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IB Maths AA SL · Paper 1

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2

Session: May 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 May 2022 TZ1 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 · 36 marks

Question 1

5 marks

Line perpendicular to BC through A

Geometry and trigonometry

(a)

3 marks

Gradient of BC:
The perpendicular gradient is 3. Use the point $A (- 2, 20)$ .

m_{B C} = \frac{12 - 6}{- 14 - 4} = - \frac{1}{3} y - 20 = 3 (x + 2) \Rightarrow y = 3 x + 26

(b)

2 marks

Put $(k, 2)$ into the equation.

2 = 3 k + 26 \Rightarrow k = - 8

Question 2

5 marks

Rewriting a quotient with a square root, then integrating

Calculus

(a)

1 mark

Divide each term by $x$ .

\frac{3 x - 5}{x} = 3 - 5 x^{- 1/2} \Rightarrow p = - \frac{1}{2}

(b)

4 marks

Integrate the rewritten form.
Substitute the limits.

\int (3 - 5 x^{- 1/2}) d x = 3 x - 10 x^{1/2} + c [3 x - 10 x]_{1}^{9} = (27 - 30) - (3 - 10) = 4

Question 3

7 marks

A box plot, outliers and two regression lines

Statistics and probability

(a)

3 marks

The quartiles are $Q_{1} = 6$ and $Q_{3} = 10$ , so the IQR is 4. A value is an outlier above $Q_{3} + 1.5 \times IQR$ .
So the largest value that is not an outlier is $c = 16$ .

10 + 1.5 \times 4 = 16

(b)

4 marks

(i) To estimate the child's age from the adult's, use the regression line of $c$ on $a$ .

(ii) Both lines pass through $(\overset{c}{ˉ}, \overset{a}{ˉ})$ . Substitute $c = \frac{1}{2} a - 9$ into $a = \frac{7}{4} c + 20$ .

The mean age of the adults is 34 years.

c = \frac{1}{2} (42) - 9 = 12 years a = \frac{7}{4} (\frac{1}{2} a - 9) + 20 = \frac{7}{8} a + 4.25 \Rightarrow \frac{1}{8} a = 4.25 \Rightarrow a = 34

Question 4

7 marks

A composite of trig functions and an equation

Functions

(a)

2 marks

Replace $x$ by $2 x$ in $f$ .

(f \circ g) (x) = f (2 x) = 3 sin 2 x + cos 2 x

(b)

5 marks

Set $3 sin 2 x + cos 2 x = 2 cos 2 x$ , so $3 sin 2 x = cos 2 x$ .
For $0 \leq x \leq π$ we need $2 x$ in $[0, 2 π]$ : the tangent equals $\frac{1}{3}$ at $\frac{π}{6}$ and $\frac{7 π}{6}$ .

tan 2 x = \frac{1}{3} 2 x = \frac{π}{6}, \frac{7 π}{6} \Rightarrow x = \frac{π}{12}, \frac{7 π}{12}

Question 5

5 marks

A tangent parallel to a given line

Number and algebra

5 marks

Use the product rule on $y = (2 x - 1) e^{k x}$ .
At $x = 1$ the gradient is $2 e^{k} + k e^{k} = (k + 2) e^{k}$ . The line is $y = 5 e^{k} x$ , with gradient $5 e^{k}$ . Set them equal and divide by $e^{k}$ .

\frac{d y}{d x} = 2 e^{k x} + k (2 x - 1) e^{k x} (k + 2) e^{k} = 5 e^{k} \Rightarrow k + 2 = 5 \Rightarrow k = 3

Question 6

7 marks

Describing two transformations, then the smallest $y$ -intercept

Functions

(a)

2 marks

The graph of $g (x) = 4 sin (x - \frac{3 π}{2}) + 2.5 + q$ comes from $f$ by a translation of $\frac{3 π}{2}$ to the right (positive $x$ direction), and a translation of $q$ upwards (positive $y$ direction).

