Loading…

Worked solutions

IB Maths AI HL · Paper 1

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2

Session: May 2023
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 2023 TZ1 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

5 marks

Savings with a deposit at the end of each month

Number and algebra

(a)

3 marks

1 January 2024 is 24 months after 1 January 2022. Monthly rate $\frac{4}{12} %$ .

A = 1000 (1 + \frac{0.04}{12})^{24} + 100 \cdot \frac{( 1 + \frac{0.04}{12} ) ^{24} - 1}{0.04/12} = $3577.43

(b)

2 marks

Test the balance after 36 and 37 months.
So it takes 37 complete months.

A_{36} = $4945.43 < 5000, A_{37} = $5061.91 > 5000

Question 2

6 marks

Memory scores: a one-tailed two-sample t-test

Statistics and probability

(a)

2 marks

The alternative hypothesis is one-sided because Carys believes bilingual people will score higher.

H_{0} : μ_{b} = μ_{m}, H_{1} : μ_{b} > μ_{m}

(b)

2 marks

Enter both samples into a pooled two-sample $t$ -test (10 and 8 values, so $ν = 16$ ).

p \approx 0.119

(c)

2 marks

Compare with 0.05.
Do not reject $H_{0}$ : there is insufficient evidence at the 5% level that the mean memory score of bilingual people is higher than that of monolingual people.

0.119 > 0.05

Question 3

5 marks

Nearest neighbour interpolation of a temperature

Geometry and trigonometry

(a)

2 marks

Distance between $A (35, 26)$ and $D (24, 19)$ .

A D = (35 - 24)^{2} + (26 - 19)^{2} = 170 \approx 13.0 miles

(b)

3 marks

Find the distance from D to each town.
C is the closest town, so the estimate of the temperature at the farm is the temperature at C.

D A = 13.04, D B = 1 3^{2} + 5^{2} = 13.93, D C = 4^{2} + 1 2^{2} = 12.65 30^{\circ} C

Question 4

8 marks

A hill profile: trapezoidal rule, exact area and percentage error

Calculus

(a)

2 marks

Trapezoidal rule with $h = 10$ and the five heights.

A \approx \frac{10}{2} [0 + 0 + 2 (3 + 8 + 9)] = 200 cm^{2}

(b)

4 marks

The exact area is the integral of the curve.
Evaluate.

A = \int_{0}^{40} (0.04 x^{2} - 0.001 x^{3}) d x A = [\frac{0.04 x ^{3}}{3} - \frac{0.001 x ^{4}}{4}]_{0}^{40} = 853.33 - 640 = 213.33 cm^{2}

(c)

2 marks

Percentage error compares the trapezoidal estimate with the exact area.

\frac{∣200 - 213.33∣}{213.33} \times 100 = 6.25%

Question 5

5 marks

A boat's return bearing

Number and algebra

5 marks

Take the starting point as the origin. After 8 km on $31 5^{\circ}$ and 6 km on $04 5^{\circ}$ :
The boat is 1.414 km west and 9.899 km north of the start, so to return it must go east 1.414 km and south 9.899 km.
The direction is south and slightly east, so the bearing is

east = - 8 sin 4 5^{\circ} + 6 sin 4 5^{\circ} = - 1.414, north = 8 cos 4 5^{\circ} + 6 cos 4 5^{\circ} = 9.899 tan α = \frac{1.414}{9.899} \Rightarrow α = 8.1 3^{\circ} 18 0^{\circ} - 8.1 3^{\circ} \approx 17 2^{\circ}

Question 6

7 marks

Baseball: a strike and a ball against the wall

Functions

(a)

4 marks

The gradient of the path is $- 0.045$ .
The ball is 1.25 m high at the front of the plate and about $1.25 - 0.0194 = 1.2307$ m at the back. The back height is between 0.53 m and 1.24 m, so at some point over the plate the ball is in the strike zone: it is a strike.

