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IB Maths AI HL · Paper 1

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2

Session: May 2022
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 17

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2022 TZ2 paper, which your school or the IB can provide.

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Questions

Questions 1–17 · 110 marks

Question 1

6 marks

University applicants: table probabilities

Statistics and probability

(a)

1 mark

There are $17 + 25 = 42$ admitted out of 130.

P (admitted) = \frac{42}{130} = \frac{21}{65} \approx 0.323

(b)

2 marks

Only the 42 admitted applicants count, and 17 of them chose Arts.

P (Arts ∣ admitted) = \frac{17}{42} \approx 0.405

(c)

3 marks

There are $17 + 24 = 41$ Arts applicants.

P (both Arts) = \frac{41}{130} \times \frac{40}{129} = \frac{1640}{16770} \approx 0.0978

Question 2

8 marks

Security cameras: angle of depression and distances

Geometry and trigonometry

(a)

2 marks

The cameras are 3.1 m up and the register is 1.0 m up, so the vertical drop is $3.1 - 1.0 = 2.1$ m.

sin θ = \frac{2.1}{2.8} \Rightarrow θ = 48. 6^{\circ}

(b)

4 marks

Horizontal distance from Camera 1 to the register:
Camera 2 is $6.4 - 1.852 = 4.548$ m away horizontally and 2.1 m above the register.

2. 8^{2} - 2. 1^{2} = 1.852 m d = 4.54 8^{2} + 2. 1^{2} = 5.01 m

(c)

2 marks

Both cameras are at the same height, so the angle of depression is larger for the one that is horizontally closer.
Camera 1 is closer to the register (1.85 m against 4.55 m), so Camera 1 has the larger angle.

Mentor tip

This works because $tan (angle) = \frac{2.1}{horizontal distance}$ gets bigger as the horizontal distance shrinks.

Question 3

7 marks

Polygraph test: binomial probabilities

Statistics and probability

(a)

2 marks

Each of the 10 people passes with probability $0.8$ . Let $X$ be the number who pass: $X \sim B (10, 0.8)$ .

E (X) = 10 \times 0.8 = 8

(b)

2 marks

Four fail means six pass.

P (X = 6) = (6 10) (0.8)^{6} (0.2)^{4} \approx 0.0881

(c)

3 marks

Fewer than 7 pass is $X \leq 6$ .

P (X \leq 6) \approx 0.121

Mentor tip

Same answer from the failures: $Y \sim B (10, 0.2)$ and $P (Y \geq 4)$ .

Question 4

7 marks

Area under a line and a curve, and the area between them

Calculus

(a)

2 marks

The region is a trapezoid with parallel sides 7 and 4 and width 3.
Or by integration: $\int_{- 1}^{2} (6 - x) d x = 16.5$ .

A = \frac{1}{2} (7 + 4) (3) = 16.5

(b)

3 marks

Integral for the area under the curve between $x = - 1$ and $x = 2$ :
Evaluate.

A = \int_{- 1}^{2} (1.5 x^{2} - 2.5 x + 3) d x [0.5 x^{3} - 1.25 x^{2} + 3 x]_{- 1}^{2} = 5 - (- 4.75) = 9.75

(c)

2 marks

The area between the line and the curve is the difference of the two areas.

16.5 - 9.75 = 6.75

Question 5

8 marks

Graduate savings: fitting a cubic model

Functions

(a)

1 mark

The curve has one local maximum and one local minimum (and crosses the horizontal axis three times).

(b)

4 marks

At $t = 0$ the value is $- 5$ .
Substitute $t = 1, 2, 3$ into $S = a t^{3} + b t^{2} + c t - 5$ .
Solve the system.

d = - 5 a + b + c = 8, 8 a + 4 b + 2 c = 4, 27 a + 9 b + 3 c = 0 a = 2, b = - 12, c = 18

(c)

3 marks

The model is $S = 2 t^{3} - 12 t^{2} + 18 t - 5$ . Solve $S = 0$ .
$S$ is negative on $0 \leq t < 0.358$ and again on $1.832 < t < 3.810$ .

t \approx 0.358, 1.832, 3.810 time in debt = 0.358 + (3.810 - 1.832) \approx 2.34 years

Mentor tip

The graduate starts in debt ( $S = - 5$ ), so the first interval starts at $t = 0$ .

