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IB Maths AA SL · Paper 1

Mathematics: Analysis & Approaches, Standard Level

Session: November 2022
Marks: 80
Duration: 1 hour 30 min
Calculator: Not allowed
Questions: 9

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

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Section A · Short response

Questions 1–6 · 34 marks

Question 1

7 marks

Parallel and perpendicular linear functions, then a composite

Functions

(a)

2 marks

Gradient $- 2$ and passing through the origin.

g (x) = - 2 x

(b)

3 marks

Gradient $\frac{1}{2}$ through $(- 1, 2)$ .

h (x) - 2 = \frac{1}{2} (x + 1) \Rightarrow h (x) = \frac{1}{2} x + \frac{5}{2}

(c)

2 marks

Work from the inside out: $h (0) = \frac{5}{2}$ .

(g \circ h) (0) = g (\frac{5}{2}) = - 5

Question 2

4 marks

Derivative of an exponential with a squared exponent

Calculus

4 marks

Differentiate using the chain rule.
Substitute $x = - 1$ .

g^{'} (x) = 2 x e^{x^{2} + 1} g^{'} (- 1) = 2 (- 1) e^{2} = - 2 e^{2}

Question 3

7 marks

A circle on a diameter and the volume of a cone

Geometry and trigonometry

(a)

4 marks

The centre is the midpoint of the diameter AB.
The radius is half of AB.

(\frac{1 + ( - 3 )}{2}, \frac{4 + 2}{2}, \frac{0 + ( - 4 )}{2}) = (- 1, 3, - 2) A B = 4^{2} + 2^{2} + 4^{2} = 6 \Rightarrow r = 3

(b)

3 marks

The height is the distance from the centre $(- 1, 3, - 2)$ to $V (- 1, - 1, 0)$ .
Volume of a cone:

h = 0^{2} + 4^{2} + 2^{2} = 20 = 25 V = \frac{1}{3} π (3)^{2} (25) = 65 π

Question 4

5 marks

A right-angled triangle with an algebraic identity

Number and algebra

(a)

3 marks

Start from the left-hand side and expand the bracket.
Put it over a common denominator.

LHS = a^{2} + (\frac{a ^{2} - 1}{2})^{2} = a^{2} + \frac{a ^{4} - 2 a ^{2} + 1}{4} \frac{4 a ^{2} + a ^{4} - 2 a ^{2} + 1}{4} = \frac{a ^{4} + 2 a ^{2} + 1}{4} = (\frac{a ^{2} + 1}{2})^{2} = RHS

(b)

2 marks

The two shorter sides are $a$ and $\frac{a ^{2} - 1}{2}$ , because the longest side is $\frac{a ^{2} + 1}{2}$ .

Area = \frac{1}{2} \cdot a \cdot \frac{a ^{2} - 1}{2} = \frac{a ( a ^{2} - 1 )}{4}

Question 5

5 marks

Recovering f from f' = 6x/(x^2 + 1)

Calculus

5 marks

Let $u = x^{2} + 1$ , so $d u = 2 x d x$ and $6 x d x = 3 d u$ .
Use $f (1) = 5$ .

f (x) = \int \frac{6 x}{x ^{2} + 1} d x = 3 ln (x^{2} + 1) + c 5 = 3 ln 2 + c \Rightarrow c = 5 - 3 ln 2 f (x) = 3 ln (x^{2} + 1) + 5 - 3 ln 2

Question 6

6 marks

Bounds on P(A and B) for two events

Statistics and probability

(a)

1 mark

For independent events the probability of both is the product.

P (A \cap B) = 0.3 \times 0.8 = 0.24

(b)

3 marks

Rearrange the addition rule.
The union cannot be more than 1, so
The minimum is 0.1.

P (A \cap B) = P (A) + P (B) - P (A \cup B) = 1.1 - P (A \cup B) P (A \cap B) \geq 1.1 - 1 = 0.1

(c)

2 marks

The intersection cannot be larger than the smaller event. If $A$ is entirely inside $B$ , then $A \cap B = A$ .

max P (A \cap B) = P (A) = 0.3

Section B · Extended response

Questions 7–9 · 46 marks

Question 7

16 marks

A quadratic from its vertex, a family of quadratics, and intersections

Functions

(a)

3 marks

The form is $f (x) = a (x - 3)^{2} + 2$ . Substitute $(5, 0)$ .

