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IB Maths AA HL · Paper 1

Mathematics: Analysis & Approaches, Higher Level

Session: November 2024
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 2024 Global paper, which your school or the IB can provide.

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Section A · Short response

Questions 1–9 · 56 marks

Question 1

5 marks

A sector with a given area and angle

Geometry and trigonometry

(a)

3 marks

Use the area formula.

\frac{1}{2} r^{2} (1.5) = 48 \Rightarrow r^{2} = 64 r = 8 cm

(b)

2 marks

The perimeter is two radii plus the arc length $r θ$ .

2 r + r θ = 16 + 8 (1.5) = 28 cm

Question 2

6 marks

Two events from a union: intersection and a conditional probability

Statistics and probability

(a)

3 marks

Rearrange the union formula.

P (A \cap B) = 0.65 + 0.45 - 0.85 = 0.25

(b)

3 marks

The probability of neither event is $P (A^{'} \cap B^{'}) = 1 - 0.85 = 0.15$ , and $P (B^{'}) = 1 - 0.45 = 0.55$ .

P (A^{'} ∣ B^{'}) = \frac{0.15}{0.55} = \frac{3}{11}

Question 3

4 marks

Proving a difference of squares is a multiple of 12

Number and algebra

4 marks

Use the difference of squares with $a = 3 n + 2$ and $b = 3 n - 2$ .
Simplify each bracket.
Factor out 12.
Since $2 n$ is a positive integer, the expression is a multiple of 12 for every $n \in Z^{+}$ .

(3 n + 2)^{2} - (3 n - 2)^{2} = [(3 n + 2) - (3 n - 2)] [(3 n + 2) + (3 n - 2)] 4 \times 6 n = 24 n 24 n = 12 (2 n)

Question 4

6 marks

The exact value of cos(theta) from a triangle with angle 2 theta

Calculus

6 marks

Use the cosine rule for the angle $2 θ$ at $B$ , opposite the side $A C = 5$ .
Use the double angle formula.
The angle $θ$ is acute, so take the positive root.
So $p = 5$ and $q = 8$ .

cos 2 θ = \frac{4 ^{2} + 6 ^{2} - 5 ^{2}}{2 ( 4 ) ( 6 )} = \frac{27}{48} = \frac{9}{16} 2 cos^{2} θ - 1 = \frac{9}{16} \Rightarrow cos^{2} θ = \frac{25}{32} cos θ = \frac{5}{32} = \frac{5}{4 2} = \frac{5 2}{8}

Question 5

6 marks

An arithmetic sequence from u10 and S25

Number and algebra

6 marks

From $u_{10} = 16$ .
From $S_{25} = 100$ .
Subtract the first equation from the second.
Solve $u_{k} = 0$ .

u_{1} + 9 d = 16 \frac{25}{2} (2 u_{1} + 24 d) = 100 \Rightarrow u_{1} + 12 d = 4 3 d = - 12 \Rightarrow d = - 4, u_{1} = 16 + 36 = 52 52 - 4 (k - 1) = 0 \Rightarrow k - 1 = 13 k = 14

Question 6

5 marks

A quadratic inequality and the largest domain for an inverse

Functions

(a)

3 marks

Find the critical values by factorising.
The parabola opens upwards, so it is negative between the roots.

2 x^{2} - 15 x + 18 = (2 x - 3) (x - 6) = 0 \Rightarrow x = \frac{3}{2}, x = 6 \frac{3}{2} < x < 6

(b)

2 marks

For $f (x) = 2 x^{2} - 15 x + 18$ to be real we need $2 x^{2} - 15 x + 18 \geq 0$ , which means $x \leq \frac{3}{2}$ or $x \geq 6$ . The domain $x \leq k$ has no gaps, so $k \leq \frac{3}{2}$ .
For $x \leq \frac{3}{2}$ the inside is decreasing (the vertex is at $x = 3.75$ ), so $f$ is one-to-one and has an inverse.

k = \frac{3}{2}

Question 7

7 marks

A secant function: its range and a volume of revolution

Functions

(a)

3 marks

The function $f (x) = sec (x - \frac{π}{4})$ has its minimum value where $x = \frac{π}{4}$ , because $cos 0 = 1$ .
The ends of the interval.

f (\frac{π}{4}) = 1 f (0) = sec (- \frac{π}{4}) = 2, f (\frac{π}{2}) = sec \frac{π}{4} = 2 1 \leq f (x) \leq 2

(b)

4 marks

Use the volume formula.
Substitute the limits.

