A2 June 2019 Q3
3. The number of visits to a website, in any particular month, is modelled as the number of visits received in the previous month plus \(k\) times the number of visits received in the month before that, where \(k\) is a positive constant.
Given that \(V_n\) is the number of visits to the website in month \(n\),
For a particular website you are given that
- \(k = 0.24\)
- In month 1, there were 65 visits to the website.
- In month 2, there were 71 visits to the website.
This model predicts that the number of visits to this website will exceed one million for the first time in month \(N\).
| Scheme | Marks | AO |
|---|---|---|
| \(V_{n+2} = V_{n+1} + kV_n\) | B1 | 3.3 |
| (1) |
Notes
B1: A correct expression for the model using the information given
| Scheme | Marks | AO |
|---|---|---|
| \(\lambda^2 - \lambda - 0.24 = 0 \Rightarrow \lambda = \ldots(1.2, -0.2)\) | M1 | 1.1b |
| \(V_n = a(1.2)^n + b(-0.2)^n\) | A1 | 2.2a |
| \(65 = a(1.2)^1 + b(-0.2)^1\) and \(71 = a(1.2)^2 + b(-0.2)^2\) | B1ft | 3.4 |
| E.g. \(\left.\begin{aligned} 78 &= 1.44a - 0.24b \\ 71 &= 1.44a + 0.04b \end{aligned}\right\} \Rightarrow 7 = -0.28b \Rightarrow b = \ldots\) | M1 | 2.1 |
| \(a = 50, b = -25 \Rightarrow V_n = 50(1.2)^n - 25(-0.2)^n\) * | A1* | 1.1b |
| (5) |
Notes
M1: Forms and solves the auxiliary equation for their answer to (a) with \(k = 0.24\)
A1: The correct closed form deduced from their solutions. This must be consistent with their equation. Note the answer is given so check carefully. This is not a follow through mark.
B1ft: Applies initial conditions to their general equation – correct two equations for their general form with \(V_1 = 65\) and \(V_2 = 71\)
M1: Attempts to solve their equations showing a correct method, reaching a value for at least one variable. It is a show that question and answers are on the paper, so method is needed. Look for one equation multiplied through to give same coefficients before attempting eliminating or substitution. If a matrix system is used the inverse must be found, not just solutions stated.
A1*: Correct expression formed following suitable working with no errors seen. With fractions instead of decimals is fine.
Alternative (b): \(V_n = 50(1.2)^n - 25(-0.2)^n\), \(V_1 = 65\), \(V_2 = 71\)
| Scheme | Marks | AO |
|---|---|---|
| \(V_1 = 50 \times 1.2 - 25 \times -0.2 = 60 + 5 = 65\) \(V_2 = 50 \times (1.2)^2 - 25 \times (-0.2)^2 = 72 - 1 = 71\) | M1 | 1.1b |
| Hence true for \(n = 1\) and \(n = 2\) Assume true for \(n = k\) and \(n = k + 1\) (for some \(k \gt 0\)) | A1 | 2.2a |
| \(V_{k+2} = V_{k+1} + 0.24V_k\) \(= 50(1.2)^{k+1} - 25(-0.2)^{k+1} + 0.24\left(50(1.2)^k - 25(-0.2)^k\right)\) | B1ft | 3.4 |
| \(\displaystyle \begin{aligned} &= \frac{50}{1.2}(1.2)^{k+2} - \frac{25}{-0.2}(-0.2)^{k+2} + \frac{12}{1.2^2}(1.2)^{k+2} - \frac{6}{(-0.2)^2}(-0.2)^{k+2} \\[10pt] &= \frac{125}{3}(1.2)^{k+2} + 125(-0.2)^{k+2} + \frac{25}{3}(1.2)^{k+2} - 150(-0.2)^{k+2} = \ldots \end{aligned}\) | M1 | 2.1 |
| So \(V_{k+2} = 50(1.2)^{k+2} - 25(-0.2)^{k+2}\) Hence true for \(n = k + 2\). So the result is true for \(n = 1\) and \(n = 2\), and if true for \(n = k\) and \(n = k + 1\) then it is true for \(n = k + 2\). Hence by mathematical induction, for all \(n \in \mathbb{N}\) \(V_n = 50(1.2)^n - 25(-0.2)^n\) * | A1* | 1.1b |
| (5) |
M1: Substitutes into equation for \(n = 1\) and \(n = 2\) to verify true for these cases.
A1: Deduces true for base cases and makes a correct assumption statement. This must include two successive cases assumed true, so e.g. as in scheme, or with \(k - 2\) and \(k - 1\) etc, or may assume true for all (integers) \(k \leqslant n\). But do not allow if assumed true for just \(k\).
B1ft: Substitutes the formula for \(k\) and \(k + 1\) (or their successive values) into the recurrence formula, follow through their equation from part (a).
M1: Rearranges to the form \(a(1.2)^{(k+2)} + b(-0.2)^{k+2}\)
A1*: Correct work leading to the correct equation for \(V_{k+2}\) and makes suitable inductive conclusion, including the ideas of “true for \(n = 1\) and \(n = 2\)”, “if true for \(n = k\) and \(n = k + 1\) then true for \(n = k + 2\)” and “hence true for all integers”.
| Scheme | Marks | AO |
|---|---|---|
| \(50(1.2)^N \gt 10^6 \Rightarrow N = \ldots\) | M1 | 3.1b |
| \(\Rightarrow N = 55\) i.e. month 55 | A1 | 3.2a |
| (2) | ||
| (8 marks) |
Notes
M1: Selects a suitable method to solve the problem. For example, realises that in the model, \((-0.2)^n\) is negligible for large \(n\) and so attempts to solve e.g. \(50(1.2)^N = 10^6\), or tries at least one value either side of \(N = 55\) as a process of trial and improvement, or uses a calculator/graphical approach – implied by a value of \(N = 55\) or \(N = 54\) stated.
A1: \(N = 55\).
The correct answer will imply both marks for this part. Ignore erroneous working if correct answer is stated as a restart.