What is 2.11: Functions in modelling?

A function model links clearly defined variables over a stated domain; it may use polynomial, exponential, logarithmic, trigonometric or reciprocal functions.

How many marks are 2.11 questions worth?

2.11 practice drills on this page are worth 2–7 marks each; marks in a real exam vary by question.

Which past exam papers has 2.11 appeared in?

We have not yet indexed a verified past-paper appearance for 2.11 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

2.11

Use of functions in modelling, including consideration of limitations and refinements of the models.

Draft — not yet indexed

Functions in modelling

Worked answers and methods for 2.11 on Edexcel A-level Maths 9MA0.

Explanation

A function model links clearly defined variables over a stated domain; it may use polynomial, exponential, logarithmic, trigonometric or reciprocal functions.
Use data to determine parameters, then calculate and interpret the result with units and appropriate rounding.
For example, $P(t)=P_0r^t$ models repeated growth, while a sine function can model periodic height and a reciprocal function can model inverse variation.
Evaluate assumptions such as fixed rates, exact periodicity, ignored constraints or extrapolation.
A refinement must answer the weakness: use a piecewise rate when growth changes, add a seasonal term when residuals are periodic, or restrict the domain when extrapolation is unreliable.

Worked example

A cooling model is $T(t)=18+62e^{-0.15t}$ , where $T$ is in degrees Celsius and $t$ is in minutes. Find the first whole minute for which the model gives $T<30$ , and state one limitation.

1.Solve $18+62e^{-0.15t}<30$ .
2.This gives $e^{-0.15t}<6/31$ , so $t>-\ln(6/31)/0.15=10.95\ldots$ .
3.The first whole minute is therefore $11$ .
4.One limitation is that the fixed ambient temperature may not remain exactly $18$ degrees Celsius.

Answer: $11$ minutes; for example, the model assumes a constant surrounding temperature of $18$ degrees Celsius.

Common mistakes

Don't report a decimal model output as an exact real-world count without appropriate contextual rounding.
Don't extrapolate beyond the stated domain without discussing whether the model assumptions remain plausible.

Exam tip

After interpreting the value, link each limitation to a concrete refinement that changes the model or restricts its domain.

Worked practice

Tier 1 · Easy

A colony is modelled by

P(t)=120(1.08)^t

, where

t

is measured in days. Estimate the number of organisms after

3

days.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$151$ organisms	2
Notes
Substitute $t=3$ : $P(3)=120(1.08)^3=151.16544$ . Since the output counts organisms, round to the nearest whole number to obtain $151$ .

(2 marks)

Tier 2 · Standard

A company uses

V=Ab^t

to estimate the value of a car, where £

V

is its value

t

years after purchase. The constants

A

and

b

satisfy

b>0

. According to this model, the car is worth £

18\,000

initially and £

14\,580

after two years. Find

A

and

b

. Hence estimate the value of the car after

5

years, giving your answer to the nearest pound. State one limitation of the model.

(5)

(Total for Question 2 is 5 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$A=18\,000$ and $b=0.9$ £ $10\,629$ For example, the model does not allow for sudden changes in value caused by damage or repairs.	5
Notes
At $t=0$ , $A=18\,000$ . Using $t=2$ gives $14\,580=18\,000b^2$ , so $b^2=0.81$ and, since $b>0$ , $b=0.9$ . Then $V(5)=18\,000(0.9)^5=10\,628.82$ , giving £ $10\,629$ to the nearest pound. A limitation is that a constant annual multiplier cannot represent unexpected damage, repairs or changes in the second-hand market.

(5 marks)

Tier 3 · Hard

The cross-section of a greenhouse roof is modelled by

h(x)=kx(30-x)

for

0\leq x\leq30

, with distances in metres. The point

(5,10)

lies on the model. Find

k

, the maximum modelled height, and the horizontal width over which

h(x)\geq12

. Give one limitation of the model.

(7)

(Total for Question 3 is 7 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$k=0.08$ Maximum height $18$ m Width $10\sqrt3$ m For example, a real roof cross-section may not be exactly parabolic.	7
Notes
Using $(5,10)$ gives $10=k(5)(25)$ , so $k=0.08$ . The quadratic is symmetric about $x=15$ , where $h(15)=0.08(15)(15)=18$ . For $h(x)\geq12$ , solve $0.08x(30-x)\geq12$ , equivalent to $x^2-30x+150\leq0$ . The roots are $15\pm5\sqrt3$ , so the allowed interval has width $(15+5\sqrt3)-(15-5\sqrt3)=10\sqrt3$ metres. The exact-parabola assumption is a limitation because manufactured or loaded roofs may have a different profile.

(7 marks)

Tier 1 · Easy

A battery's charge percentage is modelled by

C(t)=96-7t

for

0\leq t\leq10

, where

t

is measured in hours. Find

C(6)

and explain why the model should not be used to predict

C(14)

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$C(6)=54$ $t=14$ is outside the stated domain of the model.	2
Notes
Substitution gives $C(6)=96-7(6)=54$ . The model is stated only for $0\leq t\leq10$ , so using $t=14$ would be an unsupported extrapolation.

(2 marks)

Tier 2 · Standard

A concentration is modelled by

C(t)=\dfrac{a}{t+b}

for

t\geq0

, where

a

and

b

are positive constants. Given

C(0)=24

and

C(4)=12

, find

a

and

b

. Hence find the time when

C(t)=8

(5)

(Total for Question 5 is 5 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$a=96$ , $b=4$ $t=8$	5
Notes
From $C(0)=24$ , $a/b=24$ , so $a=24b$ . From $C(4)=12$ , $a=12(b+4)$ . Equating these gives $24b=12b+48$ , hence $b=4$ and $a=96$ . Finally, $96/(t+4)=8$ gives $t+4=12$ , so $t=8$ .

