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Applications of Functions

Practice exam-style IB Math AI questions for Applications of Functions, aligned with the syllabus and grouped by topic.

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Question 1
SL • Paper 1
Easy
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SL • Paper 1
Easy
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A taxi company charges a fixed fee of €4.804.80 and €1.651.65 for each kilometre travelled. The total cost, CC euros, of a journey of dd kilometres is modelled by

C(d)=4.80+1.65dC(d)=4.80+1.65d

where 0≤d≤200\leq d\leq 20.

A

the cost of a journey of 1212 kilometres.

[2]
B

the greatest distance that can be travelled for €3535.

[2]
C

the value 1.651.65 in the context of the model.

[1]
Question 2
SL • Paper 1
Easy
Calculator Permitted
SL • Paper 1
Easy
Calculator Permitted

The intensity, II, of light from a lamp varies inversely with the square of the distance, dd metres, from the lamp. The model has the form

I=kd−2I=kd^{-2}

At a distance of 33 metres, the intensity is 8080 units.

A

the value of kk.

[2]
B

the intensity at a distance of 7.57.5 metres.

[2]
C

the effect on the intensity when the distance from the lamp is doubled.

[1]
Question 3
SL • Paper 1
Easy
Calculator Permitted
SL • Paper 1
Easy
Calculator Permitted

The depth, DD metres, of water in a harbour is modelled by

D(t)=2.7sin⁡(30t∘)+5.1D(t)=2.7\sin(30t^\circ)+5.1

where tt is the number of hours after midnight and 0≤t≤240\leq t\leq24.

A

the amplitude, period and equation of the principal axis of the model.

[3]
B

the maximum depth of the water.

[1]
C

the first time after midnight at which the maximum depth occurs.

[1]
Question 4
SL • Paper 1
Medium
Calculator Permitted
SL • Paper 1
Medium
Calculator Permitted

A ball is thrown upwards from a platform. Its height, hh metres, after tt seconds is modelled by

h(t)=−0.8t2+4.8t+1.2h(t)=-0.8t^2+4.8t+1.2

where t≥0t\geq0.

A

the maximum height reached by the ball and the time at which it is reached.

[3]
B

the time at which the ball first reaches the ground.

[2]

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Question 5
SL • Paper 1
Medium
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SL • Paper 1
Medium
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The temperature, TT in degrees Celsius, of a drink tt minutes after it is placed in a room is modelled by

T(t)=18+72e−0.18tT(t)=18+72e^{-0.18t}

where t≥0t\geq0.

A

the horizontal asymptote of the graph of TT.

[1]
B

the temperature of the drink after 2020 minutes.

[2]
C

the time at which the temperature first reaches 25∘C25^\circ\text{C}.

[2]
Question 6
SL • Paper 1
Medium
Calculator Permitted
SL • Paper 1
Medium
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The number of subscribers, SS, to an online service was recorded at times t=0,1,2,3t=0,1,2,3 years.

Time, tt [years]

Subscribers, SS [subscribers]

0

240

1

300

2

375

3

469

A

why an exponential model is more appropriate than a linear model for these data.

[1]
B

an exponential model of the form S(t)=katS(t)=ka^t for these data.

[2]
C

Use the model to the number of subscribers at the end of year 77.

[1]
D

one reason why this prediction may be unreliable.

[1]
Question 7
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
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The concentration, AA milligrams per litre, of a chemical in a tank is modelled by

A(t)=15+85e−0.14tA(t)=15+85e^{-0.14t}

where tt is measured in hours.

A

the initial concentration and the long-term concentration predicted by the model.

[2]
B

the half-life of the concentration above its long-term value.

[2]
C

the concentration after two half-lives.

[1]
Question 8
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

The vertical displacement, HH metres, of a floating platform is modelled by

H(t)=1.8sin⁡(π6(t−2))+4.5H(t)=1.8\sin\left(\frac{\pi}{6}(t-2)\right)+4.5

where tt is measured in seconds.

A

the amplitude, period and phase shift of the model.

[3]
B

the maximum height of the platform and the first positive time at which this maximum occurs.

