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IB Maths AA HL · Paper 3

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2024
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

When the product rule for f times g also equals f' times g'

Calculus

(a)(i)

2 marks

Write $f (x) = (2 - x)^{- 2}$ and differentiate.

f^{'} (x) = - 2 (2 - x)^{- 3} (- 1) = \frac{2}{( 2 - x ) ^{3}}

(a)(ii)

2 marks

Multiply by $g^{'} (x) = 2 x$ .

f^{'} (x) g^{'} (x) = \frac{2}{( 2 - x ) ^{3}} \times 2 x = \frac{4 x}{( 2 - x ) ^{3}}

(a)(iii)

4 marks

Evaluate the product rule expression.
Use the common denominator $(2 - x)^{3}$ .

f (x) g^{'} (x) + g (x) f^{'} (x) = \frac{2 x}{( 2 - x ) ^{2}} + \frac{2 x ^{2}}{( 2 - x ) ^{3}} \frac{2 x ( 2 - x ) + 2 x ^{2}}{( 2 - x ) ^{3}} = \frac{4 x}{( 2 - x ) ^{3}}

(b)

2 marks

Start from $f g^{'} + g f^{'} = f^{'} g^{'}$ and collect the $f^{'}$ terms on one side.
Divide by $f (g^{'} - g)$ , which is allowed since $f > 0$ and $g \neq = g^{'}$ .

f g^{'} = f^{'} g^{'} - g f^{'} = f^{'} (g^{'} - g) \frac{f ^{'}}{f} = \frac{g ^{'}}{g ^{'} - g}

(c)

2 marks

Integrate both sides with respect to $x$ .
Take the exponential and put $A = e^{C}$ (positive).

ln f (x) = \int \frac{g ^{'} ( x )}{g ^{'} ( x ) - g ( x )} d x + C f (x) = A e^{\int \frac{g ^{'} ( x )}{g ^{'} ( x ) - g ( x )} d x}

(d)

5 marks

Differentiate $g = x e^{x}$ and form the combination.
Integrate and set $A = 1$ .

g^{'} = e^{x} (1 + x), g^{'} - g = e^{x} \frac{g ^{'}}{g ^{'} - g} = 1 + x \int (1 + x) d x = x + \frac{x ^{2}}{2} \Rightarrow f (x) = e^{x + \frac{x ^{2}}{2}}

(e)

7 marks

Differentiate $g = sin x + cos x$ .
Integrate. On $0 < x < π$ we have $sin x > 0$ , so no modulus is needed.
With $A = 1$ , the answer in the form $e^{h (x)}$ is

g^{'} = cos x - sin x, g^{'} - g = - 2 sin x \frac{g ^{'}}{g ^{'} - g} = \frac{cos x - sin x}{- 2 sin x} = \frac{1}{2} - \frac{1}{2} cot x \int (\frac{1}{2} - \frac{1}{2} cot x) d x = \frac{x}{2} - \frac{1}{2} ln (sin x) f (x) = e^{\frac{x}{2} - \frac{1}{2} l n (s i n x)} = \frac{e ^{x /2}}{sin x}, h (x) = \frac{x}{2} - \frac{1}{2} ln (sin x)

Question 2

31 marks

Random quadratics from three dice rolls: discriminants and probabilities

Statistics and probability

(a)

1 mark

Each of $a$ , $b$ and $c$ has 6 possible values, and the rolls are independent.

6 \times 6 \times 6 = 216

(b)

2 marks

Find the discriminant for $x^{2} + 4 x + 4$ .
A zero discriminant means one repeated root, so one $x$ -intercept.

Δ = 4^{2} - 4 (1) (4) = 0

(c)

6 marks

One intercept needs $b^{2} = 4 a c$ , so $b$ must be even, and $a c = \frac{b ^{2}}{4}$ .
$b = 2$ : $a c = 1$ , so $(a, c) = (1, 1)$ , 1 way.
$b = 4$ : $a c = 4$ , so $(1, 4), (2, 2), (4, 1)$ , 3 ways.
$b = 6$ : $a c = 9$ , so $(3, 3)$ , 1 way.

total = 1 + 3 + 1 = 5 \Rightarrow P = \frac{5}{216}

(d)

3 marks

Two intercepts need $b^{2} > 4 a c$ for some $b \leq 6$ , so $4 a c < 36$ .
The products of two dice that are less than 9 are

a c < 9 a c \in {1, 2, 3, 4, 5, 6, 8}

(e)(i)

1 mark

With $a c = 1$ we need $a = c = 1$ , and $b^{2} > 4$ gives $b = 3, 4, 5, 6$ .

4 functions

(e)(ii)

2 marks

With $a c = 2$ the pairs are $(1, 2)$ and $(2, 1)$ , and $b^{2} > 8$ gives $b = 3, 4, 5, 6$ for each pair.

