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IB Maths AA HL · Paper 1

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ3

Session: November 2025
Marks: 110
Duration: 2 hours
Calculator: Not allowed
Questions: 11

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11

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.

Section A · Short response

Questions 1–8 · 56 marks

Question 1

5 marks

The first term and common difference from two conditions

Number and algebra

5 marks

The 7th term is 6, and the 6th and 12th terms add to 24.
From the second equation $u_{1} + 8 d = 12$ . Subtract the first equation.

u_{1} + 6 d = 6, (u_{1} + 5 d) + (u_{1} + 11 d) = 24 \Rightarrow 2 u_{1} + 16 d = 24 2 d = 6 \Rightarrow d = 3 u_{1} = 6 - 18 = - 12

Question 2

6 marks

Two events with a given conditional probability

Statistics and probability

(a)

2 marks

Use the multiplication rule with $P (B) = \frac{1}{3}$ and $P (A ∣ B) = \frac{1}{4}$ .

P (A \cap B) = \frac{1}{4} \times \frac{1}{3} = \frac{1}{12}

(b)

4 marks

The complement of $B$ has probability $P (B^{'}) = 1 - \frac{1}{3} = \frac{2}{3}$ .
The part of $A$ outside $B$ .
Divide.

P (A \cap B^{'}) = \frac{1}{2} - \frac{1}{12} = \frac{5}{12} P (A ∣ B^{'}) = \frac{5/12}{2/3} = \frac{5}{12} \times \frac{3}{2} = \frac{5}{8}

Question 3

8 marks

A rational graph through two points and the transformations from 1/x

Functions

(a)

3 marks

Substitute $P (- 10, 1)$ .
Substitute $Q (3, - 12)$ .
Subtract the first equation from the second.

1 = \frac{- 10 A + B}{- 14} \Rightarrow - 10 A + B = - 14 - 12 = \frac{3 A + B}{- 1} \Rightarrow 3 A + B = 12 13 A = 26 \Rightarrow A = 2, B = 12 - 6 = 6

(b)

5 marks

Divide $2 x + 6$ by $x - 4$ .
Start from $y = \frac{1}{x}$ . Apply a vertical stretch with scale factor 14, to get $y = \frac{14}{x}$ .
Translate 4 units to the right, to get $y = \frac{14}{x - 4}$ .
Translate 2 units up. The vertical asymptote is then $x = 4$ and the horizontal asymptote is $y = 2$ .

y = \frac{2 x + 6}{x - 4} = 2 + \frac{14}{x - 4} y = \frac{14}{x - 4} + 2

Question 4

5 marks

A logarithmic equation with bases 8 and 4

Functions

5 marks

Change both logarithms to base 2.
Combine, using $\frac{1}{2} lo g_{2} x = lo g_{2} x$ .
Solve for $x$ .

3 lo g_{8} (10 x) - lo g_{4} x = lo g_{2} (10 x) - \frac{1}{2} lo g_{2} x = 1 lo g_{2} \frac{10 x}{x} = 1 \Rightarrow 10 x = 2 x = \frac{1}{5} \Rightarrow x = \frac{1}{25}

Question 5

8 marks

Sketching f(|x|) and |f(x)| from a graph, and an inequality

Functions

(a)

4 marks

(i) Keep the part of the graph with $x \geq 0$ : it starts at about $(0, - 1.2)$ and rises through $(2, 0)$ . Reflect it in the $y$ -axis. The result is symmetric, with a cusp (a sharp point) at the $y$ -intercept, about $(0, - 1.2)$ , and $x$ -intercepts at $x = - 2$ and $x = 2$ .

(ii) Keep the parts of the graph on or above the $x$ -axis, and reflect the part below it. The $y$ -intercept becomes about $(0, 1.2)$ , the minimum point $(- 2.7, - 3)$ becomes a maximum at about $(- 2.7, 3)$ , and there are two cusps on the $x$ -axis: one near $x = - 4.6$ and one at $x = 2$ .

(b)

4 marks

Rewrite the inequality.
This holds when $f (x) = 0$ or $∣ f (x) ∣ \leq 1$ , which means $- 1 \leq f (x) \leq 1$ .
From the graph, $f (x)$ is between $- 1$ and 1 for the following ranges of $x$ (the zero at $x = 2$ lies inside the second range).

