Loading…

Worked solutions

IB Maths AI HL · Paper 1

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2 TZ3

Session: May 2025
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 15

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

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2025 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–15 · 110 marks

Question 1

6 marks

Snow paths: a sector area and a path length

Geometry and trigonometry

(a)

3 marks

The shaded region is one fifth of the area between the circles of radius 4 and 2.8.
Since $5.13 < 6$ , Maureen has enough dye.

A = \frac{1}{5} (π (4)^{2} - π (2.8)^{2}) = \frac{1}{5} π (16 - 7.84) \approx 5.13 m^{2}

(b)

3 marks

The route has three fifths of the outer circle, plus two radii of length 4.

\frac{3}{5} \times 2 π (4) + 2 \times 4 \approx 15.08 + 8 \approx 23.1 m

Question 2

7 marks

A dolphin's jump modelled by a quadratic

Functions

(a)

2 marks

The axis of symmetry is halfway between $d = 3$ and $d = 8.5$ .

d = \frac{3 + 8.5}{2} = 5.75

(b)

2 marks

The axis of symmetry of $- 0.2 d^{2} + b d$ is $d = - \frac{b}{2 ( - 0.2 )}$ .

\frac{b}{0.4} = 5.75 \Rightarrow b = 2.3

(c)

2 marks

The dolphin re-enters the water when $h = 0$ .

- 0.2 d^{2} + 2.3 d = 0 \Rightarrow d (- 0.2 d + 2.3) = 0 \Rightarrow d = 11.5 m

(d)

1 mark

A negative $h$ means the dolphin is below the surface of the water, which is outside the jump the model describes.

Question 3

8 marks

Two loan options for a truck

Number and algebra

(a)

6 marks

(i) Option A borrows the whole price.

(ii) Option B borrows 80% of the price.

N = 72, I = 7.15%, P V = 42695, F V = 0 \Rightarrow P M T \approx 730.99 USD 0.8 \times 42695 = 34156 N = 72, I = 6.40%, P V = 34156, F V = 0 \Rightarrow P M T \approx 572.54 USD

(b)

2 marks

(i) Option A needs no initial payment, so it suits Raul if he does not have enough money for a deposit.

(ii) Option B has lower monthly payments, and less interest is paid over the life of the loan.

Question 4

7 marks

Hat prices with a discount after the first 15

Functions

(a)

1 mark

Fifteen hats at $p$ euros each.

15 p

(b)

3 marks

The remaining $n - 15$ hats cost $p - 5$ each.

C (n) = 15 p + (p - 5) (n - 15)

(c)

3 marks

(i) The total is the mean price times the number of hats.

(ii) Substitute $n = 100$ in the cost equation.

25.73 \times 100 = 2573 EUR 2573 = 15 p + (p - 5) (85) \Rightarrow 2573 = 100 p - 425 \Rightarrow p = 29.98

Question 5

9 marks

A boat off course: the cosine rule and bearings

Geometry and trigonometry

(a)

7 marks

(i) In triangle $D B C$ we have $D B = 2$ , $B C = 13$ and $D B C = 17 0^{\circ}$ . Use the cosine rule.

(ii) Use the sine rule to find the angle $B D C$ .

Moving from the due east direction at $D$ (bearing $09 0^{\circ}$ ) round towards the south by this angle gives

D C^{2} = 2^{2} + 1 3^{2} - 2 (2) (13) cos 17 0^{\circ} \Rightarrow D C \approx 14.97 \approx 15.0 km \frac{sin B D C}{13} = \frac{sin 17 0 ^{\circ}}{14.97} \Rightarrow B D C \approx 8.6 7^{\circ} bearing = 90 + 8.67 \approx 098. 7^{\circ}

(b)

2 marks

The points $D$ and $B$ are both 2 km west of $E$ and $C$ respectively (so $D B$ and $E C$ are parallel and equal in length). So $D B C E$ is a parallelogram and the opposite sides $D E$ and $B C$ are equal.