0 = 4 a + 2 \Rightarrow a = - \frac{1}{2} f (x) = - \frac{1}{2} (x - 3)^{2} + 2

(b)

7 marks

$g (1) = p + (t - 1) - p = t - 1$ .
Then $g (- 3) = 9 p - 3 (4) - p = 8 p - 12 = 4$ .
So $g (x) = 2 x^{2} + 4 x - 2$ , with vertex at $x = - 1$ .

t - 1 = 4 \Rightarrow t = 5 p = 2 g (- 1) = 2 - 4 - 2 = - 4 \Rightarrow range: g (x) \geq - 4

(c)

6 marks

Set $j (x) = g (x)$ .
The discriminant is
Since $p \neq = 0$ , $16 p^{2} > 0$ and $t^{2} \geq 0$ , so $Δ > 0$ for every $p$ and $t$ . The equation always has two distinct roots, so the graphs meet at two distinct points.

- x + 3 p = p x^{2} + (t - 1) x - p \Rightarrow p x^{2} + t x - 4 p = 0 Δ = t^{2} + 16 p^{2}

Question 8

15 marks

Logarithm values and laws with a base ab

Functions

(a)

7 marks

Write each argument as a power of the base.

lo g_{2} \frac{1}{16} = lo g_{2} 2^{- 4} = - 4 lo g_{9} 3 = lo g_{9} 9^{1/2} = \frac{1}{2} lo g_{3} 81 = lo g_{3} (3)^{8} = 8

(b)

8 marks

Since $lo g_{ab} (ab) = 1$ , the log law for products gives $lo g_{ab} a + lo g_{ab} b = 1$ .
Now use the laws for powers and quotients.

3 + lo g_{ab} b = 1 \Rightarrow lo g_{ab} b = - 2 lo g_{ab} \frac{3 a}{b} = \frac{1}{3} lo g_{ab} a - \frac{1}{2} lo g_{ab} b = \frac{1}{3} (3) - \frac{1}{2} (- 2) = 2

Question 9

15 marks

f(x) = cos^2 x - 3 sin^2 x: roots, turning points and a sketch

Functions
Calculus

(a)

5 marks

Set $f (x) = 0$ .

cos^{2} x = 3 sin^{2} x \Rightarrow tan^{2} x = \frac{1}{3} \Rightarrow tan x = \pm \frac{1}{3} x = \frac{π}{6}, \frac{5 π}{6}

(b)

7 marks

Differentiate.
Set $f^{'} (x) = 0$ so $sin 2 x = 0$ for $0 \leq x \leq π$ .
Substitute into $f$ .

f^{'} (x) = - 2 cos x sin x - 6 sin x cos x = - 8 sin x cos x = - 4 sin 2 x x = 0, \frac{π}{2}, π (0, 1), (\frac{π}{2}, - 3), (π, 1)

(c)

3 marks

Sketch: the graph starts at $(0, 1)$ , falls through $(\frac{π}{6}, 0)$ to the minimum at $(\frac{π}{2}, - 3)$ , rises through $(\frac{5 π}{6}, 0)$ and finishes at $(π, 1)$ . The $y$ -intercept is $(0, 1)$ .

Our analysis

Where the marks are

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

Functions46 marks · 57%
Q1, Q7, Q8, Q9
Calculus16 marks · 20%
Q2, Q5, Q9
Geometry and trigonometry7 marks · 9%
Q3
Statistics and probability6 marks · 8%
Q6
Number and algebra5 marks · 6%
Q4

Mentor insights

Traps to watch for

01
Parallel means equal gradient · Q1
A parallel line to $f (x) = - 2 x + 3$ has gradient $- 2$ .
02
Height of the cone · Q3
The cone's height is the distance from the centre of the base to the vertex, not the distance from A to V.
03
Use the vertex form · Q7
With the vertex $(3, 2)$ and a root at 5, substitute the root to find $a$ .
04
Write numbers as powers of the base · Q8
$\frac{1}{16} = 2^{- 4}$ , $3 = 9^{1/2}$ and $81 = (3)^{8}$ .
05
Include both roots in the domain · Q9
$tan^{2} x = \frac{1}{3}$ gives $tan x = \pm \frac{1}{3}$ , so there are two solutions in $[0, π]$ .

