A2 June 2022 Q8

EdexcelCurrent spec9 marksRecurrence Relations

8. The owner of a new company models the number of customers that the company will have at the end of each month. The owner assumes that

a constant proportion, \(p\) (where \(0 < p < 1\)), of the previous month’s customers will be retained for the next month
a constant number of new customers, \(k\), will be added each month.

Let \(u_n\) (where \(n \geqslant 1\)) represent the number of customers that the company will have at the end of \(n\) months.

The company has 5000 customers at the end of the first month.

(a) By setting up a first order recurrence relation for \(u_{n+1}\) in terms of \(u_n\), determine an expression for \(u_n\) in terms of \(n\), \(p\) and \(k\). (6)

The owner believes that 95% of the previous month’s customers will be retained each month and that there will be 10 000 new customers each month.

According to the model, the company will first have at least 135 000 customers by the end of the \(m\)th month.

(b) Using logarithms, determine the value of \(m\). (3)

Mark scheme (a)

Scheme	Marks	AO
\(u_{n+1} = pu_n + k\)	B1	3.3
(aux equation \(m - p = 0 \Rightarrow\)) complementary function is \(A(p)^n\)	B1	1.1b
Consider a trial solution of the form \(u_n = \lambda\) so \(\lambda - p\lambda = k\)	M1	1.1b
General solution is \(u_n = A(p)^n + \dfrac{k}{1-p}\)	A1	1.1b
\(u_1 = 5000 \Rightarrow 5000 = A(p) + \dfrac{k}{1-p}\) and solve for \(A\)	M1	3.4
\(u_n = \left(5000 - \dfrac{k}{1-p}\right)p^{n-1} + \dfrac{k}{1-p}\)	A1	2.2a
	(6)

Notes

B1: CAO

M1: substituting their trial solution into the recurrence relation in an attempt to find their \(\lambda\) (which if correct is \(\dfrac{k}{1-p}\))

A1: CAO for the general solution

M1: using the conditions in the model to calculate \(A\) (which if correct is \(p^{-1}\left(5000 - \dfrac{k}{1-p}\right)\))

A1: CAO for the particular solution (oe)

Mark scheme (b)

Scheme	Marks	AO
Set \(k = 10\,000\), \(p = 0.95\) and \(u_m \geqslant 135\,000\) \(\left(5000 - \dfrac{10000}{1-0.95}\right)(0.95)^{m-1} + \dfrac{10000}{1-0.95} \geqslant 135000\)	B1	3.1b
\((0.95)^{m-1} \leqslant \dfrac{1}{3} \Rightarrow (m-1)\log(0.95) \leqslant \log\left(\dfrac{1}{3}\right) \Rightarrow m \geqslant \ldots\)	M1	1.1b
\(m \geqslant 22.418\ldots\) so 23 months after the company was first up	A1	3.2a
	(3)
	(9 marks)

Notes

B1: Applying \(u_m \geqslant 135000\) (or equality or strict inequality) to their general solution together with correct values for \(k\) and \(p\) (dependent on both M marks in (a))

M1: dependent on previous B mark – solving their equation using logarithms

A1: CAO – must be rounded to 23

Special case: A common misread is 500 for 5000. Mark as a misread so final A marks in (a) and (b) deducted.

(corrected from the printed mark scheme: the inequality signs \(\geqslant\) and \(\leqslant\) in this part are printed as “…” and “,,” because of a font error)