D E = 13 km

Question 6

6 marks

Discounts at furniture stores and composite functions

Functions

(a)

2 marks

(i) A 30% discount leaves 70% of the price.

(ii) A reduction of 150 CAD.

f (x) = 0.7 x g (x) = x - 150

(b)

2 marks

$(f \circ g) (x) = f (g (x))$ means the 150 CAD reduction is applied first, and then the 30% discount is applied to the reduced price.

(f \circ g) (x) = 0.7 (x - 150) = 0.7 x - 105

(c)

2 marks

Compare the two orders at 450 CAD.
The price 165 is lower, so $(g \circ f) (x)$ is better for the customer: the 30% is taken off the higher original price first.

(f \circ g) (450) = 0.7 (300) = 210, (g \circ f) (450) = 0.7 (450) - 150 = 165

Question 7

8 marks

A particle's velocity 2 cos t + sin 2t - 0.2: sketch, turning points and distance

Calculus

(a)

5 marks

(i) Sketch: the curve starts at $(0, 1.8)$ and rises to a maximum, falls through zero velocity to a minimum, and rises again, ending at $t = 6$ .

(ii) The points of zero acceleration come from $v^{'} (t) = 0$ , found on a graph.

(0.524, 2.40), (2.62, - 2.80), (4.71, - 0.2)

(b)

3 marks

(i) The particle changes direction (where $v = 0$ at $t \approx 1.52$ and $t \approx 5.32$ ), so the distance is the integral of the modulus.

(ii) Evaluate with the calculator.

d = \int_{0}^{6} ∣ 2 cos t + sin 2 t - 0.2 ∣ d t d \approx 7.76 m

Question 8

7 marks

Area under a cubic and equal areas with a quadratic

Calculus

(a)

2 marks

The roots are 1, 5 and $a$ , and their sum is 9.

1 + 5 + a = 9 \Rightarrow a = 3

(b)

2 marks

The curve is above the axis between $x = 1$ and $x = 3$ .

\int_{1}^{3} (x^{3} - 9 x^{2} + 23 x - 15) d x = 4

(c)

3 marks

Set the two areas equal.
Solve with the calculator on $1 < b < 3$ .

\int_{1}^{b} f (x) d x = \int_{1}^{b} g (x) d x b \approx 2.57

Question 9

7 marks

Travel time to school: normal, binomial and a detention

Statistics and probability

(a)

4 marks

The probability that he is late on one day.
Let $X$ be the number of late days in 180 days, $X \sim B (180, 0.03386)$ .

P (T > 45) \approx 0.03386, T \sim N (35.5, 5. 2^{2}) P (X \geq 10) = 1 - P (X \leq 9) \approx 0.0869

(b)

3 marks

On a late day he is marked late with probability $1 - 0.35 = 0.65$ .
Let $Y$ be the number of days marked late, $Y \sim B (180, 0.02201)$ .

0.03386 \times 0.65 \approx 0.02201 P (Y \geq 4) = 1 - P (Y \leq 3) \approx 0.561

Question 10

9 marks

Combining logarithms, an inverse function and finding a constant

Functions

(a)

2 marks

Combine the logarithms.

f (x) = ln \frac{( 2 - x ) ( - 2 - x )}{x ^{2}} = ln \frac{x ^{2} - 4}{x ^{2}} g (x) = \frac{x ^{2} - 4}{x ^{2}} = 1 - \frac{4}{x ^{2}}

(b)

2 marks

The domain of $f$ is $x < - 2$ , which is the range of $f^{- 1}$ .

f^{- 1} (x) < - 2

(c)

5 marks

Write $y = ln (1 - \frac{4}{x ^{2}})$ and swap the variables.
The range of $f^{- 1}$ is $y < - 2$ , so take the negative root.

x = ln (1 - \frac{4}{y ^{2}}) \Rightarrow e^{x} = 1 - \frac{4}{y ^{2}} \Rightarrow y^{2} = \frac{4}{1 - e ^{x}} f^{- 1} (x) = - \frac{2}{1 - e ^{x}} a = - 2