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 SL · Paper 1

Mathematics: Analysis & Approaches, Standard Level

Session: November 2022
Marks: 80
Duration: 1 hour 30 min
Calculator: Not allowed
Questions: 9

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2022 Global 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 1–6 · 34 marks

Question 1

7 marks

Parallel and perpendicular linear functions, then a composite

Functions

(a)

2 marks

Gradient $- 2$ and passing through the origin.

g (x) = - 2 x

(b)

3 marks

Gradient $\frac{1}{2}$ through $(- 1, 2)$ .

h (x) - 2 = \frac{1}{2} (x + 1) \Rightarrow h (x) = \frac{1}{2} x + \frac{5}{2}

(c)

2 marks

Work from the inside out: $h (0) = \frac{5}{2}$ .

(g \circ h) (0) = g (\frac{5}{2}) = - 5

Question 2

4 marks

Derivative of an exponential with a squared exponent

Calculus

4 marks

Differentiate using the chain rule.
Substitute $x = - 1$ .

g^{'} (x) = 2 x e^{x^{2} + 1} g^{'} (- 1) = 2 (- 1) e^{2} = - 2 e^{2}

Question 3

7 marks

A circle on a diameter and the volume of a cone

Geometry and trigonometry

(a)

4 marks

The centre is the midpoint of the diameter AB.
The radius is half of AB.

(\frac{1 + ( - 3 )}{2}, \frac{4 + 2}{2}, \frac{0 + ( - 4 )}{2}) = (- 1, 3, - 2) A B = 4^{2} + 2^{2} + 4^{2} = 6 \Rightarrow r = 3

(b)

3 marks

The height is the distance from the centre $(- 1, 3, - 2)$ to $V (- 1, - 1, 0)$ .
Volume of a cone:

h = 0^{2} + 4^{2} + 2^{2} = 20 = 25 V = \frac{1}{3} π (3)^{2} (25) = 65 π

Question 4

5 marks

A right-angled triangle with an algebraic identity

Number and algebra

(a)

3 marks

Start from the left-hand side and expand the bracket.
Put it over a common denominator.

LHS = a^{2} + (\frac{a ^{2} - 1}{2})^{2} = a^{2} + \frac{a ^{4} - 2 a ^{2} + 1}{4} \frac{4 a ^{2} + a ^{4} - 2 a ^{2} + 1}{4} = \frac{a ^{4} + 2 a ^{2} + 1}{4} = (\frac{a ^{2} + 1}{2})^{2} = RHS

(b)

2 marks

The two shorter sides are $a$ and $\frac{a ^{2} - 1}{2}$ , because the longest side is $\frac{a ^{2} + 1}{2}$ .

Area = \frac{1}{2} \cdot a \cdot \frac{a ^{2} - 1}{2} = \frac{a ( a ^{2} - 1 )}{4}

Question 5

5 marks

Recovering f from f' = 6x/(x^2 + 1)

Calculus

5 marks

Let $u = x^{2} + 1$ , so $d u = 2 x d x$ and $6 x d x = 3 d u$ .
Use $f (1) = 5$ .

f (x) = \int \frac{6 x}{x ^{2} + 1} d x = 3 ln (x^{2} + 1) + c 5 = 3 ln 2 + c \Rightarrow c = 5 - 3 ln 2 f (x) = 3 ln (x^{2} + 1) + 5 - 3 ln 2

Question 6

6 marks

Bounds on P(A and B) for two events

Statistics and probability

(a)

1 mark

For independent events the probability of both is the product.

P (A \cap B) = 0.3 \times 0.8 = 0.24

(b)

3 marks

Rearrange the addition rule.
The union cannot be more than 1, so
The minimum is 0.1.

P (A \cap B) = P (A) + P (B) - P (A \cup B) = 1.1 - P (A \cup B) P (A \cap B) \geq 1.1 - 1 = 0.1

(c)

2 marks

The intersection cannot be larger than the smaller event. If $A$ is entirely inside $B$ , then $A \cap B = A$ .

max P (A \cap B) = P (A) = 0.3

Section B · Extended response

Questions 7–9 · 46 marks

Question 7

16 marks

A quadratic from its vertex, a family of quadratics, and intersections

Functions

(a)

3 marks

The form is $f (x) = a (x - 3)^{2} + 2$ . Substitute $(5, 0)$ .