V = π \int_{0}^{π /2} sec^{2} (x - \frac{π}{4}) d x = π [tan (x - \frac{π}{4})]_{0}^{π /2} π [tan \frac{π}{4} - tan (- \frac{π}{4})] = π (1 + 1) V = 2 π

Question 8

8 marks

A point on a line and when its position vector is perpendicular to AB

Number and algebra

(a)

1 mark

Write $O P = O A + A P$ with $A B = b - a$ .

O P = a + λ (b - a) = (1 - λ) a + λ b

(b)

7 marks

If $O P ⊥ A B$ , then $O P \cdot A B = 0$ .
Expand.
Use $∣ a ∣ = 1$ , $∣ b ∣ = 2$ and $a \cdot b = \frac{1}{4}$ .
Collect the terms.

[(1 - λ) a + λ b] \cdot (b - a) = 0 (1 - λ) a \cdot b - (1 - λ) ∣ a ∣^{2} + λ ∣ b ∣^{2} - λ a \cdot b = 0 \frac{1 - λ}{4} - (1 - λ) + 4 λ - \frac{λ}{4} = 0 - \frac{3}{4} (1 - λ) + \frac{15}{4} λ = 0 \Rightarrow - 3 + 18 λ = 0 λ = \frac{1}{6}

Question 9

9 marks

A tangent identity and a trigonometric equation

Functions

(a)

6 marks

Use the subtraction formula with $tan \frac{π}{4} = 1$ .
Multiply the top and bottom by $sin θ - cos θ$ .
Use $sin^{2} θ + cos^{2} θ = 1$ , $2 sin θ cos θ = sin 2 θ$ and $cos^{2} θ - sin^{2} θ = cos 2 θ$ .

tan (θ - \frac{π}{4}) = \frac{tan θ - 1}{1 + tan θ} = \frac{sin θ - cos θ}{cos θ + sin θ} \frac{( sin θ - cos θ ) ^{2}}{( sin θ + cos θ ) ( sin θ - cos θ )} = \frac{sin ^{2} θ - 2 sin θ cos θ + cos ^{2} θ}{sin ^{2} θ - cos ^{2} θ} \frac{1 - sin 2 θ}{- cos 2 θ} = \frac{sin 2 θ - 1}{cos 2 θ}

(b)

3 marks

Put $2 θ = x$ , so $θ = \frac{x}{2}$ . By part (a) the equation is
Solve for the angle.
In the interval $0 \leq x \leq 2 π$ the only solution is

tan (\frac{x}{2} - \frac{π}{4}) = 3 \frac{x}{2} - \frac{π}{4} = \frac{π}{3} + nπ \Rightarrow x = \frac{7 π}{6} + 2 nπ x = \frac{7 π}{6}

Section B · Extended response

Questions 10–12 · 54 marks

Question 10

17 marks

A hollow cylinder: surface area, volume and the maximum volume

Calculus

(a)

3 marks

The outer curved surface (radius $4 r$ ) and the inner curved surface (radius $r$ ).
The two ends are rings with area $π (4 r)^{2} - π r^{2} = 15 π r^{2}$ each.

2 π (4 r) h + 2 π r h = 8 π r h + 2 π r h = 10 π r h 2 \times 15 π r^{2} = 30 π r^{2} S = 30 π r^{2} + 10 π r h

(b)

6 marks

Set $S = 240 π$ and solve for $r h$ .
The volume is the large cylinder minus the small one.
Substitute.

30 π r^{2} + 10 π r h = 240 π \Rightarrow 3 r^{2} + r h = 24 \Rightarrow r h = 24 - 3 r^{2} V = π (4 r)^{2} h - π r^{2} h = 15 π r^{2} h = 15 π r (r h) V = 15 π r (24 - 3 r^{2}) = 360 π r - 45 π r^{3}

(c)

2 marks

Differentiate.

\frac{d V}{d r} = 360 π - 135 π r^{2}

(d)

3 marks

At the maximum $\frac{d V}{d r} = 0$ .
Compare with $r = p \frac{2}{3}$ , for which $r^{2} = \frac{2 p ^{2}}{3}$ .