a = - 0.045 \times 0.43 = - 0.01935 (\approx - 0.0194)

(b)

3 marks

Find the height at the wall, $d = 96$ .
Since $8.34 > 5$ , the ball is above the 5 m wall when it reaches it, so it goes over the wall.

h (96) = - 0.01 (96)^{2} + 1.04 (96) + 0.66 = 8.34 m

Question 7

9 marks

A triangular shade cover: vector product, area, distance and slope

Geometry and trigonometry

(a)

2 marks

Compute the components: $6 (0) - 1 (3) = - 3$ , $1 (7) - 0 (0) = 7$ , $0 (3) - 6 (7) = - 42$ .

A B \times A C = 061 \times 730 = - 3 7 - 42

(b)

2 marks

Area of the triangle is half the length of the vector product.

Area = \frac{1}{2} 9 + 49 + 1764 = \frac{1}{2} 1822 \approx 21.3 m^{2}

(c)

3 marks

BX is the height of the triangle on the base AC, so $\frac{1}{2} ∣ A C ∣ \cdot B X = Area$ , with $∣ A C ∣ = 58$ .

B X = \frac{1822}{58} \approx 5.60 m

(d)

2 marks

The angle between the cover and the horizontal equals the angle between their normals, $(- 3, 7, - 42)$ and $(0, 0, 1)$ .

cos φ = \frac{42}{1822} \Rightarrow φ \approx 10. 3^{\circ}

Question 8

7 marks

Testing for linear correlation with six observations

Statistics and probability

(a)

2 marks

The null hypothesis says there is no linear correlation between $X$ and $Y$ .

H_{0} : ρ = 0, H_{1} : ρ \neq = 0

(b)

3 marks

Enter the data and run a linear regression and a test for correlation.

r = 0.849, p = 0.0325

(c)

2 marks

Compare the $p$ -value with 0.05.
Yes, there is sufficient evidence to reject $H_{0}$ : at the 5% level there is evidence of a linear correlation between $X$ and $Y$ , so the claim of no linear correlation is not supported.

0.0325 < 0.05

Question 9

5 marks

Speed of cars: standard deviation and an asymmetric interval

Statistics and probability

(a)

2 marks

The difference from the mean is two standard deviations.

2 σ = 75.7 - 67.3 = 8.4 \Rightarrow σ = 4.2

(b)

3 marks

A central region containing 82% leaves 9% in each tail, so it is $μ \pm z σ$ with $z = 1.341$ .
The speed 74 km/h is in the 82% interval but lies above 72.9. So the interval from $p$ to $q$ is not this symmetric interval around the mean: it is not symmetrical about the mean.

67.3 \pm 1.341 (4.2) = 67.3 \pm 5.63 \Rightarrow (61.7, 72.9)

Question 10

9 marks

A decorative hook built from a line and two parabolas

Functions

(a)

1 mark

The line goes from $(0, 0)$ to $(3, 3)$ .

y = x

(b)

3 marks

Write $y = a x^{2} + b x$ since it passes through the origin. The gradient at 0 is $b = 1$ .

4 a - 2 b = 0 \Rightarrow a = \frac{1}{2} y = \frac{1}{2} x^{2} + x

(c)

4 marks

The curve through $(3, 3)$ with gradient 1 there has the form $y = a (x - 3)^{2} + (x - 3) + 3$ . It passes through $(6, 2)$ .

9 a + 3 + 3 = 2 \Rightarrow a = - \frac{4}{9} y = - \frac{4}{9} (x - 3)^{2} + x

(d)

1 mark

Put the three pieces together.

f (x) = ⎩ ⎨ ⎧ \frac{1}{2} x^{2} + x, x, - \frac{4}{9} (x - 3)^{2} + x, - 2 \leq x < 0 0 \leq x \leq 3 3 < x \leq 6

Question 11

6 marks

An orange heavier than two lemons: combining normal variables

Number and algebra

6 marks

Let $O \sim N (205, 5^{2})$ and $L_{1}, L_{2} \sim N (105, 3^{2})$ be independent. The combined weight of the lemons has mean 210 and variance $9 + 9 = 18$ .
Let $D = O - (L_{1} + L_{2})$ .
The orange is heavier than the lemons when $D > 0$ .