(5 marks)

Tier 3 · Hard

For

t\geq0

, a storm collection model is

R(t)=\dfrac{at}{t+b}

, with positive parameters

a

and

b

. Given

R(2)=18

and

R(6)=36

, find

a

and

b

and the limiting amount predicted by the model. A gauge later records

78

cubic metres from this storm. Decide whether the model can represent the whole storm, and state a specific refinement.

(7)

(Total for Question 6 is 7 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$a=72$ , $b=6$ The limiting amount is $72$ cubic metres. No; the model never reaches $78$ cubic metres. Refit the model with a limiting parameter greater than $78$ cubic metres using the later data.	7
Notes
Using $R(2)=18$ gives $18=2a/(2+b)$ , so $a=18+9b$ . Using $R(6)=36$ gives $36=6a/(6+b)$ , so $a=36+6b$ . Equating these expressions gives $b=6$ and then $a=72$ . As $t\to\infty$ , $at/(t+b)\to a$ , so the predicted limiting amount is $72$ cubic metres. For every finite $t\geq0$ , the model gives $R(t)<72$ , so it cannot represent a recorded total of $78$ cubic metres. A specific refinement is to refit the asymptote parameter using the later observation, making the limiting value exceed $78$ cubic metres.

(7 marks)

Tier 2 · Standard

A water level is modelled by

H(t)=a+b\cos(\pi t/6)

for

0\leq t\leq12

, where

H

is measured in metres and

t

in hours. The model gives

H(0)=8

and

H(6)=2

. Find

a

and

b

, then find the first time after

t=0

when

H(t)=5

. State the periodicity assumption made by the model.

(5)

(Total for Question 7 is 5 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$a=5$ , $b=3$ $t=3$ hours The water-level pattern is assumed to repeat exactly every $12$ hours.	5
Notes
Since $\cos0=1$ and $\cos\pi=-1$ , the two observations give $a+b=8$ and $a-b=2$ . Hence $a=5$ and $b=3$ . Solving $5+3\cos(\pi t/6)=5$ gives $\cos(\pi t/6)=0$ ; the first positive solution is $\pi t/6=\pi/2$ , so $t=3$ . The cosine model assumes an exactly repeating $12$ -hour cycle.

(5 marks)

Tier 3 · Hard

A quantity is modelled at whole-number times

t\geq0

P(t)=L-Ar^t

, where

A>0

L>0

and

0<r<1

. Given

P(0)=20

P(1)=44

and

P(2)=56

, find

L

A

and

r

. Hence find the first whole-number time for which

P(t)>65

. A later measurement is lower than the model because the surrounding conditions changed at

t=3

; state a specific refinement that addresses this information.

(7)

(Total for Question 8 is 7 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$L=68$ , $A=48$ , $r=\dfrac12$ The first whole-number time is $t=5$ . Use a piecewise model with new parameter values from $t=3$ onwards.	7
Notes
The first two increases are $44-20=24$ and $56-44=12$ . For this model successive increases have ratio $r$ , so $r=12/24=1/2$ . Also $P(1)-P(0)=A(1-r)=24$ , giving $A=48$ . Then $P(0)=L-A=20$ gives $L=68$ . The inequality $68-48(1/2)^t>65$ is equivalent to $(1/2)^t<1/16$ , so $t>4$ and the first whole-number time is $5$ . A piecewise model, refitted with different parameters for $t\geq3$ , directly represents the changed conditions.

(7 marks)

Tier 3 · Hard

A braking-distance model is

D=kv^n

, where

D

is measured in metres,

v

in metres per second, and

k>0

. The model gives

D=8

when

v=10

and

D=32

when

v=20

. Find

k

and

n

, then predict

D

when

v=30

. A measured braking distance at

30

metres per second is

78

metres. State the model error and give a specific refinement.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$n=2$ and $k=0.08$ The model predicts $72$ metres. It underestimates by $6$ metres; for example, fit a piecewise power model using additional high-speed data.	6
Notes
Dividing the two model equations gives $32/8=(20/10)^n$ , so $4=2^n$ and $n=2$ . Then $8=100k$ , giving $k=0.08$ . At $v=30$ , the model gives $D=0.08(30)^2=72$ metres. Compared with the measured $78$ metres, this is an underestimate of $6$ metres. A specific refinement is to collect further high-speed observations and fit a second power law above a stated speed threshold.

(6 marks)

Q10

Tier 3 · Hard

10.

For

t\geq0

, a population is modelled by

P(t)=50(2^{t/2})

for

0\leq t\leq4

and by

P(t)=A\left(\dfrac32\right)^{t-4}

for

t>4

. The model is continuous at

t=4

. Find

A

and the first whole-number time for which

P(t)>600

. State one limitation of using the second branch indefinitely and give a specific refinement.

(7)

(Total for Question 10 is 7 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$A=200$ The first whole-number time is $t=7$ . A constant $50\%$ increase per time interval cannot remain realistic indefinitely; for example, replace the second branch by a model with a finite limiting population.	7
Notes
At $t=4$ , the first branch gives $P(4)=50(2^2)=200$ . Continuity therefore requires $A=200$ . The first branch never exceeds $200$ . On the second branch, $P(6)=200(3/2)^2=450$ , while $P(7)=200(3/2)^3=675$ . Since the branch is increasing, the first whole-number time with $P(t)>600$ is $t=7$ . Indefinite fixed-percentage growth ignores limited resources, so a model with a carrying-capacity parameter would be a specific refinement.

(7 marks)

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