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IB Maths AA HL · Paper 1

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: November 2023
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 2023 TZ2 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 · 57 marks

Question 1

5 marks

A composite function with a constant, solved for k

Functions

(a)

2 marks

Substitute $f (x) = x - 3$ into $g$ .

(g \circ f) (x) = (x - 3)^{2} + k^{2}

(b)

3 marks

Put $x = 2$ .

(2 - 3)^{2} + k^{2} = 10 \Rightarrow k^{2} = 9 k = \pm 3

Question 2

4 marks

Union and complements of two events

Statistics and probability

(a)

2 marks

Use the addition rule.

P (A \cup B) = 0.65 + 0.75 - 0.6 = 0.8

(b)

2 marks

The event "neither A nor B" is the complement of the union.

P (A^{'} \cap B^{'}) = 1 - 0.8 = 0.2

Question 3

7 marks

An arithmetic sequence from its sum formula

Number and algebra

(a)

5 marks

Use the two given sums.
Subtract.

S_{4} = 16 p - 4 q = 40 \Rightarrow 4 p - q = 10 S_{5} = 25 p - 5 q = 65 \Rightarrow 5 p - q = 13 p = 3, q = 2

(b)

2 marks

The fifth term is the difference between consecutive sums.

u_{5} = S_{5} - S_{4} = 65 - 40 = 25

Question 4

6 marks

Area of a triangle from two sides and the cosine of the included angle

Number and algebra

6 marks

With $cos B A C = \frac{1}{5}$ and the angle acute:
Area of the triangle.

sin B A C = 1 - \frac{1}{25} = \frac{2 6}{5} Area = \frac{1}{2} (6) (10) sin B A C = 56 \cdot \frac{2 6}{5} = 2 (6) = 12 cm^{2}

Question 5

6 marks

Finding n and k from two terms of a binomial expansion

Number and algebra

6 marks

The coefficient of $x$ in $(1 + k x)^{n}$ is $nk$ , and of $x^{2}$ is $(2 n) k^{2}$ .
Cancel $k^{2}$ (since $k \neq = 0$ ) in the second equation.

nk = 12, (2 n) k^{2} = 28 k^{2} \frac{n ( n - 1 )}{2} = 28 \Rightarrow n^{2} - n - 56 = 0 \Rightarrow (n - 8) (n + 7) = 0 \Rightarrow n = 8 k = \frac{12}{8} = 1.5

Question 6

7 marks

Induction: 5^(2n) - 2^(3n) is divisible by 17

Number and algebra

7 marks

Base case $n = 1$ : $5^{2} - 2^{3} = 25 - 8 = 17$ , which is divisible by 17.
Assume $5^{2 k} - 2^{3 k} = 17 m$ for some integer $m$ .
Consider $n = k + 1$ .
Write $25 \cdot 5^{2 k} = 25 (5^{2 k} - 2^{3 k}) + 25 \cdot 2^{3 k}$ .
This is divisible by 17.
The statement is true for $n = 1$ and, if true for $n = k$ , it is true for $n = k + 1$ . By mathematical induction $5^{2 n} - 2^{3 n}$ is divisible by 17 for all $n \in Z^{+}$ .

5^{2 k + 2} - 2^{3 k + 3} = 25 \cdot 5^{2 k} - 8 \cdot 2^{3 k} 25 (17 m) + 25 \cdot 2^{3 k} - 8 \cdot 2^{3 k} = 17 (25 m) + 17 \cdot 2^{3 k} = 17 (25 m + 2^{3 k})

Question 7

5 marks

A complex equation: real and imaginary parts

Number and algebra

5 marks

Substitute $z = 5 + q i$ .
Compare imaginary parts.
Compare real parts.

z^{2} = 25 - q^{2} + 10 q i, i z = 5 i - q z^{2} + i z = (25 - q^{2} - q) + (10 q + 5) i 10 q + 5 = 25 \Rightarrow q = 2 25 - 4 - 2 = - p \Rightarrow p = - 19

Question 8

9 marks

Integrating x (ln x)^2 by parts and a definite integral

Calculus

(a)

6 marks

Integrate by parts with $u = (ln x)^{2}$ and $d v = x d x$ , so $d u = \frac{2 ln x}{x} d x$ and $v = \frac{x ^{2}}{2}$ .
Integrate $\int x ln x d x$ by parts: $\frac{x ^{2}}{2} ln x - \frac{x ^{2}}{4}$ .