135 r^{2} = 360 \Rightarrow r^{2} = \frac{8}{3} \Rightarrow r = 2 \frac{2}{3} \frac{2 p ^{2}}{3} = \frac{8}{3} \Rightarrow p^{2} = 4 \Rightarrow p = 2

(e)

3 marks

Substitute into $V = π r (360 - 45 r^{2})$ .
So $q = 480$ .

V = π \cdot 2 \frac{2}{3} (360 - 45 \cdot \frac{8}{3}) = 2 π \frac{2}{3} (360 - 120) V = 480 π \frac{2}{3}

Question 11

17 marks

A curve with an exponential quotient: a limit, derivatives and a Maclaurin series

Calculus

(a)

2 marks

Divide the top and bottom by $e^{2 x}$ .

x \to \infty lim \frac{e ^{2 x} - 1}{e ^{2 x} + 1} = x \to \infty lim \frac{1 - e ^{- 2 x}}{1 + e ^{- 2 x}} = \frac{1 - 0}{1 + 0} = 1

(b)

6 marks

(i) Use the quotient rule with $u = e^{2 x} - 1$ and $v = e^{2 x} + 1$ .

(ii) Work out $1 - y^{2}$ using a common denominator.

The numerator is a difference of squares: $(a + b)^{2} - (a - b)^{2} = 4 ab$ .

\frac{d y}{d x} = \frac{2 e ^{2 x} ( e ^{2 x} + 1 ) - 2 e ^{2 x} ( e ^{2 x} - 1 )}{( e ^{2 x} + 1 ) ^{2}} = \frac{4 e ^{2 x}}{( e ^{2 x} + 1 ) ^{2}} 1 - y^{2} = \frac{( e ^{2 x} + 1 ) ^{2} - ( e ^{2 x} - 1 ) ^{2}}{( e ^{2 x} + 1 ) ^{2}} 1 - y^{2} = \frac{4 e ^{2 x}}{( e ^{2 x} + 1 ) ^{2}} = \frac{d y}{d x}

(c)

5 marks

(i) Differentiate $\frac{d y}{d x} = 1 - y^{2}$ implicitly.

(ii) Differentiate $\frac{d ^{2} y}{d x ^{2}} = 2 y^{3} - 2 y$ implicitly.

\frac{d ^{2} y}{d x ^{2}} = - 2 y \frac{d y}{d x} = - 2 y (1 - y^{2}) = 2 y^{3} - 2 y \frac{d ^{3} y}{d x ^{3}} = (6 y^{2} - 2) \frac{d y}{d x} = (6 y^{2} - 2) (1 - y^{2})

(d)

4 marks

At $x = 0$ : $y = 0$ .
Use the Maclaurin series $y \approx y (0) + y^{'} (0) x + \frac{y ^{''} ( 0 )}{2 !} x^{2} + \frac{y ^{'''} ( 0 )}{3 !} x^{3}$ .

y = 0, y^{'} = 1 - 0 = 1, y^{''} = 0, y^{'''} = (- 2) (1) = - 2 \frac{e ^{2 x} - 1}{e ^{2 x} + 1} \approx x - \frac{x ^{3}}{3}

Question 12

20 marks

Roots of z to the fourth equal 16i, a geometric sequence and midpoints

Number and algebra

(a)

6 marks

Write $16 i = 16 cis \frac{π}{2}$ . The fourth roots have modulus $1 6^{1/4} = 2$ and arguments $\frac{π /2 + 2 k π}{4} = \frac{π}{8} + \frac{k π}{2}$ .

z_{1} = 2 cis \frac{π}{8}, z_{2} = 2 cis \frac{5 π}{8} z_{3} = 2 cis \frac{9 π}{8}, z_{4} = 2 cis \frac{13 π}{8}

(b)

3 marks

The common ratio is $\frac{z _{2}}{z _{1}}$ . The modulus is 1 and the argument is $\frac{5 π}{8} - \frac{π}{8} = \frac{π}{2}$ .

r = cis \frac{π}{2} = i

(c)

3 marks

All four points are 2 units from the origin, so they lie on a circle of radius 2 centred at the origin. $A$ is at an angle of $22. 5^{\circ}$ (in the first quadrant), $B$ at $112. 5^{\circ}$ , $C$ at $202. 5^{\circ}$ and $D$ at $292. 5^{\circ}$ , each $9 0^{\circ}$ round from the last, so they form a square.