(b)

5 marks

Every value of $g$ is at least $7$ . The least value of the sine term is $- 4$ , so
The $y$ -intercept is $g (0)$ . Since $sin (- \frac{3 π}{2}) = 1$ :
The smallest value of $r$ comes from the smallest $q$ .

- 4 + 2.5 + q \geq 7 \Rightarrow q \geq 8.5 r = 4 (1) + 2.5 + q = 6.5 + q r_{m i n} = 6.5 + 8.5 = 15

Section B · Extended response

Questions 7–9 · 44 marks

Question 7

14 marks

Quadratic derivative: symmetry, discriminant and inflexion

Calculus

(a)

2 marks

The axis of symmetry of $a x^{2} + b x + c$ is $x = - \frac{b}{2 a}$ .

q = \frac{12}{6} = 2

(b)

4 marks

The discriminant of $3 x^{2} - 12 x + p$ is
Set $(- 12)^{2} - 4 (3) p = 0$ .

Δ = 0 144 - 12 p = 0 \Rightarrow p = 12

(c)

3 marks

The gradient of $f^{'}$ is $f^{''}$ .

f^{''} (x) = 6 x - 12 \Rightarrow f^{''} (0) = - 12

(d)

2 marks

$f^{''}$ is a straight line with gradient 6.
Sketch: a line rising through $(0, - 12)$ and $(2, 0)$ .

(e)

3 marks

A point of inflexion is where $f^{''}$ changes sign.
The graph of $f$ is concave down where $f^{''} (x) < 0$ . The line $f^{''}$ is below the axis for $x < 2$ , so $f$ is concave down for $x < 2$ .

f^{''} (x) = 0 \Rightarrow a = 2

Question 8

15 marks

A series that is geometric, then arithmetic

Number and algebra

(a)

5 marks

The terms are $ln x, p ln x, \frac{1}{3} ln x$ . In a geometric series the ratio is constant.
With $p > 0$ the ratio is $r = \frac{1}{3}$ .

\frac{p ln x}{ln x} = \frac{\frac{1}{3} ln x}{p ln x} \Rightarrow p = \frac{1}{3 p} \Rightarrow p^{2} = \frac{1}{3} \Rightarrow p = \pm \frac{1}{3} S_{\infty} = \frac{ln x}{1 - \frac{1}{3}} = 3 + 3 \Rightarrow ln x = (3 + 3) \cdot \frac{3 - 1}{3} = 2 x = e^{2}

(b)

10 marks

In an arithmetic series the middle term is the mean of its neighbours: $p ln x = \frac{1}{2} (ln x + \frac{1}{3} ln x)$ .
The common difference is the second term minus the first.
The sum of the first $n$ terms is $S_{n} = \frac{n}{2} [2 ln x + (n - 1) (- \frac{1}{3} ln x)]$ . Set $S_{n} = - 3 ln x$ and cancel $ln x$ .

p = \frac{2}{3} d = \frac{2}{3} ln x - ln x = - \frac{1}{3} ln x n [2 - \frac{n - 1}{3}] = - 6 \Rightarrow n (7 - n) = - 18 \Rightarrow n^{2} - 7 n - 18 = 0 (n - 9) (n + 2) = 0 \Rightarrow n = 9

Question 9

15 marks

Cubic expansion, a trigonometric identity and a definite integral

Number and algebra
Calculus

(a)

6 marks

Expand using the binomial theorem.
Put $a = cos 2 x$ .
Use $cos 2 x = 1 - 2 sin^{2} x$ , so $1 - cos 2 x = 2 sin^{2} x$ .

(1 - a)^{3} = 1 - 3 a + 3 a^{2} - a^{3} 1 - 3 cos 2 x + 3 cos^{2} 2 x - cos^{3} 2 x = (1 - cos 2 x)^{3} (2 sin^{2} x)^{3} = 8 sin^{6} x

(b)

9 marks

Using part (a), $f (x) = 4 cos x \cdot 8 sin^{6} x = 32 sin^{6} x cos x$ . Let $u = sin x$ , so $d u = cos x d x$ .
For the second integral use the same antiderivative from $m$ to $\frac{π}{2}$ .