∣ f (x) ∣ - (f (x))^{2} \geq 0 \Rightarrow ∣ f (x) ∣ (1 - ∣ f (x) ∣) \geq 0 - 4.93 \leq x \leq - 4.27 and 0.292 \leq x \leq 5

Question 6

7 marks

A first order differential equation with an integrating factor

Calculus

7 marks

Divide by $x$ to get the standard form.
The integrating factor is $e^{3 l n x} = x^{3}$ . Multiply through.
Integrate, using the given result.
Use $y = \frac{8}{π ^{2}}$ when $x = \frac{π}{2}$ .
Divide by $x^{3}$ .

\frac{d y}{d x} + \frac{3}{x} y = \frac{cosec ^{2} x}{x ^{2}} x^{3} \frac{d y}{d x} + 3 x^{2} y = x cosec^{2} x \Rightarrow \frac{d}{d x} (x^{3} y) = x cosec^{2} x x^{3} y = ln ∣ sin x ∣ - x cot x + c \frac{π ^{3}}{8} \times \frac{8}{π ^{2}} = π = ln 1 - 0 + c \Rightarrow c = π y = \frac{ln ( sin x ) - x cot x + π}{x ^{3}}

Question 7

8 marks

A probability distribution with a parameter

Statistics and probability

(a)

2 marks

The three probabilities must each be non-negative.

0.6 - 2 a \geq 0, 3 a \geq 0, 0.4 - a \geq 0 a \leq 0.3, a \geq 0, a \leq 0.4 \Rightarrow 0 \leq a \leq 0.3

(b)

6 marks

With $a = 0.2$ the probabilities for $x = 1, 2, 3$ are $0.2$ , $0.6$ and $0.2$ .
Since $2 - X$ is $- X$ shifted, $Var (2 - X) = (- 1)^{2} Var (X) = Var (X)$ .
The mean is 2 by symmetry.
The mean of the squares.
The variance.

E (X) = 0.2 + 1.2 + 0.6 = 2 E (X^{2}) = 0.2 (1) + 0.6 (4) + 0.2 (9) = 4.4 Var (X) = 4.4 - 2^{2} = 0.4 Var (2 - X) = 0.4

Question 8

9 marks

The quickest route across a beach and along a promenade

Calculus

(a)

2 marks

Time on the sand is $\frac{400 sec θ}{0.8} = 500 sec θ$ . Time on the promenade is $\frac{1000 - 400 tan θ}{1.2}$ .

T = 500 sec θ + \frac{1000 - 400 tan θ}{1.2} = 500 sec θ + \frac{2500 - 1000 tan θ}{3}

(b)

3 marks

Differentiate, using $\frac{d}{d θ} sec θ = sec θ tan θ$ and $\frac{d}{d θ} tan θ = sec^{2} θ$ .

\frac{d T}{d θ} = 500 sec θ tan θ - \frac{1000}{3} sec^{2} θ

(c)

4 marks

Set the derivative to zero and divide by $sec θ$ .
Then $cos θ = 1 - \frac{4}{9} = \frac{5}{3}$ , so $tan θ = \frac{2}{5}$ .
In the right triangle $P R X$ , $P X = 400 tan θ$ .

500 tan θ = \frac{1000}{3} sec θ \Rightarrow \frac{sin θ}{cos θ} = \frac{2}{3 cos θ} \Rightarrow sin θ = \frac{2}{3} P X = 400 \times \frac{2}{5} = \frac{800}{5} = 1605 m

Section B · Extended response

Questions 9–11 · 54 marks

Question 9

15 marks

A quadratic with no real roots, its minimum, a normal line and an area

Functions
Calculus

(a)

2 marks

The discriminant of $\frac{1}{2} x^{2} + k x + 13$ is $k^{2} - 4 \cdot \frac{1}{2} \cdot 13 = k^{2} - 26$ . It must be negative.
The greatest positive integer is $k = 5$ .

k^{2} < 26 \Rightarrow k < 26 \approx 5.10

(b)

3 marks

(i) With $k = 5$ , $f (x) = \frac{1}{2} x^{2} + 5 x + 13$ . The axis is at $x = - \frac{5}{2 \times \frac{1}{2}}$ .