Question 6

5 marks

Directed network: adjacency matrix and closed walks

Number and algebra

(a)

2 marks

Read rows as "from" and columns as "to". The edges are A to A, A to B, B to A, B to E, C to B, C to D, D to A, E to A, E to C and E to D.

M = 1101110100000010010101000

(b)

3 marks

The entry in row $i$ , column $i$ of $M^{5}$ counts the walks of length 5 from vertex $i$ back to itself. Raise the matrix to the fifth power.
Add the diagonal.

M^{5} = 1717138189106511232123422454425 17 + 10 + 2 + 2 + 5 = 36

Question 7

5 marks

Infinite geometric series: find the third term

Number and algebra

5 marks

Let the first term be $u_{1}$ and the ratio $r$ . The sum is 9, so $u_{1} = 9 (1 - r)$ .
The first term is 4 more than the second term, so $u_{1} - u_{1} r = 4$ , which is $u_{1} (1 - r) = 4$ .
The sum to infinity exists only if $∣ r ∣ < 1$ , so $r = \frac{1}{3}$ .

9 (1 - r)^{2} = 4 \Rightarrow 1 - r = \pm \frac{2}{3} \Rightarrow r = \frac{1}{3} or r = \frac{5}{3} u_{1} = 9 (1 - \frac{1}{3}) = 6, u_{3} = 6 (\frac{1}{3})^{2} = \frac{2}{3}

Mentor tip

Check: $\frac{6}{1 - 1/3} = 9$ and $6 - 2 = 4$ .

Question 8

8 marks

Sector area with measurement bounds and percentage error

Geometry and trigonometry

(a)

2 marks

Convert 30 degrees to $\frac{π}{6}$ radians and use the sector formula.

A = \frac{1}{2} r^{2} θ = \frac{1}{2} (2)^{2} (\frac{30 π}{180}) = \frac{π}{3} \approx 1.047 cm^{2}

(b)

3 marks

The area is largest with the biggest radius and angle, and smallest with the smallest.

A_{m a x} = \frac{1}{2} (2.5)^{2} (\frac{35 π}{180}) \approx 1.91 cm^{2} A_{m i n} = \frac{1}{2} (1.5)^{2} (\frac{25 π}{180}) \approx 0.491 cm^{2}

(c)

3 marks

If the true area were $A_{m a x}$ :
If the true area were $A_{m i n}$ :
The largest possible percentage error is about $113%$ .

\frac{1.047 - 1.909}{1.909} \times 100 \approx - 45.2% \frac{1.047 - 0.4909}{0.4909} \times 100 \approx 113%

Question 9

8 marks

Digits of pi remembered: estimates and a one-sided test

Statistics and probability

(a)

1 mark

The 67 values are 2 (twice), 3 (six times), and so on.

\overset{x}{ˉ} = 4.63

(b)

2 marks

The sample standard deviation with divisor $n - 1$ is $s_{n - 1} = 1.0987$ .

s_{n - 1}^{2} = 1.098 7^{2} \approx 1.21

(c)

5 marks

The researcher wants to know whether the students remember more than the general population.
The population variance is not known, so use a one-sample $t$ -test with $ν = 66$ .
Compare with 0.05.
Reject $H_{0}$ : there is significant evidence that IB HL students recall more digits of $π$ than the general population.

H_{1} : μ > 4.4 p \approx 0.0479 0.0479 < 0.05

Mentor tip

A $z$ -test with $s$ in place of $σ$ gives $p = 0.0455$ and the same conclusion.

Question 10

5 marks

Inverse of a log function, then f(x) = f inverse of x

Functions

(a)

3 marks

Write $f (x) = - ln (x - 2)$ .
Swap $x$ and $y$ .

y = ln (\frac{1}{x - 2}) \Rightarrow e^{y} = \frac{1}{x - 2} \Rightarrow x = 2 + e^{- y} f^{- 1} (x) = e^{- x} + 2

(b)

2 marks

Solve $ln (\frac{1}{x - 2}) = e^{- x} + 2$ by graphing both sides.

x \approx 2.12

Mentor tip

$f$ and $f^{- 1}$ are reflections in $y = x$ , so any intersection with $y = x$ is also a solution; here the curves cross at a point just above $x = 2$ .

Question 11

6 marks

Saving for a real value of 4000: lump sum or yearly payments

Number and algebra

(a)

3 marks

The target in ten years' money is $4000 \times 1.0 2^{10} = 4875.98$ .
The minimum single investment is $3457.