[2]

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Question 9
SL • Paper 2
Medium
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SL • Paper 2
Medium
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The power, PP kilowatts, generated by a small wind turbine at wind speed vv metres per second is modelled by

P(v)=0.12v3−1.8v2+8vP(v)=0.12v^3-1.8v^2+8v

where 0≤v≤150\leq v\leq15.

A

the power generated when v=10v=10.

[2]
B

the wind speed at which the turbine generates 1818 kilowatts.

[2]
C

why any solution outside the interval 0≤v≤150\leq v\leq15 should be rejected.

[1]
Question 10
SL • Paper 1
Medium
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SL • Paper 1
Medium
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The mass, MM grams, of leaves on a plant tt days after an experiment begins is modelled by

M(t)=at2+bt+cM(t)=at^2+bt+c

The measured masses are 1212 grams when t=0t=0, 2828 grams when t=4t=4, and 2222 grams when t=10t=10.

A

the values of aa, bb and cc.

[3]
B

the maximum mass predicted by the model and the day on which it occurs.

[2]
C

The model predicts M(16)=−20M(16)=-20. why the model should not be used to the mass on day 1616.

[1]
Question 11
HL • Paper 1
Medium
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HL • Paper 1
Medium
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The sound level, SS decibels, produced by a machine operating at power xx kilowatts is modelled by

S(x)=a+bln⁡xS(x)=a+b\ln x

where x>0x>0. The sound level is 1818 decibels when x=2x=2 and 4242 decibels when x=10x=10.

A

the values of aa and bb.

[3]
B

the power at which the predicted sound level is 5050 decibels.

[2]
C

the domain restriction required by this model.

[1]
Question 12
HL • Paper 1
Medium
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HL • Paper 1
Medium
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The population, NN, of insects in a greenhouse is modelled by

N(t)=12001+Ce−0.35tN(t)=\frac{1200}{1+Ce^{-0.35t}}

where tt is the number of weeks after observations begin. Initially there are 150150 insects.

A

the value of CC.

[2]
B

the carrying capacity of the greenhouse.

[1]
C

the time at which the population first reaches 900900 insects.

[2]
D

why a logistic model may be more appropriate than an exponential model for this population.

[1]

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Question 13
HL • Paper 1
Medium
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HL • Paper 1
Medium
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The cost, C(x)C(x) euros, of delivering a package a distance of xx kilometres is modelled by

C(x)={p+1.8x,0≤x<10,q+1.2x,x≥10.C(x)= \begin{cases} p+1.8x, & 0\leq x<10,\\ q+1.2x, & x\geq10. \end{cases}

The initial delivery charge is €55, and the model is continuous at x=10x=10.

A

the value of pp.

[1]
B

the value of qq.

[2]
C

the cost of a delivery over a distance of 1616 kilometres.

[2]
Question 14
HL • Paper 1
Medium
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HL • Paper 1
Medium
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The number, yy thousands, of daily visits to a website was recorded over time. A best-fit straight line for a of ln⁡y\ln y against time tt days is

ln⁡y=1.47+0.082t\ln y=1.47+0.082t

A

an exponential model of the form y=kerty=ke^{rt}.

[2]
B

Use the model to the number of daily visits when t=20t=20.

[1]
C

the doubling time predicted by the model.

[2]
Question 15
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
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Data for positive variables xx and yy were analysed using three transformations. The following correlation coefficients were obtained:

  • a of yy against xx: r=0.941r=0.941;
  • a of ln⁡y\ln y against xx: r=0.998r=0.998;
  • a of ln⁡y\ln y against ln⁡x\ln x: r=0.972r=0.972.

For the second , the best-fit line is

ln⁡y=2.20+0.315x\ln y=2.20+0.315x

The observed values of xx were between 11 and 66.

A

the most appropriate of the linear, exponential and power models. your answer.

[2]
B

Use the best-fit line to yy when x=8x=8.

[2]
C

why the prediction in part (b) should be treated with caution.

[1]
Question 16
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
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A force FF newtons and speed vv metres per second are believed to satisfy a power model

F=KvpF=Kv^p

A best-fit line for a of log⁡10F\log_{10}F against log⁡10v\log_{10}v is

log⁡10F=−0.602+1.75log⁡10v\log_{10}F=-0.602+1.75\log_{10}v

A

the values of KK and pp, and write the power model.

[2]
B

the factor by which the force changes when the speed is multiplied by 33.

