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IB Maths AA HL · Paper 1

Mathematics: Analysis & Approaches, Higher Level

Session: November 2022
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 2022 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

4 marks

Derivative of an exponential with a squared exponent

Calculus

4 marks

Differentiate using the chain rule.
Substitute $x = - 1$ .

g^{'} (x) = 2 x e^{x^{2} + 1} g^{'} (- 1) = 2 (- 1) e^{2} = - 2 e^{2}

Question 2

7 marks

A circle on a diameter and the volume of a cone

Geometry and trigonometry

(a)

4 marks

The centre is the midpoint of the diameter AB.
The radius is half of AB.

(\frac{1 + ( - 3 )}{2}, \frac{4 + 2}{2}, \frac{0 + ( - 4 )}{2}) = (- 1, 3, - 2) A B = 4^{2} + 2^{2} + 4^{2} = 6 \Rightarrow r = 3

(b)

3 marks

The height is the distance from the centre $(- 1, 3, - 2)$ to $V (- 1, - 1, 0)$ .
Volume of a cone:

h = 0^{2} + 4^{2} + 2^{2} = 20 = 25 V = \frac{1}{3} π (3)^{2} (25) = 65 π

Question 3

5 marks

A right-angled triangle with an algebraic identity

Number and algebra

(a)

3 marks

Start from the left-hand side and expand the bracket.
Put it over a common denominator.

LHS = a^{2} + (\frac{a ^{2} - 1}{2})^{2} = a^{2} + \frac{a ^{4} - 2 a ^{2} + 1}{4} \frac{4 a ^{2} + a ^{4} - 2 a ^{2} + 1}{4} = \frac{a ^{4} + 2 a ^{2} + 1}{4} = (\frac{a ^{2} + 1}{2})^{2} = RHS

(b)

2 marks

The two shorter sides are $a$ and $\frac{a ^{2} - 1}{2}$ , because the longest side is $\frac{a ^{2} + 1}{2}$ .

Area = \frac{1}{2} \cdot a \cdot \frac{a ^{2} - 1}{2} = \frac{a ( a ^{2} - 1 )}{4}

Question 4

5 marks

Recovering f from f' = 6x/(x^2 + 1)

Calculus

5 marks

Let $u = x^{2} + 1$ , so $d u = 2 x d x$ and $6 x d x = 3 d u$ .
Use $f (1) = 5$ .

f (x) = \int \frac{6 x}{x ^{2} + 1} d x = 3 ln (x^{2} + 1) + c 5 = 3 ln 2 + c \Rightarrow c = 5 - 3 ln 2 f (x) = 3 ln (x^{2} + 1) + 5 - 3 ln 2

Question 5

7 marks

A quartic with a complex root: product and sum of roots

Number and algebra

(a)

4 marks

The four roots are $3 + i$ , $3 - i$ , $a$ and $a^{2}$ , and their product is the constant term 80.

(3 + i) (3 - i) \cdot a \cdot a^{2} = 10 a^{3} = 80 a^{3} = 8 \Rightarrow a = 2

(b)

3 marks

The roots are $3 + i$ , $3 - i$ , $2$ and $4$ . Their sum is $- p$ .

3 + i + 3 - i + 2 + 4 = 12 = - p \Rightarrow p = - 12

Mentor tip

Check by multiplying out $(z^{2} - 6 z + 10) (z^{2} - 6 z + 8)$ : the $z^{2}$ coefficient is $54$ .

Question 6

6 marks

Bounds on P(A and B) for two events

Statistics and probability

(a)

1 mark

For independent events the probability of both is the product.

P (A \cap B) = 0.3 \times 0.8 = 0.24

(b)

3 marks

Rearrange the addition rule.
The union cannot be more than 1, so
The minimum is 0.1.

