 2$ and $v < 0$ for $2 < t \leq 3$ . Integrate each piece.

\int_{0}^{2} v d t = 8, \int_{2}^{3} v d t = - 5 total distance = 8 + 5 = 13 m

Question 11

14 marks

Induction on repeated derivatives of $x^{2} e^{x}$ , then a limit

Calculus

(a)

7 marks

Base case $n = 1$ : by the product rule $\frac{d}{d x} (x^{2} e^{x}) = e^{x} (x^{2} + 2 x)$ . The formula gives $e^{x} [x^{2} + 2 x + 0]$ , the same.
Assume true for $n = k$ :
Differentiate again with the product rule.
This is the formula for $n = k + 1$ , since $(k + 1) ((k + 1) - 1) = k (k + 1)$ . It holds for $n = 1$ , and if it holds for $k$ then it holds for $k + 1$ , so it holds for all positive integers $n$ .

\frac{d ^{k}}{d x ^{k}} (x^{2} e^{x}) = [x^{2} + 2 k x + k (k - 1)] e^{x} \frac{d ^{k + 1}}{d x ^{k + 1}} (x^{2} e^{x}) = [2 x + 2 k] e^{x} + [x^{2} + 2 k x + k (k - 1)] e^{x} [x^{2} + 2 (k + 1) x + k (k + 1)] e^{x}

(b)

3 marks

Evaluate the derivatives at $x = 0$ using part (a): $f (0) = 0$ , $f^{'} (0) = 0$ , $f^{''} (0) = 2$ , $f^{'''} (0) = 6$ , $f^{(4)} (0) = 12$ .

f (x) \approx 0 + 0 + \frac{2}{2 !} x^{2} + \frac{6}{3 !} x^{3} + \frac{12}{4 !} x^{4} = x^{2} + x^{3} + \frac{1}{2} x^{4}

Mentor tip

Alternatively multiply the series for $e^{x}$ by $x^{2}$ .

(c)

4 marks

Factor the numerator: $x^{2} e^{x} - x^{2} = x^{2} (e^{x} - 1)$ . By part (b) it equals $x^{3} + \frac{1}{2} x^{4} + \dots = x^{3} (1 + \frac{x}{2} + \dots)$ .
Let $x \to 0$ .

\frac{( x ^{2} e ^{x} - x ^{2} ) ^{3}}{x ^{9}} = \frac{x ^{9} ( 1 + \frac{x}{2} + \dots ) ^{3}}{x ^{9}} = (1 + \frac{x}{2} + \dots)^{3} 1

Question 12

22 marks

Roots of $(z - 1)^{3} = i$ and a second equation

Number and algebra

(a)

6 marks

(i) Substitute $ω_{1} = 1 + e^{i π /6}$ .

(ii) The other roots have $ω - 1$ turned through $\frac{2 π}{3}$ and $\frac{4 π}{3}$ .

Check the signs: $Im ω_{2} = \frac{1}{2} > 0$ and $ω_{3} = 1 - i$ has a negative imaginary part.

(ω_{1} - 1)^{3} = (e^{i π /6})^{3} = e^{i π /2} = cos \frac{π}{2} + i sin \frac{π}{2} = i ω_{2} = 1 + e^{i (π /6 + 2 π /3)} = 1 + e^{5 π i /6}, ω_{3} = 1 + e^{i (π /6 + 4 π /3)} = 1 + e^{9 π i /6}

(b)

4 marks

Write each in Cartesian form: A is $1 + (\frac{3}{2} + \frac{1}{2} i) \approx (1.87, 0.5)$ , B is $1 + (- \frac{3}{2} + \frac{1}{2} i) \approx (0.13, 0.5)$ , and C is $(1, - 1)$ . They lie on a circle of radius 1 centred at $(1, 0)$ .

(c)

3 marks

Use the distance formula between $A \approx (1.866, 0.5)$ and $C = (1, - 1)$ .

AC = (\frac{3}{2})^{2} + (\frac{3}{2})^{2} = \frac{3}{4} + \frac{9}{4} = 3

(d)

3 marks

With $α = \frac{1}{1 - e ^{i π /6}}$ we get $\frac{1}{α} = 1 - e^{i π /6}$ , so $\frac{α - 1}{α} = 1 - \frac{1}{α} = e^{i π /6}$ .
Cube both sides, using de Moivre's theorem.