P V = \frac{4875.98}{1.03 5 ^{10}} = 3456.67

(b)

3 marks

The $1000 grows to
The yearly payments must make up the rest.
So about $295 p er y e a r;$ 296 is the smallest whole-dollar payment that reaches the target.

1000 \times 1.03 5^{10} = 1410.60 x \cdot \frac{1.03 5 ^{10} - 1}{0.035} = 4875.98 - 1410.60 = 3465.38 x = \frac{3465.38}{11.7314} \approx 295.39

Mentor tip

Use a finance solver with N = 10, I = 3.5, PV = $- 1000$ , FV = 4875.98 and 1 payment per year to check.

Question 12

6 marks

Cuttlefish weight: Type I and Type II errors

Statistics and probability

(a)

2 marks

Type I: a male is classified as female, so $W > 11.5$ with $W \sim N (10, 0. 5^{2})$ .

P (W > 11.5) = P (Z > 3) \approx 0.00135

(b)

2 marks

Type II: a female is classified as male, so $W < 11.5$ with $W \sim N (12, 1^{2})$ .

P (W < 11.5) = P (Z < - 0.5) \approx 0.309

(c)

2 marks

Use total probability.

P (error) = 0.9 (0.00135) + 0.1 (0.3085) \approx 0.0321

Question 13

5 marks

Adding two sinusoidal voltages with complex numbers

Number and algebra

(a)

4 marks

Both terms share $e^{3 t i}$ .
Evaluate the bracket.
So $V_{T} = Re (3.99 e^{i (3 t - 1.894)})$ .

V_{T} = Re [e^{3 t i} (2 + 5 e^{4 i})] 2 + 5 e^{4 i} = 2 + 5 cos 4 + 5 i sin 4 \approx - 1.268 - 3.784 i = 3.99 e^{- 1.894 i} V_{T} = 3.99 cos (3 t - 1.89)

(b)

1 mark

The cosine reaches 1.

maximum voltage = 3.99

Question 14

4 marks

Vase volume: the trapezoidal rule

Number and algebra

4 marks

Rotating about the $y$ -axis, the volume is $V = π \int_{0}^{10} x^{2} d y$ , where $x$ is the radius.
The squared radii at heights 0, 2, 4, 6, 8, 10 are 16, 36, 64, 49, 9, 25. The strip width is $h = 2$ .

V \approx π \times \frac{2}{2} [16 + 25 + 2 (36 + 64 + 49 + 9)] = 357 π V \approx 1120 cm^{3}

Mentor tip

Common slip: using the radii instead of the squared radii. That gives an area-like number and loses the $π$ factor.

Question 15

7 marks

A line, a singular matrix and why the image does not depend on m

Geometry and trigonometry

(a)

2 marks

The line passes through $(0, c)$ and goes up $m$ for every 1 across.

a = (0 c), b = (1 m)

(b)

1 mark

Cross-multiply the entries of $(6432)$ .

det M = 6 \times 2 - 3 \times 4 = 0

(c)

4 marks

A point of the line is $(x, m x + c)$ . Apply $M$ .
Since $m \neq = - 2$ , the factor $(2 + m) x + c$ takes every real value as $x$ varies. So the image is the set of all multiples of $(32)$ .
This line contains no $m$ or $c$ .

M (x m x + c) = (6 x + 3 m x + 3 c 4 x + 2 m x + 2 c) = ((2 + m) x + c) (32) y = \frac{2}{3} x

Question 16

7 marks

A particle on a circle: velocity and acceleration are never parallel to the position

Calculus

(a)

2 marks

Chain rule on each component.

v = (2 t cos (t^{2}) - 2 t sin (t^{2}))

(b)

5 marks

Differentiate again with the product rule.
Split $a$ into two pieces.
The second piece is a multiple of the position vector $O P$ . The first piece is perpendicular to $O P$ , because the scalar product is $sin (t^{2}) cos (t^{2}) - cos (t^{2}) sin (t^{2}) = 0$ , and it has length 2.
So $a$ always has a component of size 2 at right angles to $O P$ . A vector parallel to $O P$ has no such component, so $a$ is never parallel to $O P$ .

a = (2 cos (t^{2}) - 4 t^{2} sin (t^{2}) - 2 sin (t^{2}) - 4 t^{2} cos (t^{2})) a = 2 (cos (t^{2}) - sin (t^{2})) - 4 t^{2} (sin (t^{2}) cos (t^{2}))

Mentor tip

The 2D cross product gives the same result: $x_{P} a_{y} - y_{P} a_{x} = - 2$ , which is never 0.

