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IB Maths AI HL · Paper 1

Mathematics: Applications & Interpretation, Higher Level

Session: November 2021
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 November 2021 Global 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

A kite on coordinate axes: gradients and a perpendicular diagonal

Geometry and trigonometry

(a)

2 marks

The gradient of AC.

m_{A C} = \frac{6 - 0}{4 - 2} = 3

(b)

1 mark

BD is perpendicular to AC.

m_{B D} = - \frac{1}{3}

(c)

2 marks

Use the point B $(0, 4)$ with gradient $- \frac{1}{3}$ .

y = - \frac{1}{3} x + 4 \Rightarrow x + 3 y - 12 = 0

(d)

1 mark

D is on the $x$ -axis, so $y = 0$ .

x + 0 - 12 = 0 \Rightarrow x = 12

Question 2

5 marks

Carbon dioxide emissions: the trapezoidal rule and percentage error

Calculus

(a)

3 marks

Use the trapezoidal rule with width 0.4: half the width times (first + last + twice the middle values).

\frac{1}{2} (0.4) [30 + 50 + 2 (50 + 60 + 40 + 20)] = 0.2 (80 + 340) = 84 tonnes

(b)

2 marks

Percentage error compared with the real value, 72 tonnes.

\frac{84 - 72}{72} \times 100% \approx 16.7%

Question 3

7 marks

A tin can's height as a function of diameter, and its inverse

Functions

(a)

3 marks

$h$ is decreasing on its domain, so the largest value is at $x = 4$ and the smallest at $x = 14$ .

h (4) = \frac{640}{16} + 0.5 = 40.5, h (14) = \frac{640}{196} + 0.5 \approx 3.77 3.77 \leq h (x) \leq 40.5

(b)

4 marks

(i) Solve $h (x) = 10$ .

(ii) A tin that is 10 cm high has a diameter of about 8.21 cm.

(iii) The range of the inverse is the domain of $h$ .

\frac{640}{x ^{2}} + 0.5 = 10 \Rightarrow x^{2} = \frac{640}{9.5} \Rightarrow x \approx 8.21 4 \leq h^{- 1} (x) \leq 14

Question 4

6 marks

Fire stations and a Voronoi diagram: where to put a new station

Geometry and trigonometry

(a)

3 marks

The new station should be as far as possible from the nearest existing one, which happens at a vertex of the Voronoi diagram, P or Q. Compare the distance from each to its nearest stations.
P is farther from its nearest stations than Q is, so the new station should be at P.

AP = (10 - 6)^{2} + (6 - 2)^{2} = 32 \approx 5.66, BQ = (13 - 14)^{2} + (7 - 2)^{2} = 26 \approx 5.10

(b)

3 marks

(i) P is $(10, 6)$ and C is $(18, 6)$ . They have the same $y$ , so the bisector is vertical through the midpoint $(14, 6)$ .

(ii) The missing boundaries are the perpendicular bisectors of PC and PD, joined to the existing boundaries. The one for PD passes through the midpoint $(10.4, 8.8)$ and has gradient $- \frac{1}{7}$ , since PD has gradient 7.

x = 14 y - 8.8 = - \frac{1}{7} (x - 10.4)

Question 5

9 marks

A 30-year home loan: the payment, the balance and the interest paid

Number and algebra

(a)

3 marks

Use the finance solver with $N = 360$ , $I = 3.8%$ , $PV = 170000$ , $FV = 0$ , 12 payments and 12 compounding periods per year.

PMT \approx 792.13 AUD per month

(b)

6 marks

(i) After 10 years ( $N = 120$ ) the balance is the future value with the payment of $- 792.13$ .

(ii) They paid $120 \times 792.13 = 95 055.60$ . Of this, $170000 - 133019.94 = 36 980.06$ went to repaying the loan. The rest is interest.

balance \approx 133 019.94 AUD 95 055.60 - 36 980.06 = 58 075.54 AUD

Mentor tip

Using the unrounded payment gives 58 075.60. Both are fine to the cent, as long as you say which one you used.