Question 11

8 marks

A geometric sequence with a complex ratio

Number and algebra

(a)

2 marks

Multiply by $r = \frac{2}{3} + \frac{2}{3} i$ .

u_{2} = 9 r = 6 + 6 i u_{3} = u_{2} r = (6 + 6 i) (\frac{2}{3} + \frac{2}{3} i) = 4 + 4 i + 4 i + 4 i^{2} = 8 i

(b)

4 marks

The modulus of the common ratio is
Since $∣ u_{n + 1} ∣ = ∣ u_{n} ∣∣ r ∣$ , each modulus is $\frac{2 2}{3}$ times the one before. So $∣ u_{1} ∣, ∣ u_{2} ∣, ∣ u_{3} ∣, \dots$ form a geometric sequence with first term 9 and common ratio $\frac{2 2}{3} \approx 0.943$ .

∣ r ∣ = (\frac{2}{3})^{2} + (\frac{2}{3})^{2} = \frac{2 2}{3}

(c)

2 marks

The ratio satisfies $∣ r ∣ < 1$ , so the sum to infinity exists.

S_{\infty} = \frac{9}{1 - \frac{2 2}{3}} = \frac{27}{3 - 2 2} = 81 + 542 \approx 157

Question 12

7 marks

A slope field for dy/dx = x^4 + 4x^2 y + 4y^2

Calculus

(a)

2 marks

The right side is a square, so every gradient is zero or positive and the solution through $(0, 1)$ is increasing. Starting from $(0, 1)$ , follow the short slope marks to the right and left.
Sketch: a curve that rises from the left through $(0, 1)$ and steepens to the right, crossing the $x$ -axis near $x \approx - 0.9$ .

(b)

2 marks

Sketch: a downward parabola $y = a x^{2}$ with its maximum at $(0, 0)$ , going through about $(- 2, - 2)$ and $(2, - 2)$ .

(c)

3 marks

Substitute $y = a x^{2}$ into the right side and set it equal to 0.

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

Question 13

7 marks

A rotation followed by a reflection matrix, replaced by a single reflection

Number and algebra

(a)

2 marks

The matrix $A = (cos \frac{π}{3} sin \frac{π}{3} - sin \frac{π}{3} cos \frac{π}{3})$ has the form of a rotation matrix.
It is a rotation about the origin through $\frac{π}{3}$ radians anticlockwise.

(b)

5 marks

With $B = (- 0.8 0.6 0.6 0.8)$ , the combined matrix is $B A$ .
Compare with the reflection matrix.
The line is $y = x tan θ$ .

B A = (- 0.8 0.6 0.6 0.8) (0.5 0.8660 - 0.8660 0.5) = (0.1196 0.9928 0.9928 - 0.1196) cos 2 θ = 0.1196, sin 2 θ = 0.9928 \Rightarrow tan 2 θ = \frac{0.9928}{0.1196} 2 θ = 1.4509 \Rightarrow θ = 0.7254 a = tan 0.7254 \approx 0.887

Question 14

6 marks

Adding two sine waves with complex exponentials

Number and algebra

(a)

2 marks

Write each wave as the imaginary part of a complex exponential.

L_{1} (x) = Im (14.8 e^{0.35 x i} e^{1.27 i}), L_{2} (x) = Im (18.5 e^{0.35 x i} e^{- 2.36 i})

(b)

4 marks

Add the waves and factor out $e^{0.35 x i}$ .
Evaluate the bracket on the calculator in the form $q e^{r i}$ .