Question 17

8 marks

A snail and the incoming tide on a parabolic beach

Calculus

(a)

3 marks

The water level rises at $\frac{d y}{d t} = 0.2$ , and $y = 0.02 x^{2}$ gives $\frac{d y}{d x} = 0.04 x$ .
At $x = 1$ :

0.2 = 0.04 x \frac{d x}{d t} \frac{d x}{d t} = \frac{0.2}{0.04} = 5 m h^{- 1}

(b)

5 marks

Speed 1 along the curve means arc length equals time. The arc length from $x = 1$ to $x = 10$ of $y = 0.02 x^{2}$ is
The water reaches $y = 2$ when $0.2 t = 2$ , that is after 10 hours.
So the snail gets there first.

T = \int_{1}^{10} 1 + (0.04 x)^{2} d x \approx 9.26 hours 9.26 < 10

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 probability27 marks · 25%
Q1, Q3, Q9, Q12
Number and algebra25 marks · 23%
Q6, Q7, Q11, Q13, Q14
Geometry and trigonometry23 marks · 21%
Q2, Q8, Q15
Calculus22 marks · 20%
Q4, Q16, Q17
Functions13 marks · 12%
Q5, Q10

Mentor insights

Traps to watch for

01
Angle of depression is measured from the horizontal · Q2
In the right triangle the 2.1 m drop is opposite the angle, and the 2.8 m sightline is the hypotenuse.
02
Use the first point for d · Q5
The point $(0, - 5)$ gives $d = - 5$ immediately, which makes the other three equations simpler.
03
Bounds from one significant figure · Q8
A radius of 2 cm to 1 s.f. lies between 1.5 and 2.5; 30 degrees lies between 25 and 35.
04
Divide by n-1 for the variance estimate · Q9
The unbiased estimate of the population variance uses $s_{n - 1}^{2}$ , not the biased $s_{n}^{2}$ (which gives 1.19).
05
Chain rule links the two rates · Q17
$\frac{d y}{d t} = \frac{d y}{d x} \cdot \frac{d x}{d t}$ , with $\frac{d y}{d x} = 0.04 x$ .

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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Worked solutions

IB Maths AI HL · Paper 1

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2

Session: May 2022
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 17

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2022 TZ2 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–17 · 110 marks

Question 1

6 marks

University applicants: table probabilities

Statistics and probability

(a)

1 mark

There are $17 + 25 = 42$ admitted out of 130.

P (admitted) = \frac{42}{130} = \frac{21}{65} \approx 0.323

(b)

2 marks

Only the 42 admitted applicants count, and 17 of them chose Arts.

P (Arts ∣ admitted) = \frac{17}{42} \approx 0.405

(c)

3 marks

There are $17 + 24 = 41$ Arts applicants.

P (both Arts) = \frac{41}{130} \times \frac{40}{129} = \frac{1640}{16770} \approx 0.0978

Question 2

8 marks

Security cameras: angle of depression and distances

Geometry and trigonometry

(a)

2 marks

The cameras are 3.1 m up and the register is 1.0 m up, so the vertical drop is $3.1 - 1.0 = 2.1$ m.

sin θ = \frac{2.1}{2.8} \Rightarrow θ = 48. 6^{\circ}

(b)

4 marks

Horizontal distance from Camera 1 to the register:
Camera 2 is $6.4 - 1.852 = 4.548$ m away horizontally and 2.1 m above the register.

2. 8^{2} - 2. 1^{2} = 1.852 m d = 4.54 8^{2} + 2. 1^{2} = 5.01 m

(c)

2 marks

Both cameras are at the same height, so the angle of depression is larger for the one that is horizontally closer.
Camera 1 is closer to the register (1.85 m against 4.55 m), so Camera 1 has the larger angle.

Mentor tip

This works because $tan (angle) = \frac{2.1}{horizontal distance}$ gets bigger as the horizontal distance shrinks.

Question 3

7 marks

Polygraph test: binomial probabilities

Statistics and probability

(a)

2 marks

Each of the 10 people passes with probability $0.8$ . Let $X$ be the number who pass: $X \sim B (10, 0.8)$ .

E (X) = 10 \times 0.8 = 8

(b)

2 marks

Four fail means six pass.