Question 6

5 marks

An infinite geometric series: its ratio and when the terms become small

Number and algebra

(a)

2 marks

For an infinite geometric series $S_{\infty} = \frac{u _{1}}{1 - r}$ .

10 = \frac{2}{1 - r} \Rightarrow r = 0.8

(b)

3 marks

The terms are $u_{n} = 2 (0.8)^{n - 1}$ . Solve $u_{n} < \frac{1}{2}$ .
So $n > 7.21$ , and the least whole value is

2 (0.8)^{n - 1} < 0.5 \Rightarrow (0.8)^{n - 1} < 0.25 \Rightarrow n - 1 > 6.21 n = 8

Question 7

8 marks

History exam scores: a cumulative frequency graph and grade boundaries

Statistics and probability

(a)

1 mark

The median is at the half-way cumulative frequency, $\frac{120}{2} = 60$ . Read across to the curve and down to the score axis.

median = 75

(b)

2 marks

The grades account for all 120 students.

6 + 13 + 26 + a + b = 120 \Rightarrow a = 75 - b

(c)

5 marks

(i) The mean grade is the weighted total of the grades divided by 120.

(ii) 28 students got grade 5, so the other $120 - 28 = 92$ scored lower. Read the graph at a cumulative frequency of 92.

\frac{6 ( 1 ) + 13 ( 2 ) + 26 ( 3 ) + 4 ( 75 - b ) + 5 b}{120} = 3.65 \Rightarrow \frac{410 + b}{120} = 3.65 410 + b = 438 \Rightarrow b = 28 minimum score for grade 5 = 84

Question 8

5 marks

A ski slope from the graph of its gradient

Calculus

(a)

2 marks

(i) $∣ h^{'} (x) ∣$ is largest at the lowest point of the graph of $h^{'}$ .

(ii) This is where the hill is steepest, where the skier descends fastest.

x = a

(b)

3 marks

Where $h^{'} < 0$ the hill goes down, and where $h^{'} > 0$ it goes up. The graph of $h^{'}$ is below the axis from 0 to $b$ , above from $b$ to $d$ , and below again from $d$ to $f$ .
So the sketch starts at height 60 when $x = 0$ and falls steeply (steepest at $a$ ) to a local minimum at $x = b$ , then rises to a local maximum at $x = d$ , and finally falls (steepest at $e$ ) to height 0 at $x = f$ .

Mentor tip

Inflexion points of $h$ come at $a$ , $c$ and $e$ , where $h^{'}$ has its extreme values.

Question 9

5 marks

Katie's commute: a two-state transition diagram and the long run

Number and algebra

(a)

2 marks

From "cycles", the chance of not cycling next day is 0.2, so the chance of cycling again is 0.8. From "does not cycle", the chance of not cycling again is 0.4, so the chance of cycling is 0.6.

Cycles \to Cycles 0.8, Cycles \to Does not cycle 0.2, Does not \to Cycles 0.6, Does not \to Does not 0.4

(b)

3 marks

The transition matrix (columns are "from") is $(0.8 0.2 0.6 0.4)$ . After many days the probabilities settle to the steady state, and 180 days is long enough.
So the probability that she cycles on the last working day is 0.75.

(0.8 0.2 0.6 0.4)^{180} \approx (0.75 0.25 0.75 0.25)

Mentor tip

Solving $0.2 c = 0.6 (1 - c)$ for the steady state gives the same value, $c = 0.75$ .

Question 10

4 marks

Transforming a graph: a stretch, a shift and a lift

Functions

(a)

2 marks

Substitute $x = 0$ , so $f (0 - 2) = f (- 2) = 6$ .

g (0) = 2 f (- 2) + 4 = 2 (6) + 4 = 16

(b)

2 marks

The graph of $y = g (x)$ is the graph of $f$ moved 2 to the right, stretched by a factor of 2 vertically, and then lifted 4 units. The horizontal asymptote $y = 0$ becomes $y = 4$ . The point $(- 2, 6)$ goes to $(0, 16)$ and the point $(0, 1)$ goes to $(2, 6)$ . The curve is decreasing and concave up, always above $y = 4$ , so there are no $x$ -intercepts, and it crosses the $y$ -axis at $(0, 16)$ .