\int_{0}^{m} f (x) d x = \int 32 u^{6} d u = [\frac{32}{7} sin^{7} x]_{0}^{m} = \frac{32}{7} sin^{7} m \frac{32}{7} (1 - sin^{7} m) = \frac{127}{28} \Rightarrow 1 - sin^{7} m = \frac{127}{128} sin^{7} m = \frac{1}{128} = (\frac{1}{2})^{7} \Rightarrow sin m = \frac{1}{2} \Rightarrow m = \frac{π}{6}

Our analysis

Where the marks are

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

Calculus28 marks · 35%
Q2, Q7, Q9
Number and algebra26 marks · 33%
Q5, Q8, Q9
Functions14 marks · 18%
Q4, Q6
Statistics and probability7 marks · 9%
Q3
Geometry and trigonometry5 marks · 6%
Q1

Mentor insights

Traps to watch for

01
The two lines cross at the means · Q3
The regression line of $a$ on $c$ and the regression line of $c$ on $a$ meet at $(\overset{c}{ˉ}, \overset{a}{ˉ})$ . Solving them together gives the mean ages.
02
Divide by cosine · Q4
In $3 sin 2 x = cos 2 x$ divide both sides by $cos 2 x$ to get a tangent. Then list all solutions for $2 x$ in $0 \leq 2 x \leq 2 π$ .
03
Vertex on the x-axis means a double root · Q7
The discriminant is 0. That fixes $p$ using $b^{2} - 4 a c = 0$ .
04
Use consecutive terms · Q8
For a geometric series the ratio is the same between neighbouring terms. For an arithmetic series the difference is the same.
05
Choose the substitution that matches the pattern · Q9
The sum $1 - 3 a + 3 a^{2} - a^{3}$ is $(1 - a)^{3}$ . Set $a = cos 2 x$ and use $1 - cos 2 x = 2 sin^{2} x$ .

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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Our maths mentors took these exams recently. One-to-one, they'll find where your marks are going and fix it before May.

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Worked solutions

IB Maths AA SL · Paper 1

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2

Session: May 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 May 2022 TZ1 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 · 36 marks

Question 1

5 marks

Line perpendicular to BC through A

Geometry and trigonometry

(a)

3 marks

Gradient of BC:
The perpendicular gradient is 3. Use the point $A (- 2, 20)$ .

m_{B C} = \frac{12 - 6}{- 14 - 4} = - \frac{1}{3} y - 20 = 3 (x + 2) \Rightarrow y = 3 x + 26

(b)

2 marks

Put $(k, 2)$ into the equation.

2 = 3 k + 26 \Rightarrow k = - 8

Question 2

5 marks

Rewriting a quotient with a square root, then integrating

Calculus

(a)

1 mark

Divide each term by $x$ .

\frac{3 x - 5}{x} = 3 - 5 x^{- 1/2} \Rightarrow p = - \frac{1}{2}

(b)

4 marks

Integrate the rewritten form.
Substitute the limits.

\int (3 - 5 x^{- 1/2}) d x = 3 x - 10 x^{1/2} + c [3 x - 10 x]_{1}^{9} = (27 - 30) - (3 - 10) = 4

Question 3

7 marks

A box plot, outliers and two regression lines

Statistics and probability

(a)

3 marks

The quartiles are $Q_{1} = 6$ and $Q_{3} = 10$ , so the IQR is 4. A value is an outlier above $Q_{3} + 1.5 \times IQR$ .
So the largest value that is not an outlier is $c = 16$ .

10 + 1.5 \times 4 = 16

(b)

4 marks

(i) To estimate the child's age from the adult's, use the regression line of $c$ on $a$ .

(ii) Both lines pass through $(\overset{c}{ˉ}, \overset{a}{ˉ})$ . Substitute $c = \frac{1}{2} a - 9$ into $a = \frac{7}{4} c + 20$ .