E (D) = 205 - 210 = - 5, Var (D) = 25 + 18 = 43 D \sim N (- 5, 43) P (D > 0) = P (Z > \frac{5}{43}) \approx 0.223

Question 12

5 marks

Adding two AC voltages of the same frequency

Number and algebra

5 marks

Use the sum-to-product identity $sin A + sin B = 2 sin \frac{A + B}{2} cos \frac{A - B}{2}$ .
Since $cos 3 0^{\circ} = \frac{3}{2}$ :

V_{1} + V_{2} = 6 [sin (a t + 3 0^{\circ}) + sin (a t + 9 0^{\circ})] = 12 sin (a t + 6 0^{\circ}) cos (3 0^{\circ}) V_{1} + V_{2} = 63 sin (a t + 6 0^{\circ}) V = 63 \approx 10.4, θ = 60

Question 13

6 marks

Euler's method for a damped spring

Number and algebra

6 marks

Put $y = \frac{d x}{d t}$ , so $\frac{d x}{d t} = y$ and $\frac{d y}{d t} = - 2 y - 10 x$ , with $x_{0} = 0.75$ , $y_{0} = 0$ and $h = 0.1$ .
Five steps give

x_{n + 1} = x_{n} + 0.1 y_{n}, y_{n + 1} = y_{n} + 0.1 (- 2 y_{n} - 10 x_{n}) (x, y) = (0.75, - 0.75), (0.675, - 1.35), (0.54, - 1.755), (0.3645, - 1.944), (0.1701, - 1.9197) x (0.5) \approx 0.170 cm

Question 14

6 marks

A straight line on a log-log graph gives a power law

Number and algebra

6 marks

A straight line on a log-log graph has the form $ln y = b ln x + ln a$ . Use the two given points.
Find $a$ from $(2, 13.1951)$ .

b = \frac{ln 34.822 - ln 13.1951}{ln 4 - ln 2} = \frac{ln ( 2.6389 )}{ln 2} = 1.4 a = \frac{13.1951}{2 ^{1.4}} = 5 y = 5 x^{1.4}

Question 15

6 marks

Volume of a solid formed by rotating a hexagon about the y-axis

Number and algebra

6 marks

The right-hand edge is $x = 2 y$ for $1 \leq y \leq 2$ , $x = 4$ for $2 \leq y \leq 4$ , and $x = 12 - 2 y$ for $4 \leq y \leq 5$ . The left edge is the $y$ -axis.
The volume is about 159 cubic units.

V = π \int_{1}^{2} (2 y)^{2} d y + π \int_{2}^{4} 16 d y + π \int_{4}^{5} (12 - 2 y)^{2} d y V = π (\frac{28}{3} + 32 + \frac{28}{3}) = \frac{152 π}{3} \approx 159

Question 16

6 marks

Sum of square residuals for the model I = k/d^2

Calculus

(a)

4 marks

The sum of squares is
The constant term (when $k = 0$ ) is the sum of the squares of the observed values.

S = (42 - k)^{2} + (11 - \frac{k}{4})^{2} + (1.5 - \frac{k}{25})^{2} c = 4 2^{2} + 1 1^{2} + 1. 5^{2} = 1764 + 121 + 2.25 = 1887.25

(b)

2 marks

The minimum of $1.0641 k^{2} - 89.62 k + 1887.25$ is at the vertex.

k = \frac{89.62}{2 ( 1.0641 )} = 42.1 I = \frac{42.1}{d ^{2}}

Question 17

9 marks

A plane approaching a tracking station: related rates

Calculus
Geometry and trigonometry

(a)

2 marks

The plane flies towards Q, so $x$ is decreasing at 250 m/s, and $θ$ is increasing at 0.075 rad/s.

