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IB Maths AA HL · Paper 3

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2023
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

24 marks

Where does log base a of x meet the line y = x?

Functions
Calculus

(a)

4 marks

Both logarithm graphs pass through $(1, 0)$ , and they are increasing and concave down. $y = lo g_{10} x$ is flatter than $y = lo g_{2} x$ and both lie below the line $y = x$ .
Sketch: $y = x$ through the origin; $y = lo g_{2} x$ and $y = lo g_{10} x$ each labelled and crossing the $x$ -axis at $x = 1$ .

non-zero x -intercept: x = 1 for both logs

(b)

5 marks

Let $h (x) = x - ln x$ for $x > 0$ .
The second derivative is positive, so this is a minimum.

h^{'} (x) = 1 - \frac{1}{x} = 0 \Rightarrow x = 1 h^{''} (x) = \frac{1}{x ^{2}} \Rightarrow h^{''} (1) = 1 > 0 h (1) = 1 - 0 = 1

(c)

1 mark

The minimum value of $x - ln x$ is 1, so $x - ln x \geq 1 > 0$ for all $x > 0$ .

x > ln x

(d)

4 marks

For $0 < a < 1$ the graph of $lo g_{a} x$ falls and crosses $y = x$ exactly once. For $1 < a < 1.4$ the curve rises more steeply at first and cuts the line twice. For $1.5 < a < 2$ it stays below the line.

p = 1, q = 2, r = 0

(e)

8 marks

At the tangent point the curve and the line have the same height and the same gradient.
The second equation gives $x ln a = 1$ . Then the first gives $ln x = x ln a = 1$ , so $x = e$ .

\frac{ln x}{ln a} = x, \frac{1}{x ln a} = 1 ln a = \frac{1}{e} \Rightarrow a = e^{1/ e} P = (e, e), a = e^{1/ e} \approx 1.44

(f)(i)

1 mark

Two intersection points occur for values of $a$ between 1 and the tangent case.

1 < a < e^{1/ e}

(f)(ii)

1 mark

No intersection points occur once $a$ is above the tangent case.

a > e^{1/ e}

Question 2

31 marks

Arithmetic sequences hidden in linear and quadratic functions

Number and algebra

(a)

2 marks

For $L (x) = 2 x - 1$ we have $m = 2$ and $c = - 1$ . The root is where $2 x - 1 = 0$ .
The numbers $m, r, c$ are $2, \frac{1}{2}, - 1$ . The differences are $\frac{1}{2} - 2 = - \frac{3}{2}$ and $- 1 - \frac{1}{2} = - \frac{3}{2}$ , which are equal, so it is an AS-linear function.

r = \frac{1}{2}

(b)(i)

1 mark

The root of $m x + c = 0$ .

r = - \frac{c}{m}

(b)(ii)

4 marks

$m, r, c$ in arithmetic sequence means $2 r = m + c$ .

- \frac{2 c}{m} = m + c \Rightarrow - 2 c = m^{2} + m c \Rightarrow c (m + 2) = - m^{2} c = - \frac{m ^{2}}{m + 2} L (x) = m x - \frac{m ^{2}}{m + 2}

(b)(iii)

1 mark

The denominator must not be zero.

m \neq = - 2

(c)

3 marks

We need $c = - \frac{m ^{2}}{m + 2}$ and $r = - \frac{c}{m}$ to be integers. Since $m^{2} = (m + 2) (m - 2) + 4$ , the number $m + 2$ must divide 4: $m + 2 \in {\pm 1, \pm 2, \pm 4}$ , with $m \neq = 0$ .
Check each: $m = - 3$ gives $c = 9$ , $r = 3$ . $m = - 4$ gives $c = 8$ , $r = 2$ . $m = - 1$ gives $c = - 1$ , $r = - 1$ (the given one). The cases $m = 2$ ( $r = \frac{1}{2}$ ) and $m = - 6$ ( $r = \frac{3}{2}$ ) do not give integer roots.