P (A \cap B) = P (A) + P (B) - P (A \cup B) = 1.1 - P (A \cup B) P (A \cap B) \geq 1.1 - 1 = 0.1

(c)

2 marks

The intersection cannot be larger than the smaller event. If $A$ is entirely inside $B$ , then $A \cap B = A$ .

max P (A \cap B) = P (A) = 0.3

Question 7

7 marks

A curve with no local maximum or minimum for x > 0

Number and algebra

7 marks

Expand first: $x^{2} y^{2} + y^{4} = 4 x^{2}$ . Differentiate with respect to $x$ .
Collect the $\frac{d y}{d x}$ terms.
A local maximum or minimum needs $\frac{d y}{d x} = 0$ , so $x (4 - y^{2}) = 0$ . With $x > 0$ this means $y^{2} = 4$ , so $y = \pm 2$ .
That is excluded by $- 2 < y < 2$ . So there are no local maximum or minimum points for $x > 0$ .

2 x y^{2} + 2 x^{2} y \frac{d y}{d x} + 4 y^{3} \frac{d y}{d x} = 8 x \frac{d y}{d x} = \frac{8 x - 2 x y ^{2}}{2 y ( x ^{2} + 2 y ^{2} )} = \frac{x ( 4 - y ^{2} )}{y ( x ^{2} + 2 y ^{2} )}

Question 8

5 marks

When does the inverse of cos(x - k) exist?

Functions

(a)

2 marks

When $k = \frac{π}{2}$ , $f (x) = cos (x - \frac{π}{2}) = sin x$ , which increases from $x = 0$ to its maximum at $x = \frac{π}{2}$ .

a = \frac{π}{2}

(b)

1 mark

When $k = π$ , $f (x) = cos (x - π) = - cos x$ , which increases until $x = π$ .

a = π

(c)

2 marks

The turning points of $cos (x - k)$ occur when $x - k$ is a multiple of $π$ . For $π < k < 2 π$ the first one for $x \geq 0$ is $x - k = - π$ .
At $x = 0$ the argument is $- k$ , between $- 2 π$ and $- π$ , where cosine is decreasing. It reaches its minimum at $x = k - π$ (which lies between 0 and $π$ ).

a = k - π

Question 9

10 marks

A homogeneous differential equation and its zero-gradient points

Calculus

(a)

8 marks

Put $y = v x$ into $\frac{d y}{d x} = \frac{y ^{2} - 2 x ^{2}}{x y}$ .
Separate the variables.
Use $y = 2$ when $x = 1$ .

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

(b)

2 marks

Zero gradient means the numerator of $\frac{y ^{2} - 2 x ^{2}}{x y}$ is 0.

y^{2} = 2 x^{2} \Rightarrow y = \pm 2 x m = \pm 2

Section B · Extended response

Questions 10–12 · 54 marks

Question 10

20 marks

f(x) = cos^2 x - 3 sin^2 x: roots, turning points and an inequality

Functions
Calculus

(a)

5 marks

Set $f (x) = 0$ .

cos^{2} x = 3 sin^{2} x \Rightarrow tan^{2} x = \frac{1}{3} \Rightarrow tan x = \pm \frac{1}{3} x = \frac{π}{6}, \frac{5 π}{6}

(b)

7 marks

Differentiate with the chain rule.
Set $f^{'} (x) = 0$ , so $sin 2 x = 0$ for $0 \leq x \leq π$ .
Substitute into $f$ .

f^{'} (x) = - 2 cos x sin x - 6 sin x cos x = - 8 sin x cos x = - 4 sin 2 x x = 0, \frac{π}{2}, π (0, 1), (\frac{π}{2}, - 3), (π, 1)

(c)

4 marks

Reflect the part of $y = f (x)$ below the axis in the axis.
Sketch: starts at $(0, 1)$ , falls to a sharp point at $(\frac{π}{6}, 0)$ , rises to a smooth maximum at $(\frac{π}{2}, 3)$ , falls to a sharp point at $(\frac{5 π}{6}, 0)$ and rises to $(π, 1)$ . The $y$ -axis intercept is $(0, 1)$ .