(ii) Substitute $x = - 5$ .

x = - 5 f (- 5) = \frac{25}{2} - 25 + 13 = \frac{1}{2} \Rightarrow (- 5, \frac{1}{2})

(c)

5 marks

Evaluate at $x = - 3$ .
The gradient of the normal is $- \frac{1}{2}$ .

f (- 3) = \frac{9}{2} - 15 + 13 = \frac{5}{2}, f^{'} (x) = x + 5 \Rightarrow f^{'} (- 3) = 2 y - \frac{5}{2} = - \frac{1}{2} (x + 3) \Rightarrow y = - \frac{1}{2} x + 1

(d)

5 marks

Between $x = - 3$ and $x = 0$ the curve is above the line $L$ .
Integrate.

Area = \int_{- 3}^{0} [(\frac{1}{2} x^{2} + 5 x + 13) - (- \frac{1}{2} x + 1)] d x = \int_{- 3}^{0} (\frac{1}{2} x^{2} + \frac{11}{2} x + 12) d x [\frac{x ^{3}}{6} + \frac{11 x ^{2}}{4} + 12 x]_{- 3}^{0} = 0 - (- \frac{9}{2} + \frac{99}{4} - 36) = \frac{63}{4}

Question 10

19 marks

Two lines in 3D: skew lines, a closest point and a plane

Number and algebra

(a)

1 mark

Use the point $P$ as the position vector and the given direction vector.

r = - 1 1 - 13 + λ 712

(b)

2 marks

The direction is $A B = 5 - 2 - 1$ .

s = 2 - 4 2 + μ 5 - 2 - 1

(c)

5 marks

The directions $(7, 1, 2)$ and $(5, - 2, - 1)$ are not multiples of each other, so the lines are not parallel.
Look for an intersection. Equate the first two components.
From the second, $λ = - 5 - 2 μ$ . Substitute in the first.
Check the third component.
These are different, so there is no common point. The lines are not parallel and do not intersect, so they are skew.

- 1 + 7 λ = 2 + 5 μ, 1 + λ = - 4 - 2 μ - 1 - 35 - 14 μ = 2 + 5 μ \Rightarrow μ = - 2, λ = - 1 - 13 + 2 (- 1) = - 15 but 2 - (- 2) = 4

(d)

4 marks

A point on $L_{2}$ is $N = (2 + 5 μ, - 4 - 2 μ, 2 - μ)$ .
Take the dot product with $A B$ .

P N = 3 + 5 μ - 5 - 2 μ 15 - μ P N \cdot A B = 5 (3 + 5 μ) - 2 (- 5 - 2 μ) - (15 - μ) = 15 + 25 μ + 10 + 4 μ - 15 + μ = 10 + 30 μ

(e)

3 marks

At the closest point $P N ⊥ A B$ , so the dot product is zero.
Substitute.

10 + 30 μ = 0 \Rightarrow μ = - \frac{1}{3} N = (2 - \frac{5}{3}, - 4 + \frac{2}{3}, 2 + \frac{1}{3}) = (\frac{1}{3}, - \frac{10}{3}, \frac{7}{3})

(f)

4 marks

Two vectors in the plane are $O P = (- 1, 1, - 13)$ and, scaling $O N$ by 3, $(1, - 10, 7)$ . Take their cross product.
The plane passes through the origin, so $δ = 0$ .
Dividing by 3 gives the equivalent $- 41 x - 2 y + 3 z = 0$ .

- 1 1 - 13 \times 1 - 10 7 = 7 - 130 - 13 + 7 10 - 1 = - 123 - 6 9 - 123 x - 6 y + 9 z = 0

Question 11

20 marks

De Moivre, the tangent identity and the exact value of tan(pi/8)

Number and algebra

(a)

6 marks

(i) The modulus is $1 + 3 = 2$ and the argument is $arctan 3 = \frac{π}{3}$ .

(ii) Use de Moivre's theorem on $w$ and its conjugate.

