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Date November 2011 Marks available 2 Reference code 11N.2.sl.TZ0.6
Level SL only Paper 2 Time zone TZ0
Command term Find Question number 6 Adapted from N/A

Question

Jose takes medication. After t minutes, the concentration of medication left in his bloodstream is given by \(A(t) = 10{(0.5)^{0.014t}}\) , where A is in milligrams per litre.

Write down \(A(0)\) .

[1]
a.

Find the concentration of medication left in his bloodstream after 50 minutes.

[2]
b.

At 13:00, when there is no medication in Jose’s bloodstream, he takes his first dose of medication. He can take his medication again when the concentration of medication reaches 0.395 milligrams per litre. What time will Jose be able to take his medication again?

[5]
c.

Markscheme

\(A(0) = 10\)     A1     N1

[1 mark]

a.

substitution into formula     (A1)

e.g. \(10{(0.5)^{0.014(50)}}\) , \(A(50)\)

\(A(50) = 6.16\)     A1     N2

[2 marks]

b.

set up equation     (M1)

e.g. \(A(t) = 0.395\)

attempting to solve     (M1)

e.g. graph, use of logs

correct working    (A1)

e.g. sketch of intersection, \(0.014t\log 0.5 = \log 0.0395\)

\(t = 333.00025 \ldots \)     A1

correct time 18:33 or 18:34 (accept 6:33 or 6:34 but nothing else)     A1     N3

[5 marks]

c.

Examiners report

For a later question in Section A, a pleasing number of candidates made good progress. Some candidates believed that raising a base to the zero power gave zero which indicated that they most likely did not begin by analysing the function with their GDC. For part (c), many candidates could set up the equation correctly and had some idea to apply logarithms but became lost in the algebra. Those who used their GDC to find when the function equalled 0.395 typically did so successfully. A common error for those who obtained a correct value for time in minutes was to treat 5.55 hours as 5 hours and 55 minutes after 13:00.

a.

For a later question in Section A, a pleasing number of candidates made good progress. Some candidates believed that raising a base to the zero power gave zero which indicated that they most likely did not begin by analysing the function with their GDC. For part (c), many candidates could set up the equation correctly and had some idea to apply logarithms but became lost in the algebra. Those who used their GDC to find when the function equalled 0.395 typically did so successfully. A common error for those who obtained a correct value for time in minutes was to treat 5.55 hours as 5 hours and 55 minutes after 13:00.

b.

For a later question in Section A, a pleasing number of candidates made good progress. Some candidates believed that raising a base to the zero power gave zero which indicated that they most likely did not begin by analysing the function with their GDC. For part (c), many candidates could set up the equation correctly and had some idea to apply logarithms but became lost in the algebra. Those who used their GDC to find when the function equalled 0.395 typically did so successfully. A common error for those who obtained a correct value for time in minutes was to treat 5.55 hours as 5 hours and 55 minutes after 13:00.

c.

Syllabus sections

Topic 2 - Functions and equations » 2.6 » Exponential functions and their graphs: \(x \mapsto {a^x}\) , \(a > 0\) , \(x \mapsto {{\text{e}}^x}\) .
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