(d)

3 marks

The roots of $v^{4} = a + bi$ are $z_{k}^{*}$ , the conjugates, so $v^{4} = (z^{4})^{*} = (16 i)^{*} = - 16 i$ .

a = 0, b = - 16

(e)

5 marks

$A^{'}$ is the midpoint of $A B$ . The chord $A B$ subtends $9 0^{\circ}$ at the centre, so $∣ O A^{'} ∣ = 2 cos 4 5^{\circ} = 2$ , and its argument is the average $\frac{1}{2} (\frac{π}{8} + \frac{5 π}{8}) = \frac{3 π}{8}$ .
The four midpoints are $2 cis (\frac{3 π}{8} + \frac{k π}{2})$ . They are roots of $w^{p} = 2^{q}$ when the modulus matches and $p$ times the argument is a multiple of $2 π$ .
So $p$ must be a multiple of 16, and the least value is 16. The spacing of the roots is then $\frac{2 π}{16} = \frac{π}{8}$ , which includes the four points spaced by $\frac{π}{2}$ .

A^{'} = 2 cis \frac{3 π}{8} (2)^{p} = 2^{q} \Rightarrow p = 2 q p \cdot \frac{3 π}{8} = 2 mπ \Rightarrow p = \frac{16 m}{3} p = 16, q = 8

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 · 36%
Q4, Q10, Q11
Number and algebra38 marks · 35%
Q3, Q5, Q8, Q12
Functions21 marks · 19%
Q6, Q7, Q9
Statistics and probability6 marks · 5%
Q2
Geometry and trigonometry5 marks · 5%
Q1

Mentor insights

Traps to watch for

01
Parametrise · Q8
$A P = λ A B$ .
02
Subtraction formula · Q9
$tan (A - B) = \frac{tan A - tan B}{1 + tan A tan B}$ .
03
Surface area · Q10
Outer curved, inner curved and two ring-shaped ends.
04
Limit at infinity · Q11
Divide the top and bottom by the fastest growing term.
05
De Moivre · Q12
The roots of $z^{n} = r e^{i α}$ have modulus $r^{1/ n}$ and arguments $\frac{α + 2 k π}{n}$ .

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Worked solutions

IB Maths AA HL · Paper 1

Mathematics: Analysis & Approaches, Higher Level

Session: November 2024
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 2024 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 · 56 marks

Question 1

5 marks

A sector with a given area and angle

Geometry and trigonometry

(a)

3 marks

Use the area formula.

\frac{1}{2} r^{2} (1.5) = 48 \Rightarrow r^{2} = 64 r = 8 cm

(b)

2 marks

The perimeter is two radii plus the arc length $r θ$ .

2 r + r θ = 16 + 8 (1.5) = 28 cm

Question 2

6 marks

Two events from a union: intersection and a conditional probability

Statistics and probability

(a)

3 marks

Rearrange the union formula.

P (A \cap B) = 0.65 + 0.45 - 0.85 = 0.25

(b)

3 marks

The probability of neither event is $P (A^{'} \cap B^{'}) = 1 - 0.85 = 0.15$ , and $P (B^{'}) = 1 - 0.45 = 0.55$ .

P (A^{'} ∣ B^{'}) = \frac{0.15}{0.55} = \frac{3}{11}

Question 3

4 marks

Proving a difference of squares is a multiple of 12

Number and algebra

4 marks

Use the difference of squares with $a = 3 n + 2$ and $b = 3 n - 2$ .
Simplify each bracket.
Factor out 12.
Since $2 n$ is a positive integer, the expression is a multiple of 12 for every $n \in Z^{+}$ .