L (x) = - 3 x + 9 and L (x) = - 4 x + 8

(d)(i)

1 mark

Use the sum of roots of $a x^{2} + b x + c$ .

r_{1} + r_{2} = - \frac{b}{a}

(d)(ii)

1 mark

Use the product of roots.

r_{1} r_{2} = \frac{c}{a}

(e)(i)

1 mark

In $a, r_{1}, b, r_{2}, c$ with common difference $d$ : $b = a + 2 d$ and $r_{2} - r_{1} = 2 d$ .

r_{2} - r_{1} = b - a

(e)(ii)

2 marks

Solve the two equations $r_{1} + r_{2} = - \frac{b}{a}$ and $r_{2} - r_{1} = b - a$ for $r_{1}$ .

2 r_{1} = - \frac{b}{a} - (b - a) \Rightarrow r_{1} = \frac{a ^{2} - ab - b}{2 a}

(e)(iii)

3 marks

In an arithmetic sequence the middle term is the mean of the two terms on either side of it: $b = \frac{r _{1} + r _{2}}{2}$ , so $r_{1} + r_{2} = 2 b$ .
Also $r_{1} + r_{2} = - \frac{b}{a}$ .

2 b = - \frac{b}{a} \Rightarrow b (2 + \frac{1}{a}) = 0 \Rightarrow b = 0 or a = - \frac{1}{2}

(f)

5 marks

With $b = 0$ the sequence is $a, r_{1}, 0, r_{2}, c$ , so $r_{2} = - r_{1}$ and $r_{1} = \frac{a}{2}$ , $c = - a$ .
The product of roots is $\frac{c}{a} = - 1$ .
So the two functions are
Check: $2, 1, 0, - 1, - 2$ and $- 2, - 1, 0, 1, 2$ are arithmetic.

r_{1} r_{2} = - r_{1}^{2} = - 1 \Rightarrow r_{1} = \pm 1 \Rightarrow a = \pm 2 Q (x) = 2 x^{2} - 2 and Q (x) = - 2 x^{2} + 2

(g)(i)

2 marks

Put $a = - \frac{1}{2}$ in the formula from (e)(ii).

r_{1} = \frac{\frac{1}{4} + \frac{b}{2} - b}{- 1} = \frac{b}{2} - \frac{1}{4}

(g)(ii)

5 marks

The sequence has $c = 2 b - a = 2 b + \frac{1}{2}$ . The other root is $r_{2} = 2 b - r_{1} = \frac{3 b}{2} + \frac{1}{4}$ .
The product of roots is $\frac{c}{a} = - 2 (2 b + \frac{1}{2}) = - 4 b - 1$ .
Then $c = 2 b + \frac{1}{2}$ .

(\frac{b}{2} - \frac{1}{4}) (\frac{3 b}{2} + \frac{1}{4}) = - 4 b - 1 \frac{3 b ^{2}}{4} - \frac{b}{4} - \frac{1}{16} + 4 b + 1 = 0 \Rightarrow 4 b^{2} + 20 b + 5 = 0 b = \frac{- 20 \pm 320}{8} = - \frac{5}{2} \pm 5 b = - \frac{5}{2} \pm 5, c = - \frac{9}{2} \pm 25 (same sign in both)

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 algebra31 marks · 56%
Q2
Calculus14 marks · 25%
Q1
Functions10 marks · 18%
Q1

Mentor insights

Traps to watch for

01
Second derivative test · Q1
To justify a minimum, show $\frac{d ^{2}}{d x ^{2}} (x - ln x) > 0$ at the stationary point.
02
Tangent condition · Q1
The tangent case has both the curve and the line with the same value and gradient at P.
03
Arithmetic means the middle term is the mean · Q2
In $a, r_{1}, b, r_{2}, c$ the middle term is $b = \frac{r _{1} + r _{2}}{2}$ , so $r_{1} + r_{2} = 2 b$ .
04
Use the sum of roots twice · Q2
Compare $r_{1} + r_{2} = - \frac{b}{a}$ with $r_{1} + r_{2} = 2 b$ to get $b = 0$ or $a = - \frac{1}{2}$ .
05
Integer divisibility · Q2
$m + 2$ must divide $m^{2}$ , and $m^{2} = (m + 2) (m - 2) + 4$ .