\frac{d x}{d θ} = \frac{d x / d t}{d θ / d t} = \frac{- 250}{0.075} = - \frac{10000}{3}

(b)

1 mark

In the right triangle SQP: $tan θ = \frac{3000}{x}$ .

x = \frac{3000}{tan θ} = 3000 cot θ

(c)

3 marks

Differentiate $x = 3000 cot θ$ .

\frac{d x}{d θ} = - 3000 csc^{2} θ = - \frac{3000}{sin ^{2} θ}

(d)

3 marks

Equate with the value from part (a).
Then $cos^{2} θ = 0.1$ and $cot θ = \frac{0.1}{0.9} = \frac{1}{3}$ .

\frac{3000}{sin ^{2} θ} = \frac{10000}{3} \Rightarrow sin^{2} θ = 0.9 x = 3000 cot θ = 1000 m

Our analysis

Where the marks are

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

Number and algebra39 marks · 35%
Q1, Q5, Q11, Q12, Q13, Q14, Q15
Calculus22 marks · 20%
Q4, Q16, Q17
Statistics and probability18 marks · 16%
Q2, Q8, Q9
Functions16 marks · 15%
Q6, Q10
Geometry and trigonometry15 marks · 14%
Q3, Q7, Q17

Mentor insights

Traps to watch for

01
Exact area uses the equation · Q4
Integrate the given cubic from 0 to 40.
02
Height change over the plate · Q6
The ball is over the plate for 0.43 m of horizontal distance, so the height changes by gradient times 0.43.
03
Vector product gives the normal · Q7
$A B \times A C$ is perpendicular to the cover and its length is twice the triangle's area.
04
Match gradients at joins · Q10
The quadratic pieces have the same gradient as the line at the join, which is 1.
05
The plane is approaching Q · Q17
The distance $x$ is decreasing, so $\frac{d x}{d t} = - 250$ .

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

Session: May 2023
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 2023 TZ1 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

5 marks

Savings with a deposit at the end of each month

Number and algebra

(a)

3 marks

1 January 2024 is 24 months after 1 January 2022. Monthly rate $\frac{4}{12} %$ .

A = 1000 (1 + \frac{0.04}{12})^{24} + 100 \cdot \frac{( 1 + \frac{0.04}{12} ) ^{24} - 1}{0.04/12} = $3577.43

(b)

2 marks

Test the balance after 36 and 37 months.
So it takes 37 complete months.

A_{36} = $4945.43 < 5000, A_{37} = $5061.91 > 5000

Question 2

6 marks

Memory scores: a one-tailed two-sample t-test

Statistics and probability

(a)

2 marks

The alternative hypothesis is one-sided because Carys believes bilingual people will score higher.

H_{0} : μ_{b} = μ_{m}, H_{1} : μ_{b} > μ_{m}

(b)

2 marks

Enter both samples into a pooled two-sample $t$ -test (10 and 8 values, so $ν = 16$ ).

p \approx 0.119

(c)

2 marks

Compare with 0.05.
Do not reject $H_{0}$ : there is insufficient evidence at the 5% level that the mean memory score of bilingual people is higher than that of monolingual people.

0.119 > 0.05

Question 3

5 marks

Nearest neighbour interpolation of a temperature

Geometry and trigonometry

(a)

2 marks

Distance between $A (35, 26)$ and $D (24, 19)$ .

A D = (35 - 24)^{2} + (26 - 19)^{2} = 170 \approx 13.0 miles

(b)

3 marks

Find the distance from D to each town.
C is the closest town, so the estimate of the temperature at the farm is the temperature at C.

D A = 13.04, D B = 1 3^{2} + 5^{2} = 13.93, D C = 4^{2} + 1 2^{2} = 12.65 30^{\circ} C

Question 4

8 marks

A hill profile: trapezoidal rule, exact area and percentage error

Calculus

(a)

2 marks

Trapezoidal rule with $h = 10$ and the five heights.

