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IB Maths AA HL · Paper 3

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2022
Marks: 55
Duration: 1 hour 15 min
Calculator: Allowed
Questions: 2

Q1 Q2

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Questions

Questions 1–2 · 55 marks

Question 1

27 marks

Polygonal numbers: triangular, square and pentagonal, and a proof by induction

Number and algebra

(a)(i)

2 marks

With $r = 3$ the formula is $P_{3} (n) = \frac{( 3 - 2 ) n ^{2} - ( 3 - 4 ) n}{2}$ .

P_{3} (n) = \frac{n ^{2} + n}{2} = \frac{n ( n + 1 )}{2}

(a)(ii)

2 marks

Set $\frac{n ( n + 1 )}{2} = 351$ .
So 351 is the 26th triangular number.

n^{2} + n - 702 = 0 \Rightarrow (n - 26) (n + 27) = 0 \Rightarrow n = 26

(b)(i)

2 marks

Add the two consecutive triangular numbers.

P_{3} (n) + P_{3} (n + 1) = \frac{n ( n + 1 )}{2} + \frac{( n + 1 ) ( n + 2 )}{2} = \frac{( n + 1 ) ( 2 n + 2 )}{2} = (n + 1)^{2}

(b)(ii)

1 mark

Two consecutive triangular numbers add up to a square number.

(b)(iii)

1 mark

For $n = 4$ : $P_{3} (4) = 10$ and $P_{3} (5) = 15$ . Draw a staircase triangle of 10 dots and fit a staircase triangle of 15 dots against it, upside down. Together they fill a $5 \times 5$ grid of 25 dots.

10 + 15 = 25 = 5^{2}

(c)

3 marks

Use $P_{3} (n) = \frac{n ( n + 1 )}{2}$ .
Since $2 n + 1$ is odd for every integer $n$ , this is the square of an odd number.

8 P_{3} (n) + 1 = 4 n (n + 1) + 1 = 4 n^{2} + 4 n + 1 = (2 n + 1)^{2}

(d)

3 marks

The series $1 + 4 + 7 + \dots + (3 n - 2)$ is arithmetic with first term 1, last term $3 n - 2$ and $n$ terms.

P_{5} (n) = \frac{n}{2} (1 + (3 n - 2)) = \frac{n ( 3 n - 1 )}{2}

(e)

5 marks

List the triangular numbers $\frac{m ( m + 1 )}{2}$ and the pentagonal numbers $\frac{n ( 3 n - 1 )}{2}$ until a value above 1 appears in both.
The first common value above 1 is

triangular: 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, \dots pentagonal: 1, 5, 12, 22, 35, 51, 70, 92, 117, 145, 176, 210, \dots 210 (the 20th triangular and the 12th pentagonal number)

(f)

8 marks

For $n = 1$ the series gives $P_{r} (1) = 1$ . The formula gives $\frac{( r - 2 ) - ( r - 4 )}{2} = \frac{2}{2} = 1$ . So it is true for $n = 1$ .
Assume it is true for $n = k$ : $P_{r} (k) = \frac{( r - 2 ) k ^{2} - ( r - 4 ) k}{2}$ .
The polygonal number is the sum $P_{r} (n) = \sum_{m = 1}^{n} (1 + (m - 1) (r - 2))$ . For $n = k + 1$ there is one more term, the one with $m = k + 1$ , which is $1 + k (r - 2)$ .
Put everything over 2 and collect the powers of $k$ . The coefficient of $k$ is $- (r - 4) + 2 (r - 2) = r$ .
Compare with the formula at $n = k + 1$ .
The two expressions agree. So if the formula holds for $n = k$ it holds for $n = k + 1$ . It holds for $n = 1$ , so by induction it is true for all positive integers $n$ .

P_{r} (k + 1) = P_{r} (k) + 1 + k (r - 2) = \frac{( r - 2 ) k ^{2} - ( r - 4 ) k}{2} + 1 + k (r - 2) P_{r} (k + 1) = \frac{( r - 2 ) k ^{2} - ( r - 4 ) k + 2 + 2 ( r - 2 ) k}{2} = \frac{( r - 2 ) k ^{2} + r k + 2}{2} \frac{( r - 2 ) ( k + 1 ) ^{2} - ( r - 4 ) ( k + 1 )}{2} = \frac{( r - 2 ) k ^{2} + 2 ( r - 2 ) k + ( r - 2 ) - ( r - 4 ) k - ( r - 4 )}{2} = \frac{( r - 2 ) k ^{2} + r k + 2}{2}

Question 2

28 marks

Cubics with one real and two complex roots, and a tangent that reaches the real root

Number and algebra
Calculus
Functions

(a)(i)

1 mark

Complex roots of a real polynomial come in conjugate pairs.