P (X = 6) = (6 10) (0.8)^{6} (0.2)^{4} \approx 0.0881

(c)

3 marks

Fewer than 7 pass is $X \leq 6$ .

P (X \leq 6) \approx 0.121

Mentor tip

Same answer from the failures: $Y \sim B (10, 0.2)$ and $P (Y \geq 4)$ .

Question 4

7 marks

Area under a line and a curve, and the area between them

Calculus

(a)

2 marks

The region is a trapezoid with parallel sides 7 and 4 and width 3.
Or by integration: $\int_{- 1}^{2} (6 - x) d x = 16.5$ .

A = \frac{1}{2} (7 + 4) (3) = 16.5

(b)

3 marks

Integral for the area under the curve between $x = - 1$ and $x = 2$ :
Evaluate.

A = \int_{- 1}^{2} (1.5 x^{2} - 2.5 x + 3) d x [0.5 x^{3} - 1.25 x^{2} + 3 x]_{- 1}^{2} = 5 - (- 4.75) = 9.75

(c)

2 marks

The area between the line and the curve is the difference of the two areas.

16.5 - 9.75 = 6.75

Question 5

8 marks

Graduate savings: fitting a cubic model

Functions

(a)

1 mark

The curve has one local maximum and one local minimum (and crosses the horizontal axis three times).

(b)

4 marks

At $t = 0$ the value is $- 5$ .
Substitute $t = 1, 2, 3$ into $S = a t^{3} + b t^{2} + c t - 5$ .
Solve the system.

d = - 5 a + b + c = 8, 8 a + 4 b + 2 c = 4, 27 a + 9 b + 3 c = 0 a = 2, b = - 12, c = 18

(c)

3 marks

The model is $S = 2 t^{3} - 12 t^{2} + 18 t - 5$ . Solve $S = 0$ .
$S$ is negative on $0 \leq t < 0.358$ and again on $1.832 < t < 3.810$ .

t \approx 0.358, 1.832, 3.810 time in debt = 0.358 + (3.810 - 1.832) \approx 2.34 years

Mentor tip

The graduate starts in debt ( $S = - 5$ ), so the first interval starts at $t = 0$ .

Question 6

5 marks

Directed network: adjacency matrix and closed walks

Number and algebra

(a)

2 marks

Read rows as "from" and columns as "to". The edges are A to A, A to B, B to A, B to E, C to B, C to D, D to A, E to A, E to C and E to D.

M = 1101110100000010010101000

(b)

3 marks

The entry in row $i$ , column $i$ of $M^{5}$ counts the walks of length 5 from vertex $i$ back to itself. Raise the matrix to the fifth power.
Add the diagonal.

M^{5} = 1717138189106511232123422454425 17 + 10 + 2 + 2 + 5 = 36

Question 7

5 marks

Infinite geometric series: find the third term

Number and algebra

5 marks

Let the first term be $u_{1}$ and the ratio $r$ . The sum is 9, so $u_{1} = 9 (1 - r)$ .
The first term is 4 more than the second term, so $u_{1} - u_{1} r = 4$ , which is $u_{1} (1 - r) = 4$ .
The sum to infinity exists only if $∣ r ∣ < 1$ , so $r = \frac{1}{3}$ .

9 (1 - r)^{2} = 4 \Rightarrow 1 - r = \pm \frac{2}{3} \Rightarrow r = \frac{1}{3} or r = \frac{5}{3} u_{1} = 9 (1 - \frac{1}{3}) = 6, u_{3} = 6 (\frac{1}{3})^{2} = \frac{2}{3}

Mentor tip

Check: $\frac{6}{1 - 1/3} = 9$ and $6 - 2 = 4$ .

Question 8

8 marks

Sector area with measurement bounds and percentage error

Geometry and trigonometry

(a)

2 marks

Convert 30 degrees to $\frac{π}{6}$ radians and use the sector formula.

A = \frac{1}{2} r^{2} θ = \frac{1}{2} (2)^{2} (\frac{30 π}{180}) = \frac{π}{3} \approx 1.047 cm^{2}

(b)

3 marks

The area is largest with the biggest radius and angle, and smallest with the smallest.

A_{m a x} = \frac{1}{2} (2.5)^{2} (\frac{35 π}{180}) \approx 1.91 cm^{2} A_{m i n} = \frac{1}{2} (1.5)^{2} (\frac{25 π}{180}) \approx 0.491 cm^{2}

(c)

3 marks

If the true area were $A_{m a x}$ :
If the true area were $A_{m i n}$ :
The largest possible percentage error is about $113%$ .

