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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:nZ+} has first term u1=1.6 and common difference d = 1.5. The geometric sequence {vn:nZ+} has first term v1=3 and common ratio r = 1.2.

Find an expression for unvn in terms of n.

[2]
a.

Determine the set of values of n for which un>vn.

[3]
b.

Determine the greatest value of unvn. Give your answer correct to four significant figures.

[1]
c.

Markscheme

unvn=1.6+(n1)×1.53×1.2n1 (=1.5n+0.13×1.2n1)     A1A1

[2 marks]

a.

attempting to solve un>vn numerically or graphically.     (M1)

n=2.621,9.695     (A1)

So 3     A1

[3 marks]

b.

The greatest value of {u_n} - {v_n} is 1.642.     A1

Note: Do not accept 1.64.

 

[1 mark]

c.

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}.

a.

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.

b.

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}.

c.

Syllabus sections

Topic 2 - Core: Functions and equations » 2.7 » Use of technology for these and other functions.

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