4 - i

(a)(ii)

1 mark

The mean of the two complex roots.

\frac{( 4 + i ) + ( 4 - i )}{2} = 4

(b)

4 marks

Differentiate $f (x) = (x - 1) (x^{2} - 8 x + 17)$ with the product rule.
At $x = 4$ : the first bracket is $16 - 32 + 17 = 1$ and the second term is $3 \times 0 = 0$ , so $f^{'} (4) = 1$ . Also $f (4) = 3 \times 1 = 3$ .
The tangent at $A (4, 3)$ has gradient 1.

f^{'} (x) = (x^{2} - 8 x + 17) + (x - 1) (2 x - 8) y - 3 = 1 (x - 4) \Rightarrow y = x - 1

(c)

2 marks

The curve $f (x) = (x - 1) ((x - 4)^{2} + 1)$ crosses the $x$ -axis only at $x = 1$ and is increasing throughout. The tangent $y = x - 1$ touches the curve at $A (4, 3)$ and crosses the $x$ -axis at $(1, 0)$ , the same point where the curve crosses.

(d)(i)

2 marks

Use the product rule on $g (x) = (x - r) (x^{2} - 2 a x + a^{2} + b^{2})$ , noting that $x^{2} - 2 a x + a^{2} = (x - a)^{2}$ has derivative $2 (x - a)$ .

g^{'} (x) = (x^{2} - 2 a x + a^{2} + b^{2}) + (x - r) (2 x - 2 a) = 2 (x - r) (x - a) + x^{2} - 2 a x + a^{2} + b^{2}

(d)(ii)

6 marks

At $x = a$ : $g (a) = (a - r) b^{2}$ and $g^{'} (a) = 0 + a^{2} - 2 a^{2} + a^{2} + b^{2} = b^{2}$ .
The tangent at $A (a, g (a))$ is
This is zero when $x = r$ (since $b > 0$ ). So the tangent crosses the $x$ -axis at $R (r, 0)$ .

y - (a - r) b^{2} = b^{2} (x - a) \Rightarrow y = b^{2} (x - r)

(e)

1 mark

From part (d)(ii), $g^{'} (a) = b^{2}$ , and $b > 0$ , so $b = g^{'} (a)$ .

a \pm b i = a \pm i g^{'} (a)

(f)(i)

4 marks

In $y = (x - r) (x^{2} - 2 a x + a^{2} + 16)$ we have $b^{2} = 16$ , so $b = 4$ and $r = - 2$ (R is at $(- 2, 0)$ ). The tangent at A has gradient $b^{2} = 16$ and passes through R and A $(a, 80)$ .
The roots are $r$ and $a \pm b i$ .

80 = 16 (a - (- 2)) \Rightarrow a + 2 = 5 \Rightarrow a = 3 z = - 2, z = 3 \pm 4 i

(f)(ii)

1 mark

C $_{2}$ has coordinates $(a, - g^{'} (a))$ .

C_{2} = (3, - 4)

(g)(i)

2 marks

Differentiate $g^{'} (x)$ again: $g^{''} (x) = 2 (x - a) + 2 (x - r) + 2 x - 2 a = 6 x - 4 a - 2 r$ . A point of inflexion has $g^{''} (x) = 0$ .

6 x = 4 a + 2 r \Rightarrow x = \frac{2 a + r}{3} = \frac{1}{3} (2 a + r)

(g)(ii)

1 mark

$\frac{2 a + r}{3} = r + \frac{2}{3} (a - r)$ . So P is two thirds of the way horizontally from R to A.

(h)(i)

2 marks

With $a = r = 1$ , $b = 2$ : $y = (x - 1) (x^{2} - 2 x + 5) = (x - 1) ((x - 1)^{2} + 4)$ . It is a cubic with a positive leading coefficient, always increasing, with no turning points, and a point of inflexion on the $x$ -axis at $(1, 0)$ .

(h)(ii)

1 mark

When $a = r$ , then $g (a) = (a - r) b^{2} = 0$ and the inflexion is at $x = \frac{2 a + r}{3} = r$ .

P = (r, 0), A = (r, 0)

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 algebra35 marks · 64%
Q1, Q2
Calculus16 marks · 29%
Q2
Functions4 marks · 7%
Q2

Mentor insights

Traps to watch for

01
Verify, don't assume · Q1
In (a)(i) put $r = 3$ into the given formula and simplify. Say that the result matches. In (e) list both sequences until a value appears in both.
02
Induction needs the base case and the link · Q1
In (f) show the formula holds for $n = 1$ , assume it for $n = k$ , and use the series definition to go from $k$ to $k + 1$ . End with all three parts of the conclusion.
03
The tangent at A hits the real root · Q2
The surprising fact is that the tangent at $(a, g (a))$ passes through $(r, 0)$ . Show it by finding $g (a)$ and $g^{'} (a)$ and writing the tangent in point-gradient form.
04
Complex roots are conjugates · Q2
The roots $a \pm b i$ come as a pair, and their real part is the mean of the two. The imaginary part is $b = g^{'} (a)$ .
05
Two thirds along · Q2
The point of inflexion lies at $x = \frac{2 a + r}{3}$ , so two thirds of the way horizontally from R to A. Use that as the numerical statement.