L_{3} (x) = Im (e^{0.35 x i} (14.8 e^{1.27 i} + 18.5 e^{- 2.36 i})) 14.8 e^{1.27 i} + 18.5 e^{- 2.36 i} \approx 8.816 e^{3.016 i} L_{3} (x) = 8.816 sin (0.35 x + 3.016) q \approx 8.82, r \approx 3.02

Question 15

8 marks

A mass on a spring: Euler's method for a second-order equation

Calculus

(a)

5 marks

Divide the equation by 2 to get $\frac{d ^{2} x}{d t ^{2}} = 4.9 - 1.2 \frac{d x}{d t} - 4 x$ . Use Euler's method with $h = 0.1$ for $x$ and $v$ .
Start from $x_{0} = 0.5$ , $v_{0} = 9.8$ and take 10 steps.
Continue the same steps. The first estimated displacement at or below zero is at $t = 2.0$ seconds (where the estimate is $- 0.19$ ).

x_{n + 1} = x_{n} + 0.1 v_{n}, v_{n + 1} = v_{n} + 0.1 (4.9 - 1.2 v_{n} - 4 x_{n}) x (1) \approx 4.36 m t = 2 seconds

(b)

3 marks

The exact value is 385 cm to the nearest centimetre, so it lies between 3.845 m and 3.855 m. The largest error is with 3.845.

\frac{4.3633 - 3.845}{3.845} \times 100 \approx 13.5%

Our analysis

Where the marks are

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

Calculus30 marks · 27%
Q7, Q8, Q12, Q15
Number and algebra29 marks · 26%
Q3, Q11, Q13, Q14
Functions29 marks · 26%
Q2, Q4, Q6, Q10
Geometry and trigonometry15 marks · 14%
Q1, Q5
Statistics and probability7 marks · 6%
Q9

Mentor insights

Traps to watch for

01
Finance solver · Q3
$N = 72$ monthly payments, $F V = 0$ , 12 payments a year.
02
Angle at B · Q5
The bearing of $C$ from $B$ is $10 0^{\circ}$ , and $D$ is due west of $B$ (bearing $27 0^{\circ}$ ), so the angle $D B C$ is $27 0^{\circ} - 10 0^{\circ} = 17 0^{\circ}$ .
03
Zero acceleration · Q7
The acceleration is the gradient of the velocity graph, so zero acceleration is where $v$ has a maximum or minimum.
04
Log laws · Q10
$ln a + ln b = ln (ab)$ and $2 ln x = ln x^{2}$ .
05
Modulus of a product · Q11
$∣ z w ∣ = ∣ z ∣∣ w ∣$ , so the moduli of the terms form a geometric sequence with ratio $∣ r ∣$ .

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

Worked solutions

IB Maths AI HL · Paper 1

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2 TZ3

Session: May 2025
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 15

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

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2025 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–15 · 110 marks

Question 1

6 marks

Snow paths: a sector area and a path length

Geometry and trigonometry

(a)

3 marks

The shaded region is one fifth of the area between the circles of radius 4 and 2.8.
Since $5.13 < 6$ , Maureen has enough dye.

A = \frac{1}{5} (π (4)^{2} - π (2.8)^{2}) = \frac{1}{5} π (16 - 7.84) \approx 5.13 m^{2}

(b)

3 marks

The route has three fifths of the outer circle, plus two radii of length 4.

\frac{3}{5} \times 2 π (4) + 2 \times 4 \approx 15.08 + 8 \approx 23.1 m

Question 2

7 marks

A dolphin's jump modelled by a quadratic

Functions

(a)

2 marks

The axis of symmetry is halfway between $d = 3$ and $d = 8.5$ .

d = \frac{3 + 8.5}{2} = 5.75

(b)

2 marks

The axis of symmetry of $- 0.2 d^{2} + b d$ is $d = - \frac{b}{2 ( - 0.2 )}$ .

\frac{b}{0.4} = 5.75 \Rightarrow b = 2.3

(c)

2 marks

The dolphin re-enters the water when $h = 0$ .

- 0.2 d^{2} + 2.3 d = 0 \Rightarrow d (- 0.2 d + 2.3) = 0 \Rightarrow d = 11.5 m

(d)

1 mark

A negative $h$ means the dolphin is below the surface of the water, which is outside the jump the model describes.

Question 3

8 marks

Two loan options for a truck

Number and algebra

(a)

6 marks

(i) Option A borrows the whole price.

(ii) Option B borrows 80% of the price.