[2]
C

the speed at which the model predicts a force of 2020 newtons.

[2]

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Question 17
SL • Paper 2
Medium
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SL • Paper 2
Medium
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An electric bicycle company charges according to the duration, tt minutes, for which a bicycle is hired. The cost, C(t)C(t) dollars, is modelled by

C(t)={6+0.18t,0≤t≤30,p+0.10t,30<t≤120.C(t)= \begin{cases} 6+0.18t, & 0\leq t\leq 30,\\ p+0.10t, & 30<t\leq120. \end{cases}

The cost model has no jump at t=30t=30. A competing company charges D(t)=9+0.07tD(t)=9+0.07t dollars for 0≤t≤1200\leq t\leq120.

A
I.

the cost of hiring a bicycle for 2525 minutes.

[1]
II.

the value of pp.

[2]
B

the duration of a hire costing $15 under the model CC.

[2]
C

the durations for which hiring from the electric bicycle company is cheaper than hiring from the competing company.

[3]
Question 18
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

The value, V(t)V(t) dollars, of a specialized camera tt years after purchase is modelled by

V(t)=ke−rt+120V(t)=ke^{-rt}+120

where k>0k>0, r>0r>0 and t≥0t\geq0. The camera initially costs $1320 and has a value of $780 after three years.

A
I.

the value of kk.

[1]
II.

the value of rr.

[3]
B

the predicted value of the camera after eight years.

[2]
C

A linear model through the two given values predicts that the camera will be worth a negative amount after eight years. why the exponential model is more reasonable for long-term use.

[3]
Question 19
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

On a particular wet road, the stopping distance, ss metres, of a car varies directly with the square of its speed, vv kilometres per hour. The model is

s=kv2s=kv^2

At 50 km h−150\ \text{km h}^{-1}, the stopping distance is 2828 m.

A
I.

the value of kk.

[2]
II.

the factor by which the stopping distance changes when the speed is doubled.

[1]
B

the stopping distance predicted at 80 km h−180\ \text{km h}^{-1}.

[2]
C

A driver needs the stopping distance to be no more than 4545 m. the greatest speed predicted by the model and one limitation of using this result.

[3]
Question 20
SL • Paper 2
Hard
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SL • Paper 2
Hard
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The cross-section of a decorative entrance is modelled by the quadratic function

h(x)=ax2+bx+ch(x)=ax^2+bx+c

where hh is the height in metres and xx is the horizontal distance in metres from the left-hand end. The curve passes through (0,1.2)(0,1.2), (4,5.6)(4,5.6) and (10,1.2)(10,1.2).

A
I.

the value of cc.

[1]
II.

the values of aa and bb.

[3]
B

the maximum height of the entrance and the horizontal position at which it occurs.

[3]
C

A rectangular vehicle of height 2.002.00 m is to pass centrally through the entrance. the greatest possible width of the vehicle.

[3]

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Question 21
SL • Paper 2
Hard
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SL • Paper 2
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An open rectangular box is made from a 3030 cm by 2020 cm sheet of card by cutting squares of side xx cm from each corner and folding up the sides. The volume is modelled by

V(x)=x(30−2x)(20−2x)V(x)=x(30-2x)(20-2x)

where 0<x<100<x<10.

A
I.

V(x)=4x3−100x2+600xV(x)=4x^3-100x^2+600x.

[2]
II.

the volume when x=3x=3.

[1]
B

Use your GDC to the value of xx that gives the maximum volume, and this maximum volume.

[3]
C

A volume of exactly 900 cm3900\ \text{cm}^3 is required. all possible values of xx and why there are two possible box designs.

[4]
Question 22
SL • Paper 2
Hard
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SL • Paper 2
Hard
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The temperature, T(t)T(t) in degrees Celsius, inside a greenhouse is modelled over a repeating daily cycle. The variable tt is the number of hours after 15:00. The model is

T(t)=acos⁡(bt∘)+dT(t)=a\cos(bt^\circ)+d

The maximum temperature is 32∘C32^\circ\text{C} at t=0t=0, the minimum temperature is 14∘C14^\circ\text{C}, and the period is 2424 hours. Assume that b>0b>0.

A
I.

the values of aa and dd.

[3]
II.

the value of bb.