Question 11

6 marks

The smallest fenced triangle: the ambiguous case

Number and algebra

6 marks

Let $A C = x$ . The 7 m fence BC is opposite the $4 0^{\circ}$ angle at A. By the cosine rule:
Both roots give a triangle.
The area is $\frac{1}{2} (10) (x) sin 4 0^{\circ}$ , which is smaller for the shorter $x$ .

7^{2} = 1 0^{2} + x^{2} - 2 (10) (x) cos 4 0^{\circ} \Rightarrow x^{2} - 15.32 x + 51 = 0 x \approx 4.888 or x \approx 10.43 minimum area = \frac{1}{2} (10) (4.888) sin 4 0^{\circ} \approx 15.7 m^{2}

Question 12

5 marks

Christmas trees in stock: a log–log regression and a prediction

Statistics and probability

(a)

2 marks

Regression of $ln x$ on $ln d$ using the second table.

m \approx - 0.695, b \approx 4.63

(b)

3 marks

For $d = 25$ , $ln d = ln 25$ .
Now undo the logarithm.

ln x = - 0.695 ln 25 + 4.63 \approx 2.393 x = e^{2.393} \approx 10.9%

Mentor tip

Using the unrounded regression values also gives about 10.9%.

Question 13

7 marks

A slope field: a value, the zero-slope curve and solution curves

Calculus

(a)

1 mark

Substitute $x = 0$ , $y = 1$ .

\frac{d y}{d x} = e^{0} - 1 = 0

(b)

2 marks

$\frac{d y}{d x} = 0$ where $y = e^{- x^{2}}$ . Sketch this bell-shaped curve: symmetric about the $y$ -axis, with a maximum at $(0, 1)$ and approaching $y = 0$ on each side.

(c)

4 marks

(i) The solution through $(0, 0)$ has gradient $e^{0} - 0 = 1$ there, so it rises from the origin. It climbs to a single maximum in the first quadrant, where it meets the bell curve from part (b), and then falls back towards the $x$ -axis.

(ii) The solution through $(0, 0.75)$ has gradient $1 - 0.75 = 0.25 > 0$ . Moving left it reaches a single minimum in the second quadrant, where it crosses the zero-slope curve. Moving right it rises to a single maximum in the first quadrant and then falls towards the $x$ -axis.

Question 14

7 marks

Sacks of potatoes: one sack, the quartile and a crate of ten

Statistics and probability

(a)

2 marks

With $X \sim N (49.8, 0. 9^{2})$ :

P (X < 50) \approx 0.588

(b)

2 marks

The lower quartile is the weight below which 25% of sacks fall.

P (X < l) = 0.25 \Rightarrow l \approx 49.2 kg

(c)

3 marks

The total of 10 independent sacks has mean $10 \times 49.8 = 498$ and variance $10 \times 0. 9^{2} = 8.1$ .

T \sim N (498, 8.1), P (T > 500) \approx 0.241

Question 15

9 marks

A wire frame: its radius, the volume and the angle that maximises it

Calculus

(a)

2 marks

The wire has two arcs of length $r θ$ , four edges of length $r$ and three connecting pieces of 1 cm each.

15 = 2 r θ + 4 r + 3 \Rightarrow 12 = 2 r (θ + 2) \Rightarrow r = \frac{6}{2 + θ}

(b)

7 marks

(i) The volume is the area of one sector times the 1 cm height.

(ii) Quotient rule.

(iii) Set the derivative to zero.

V = \frac{1}{2} r^{2} θ \times 1 = \frac{1}{2} \cdot \frac{36}{( 2 + θ ) ^{2}} \cdot θ = \frac{18 θ}{( 2 + θ ) ^{2}} \frac{d V}{d θ} = \frac{18 ( 2 + θ ) ^{2} - 18 θ \cdot 2 ( 2 + θ )}{( 2 + θ ) ^{4}} = \frac{18 ( 2 + θ ) - 36 θ}{( 2 + θ ) ^{3}} = \frac{36 - 18 θ}{( 2 + θ ) ^{3}} 36 - 18 θ = 0 \Rightarrow θ = 2

Question 16

9 marks

A ship and a lighthouse: closest approach and an alarm

Geometry and trigonometry

(a)

1 mark

The position is the start plus velocity times time.