\int x (ln x)^{2} d x = \frac{x ^{2}}{2} (ln x)^{2} - \int x ln x d x \int x (ln x)^{2} d x = \frac{x ^{2} ( ln x ) ^{2}}{2} - \frac{x ^{2} ln x}{2} + \frac{x ^{2}}{4} + c

(b)

3 marks

Evaluate between 1 and 4, using $ln 4 = 2 ln 2$ .

[\frac{x ^{2} ( ln x ) ^{2}}{2} - \frac{x ^{2} ln x}{2} + \frac{x ^{2}}{4}]_{1}^{4} = (8 (2 ln 2)^{2} - 8 (2 ln 2) + 4) - \frac{1}{4} 32 (ln 2)^{2} - 16 ln 2 + \frac{15}{4}

Question 9

8 marks

An even function and a limit that needs l'Hopital or a series

Functions
Calculus

(a)

2 marks

The function is $f (x) = \frac{sin ^{2} ( k x )}{x ^{2}}$ . Replace $x$ by $- x$ .
So $f$ is an even function.

f (- x) = \frac{sin ^{2} ( - k x )}{( - x ) ^{2}} = \frac{( - sin k x ) ^{2}}{x ^{2}} = \frac{sin ^{2} ( k x )}{x ^{2}} = f (x)

(b)

6 marks

Use the series $sin (k x) = k x - \dots$ , so $sin^{2} (k x) = k^{2} x^{2} - \dots$ .
Set this equal to 16.

x \to 0 lim \frac{sin ^{2} ( k x )}{x ^{2}} = x \to 0 lim \frac{k ^{2} x ^{2} + \dots}{x ^{2}} = k^{2} k^{2} = 16 \Rightarrow k = 4 (k > 0)

Mentor tip

With l'Hopital's rule: differentiate top and bottom to get $\frac{2 k sin k x cos k x}{2 x} = \frac{k sin 2 k x}{2 x}$ , still $\frac{0}{0}$ , then once more to get $k^{2} cos 2 k x \to k^{2}$ .

Section B · Extended response

Questions 10–12 · 53 marks

Question 10

15 marks

Two logarithmic graphs that meet twice

Functions

(a)

1 mark

$g (x) = 2 ln x - ln d$ is undefined at $x = 0$ .

x = 0

(b)

9 marks

At an intersection $ln (2 x - 9) = 2 ln x - ln d = ln \frac{x ^{2}}{d}$ .
Two distinct intersection points mean the discriminant is positive.
Factorise: $d (d - 9) > 0$ . Since $d > 0$ this means

2 x - 9 = \frac{x ^{2}}{d} \Rightarrow x^{2} - 2 d x + 9 d = 0 (- 2 d)^{2} - 4 (1) (9 d) > 0 \Rightarrow 4 d^{2} - 36 d > 0 \Rightarrow d^{2} - 9 d > 0 d > 9

(c)

5 marks

With $d = 10$ : $x^{2} - 20 x + 90 = 0$ .
The difference between the roots:

x = \frac{20 \pm 400 - 360}{2} = 10 \pm 10 q - p = 210

Question 11

21 marks

f(x) = e^(cos 2x): turning points, a sketch and a Maclaurin series

Calculus

(a)

5 marks

Differentiate with the chain rule.
On $0 \leq x \leq π$ this gives $x = 0, \frac{π}{2}, π$ .

f^{'} (x) = - 2 sin 2 x e^{c o s 2 x} = 0 \Rightarrow sin 2 x = 0 (0, e), (\frac{π}{2}, e^{- 1}), (π, e)

(b)

4 marks

Differentiate again with the product rule.
At $x = 0$ and $x = π$ the value is $- 4 e < 0$ (maximum). At $x = \frac{π}{2}$ it is $\frac{4}{e} > 0$ (minimum).
So there are two local maxima and one local minimum.

f^{''} (x) = [4 sin^{2} 2 x - 4 cos 2 x] e^{c o s 2 x}

(c)

3 marks

Sketch: a smooth wave starting at the maximum $(0, e)$ , falling to the minimum $(\frac{π}{2}, \frac{1}{e})$ and rising to the maximum $(π, e)$ . The curve is sharply curved at the maxima (second derivative $- 4 e \approx - 10.9$ ) and much flatter at the minimum (second derivative $\frac{4}{e} \approx 1.47$ ).

