Lesson 4D.6.1
4D.6.1 Modelling with recurrence relations Quiz: Pearson Edexcel Further Maths, Unit 45
20 questions
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Lesson 4D.6.1, Modelling with recurrence relations: 20 multiple choice questions for the Pearson Edexcel Further Maths (9FM0), Unit 45: Recurrence relations, written with Revision Ninja.
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The 20 questions
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What does a first-order recurrence relation u_(n+1) = a u_n + b model?
- A process in which each term is determined from the previous term.
- A random sequence with no rule.
- Only geometric series with a = 1.
- A set of simultaneous equations solved all at once.
-
What is the initial condition in a recurrence model?
- The limit of the terms as n tends to infinity.
- The sum of all the terms so far.
- The value of the constant b only.
- The given starting value, such as u_0 or u_1, needed to fix the solution.
-
A population grows by 5% a year and 200 are removed each year. Which recurrence models this, with P_n the population after n years?
- P_(n+1) = 1.05 P_n - 200
- P_(n+1) = 0.05 P_n - 200
- P_(n+1) = P_n + 0.05 - 200
- P_(n+1) = 1.05 P_n + 200
-
For u_(n+1) = a u_n + b, what equation does a steady state u satisfy?
- u = b - a
- u = a u + b
- u = a + b
- u = u + 1
-
What is the complementary function of a non-homogeneous recurrence?
- The sum of the initial conditions.
- The particular solution found by trying a constant.
- The general solution of the homogeneous recurrence with the constant term set to zero.
- The solution of the auxiliary equation with b added.
-
What is the auxiliary equation of u_(n+2) - 5u_(n+1) + 6u_n = 0?
- m^2 - 5m + 6 = 0
- m^2 - 6m + 5 = 0
- m^2 + 5m + 6 = 0
- m - 5 + 6 = 0
-
Why is modelling with recurrence relations useful?
- It replaces all data collection.
- It works only for constant growth.
- It describes how a quantity changes from one period to the next, such as yearly population.
- It gives the exact value at any time without starting values.
-
A population P_0 = 1000 follows P_(n+1) = 1.05 P_n - 200. What is P_1?
- 950
- 800
- 850
- 1050
-
For the same model, what is P_2?
- 700
- 722.5
- 692.5
- 850
-
For P_(n+1) = 1.05 P_n - 200, what is the steady state?
- 1000
- 5000
- 200
- 4000
-
A balance follows B_(n+1) = 1.02 B_n - 150 with B_0 = 2000. What is B_1?
- 2040
- 1890
- 1950
- 1850
-
For the population model P_(n+1) = 1.05 P_n - 200 with P_0 = 1000, what is P_3?
- 527.125
- 527.5
- 600
- 692.5
-
A savings account earns 4% a year and £100 is added at the end of each year. Which recurrence models the balance u_n?
- u_(n+1) = 1.04 u_n + 100
- u_(n+1) = 1.4 u_n + 100
- u_(n+1) = 1.04 u_n - 100
- u_(n+1) = u_n + 4 + 100
-
For the savings model u_(n+1) = 1.04 u_n + 100 with u_0 = 1000, what is u_2?
- 1285.6
- 1240
- 1300
- 1180
-
For the model u_(n+1) = 0.8 u_n + 50, what is the steady state?
- 62.5
- 250
- 200
- 50
-
For the population model P_(n+1) = 1.05 P_n - 200 starting at 1000, which way does the population move in the first year?
- It rises and then returns to 1000.
- It falls, since P_1 = 850 is below 1000.
- It stays at 1000.
- It rises, since 5% growth outweighs the removal.
-
Why is the steady state 4000 of P_(n+1) = 1.05 P_n - 200 unstable?
- Above 4000 the population keeps rising and below 4000 it keeps falling, so the sequence moves away from 4000.
- It is stable because 1.05 is less than 1.
- It is the first term of the sequence.
- Because b is negative, it attracts all sequences.
-
Does the sequence u_(n+1) = 0.9 u_n + 30 converge from any starting value?
- Yes, it converges to 300 from any starting value, since the multiplier 0.9 has absolute value below 1.
- No, it diverges from any starting value above 300.
- No, it oscillates between two values for every start.
- Only if the starting value is exactly 300.
-
Why must a recurrence model include an initial condition?
- It removes the need for a steady state.
- It is needed only when a is negative.
- It makes b equal to zero.
- Without one, infinitely many sequences satisfy the recurrence, so the actual quantity cannot be determined.
-
Why does a steady state fail to exist when a = 1 and b is not zero?
- The equation u = u + b has no solution, so there is no fixed value for the sequence to settle at.
- Because b must be zero when a is 1, so the steady state is always zero.
- Because the population must double each year.
- Because a = 1 always gives an oscillating sequence.
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