(3 n + 2)^{2} - (3 n - 2)^{2} = [(3 n + 2) - (3 n - 2)] [(3 n + 2) + (3 n - 2)] 4 \times 6 n = 24 n 24 n = 12 (2 n)

Question 4

6 marks

The exact value of cos(theta) from a triangle with angle 2 theta

Calculus

6 marks

Use the cosine rule for the angle $2 θ$ at $B$ , opposite the side $A C = 5$ .
Use the double angle formula.
The angle $θ$ is acute, so take the positive root.
So $p = 5$ and $q = 8$ .

cos 2 θ = \frac{4 ^{2} + 6 ^{2} - 5 ^{2}}{2 ( 4 ) ( 6 )} = \frac{27}{48} = \frac{9}{16} 2 cos^{2} θ - 1 = \frac{9}{16} \Rightarrow cos^{2} θ = \frac{25}{32} cos θ = \frac{5}{32} = \frac{5}{4 2} = \frac{5 2}{8}

Question 5

6 marks

An arithmetic sequence from u10 and S25

Number and algebra

6 marks

From $u_{10} = 16$ .
From $S_{25} = 100$ .
Subtract the first equation from the second.
Solve $u_{k} = 0$ .

u_{1} + 9 d = 16 \frac{25}{2} (2 u_{1} + 24 d) = 100 \Rightarrow u_{1} + 12 d = 4 3 d = - 12 \Rightarrow d = - 4, u_{1} = 16 + 36 = 52 52 - 4 (k - 1) = 0 \Rightarrow k - 1 = 13 k = 14

Question 6

5 marks

A quadratic inequality and the largest domain for an inverse

Functions

(a)

3 marks

Find the critical values by factorising.
The parabola opens upwards, so it is negative between the roots.

2 x^{2} - 15 x + 18 = (2 x - 3) (x - 6) = 0 \Rightarrow x = \frac{3}{2}, x = 6 \frac{3}{2} < x < 6

(b)

2 marks

For $f (x) = 2 x^{2} - 15 x + 18$ to be real we need $2 x^{2} - 15 x + 18 \geq 0$ , which means $x \leq \frac{3}{2}$ or $x \geq 6$ . The domain $x \leq k$ has no gaps, so $k \leq \frac{3}{2}$ .
For $x \leq \frac{3}{2}$ the inside is decreasing (the vertex is at $x = 3.75$ ), so $f$ is one-to-one and has an inverse.

k = \frac{3}{2}

Question 7

7 marks

A secant function: its range and a volume of revolution

Functions

(a)

3 marks

The function $f (x) = sec (x - \frac{π}{4})$ has its minimum value where $x = \frac{π}{4}$ , because $cos 0 = 1$ .
The ends of the interval.

f (\frac{π}{4}) = 1 f (0) = sec (- \frac{π}{4}) = 2, f (\frac{π}{2}) = sec \frac{π}{4} = 2 1 \leq f (x) \leq 2

(b)

4 marks

Use the volume formula.
Substitute the limits.

V = π \int_{0}^{π /2} sec^{2} (x - \frac{π}{4}) d x = π [tan (x - \frac{π}{4})]_{0}^{π /2} π [tan \frac{π}{4} - tan (- \frac{π}{4})] = π (1 + 1) V = 2 π

Question 8

8 marks

A point on a line and when its position vector is perpendicular to AB

Number and algebra

(a)

1 mark

Write $O P = O A + A P$ with $A B = b - a$ .

O P = a + λ (b - a) = (1 - λ) a + λ b

(b)

7 marks

If $O P ⊥ A B$ , then $O P \cdot A B = 0$ .
Expand.
Use $∣ a ∣ = 1$ , $∣ b ∣ = 2$ and $a \cdot b = \frac{1}{4}$ .
Collect the terms.

[(1 - λ) a + λ b] \cdot (b - a) = 0 (1 - λ) a \cdot b - (1 - λ) ∣ a ∣^{2} + λ ∣ b ∣^{2} - λ a \cdot b = 0 \frac{1 - λ}{4} - (1 - λ) + 4 λ - \frac{λ}{4} = 0 - \frac{3}{4} (1 - λ) + \frac{15}{4} λ = 0 \Rightarrow - 3 + 18 λ = 0 λ = \frac{1}{6}

Question 9

9 marks

A tangent identity and a trigonometric equation

Functions

(a)

6 marks

Use the subtraction formula with $tan \frac{π}{4} = 1$ .
Multiply the top and bottom by $sin θ - cos θ$ .
Use $sin^{2} θ + cos^{2} θ = 1$ , $2 sin θ cos θ = sin 2 θ$ and $cos^{2} θ - sin^{2} θ = cos 2 θ$ .