Add the two. The imaginary parts cancel.

w = 2 e^{iπ /3} (1 + i 3)^{8} = 2^{8} e^{8 π i /3}, (1 - i 3)^{8} = 2^{8} e^{- 8 π i /3} 256 \cdot 2 cos \frac{8 π}{3} = 512 cos \frac{2 π}{3} = 512 (- \frac{1}{2}) = - 256

(b)

3 marks

Since $1 - i tan θ$ is the conjugate of $1 + i tan θ$ , apply the given result to each.

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

(c)

4 marks

Put $z = i tan θ$ . The equation becomes $\frac{2 cos 4 θ}{cos ^{4} θ} = 0$ , so $cos 4 θ = 0$ .
The four roots.

4 θ = \frac{π}{2}, \frac{3 π}{2}, \frac{5 π}{2}, \frac{7 π}{2} \Rightarrow θ = \frac{π}{8}, \frac{3 π}{8}, \frac{5 π}{8}, \frac{7 π}{8} z = i tan \frac{k π}{8}, k = 1, 3, 5, 7

(d)

7 marks

Expand.
Set it to zero and solve for $z^{2}$ .
For $z = i tan \frac{π}{8}$ we have $z^{2} = - tan^{2} \frac{π}{8}$ .
Since $tan \frac{π}{8} < tan \frac{π}{4} = 1$ , we need $tan^{2} \frac{π}{8} < 1$ . The value $3 + 22$ is bigger than 1, so it is ruled out.

(1 + z)^{4} + (1 - z)^{4} = 2 + 12 z^{2} + 2 z^{4} z^{4} + 6 z^{2} + 1 = 0 \Rightarrow z^{2} = \frac{- 6 \pm 32}{2} = - 3 \pm 22 tan^{2} \frac{π}{8} = 3 \mp 22 tan^{2} \frac{π}{8} = 3 - 22 \Rightarrow tan \frac{π}{8} = 3 - 22

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 algebra44 marks · 40%
Q1, Q10, Q11
Functions26 marks · 24%
Q3, Q4, Q5, Q9
Calculus26 marks · 24%
Q6, Q8, Q9
Statistics and probability14 marks · 13%
Q2, Q7

Mentor insights

Traps to watch for

01
Two equations · Q3
Substitute each point into $y = \frac{A x + B}{x - 4}$ to get two linear equations.
02
Time = distance / speed · Q8
Time on the sand plus time on the promenade.
03
Discriminant · Q9
No real roots means $b^{2} - 4 a c < 0$ .
04
Skew lines · Q10
Not parallel and no common point.
05
Polar form · Q11
$1 + i 3 = 2 e^{iπ /3}$ .

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 1

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ3

Session: November 2025
Marks: 110
Duration: 2 hours
Calculator: Not allowed
Questions: 11

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11

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.

Section A · Short response

Questions 1–8 · 56 marks

Question 1

5 marks

The first term and common difference from two conditions

Number and algebra

5 marks

The 7th term is 6, and the 6th and 12th terms add to 24.
From the second equation $u_{1} + 8 d = 12$ . Subtract the first equation.

u_{1} + 6 d = 6, (u_{1} + 5 d) + (u_{1} + 11 d) = 24 \Rightarrow 2 u_{1} + 16 d = 24 2 d = 6 \Rightarrow d = 3 u_{1} = 6 - 18 = - 12

Question 2

6 marks

Two events with a given conditional probability

Statistics and probability

(a)

2 marks

Use the multiplication rule with $P (B) = \frac{1}{3}$ and $P (A ∣ B) = \frac{1}{4}$ .

P (A \cap B) = \frac{1}{4} \times \frac{1}{3} = \frac{1}{12}

(b)

4 marks

The complement of $B$ has probability $P (B^{'}) = 1 - \frac{1}{3} = \frac{2}{3}$ .
The part of $A$ outside $B$ .
Divide.

P (A \cap B^{'}) = \frac{1}{2} - \frac{1}{12} = \frac{5}{12} P (A ∣ B^{'}) = \frac{5/12}{2/3} = \frac{5}{12} \times \frac{3}{2} = \frac{5}{8}

Question 3

8 marks

A rational graph through two points and the transformations from 1/x

Functions

(a)

3 marks

Substitute $P (- 10, 1)$ .
Substitute $Q (3, - 12)$ .
Subtract the first equation from the second.