O S = (300100) + t (- 12 15)

(b)

6 marks

The vector from the lighthouse L $(129, 283)$ to the ship is
It is perpendicular to the velocity when the dot product is zero.

L S = (171 - 12 t - 183 + 15 t) (171 - 12 t) (- 12) + (- 183 + 15 t) (15) = 0 \Rightarrow - 2052 + 144 t - 2745 + 225 t = 0 369 t = 4797 \Rightarrow t = 13

(c)

2 marks

At $t = 13$ the vector is $L S = (15, 12)$ .
This is less than 20 km, so the alarm will sound.

∣ L S ∣ = 1 5^{2} + 1 2^{2} = 369 \approx 19.2 km

Question 17

7 marks

The volume of a bowl as a solid of revolution about the $y$ -axis

Calculus

(a)

5 marks

From $y = 6 ln x$ we get $x = e^{y /6}$ , so $x^{2} = e^{y /3}$ .

V = π \int_{0}^{h} e^{y /3} d y = π [3 e^{y /3}]_{0}^{h} = 3 π (e^{h /3} - 1)

(b)

2 marks

The bowl is full when $h = 9$ .

V = 3 π (e^{3} - 1) \approx 180 cm^{3}

Our analysis

Where the marks are

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

Calculus33 marks · 30%
Q2, Q8, Q13, Q15, Q17
Number and algebra25 marks · 23%
Q5, Q6, Q9, Q11
Geometry and trigonometry21 marks · 19%
Q1, Q4, Q16
Statistics and probability20 marks · 18%
Q7, Q12, Q14
Functions11 marks · 10%
Q3, Q10

Mentor insights

Traps to watch for

01
Interpret the inverse · Q3
The inverse takes a height and gives the diameter. So $h^{- 1} (10)$ is the diameter of a tin that is 10 cm high.
02
Use the finance solver correctly · Q5
PV and the payment must have opposite signs, and the payment is at the end of each month. Round the payment to the cent before using it, as banks do.
03
Use the total of 120 · Q7
Read the median at 60 on the cumulative axis. For the minimum score for grade 5, count down from 120 by the number of grade 5 students and read the graph there.
04
Differentiate with the quotient rule · Q15
The volume $\frac{18 θ}{( 2 + θ ) ^{2}}$ is a quotient, so use the quotient rule and simplify before setting the derivative to zero.
05
Closest when perpendicular · Q16
The ship is closest to the lighthouse when the line from the lighthouse to the ship is perpendicular to the ship's velocity. Set the dot product to zero.

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

Session: November 2021
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 November 2021 Global 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

A kite on coordinate axes: gradients and a perpendicular diagonal

Geometry and trigonometry

(a)

2 marks

The gradient of AC.

m_{A C} = \frac{6 - 0}{4 - 2} = 3

(b)

1 mark

BD is perpendicular to AC.

m_{B D} = - \frac{1}{3}

(c)

2 marks

Use the point B $(0, 4)$ with gradient $- \frac{1}{3}$ .

y = - \frac{1}{3} x + 4 \Rightarrow x + 3 y - 12 = 0

(d)

1 mark

D is on the $x$ -axis, so $y = 0$ .

x + 0 - 12 = 0 \Rightarrow x = 12

Question 2

5 marks

Carbon dioxide emissions: the trapezoidal rule and percentage error

Calculus

(a)

3 marks

Use the trapezoidal rule with width 0.4: half the width times (first + last + twice the middle values).

\frac{1}{2} (0.4) [30 + 50 + 2 (50 + 60 + 40 + 20)] = 0.2 (80 + 340) = 84 tonnes

(b)

2 marks

Percentage error compared with the real value, 72 tonnes.