[2]
B

the temperature at 21:00.

[1]
C

During each cycle, ventilation is required whenever the temperature is above 28∘C28^\circ\text{C}. the total time for which ventilation is required.

[3]
Question 23
SL • Paper 2
Hard
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SL • Paper 2
Hard
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The yield, YY tonnes per hectare, of a crop is modelled by

Y(x)=ax2+bx+cY(x)=ax^2+bx+c

where xx is the mass of fertilizer applied in hundreds of kilograms per hectare. Measurements give Y(0)=2.4Y(0)=2.4, Y(4)=5.8Y(4)=5.8 and Y(10)=4.0Y(10)=4.0.

A
I.

the value of cc.

[1]
II.

aa and bb.

[3]
B

the fertilizer amount that maximizes the predicted yield, and the maximum yield.

[3]
C

The measured fertilizer amounts were between 00 and 1010. The model predicts a negative yield when x=14x=14. why this does not invalidate the model over the measured interval.

[3]
Question 24
SL • Paper 2
Hard
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SL • Paper 2
Hard
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The surface area, AA square metres, covered by a water plant was recorded every two days from day 00 to day 66. The results are shown in the table.

tt / days

AA / m2\text{m}^2

0

40

2

56

4

78.4

6

109.76

A
I.

why an exponential model is appropriate for these data.

[2]
II.

an exponential model for AA in terms of tt, the time in days.

[2]
B

when the model first predicts that the plant will cover 500 m2500\ \text{m}^2.

[2]
C
I.

A survey on day 1212 found an actual area of 290 m2290\ \text{m}^2. the model's prediction and on its accuracy.

[2]
II.

one reason why predictions far beyond day 1212 should be treated with caution.

[1]

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Question 25
HL • Paper 2
Hard
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HL • Paper 2
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The count rate, R(t)R(t) counts per minute, from a sample is modelled by

R(t)=12+Ae−ktR(t)=12+Ae^{-kt}

where tt is measured in hours. The constant value 1212 represents background radiation. Initially the count rate is 212212, and after six hours it is 8282.

A
I.

AA.

[1]
II.

kk.

[3]
B

the half-life of the count rate above background radiation.

[2]
C

when the total count rate first reaches 2020 counts per minute, and why the total count rate itself does not halve every half-life.

[3]
Question 26
HL • Paper 2
Hard
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HL • Paper 2
Hard
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The number of bird species, SS, observed in a protected region of area AA square kilometres is modelled by

S(A)=a+bln⁡A,A>0S(A)=a+b\ln A,\qquad A>0

A region of area 5 km25\ \text{km}^2 contains 4242 species, while a region of area 80 km280\ \text{km}^2 contains 7070 species.

A
I.

Form two equations involving aa and bb.

[1]
II.

aa and bb.

[3]
B

the increase in the predicted number of species whenever the area is doubled.

[2]
C

the area for which the model predicts 8585 species, and on the reliability of this prediction.

[3]
Question 27
HL • Paper 2
Hard
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HL • Paper 2
Hard
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The charge, Q(t)Q(t) percent, of a battery during a rapid-charging process is modelled by

Q(t)={6t,0≤t<8,80−Ce−0.25(t−8),t≥8,Q(t)= \begin{cases} 6t, & 0\leq t<8,\\ 80-Ce^{-0.25(t-8)}, & t\geq8, \end{cases}

where tt is measured in minutes. The model is continuous at t=8t=8.

A
I.

the charge predicted immediately before t=8t=8.

[1]
II.

CC.

[2]
B

when the battery first reaches 70%70\% charge.

[2]
C

the charge predicted after 1414 minutes. why extending the first linear rule to t=14t=14 would be inappropriate.

[3]
Question 28
HL • Paper 3
Hard
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HL • Paper 3
Hard
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A streaming company models the cumulative number of subscribers, N(t)N(t), by
N(t)=800001+Ce−ktN(t)=\frac{80000}{1+Ce^{-kt}}
where tt is the number of months after launch and C,k>0C,k>0. At launch there are 40004000 subscribers. After six months there are 3000030000 subscribers.

A
I.

the value of CC.

[1]
II.

Using the observation at six months, kk.

[3]
B
I.

the number of subscribers predicted after 1212 months.

[2]
II.