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 2023
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 2023 TZ2 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

24 marks

Where does log base a of x meet the line y = x?

Functions
Calculus

(a)

4 marks

Both logarithm graphs pass through $(1, 0)$ , and they are increasing and concave down. $y = lo g_{10} x$ is flatter than $y = lo g_{2} x$ and both lie below the line $y = x$ .
Sketch: $y = x$ through the origin; $y = lo g_{2} x$ and $y = lo g_{10} x$ each labelled and crossing the $x$ -axis at $x = 1$ .

non-zero x -intercept: x = 1 for both logs

(b)

5 marks

Let $h (x) = x - ln x$ for $x > 0$ .
The second derivative is positive, so this is a minimum.

h^{'} (x) = 1 - \frac{1}{x} = 0 \Rightarrow x = 1 h^{''} (x) = \frac{1}{x ^{2}} \Rightarrow h^{''} (1) = 1 > 0 h (1) = 1 - 0 = 1

(c)

1 mark

The minimum value of $x - ln x$ is 1, so $x - ln x \geq 1 > 0$ for all $x > 0$ .

x > ln x

(d)

4 marks

p = 1, q = 2, r = 0

(e)

8 marks

At the tangent point the curve and the line have the same height and the same gradient.
The second equation gives $x ln a = 1$ . Then the first gives $ln x = x ln a = 1$ , so $x = e$ .

\frac{ln x}{ln a} = x, \frac{1}{x ln a} = 1 ln a = \frac{1}{e} \Rightarrow a = e^{1/ e} P = (e, e), a = e^{1/ e} \approx 1.44

(f)(i)

1 mark

Two intersection points occur for values of $a$ between 1 and the tangent case.

1 < a < e^{1/ e}

(f)(ii)

1 mark

No intersection points occur once $a$ is above the tangent case.

a > e^{1/ e}

Question 2

31 marks

Arithmetic sequences hidden in linear and quadratic functions

Number and algebra

(a)

2 marks

For $L (x) = 2 x - 1$ we have $m = 2$ and $c = - 1$ . The root is where $2 x - 1 = 0$ .
The numbers $m, r, c$ are $2, \frac{1}{2}, - 1$ . The differences are $\frac{1}{2} - 2 = - \frac{3}{2}$ and $- 1 - \frac{1}{2} = - \frac{3}{2}$ , which are equal, so it is an AS-linear function.

r = \frac{1}{2}

(b)(i)

1 mark

The root of $m x + c = 0$ .

r = - \frac{c}{m}

(b)(ii)

4 marks

$m, r, c$ in arithmetic sequence means $2 r = m + c$ .

- \frac{2 c}{m} = m + c \Rightarrow - 2 c = m^{2} + m c \Rightarrow c (m + 2) = - m^{2} c = - \frac{m ^{2}}{m + 2} L (x) = m x - \frac{m ^{2}}{m + 2}

(b)(iii)

1 mark

The denominator must not be zero.

m \neq = - 2

(c)

3 marks

We need $c = - \frac{m ^{2}}{m + 2}$ and $r = - \frac{c}{m}$ to be integers. Since $m^{2} = (m + 2) (m - 2) + 4$ , the number $m + 2$ must divide 4: $m + 2 \in {\pm 1, \pm 2, \pm 4}$ , with $m \neq = 0$ .
Check each: $m = - 3$ gives $c = 9$ , $r = 3$ . $m = - 4$ gives $c = 8$ , $r = 2$ . $m = - 1$ gives $c = - 1$ , $r = - 1$ (the given one). The cases $m = 2$ ( $r = \frac{1}{2}$ ) and $m = - 6$ ( $r = \frac{3}{2}$ ) do not give integer roots.

