Date | May 2013 | Marks available | 1 | Reference code | 13M.2.hl.TZ2.5 |
Level | HL only | Paper | 2 | Time zone | TZ2 |
Command term | Determine | Question number | 5 | Adapted from | N/A |
Question
The arithmetic sequence {un:n∈Z+} has first term u1=1.6 and common difference d = 1.5. The geometric sequence {vn:n∈Z+} has first term v1=3 and common ratio r = 1.2.
Find an expression for un−vn in terms of n.
Determine the set of values of n for which un>vn.
Determine the greatest value of un−vn. Give your answer correct to four significant figures.
Markscheme
un−vn=1.6+(n−1)×1.5−3×1.2n−1 (=1.5n+0.1−3×1.2n−1) A1A1
[2 marks]
attempting to solve un>vn numerically or graphically. (M1)
n=2.621…,9.695… (A1)
So 3⩽ A1
[3 marks]
The greatest value of {u_n} - {v_n} is 1.642. A1
Note: Do not accept 1.64.
[1 mark]
Examiners report
In part (a), most candidates were able to express {u_n} and {v_n} correctly and hence obtain a correct expression for {u_n} - {v_n}. Some candidates made careless algebraic errors when unnecessarily simplifying {u_n} while other candidates incorrectly stated {v_n} as 3{(1.2)^n}.
In parts (b) and (c), most candidates treated n as a continuous variable rather than as a discrete variable. Candidates should be aware that a GDC’s table feature can be extremely useful when attempting such question types.
In parts (b) and (c), most candidates treated n as a continuous variable rather than as a discrete variable. Candidates should be aware that a GDC’s table feature can be extremely useful when attempting such question types. In part (c), a number of candidates attempted to find the maximum value of n rather than attempting to find the maximum value of {u_n} - {v_n}.