(d)

4 marks

Write $f (x) = 2 cos 2 x - 1$ , since $cos^{2} x - 3 sin^{2} x = \frac{1 + cos 2 x}{2} - \frac{3 ( 1 - cos 2 x )}{2}$ .
$∣ f (x) ∣ > 1$ means $f (x) > 1$ or $f (x) < - 1$ . The first needs $cos 2 x > 1$ , which never happens. The second:
This matches the sketch: $∣ f ∣$ is above the line $y = 1$ between the two points where it crosses 1.

2 cos 2 x - 1 < - 1 \Rightarrow cos 2 x < 0 \Rightarrow \frac{π}{4} < x < \frac{3 π}{4}

Question 11

16 marks

Three-digit codes and a cubic with a given quadratic factor

Number and algebra

(a)

4 marks

With repeats: $5 \times 5 \times 5$ . Without repeats: $5 \times 4 \times 3$ .

125 and 60

(b)

12 marks

The factor $(x + 1) (x + 2)$ means $P (- 1) = 0$ and $P (- 2) = 0$ .
Subtract the first from the second to eliminate $c$ .
Now test the values of $a$ from 1 to 5. The value $b$ must also be between 1 and 5: $a = 1, 2$ give $b \leq - 1$ , and $a = 5$ gives $b = 8$ .
With $a = 3$ : $b = 2$ and $c = 1 - a + b = 0$ , which is not allowed. With $a = 4$ : $b = 5$ and $c = 1 - 4 + 5 = 2$ , all different and in range.
So $P (x) = x^{3} + 4 x^{2} + 5 x + 2$ . Divide by $(x + 1) (x + 2) = x^{2} + 3 x + 2$ to get the third factor $x + 1$ .
Sketch: a cubic with positive leading coefficient. It crosses the $x$ -axis at $(- 2, 0)$ and touches it at $(- 1, 0)$ , where it has a local minimum, and it crosses the $y$ -axis at $(0, 2)$ . There is a local maximum between $x = - 2$ and $x = - 1$ .

- 1 + a - b + c = 0, - 8 + 4 a - 2 b + c = 0 - 7 + 3 a - b = 0 \Rightarrow b = 3 a - 7 a = 4, b = 5, c = 2 P (x) = (x + 1)^{2} (x + 2)

Question 12

18 marks

The argument of a product of complex numbers, proved by induction

Number and algebra

(a)

3 marks

$z_{0} = 1 + i$ .
For general $n$ , the real part is $n^{2} + n + 1$ and the imaginary part is 1.

ar g z_{0} = arctan 1 = \frac{π}{4} ar g z_{n} = arctan (\frac{1}{n ^{2} + n + 1})

(b)

5 marks

Use $tan (A + B) = \frac{tan A + tan B}{1 - tan A tan B}$ with $A = arctan a$ and $B = arctan b$ .
Since $a, b > 0$ and $ab < 1$ , the sum is an angle between 0 and $\frac{π}{2}$ , so
For $w_{1} = z_{0} z_{1}$ we add the arguments.

tan (arctan a + arctan b) = \frac{a + b}{1 - ab} arctan a + arctan b = arctan (\frac{a + b}{1 - ab}) ar g w_{1} = arctan 1 + arctan \frac{1}{3} = arctan (\frac{1 + \frac{1}{3}}{1 - \frac{1}{3}}) = arctan 2

(c)

10 marks

Base case $n = 0$ : $ar g w_{0} = ar g z_{0} = \frac{π}{4} = arctan 1 = arctan (0 + 1)$ , so it holds.
Assume it holds for $n = k$ : $ar g w_{k} = arctan (k + 1)$ .
Then $w_{k + 1} = w_{k} z_{k + 1}$ , so the arguments add.
The product of the two numbers is $\frac{k + 1}{k ^{2} + 3 k + 3} < 1$ , so the formula from (b) applies.
Expand the numerator: $(k + 1) (k^{2} + 3 k + 3) + 1 = k^{3} + 4 k^{2} + 6 k + 4 = (k + 2) (k^{2} + 2 k + 2)$ .
The statement is true for $n = 0$ and, if true for $n = k$ , it is true for $n = k + 1$ . By mathematical induction it is true for all $n \in N$ .

