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IB Maths AA HL · Paper 3

Mathematics: Analysis & Approaches, Higher Level

Session: November 2023
Marks: 55
Duration: 1 hour 15 min
Calculator: Allowed
Questions: 2

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Questions

Questions 1–2 · 55 marks

Question 1

24 marks

The family of cubics y = x^3 + ax^2 + b: how many x-intercepts?

Functions
Calculus

(a)

2 marks

For $y = x^{3} + 3 x^{2} + b$ the turning points are at $x = 0$ (a local minimum with $y = b$ ) and at $x = - 2$ (a local maximum with $y = 4 + b$ ). The curve has exactly two x-intercepts when one of these turning points is on the axis.

b = 0 or b = - 4

(b)(i)

1 mark

One intercept occurs when both turning points are above the axis ( $b > 0$ ) or both below ( $4 + b < 0$ ).

b < - 4 or b > 0

(b)(ii)

1 mark

Three intercepts occur when the local maximum is above and the local minimum below the axis.

- 4 < b < 0

(c)(i)

1 mark

For $y = x^{3} - 3 x^{2} + b$ the local maximum is at $x = 0$ with $y = b$ and the local minimum is at $x = 2$ with $y = b - 4$ . Two intercepts occur when a turning point is on the axis.

b = 0 or b = 4

(c)(ii)

1 mark

One intercept: both turning points on the same side of the axis.

b < 0 or b > 4

(c)(iii)

1 mark

Three intercepts: the maximum $b$ is positive and the minimum $b - 4$ is negative.

0 < b < 4

(d)

1 mark

With three intercepts the curve rises through an axis crossing to a local maximum, falls through a second crossing to a local minimum and rises through a third crossing. So the local maximum is above the axis and the local minimum is below it.

(e)

5 marks

Differentiate $y = x^{3} + a x^{2} + b$ .
Find the $y$ -coordinates.
So the points are $P (0, b)$ and $Q (- \frac{2 a}{3}, \frac{4 a ^{3}}{27} + b)$ .

\frac{d y}{d x} = 3 x^{2} + 2 a x = x (3 x + 2 a) = 0 \Rightarrow x = 0 or x = - \frac{2 a}{3} x = 0 : y = b x = - \frac{2 a}{3} : y = - \frac{8 a ^{3}}{27} + \frac{4 a ^{3}}{9} + b = \frac{4 a ^{3}}{27} + b

(f)(i)

3 marks

The second derivative.
At P, $\frac{d ^{2} y}{d x ^{2}} = 2 a > 0$ so P is a local minimum. At Q, $\frac{d ^{2} y}{d x ^{2}} = - 4 a + 2 a = - 2 a < 0$ so Q is a local maximum.

\frac{d ^{2} y}{d x ^{2}} = 6 x + 2 a

(f)(ii)

1 mark

With $a > 0$ and $b > 0$ : P has $y = b > 0$ , and Q has $y = \frac{4 a ^{3}}{27} + b > 0$ . Both are above the x-axis.

(g)(i)

1 mark

With $a < 0$ : $\frac{d ^{2} y}{d x ^{2}}$ at P is $2 a < 0$ , so P is a local maximum, and since $b > 0$ it is above the x-axis.

(g)(ii)

1 mark

With $a < 0$ the point Q is a local minimum, with $y = \frac{4 a ^{3}}{27} + b$ . It is below the axis when this is negative.

4 a^{3} + 27 b < 0

(h)

5 marks

The condition $4 a^{3} b + 27 b^{2} = b (4 a^{3} + 27 b) < 0$ means $b$ and $4 a^{3} + 27 b$ have opposite signs.
Case 1: $b > 0$ and $4 a^{3} + 27 b < 0$ . The second inequality needs $a^{3} < 0$ , so $a < 0$ . By (g), P is a local maximum above the axis and Q is a local minimum below the axis, since $\frac{4 a ^{3} + 27 b}{27} < 0$ .
Case 2: $b < 0$ and $4 a^{3} + 27 b > 0$ . This needs $a^{3} > 0$ so $a > 0$ . By (f), P is a local minimum, below the axis since $b < 0$ , and Q is a local maximum above the axis, since $\frac{4 a ^{3} + 27 b}{27} > 0$ .
In both cases the local maximum is above the axis and the local minimum is below it. The cubic tends to $- \infty$ on the left and $+ \infty$ on the right, so the curve crosses the axis once before the maximum, once between the two turning points and once after the minimum. That gives exactly three x-intercepts.