(\frac{α - 1}{α})^{3} = e^{i π /2} = i \Rightarrow (α - 1)^{3} = i α^{3}

(e)

6 marks

Multiply the top and bottom by the conjugate of the denominator. With $θ = \frac{π}{6}$ :
The denominator simplifies, using $cos^{2} θ + sin^{2} θ = 1$ .

α = \frac{1}{( 1 - cos θ ) - i sin θ} = \frac{( 1 - cos θ ) + i sin θ}{( 1 - cos θ ) ^{2} + sin ^{2} θ} (1 - cos θ)^{2} + sin^{2} θ = 2 - 2 cos θ Re (α) = \frac{1 - cos θ}{2 ( 1 - cos θ )} = \frac{1}{2}

Our analysis

Where the marks are

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

Number and algebra50 marks · 45%
Q3, Q6, Q7, Q8, Q9, Q12
Calculus41 marks · 37%
Q1, Q5, Q10, Q11
Functions9 marks · 8%
Q2
Geometry and trigonometry5 marks · 5%
Q6
Statistics and probability5 marks · 5%
Q4

Mentor insights

Traps to watch for

01
Self-inverse means $g = g^{- 1}$ · Q2
Find the inverse by swapping $x$ and $y$ , then compare it with $g$ and match coefficients.
02
Chain rule for the composite · Q5
$h = f \circ g$ , so $h^{'} (x) = f^{'} (g (x)) g^{'} (x)$ . At $x = 4$ we need $g (4)$ and $g^{'} (4)$ .
03
Total distance needs the sign change · Q10
The velocity changes sign at $t = 2$ . Integrate each side separately and add the sizes.
04
Differentiate the right-hand side · Q11
In the inductive step, differentiate the formula for $n = k$ once more and check it matches the formula with $k + 1$ in place of $n$ .
05
Cube roots are equally spaced · Q12
The three roots of $(z - 1)^{3} = i$ are $1$ plus the three cube roots of $i$ . They are spaced $\frac{2 π}{3}$ apart on a circle.

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 1

Mathematics: Analysis & Approaches, Higher Level

Session: November 2021
Marks: 110
Duration: 2 hours
Calculator: Not allowed
Questions: 12

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12

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.

Section A · Short response

Questions 1–9 · 58 marks

Question 1

4 marks

A simple differential equation with an initial condition

Calculus

4 marks

Integrate $cos (x - \frac{π}{4})$ with respect to $x$ .
Use $y = 2$ at $x = \frac{3 π}{4}$ , where $sin \frac{π}{2} = 1$ .

y = sin (x - \frac{π}{4}) + c 2 = 1 + c \Rightarrow c = 1 y = sin (x - \frac{π}{4}) + 1

Question 2

9 marks

A rational function: asymptotes, sketch, and a self-inverse member of a family

Functions

(a)

2 marks

(i) The denominator is zero at $x = 3$ .

(ii) As $x \to \infty$ , $f (x) \to \frac{2 x}{- x}$ .

x = 3 y = - 2

(b)

2 marks

(i) The graph crosses the $x$ -axis where $2 x + 4 = 0$ .

(ii) At $x = 0$ , $f (0) = \frac{4}{3}$ .

(- 2, 0) (0, \frac{4}{3})

(c)

1 mark

(d)

4 marks

Write $y = \frac{a x + 4}{3 - x}$ , swap $x$ and $y$ , and solve for $y$ .
For $g = g^{- 1}$ , compare $\frac{a x + 4}{3 - x}$ with $\frac{3 x - 4}{x + a}$ . Writing the first as $\frac{- a x - 4}{x - 3}$ and matching with a denominator of $x - 3$ requires $a = - 3$ .

x (3 - y) = a y + 4 \Rightarrow 3 x - 4 = y (x + a) \Rightarrow g^{- 1} (x) = \frac{3 x - 4}{x + a} \frac{- 3 x + 4}{3 - x} = \frac{3 x - 4}{x - 3} = g^{- 1} (x) \Rightarrow a = - 3

Question 3

5 marks

A logarithm equation with mixed bases

Number and algebra

5 marks

Rewrite $lo g_{3} x = \frac{1}{2} lo g_{3} x$ , and use $\frac{1}{lo g _{2} 3} = lo g_{3} 2$ for the fraction. Expand $lo g_{3} (4 x^{3}) = 2 lo g_{3} 2 + 3 lo g_{3} x$ .
Collect the terms in $lo g_{3} x$ on one side.