ar g w_{k + 1} = arctan (k + 1) + arctan (\frac{1}{( k + 1 ) ^{2} + ( k + 1 ) + 1}) = arctan (k + 1) + arctan (\frac{1}{k ^{2} + 3 k + 3}) arctan (\frac{( k + 1 ) + \frac{1}{k ^{2} + 3 k + 3}}{1 - \frac{k + 1}{k ^{2} + 3 k + 3}}) = arctan (\frac{( k + 1 ) ( k ^{2} + 3 k + 3 ) + 1}{k ^{2} + 2 k + 2}) ar g w_{k + 1} = arctan (k + 2) = arctan ((k + 1) + 1)

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 algebra53 marks · 48%
Q3, Q5, Q7, Q11, Q12
Calculus26 marks · 24%
Q1, Q4, Q9, Q10
Functions18 marks · 16%
Q8, Q10
Geometry and trigonometry7 marks · 6%
Q2
Statistics and probability6 marks · 5%
Q6

Mentor insights

Traps to watch for

01
Height of the cone · Q2
The cone's height is the distance from the centre of the base to the vertex, not the distance from A to V.
02
Substitution y = vx · Q9
Differentiate with the product rule: $\frac{d y}{d x} = v + x \frac{d v}{d x}$ .
03
Include both roots in the domain · Q10
$tan^{2} x = \frac{1}{3}$ gives $tan x = \pm \frac{1}{3}$ , so there are two solutions in $[0, π]$ .
04
Repeats allowed or not · Q11
With repeats each digit has 5 choices. Without repeats there are 5, then 4, then 3.
05
Base case at n = 0 · Q12
The statement is claimed for $n \in N$ starting from 0, so check $n = 0$ first.

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Worked solutions

IB Maths AA HL · Paper 1

Mathematics: Analysis & Approaches, Higher Level

Session: November 2022
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 2022 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

4 marks

Derivative of an exponential with a squared exponent

Calculus

4 marks

Differentiate using the chain rule.
Substitute $x = - 1$ .

g^{'} (x) = 2 x e^{x^{2} + 1} g^{'} (- 1) = 2 (- 1) e^{2} = - 2 e^{2}

Question 2

7 marks

A circle on a diameter and the volume of a cone

Geometry and trigonometry

(a)

4 marks

The centre is the midpoint of the diameter AB.
The radius is half of AB.

(\frac{1 + ( - 3 )}{2}, \frac{4 + 2}{2}, \frac{0 + ( - 4 )}{2}) = (- 1, 3, - 2) A B = 4^{2} + 2^{2} + 4^{2} = 6 \Rightarrow r = 3

(b)

3 marks

The height is the distance from the centre $(- 1, 3, - 2)$ to $V (- 1, - 1, 0)$ .
Volume of a cone:

h = 0^{2} + 4^{2} + 2^{2} = 20 = 25 V = \frac{1}{3} π (3)^{2} (25) = 65 π

Question 3

5 marks

A right-angled triangle with an algebraic identity

Number and algebra

(a)

3 marks

Start from the left-hand side and expand the bracket.
Put it over a common denominator.

LHS = a^{2} + (\frac{a ^{2} - 1}{2})^{2} = a^{2} + \frac{a ^{4} - 2 a ^{2} + 1}{4} \frac{4 a ^{2} + a ^{4} - 2 a ^{2} + 1}{4} = \frac{a ^{4} + 2 a ^{2} + 1}{4} = (\frac{a ^{2} + 1}{2})^{2} = RHS

(b)

2 marks

The two shorter sides are $a$ and $\frac{a ^{2} - 1}{2}$ , because the longest side is $\frac{a ^{2} + 1}{2}$ .