1 = \frac{- 10 A + B}{- 14} \Rightarrow - 10 A + B = - 14 - 12 = \frac{3 A + B}{- 1} \Rightarrow 3 A + B = 12 13 A = 26 \Rightarrow A = 2, B = 12 - 6 = 6

(b)

5 marks

Divide $2 x + 6$ by $x - 4$ .
Start from $y = \frac{1}{x}$ . Apply a vertical stretch with scale factor 14, to get $y = \frac{14}{x}$ .
Translate 4 units to the right, to get $y = \frac{14}{x - 4}$ .
Translate 2 units up. The vertical asymptote is then $x = 4$ and the horizontal asymptote is $y = 2$ .

y = \frac{2 x + 6}{x - 4} = 2 + \frac{14}{x - 4} y = \frac{14}{x - 4} + 2

Question 4

5 marks

A logarithmic equation with bases 8 and 4

Functions

5 marks

Change both logarithms to base 2.
Combine, using $\frac{1}{2} lo g_{2} x = lo g_{2} x$ .
Solve for $x$ .

3 lo g_{8} (10 x) - lo g_{4} x = lo g_{2} (10 x) - \frac{1}{2} lo g_{2} x = 1 lo g_{2} \frac{10 x}{x} = 1 \Rightarrow 10 x = 2 x = \frac{1}{5} \Rightarrow x = \frac{1}{25}

Question 5

8 marks

Sketching f(|x|) and |f(x)| from a graph, and an inequality

Functions

(a)

4 marks

(b)

4 marks

Rewrite the inequality.
This holds when $f (x) = 0$ or $∣ f (x) ∣ \leq 1$ , which means $- 1 \leq f (x) \leq 1$ .
From the graph, $f (x)$ is between $- 1$ and 1 for the following ranges of $x$ (the zero at $x = 2$ lies inside the second range).

∣ f (x) ∣ - (f (x))^{2} \geq 0 \Rightarrow ∣ f (x) ∣ (1 - ∣ f (x) ∣) \geq 0 - 4.93 \leq x \leq - 4.27 and 0.292 \leq x \leq 5

Question 6

7 marks

A first order differential equation with an integrating factor

Calculus

7 marks

Divide by $x$ to get the standard form.
The integrating factor is $e^{3 l n x} = x^{3}$ . Multiply through.
Integrate, using the given result.
Use $y = \frac{8}{π ^{2}}$ when $x = \frac{π}{2}$ .
Divide by $x^{3}$ .

\frac{d y}{d x} + \frac{3}{x} y = \frac{cosec ^{2} x}{x ^{2}} x^{3} \frac{d y}{d x} + 3 x^{2} y = x cosec^{2} x \Rightarrow \frac{d}{d x} (x^{3} y) = x cosec^{2} x x^{3} y = ln ∣ sin x ∣ - x cot x + c \frac{π ^{3}}{8} \times \frac{8}{π ^{2}} = π = ln 1 - 0 + c \Rightarrow c = π y = \frac{ln ( sin x ) - x cot x + π}{x ^{3}}

Question 7

8 marks

A probability distribution with a parameter

Statistics and probability

(a)

2 marks

The three probabilities must each be non-negative.

0.6 - 2 a \geq 0, 3 a \geq 0, 0.4 - a \geq 0 a \leq 0.3, a \geq 0, a \leq 0.4 \Rightarrow 0 \leq a \leq 0.3

(b)

6 marks

With $a = 0.2$ the probabilities for $x = 1, 2, 3$ are $0.2$ , $0.6$ and $0.2$ .
Since $2 - X$ is $- X$ shifted, $Var (2 - X) = (- 1)^{2} Var (X) = Var (X)$ .
The mean is 2 by symmetry.
The mean of the squares.
The variance.

E (X) = 0.2 + 1.2 + 0.6 = 2 E (X^{2}) = 0.2 (1) + 0.6 (4) + 0.2 (9) = 4.4 Var (X) = 4.4 - 2^{2} = 0.4 Var (2 - X) = 0.4

Question 8

9 marks

The quickest route across a beach and along a promenade

Calculus

(a)

2 marks

Time on the sand is $\frac{400 sec θ}{0.8} = 500 sec θ$ . Time on the promenade is $\frac{1000 - 400 tan θ}{1.2}$ .