\frac{84 - 72}{72} \times 100% \approx 16.7%

Question 3

7 marks

A tin can's height as a function of diameter, and its inverse

Functions

(a)

3 marks

$h$ is decreasing on its domain, so the largest value is at $x = 4$ and the smallest at $x = 14$ .

h (4) = \frac{640}{16} + 0.5 = 40.5, h (14) = \frac{640}{196} + 0.5 \approx 3.77 3.77 \leq h (x) \leq 40.5

(b)

4 marks

(i) Solve $h (x) = 10$ .

(ii) A tin that is 10 cm high has a diameter of about 8.21 cm.

(iii) The range of the inverse is the domain of $h$ .

\frac{640}{x ^{2}} + 0.5 = 10 \Rightarrow x^{2} = \frac{640}{9.5} \Rightarrow x \approx 8.21 4 \leq h^{- 1} (x) \leq 14

Question 4

6 marks

Fire stations and a Voronoi diagram: where to put a new station

Geometry and trigonometry

(a)

3 marks

The new station should be as far as possible from the nearest existing one, which happens at a vertex of the Voronoi diagram, P or Q. Compare the distance from each to its nearest stations.
P is farther from its nearest stations than Q is, so the new station should be at P.

AP = (10 - 6)^{2} + (6 - 2)^{2} = 32 \approx 5.66, BQ = (13 - 14)^{2} + (7 - 2)^{2} = 26 \approx 5.10

(b)

3 marks

(i) P is $(10, 6)$ and C is $(18, 6)$ . They have the same $y$ , so the bisector is vertical through the midpoint $(14, 6)$ .

x = 14 y - 8.8 = - \frac{1}{7} (x - 10.4)

Question 5

9 marks

A 30-year home loan: the payment, the balance and the interest paid

Number and algebra

(a)

3 marks

Use the finance solver with $N = 360$ , $I = 3.8%$ , $PV = 170000$ , $FV = 0$ , 12 payments and 12 compounding periods per year.

PMT \approx 792.13 AUD per month

(b)

6 marks

(i) After 10 years ( $N = 120$ ) the balance is the future value with the payment of $- 792.13$ .

(ii) They paid $120 \times 792.13 = 95 055.60$ . Of this, $170000 - 133019.94 = 36 980.06$ went to repaying the loan. The rest is interest.

balance \approx 133 019.94 AUD 95 055.60 - 36 980.06 = 58 075.54 AUD

Mentor tip

Using the unrounded payment gives 58 075.60. Both are fine to the cent, as long as you say which one you used.

Question 6

5 marks

An infinite geometric series: its ratio and when the terms become small

Number and algebra

(a)

2 marks

For an infinite geometric series $S_{\infty} = \frac{u _{1}}{1 - r}$ .

10 = \frac{2}{1 - r} \Rightarrow r = 0.8

(b)

3 marks

The terms are $u_{n} = 2 (0.8)^{n - 1}$ . Solve $u_{n} < \frac{1}{2}$ .
So $n > 7.21$ , and the least whole value is

2 (0.8)^{n - 1} < 0.5 \Rightarrow (0.8)^{n - 1} < 0.25 \Rightarrow n - 1 > 6.21 n = 8

Question 7

8 marks

History exam scores: a cumulative frequency graph and grade boundaries

Statistics and probability

(a)

1 mark

The median is at the half-way cumulative frequency, $\frac{120}{2} = 60$ . Read across to the curve and down to the score axis.

median = 75

(b)

2 marks

The grades account for all 120 students.

6 + 13 + 26 + a + b = 120 \Rightarrow a = 75 - b

(c)

5 marks

(i) The mean grade is the weighted total of the grades divided by 120.

(ii) 28 students got grade 5, so the other $120 - 28 = 92$ scored lower. Read the graph at a cumulative frequency of 92.

\frac{6 ( 1 ) + 13 ( 2 ) + 26 ( 3 ) + 4 ( 75 - b ) + 5 b}{120} = 3.65 \Rightarrow \frac{410 + b}{120} = 3.65 410 + b = 438 \Rightarrow b = 28 minimum score for grade 5 = 84

Question 8

5 marks

A ski slope from the graph of its gradient

Calculus

(a)

2 marks

(i) $∣ h^{'} (x) ∣$ is largest at the lowest point of the graph of $h^{'}$ .