\frac{1}{2} lo g_{3} x = \frac{1}{2} lo g_{3} 2 + 2 lo g_{3} 2 + 3 lo g_{3} x - \frac{5}{2} lo g_{3} x = \frac{5}{2} lo g_{3} 2 \Rightarrow lo g_{3} x = - lo g_{3} 2 = lo g_{3} \frac{1}{2} x = \frac{1}{2}

Question 4

5 marks

Two boxes of balls: a total probability and an independence test

Statistics and probability

(a)

3 marks

Either box is chosen with probability $\frac{1}{2}$ , then a red ball is drawn.

P (R) = \frac{1}{2} \cdot \frac{5}{7} + \frac{1}{2} \cdot \frac{4}{7} = \frac{9}{14}

(b)

2 marks

Compare $P (A) P (R)$ with $P (A \cap R)$ .
They are not equal, so $A$ and $R$ are not independent. (Box 1 has a higher proportion of red balls.)

P (A) P (R) = \frac{1}{2} \cdot \frac{9}{14} = \frac{9}{28}, P (A \cap R) = \frac{1}{2} \cdot \frac{5}{7} = \frac{5}{14} = \frac{10}{28}

Question 5

7 marks

A tangent to $f$ and the tangent to a composite

Calculus

(a)

1 mark

The tangent has gradient 6.

f^{'} (4) = 6

(b)

1 mark

The tangent passes through the graph at $x = 4$ .

f (4) = 6 (4) - 1 = 23

(c)

2 marks

First $g (4) = 16 - 12 = 4$ .

h (4) = f (g (4)) = f (4) = 23

(d)

3 marks

Use the chain rule with $g^{'} (x) = 2 x - 3$ , so $g^{'} (4) = 5$ .
The tangent at $x = 4$ passes through $(4, 23)$ with gradient 30.

h^{'} (4) = f^{'} (g (4)) g^{'} (4) = 6 \times 5 = 30 y - 23 = 30 (x - 4) \Rightarrow y = 30 x - 97

Question 6

7 marks

A rational identity, used to solve a trig equation

Number and algebra
Geometry and trigonometry

(a)

2 marks

Put the right-hand side over the common denominator.

2 x - 3 - \frac{6}{x - 1} = \frac{( 2 x - 3 ) ( x - 1 ) - 6}{x - 1} = \frac{2 x ^{2} - 5 x + 3 - 6}{x - 1} = \frac{2 x ^{2} - 5 x - 3}{x - 1}

(b)

5 marks

Put $x = sin 2 θ$ in the equation $2 sin 2 θ - 3 - \frac{6}{sin 2 θ - 1} = 0$ . By part (a) it becomes a quadratic.
$sin 2 θ = 3$ is impossible, so $sin 2 θ = - \frac{1}{2}$ for $0 \leq 2 θ \leq 2 π$ .

\frac{2 sin ^{2} 2 θ - 5 sin 2 θ - 3}{sin 2 θ - 1} = 0 \Rightarrow (2 sin 2 θ + 1) (sin 2 θ - 3) = 0 2 θ = \frac{7 π}{6}, \frac{11 π}{6} \Rightarrow θ = \frac{7 π}{12}, \frac{11 π}{12}

Mentor tip

Neither answer is $\frac{π}{4}$ , the excluded value, where $sin 2 θ = 1$ would make the denominator zero.

Question 7

7 marks

A quadratic with two distinct roots

Number and algebra

(a)

5 marks

Rewrite as $3 p x^{2} + 2 p x + (1 - p) = 0$ and take the discriminant.
Factorise and find where it is positive.

Δ = (2 p)^{2} - 4 (3 p) (1 - p) = 16 p^{2} - 12 p > 0 4 p (4 p - 3) > 0 \Rightarrow p < 0 or p > \frac{3}{4}

(b)

2 marks

For $p = 4$ the equation is $12 x^{2} + 8 x - 3 = 0$ .

x = \frac{- 8 \pm 64 + 144}{24} = \frac{- 8 \pm 4 13}{24} = \frac{- 2 \pm 13}{6} a = - 2

Question 8

7 marks

A first-order linear differential equation with an integrating factor

Number and algebra

7 marks

Rewrite as $\frac{d y}{d x} + \frac{2}{x} y = \frac{ln 2 x}{x ^{2}}$ . The integrating factor is $e^{\int 2/ x d x} = e^{2 l n x} = x^{2}$ .
Integrate the right side, with $\int ln 2 x d x = x ln 2 x - x$ by parts.
Use $y = 4$ at $x = \frac{1}{2}$ : the log term is $ln 1 = 0$ .