tan (θ - \frac{π}{4}) = \frac{tan θ - 1}{1 + tan θ} = \frac{sin θ - cos θ}{cos θ + sin θ} \frac{( sin θ - cos θ ) ^{2}}{( sin θ + cos θ ) ( sin θ - cos θ )} = \frac{sin ^{2} θ - 2 sin θ cos θ + cos ^{2} θ}{sin ^{2} θ - cos ^{2} θ} \frac{1 - sin 2 θ}{- cos 2 θ} = \frac{sin 2 θ - 1}{cos 2 θ}

(b)

3 marks

Put $2 θ = x$ , so $θ = \frac{x}{2}$ . By part (a) the equation is
Solve for the angle.
In the interval $0 \leq x \leq 2 π$ the only solution is

tan (\frac{x}{2} - \frac{π}{4}) = 3 \frac{x}{2} - \frac{π}{4} = \frac{π}{3} + nπ \Rightarrow x = \frac{7 π}{6} + 2 nπ x = \frac{7 π}{6}

Section B · Extended response

Questions 10–12 · 54 marks

Question 10

17 marks

A hollow cylinder: surface area, volume and the maximum volume

Calculus

(a)

3 marks

The outer curved surface (radius $4 r$ ) and the inner curved surface (radius $r$ ).
The two ends are rings with area $π (4 r)^{2} - π r^{2} = 15 π r^{2}$ each.

2 π (4 r) h + 2 π r h = 8 π r h + 2 π r h = 10 π r h 2 \times 15 π r^{2} = 30 π r^{2} S = 30 π r^{2} + 10 π r h

(b)

6 marks

Set $S = 240 π$ and solve for $r h$ .
The volume is the large cylinder minus the small one.
Substitute.

30 π r^{2} + 10 π r h = 240 π \Rightarrow 3 r^{2} + r h = 24 \Rightarrow r h = 24 - 3 r^{2} V = π (4 r)^{2} h - π r^{2} h = 15 π r^{2} h = 15 π r (r h) V = 15 π r (24 - 3 r^{2}) = 360 π r - 45 π r^{3}

(c)

2 marks

Differentiate.

\frac{d V}{d r} = 360 π - 135 π r^{2}

(d)

3 marks

At the maximum $\frac{d V}{d r} = 0$ .
Compare with $r = p \frac{2}{3}$ , for which $r^{2} = \frac{2 p ^{2}}{3}$ .

135 r^{2} = 360 \Rightarrow r^{2} = \frac{8}{3} \Rightarrow r = 2 \frac{2}{3} \frac{2 p ^{2}}{3} = \frac{8}{3} \Rightarrow p^{2} = 4 \Rightarrow p = 2

(e)

3 marks

Substitute into $V = π r (360 - 45 r^{2})$ .
So $q = 480$ .

V = π \cdot 2 \frac{2}{3} (360 - 45 \cdot \frac{8}{3}) = 2 π \frac{2}{3} (360 - 120) V = 480 π \frac{2}{3}

Question 11

17 marks

A curve with an exponential quotient: a limit, derivatives and a Maclaurin series

Calculus

(a)

2 marks

Divide the top and bottom by $e^{2 x}$ .

x \to \infty lim \frac{e ^{2 x} - 1}{e ^{2 x} + 1} = x \to \infty lim \frac{1 - e ^{- 2 x}}{1 + e ^{- 2 x}} = \frac{1 - 0}{1 + 0} = 1

(b)

6 marks

(i) Use the quotient rule with $u = e^{2 x} - 1$ and $v = e^{2 x} + 1$ .

(ii) Work out $1 - y^{2}$ using a common denominator.

The numerator is a difference of squares: $(a + b)^{2} - (a - b)^{2} = 4 ab$ .

\frac{d y}{d x} = \frac{2 e ^{2 x} ( e ^{2 x} + 1 ) - 2 e ^{2 x} ( e ^{2 x} - 1 )}{( e ^{2 x} + 1 ) ^{2}} = \frac{4 e ^{2 x}}{( e ^{2 x} + 1 ) ^{2}} 1 - y^{2} = \frac{( e ^{2 x} + 1 ) ^{2} - ( e ^{2 x} - 1 ) ^{2}}{( e ^{2 x} + 1 ) ^{2}} 1 - y^{2} = \frac{4 e ^{2 x}}{( e ^{2 x} + 1 ) ^{2}} = \frac{d y}{d x}

(c)

5 marks

(i) Differentiate $\frac{d y}{d x} = 1 - y^{2}$ implicitly.