0 = 4 a + 2 \Rightarrow a = - \frac{1}{2} f (x) = - \frac{1}{2} (x - 3)^{2} + 2

(b)

7 marks

$g (1) = p + (t - 1) - p = t - 1$ .
Then $g (- 3) = 9 p - 3 (4) - p = 8 p - 12 = 4$ .
So $g (x) = 2 x^{2} + 4 x - 2$ , with vertex at $x = - 1$ .

t - 1 = 4 \Rightarrow t = 5 p = 2 g (- 1) = 2 - 4 - 2 = - 4 \Rightarrow range: g (x) \geq - 4

(c)

6 marks

Set $j (x) = g (x)$ .
The discriminant is
Since $p \neq = 0$ , $16 p^{2} > 0$ and $t^{2} \geq 0$ , so $Δ > 0$ for every $p$ and $t$ . The equation always has two distinct roots, so the graphs meet at two distinct points.

- x + 3 p = p x^{2} + (t - 1) x - p \Rightarrow p x^{2} + t x - 4 p = 0 Δ = t^{2} + 16 p^{2}

Question 8

15 marks

Logarithm values and laws with a base ab

Functions

(a)

7 marks

Write each argument as a power of the base.

lo g_{2} \frac{1}{16} = lo g_{2} 2^{- 4} = - 4 lo g_{9} 3 = lo g_{9} 9^{1/2} = \frac{1}{2} lo g_{3} 81 = lo g_{3} (3)^{8} = 8

(b)

8 marks

Since $lo g_{ab} (ab) = 1$ , the log law for products gives $lo g_{ab} a + lo g_{ab} b = 1$ .
Now use the laws for powers and quotients.

3 + lo g_{ab} b = 1 \Rightarrow lo g_{ab} b = - 2 lo g_{ab} \frac{3 a}{b} = \frac{1}{3} lo g_{ab} a - \frac{1}{2} lo g_{ab} b = \frac{1}{3} (3) - \frac{1}{2} (- 2) = 2

Question 9

15 marks

f(x) = cos^2 x - 3 sin^2 x: roots, turning points and a sketch

Functions
Calculus

(a)

5 marks

Set $f (x) = 0$ .

cos^{2} x = 3 sin^{2} x \Rightarrow tan^{2} x = \frac{1}{3} \Rightarrow tan x = \pm \frac{1}{3} x = \frac{π}{6}, \frac{5 π}{6}

(b)

7 marks

Differentiate.
Set $f^{'} (x) = 0$ so $sin 2 x = 0$ for $0 \leq x \leq π$ .
Substitute into $f$ .

f^{'} (x) = - 2 cos x sin x - 6 sin x cos x = - 8 sin x cos x = - 4 sin 2 x x = 0, \frac{π}{2}, π (0, 1), (\frac{π}{2}, - 3), (π, 1)

(c)

3 marks

Our analysis

Where the marks are

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

Functions46 marks · 57%
Q1, Q7, Q8, Q9
Calculus16 marks · 20%
Q2, Q5, Q9
Geometry and trigonometry7 marks · 9%
Q3
Statistics and probability6 marks · 8%
Q6
Number and algebra5 marks · 6%
Q4

Mentor insights

Traps to watch for

01
Parallel means equal gradient · Q1
A parallel line to $f (x) = - 2 x + 3$ has gradient $- 2$ .
02
Height of the cone · Q3
The cone's height is the distance from the centre of the base to the vertex, not the distance from A to V.
03
Use the vertex form · Q7
With the vertex $(3, 2)$ and a root at 5, substitute the root to find $a$ .
04
Write numbers as powers of the base · Q8
$\frac{1}{16} = 2^{- 4}$ , $3 = 9^{1/2}$ and $81 = (3)^{8}$ .
05
Include both roots in the domain · Q9
$tan^{2} x = \frac{1}{3}$ gives $tan x = \pm \frac{1}{3}$ , so there are two solutions in $[0, π]$ .

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