\frac{d}{d x} (x^{2} y) = ln 2 x x^{2} y = x ln 2 x - x + c \frac{1}{4} (4) = 0 - \frac{1}{2} + c \Rightarrow c = \frac{3}{2} y = \frac{ln 2 x - 1}{x} + \frac{3}{2 x ^{2}}

Question 9

7 marks

A binomial expansion with a parameter, and when it is valid

Number and algebra

(a)

6 marks

Expand each part to the $x^{2}$ term.
Subtract the series.
Compare with $4 b x + b x^{2}$ . From the $x^{2}$ coefficient $b = \frac{3 a ^{2} + 1}{8}$ , and from the $x$ coefficient $4 b = \frac{1 - a}{2}$ .

(1 + a x)^{- 1/2} = 1 - \frac{a}{2} x + \frac{3}{8} a^{2} x^{2} + \dots, 1 - x = 1 - \frac{1}{2} x - \frac{1}{8} x^{2} - \dots \frac{1}{1 + a x} - 1 - x = \frac{1 - a}{2} x + \frac{3 a ^{2} + 1}{8} x^{2} + \dots \frac{3 a ^{2} + 1}{2} = \frac{1 - a}{2} \Rightarrow 3 a^{2} + a = 0 \Rightarrow a = - \frac{1}{3} (a \neq = 0) b = \frac{3 \cdot \frac{1}{9} + 1}{8} = \frac{1}{6}

(b)

1 mark

The series for $(1 + a x)^{- 1/2}$ needs $∣ a x ∣ < 1$ , which is $∣ x ∣ < 3$ . The series for $1 - x$ needs $∣ x ∣ < 1$ . Both hold only if

∣ x ∣ < 1

Section B · Extended response

Questions 10–12 · 52 marks

Question 10

16 marks

A particle's velocity: maximum speed, distance, graph and total distance

Calculus

(a)

7 marks

(i) The velocity is largest where $v^{'} = 0$ .

(ii) Integrate $v$ from 0 (since $P$ starts at the origin) up to $t = \frac{2}{3}$ .

v^{'} = 4 - 6 t = 0 \Rightarrow t = \frac{2}{3} x = \int_{0}^{2/3} (4 + 4 t - 3 t^{2}) d t = [4 t + 2 t^{2} - t^{3}]_{0}^{2/3} \frac{8}{3} + \frac{8}{9} - \frac{8}{27} = \frac{72 + 24 - 8}{27} = \frac{88}{27} m

(b)

4 marks

The graph is a downward-opening parabola for $0 \leq t \leq 3$ . It starts at $(0, 4)$ and has a maximum at $t = \frac{2}{3}$ with $v = \frac{16}{3}$ . It crosses the $t$ -axis where $3 t^{2} - 4 t - 4 = 0$ .
It ends at $t = 3$ with $v = 4 + 12 - 27 = - 11$ .

(3 t + 2) (t - 2) = 0 \Rightarrow t = 2

(c)

5 marks

$v > 0$ for $0 < t < 2$ and $v < 0$ for $2 < t \leq 3$ . Integrate each piece.

\int_{0}^{2} v d t = 8, \int_{2}^{3} v d t = - 5 total distance = 8 + 5 = 13 m

Question 11

14 marks

Induction on repeated derivatives of $x^{2} e^{x}$ , then a limit

Calculus

(a)

7 marks

Base case $n = 1$ : by the product rule $\frac{d}{d x} (x^{2} e^{x}) = e^{x} (x^{2} + 2 x)$ . The formula gives $e^{x} [x^{2} + 2 x + 0]$ , the same.
Assume true for $n = k$ :
Differentiate again with the product rule.
This is the formula for $n = k + 1$ , since $(k + 1) ((k + 1) - 1) = k (k + 1)$ . It holds for $n = 1$ , and if it holds for $k$ then it holds for $k + 1$ , so it holds for all positive integers $n$ .

\frac{d ^{k}}{d x ^{k}} (x^{2} e^{x}) = [x^{2} + 2 k x + k (k - 1)] e^{x} \frac{d ^{k + 1}}{d x ^{k + 1}} (x^{2} e^{x}) = [2 x + 2 k] e^{x} + [x^{2} + 2 k x + k (k - 1)] e^{x} [x^{2} + 2 (k + 1) x + k (k + 1)] e^{x}

(b)

3 marks

Evaluate the derivatives at $x = 0$ using part (a): $f (0) = 0$ , $f^{'} (0) = 0$ , $f^{''} (0) = 2$ , $f^{'''} (0) = 6$ , $f^{(4)} (0) = 12$ .

f (x) \approx 0 + 0 + \frac{2}{2 !} x^{2} + \frac{6}{3 !} x^{3} + \frac{12}{4 !} x^{4} = x^{2} + x^{3} + \frac{1}{2} x^{4}

Mentor tip

Alternatively multiply the series for $e^{x}$ by $x^{2}$ .