N = 72, I = 7.15%, P V = 42695, F V = 0 \Rightarrow P M T \approx 730.99 USD 0.8 \times 42695 = 34156 N = 72, I = 6.40%, P V = 34156, F V = 0 \Rightarrow P M T \approx 572.54 USD

(b)

2 marks

(i) Option A needs no initial payment, so it suits Raul if he does not have enough money for a deposit.

(ii) Option B has lower monthly payments, and less interest is paid over the life of the loan.

Question 4

7 marks

Hat prices with a discount after the first 15

Functions

(a)

1 mark

Fifteen hats at $p$ euros each.

15 p

(b)

3 marks

The remaining $n - 15$ hats cost $p - 5$ each.

C (n) = 15 p + (p - 5) (n - 15)

(c)

3 marks

(i) The total is the mean price times the number of hats.

(ii) Substitute $n = 100$ in the cost equation.

25.73 \times 100 = 2573 EUR 2573 = 15 p + (p - 5) (85) \Rightarrow 2573 = 100 p - 425 \Rightarrow p = 29.98

Question 5

9 marks

A boat off course: the cosine rule and bearings

Geometry and trigonometry

(a)

7 marks

(i) In triangle $D B C$ we have $D B = 2$ , $B C = 13$ and $D B C = 17 0^{\circ}$ . Use the cosine rule.

(ii) Use the sine rule to find the angle $B D C$ .

Moving from the due east direction at $D$ (bearing $09 0^{\circ}$ ) round towards the south by this angle gives

D C^{2} = 2^{2} + 1 3^{2} - 2 (2) (13) cos 17 0^{\circ} \Rightarrow D C \approx 14.97 \approx 15.0 km \frac{sin B D C}{13} = \frac{sin 17 0 ^{\circ}}{14.97} \Rightarrow B D C \approx 8.6 7^{\circ} bearing = 90 + 8.67 \approx 098. 7^{\circ}

(b)

2 marks

D E = 13 km

Question 6

6 marks

Discounts at furniture stores and composite functions

Functions

(a)

2 marks

(i) A 30% discount leaves 70% of the price.

(ii) A reduction of 150 CAD.

f (x) = 0.7 x g (x) = x - 150

(b)

2 marks

$(f \circ g) (x) = f (g (x))$ means the 150 CAD reduction is applied first, and then the 30% discount is applied to the reduced price.

(f \circ g) (x) = 0.7 (x - 150) = 0.7 x - 105

(c)

2 marks

Compare the two orders at 450 CAD.
The price 165 is lower, so $(g \circ f) (x)$ is better for the customer: the 30% is taken off the higher original price first.

(f \circ g) (450) = 0.7 (300) = 210, (g \circ f) (450) = 0.7 (450) - 150 = 165

Question 7

8 marks

A particle's velocity 2 cos t + sin 2t - 0.2: sketch, turning points and distance

Calculus

(a)

5 marks

(i) Sketch: the curve starts at $(0, 1.8)$ and rises to a maximum, falls through zero velocity to a minimum, and rises again, ending at $t = 6$ .

(ii) The points of zero acceleration come from $v^{'} (t) = 0$ , found on a graph.

(0.524, 2.40), (2.62, - 2.80), (4.71, - 0.2)

(b)

3 marks

(i) The particle changes direction (where $v = 0$ at $t \approx 1.52$ and $t \approx 5.32$ ), so the distance is the integral of the modulus.

(ii) Evaluate with the calculator.

d = \int_{0}^{6} ∣ 2 cos t + sin 2 t - 0.2 ∣ d t d \approx 7.76 m

Question 8

7 marks

Area under a cubic and equal areas with a quadratic

Calculus

(a)

2 marks

The roots are 1, 5 and $a$ , and their sum is 9.

1 + 5 + a = 9 \Rightarrow a = 3

(b)

2 marks

The curve is above the axis between $x = 1$ and $x = 3$ .

\int_{1}^{3} (x^{3} - 9 x^{2} + 23 x - 15) d x = 4

(c)

3 marks

Set the two areas equal.
Solve with the calculator on $1 < b < 3$ .