(ii) Differentiate $\frac{d ^{2} y}{d x ^{2}} = 2 y^{3} - 2 y$ implicitly.

\frac{d ^{2} y}{d x ^{2}} = - 2 y \frac{d y}{d x} = - 2 y (1 - y^{2}) = 2 y^{3} - 2 y \frac{d ^{3} y}{d x ^{3}} = (6 y^{2} - 2) \frac{d y}{d x} = (6 y^{2} - 2) (1 - y^{2})

(d)

4 marks

At $x = 0$ : $y = 0$ .
Use the Maclaurin series $y \approx y (0) + y^{'} (0) x + \frac{y ^{''} ( 0 )}{2 !} x^{2} + \frac{y ^{'''} ( 0 )}{3 !} x^{3}$ .

y = 0, y^{'} = 1 - 0 = 1, y^{''} = 0, y^{'''} = (- 2) (1) = - 2 \frac{e ^{2 x} - 1}{e ^{2 x} + 1} \approx x - \frac{x ^{3}}{3}

Question 12

20 marks

Roots of z to the fourth equal 16i, a geometric sequence and midpoints

Number and algebra

(a)

6 marks

Write $16 i = 16 cis \frac{π}{2}$ . The fourth roots have modulus $1 6^{1/4} = 2$ and arguments $\frac{π /2 + 2 k π}{4} = \frac{π}{8} + \frac{k π}{2}$ .

z_{1} = 2 cis \frac{π}{8}, z_{2} = 2 cis \frac{5 π}{8} z_{3} = 2 cis \frac{9 π}{8}, z_{4} = 2 cis \frac{13 π}{8}

(b)

3 marks

The common ratio is $\frac{z _{2}}{z _{1}}$ . The modulus is 1 and the argument is $\frac{5 π}{8} - \frac{π}{8} = \frac{π}{2}$ .

r = cis \frac{π}{2} = i

(c)

3 marks

(d)

3 marks

The roots of $v^{4} = a + bi$ are $z_{k}^{*}$ , the conjugates, so $v^{4} = (z^{4})^{*} = (16 i)^{*} = - 16 i$ .

a = 0, b = - 16

(e)

5 marks

$A^{'}$ is the midpoint of $A B$ . The chord $A B$ subtends $9 0^{\circ}$ at the centre, so $∣ O A^{'} ∣ = 2 cos 4 5^{\circ} = 2$ , and its argument is the average $\frac{1}{2} (\frac{π}{8} + \frac{5 π}{8}) = \frac{3 π}{8}$ .
The four midpoints are $2 cis (\frac{3 π}{8} + \frac{k π}{2})$ . They are roots of $w^{p} = 2^{q}$ when the modulus matches and $p$ times the argument is a multiple of $2 π$ .
So $p$ must be a multiple of 16, and the least value is 16. The spacing of the roots is then $\frac{2 π}{16} = \frac{π}{8}$ , which includes the four points spaced by $\frac{π}{2}$ .

A^{'} = 2 cis \frac{3 π}{8} (2)^{p} = 2^{q} \Rightarrow p = 2 q p \cdot \frac{3 π}{8} = 2 mπ \Rightarrow p = \frac{16 m}{3} p = 16, q = 8

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 · 36%
Q4, Q10, Q11
Number and algebra38 marks · 35%
Q3, Q5, Q8, Q12
Functions21 marks · 19%
Q6, Q7, Q9
Statistics and probability6 marks · 5%
Q2
Geometry and trigonometry5 marks · 5%
Q1

Mentor insights

Traps to watch for

01
Parametrise · Q8
$A P = λ A B$ .
02
Subtraction formula · Q9
$tan (A - B) = \frac{tan A - tan B}{1 + tan A tan B}$ .
03
Surface area · Q10
Outer curved, inner curved and two ring-shaped ends.
04
Limit at infinity · Q11
Divide the top and bottom by the fastest growing term.
05
De Moivre · Q12
The roots of $z^{n} = r e^{i α}$ have modulus $r^{1/ n}$ and arguments $\frac{α + 2 k π}{n}$ .

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