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 3

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2022
Marks: 55
Duration: 1 hour 15 min
Calculator: Allowed
Questions: 2

Q1 Q2

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2022 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–2 · 55 marks

Question 1

27 marks

Polygonal numbers: triangular, square and pentagonal, and a proof by induction

Number and algebra

(a)(i)

2 marks

With $r = 3$ the formula is $P_{3} (n) = \frac{( 3 - 2 ) n ^{2} - ( 3 - 4 ) n}{2}$ .

P_{3} (n) = \frac{n ^{2} + n}{2} = \frac{n ( n + 1 )}{2}

(a)(ii)

2 marks

Set $\frac{n ( n + 1 )}{2} = 351$ .
So 351 is the 26th triangular number.

n^{2} + n - 702 = 0 \Rightarrow (n - 26) (n + 27) = 0 \Rightarrow n = 26

(b)(i)

2 marks

Add the two consecutive triangular numbers.

P_{3} (n) + P_{3} (n + 1) = \frac{n ( n + 1 )}{2} + \frac{( n + 1 ) ( n + 2 )}{2} = \frac{( n + 1 ) ( 2 n + 2 )}{2} = (n + 1)^{2}

(b)(ii)

1 mark

Two consecutive triangular numbers add up to a square number.

(b)(iii)

1 mark

10 + 15 = 25 = 5^{2}

(c)

3 marks

Use $P_{3} (n) = \frac{n ( n + 1 )}{2}$ .
Since $2 n + 1$ is odd for every integer $n$ , this is the square of an odd number.

8 P_{3} (n) + 1 = 4 n (n + 1) + 1 = 4 n^{2} + 4 n + 1 = (2 n + 1)^{2}

(d)

3 marks

The series $1 + 4 + 7 + \dots + (3 n - 2)$ is arithmetic with first term 1, last term $3 n - 2$ and $n$ terms.

P_{5} (n) = \frac{n}{2} (1 + (3 n - 2)) = \frac{n ( 3 n - 1 )}{2}

(e)

5 marks

List the triangular numbers $\frac{m ( m + 1 )}{2}$ and the pentagonal numbers $\frac{n ( 3 n - 1 )}{2}$ until a value above 1 appears in both.
The first common value above 1 is

triangular: 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, \dots pentagonal: 1, 5, 12, 22, 35, 51, 70, 92, 117, 145, 176, 210, \dots 210 (the 20th triangular and the 12th pentagonal number)

(f)

8 marks

For $n = 1$ the series gives $P_{r} (1) = 1$ . The formula gives $\frac{( r - 2 ) - ( r - 4 )}{2} = \frac{2}{2} = 1$ . So it is true for $n = 1$ .
Assume it is true for $n = k$ : $P_{r} (k) = \frac{( r - 2 ) k ^{2} - ( r - 4 ) k}{2}$ .
The polygonal number is the sum $P_{r} (n) = \sum_{m = 1}^{n} (1 + (m - 1) (r - 2))$ . For $n = k + 1$ there is one more term, the one with $m = k + 1$ , which is $1 + k (r - 2)$ .
Put everything over 2 and collect the powers of $k$ . The coefficient of $k$ is $- (r - 4) + 2 (r - 2) = r$ .
Compare with the formula at $n = k + 1$ .
The two expressions agree. So if the formula holds for $n = k$ it holds for $n = k + 1$ . It holds for $n = 1$ , so by induction it is true for all positive integers $n$ .

P_{r} (k + 1) = P_{r} (k) + 1 + k (r - 2) = \frac{( r - 2 ) k ^{2} - ( r - 4 ) k}{2} + 1 + k (r - 2) P_{r} (k + 1) = \frac{( r - 2 ) k ^{2} - ( r - 4 ) k + 2 + 2 ( r - 2 ) k}{2} = \frac{( r - 2 ) k ^{2} + r k + 2}{2} \frac{( r - 2 ) ( k + 1 ) ^{2} - ( r - 4 ) ( k + 1 )}{2} = \frac{( r - 2 ) k ^{2} + 2 ( r - 2 ) k + ( r - 2 ) - ( r - 4 ) k - ( r - 4 )}{2} = \frac{( r - 2 ) k ^{2} + r k + 2}{2}

Question 2

28 marks

Cubics with one real and two complex roots, and a tangent that reaches the real root

Number and algebra
Calculus
Functions

(a)(i)

1 mark

Complex roots of a real polynomial come in conjugate pairs.