\frac{1.047 - 1.909}{1.909} \times 100 \approx - 45.2% \frac{1.047 - 0.4909}{0.4909} \times 100 \approx 113%

Question 9

8 marks

Digits of pi remembered: estimates and a one-sided test

Statistics and probability

(a)

1 mark

The 67 values are 2 (twice), 3 (six times), and so on.

\overset{x}{ˉ} = 4.63

(b)

2 marks

The sample standard deviation with divisor $n - 1$ is $s_{n - 1} = 1.0987$ .

s_{n - 1}^{2} = 1.098 7^{2} \approx 1.21

(c)

5 marks

The researcher wants to know whether the students remember more than the general population.
The population variance is not known, so use a one-sample $t$ -test with $ν = 66$ .
Compare with 0.05.
Reject $H_{0}$ : there is significant evidence that IB HL students recall more digits of $π$ than the general population.

H_{1} : μ > 4.4 p \approx 0.0479 0.0479 < 0.05

Mentor tip

A $z$ -test with $s$ in place of $σ$ gives $p = 0.0455$ and the same conclusion.

Question 10

5 marks

Inverse of a log function, then f(x) = f inverse of x

Functions

(a)

3 marks

Write $f (x) = - ln (x - 2)$ .
Swap $x$ and $y$ .

y = ln (\frac{1}{x - 2}) \Rightarrow e^{y} = \frac{1}{x - 2} \Rightarrow x = 2 + e^{- y} f^{- 1} (x) = e^{- x} + 2

(b)

2 marks

Solve $ln (\frac{1}{x - 2}) = e^{- x} + 2$ by graphing both sides.

x \approx 2.12

Mentor tip

$f$ and $f^{- 1}$ are reflections in $y = x$ , so any intersection with $y = x$ is also a solution; here the curves cross at a point just above $x = 2$ .

Question 11

6 marks

Saving for a real value of 4000: lump sum or yearly payments

Number and algebra

(a)

3 marks

The target in ten years' money is $4000 \times 1.0 2^{10} = 4875.98$ .
The minimum single investment is $3457.

P V = \frac{4875.98}{1.03 5 ^{10}} = 3456.67

(b)

3 marks

The $1000 grows to
The yearly payments must make up the rest.
So about $295 p er y e a r;$ 296 is the smallest whole-dollar payment that reaches the target.

1000 \times 1.03 5^{10} = 1410.60 x \cdot \frac{1.03 5 ^{10} - 1}{0.035} = 4875.98 - 1410.60 = 3465.38 x = \frac{3465.38}{11.7314} \approx 295.39

Mentor tip

Use a finance solver with N = 10, I = 3.5, PV = $- 1000$ , FV = 4875.98 and 1 payment per year to check.

Question 12

6 marks

Cuttlefish weight: Type I and Type II errors

Statistics and probability

(a)

2 marks

Type I: a male is classified as female, so $W > 11.5$ with $W \sim N (10, 0. 5^{2})$ .

P (W > 11.5) = P (Z > 3) \approx 0.00135

(b)

2 marks

Type II: a female is classified as male, so $W < 11.5$ with $W \sim N (12, 1^{2})$ .

P (W < 11.5) = P (Z < - 0.5) \approx 0.309

(c)

2 marks

Use total probability.

P (error) = 0.9 (0.00135) + 0.1 (0.3085) \approx 0.0321

Question 13

5 marks

Adding two sinusoidal voltages with complex numbers

Number and algebra

(a)

4 marks

Both terms share $e^{3 t i}$ .
Evaluate the bracket.
So $V_{T} = Re (3.99 e^{i (3 t - 1.894)})$ .

V_{T} = Re [e^{3 t i} (2 + 5 e^{4 i})] 2 + 5 e^{4 i} = 2 + 5 cos 4 + 5 i sin 4 \approx - 1.268 - 3.784 i = 3.99 e^{- 1.894 i} V_{T} = 3.99 cos (3 t - 1.89)

(b)

1 mark

The cosine reaches 1.

maximum voltage = 3.99

Question 14

4 marks

Vase volume: the trapezoidal rule

Number and algebra

4 marks

Rotating about the $y$ -axis, the volume is $V = π \int_{0}^{10} x^{2} d y$ , where $x$ is the radius.
The squared radii at heights 0, 2, 4, 6, 8, 10 are 16, 36, 64, 49, 9, 25. The strip width is $h = 2$ .