2 \times 4 = 8 functions

(f)

6 marks

For each value of $a c$ , count the pairs $(a, c)$ and the values of $b$ with $b^{2} > 4 a c$ .
The total is $4 + 8 + 6 + 6 + 4 + 8 + 2 = 38$ .

a c = 1 : 1 \times 4 = 4, a c = 2 : 2 \times 4 = 8, a c = 3 : 2 \times 3 = 6, a c = 4 : 3 \times 2 = 6 a c = 5 : 2 \times 2 = 4, a c = 6 : 4 \times 2 = 8, a c = 8 : 2 \times 1 = 2 p = \frac{38}{216} = \frac{19}{108} \approx 0.176

(g)

3 marks

$f (x) = x^{2} + 2 Z x + 1$ has two intercepts when the discriminant $4 Z^{2} - 4 > 0$ , so $∣ Z ∣ > 1$ .

P (∣ Z ∣ > 1) = 2 P (Z > 1) = 1 - 0.6827 \approx 0.317

(h)

7 marks

The roots are $X_{1} = - Z - Z^{2} - 1$ (the smaller) and $X_{2} = - Z + Z^{2} - 1$ . Both are greater than 0.5 exactly when the smaller one is.
Solve $- Z - Z^{2} - 1 > 0.5$ . This needs $- Z - 0.5 > 0$ , that is $Z < - 0.5$ , and then squaring $(- Z - 0.5)^{2} > Z^{2} - 1$ gives
With $∣ Z ∣ > 1$ and $Z < - 0.5$ we need
The probability that both roots exceed 0.5 and there are two intercepts is
Conditional on two intercepts:

Z^{2} + Z + 0.25 > Z^{2} - 1 \Rightarrow Z > - 1.25 - 1.25 < Z < - 1 P (- 1.25 < Z < - 1) \approx 0.0530 \frac{0.0530}{0.3173} \approx 0.167

Our analysis

Where the marks are

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

Statistics and probability31 marks · 56%
Q2
Calculus24 marks · 44%
Q1

Mentor insights

Traps to watch for

01
Rearrange · Q1
$f g^{'} + g f^{'} = f^{'} g^{'}$ gives $f g^{'} = f^{'} (g^{'} - g)$ , a separable equation in $f$ .
02
Integrate the log derivative · Q1
$\frac{f ^{'}}{f} = \frac{g ^{'}}{g ^{'} - g}$ gives $ln f = \int \frac{g ^{'}}{g ^{'} - g} d x$ .
03
Discriminant · Q2
The graph of $a x^{2} + b x + c$ has two intercepts when $b^{2} > 4 a c$ and one when $b^{2} = 4 a c$ .
04
Count by cases · Q2
List the values of $a c$ allowed, and for each $a c$ count the values of $b$ with $b^{2} > 4 a c$ .

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 2024
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 2024 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

When the product rule for f times g also equals f' times g'

Calculus

(a)(i)

2 marks

Write $f (x) = (2 - x)^{- 2}$ and differentiate.

f^{'} (x) = - 2 (2 - x)^{- 3} (- 1) = \frac{2}{( 2 - x ) ^{3}}

(a)(ii)

2 marks

Multiply by $g^{'} (x) = 2 x$ .

f^{'} (x) g^{'} (x) = \frac{2}{( 2 - x ) ^{3}} \times 2 x = \frac{4 x}{( 2 - x ) ^{3}}

(a)(iii)

4 marks

Evaluate the product rule expression.
Use the common denominator $(2 - x)^{3}$ .

f (x) g^{'} (x) + g (x) f^{'} (x) = \frac{2 x}{( 2 - x ) ^{2}} + \frac{2 x ^{2}}{( 2 - x ) ^{3}} \frac{2 x ( 2 - x ) + 2 x ^{2}}{( 2 - x ) ^{3}} = \frac{4 x}{( 2 - x ) ^{3}}

(b)

2 marks

Start from $f g^{'} + g f^{'} = f^{'} g^{'}$ and collect the $f^{'}$ terms on one side.
Divide by $f (g^{'} - g)$ , which is allowed since $f > 0$ and $g \neq = g^{'}$ .

f g^{'} = f^{'} g^{'} - g f^{'} = f^{'} (g^{'} - g) \frac{f ^{'}}{f} = \frac{g ^{'}}{g ^{'} - g}

(c)

2 marks

Integrate both sides with respect to $x$ .
Take the exponential and put $A = e^{C}$ (positive).

ln f (x) = \int \frac{g ^{'} ( x )}{g ^{'} ( x ) - g ( x )} d x + C f (x) = A e^{\int \frac{g ^{'} ( x )}{g ^{'} ( x ) - g ( x )} d x}

(d)

5 marks

Differentiate $g = x e^{x}$ and form the combination.
Integrate and set $A = 1$ .

g^{'} = e^{x} (1 + x), g^{'} - g = e^{x} \frac{g ^{'}}{g ^{'} - g} = 1 + x \int (1 + x) d x = x + \frac{x ^{2}}{2} \Rightarrow f (x) = e^{x + \frac{x ^{2}}{2}}

(e)