4 - i

(a)(ii)

1 mark

The mean of the two complex roots.

\frac{( 4 + i ) + ( 4 - i )}{2} = 4

(b)

4 marks

Differentiate $f (x) = (x - 1) (x^{2} - 8 x + 17)$ with the product rule.
At $x = 4$ : the first bracket is $16 - 32 + 17 = 1$ and the second term is $3 \times 0 = 0$ , so $f^{'} (4) = 1$ . Also $f (4) = 3 \times 1 = 3$ .
The tangent at $A (4, 3)$ has gradient 1.

f^{'} (x) = (x^{2} - 8 x + 17) + (x - 1) (2 x - 8) y - 3 = 1 (x - 4) \Rightarrow y = x - 1

(c)

2 marks

(d)(i)

2 marks

Use the product rule on $g (x) = (x - r) (x^{2} - 2 a x + a^{2} + b^{2})$ , noting that $x^{2} - 2 a x + a^{2} = (x - a)^{2}$ has derivative $2 (x - a)$ .

g^{'} (x) = (x^{2} - 2 a x + a^{2} + b^{2}) + (x - r) (2 x - 2 a) = 2 (x - r) (x - a) + x^{2} - 2 a x + a^{2} + b^{2}

(d)(ii)

6 marks

At $x = a$ : $g (a) = (a - r) b^{2}$ and $g^{'} (a) = 0 + a^{2} - 2 a^{2} + a^{2} + b^{2} = b^{2}$ .
The tangent at $A (a, g (a))$ is
This is zero when $x = r$ (since $b > 0$ ). So the tangent crosses the $x$ -axis at $R (r, 0)$ .

y - (a - r) b^{2} = b^{2} (x - a) \Rightarrow y = b^{2} (x - r)

(e)

1 mark

From part (d)(ii), $g^{'} (a) = b^{2}$ , and $b > 0$ , so $b = g^{'} (a)$ .

a \pm b i = a \pm i g^{'} (a)

(f)(i)

4 marks

In $y = (x - r) (x^{2} - 2 a x + a^{2} + 16)$ we have $b^{2} = 16$ , so $b = 4$ and $r = - 2$ (R is at $(- 2, 0)$ ). The tangent at A has gradient $b^{2} = 16$ and passes through R and A $(a, 80)$ .
The roots are $r$ and $a \pm b i$ .

80 = 16 (a - (- 2)) \Rightarrow a + 2 = 5 \Rightarrow a = 3 z = - 2, z = 3 \pm 4 i

(f)(ii)

1 mark

C $_{2}$ has coordinates $(a, - g^{'} (a))$ .

C_{2} = (3, - 4)

(g)(i)

2 marks

Differentiate $g^{'} (x)$ again: $g^{''} (x) = 2 (x - a) + 2 (x - r) + 2 x - 2 a = 6 x - 4 a - 2 r$ . A point of inflexion has $g^{''} (x) = 0$ .

6 x = 4 a + 2 r \Rightarrow x = \frac{2 a + r}{3} = \frac{1}{3} (2 a + r)

(g)(ii)

1 mark

$\frac{2 a + r}{3} = r + \frac{2}{3} (a - r)$ . So P is two thirds of the way horizontally from R to A.

(h)(i)

2 marks

(h)(ii)

1 mark

When $a = r$ , then $g (a) = (a - r) b^{2} = 0$ and the inflexion is at $x = \frac{2 a + r}{3} = r$ .

P = (r, 0), A = (r, 0)

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 algebra35 marks · 64%
Q1, Q2
Calculus16 marks · 29%
Q2
Functions4 marks · 7%
Q2

Mentor insights

Traps to watch for

01
Verify, don't assume · Q1
In (a)(i) put $r = 3$ into the given formula and simplify. Say that the result matches. In (e) list both sequences until a value appears in both.
02
Induction needs the base case and the link · Q1
In (f) show the formula holds for $n = 1$ , assume it for $n = k$ , and use the series definition to go from $k$ to $k + 1$ . End with all three parts of the conclusion.
03
The tangent at A hits the real root · Q2
The surprising fact is that the tangent at $(a, g (a))$ passes through $(r, 0)$ . Show it by finding $g (a)$ and $g^{'} (a)$ and writing the tangent in point-gradient form.
04
Complex roots are conjugates · Q2
The roots $a \pm b i$ come as a pair, and their real part is the mean of the two. The imaginary part is $b = g^{'} (a)$ .
05
Two thirds along · Q2
The point of inflexion lies at $x = \frac{2 a + r}{3}$ , so two thirds of the way horizontally from R to A. Use that as the numerical statement.

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