T = 500 sec θ + \frac{1000 - 400 tan θ}{1.2} = 500 sec θ + \frac{2500 - 1000 tan θ}{3}

(b)

3 marks

Differentiate, using $\frac{d}{d θ} sec θ = sec θ tan θ$ and $\frac{d}{d θ} tan θ = sec^{2} θ$ .

\frac{d T}{d θ} = 500 sec θ tan θ - \frac{1000}{3} sec^{2} θ

(c)

4 marks

Set the derivative to zero and divide by $sec θ$ .
Then $cos θ = 1 - \frac{4}{9} = \frac{5}{3}$ , so $tan θ = \frac{2}{5}$ .
In the right triangle $P R X$ , $P X = 400 tan θ$ .

500 tan θ = \frac{1000}{3} sec θ \Rightarrow \frac{sin θ}{cos θ} = \frac{2}{3 cos θ} \Rightarrow sin θ = \frac{2}{3} P X = 400 \times \frac{2}{5} = \frac{800}{5} = 1605 m

Section B · Extended response

Questions 9–11 · 54 marks

Question 9

15 marks

A quadratic with no real roots, its minimum, a normal line and an area

Functions
Calculus

(a)

2 marks

The discriminant of $\frac{1}{2} x^{2} + k x + 13$ is $k^{2} - 4 \cdot \frac{1}{2} \cdot 13 = k^{2} - 26$ . It must be negative.
The greatest positive integer is $k = 5$ .

k^{2} < 26 \Rightarrow k < 26 \approx 5.10

(b)

3 marks

(i) With $k = 5$ , $f (x) = \frac{1}{2} x^{2} + 5 x + 13$ . The axis is at $x = - \frac{5}{2 \times \frac{1}{2}}$ .

(ii) Substitute $x = - 5$ .

x = - 5 f (- 5) = \frac{25}{2} - 25 + 13 = \frac{1}{2} \Rightarrow (- 5, \frac{1}{2})

(c)

5 marks

Evaluate at $x = - 3$ .
The gradient of the normal is $- \frac{1}{2}$ .

f (- 3) = \frac{9}{2} - 15 + 13 = \frac{5}{2}, f^{'} (x) = x + 5 \Rightarrow f^{'} (- 3) = 2 y - \frac{5}{2} = - \frac{1}{2} (x + 3) \Rightarrow y = - \frac{1}{2} x + 1

(d)

5 marks

Between $x = - 3$ and $x = 0$ the curve is above the line $L$ .
Integrate.

Area = \int_{- 3}^{0} [(\frac{1}{2} x^{2} + 5 x + 13) - (- \frac{1}{2} x + 1)] d x = \int_{- 3}^{0} (\frac{1}{2} x^{2} + \frac{11}{2} x + 12) d x [\frac{x ^{3}}{6} + \frac{11 x ^{2}}{4} + 12 x]_{- 3}^{0} = 0 - (- \frac{9}{2} + \frac{99}{4} - 36) = \frac{63}{4}

Question 10

19 marks

Two lines in 3D: skew lines, a closest point and a plane

Number and algebra

(a)

1 mark

Use the point $P$ as the position vector and the given direction vector.

r = - 1 1 - 13 + λ 712

(b)

2 marks

The direction is $A B = 5 - 2 - 1$ .

s = 2 - 4 2 + μ 5 - 2 - 1

(c)

5 marks

The directions $(7, 1, 2)$ and $(5, - 2, - 1)$ are not multiples of each other, so the lines are not parallel.
Look for an intersection. Equate the first two components.
From the second, $λ = - 5 - 2 μ$ . Substitute in the first.
Check the third component.
These are different, so there is no common point. The lines are not parallel and do not intersect, so they are skew.

- 1 + 7 λ = 2 + 5 μ, 1 + λ = - 4 - 2 μ - 1 - 35 - 14 μ = 2 + 5 μ \Rightarrow μ = - 2, λ = - 1 - 13 + 2 (- 1) = - 15 but 2 - (- 2) = 4

(d)

4 marks

A point on $L_{2}$ is $N = (2 + 5 μ, - 4 - 2 μ, 2 - μ)$ .
Take the dot product with $A B$ .