The subscriber population grows fastest when it reaches half its limiting value. when this occurs. If you did not obtain a value for kk, use k=0.406k=0.406.

[2]
C

For a logistic model P(t)=L1+Ce−ktP(t)=\dfrac{L}{1+Ce^{-kt}}, derive an expression for the time at which P(t)=pLP(t)=pL, where 0<p<10<p<1. the effect on this time as pp approaches 11.

[4]

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Question 29
HL • Paper 3
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HL • Paper 3
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The mass, MM grams, of a biodegradable coating remaining on an object was measured at different times. A linear regression of ln⁡M\ln M on time tt, measured in days, gives
ln⁡M=4.35−0.165t\ln M=4.35-0.165t
The recorded times lie between 00 and 1515 days.

A
I.

an exponential model of the form M(t)=AertM(t)=Ae^{rt}.

[2]
II.

the values of AA and rr in context.

[2]
B

the half-life of the coating.

[2]
C
I.

The measured mass on day 1818 is 4.404.40 grams. the percentage by which this measurement exceeds the model prediction.

[2]
II.

two reasons why the day 1818 prediction should be treated cautiously.

[2]
Question 30
HL • Paper 3
Hard
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HL • Paper 3
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Engineers how the heat-loss rate, HH kilowatts, from a building depends on wind speed, vv metres per second. The data were collected for 2≤v≤92\leq v\leq9. A regression of ln⁡H\ln H on ln⁡v\ln v gives
ln⁡H=−1.10+1.60ln⁡v\ln H=-1.10+1.60\ln v

A
I.

why the regression supports a power model H=KvpH=Kv^p.

[1]
II.

KK and pp.

[3]
B

the factor by which the model predicts HH will change when the wind speed is doubled.

[2]
C
I.

the wind speed at which the predicted heat-loss rate is 2020 kilowatts.

[2]
II.

Evaluate the reliability of this result.

[2]
Question 31
HL • Paper 3
Hard
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HL • Paper 3
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A factory models the daily cost, C(x)C(x) dollars, of producing xx units by
C(x)={120+8x,0≤x<20,ax+b,20≤x<50,460+4(x−50),x≥50.C(x)=\begin{cases}120+8x,&0\leq x<20,\\ax+b,&20\leq x<50,\\460+4(x-50),&x\geq50.\end{cases}
The cost model is continuous at both production thresholds.

Production interval

Daily cost C(x)C(x) / dollars

0≤x<200\leq x<20

120+8x120+8x

20≤x<5020\leq x<50

ax+bax+b

x≥50x\geq50

460+4(x−50)460+4(x-50)

A
I.

Use continuity at x=20x=20 and x=50x=50 to form two equations in aa and bb.

[2]
II.

aa and bb.

[3]
B

the cost of producing 3838 units and the value of aa in context.

[2]
C
I.

the smallest integer production level for which the daily cost is at least 500500 dollars.

[2]
II.

one advantage of this piecewise model over a single linear cost model.

[1]
Question 32
HL • Paper 3
Hard
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HL • Paper 3
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The yield, YY tonnes per hectare, of a crop is modelled in terms of fertilizer application, xx kilograms per hectare, by
Y(x)=a+bln⁡x,x>0Y(x)=a+b\ln x,\qquad x>0
Field trials give Y(20)=4.5Y(20)=4.5 and Y(80)=7.3Y(80)=7.3.

A
I.

b=2.8ln⁡4b=\dfrac{2.8}{\ln4}.

[2]
II.

aa and bb, giving each value correct to three significant figures.

[2]
B

the fertilizer application predicted to produce a yield of 88 tonnes per hectare.

[2]
C
I.

the increase in predicted yield when the fertilizer application is multiplied by 1.51.5.

[2]
II.

why the model exhibits diminishing increases in yield for equal additions of fertilizer, and one contextual limitation.

[2]

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Question 33
HL • Paper 3
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HL • Paper 3
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Radiation detected near a sealed storage unit is modelled by
R(t)=B+Ae−ktR(t)=B+Ae^{-kt}
where RR is measured in counts per minute and tt is measured in hours. The background radiation level is 1212 counts per minute. Initially, R=92R=92, and after 55 hours, R=42R=42.

A
I.

AA and BB.

[2]
II.

kk.