V \approx π \times \frac{2}{2} [16 + 25 + 2 (36 + 64 + 49 + 9)] = 357 π V \approx 1120 cm^{3}

Mentor tip

Common slip: using the radii instead of the squared radii. That gives an area-like number and loses the $π$ factor.

Question 15

7 marks

A line, a singular matrix and why the image does not depend on m

Geometry and trigonometry

(a)

2 marks

The line passes through $(0, c)$ and goes up $m$ for every 1 across.

a = (0 c), b = (1 m)

(b)

1 mark

Cross-multiply the entries of $(6432)$ .

det M = 6 \times 2 - 3 \times 4 = 0

(c)

4 marks

A point of the line is $(x, m x + c)$ . Apply $M$ .
Since $m \neq = - 2$ , the factor $(2 + m) x + c$ takes every real value as $x$ varies. So the image is the set of all multiples of $(32)$ .
This line contains no $m$ or $c$ .

M (x m x + c) = (6 x + 3 m x + 3 c 4 x + 2 m x + 2 c) = ((2 + m) x + c) (32) y = \frac{2}{3} x

Question 16

7 marks

A particle on a circle: velocity and acceleration are never parallel to the position

Calculus

(a)

2 marks

Chain rule on each component.

v = (2 t cos (t^{2}) - 2 t sin (t^{2}))

(b)

5 marks

Differentiate again with the product rule.
Split $a$ into two pieces.
The second piece is a multiple of the position vector $O P$ . The first piece is perpendicular to $O P$ , because the scalar product is $sin (t^{2}) cos (t^{2}) - cos (t^{2}) sin (t^{2}) = 0$ , and it has length 2.
So $a$ always has a component of size 2 at right angles to $O P$ . A vector parallel to $O P$ has no such component, so $a$ is never parallel to $O P$ .

a = (2 cos (t^{2}) - 4 t^{2} sin (t^{2}) - 2 sin (t^{2}) - 4 t^{2} cos (t^{2})) a = 2 (cos (t^{2}) - sin (t^{2})) - 4 t^{2} (sin (t^{2}) cos (t^{2}))

Mentor tip

The 2D cross product gives the same result: $x_{P} a_{y} - y_{P} a_{x} = - 2$ , which is never 0.

Question 17

8 marks

A snail and the incoming tide on a parabolic beach

Calculus

(a)

3 marks

The water level rises at $\frac{d y}{d t} = 0.2$ , and $y = 0.02 x^{2}$ gives $\frac{d y}{d x} = 0.04 x$ .
At $x = 1$ :

0.2 = 0.04 x \frac{d x}{d t} \frac{d x}{d t} = \frac{0.2}{0.04} = 5 m h^{- 1}

(b)

5 marks

Speed 1 along the curve means arc length equals time. The arc length from $x = 1$ to $x = 10$ of $y = 0.02 x^{2}$ is
The water reaches $y = 2$ when $0.2 t = 2$ , that is after 10 hours.
So the snail gets there first.

T = \int_{1}^{10} 1 + (0.04 x)^{2} d x \approx 9.26 hours 9.26 < 10

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 probability27 marks · 25%
Q1, Q3, Q9, Q12
Number and algebra25 marks · 23%
Q6, Q7, Q11, Q13, Q14
Geometry and trigonometry23 marks · 21%
Q2, Q8, Q15
Calculus22 marks · 20%
Q4, Q16, Q17
Functions13 marks · 12%
Q5, Q10

Mentor insights

Traps to watch for

01
Angle of depression is measured from the horizontal · Q2
In the right triangle the 2.1 m drop is opposite the angle, and the 2.8 m sightline is the hypotenuse.
02
Use the first point for d · Q5
The point $(0, - 5)$ gives $d = - 5$ immediately, which makes the other three equations simpler.
03
Bounds from one significant figure · Q8
A radius of 2 cm to 1 s.f. lies between 1.5 and 2.5; 30 degrees lies between 25 and 35.
04
Divide by n-1 for the variance estimate · Q9
The unbiased estimate of the population variance uses $s_{n - 1}^{2}$ , not the biased $s_{n}^{2}$ (which gives 1.19).
05
Chain rule links the two rates · Q17
$\frac{d y}{d t} = \frac{d y}{d x} \cdot \frac{d x}{d t}$ , with $\frac{d y}{d x} = 0.04 x$ .

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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