(ii) This is where the hill is steepest, where the skier descends fastest.

x = a

(b)

3 marks

Where $h^{'} < 0$ the hill goes down, and where $h^{'} > 0$ it goes up. The graph of $h^{'}$ is below the axis from 0 to $b$ , above from $b$ to $d$ , and below again from $d$ to $f$ .
So the sketch starts at height 60 when $x = 0$ and falls steeply (steepest at $a$ ) to a local minimum at $x = b$ , then rises to a local maximum at $x = d$ , and finally falls (steepest at $e$ ) to height 0 at $x = f$ .

Mentor tip

Inflexion points of $h$ come at $a$ , $c$ and $e$ , where $h^{'}$ has its extreme values.

Question 9

5 marks

Katie's commute: a two-state transition diagram and the long run

Number and algebra

(a)

2 marks

From "cycles", the chance of not cycling next day is 0.2, so the chance of cycling again is 0.8. From "does not cycle", the chance of not cycling again is 0.4, so the chance of cycling is 0.6.

Cycles \to Cycles 0.8, Cycles \to Does not cycle 0.2, Does not \to Cycles 0.6, Does not \to Does not 0.4

(b)

3 marks

The transition matrix (columns are "from") is $(0.8 0.2 0.6 0.4)$ . After many days the probabilities settle to the steady state, and 180 days is long enough.
So the probability that she cycles on the last working day is 0.75.

(0.8 0.2 0.6 0.4)^{180} \approx (0.75 0.25 0.75 0.25)

Mentor tip

Solving $0.2 c = 0.6 (1 - c)$ for the steady state gives the same value, $c = 0.75$ .

Question 10

4 marks

Transforming a graph: a stretch, a shift and a lift

Functions

(a)

2 marks

Substitute $x = 0$ , so $f (0 - 2) = f (- 2) = 6$ .

g (0) = 2 f (- 2) + 4 = 2 (6) + 4 = 16

(b)

2 marks

Question 11

6 marks

The smallest fenced triangle: the ambiguous case

Number and algebra

6 marks

Let $A C = x$ . The 7 m fence BC is opposite the $4 0^{\circ}$ angle at A. By the cosine rule:
Both roots give a triangle.
The area is $\frac{1}{2} (10) (x) sin 4 0^{\circ}$ , which is smaller for the shorter $x$ .

7^{2} = 1 0^{2} + x^{2} - 2 (10) (x) cos 4 0^{\circ} \Rightarrow x^{2} - 15.32 x + 51 = 0 x \approx 4.888 or x \approx 10.43 minimum area = \frac{1}{2} (10) (4.888) sin 4 0^{\circ} \approx 15.7 m^{2}

Question 12

5 marks

Christmas trees in stock: a log–log regression and a prediction

Statistics and probability

(a)

2 marks

Regression of $ln x$ on $ln d$ using the second table.

m \approx - 0.695, b \approx 4.63

(b)

3 marks

For $d = 25$ , $ln d = ln 25$ .
Now undo the logarithm.

ln x = - 0.695 ln 25 + 4.63 \approx 2.393 x = e^{2.393} \approx 10.9%

Mentor tip

Using the unrounded regression values also gives about 10.9%.

Question 13

7 marks

A slope field: a value, the zero-slope curve and solution curves

Calculus

(a)

1 mark

Substitute $x = 0$ , $y = 1$ .

\frac{d y}{d x} = e^{0} - 1 = 0

(b)

2 marks

$\frac{d y}{d x} = 0$ where $y = e^{- x^{2}}$ . Sketch this bell-shaped curve: symmetric about the $y$ -axis, with a maximum at $(0, 1)$ and approaching $y = 0$ on each side.

(c)

4 marks

Question 14

7 marks

Sacks of potatoes: one sack, the quartile and a crate of ten

Statistics and probability

(a)

2 marks

With $X \sim N (49.8, 0. 9^{2})$ :

P (X < 50) \approx 0.588

(b)

2 marks

The lower quartile is the weight below which 25% of sacks fall.