Question 2

31 marks

Orthogonal families of curves and curves meeting at an acute angle

Calculus
Functions

(a)(i)

1 mark

The line $y = m x$ has gradient $m = \frac{y}{x}$ .

gradient of L = \frac{y}{x}

(a)(ii)

1 mark

The curves $C$ meet every line at right angles, so their gradient is the negative reciprocal.

\frac{d y}{d x} = - \frac{x}{y}

(a)(iii)

2 marks

Separate the variables and integrate.

\int y d y = - \int x d x \Rightarrow \frac{y ^{2}}{2} = - \frac{x ^{2}}{2} + c \Rightarrow x^{2} + y^{2} = k

(b)

3 marks

The curve $y^{2} = 16 - 8 x$ is a parabola opening to the left with vertex (and $x$ -intercept) at $(2, 0)$ . The curve $y^{2} = 4 + 4 x$ is a parabola opening to the right with vertex (and $x$ -intercept) at $(- 1, 0)$ .
Sketch: both parabolas about the $x$ -axis, labelled with their equations, the left-opening one passing through $(2, 0)$ and the right-opening one through $(- 1, 0)$ , crossing each other above and below the axis.

(c)

6 marks

Equate the two expressions for $y^{2}$ .
Substitute back.

4 a^{2} - 4 a x = 4 b^{2} + 4 b x \Rightarrow 4 (a^{2} - b^{2}) = 4 x (a + b) \Rightarrow x = a - b y^{2} = 4 a^{2} - 4 a (a - b) = 4 ab \Rightarrow y = \pm 2 ab M (a - b, 2 ab), N (a - b, - 2 ab)

(d)

5 marks

Differentiate each curve implicitly.
At M, $y^{2} = 4 ab$ . The product of the gradients is
The product is $- 1$ , so the curves intersect at right angles.

2 y \frac{d y}{d x} = - 4 a \Rightarrow \frac{d y}{d x} = - \frac{2 a}{y}, 2 y \frac{d y}{d x} = 4 b \Rightarrow \frac{d y}{d x} = \frac{2 b}{y} (- \frac{2 a}{y}) (\frac{2 b}{y}) = - \frac{4 ab}{y ^{2}} = - \frac{4 ab}{4 ab} = - 1

(e)(i)

2 marks

With $α = \frac{π}{4}$ , $tan α = 1$ , and $f (x, y) = - \frac{x}{y}$ .

g (x, y) = \frac{f + 1}{1 - f} = \frac{- \frac{x}{y} + 1}{1 + \frac{x}{y}} = \frac{y - x}{y + x}

(e)(ii)

9 marks

Solve $\frac{d y}{d x} = \frac{y - x}{y + x}$ with $y = v x$ .
Separate the variables.
Replace $v = \frac{y}{x}$ and combine the logarithms.

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

(f)

2 marks

Divide the top and bottom of $g = \frac{f + tan α}{1 - f tan α}$ by $tan α$ .
As $α \to \frac{π}{2}$ , $tan α \to \infty$ , so $\frac{1}{tan α} \to 0$ and for finite $f$ :

g = \frac{\frac{f}{t a n α} + 1}{\frac{1}{t a n α} - f} α \to π /2 lim g (x, y) = \frac{0 + 1}{0 - f ( x , y )} = - \frac{1}{f ( x , y )}

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Calculus carries the most marks on this paper.

Calculus44 marks · 80%
Q1, Q2
Functions11 marks · 20%
Q1, Q2

Mentor insights

Traps to watch for

01
Intercepts depend on the turning points · Q1
A cubic has exactly two x-intercepts when a turning point sits on the axis, and three when the local maximum is above and the local minimum below.
02
Sign of the second derivative · Q1
Positive means a local minimum and negative means a local maximum.
03
Perpendicular means gradients multiply to -1 · Q2
The gradient of the second family is the negative reciprocal of the first.
04
Homogeneous equation · Q2
Use $y = v x$ and separate the variables in $v$ and $x$ .
05
A limit with tan · Q2
As $α \to \frac{π}{2}$ , $tan α \to \infty$ , so divide the top and bottom by $tan α$ .