L (x) = - 3 x + 9 and L (x) = - 4 x + 8

(d)(i)

1 mark

Use the sum of roots of $a x^{2} + b x + c$ .

r_{1} + r_{2} = - \frac{b}{a}

(d)(ii)

1 mark

Use the product of roots.

r_{1} r_{2} = \frac{c}{a}

(e)(i)

1 mark

In $a, r_{1}, b, r_{2}, c$ with common difference $d$ : $b = a + 2 d$ and $r_{2} - r_{1} = 2 d$ .

r_{2} - r_{1} = b - a

(e)(ii)

2 marks

Solve the two equations $r_{1} + r_{2} = - \frac{b}{a}$ and $r_{2} - r_{1} = b - a$ for $r_{1}$ .

2 r_{1} = - \frac{b}{a} - (b - a) \Rightarrow r_{1} = \frac{a ^{2} - ab - b}{2 a}

(e)(iii)

3 marks

In an arithmetic sequence the middle term is the mean of the two terms on either side of it: $b = \frac{r _{1} + r _{2}}{2}$ , so $r_{1} + r_{2} = 2 b$ .
Also $r_{1} + r_{2} = - \frac{b}{a}$ .

2 b = - \frac{b}{a} \Rightarrow b (2 + \frac{1}{a}) = 0 \Rightarrow b = 0 or a = - \frac{1}{2}

(f)

5 marks

With $b = 0$ the sequence is $a, r_{1}, 0, r_{2}, c$ , so $r_{2} = - r_{1}$ and $r_{1} = \frac{a}{2}$ , $c = - a$ .
The product of roots is $\frac{c}{a} = - 1$ .
So the two functions are
Check: $2, 1, 0, - 1, - 2$ and $- 2, - 1, 0, 1, 2$ are arithmetic.

r_{1} r_{2} = - r_{1}^{2} = - 1 \Rightarrow r_{1} = \pm 1 \Rightarrow a = \pm 2 Q (x) = 2 x^{2} - 2 and Q (x) = - 2 x^{2} + 2

(g)(i)

2 marks

Put $a = - \frac{1}{2}$ in the formula from (e)(ii).

r_{1} = \frac{\frac{1}{4} + \frac{b}{2} - b}{- 1} = \frac{b}{2} - \frac{1}{4}

(g)(ii)

5 marks

The sequence has $c = 2 b - a = 2 b + \frac{1}{2}$ . The other root is $r_{2} = 2 b - r_{1} = \frac{3 b}{2} + \frac{1}{4}$ .
The product of roots is $\frac{c}{a} = - 2 (2 b + \frac{1}{2}) = - 4 b - 1$ .
Then $c = 2 b + \frac{1}{2}$ .

(\frac{b}{2} - \frac{1}{4}) (\frac{3 b}{2} + \frac{1}{4}) = - 4 b - 1 \frac{3 b ^{2}}{4} - \frac{b}{4} - \frac{1}{16} + 4 b + 1 = 0 \Rightarrow 4 b^{2} + 20 b + 5 = 0 b = \frac{- 20 \pm 320}{8} = - \frac{5}{2} \pm 5 b = - \frac{5}{2} \pm 5, c = - \frac{9}{2} \pm 25 (same sign in both)

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 algebra31 marks · 56%
Q2
Calculus14 marks · 25%
Q1
Functions10 marks · 18%
Q1

Mentor insights

Traps to watch for

01
Second derivative test · Q1
To justify a minimum, show $\frac{d ^{2}}{d x ^{2}} (x - ln x) > 0$ at the stationary point.
02
Tangent condition · Q1
The tangent case has both the curve and the line with the same value and gradient at P.
03
Arithmetic means the middle term is the mean · Q2
In $a, r_{1}, b, r_{2}, c$ the middle term is $b = \frac{r _{1} + r _{2}}{2}$ , so $r_{1} + r_{2} = 2 b$ .
04
Use the sum of roots twice · Q2
Compare $r_{1} + r_{2} = - \frac{b}{a}$ with $r_{1} + r_{2} = 2 b$ to get $b = 0$ or $a = - \frac{1}{2}$ .
05
Integer divisibility · Q2
$m + 2$ must divide $m^{2}$ , and $m^{2} = (m + 2) (m - 2) + 4$ .

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