P (X < l) = 0.25 \Rightarrow l \approx 49.2 kg

(c)

3 marks

The total of 10 independent sacks has mean $10 \times 49.8 = 498$ and variance $10 \times 0. 9^{2} = 8.1$ .

T \sim N (498, 8.1), P (T > 500) \approx 0.241

Question 15

9 marks

A wire frame: its radius, the volume and the angle that maximises it

Calculus

(a)

2 marks

The wire has two arcs of length $r θ$ , four edges of length $r$ and three connecting pieces of 1 cm each.

15 = 2 r θ + 4 r + 3 \Rightarrow 12 = 2 r (θ + 2) \Rightarrow r = \frac{6}{2 + θ}

(b)

7 marks

(i) The volume is the area of one sector times the 1 cm height.

(ii) Quotient rule.

(iii) Set the derivative to zero.

V = \frac{1}{2} r^{2} θ \times 1 = \frac{1}{2} \cdot \frac{36}{( 2 + θ ) ^{2}} \cdot θ = \frac{18 θ}{( 2 + θ ) ^{2}} \frac{d V}{d θ} = \frac{18 ( 2 + θ ) ^{2} - 18 θ \cdot 2 ( 2 + θ )}{( 2 + θ ) ^{4}} = \frac{18 ( 2 + θ ) - 36 θ}{( 2 + θ ) ^{3}} = \frac{36 - 18 θ}{( 2 + θ ) ^{3}} 36 - 18 θ = 0 \Rightarrow θ = 2

Question 16

9 marks

A ship and a lighthouse: closest approach and an alarm

Geometry and trigonometry

(a)

1 mark

The position is the start plus velocity times time.

O S = (300100) + t (- 12 15)

(b)

6 marks

The vector from the lighthouse L $(129, 283)$ to the ship is
It is perpendicular to the velocity when the dot product is zero.

L S = (171 - 12 t - 183 + 15 t) (171 - 12 t) (- 12) + (- 183 + 15 t) (15) = 0 \Rightarrow - 2052 + 144 t - 2745 + 225 t = 0 369 t = 4797 \Rightarrow t = 13

(c)

2 marks

At $t = 13$ the vector is $L S = (15, 12)$ .
This is less than 20 km, so the alarm will sound.

∣ L S ∣ = 1 5^{2} + 1 2^{2} = 369 \approx 19.2 km

Question 17

7 marks

The volume of a bowl as a solid of revolution about the $y$ -axis

Calculus

(a)

5 marks

From $y = 6 ln x$ we get $x = e^{y /6}$ , so $x^{2} = e^{y /3}$ .

V = π \int_{0}^{h} e^{y /3} d y = π [3 e^{y /3}]_{0}^{h} = 3 π (e^{h /3} - 1)

(b)

2 marks

The bowl is full when $h = 9$ .

V = 3 π (e^{3} - 1) \approx 180 cm^{3}

Our analysis

Where the marks are

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

Calculus33 marks · 30%
Q2, Q8, Q13, Q15, Q17
Number and algebra25 marks · 23%
Q5, Q6, Q9, Q11
Geometry and trigonometry21 marks · 19%
Q1, Q4, Q16
Statistics and probability20 marks · 18%
Q7, Q12, Q14
Functions11 marks · 10%
Q3, Q10

Mentor insights

Traps to watch for

01
Interpret the inverse · Q3
The inverse takes a height and gives the diameter. So $h^{- 1} (10)$ is the diameter of a tin that is 10 cm high.
02
Use the finance solver correctly · Q5
PV and the payment must have opposite signs, and the payment is at the end of each month. Round the payment to the cent before using it, as banks do.
03
Use the total of 120 · Q7
Read the median at 60 on the cumulative axis. For the minimum score for grade 5, count down from 120 by the number of grade 5 students and read the graph there.
04
Differentiate with the quotient rule · Q15
The volume $\frac{18 θ}{( 2 + θ ) ^{2}}$ is a quotient, so use the quotient rule and simplify before setting the derivative to zero.
05
Closest when perpendicular · Q16
The ship is closest to the lighthouse when the line from the lighthouse to the ship is perpendicular to the ship's velocity. Set the dot product to zero.

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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