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

Session: November 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 November 2023 Global 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

The family of cubics y = x^3 + ax^2 + b: how many x-intercepts?

Functions
Calculus

(a)

2 marks

b = 0 or b = - 4

(b)(i)

1 mark

One intercept occurs when both turning points are above the axis ( $b > 0$ ) or both below ( $4 + b < 0$ ).

b < - 4 or b > 0

(b)(ii)

1 mark

Three intercepts occur when the local maximum is above and the local minimum below the axis.

- 4 < b < 0

(c)(i)

1 mark

For $y = x^{3} - 3 x^{2} + b$ the local maximum is at $x = 0$ with $y = b$ and the local minimum is at $x = 2$ with $y = b - 4$ . Two intercepts occur when a turning point is on the axis.

b = 0 or b = 4

(c)(ii)

1 mark

One intercept: both turning points on the same side of the axis.

b < 0 or b > 4

(c)(iii)

1 mark

Three intercepts: the maximum $b$ is positive and the minimum $b - 4$ is negative.

0 < b < 4

(d)

1 mark

(e)

5 marks

Differentiate $y = x^{3} + a x^{2} + b$ .
Find the $y$ -coordinates.
So the points are $P (0, b)$ and $Q (- \frac{2 a}{3}, \frac{4 a ^{3}}{27} + b)$ .

\frac{d y}{d x} = 3 x^{2} + 2 a x = x (3 x + 2 a) = 0 \Rightarrow x = 0 or x = - \frac{2 a}{3} x = 0 : y = b x = - \frac{2 a}{3} : y = - \frac{8 a ^{3}}{27} + \frac{4 a ^{3}}{9} + b = \frac{4 a ^{3}}{27} + b

(f)(i)

3 marks

The second derivative.
At P, $\frac{d ^{2} y}{d x ^{2}} = 2 a > 0$ so P is a local minimum. At Q, $\frac{d ^{2} y}{d x ^{2}} = - 4 a + 2 a = - 2 a < 0$ so Q is a local maximum.

\frac{d ^{2} y}{d x ^{2}} = 6 x + 2 a

(f)(ii)

1 mark

With $a > 0$ and $b > 0$ : P has $y = b > 0$ , and Q has $y = \frac{4 a ^{3}}{27} + b > 0$ . Both are above the x-axis.

(g)(i)

1 mark

With $a < 0$ : $\frac{d ^{2} y}{d x ^{2}}$ at P is $2 a < 0$ , so P is a local maximum, and since $b > 0$ it is above the x-axis.

(g)(ii)

1 mark

With $a < 0$ the point Q is a local minimum, with $y = \frac{4 a ^{3}}{27} + b$ . It is below the axis when this is negative.

4 a^{3} + 27 b < 0

(h)

5 marks

The condition $4 a^{3} b + 27 b^{2} = b (4 a^{3} + 27 b) < 0$ means $b$ and $4 a^{3} + 27 b$ have opposite signs.
Case 1: $b > 0$ and $4 a^{3} + 27 b < 0$ . The second inequality needs $a^{3} < 0$ , so $a < 0$ . By (g), P is a local maximum above the axis and Q is a local minimum below the axis, since $\frac{4 a ^{3} + 27 b}{27} < 0$ .
Case 2: $b < 0$ and $4 a^{3} + 27 b > 0$ . This needs $a^{3} > 0$ so $a > 0$ . By (f), P is a local minimum, below the axis since $b < 0$ , and Q is a local maximum above the axis, since $\frac{4 a ^{3} + 27 b}{27} > 0$ .
In both cases the local maximum is above the axis and the local minimum is below it. The cubic tends to $- \infty$ on the left and $+ \infty$ on the right, so the curve crosses the axis once before the maximum, once between the two turning points and once after the minimum. That gives exactly three x-intercepts.

