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Date November Example question Marks available 2 Reference code EXN.2.AHL.TZ0.4
Level Additional Higher Level Paper Paper 2 Time zone Time zone 0
Command term Show that Question number 4 Adapted from N/A

Question

Jorge is carefully observing the rise in sales of a new app he has created.

The number of sales in the first four months is shown in the table below.

Jorge believes that the increase is exponential and proposes to model the number of sales N in month t with the equation

N=Aert, A,r

Jorge plans to adapt Euler’s method to find an approximate value for r.

With a step length of one month the solution to the differential equation can be approximated using Euler’s method where

Nn+1Nn+1×N'n, n

Jorge decides to take the mean of these values as the approximation of r for his model. He also decides the graph of the model should pass through the point (2, 52).

The sum of the square residuals for these points for the least squares regression model is approximately 6.555.

Show that Jorge’s model satisfies the differential equation

dNdt=rN

[2]
a.

Show that rNn+1-NnNn

[3]
b.

Hence find three approximations for the value of r.

[3]
c.

Find the equation for Jorge’s model.

[3]
d.

Find the sum of the square residuals for Jorge’s model using the values t=1, 2, 3, 4.

[2]
e.

Comment how well Jorge’s model fits the data.

[1]
f.i.

Give two possible sources of error in the construction of his model.

[2]
f.ii.

Markscheme

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

dNdt=rAert        (M1)A1

 

Note: M1 is for an attempt to find dNdt

 

=rN        AG

 

Note: Accept solution of the differential equation by separating variables

 

[2 marks]

a.

Nn+1Nn+1×N'nN'nNn+1-Nn        M1

rNnNn+1-Nn        M1A1

rNn+1-NnNn        AG

 

Note: Do not penalize the use of the = sign.

 

[3 marks]

b.

Correct method         (M1)

r52-4040=0.3

r70-5252=0.346

r98-7070=0.4        A2

 

Note: A1 for a single error A0 for two or more errors.

 

[3 marks]

c.

r=0.349 0.34871 or 68195        A1

52=Ae0.34871×2        (M1)

A=25.8887

N=25.9e0.349t        A1

 

[3 marks]

d.

36.6904-402+0+73.6951-702+104.4435-982        (M1)

=66.1 66.126        A1

 

[2 marks]

e.

The sum of the square residuals is approximately 10 times as large as the minimum possible, so Jorge’s model is unlikely to fit the data exactly     R1

 

[1 mark]

f.i.

For example

Selecting a single point for the curve to pass through

Approximating the gradient of the curve by the gradient of a chord       R1R1

 

[2 marks]

f.ii.

Examiners report

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Syllabus sections

Topic 5—Calculus » AHL 5.9—Differentiating standard functions and derivative rules
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Topic 5—Calculus

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