P N = 3 + 5 μ - 5 - 2 μ 15 - μ P N \cdot A B = 5 (3 + 5 μ) - 2 (- 5 - 2 μ) - (15 - μ) = 15 + 25 μ + 10 + 4 μ - 15 + μ = 10 + 30 μ

(e)

3 marks

At the closest point $P N ⊥ A B$ , so the dot product is zero.
Substitute.

10 + 30 μ = 0 \Rightarrow μ = - \frac{1}{3} N = (2 - \frac{5}{3}, - 4 + \frac{2}{3}, 2 + \frac{1}{3}) = (\frac{1}{3}, - \frac{10}{3}, \frac{7}{3})

(f)

4 marks

Two vectors in the plane are $O P = (- 1, 1, - 13)$ and, scaling $O N$ by 3, $(1, - 10, 7)$ . Take their cross product.
The plane passes through the origin, so $δ = 0$ .
Dividing by 3 gives the equivalent $- 41 x - 2 y + 3 z = 0$ .

- 1 1 - 13 \times 1 - 10 7 = 7 - 130 - 13 + 7 10 - 1 = - 123 - 6 9 - 123 x - 6 y + 9 z = 0

Question 11

20 marks

De Moivre, the tangent identity and the exact value of tan(pi/8)

Number and algebra

(a)

6 marks

(i) The modulus is $1 + 3 = 2$ and the argument is $arctan 3 = \frac{π}{3}$ .

(ii) Use de Moivre's theorem on $w$ and its conjugate.

Add the two. The imaginary parts cancel.

w = 2 e^{iπ /3} (1 + i 3)^{8} = 2^{8} e^{8 π i /3}, (1 - i 3)^{8} = 2^{8} e^{- 8 π i /3} 256 \cdot 2 cos \frac{8 π}{3} = 512 cos \frac{2 π}{3} = 512 (- \frac{1}{2}) = - 256

(b)

3 marks

Since $1 - i tan θ$ is the conjugate of $1 + i tan θ$ , apply the given result to each.

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

(c)

4 marks

Put $z = i tan θ$ . The equation becomes $\frac{2 cos 4 θ}{cos ^{4} θ} = 0$ , so $cos 4 θ = 0$ .
The four roots.

4 θ = \frac{π}{2}, \frac{3 π}{2}, \frac{5 π}{2}, \frac{7 π}{2} \Rightarrow θ = \frac{π}{8}, \frac{3 π}{8}, \frac{5 π}{8}, \frac{7 π}{8} z = i tan \frac{k π}{8}, k = 1, 3, 5, 7

(d)

7 marks

Expand.
Set it to zero and solve for $z^{2}$ .
For $z = i tan \frac{π}{8}$ we have $z^{2} = - tan^{2} \frac{π}{8}$ .
Since $tan \frac{π}{8} < tan \frac{π}{4} = 1$ , we need $tan^{2} \frac{π}{8} < 1$ . The value $3 + 22$ is bigger than 1, so it is ruled out.

(1 + z)^{4} + (1 - z)^{4} = 2 + 12 z^{2} + 2 z^{4} z^{4} + 6 z^{2} + 1 = 0 \Rightarrow z^{2} = \frac{- 6 \pm 32}{2} = - 3 \pm 22 tan^{2} \frac{π}{8} = 3 \mp 22 tan^{2} \frac{π}{8} = 3 - 22 \Rightarrow tan \frac{π}{8} = 3 - 22

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 algebra44 marks · 40%
Q1, Q10, Q11
Functions26 marks · 24%
Q3, Q4, Q5, Q9
Calculus26 marks · 24%
Q6, Q8, Q9
Statistics and probability14 marks · 13%
Q2, Q7

Mentor insights

Traps to watch for

01
Two equations · Q3
Substitute each point into $y = \frac{A x + B}{x - 4}$ to get two linear equations.
02
Time = distance / speed · Q8
Time on the sand plus time on the promenade.
03
Discriminant · Q9
No real roots means $b^{2} - 4 a c < 0$ .
04
Skew lines · Q10
Not parallel and no common point.
05
Polar form · Q11
$1 + i 3 = 2 e^{iπ /3}$ .

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