Area = \frac{1}{2} \cdot a \cdot \frac{a ^{2} - 1}{2} = \frac{a ( a ^{2} - 1 )}{4}

Question 4

5 marks

Recovering f from f' = 6x/(x^2 + 1)

Calculus

5 marks

Let $u = x^{2} + 1$ , so $d u = 2 x d x$ and $6 x d x = 3 d u$ .
Use $f (1) = 5$ .

f (x) = \int \frac{6 x}{x ^{2} + 1} d x = 3 ln (x^{2} + 1) + c 5 = 3 ln 2 + c \Rightarrow c = 5 - 3 ln 2 f (x) = 3 ln (x^{2} + 1) + 5 - 3 ln 2

Question 5

7 marks

A quartic with a complex root: product and sum of roots

Number and algebra

(a)

4 marks

The four roots are $3 + i$ , $3 - i$ , $a$ and $a^{2}$ , and their product is the constant term 80.

(3 + i) (3 - i) \cdot a \cdot a^{2} = 10 a^{3} = 80 a^{3} = 8 \Rightarrow a = 2

(b)

3 marks

The roots are $3 + i$ , $3 - i$ , $2$ and $4$ . Their sum is $- p$ .

3 + i + 3 - i + 2 + 4 = 12 = - p \Rightarrow p = - 12

Mentor tip

Check by multiplying out $(z^{2} - 6 z + 10) (z^{2} - 6 z + 8)$ : the $z^{2}$ coefficient is $54$ .

Question 6

6 marks

Bounds on P(A and B) for two events

Statistics and probability

(a)

1 mark

For independent events the probability of both is the product.

P (A \cap B) = 0.3 \times 0.8 = 0.24

(b)

3 marks

Rearrange the addition rule.
The union cannot be more than 1, so
The minimum is 0.1.

P (A \cap B) = P (A) + P (B) - P (A \cup B) = 1.1 - P (A \cup B) P (A \cap B) \geq 1.1 - 1 = 0.1

(c)

2 marks

The intersection cannot be larger than the smaller event. If $A$ is entirely inside $B$ , then $A \cap B = A$ .

max P (A \cap B) = P (A) = 0.3

Question 7

7 marks

A curve with no local maximum or minimum for x > 0

Number and algebra

7 marks

Expand first: $x^{2} y^{2} + y^{4} = 4 x^{2}$ . Differentiate with respect to $x$ .
Collect the $\frac{d y}{d x}$ terms.
A local maximum or minimum needs $\frac{d y}{d x} = 0$ , so $x (4 - y^{2}) = 0$ . With $x > 0$ this means $y^{2} = 4$ , so $y = \pm 2$ .
That is excluded by $- 2 < y < 2$ . So there are no local maximum or minimum points for $x > 0$ .

2 x y^{2} + 2 x^{2} y \frac{d y}{d x} + 4 y^{3} \frac{d y}{d x} = 8 x \frac{d y}{d x} = \frac{8 x - 2 x y ^{2}}{2 y ( x ^{2} + 2 y ^{2} )} = \frac{x ( 4 - y ^{2} )}{y ( x ^{2} + 2 y ^{2} )}

Question 8

5 marks

When does the inverse of cos(x - k) exist?

Functions

(a)

2 marks

When $k = \frac{π}{2}$ , $f (x) = cos (x - \frac{π}{2}) = sin x$ , which increases from $x = 0$ to its maximum at $x = \frac{π}{2}$ .

a = \frac{π}{2}

(b)

1 mark

When $k = π$ , $f (x) = cos (x - π) = - cos x$ , which increases until $x = π$ .

a = π

(c)

2 marks

The turning points of $cos (x - k)$ occur when $x - k$ is a multiple of $π$ . For $π < k < 2 π$ the first one for $x \geq 0$ is $x - k = - π$ .
At $x = 0$ the argument is $- k$ , between $- 2 π$ and $- π$ , where cosine is decreasing. It reaches its minimum at $x = k - π$ (which lies between 0 and $π$ ).

a = k - π

Question 9

10 marks

A homogeneous differential equation and its zero-gradient points

Calculus

(a)

8 marks

Put $y = v x$ into $\frac{d y}{d x} = \frac{y ^{2} - 2 x ^{2}}{x y}$ .
Separate the variables.
Use $y = 2$ when $x = 1$ .