A \approx \frac{10}{2} [0 + 0 + 2 (3 + 8 + 9)] = 200 cm^{2}

(b)

4 marks

The exact area is the integral of the curve.
Evaluate.

A = \int_{0}^{40} (0.04 x^{2} - 0.001 x^{3}) d x A = [\frac{0.04 x ^{3}}{3} - \frac{0.001 x ^{4}}{4}]_{0}^{40} = 853.33 - 640 = 213.33 cm^{2}

(c)

2 marks

Percentage error compares the trapezoidal estimate with the exact area.

\frac{∣200 - 213.33∣}{213.33} \times 100 = 6.25%

Question 5

5 marks

A boat's return bearing

Number and algebra

5 marks

Take the starting point as the origin. After 8 km on $31 5^{\circ}$ and 6 km on $04 5^{\circ}$ :
The boat is 1.414 km west and 9.899 km north of the start, so to return it must go east 1.414 km and south 9.899 km.
The direction is south and slightly east, so the bearing is

east = - 8 sin 4 5^{\circ} + 6 sin 4 5^{\circ} = - 1.414, north = 8 cos 4 5^{\circ} + 6 cos 4 5^{\circ} = 9.899 tan α = \frac{1.414}{9.899} \Rightarrow α = 8.1 3^{\circ} 18 0^{\circ} - 8.1 3^{\circ} \approx 17 2^{\circ}

Question 6

7 marks

Baseball: a strike and a ball against the wall

Functions

(a)

4 marks

The gradient of the path is $- 0.045$ .
The ball is 1.25 m high at the front of the plate and about $1.25 - 0.0194 = 1.2307$ m at the back. The back height is between 0.53 m and 1.24 m, so at some point over the plate the ball is in the strike zone: it is a strike.

a = - 0.045 \times 0.43 = - 0.01935 (\approx - 0.0194)

(b)

3 marks

Find the height at the wall, $d = 96$ .
Since $8.34 > 5$ , the ball is above the 5 m wall when it reaches it, so it goes over the wall.

h (96) = - 0.01 (96)^{2} + 1.04 (96) + 0.66 = 8.34 m

Question 7

9 marks

A triangular shade cover: vector product, area, distance and slope

Geometry and trigonometry

(a)

2 marks

Compute the components: $6 (0) - 1 (3) = - 3$ , $1 (7) - 0 (0) = 7$ , $0 (3) - 6 (7) = - 42$ .

A B \times A C = 061 \times 730 = - 3 7 - 42

(b)

2 marks

Area of the triangle is half the length of the vector product.

Area = \frac{1}{2} 9 + 49 + 1764 = \frac{1}{2} 1822 \approx 21.3 m^{2}

(c)

3 marks

BX is the height of the triangle on the base AC, so $\frac{1}{2} ∣ A C ∣ \cdot B X = Area$ , with $∣ A C ∣ = 58$ .

B X = \frac{1822}{58} \approx 5.60 m

(d)

2 marks

The angle between the cover and the horizontal equals the angle between their normals, $(- 3, 7, - 42)$ and $(0, 0, 1)$ .

cos φ = \frac{42}{1822} \Rightarrow φ \approx 10. 3^{\circ}

Question 8

7 marks

Testing for linear correlation with six observations

Statistics and probability

(a)

2 marks

The null hypothesis says there is no linear correlation between $X$ and $Y$ .

H_{0} : ρ = 0, H_{1} : ρ \neq = 0

(b)

3 marks

Enter the data and run a linear regression and a test for correlation.

r = 0.849, p = 0.0325

(c)

2 marks

Compare the $p$ -value with 0.05.
Yes, there is sufficient evidence to reject $H_{0}$ : at the 5% level there is evidence of a linear correlation between $X$ and $Y$ , so the claim of no linear correlation is not supported.