7 marks

Differentiate $g = sin x + cos x$ .
Integrate. On $0 < x < π$ we have $sin x > 0$ , so no modulus is needed.
With $A = 1$ , the answer in the form $e^{h (x)}$ is

g^{'} = cos x - sin x, g^{'} - g = - 2 sin x \frac{g ^{'}}{g ^{'} - g} = \frac{cos x - sin x}{- 2 sin x} = \frac{1}{2} - \frac{1}{2} cot x \int (\frac{1}{2} - \frac{1}{2} cot x) d x = \frac{x}{2} - \frac{1}{2} ln (sin x) f (x) = e^{\frac{x}{2} - \frac{1}{2} l n (s i n x)} = \frac{e ^{x /2}}{sin x}, h (x) = \frac{x}{2} - \frac{1}{2} ln (sin x)

Question 2

31 marks

Random quadratics from three dice rolls: discriminants and probabilities

Statistics and probability

(a)

1 mark

Each of $a$ , $b$ and $c$ has 6 possible values, and the rolls are independent.

6 \times 6 \times 6 = 216

(b)

2 marks

Find the discriminant for $x^{2} + 4 x + 4$ .
A zero discriminant means one repeated root, so one $x$ -intercept.

Δ = 4^{2} - 4 (1) (4) = 0

(c)

6 marks

One intercept needs $b^{2} = 4 a c$ , so $b$ must be even, and $a c = \frac{b ^{2}}{4}$ .
$b = 2$ : $a c = 1$ , so $(a, c) = (1, 1)$ , 1 way.
$b = 4$ : $a c = 4$ , so $(1, 4), (2, 2), (4, 1)$ , 3 ways.
$b = 6$ : $a c = 9$ , so $(3, 3)$ , 1 way.

total = 1 + 3 + 1 = 5 \Rightarrow P = \frac{5}{216}

(d)

3 marks

Two intercepts need $b^{2} > 4 a c$ for some $b \leq 6$ , so $4 a c < 36$ .
The products of two dice that are less than 9 are

a c < 9 a c \in {1, 2, 3, 4, 5, 6, 8}

(e)(i)

1 mark

With $a c = 1$ we need $a = c = 1$ , and $b^{2} > 4$ gives $b = 3, 4, 5, 6$ .

4 functions

(e)(ii)

2 marks

With $a c = 2$ the pairs are $(1, 2)$ and $(2, 1)$ , and $b^{2} > 8$ gives $b = 3, 4, 5, 6$ for each pair.

2 \times 4 = 8 functions

(f)

6 marks

For each value of $a c$ , count the pairs $(a, c)$ and the values of $b$ with $b^{2} > 4 a c$ .
The total is $4 + 8 + 6 + 6 + 4 + 8 + 2 = 38$ .

a c = 1 : 1 \times 4 = 4, a c = 2 : 2 \times 4 = 8, a c = 3 : 2 \times 3 = 6, a c = 4 : 3 \times 2 = 6 a c = 5 : 2 \times 2 = 4, a c = 6 : 4 \times 2 = 8, a c = 8 : 2 \times 1 = 2 p = \frac{38}{216} = \frac{19}{108} \approx 0.176

(g)

3 marks

$f (x) = x^{2} + 2 Z x + 1$ has two intercepts when the discriminant $4 Z^{2} - 4 > 0$ , so $∣ Z ∣ > 1$ .

P (∣ Z ∣ > 1) = 2 P (Z > 1) = 1 - 0.6827 \approx 0.317

(h)

7 marks

The roots are $X_{1} = - Z - Z^{2} - 1$ (the smaller) and $X_{2} = - Z + Z^{2} - 1$ . Both are greater than 0.5 exactly when the smaller one is.
Solve $- Z - Z^{2} - 1 > 0.5$ . This needs $- Z - 0.5 > 0$ , that is $Z < - 0.5$ , and then squaring $(- Z - 0.5)^{2} > Z^{2} - 1$ gives
With $∣ Z ∣ > 1$ and $Z < - 0.5$ we need
The probability that both roots exceed 0.5 and there are two intercepts is
Conditional on two intercepts:

Z^{2} + Z + 0.25 > Z^{2} - 1 \Rightarrow Z > - 1.25 - 1.25 < Z < - 1 P (- 1.25 < Z < - 1) \approx 0.0530 \frac{0.0530}{0.3173} \approx 0.167

Our analysis

Where the marks are

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

Statistics and probability31 marks · 56%
Q2
Calculus24 marks · 44%
Q1

Mentor insights

Traps to watch for

01
Rearrange · Q1
$f g^{'} + g f^{'} = f^{'} g^{'}$ gives $f g^{'} = f^{'} (g^{'} - g)$ , a separable equation in $f$ .
02
Integrate the log derivative · Q1
$\frac{f ^{'}}{f} = \frac{g ^{'}}{g ^{'} - g}$ gives $ln f = \int \frac{g ^{'}}{g ^{'} - g} d x$ .
03
Discriminant · Q2
The graph of $a x^{2} + b x + c$ has two intercepts when $b^{2} > 4 a c$ and one when $b^{2} = 4 a c$ .
04
Count by cases · Q2
List the values of $a c$ allowed, and for each $a c$ count the values of $b$ with $b^{2} > 4 a c$ .

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