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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

$N = A e^{r t}, A, r \in ℝ$

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

$N (n + 1) \approx N (n) + 1 \times N' (n), n \in ℕ$

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

$\frac{d N}{d t} = r N$

[2]

Show that $r \approx \frac{N (n + 1) - N (n)}{N (n)}$

[3]

Hence find three approximations for the value of $r$ .

[3]

Find the equation for Jorge’s model.

[3]

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

[2]

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.

$\frac{d N}{d t} = r A e^{r t}$ (M1)A1

Note: M1 is for an attempt to find $\frac{d N}{d t}$

$= r N$ AG

Note: Accept solution of the differential equation by separating variables

[2 marks]

$N (n + 1) \approx N (n) + 1 \times N' (n) \Rightarrow N' (n) \approx N (n + 1) - N (n)$ M1

$\Rightarrow r N (n) \approx N (n + 1) - N (n)$ M1A1

$\Rightarrow r \approx \frac{N (n + 1) - N (n)}{N (n)}$ AG

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

[3 marks]

Correct method (M1)

$r \approx \frac{52 - 40}{40} = 0.3$

$r \approx \frac{70 - 52}{52} = 0.346$

$r \approx \frac{98 - 70}{70} = 0.4$ A2

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

[3 marks]

$r = 0.349 (0.34871 \dots)$ or $\frac{68}{195}$ A1

$52 = A e^{0.34871 \dots \times 2}$ (M1)

$A = 25.8887 \dots$

$N = 25.9 e^{0.349 t}$ A1

[3 marks]

${(36.6904 \dots - 40)}^{2} + 0 + {(73.6951 \dots - 70)}^{2} + {(104.4435 \dots - 98)}^{2}$ (M1)

$= 66.1 (66.126 \dots)$ A1

[2 marks]

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

[N/A]

f.i.

[N/A]

f.ii.

Syllabus sections

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

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16N.1.SL.TZ0.S_10b:
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17M.1.SL.TZ2.S_6b:
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16N.1.SL.TZ0.S_10c:
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18N.2.AHL.TZ0.H_9d:
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20N.2.SL.TZ0.S_10d:
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21N.1.AHL.TZ0.15a:
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21N.3.AHL.TZ0.2a.ii:
Interpret the meaning of parameter $a$ in the context of the model.
21N.3.AHL.TZ0.2a.iii:
Suggest why Eva’s use of the linear regression equation in this way could be unreliable.
21N.3.AHL.TZ0.2b.i:
Find the equation of the least squares quadratic regression curve.
21N.3.AHL.TZ0.2b.ii:
Find the value of $k$ .
21N.3.AHL.TZ0.2b.iii:
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21N.3.AHL.TZ0.2c:
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Use the general solution from part (d) and the initial condition $h (0) = 3.2$ to predict the value of $T$ .
21N.3.AHL.TZ0.2f:
Find this new height.
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21N.3.AHL.TZ0.2g.ii:
Use Euler’s method with a step length of $0.5$ minutes to estimate the maximum value of $H$ .

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