Question 2

31 marks

Orthogonal families of curves and curves meeting at an acute angle

Calculus
Functions

(a)(i)

1 mark

The line $y = m x$ has gradient $m = \frac{y}{x}$ .

gradient of L = \frac{y}{x}

(a)(ii)

1 mark

The curves $C$ meet every line at right angles, so their gradient is the negative reciprocal.

\frac{d y}{d x} = - \frac{x}{y}

(a)(iii)

2 marks

Separate the variables and integrate.

\int y d y = - \int x d x \Rightarrow \frac{y ^{2}}{2} = - \frac{x ^{2}}{2} + c \Rightarrow x^{2} + y^{2} = k

(b)

3 marks

The curve $y^{2} = 16 - 8 x$ is a parabola opening to the left with vertex (and $x$ -intercept) at $(2, 0)$ . The curve $y^{2} = 4 + 4 x$ is a parabola opening to the right with vertex (and $x$ -intercept) at $(- 1, 0)$ .
Sketch: both parabolas about the $x$ -axis, labelled with their equations, the left-opening one passing through $(2, 0)$ and the right-opening one through $(- 1, 0)$ , crossing each other above and below the axis.

(c)

6 marks

Equate the two expressions for $y^{2}$ .
Substitute back.

4 a^{2} - 4 a x = 4 b^{2} + 4 b x \Rightarrow 4 (a^{2} - b^{2}) = 4 x (a + b) \Rightarrow x = a - b y^{2} = 4 a^{2} - 4 a (a - b) = 4 ab \Rightarrow y = \pm 2 ab M (a - b, 2 ab), N (a - b, - 2 ab)

(d)

5 marks

Differentiate each curve implicitly.
At M, $y^{2} = 4 ab$ . The product of the gradients is
The product is $- 1$ , so the curves intersect at right angles.

2 y \frac{d y}{d x} = - 4 a \Rightarrow \frac{d y}{d x} = - \frac{2 a}{y}, 2 y \frac{d y}{d x} = 4 b \Rightarrow \frac{d y}{d x} = \frac{2 b}{y} (- \frac{2 a}{y}) (\frac{2 b}{y}) = - \frac{4 ab}{y ^{2}} = - \frac{4 ab}{4 ab} = - 1

(e)(i)

2 marks

With $α = \frac{π}{4}$ , $tan α = 1$ , and $f (x, y) = - \frac{x}{y}$ .

g (x, y) = \frac{f + 1}{1 - f} = \frac{- \frac{x}{y} + 1}{1 + \frac{x}{y}} = \frac{y - x}{y + x}

(e)(ii)

9 marks

Solve $\frac{d y}{d x} = \frac{y - x}{y + x}$ with $y = v x$ .
Separate the variables.
Replace $v = \frac{y}{x}$ and combine the logarithms.

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

(f)

2 marks

Divide the top and bottom of $g = \frac{f + tan α}{1 - f tan α}$ by $tan α$ .
As $α \to \frac{π}{2}$ , $tan α \to \infty$ , so $\frac{1}{tan α} \to 0$ and for finite $f$ :

g = \frac{\frac{f}{t a n α} + 1}{\frac{1}{t a n α} - f} α \to π /2 lim g (x, y) = \frac{0 + 1}{0 - f ( x , y )} = - \frac{1}{f ( x , y )}

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Calculus carries the most marks on this paper.

Calculus44 marks · 80%
Q1, Q2
Functions11 marks · 20%
Q1, Q2

Mentor insights

Traps to watch for

01
Intercepts depend on the turning points · Q1
A cubic has exactly two x-intercepts when a turning point sits on the axis, and three when the local maximum is above and the local minimum below.
02
Sign of the second derivative · Q1
Positive means a local minimum and negative means a local maximum.
03
Perpendicular means gradients multiply to -1 · Q2
The gradient of the second family is the negative reciprocal of the first.
04
Homogeneous equation · Q2
Use $y = v x$ and separate the variables in $v$ and $x$ .
05
A limit with tan · Q2
As $α \to \frac{π}{2}$ , $tan α \to \infty$ , so divide the top and bottom by $tan α$ .

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