[3]
B

the half-life of the radiation level above background.

[2]
C
I.

the time at which the detected radiation reaches 2020 counts per minute, and when it is below this level.

[2]
II.

RR after two half-lives and why the total radiation level has not been divided by four.

[2]
Question 34
HL • Paper 3
Hard
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HL • Paper 3
Hard
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The tensile strength, SS megapascals, of a polymer is measured after nn thousand loading cycles. A regression of ln⁡S\ln S on nn gives
ln⁡S=6.12−0.184n\ln S=6.12-0.184n
The data were collected for 0≤n≤50\leq n\leq5.

A
AI.

an exponential model for SS in terms of nn.

[2]
AII.

the intercept of the transformed regression line.

[2]
B

the half-life of the tensile strength according to the model, giving your answer in loading cycles.

[2]
C
CI.

A component must be replaced when its predicted strength first falls below 150150 MPa. the corresponding number of loading cycles.

[2]
CII.

why this replacement prediction may not be reliable.

[2]
Question 35
HL • Paper 3
Hard
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HL • Paper 3
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Hydrologists model river discharge, QQ cubic metres per second, in terms of mean river depth, dd metres. For measurements with 1.5≤d≤81.5\leq d\leq8, a log-log regression gives
ln⁡Q=−2.30+2.40ln⁡d\ln Q=-2.30+2.40\ln d

A
I.

a power model Q=KdpQ=Kd^p.

[3]
II.

the exponent pp in terms of a percentage change.

[1]
B

the factor by which discharge changes when depth increases by 20%20\%.

[2]
C
I.

the depth at which the model predicts a discharge of 5050 cubic metres per second.

[2]
II.

Evaluate whether this depth should be used for flood planning.

[2]
Question 36
HL • Paper 3
Hard
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HL • Paper 3
Hard
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The number of daylight hours, D(t)D(t), at a location is modelled sinusoidally, where tt is the day number of a 365365-day year. The maximum daylight is 15.415.4 hours on day 172172, and the minimum daylight is 9.09.0 hours.

A
I.

the amplitude and principal-axis value.

[2]
II.

a suitable cosine model for D(t)D(t).

[3]
B

the predicted number of daylight hours on day 300300.

[2]
C
I.

the range of day numbers during which the model predicts more than 1414 hours of daylight.

[3]
II.

one reason why actual daylight measurements may differ from this model.

[1]

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Question 37
HL • Paper 2
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HL • Paper 2
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The vertical position, H(t)H(t) metres, of a moving inspection platform is modelled by

H(t)=asin⁡(b(t−c))+dH(t)=a\sin(b(t-c))+d

where tt is measured in seconds and angles are measured in radians. The maximum height is 7.27.2 m, the minimum height is 2.82.8 m, and the platform crosses its principal axis moving upwards at t=1.5t=1.5 and at the next such crossing, t=9.5t=9.5. Take b>0b>0 and choose 0≤c<8 s0\le c<8\ \text{s}.

A
I.

aa and dd, taking a>0a>0.

[3]
II.

bb and cc.

[3]
B

During the cycle from t=1.5t=1.5 to t=9.5t=9.5, the two times at which the platform is at height 6.56.5 m.

[2]
C

the length of time during this cycle for which the platform is above 6.56.5 m.

[2]
Question 38
HL • Paper 2
Hard
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HL • Paper 2
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The number of users, N(t)N(t), of a new regional transport card is modelled by

N(t)=500001+Ce−ktN(t)=\frac{50000}{1+Ce^{-kt}}

where tt is the number of months after its introduction and C,k>0C,k>0. Initially there are 20002000 users, and after four months there are 1000010000 users.

A
I.

CC.

[2]
II.

kk.

[3]
B

when the number of users first reaches 80%80\% of the carrying capacity.

[2]
C
I.

the time at which the model's growth is fastest and this point.

[2]
II.

An exponential model fitted only to the first two observations is E(t)=2000e(ln⁡54)t=2000⋅5t/4E(t)=2000e^{(\frac{\ln 5}{4})t}=2000\cdot 5^{t/4}. why the logistic model is more appropriate for long-term prediction.