(c)

4 marks

Factor the numerator: $x^{2} e^{x} - x^{2} = x^{2} (e^{x} - 1)$ . By part (b) it equals $x^{3} + \frac{1}{2} x^{4} + \dots = x^{3} (1 + \frac{x}{2} + \dots)$ .
Let $x \to 0$ .

\frac{( x ^{2} e ^{x} - x ^{2} ) ^{3}}{x ^{9}} = \frac{x ^{9} ( 1 + \frac{x}{2} + \dots ) ^{3}}{x ^{9}} = (1 + \frac{x}{2} + \dots)^{3} 1

Question 12

22 marks

Roots of $(z - 1)^{3} = i$ and a second equation

Number and algebra

(a)

6 marks

(i) Substitute $ω_{1} = 1 + e^{i π /6}$ .

(ii) The other roots have $ω - 1$ turned through $\frac{2 π}{3}$ and $\frac{4 π}{3}$ .

Check the signs: $Im ω_{2} = \frac{1}{2} > 0$ and $ω_{3} = 1 - i$ has a negative imaginary part.

(ω_{1} - 1)^{3} = (e^{i π /6})^{3} = e^{i π /2} = cos \frac{π}{2} + i sin \frac{π}{2} = i ω_{2} = 1 + e^{i (π /6 + 2 π /3)} = 1 + e^{5 π i /6}, ω_{3} = 1 + e^{i (π /6 + 4 π /3)} = 1 + e^{9 π i /6}

(b)

4 marks

(c)

3 marks

Use the distance formula between $A \approx (1.866, 0.5)$ and $C = (1, - 1)$ .

AC = (\frac{3}{2})^{2} + (\frac{3}{2})^{2} = \frac{3}{4} + \frac{9}{4} = 3

(d)

3 marks

With $α = \frac{1}{1 - e ^{i π /6}}$ we get $\frac{1}{α} = 1 - e^{i π /6}$ , so $\frac{α - 1}{α} = 1 - \frac{1}{α} = e^{i π /6}$ .
Cube both sides, using de Moivre's theorem.

(\frac{α - 1}{α})^{3} = e^{i π /2} = i \Rightarrow (α - 1)^{3} = i α^{3}

(e)

6 marks

Multiply the top and bottom by the conjugate of the denominator. With $θ = \frac{π}{6}$ :
The denominator simplifies, using $cos^{2} θ + sin^{2} θ = 1$ .

α = \frac{1}{( 1 - cos θ ) - i sin θ} = \frac{( 1 - cos θ ) + i sin θ}{( 1 - cos θ ) ^{2} + sin ^{2} θ} (1 - cos θ)^{2} + sin^{2} θ = 2 - 2 cos θ Re (α) = \frac{1 - cos θ}{2 ( 1 - cos θ )} = \frac{1}{2}

Our analysis

Where the marks are

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

Number and algebra50 marks · 45%
Q3, Q6, Q7, Q8, Q9, Q12
Calculus41 marks · 37%
Q1, Q5, Q10, Q11
Functions9 marks · 8%
Q2
Geometry and trigonometry5 marks · 5%
Q6
Statistics and probability5 marks · 5%
Q4

Mentor insights

Traps to watch for

01
Self-inverse means $g = g^{- 1}$ · Q2
Find the inverse by swapping $x$ and $y$ , then compare it with $g$ and match coefficients.
02
Chain rule for the composite · Q5
$h = f \circ g$ , so $h^{'} (x) = f^{'} (g (x)) g^{'} (x)$ . At $x = 4$ we need $g (4)$ and $g^{'} (4)$ .
03
Total distance needs the sign change · Q10
The velocity changes sign at $t = 2$ . Integrate each side separately and add the sizes.
04
Differentiate the right-hand side · Q11
In the inductive step, differentiate the formula for $n = k$ once more and check it matches the formula with $k + 1$ in place of $n$ .
05
Cube roots are equally spaced · Q12
The three roots of $(z - 1)^{3} = i$ are $1$ plus the three cube roots of $i$ . They are spaced $\frac{2 π}{3}$ apart on a circle.

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