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

(b)

2 marks

Zero gradient means the numerator of $\frac{y ^{2} - 2 x ^{2}}{x y}$ is 0.

y^{2} = 2 x^{2} \Rightarrow y = \pm 2 x m = \pm 2

Section B · Extended response

Questions 10–12 · 54 marks

Question 10

20 marks

f(x) = cos^2 x - 3 sin^2 x: roots, turning points and an inequality

Functions
Calculus

(a)

5 marks

Set $f (x) = 0$ .

cos^{2} x = 3 sin^{2} x \Rightarrow tan^{2} x = \frac{1}{3} \Rightarrow tan x = \pm \frac{1}{3} x = \frac{π}{6}, \frac{5 π}{6}

(b)

7 marks

Differentiate with the chain rule.
Set $f^{'} (x) = 0$ , so $sin 2 x = 0$ for $0 \leq x \leq π$ .
Substitute into $f$ .

f^{'} (x) = - 2 cos x sin x - 6 sin x cos x = - 8 sin x cos x = - 4 sin 2 x x = 0, \frac{π}{2}, π (0, 1), (\frac{π}{2}, - 3), (π, 1)

(c)

4 marks

Reflect the part of $y = f (x)$ below the axis in the axis.
Sketch: starts at $(0, 1)$ , falls to a sharp point at $(\frac{π}{6}, 0)$ , rises to a smooth maximum at $(\frac{π}{2}, 3)$ , falls to a sharp point at $(\frac{5 π}{6}, 0)$ and rises to $(π, 1)$ . The $y$ -axis intercept is $(0, 1)$ .

(d)

4 marks

Write $f (x) = 2 cos 2 x - 1$ , since $cos^{2} x - 3 sin^{2} x = \frac{1 + cos 2 x}{2} - \frac{3 ( 1 - cos 2 x )}{2}$ .
$∣ f (x) ∣ > 1$ means $f (x) > 1$ or $f (x) < - 1$ . The first needs $cos 2 x > 1$ , which never happens. The second:
This matches the sketch: $∣ f ∣$ is above the line $y = 1$ between the two points where it crosses 1.

2 cos 2 x - 1 < - 1 \Rightarrow cos 2 x < 0 \Rightarrow \frac{π}{4} < x < \frac{3 π}{4}

Question 11

16 marks

Three-digit codes and a cubic with a given quadratic factor

Number and algebra

(a)

4 marks

With repeats: $5 \times 5 \times 5$ . Without repeats: $5 \times 4 \times 3$ .

125 and 60

(b)

12 marks

The factor $(x + 1) (x + 2)$ means $P (- 1) = 0$ and $P (- 2) = 0$ .
Subtract the first from the second to eliminate $c$ .
Now test the values of $a$ from 1 to 5. The value $b$ must also be between 1 and 5: $a = 1, 2$ give $b \leq - 1$ , and $a = 5$ gives $b = 8$ .
With $a = 3$ : $b = 2$ and $c = 1 - a + b = 0$ , which is not allowed. With $a = 4$ : $b = 5$ and $c = 1 - 4 + 5 = 2$ , all different and in range.
So $P (x) = x^{3} + 4 x^{2} + 5 x + 2$ . Divide by $(x + 1) (x + 2) = x^{2} + 3 x + 2$ to get the third factor $x + 1$ .
Sketch: a cubic with positive leading coefficient. It crosses the $x$ -axis at $(- 2, 0)$ and touches it at $(- 1, 0)$ , where it has a local minimum, and it crosses the $y$ -axis at $(0, 2)$ . There is a local maximum between $x = - 2$ and $x = - 1$ .