0.0325 < 0.05

Question 9

5 marks

Speed of cars: standard deviation and an asymmetric interval

Statistics and probability

(a)

2 marks

The difference from the mean is two standard deviations.

2 σ = 75.7 - 67.3 = 8.4 \Rightarrow σ = 4.2

(b)

3 marks

A central region containing 82% leaves 9% in each tail, so it is $μ \pm z σ$ with $z = 1.341$ .
The speed 74 km/h is in the 82% interval but lies above 72.9. So the interval from $p$ to $q$ is not this symmetric interval around the mean: it is not symmetrical about the mean.

67.3 \pm 1.341 (4.2) = 67.3 \pm 5.63 \Rightarrow (61.7, 72.9)

Question 10

9 marks

A decorative hook built from a line and two parabolas

Functions

(a)

1 mark

The line goes from $(0, 0)$ to $(3, 3)$ .

y = x

(b)

3 marks

Write $y = a x^{2} + b x$ since it passes through the origin. The gradient at 0 is $b = 1$ .

4 a - 2 b = 0 \Rightarrow a = \frac{1}{2} y = \frac{1}{2} x^{2} + x

(c)

4 marks

The curve through $(3, 3)$ with gradient 1 there has the form $y = a (x - 3)^{2} + (x - 3) + 3$ . It passes through $(6, 2)$ .

9 a + 3 + 3 = 2 \Rightarrow a = - \frac{4}{9} y = - \frac{4}{9} (x - 3)^{2} + x

(d)

1 mark

Put the three pieces together.

f (x) = ⎩ ⎨ ⎧ \frac{1}{2} x^{2} + x, x, - \frac{4}{9} (x - 3)^{2} + x, - 2 \leq x < 0 0 \leq x \leq 3 3 < x \leq 6

Question 11

6 marks

An orange heavier than two lemons: combining normal variables

Number and algebra

6 marks

Let $O \sim N (205, 5^{2})$ and $L_{1}, L_{2} \sim N (105, 3^{2})$ be independent. The combined weight of the lemons has mean 210 and variance $9 + 9 = 18$ .
Let $D = O - (L_{1} + L_{2})$ .
The orange is heavier than the lemons when $D > 0$ .

E (D) = 205 - 210 = - 5, Var (D) = 25 + 18 = 43 D \sim N (- 5, 43) P (D > 0) = P (Z > \frac{5}{43}) \approx 0.223

Question 12

5 marks

Adding two AC voltages of the same frequency

Number and algebra

5 marks

Use the sum-to-product identity $sin A + sin B = 2 sin \frac{A + B}{2} cos \frac{A - B}{2}$ .
Since $cos 3 0^{\circ} = \frac{3}{2}$ :

V_{1} + V_{2} = 6 [sin (a t + 3 0^{\circ}) + sin (a t + 9 0^{\circ})] = 12 sin (a t + 6 0^{\circ}) cos (3 0^{\circ}) V_{1} + V_{2} = 63 sin (a t + 6 0^{\circ}) V = 63 \approx 10.4, θ = 60

Question 13

6 marks

Euler's method for a damped spring

Number and algebra

6 marks

Put $y = \frac{d x}{d t}$ , so $\frac{d x}{d t} = y$ and $\frac{d y}{d t} = - 2 y - 10 x$ , with $x_{0} = 0.75$ , $y_{0} = 0$ and $h = 0.1$ .
Five steps give

x_{n + 1} = x_{n} + 0.1 y_{n}, y_{n + 1} = y_{n} + 0.1 (- 2 y_{n} - 10 x_{n}) (x, y) = (0.75, - 0.75), (0.675, - 1.35), (0.54, - 1.755), (0.3645, - 1.944), (0.1701, - 1.9197) x (0.5) \approx 0.170 cm