The mean age of the adults is 34 years.

c = \frac{1}{2} (42) - 9 = 12 years a = \frac{7}{4} (\frac{1}{2} a - 9) + 20 = \frac{7}{8} a + 4.25 \Rightarrow \frac{1}{8} a = 4.25 \Rightarrow a = 34

Question 4

7 marks

A composite of trig functions and an equation

Functions

(a)

2 marks

Replace $x$ by $2 x$ in $f$ .

(f \circ g) (x) = f (2 x) = 3 sin 2 x + cos 2 x

(b)

5 marks

Set $3 sin 2 x + cos 2 x = 2 cos 2 x$ , so $3 sin 2 x = cos 2 x$ .
For $0 \leq x \leq π$ we need $2 x$ in $[0, 2 π]$ : the tangent equals $\frac{1}{3}$ at $\frac{π}{6}$ and $\frac{7 π}{6}$ .

tan 2 x = \frac{1}{3} 2 x = \frac{π}{6}, \frac{7 π}{6} \Rightarrow x = \frac{π}{12}, \frac{7 π}{12}

Question 5

5 marks

A tangent parallel to a given line

Number and algebra

5 marks

Use the product rule on $y = (2 x - 1) e^{k x}$ .
At $x = 1$ the gradient is $2 e^{k} + k e^{k} = (k + 2) e^{k}$ . The line is $y = 5 e^{k} x$ , with gradient $5 e^{k}$ . Set them equal and divide by $e^{k}$ .

\frac{d y}{d x} = 2 e^{k x} + k (2 x - 1) e^{k x} (k + 2) e^{k} = 5 e^{k} \Rightarrow k + 2 = 5 \Rightarrow k = 3

Question 6

7 marks

Describing two transformations, then the smallest $y$ -intercept

Functions

(a)

2 marks

(b)

5 marks

Every value of $g$ is at least $7$ . The least value of the sine term is $- 4$ , so
The $y$ -intercept is $g (0)$ . Since $sin (- \frac{3 π}{2}) = 1$ :
The smallest value of $r$ comes from the smallest $q$ .

- 4 + 2.5 + q \geq 7 \Rightarrow q \geq 8.5 r = 4 (1) + 2.5 + q = 6.5 + q r_{m i n} = 6.5 + 8.5 = 15

Section B · Extended response

Questions 7–9 · 44 marks

Question 7

14 marks

Quadratic derivative: symmetry, discriminant and inflexion

Calculus

(a)

2 marks

The axis of symmetry of $a x^{2} + b x + c$ is $x = - \frac{b}{2 a}$ .

q = \frac{12}{6} = 2

(b)

4 marks

The discriminant of $3 x^{2} - 12 x + p$ is
Set $(- 12)^{2} - 4 (3) p = 0$ .

Δ = 0 144 - 12 p = 0 \Rightarrow p = 12

(c)

3 marks

The gradient of $f^{'}$ is $f^{''}$ .

f^{''} (x) = 6 x - 12 \Rightarrow f^{''} (0) = - 12

(d)

2 marks

$f^{''}$ is a straight line with gradient 6.
Sketch: a line rising through $(0, - 12)$ and $(2, 0)$ .

(e)

3 marks

A point of inflexion is where $f^{''}$ changes sign.
The graph of $f$ is concave down where $f^{''} (x) < 0$ . The line $f^{''}$ is below the axis for $x < 2$ , so $f$ is concave down for $x < 2$ .

f^{''} (x) = 0 \Rightarrow a = 2

Question 8

15 marks

A series that is geometric, then arithmetic

Number and algebra

(a)

5 marks

The terms are $ln x, p ln x, \frac{1}{3} ln x$ . In a geometric series the ratio is constant.
With $p > 0$ the ratio is $r = \frac{1}{3}$ .