(d)

6 marks

The series of cosine is $cos 2 x = 1 - \frac{( 2 x ) ^{2}}{2 !} + \frac{( 2 x ) ^{4}}{4 !} - \dots$ .
Then $cos 2 x - 1 = - 2 x^{2} + \frac{2}{3} x^{4}$ and use $e^{u} = 1 + u + \frac{u ^{2}}{2} + \dots$ with $u = - 2 x^{2} + \frac{2}{3} x^{4}$ .
Multiply by $e$ , since $f (x) = e \cdot e^{c o s 2 x - 1}$ .

cos 2 x = 1 - 2 x^{2} + \frac{2}{3} x^{4} - \dots e^{c o s 2 x - 1} = 1 + (- 2 x^{2} + \frac{2}{3} x^{4}) + \frac{1}{2} (4 x^{4}) + \dots = 1 - 2 x^{2} + \frac{8}{3} x^{4} f (x) = e (1 - 2 x^{2} + \frac{8}{3} x^{4})

(e)

3 marks

Use the first two terms, $e (1 - 2 x^{2})$ , and integrate from 0 to 0.1.

\int_{0}^{0.1} e (1 - 2 x^{2}) d x = e [x - \frac{2}{3} x^{3}]_{0}^{0.1} = e (0.1 - \frac{2 ( 0.001 )}{3}) = e \cdot \frac{149}{1500}

Question 12

17 marks

Using De Moivre's theorem to expand sin 5 theta

Number and algebra

(a)

4 marks

Expand with the binomial theorem, using $i^{2} = - 1$ , $i^{3} = - i$ , $i^{4} = 1$ , $i^{5} = i$ .

(cos θ + i sin θ)^{5} = (cos^{5} θ - 10 cos^{3} θ sin^{2} θ + 5 cos θ sin^{4} θ) + i (5 cos^{4} θ sin θ - 10 cos^{2} θ sin^{3} θ + sin^{5} θ)

(b)

6 marks

By De Moivre the left side equals $cos 5 θ + i sin 5 θ$ , so equate the imaginary parts.
Replace $cos^{2} θ$ with $1 - sin^{2} θ$ (let $s = sin θ$ ).

sin 5 θ = 5 cos^{4} θ sin θ - 10 cos^{2} θ sin^{3} θ + sin^{5} θ sin 5 θ = 5 (1 - s^{2})^{2} s - 10 (1 - s^{2}) s^{3} + s^{5} = 5 s - 10 s^{3} + 5 s^{5} - 10 s^{3} + 10 s^{5} + s^{5} sin 5 θ = 16 s^{5} - 20 s^{3} + 5 s

(c)

7 marks

For $θ = \frac{π}{5}$ and $θ = \frac{3 π}{5}$ , $sin 5 θ = sin π = 0$ and $sin 3 π = 0$ .
Since $sin θ \neq = 0$ for these angles, $16 sin^{4} θ - 20 sin^{2} θ + 5 = 0$ , so both angles are solutions.
This equation is a quadratic in $t = sin^{2} θ$ : $16 t^{2} - 20 t + 5 = 0$ , with two different roots $t_{1} = sin^{2} \frac{π}{5}$ and $t_{2} = sin^{2} \frac{3 π}{5}$ (the angles have different sines). The product of the roots is
The positive square root is used because both sines are positive.

16 s^{5} - 20 s^{3} + 5 s = s (16 s^{4} - 20 s^{2} + 5) = 0 t_{1} t_{2} = \frac{5}{16} \Rightarrow sin \frac{π}{5} sin \frac{3 π}{5} = \frac{5}{16} = \frac{5}{4}

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 algebra48 marks · 44%
Q3, Q4, Q5, Q6, Q7, Q12
Calculus36 marks · 33%
Q8, Q9, Q11
Functions22 marks · 20%
Q1, Q9, Q10
Statistics and probability4 marks · 4%
Q2

Mentor insights

Traps to watch for

01
Integration by parts twice · Q8
Use $u = (ln x)^{2}$ and $d v = x d x$ , then $u = ln x$ and $d v = x d x$ .
02
Even function test · Q9
Show $f (- x) = f (x)$ .
03
Combine the logs · Q10
$2 ln x - ln d = ln \frac{x ^{2}}{d}$ , so equal logs give equal arguments.
04
Second derivative test · Q11
Evaluate $f^{''}$ at each stationary point: negative is a maximum and positive is a minimum.
05
Equate imaginary parts · Q12
$(cos θ + i sin θ)^{5} = cos 5 θ + i sin 5 θ$ , so match the imaginary parts of both expansions.