Question 14

6 marks

A straight line on a log-log graph gives a power law

Number and algebra

6 marks

A straight line on a log-log graph has the form $ln y = b ln x + ln a$ . Use the two given points.
Find $a$ from $(2, 13.1951)$ .

b = \frac{ln 34.822 - ln 13.1951}{ln 4 - ln 2} = \frac{ln ( 2.6389 )}{ln 2} = 1.4 a = \frac{13.1951}{2 ^{1.4}} = 5 y = 5 x^{1.4}

Question 15

6 marks

Volume of a solid formed by rotating a hexagon about the y-axis

Number and algebra

6 marks

The right-hand edge is $x = 2 y$ for $1 \leq y \leq 2$ , $x = 4$ for $2 \leq y \leq 4$ , and $x = 12 - 2 y$ for $4 \leq y \leq 5$ . The left edge is the $y$ -axis.
The volume is about 159 cubic units.

V = π \int_{1}^{2} (2 y)^{2} d y + π \int_{2}^{4} 16 d y + π \int_{4}^{5} (12 - 2 y)^{2} d y V = π (\frac{28}{3} + 32 + \frac{28}{3}) = \frac{152 π}{3} \approx 159

Question 16

6 marks

Sum of square residuals for the model I = k/d^2

Calculus

(a)

4 marks

The sum of squares is
The constant term (when $k = 0$ ) is the sum of the squares of the observed values.

S = (42 - k)^{2} + (11 - \frac{k}{4})^{2} + (1.5 - \frac{k}{25})^{2} c = 4 2^{2} + 1 1^{2} + 1. 5^{2} = 1764 + 121 + 2.25 = 1887.25

(b)

2 marks

The minimum of $1.0641 k^{2} - 89.62 k + 1887.25$ is at the vertex.

k = \frac{89.62}{2 ( 1.0641 )} = 42.1 I = \frac{42.1}{d ^{2}}

Question 17

9 marks

A plane approaching a tracking station: related rates

Calculus
Geometry and trigonometry

(a)

2 marks

The plane flies towards Q, so $x$ is decreasing at 250 m/s, and $θ$ is increasing at 0.075 rad/s.

\frac{d x}{d θ} = \frac{d x / d t}{d θ / d t} = \frac{- 250}{0.075} = - \frac{10000}{3}

(b)

1 mark

In the right triangle SQP: $tan θ = \frac{3000}{x}$ .

x = \frac{3000}{tan θ} = 3000 cot θ

(c)

3 marks

Differentiate $x = 3000 cot θ$ .

\frac{d x}{d θ} = - 3000 csc^{2} θ = - \frac{3000}{sin ^{2} θ}

(d)

3 marks

Equate with the value from part (a).
Then $cos^{2} θ = 0.1$ and $cot θ = \frac{0.1}{0.9} = \frac{1}{3}$ .

\frac{3000}{sin ^{2} θ} = \frac{10000}{3} \Rightarrow sin^{2} θ = 0.9 x = 3000 cot θ = 1000 m

Our analysis

Where the marks are

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

Number and algebra39 marks · 35%
Q1, Q5, Q11, Q12, Q13, Q14, Q15
Calculus22 marks · 20%
Q4, Q16, Q17
Statistics and probability18 marks · 16%
Q2, Q8, Q9
Functions16 marks · 15%
Q6, Q10
Geometry and trigonometry15 marks · 14%
Q3, Q7, Q17

Mentor insights

Traps to watch for

01
Exact area uses the equation · Q4
Integrate the given cubic from 0 to 40.
02
Height change over the plate · Q6
The ball is over the plate for 0.43 m of horizontal distance, so the height changes by gradient times 0.43.
03
Vector product gives the normal · Q7
$A B \times A C$ is perpendicular to the cover and its length is twice the triangle's area.
04
Match gradients at joins · Q10
The quadratic pieces have the same gradient as the line at the join, which is 1.
05
The plane is approaching Q · Q17
The distance $x$ is decreasing, so $\frac{d x}{d t} = - 250$ .

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