\frac{p ln x}{ln x} = \frac{\frac{1}{3} ln x}{p ln x} \Rightarrow p = \frac{1}{3 p} \Rightarrow p^{2} = \frac{1}{3} \Rightarrow p = \pm \frac{1}{3} S_{\infty} = \frac{ln x}{1 - \frac{1}{3}} = 3 + 3 \Rightarrow ln x = (3 + 3) \cdot \frac{3 - 1}{3} = 2 x = e^{2}

(b)

10 marks

In an arithmetic series the middle term is the mean of its neighbours: $p ln x = \frac{1}{2} (ln x + \frac{1}{3} ln x)$ .
The common difference is the second term minus the first.
The sum of the first $n$ terms is $S_{n} = \frac{n}{2} [2 ln x + (n - 1) (- \frac{1}{3} ln x)]$ . Set $S_{n} = - 3 ln x$ and cancel $ln x$ .

p = \frac{2}{3} d = \frac{2}{3} ln x - ln x = - \frac{1}{3} ln x n [2 - \frac{n - 1}{3}] = - 6 \Rightarrow n (7 - n) = - 18 \Rightarrow n^{2} - 7 n - 18 = 0 (n - 9) (n + 2) = 0 \Rightarrow n = 9

Question 9

15 marks

Cubic expansion, a trigonometric identity and a definite integral

Number and algebra
Calculus

(a)

6 marks

Expand using the binomial theorem.
Put $a = cos 2 x$ .
Use $cos 2 x = 1 - 2 sin^{2} x$ , so $1 - cos 2 x = 2 sin^{2} x$ .

(1 - a)^{3} = 1 - 3 a + 3 a^{2} - a^{3} 1 - 3 cos 2 x + 3 cos^{2} 2 x - cos^{3} 2 x = (1 - cos 2 x)^{3} (2 sin^{2} x)^{3} = 8 sin^{6} x

(b)

9 marks

Using part (a), $f (x) = 4 cos x \cdot 8 sin^{6} x = 32 sin^{6} x cos x$ . Let $u = sin x$ , so $d u = cos x d x$ .
For the second integral use the same antiderivative from $m$ to $\frac{π}{2}$ .

\int_{0}^{m} f (x) d x = \int 32 u^{6} d u = [\frac{32}{7} sin^{7} x]_{0}^{m} = \frac{32}{7} sin^{7} m \frac{32}{7} (1 - sin^{7} m) = \frac{127}{28} \Rightarrow 1 - sin^{7} m = \frac{127}{128} sin^{7} m = \frac{1}{128} = (\frac{1}{2})^{7} \Rightarrow sin m = \frac{1}{2} \Rightarrow m = \frac{π}{6}

Our analysis

Where the marks are

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

Calculus28 marks · 35%
Q2, Q7, Q9
Number and algebra26 marks · 33%
Q5, Q8, Q9
Functions14 marks · 18%
Q4, Q6
Statistics and probability7 marks · 9%
Q3
Geometry and trigonometry5 marks · 6%
Q1

Mentor insights

Traps to watch for

01
The two lines cross at the means · Q3
The regression line of $a$ on $c$ and the regression line of $c$ on $a$ meet at $(\overset{c}{ˉ}, \overset{a}{ˉ})$ . Solving them together gives the mean ages.
02
Divide by cosine · Q4
In $3 sin 2 x = cos 2 x$ divide both sides by $cos 2 x$ to get a tangent. Then list all solutions for $2 x$ in $0 \leq 2 x \leq 2 π$ .
03
Vertex on the x-axis means a double root · Q7
The discriminant is 0. That fixes $p$ using $b^{2} - 4 a c = 0$ .
04
Use consecutive terms · Q8
For a geometric series the ratio is the same between neighbouring terms. For an arithmetic series the difference is the same.
05
Choose the substitution that matches the pattern · Q9
The sum $1 - 3 a + 3 a^{2} - a^{3}$ is $(1 - a)^{3}$ . Set $a = cos 2 x$ and use $1 - cos 2 x = 2 sin^{2} x$ .

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