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

Timezone

TZ1TZ2

Session: November 2023
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 2023 TZ2 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 · 57 marks

Question 1

5 marks

A composite function with a constant, solved for k

Functions

(a)

2 marks

Substitute $f (x) = x - 3$ into $g$ .

(g \circ f) (x) = (x - 3)^{2} + k^{2}

(b)

3 marks

Put $x = 2$ .

(2 - 3)^{2} + k^{2} = 10 \Rightarrow k^{2} = 9 k = \pm 3

Question 2

4 marks

Union and complements of two events

Statistics and probability

(a)

2 marks

Use the addition rule.

P (A \cup B) = 0.65 + 0.75 - 0.6 = 0.8

(b)

2 marks

The event "neither A nor B" is the complement of the union.

P (A^{'} \cap B^{'}) = 1 - 0.8 = 0.2

Question 3

7 marks

An arithmetic sequence from its sum formula

Number and algebra

(a)

5 marks

Use the two given sums.
Subtract.

S_{4} = 16 p - 4 q = 40 \Rightarrow 4 p - q = 10 S_{5} = 25 p - 5 q = 65 \Rightarrow 5 p - q = 13 p = 3, q = 2

(b)

2 marks

The fifth term is the difference between consecutive sums.

u_{5} = S_{5} - S_{4} = 65 - 40 = 25

Question 4

6 marks

Area of a triangle from two sides and the cosine of the included angle

Number and algebra

6 marks

With $cos B A C = \frac{1}{5}$ and the angle acute:
Area of the triangle.

sin B A C = 1 - \frac{1}{25} = \frac{2 6}{5} Area = \frac{1}{2} (6) (10) sin B A C = 56 \cdot \frac{2 6}{5} = 2 (6) = 12 cm^{2}

Question 5

6 marks

Finding n and k from two terms of a binomial expansion

Number and algebra

6 marks

The coefficient of $x$ in $(1 + k x)^{n}$ is $nk$ , and of $x^{2}$ is $(2 n) k^{2}$ .
Cancel $k^{2}$ (since $k \neq = 0$ ) in the second equation.

nk = 12, (2 n) k^{2} = 28 k^{2} \frac{n ( n - 1 )}{2} = 28 \Rightarrow n^{2} - n - 56 = 0 \Rightarrow (n - 8) (n + 7) = 0 \Rightarrow n = 8 k = \frac{12}{8} = 1.5

Question 6

7 marks

Induction: 5^(2n) - 2^(3n) is divisible by 17

Number and algebra

7 marks

Base case $n = 1$ : $5^{2} - 2^{3} = 25 - 8 = 17$ , which is divisible by 17.
Assume $5^{2 k} - 2^{3 k} = 17 m$ for some integer $m$ .
Consider $n = k + 1$ .
Write $25 \cdot 5^{2 k} = 25 (5^{2 k} - 2^{3 k}) + 25 \cdot 2^{3 k}$ .
This is divisible by 17.
The statement is true for $n = 1$ and, if true for $n = k$ , it is true for $n = k + 1$ . By mathematical induction $5^{2 n} - 2^{3 n}$ is divisible by 17 for all $n \in Z^{+}$ .

5^{2 k + 2} - 2^{3 k + 3} = 25 \cdot 5^{2 k} - 8 \cdot 2^{3 k} 25 (17 m) + 25 \cdot 2^{3 k} - 8 \cdot 2^{3 k} = 17 (25 m) + 17 \cdot 2^{3 k} = 17 (25 m + 2^{3 k})

Question 7

5 marks

A complex equation: real and imaginary parts

Number and algebra

5 marks

Substitute $z = 5 + q i$ .
Compare imaginary parts.
Compare real parts.

z^{2} = 25 - q^{2} + 10 q i, i z = 5 i - q z^{2} + i z = (25 - q^{2} - q) + (10 q + 5) i 10 q + 5 = 25 \Rightarrow q = 2 25 - 4 - 2 = - p \Rightarrow p = - 19

Question 8

9 marks

Integrating x (ln x)^2 by parts and a definite integral

Calculus

(a)

6 marks

Integrate by parts with $u = (ln x)^{2}$ and $d v = x d x$ , so $d u = \frac{2 ln x}{x} d x$ and $v = \frac{x ^{2}}{2}$ .
Integrate $\int x ln x d x$ by parts: $\frac{x ^{2}}{2} ln x - \frac{x ^{2}}{4}$ .