- 1 + a - b + c = 0, - 8 + 4 a - 2 b + c = 0 - 7 + 3 a - b = 0 \Rightarrow b = 3 a - 7 a = 4, b = 5, c = 2 P (x) = (x + 1)^{2} (x + 2)

Question 12

18 marks

The argument of a product of complex numbers, proved by induction

Number and algebra

(a)

3 marks

$z_{0} = 1 + i$ .
For general $n$ , the real part is $n^{2} + n + 1$ and the imaginary part is 1.

ar g z_{0} = arctan 1 = \frac{π}{4} ar g z_{n} = arctan (\frac{1}{n ^{2} + n + 1})

(b)

5 marks

Use $tan (A + B) = \frac{tan A + tan B}{1 - tan A tan B}$ with $A = arctan a$ and $B = arctan b$ .
Since $a, b > 0$ and $ab < 1$ , the sum is an angle between 0 and $\frac{π}{2}$ , so
For $w_{1} = z_{0} z_{1}$ we add the arguments.

tan (arctan a + arctan b) = \frac{a + b}{1 - ab} arctan a + arctan b = arctan (\frac{a + b}{1 - ab}) ar g w_{1} = arctan 1 + arctan \frac{1}{3} = arctan (\frac{1 + \frac{1}{3}}{1 - \frac{1}{3}}) = arctan 2

(c)

10 marks

Base case $n = 0$ : $ar g w_{0} = ar g z_{0} = \frac{π}{4} = arctan 1 = arctan (0 + 1)$ , so it holds.
Assume it holds for $n = k$ : $ar g w_{k} = arctan (k + 1)$ .
Then $w_{k + 1} = w_{k} z_{k + 1}$ , so the arguments add.
The product of the two numbers is $\frac{k + 1}{k ^{2} + 3 k + 3} < 1$ , so the formula from (b) applies.
Expand the numerator: $(k + 1) (k^{2} + 3 k + 3) + 1 = k^{3} + 4 k^{2} + 6 k + 4 = (k + 2) (k^{2} + 2 k + 2)$ .
The statement is true for $n = 0$ and, if true for $n = k$ , it is true for $n = k + 1$ . By mathematical induction it is true for all $n \in N$ .

ar g w_{k + 1} = arctan (k + 1) + arctan (\frac{1}{( k + 1 ) ^{2} + ( k + 1 ) + 1}) = arctan (k + 1) + arctan (\frac{1}{k ^{2} + 3 k + 3}) arctan (\frac{( k + 1 ) + \frac{1}{k ^{2} + 3 k + 3}}{1 - \frac{k + 1}{k ^{2} + 3 k + 3}}) = arctan (\frac{( k + 1 ) ( k ^{2} + 3 k + 3 ) + 1}{k ^{2} + 2 k + 2}) ar g w_{k + 1} = arctan (k + 2) = arctan ((k + 1) + 1)

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 algebra53 marks · 48%
Q3, Q5, Q7, Q11, Q12
Calculus26 marks · 24%
Q1, Q4, Q9, Q10
Functions18 marks · 16%
Q8, Q10
Geometry and trigonometry7 marks · 6%
Q2
Statistics and probability6 marks · 5%
Q6

Mentor insights

Traps to watch for

01
Height of the cone · Q2
The cone's height is the distance from the centre of the base to the vertex, not the distance from A to V.
02
Substitution y = vx · Q9
Differentiate with the product rule: $\frac{d y}{d x} = v + x \frac{d v}{d x}$ .
03
Include both roots in the domain · Q10
$tan^{2} x = \frac{1}{3}$ gives $tan x = \pm \frac{1}{3}$ , so there are two solutions in $[0, π]$ .
04
Repeats allowed or not · Q11
With repeats each digit has 5 choices. Without repeats there are 5, then 4, then 3.
05
Base case at n = 0 · Q12
The statement is claimed for $n \in N$ starting from 0, so check $n = 0$ first.

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