[2]
Question 39
HL • Paper 2
Hard
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HL • Paper 2
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The mass, MM grams, of a biodegradable material was measured over time. A best-fit straight line for a of ln⁡M\ln M against time tt days is

ln⁡M=2.91−0.128t\ln M=2.91-0.128t

The correlation coefficient for the transformed data is ρ=−0.997\rho=-0.997.

A
I.

an exponential model for MM in the form M=kertM=ke^{rt}.

[2]
II.

the value −0.128-0.128 in the transformed model.

[2]
B

the predicted mass after 1010 days. An observed mass at this time was 5.605.60 g; the residual, defined as observed minus predicted.

[3]
C

when the predicted mass first reaches 22 g, and what the correlation coefficient ρ\rho suggests about the exponential model.

[2]
Question 40
HL • Paper 2
Hard
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HL • Paper 2
Hard
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The energy use, EE kilojoules per day, of an animal is believed to be related to its mass, mm kilograms, by a power model

E=KmpE=Km^p

A best-fit straight line for a of ln⁡E\ln E against ln⁡m\ln m is

ln⁡E=1.12+0.734ln⁡m\ln E=1.12+0.734\ln m

The transformed data have correlation coefficient r=0.996r=0.996.

A
I.

KK and pp.

[2]
II.

the resulting power model.

[2]
B

the factor by which the predicted energy use changes when the animal's mass is multiplied by 55.

[2]
C

the mass for which the model predicts an energy use of 100100 kJ per day. on whether the transformed data support the proposed power relationship.

[3]

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Question 41
HL • Paper 3
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HL • Paper 3
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The speed, VV kilometres per hour, of a tidal current is assumed to vary sinusoidally over an 1818-hour monitoring period. The maximum speed is 7.27.2 kilometres per hour and the minimum speed is 1.61.6 kilometres per hour. Successive maxima occur at t=3t=3 and t=15t=15, where tt is measured in hours.

A
I.

the amplitude, principal-axis value and period of the model.

[3]
II.

a suitable model in the form V(t)=acos⁡(b(t−c))+dV(t)=a\cos(b(t-c))+d.

[2]
B

A turbine operates only when the current speed exceeds 66 kilometres per hour. the intervals during 0≤t≤180\leq t\leq18 when the turbine operates.

[4]
C
I.

the total operating time during the monitoring period.

[1]
II.

one assumption of the sinusoidal model and how its failure could affect the operating-time prediction.

[2]
Question 42
HL • Paper 3
Hard
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HL • Paper 3
Hard
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An open rectangular box is made by cutting squares of side xx centimetres from each corner of a 3030 cm by 2020 cm sheet and folding up the sides. The volume is modelled by
V(x)=x(30−2x)(20−2x)V(x)=x(30-2x)(20-2x)

A
I.

the contextual domain of xx.

[1]
II.

Expand V(x)V(x) and V′(x)V'(x).

[3]
B

the value of xx that maximizes the volume and the maximum volume.

[4]
C
I.

the range of values of xx for which the box has volume at least 1000 cm31000\ \text{cm}^3.

[3]
II.

why the largest algebraic root is rejected.

[1]
Question 43
HL • Paper 3
Hard
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HL • Paper 3
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A new product has 500500 users when it is released. After four months it has 18001800 users. Two models are considered:

  • an exponential model E(t)=500ertE(t)=500e^{rt};
  • a logistic model L(t)=100001+Ce−ktL(t)=\dfrac{10000}{1+Ce^{-kt}}.

Here tt is measured in months, and 1000010000 is the estimated size of the potential market.

A
I.

rr for the exponential model.

[2]
II.

CC and kk for the logistic model.

[3]
B

the number of users predicted after 1010 months by each model.

[3]
C
I.

Using the logistic model, when the number of users reaches 80%80\% of the potential market. If you did not obtain kk, use k=0.357k=0.357.

[2]
II.

Evaluate which model is more appropriate for long-term prediction.

[2]
Question 44
HL • Paper 3
Hard
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HL • Paper 3
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During a machine test, its internal temperature T(t)T(t), in degrees Celsius, is modelled by
T(t)={20+6t,0≤t<5,at+b,5≤t<12,86−2(t−12),t≥12,T(t)=\begin{cases}20+6t,&0\leq t<5,\\at+b,&5\leq t<12,\\86-2(t-12),&t\geq12,\end{cases}
where tt is measured in minutes. The model is continuous.