\int x (ln x)^{2} d x = \frac{x ^{2}}{2} (ln x)^{2} - \int x ln x d x \int x (ln x)^{2} d x = \frac{x ^{2} ( ln x ) ^{2}}{2} - \frac{x ^{2} ln x}{2} + \frac{x ^{2}}{4} + c

(b)

3 marks

Evaluate between 1 and 4, using $ln 4 = 2 ln 2$ .

[\frac{x ^{2} ( ln x ) ^{2}}{2} - \frac{x ^{2} ln x}{2} + \frac{x ^{2}}{4}]_{1}^{4} = (8 (2 ln 2)^{2} - 8 (2 ln 2) + 4) - \frac{1}{4} 32 (ln 2)^{2} - 16 ln 2 + \frac{15}{4}

Question 9

8 marks

An even function and a limit that needs l'Hopital or a series

Functions
Calculus

(a)

2 marks

The function is $f (x) = \frac{sin ^{2} ( k x )}{x ^{2}}$ . Replace $x$ by $- x$ .
So $f$ is an even function.

f (- x) = \frac{sin ^{2} ( - k x )}{( - x ) ^{2}} = \frac{( - sin k x ) ^{2}}{x ^{2}} = \frac{sin ^{2} ( k x )}{x ^{2}} = f (x)

(b)

6 marks

Use the series $sin (k x) = k x - \dots$ , so $sin^{2} (k x) = k^{2} x^{2} - \dots$ .
Set this equal to 16.

x \to 0 lim \frac{sin ^{2} ( k x )}{x ^{2}} = x \to 0 lim \frac{k ^{2} x ^{2} + \dots}{x ^{2}} = k^{2} k^{2} = 16 \Rightarrow k = 4 (k > 0)

Mentor tip

With l'Hopital's rule: differentiate top and bottom to get $\frac{2 k sin k x cos k x}{2 x} = \frac{k sin 2 k x}{2 x}$ , still $\frac{0}{0}$ , then once more to get $k^{2} cos 2 k x \to k^{2}$ .

Section B · Extended response

Questions 10–12 · 53 marks

Question 10

15 marks

Two logarithmic graphs that meet twice

Functions

(a)

1 mark

$g (x) = 2 ln x - ln d$ is undefined at $x = 0$ .

x = 0

(b)

9 marks

At an intersection $ln (2 x - 9) = 2 ln x - ln d = ln \frac{x ^{2}}{d}$ .
Two distinct intersection points mean the discriminant is positive.
Factorise: $d (d - 9) > 0$ . Since $d > 0$ this means

2 x - 9 = \frac{x ^{2}}{d} \Rightarrow x^{2} - 2 d x + 9 d = 0 (- 2 d)^{2} - 4 (1) (9 d) > 0 \Rightarrow 4 d^{2} - 36 d > 0 \Rightarrow d^{2} - 9 d > 0 d > 9

(c)

5 marks

With $d = 10$ : $x^{2} - 20 x + 90 = 0$ .
The difference between the roots:

x = \frac{20 \pm 400 - 360}{2} = 10 \pm 10 q - p = 210

Question 11

21 marks

f(x) = e^(cos 2x): turning points, a sketch and a Maclaurin series

Calculus

(a)

5 marks

Differentiate with the chain rule.
On $0 \leq x \leq π$ this gives $x = 0, \frac{π}{2}, π$ .

f^{'} (x) = - 2 sin 2 x e^{c o s 2 x} = 0 \Rightarrow sin 2 x = 0 (0, e), (\frac{π}{2}, e^{- 1}), (π, e)

(b)

4 marks

Differentiate again with the product rule.
At $x = 0$ and $x = π$ the value is $- 4 e < 0$ (maximum). At $x = \frac{π}{2}$ it is $\frac{4}{e} > 0$ (minimum).
So there are two local maxima and one local minimum.

f^{''} (x) = [4 sin^{2} 2 x - 4 cos 2 x] e^{c o s 2 x}

(c)

3 marks

(d)

