Variables t and N are such that when lgN is plotted against lgt, a straight line graph passing through the points (0.45, 1.2) and (1, 3.4) is obtained. [Figure 3.1]
- (i)Express the equation of the straight line graph in the form lgN = m lgt + lgc, where m and c are constants to be found.[4]
- (ii)Hence express N in terms of t.[1]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2011). The mark or mark type is shown in brackets.
- [1] (i) Gradient m = 4
- [1] (i) lg c = -0.6
- [1] (i) M1 for attempt to deal with lgc
- [1] (i) c = 0.251
- [1] (ii) N = 0.251t^4
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