A
I.

Form two equations in aa and bb using continuity.

[2]
II.

aa and bb.

[3]
B

A warning light is active whenever T(t)>70T(t)>70. the total time for which the warning light is active.

[4]
C
I.

the value of the gradient in the final interval.

[1]
II.

one limitation of extending the final linear rule indefinitely.

[1]

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Question 45
HL • Paper 3
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HL • Paper 3
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Positive variables xx and yy are measured in an experiment for 1≤x≤101\leq x\leq10. Three transformations produce the following correlation coefficients:

  • yy against xx: r=0.954r=0.954;
  • ln⁡y\ln y against xx: r=0.991r=0.991;
  • ln⁡y\ln y against ln⁡x\ln x: r=0.9996r=0.9996.

For the third , the best-fit line is
ln⁡y=0.620+1.35ln⁡x\ln y=0.620+1.35\ln x

A
I.

the most appropriate model from linear, exponential and power models. your answer.

[2]
II.

the corresponding model y=Kxpy=Kx^p.

[2]
B

Use the model to yy when x=12x=12.

[2]
C
I.

the constant factor by which yy changes when xx is doubled.

[2]
II.

Give two reasons why the prediction in part (b) should be treated cautiously.

[2]
III.

why zero values could not be included in the log-log regression.

[1]
Question 46
HL • Paper 2
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HL • Paper 2
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The height, H(t)H(t) metres, of a young tree is modelled by

H(t)=18(1−e−kt)H(t)=18(1-e^{-kt})

where tt is the number of years after planting and k>0k>0. The tree has height 88 m after four years. For a larger set of trees, a best-fit line for a of z=ln⁡(18−H)z=\ln(18-H) against tt is

z=2.87−0.139tz=2.87-0.139t

A
I.

Using the single-tree model, kk.

[3]
II.

when this model predicts that the tree will reach 1515 m.

[1]
B

the model H(t)=18(1−e−kt)H(t)=18(1-e^{-kt}) can be linearized in the form

ln⁡(18−H)=ln⁡18−kt\ln(18-H)=\ln18-kt

[2]
C
I.

Use the best-fit line to obtain a model for HH in the form H=18−Ae−ktH=18-Ae^{-kt}.

[2]
II.

the two estimates of kk and evaluate whether the fitted model is consistent with the single-tree observation.

[2]
Question 47
HL • Paper 3
Hard
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HL • Paper 3
Hard
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A population P(t)P(t) of organisms in a controlled habitat satisfies
dPdt=0.48P(1−P2400)\frac{dP}{dt}=0.48P\left(1-\frac{P}{2400}\right)
where tt is measured in weeks and P(0)=120P(0)=120. The corresponding logistic model has the form
P(t)=24001+Ce−0.48tP(t)=\frac{2400}{1+Ce^{-0.48t}}

A
I.

CC.

[1]
II.

the population after 1010 weeks.

[2]
III.

the long-term population predicted by the model.

[1]
B

the maximum growth rate occurs when P=1200P=1200, and this maximum rate.

[4]
C
I.

the two population sizes for which the growth rate is 200200 organisms per week.

[2]
II.

the two times at which these growth rates occur. If you did not obtain the populations in part (c)(i), use P=536.675P=536.675 and P=1863.325P=1863.325.

[2]
III.

why the same growth rate occurs at two different population sizes.

[1]
Question 48
HL • Paper 3
Hard
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HL • Paper 3
Hard
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A restoration project begins with 66 hectares of established tree canopy. After three years, the canopy covers 1515 hectares. Two models are proposed:
E(t)=6ertE(t)=6e^{rt}
and
L(t)=801+Ce−ktL(t)=\frac{80}{1+Ce^{-kt}}
where tt is measured in years. The value 8080 hectares is an estimated maximum area available for tree canopy.

A
I.

rr for the exponential model.

[2]
II.

CC and kk for the logistic model.

[3]
B

when each model predicts that the canopy will first cover 6060 hectares.

[4]
C
I.

why the two predictions diverge despite both models fitting the two observations exactly.

[2]
II.

Recommend one additional observation that would be useful for choosing between the models, and your answer.

[1]

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Linear Equations & Graphs