\int_{1}^{b} f (x) d x = \int_{1}^{b} g (x) d x b \approx 2.57

Question 9

7 marks

Travel time to school: normal, binomial and a detention

Statistics and probability

(a)

4 marks

The probability that he is late on one day.
Let $X$ be the number of late days in 180 days, $X \sim B (180, 0.03386)$ .

P (T > 45) \approx 0.03386, T \sim N (35.5, 5. 2^{2}) P (X \geq 10) = 1 - P (X \leq 9) \approx 0.0869

(b)

3 marks

On a late day he is marked late with probability $1 - 0.35 = 0.65$ .
Let $Y$ be the number of days marked late, $Y \sim B (180, 0.02201)$ .

0.03386 \times 0.65 \approx 0.02201 P (Y \geq 4) = 1 - P (Y \leq 3) \approx 0.561

Question 10

9 marks

Combining logarithms, an inverse function and finding a constant

Functions

(a)

2 marks

Combine the logarithms.

f (x) = ln \frac{( 2 - x ) ( - 2 - x )}{x ^{2}} = ln \frac{x ^{2} - 4}{x ^{2}} g (x) = \frac{x ^{2} - 4}{x ^{2}} = 1 - \frac{4}{x ^{2}}

(b)

2 marks

The domain of $f$ is $x < - 2$ , which is the range of $f^{- 1}$ .

f^{- 1} (x) < - 2

(c)

5 marks

Write $y = ln (1 - \frac{4}{x ^{2}})$ and swap the variables.
The range of $f^{- 1}$ is $y < - 2$ , so take the negative root.

x = ln (1 - \frac{4}{y ^{2}}) \Rightarrow e^{x} = 1 - \frac{4}{y ^{2}} \Rightarrow y^{2} = \frac{4}{1 - e ^{x}} f^{- 1} (x) = - \frac{2}{1 - e ^{x}} a = - 2

Question 11

8 marks

A geometric sequence with a complex ratio

Number and algebra

(a)

2 marks

Multiply by $r = \frac{2}{3} + \frac{2}{3} i$ .

u_{2} = 9 r = 6 + 6 i u_{3} = u_{2} r = (6 + 6 i) (\frac{2}{3} + \frac{2}{3} i) = 4 + 4 i + 4 i + 4 i^{2} = 8 i

(b)

4 marks

The modulus of the common ratio is
Since $∣ u_{n + 1} ∣ = ∣ u_{n} ∣∣ r ∣$ , each modulus is $\frac{2 2}{3}$ times the one before. So $∣ u_{1} ∣, ∣ u_{2} ∣, ∣ u_{3} ∣, \dots$ form a geometric sequence with first term 9 and common ratio $\frac{2 2}{3} \approx 0.943$ .

∣ r ∣ = (\frac{2}{3})^{2} + (\frac{2}{3})^{2} = \frac{2 2}{3}

(c)

2 marks

The ratio satisfies $∣ r ∣ < 1$ , so the sum to infinity exists.

S_{\infty} = \frac{9}{1 - \frac{2 2}{3}} = \frac{27}{3 - 2 2} = 81 + 542 \approx 157

Question 12

7 marks

A slope field for dy/dx = x^4 + 4x^2 y + 4y^2

Calculus

(a)

2 marks

The right side is a square, so every gradient is zero or positive and the solution through $(0, 1)$ is increasing. Starting from $(0, 1)$ , follow the short slope marks to the right and left.
Sketch: a curve that rises from the left through $(0, 1)$ and steepens to the right, crossing the $x$ -axis near $x \approx - 0.9$ .

(b)

2 marks

Sketch: a downward parabola $y = a x^{2}$ with its maximum at $(0, 0)$ , going through about $(- 2, - 2)$ and $(2, - 2)$ .

(c)

3 marks

Substitute $y = a x^{2}$ into the right side and set it equal to 0.