6 marks

The series of cosine is $cos 2 x = 1 - \frac{( 2 x ) ^{2}}{2 !} + \frac{( 2 x ) ^{4}}{4 !} - \dots$ .
Then $cos 2 x - 1 = - 2 x^{2} + \frac{2}{3} x^{4}$ and use $e^{u} = 1 + u + \frac{u ^{2}}{2} + \dots$ with $u = - 2 x^{2} + \frac{2}{3} x^{4}$ .
Multiply by $e$ , since $f (x) = e \cdot e^{c o s 2 x - 1}$ .

cos 2 x = 1 - 2 x^{2} + \frac{2}{3} x^{4} - \dots e^{c o s 2 x - 1} = 1 + (- 2 x^{2} + \frac{2}{3} x^{4}) + \frac{1}{2} (4 x^{4}) + \dots = 1 - 2 x^{2} + \frac{8}{3} x^{4} f (x) = e (1 - 2 x^{2} + \frac{8}{3} x^{4})

(e)

3 marks

Use the first two terms, $e (1 - 2 x^{2})$ , and integrate from 0 to 0.1.

\int_{0}^{0.1} e (1 - 2 x^{2}) d x = e [x - \frac{2}{3} x^{3}]_{0}^{0.1} = e (0.1 - \frac{2 ( 0.001 )}{3}) = e \cdot \frac{149}{1500}

Question 12

17 marks

Using De Moivre's theorem to expand sin 5 theta

Number and algebra

(a)

4 marks

Expand with the binomial theorem, using $i^{2} = - 1$ , $i^{3} = - i$ , $i^{4} = 1$ , $i^{5} = i$ .

(cos θ + i sin θ)^{5} = (cos^{5} θ - 10 cos^{3} θ sin^{2} θ + 5 cos θ sin^{4} θ) + i (5 cos^{4} θ sin θ - 10 cos^{2} θ sin^{3} θ + sin^{5} θ)

(b)

6 marks

By De Moivre the left side equals $cos 5 θ + i sin 5 θ$ , so equate the imaginary parts.
Replace $cos^{2} θ$ with $1 - sin^{2} θ$ (let $s = sin θ$ ).

sin 5 θ = 5 cos^{4} θ sin θ - 10 cos^{2} θ sin^{3} θ + sin^{5} θ sin 5 θ = 5 (1 - s^{2})^{2} s - 10 (1 - s^{2}) s^{3} + s^{5} = 5 s - 10 s^{3} + 5 s^{5} - 10 s^{3} + 10 s^{5} + s^{5} sin 5 θ = 16 s^{5} - 20 s^{3} + 5 s

(c)

7 marks

For $θ = \frac{π}{5}$ and $θ = \frac{3 π}{5}$ , $sin 5 θ = sin π = 0$ and $sin 3 π = 0$ .
Since $sin θ \neq = 0$ for these angles, $16 sin^{4} θ - 20 sin^{2} θ + 5 = 0$ , so both angles are solutions.
This equation is a quadratic in $t = sin^{2} θ$ : $16 t^{2} - 20 t + 5 = 0$ , with two different roots $t_{1} = sin^{2} \frac{π}{5}$ and $t_{2} = sin^{2} \frac{3 π}{5}$ (the angles have different sines). The product of the roots is
The positive square root is used because both sines are positive.

16 s^{5} - 20 s^{3} + 5 s = s (16 s^{4} - 20 s^{2} + 5) = 0 t_{1} t_{2} = \frac{5}{16} \Rightarrow sin \frac{π}{5} sin \frac{3 π}{5} = \frac{5}{16} = \frac{5}{4}

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 algebra48 marks · 44%
Q3, Q4, Q5, Q6, Q7, Q12
Calculus36 marks · 33%
Q8, Q9, Q11
Functions22 marks · 20%
Q1, Q9, Q10
Statistics and probability4 marks · 4%
Q2

Mentor insights

Traps to watch for

01
Integration by parts twice · Q8
Use $u = (ln x)^{2}$ and $d v = x d x$ , then $u = ln x$ and $d v = x d x$ .
02
Even function test · Q9
Show $f (- x) = f (x)$ .
03
Combine the logs · Q10
$2 ln x - ln d = ln \frac{x ^{2}}{d}$ , so equal logs give equal arguments.
04
Second derivative test · Q11
Evaluate $f^{''}$ at each stationary point: negative is a maximum and positive is a minimum.
05
Equate imaginary parts · Q12
$(cos θ + i sin θ)^{5} = cos 5 θ + i sin 5 θ$ , so match the imaginary parts of both expansions.

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