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

Question 13

7 marks

A rotation followed by a reflection matrix, replaced by a single reflection

Number and algebra

(a)

2 marks

The matrix $A = (cos \frac{π}{3} sin \frac{π}{3} - sin \frac{π}{3} cos \frac{π}{3})$ has the form of a rotation matrix.
It is a rotation about the origin through $\frac{π}{3}$ radians anticlockwise.

(b)

5 marks

With $B = (- 0.8 0.6 0.6 0.8)$ , the combined matrix is $B A$ .
Compare with the reflection matrix.
The line is $y = x tan θ$ .

B A = (- 0.8 0.6 0.6 0.8) (0.5 0.8660 - 0.8660 0.5) = (0.1196 0.9928 0.9928 - 0.1196) cos 2 θ = 0.1196, sin 2 θ = 0.9928 \Rightarrow tan 2 θ = \frac{0.9928}{0.1196} 2 θ = 1.4509 \Rightarrow θ = 0.7254 a = tan 0.7254 \approx 0.887

Question 14

6 marks

Adding two sine waves with complex exponentials

Number and algebra

(a)

2 marks

Write each wave as the imaginary part of a complex exponential.

L_{1} (x) = Im (14.8 e^{0.35 x i} e^{1.27 i}), L_{2} (x) = Im (18.5 e^{0.35 x i} e^{- 2.36 i})

(b)

4 marks

Add the waves and factor out $e^{0.35 x i}$ .
Evaluate the bracket on the calculator in the form $q e^{r i}$ .

L_{3} (x) = Im (e^{0.35 x i} (14.8 e^{1.27 i} + 18.5 e^{- 2.36 i})) 14.8 e^{1.27 i} + 18.5 e^{- 2.36 i} \approx 8.816 e^{3.016 i} L_{3} (x) = 8.816 sin (0.35 x + 3.016) q \approx 8.82, r \approx 3.02

Question 15

8 marks

A mass on a spring: Euler's method for a second-order equation

Calculus

(a)

5 marks

Divide the equation by 2 to get $\frac{d ^{2} x}{d t ^{2}} = 4.9 - 1.2 \frac{d x}{d t} - 4 x$ . Use Euler's method with $h = 0.1$ for $x$ and $v$ .
Start from $x_{0} = 0.5$ , $v_{0} = 9.8$ and take 10 steps.
Continue the same steps. The first estimated displacement at or below zero is at $t = 2.0$ seconds (where the estimate is $- 0.19$ ).

x_{n + 1} = x_{n} + 0.1 v_{n}, v_{n + 1} = v_{n} + 0.1 (4.9 - 1.2 v_{n} - 4 x_{n}) x (1) \approx 4.36 m t = 2 seconds

(b)

3 marks

The exact value is 385 cm to the nearest centimetre, so it lies between 3.845 m and 3.855 m. The largest error is with 3.845.

\frac{4.3633 - 3.845}{3.845} \times 100 \approx 13.5%

Our analysis

Where the marks are

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

Calculus30 marks · 27%
Q7, Q8, Q12, Q15
Number and algebra29 marks · 26%
Q3, Q11, Q13, Q14
Functions29 marks · 26%
Q2, Q4, Q6, Q10
Geometry and trigonometry15 marks · 14%
Q1, Q5
Statistics and probability7 marks · 6%
Q9

Mentor insights

Traps to watch for

01
Finance solver · Q3
$N = 72$ monthly payments, $F V = 0$ , 12 payments a year.
02
Angle at B · Q5
The bearing of $C$ from $B$ is $10 0^{\circ}$ , and $D$ is due west of $B$ (bearing $27 0^{\circ}$ ), so the angle $D B C$ is $27 0^{\circ} - 10 0^{\circ} = 17 0^{\circ}$ .
03
Zero acceleration · Q7
The acceleration is the gradient of the velocity graph, so zero acceleration is where $v$ has a maximum or minimum.
04
Log laws · Q10
$ln a + ln b = ln (ab)$ and $2 ln x = ln x^{2}$ .
05
Modulus of a product · Q11
$∣ z w ∣ = ∣ z ∣∣ w ∣$ , so the moduli of the terms form a geometric sequence with ratio $∣ r ∣$ .

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