An experiment was carried out and values of y for certain values of x were recorded. The table shows the values recorded. x 15 30 45 60 75 y 10 13 22 35 50 The relationship between y and x is modelled by y = Ae^(kx), where A and k are constants.
- (a)Draw a straight line graph for lny against x.[2]
- (b)Find the equation of the line in part (a) and hence find the values of A and k. Give each value correct to 1 significant figure.[5]
- (c)Find the value of x for which y = 17.[2]
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 (May/June 2024). The mark or mark type is shown in brackets.
- [B2] Points plotted at x 15 30 45 60 75 lny 2.3 2.5 3.1 3.5 3.9 or or 2.6 3.6
- [B2] B1 for at least 4 correctly plotted points
- [B2] soi
- [B2] and single, ruled, straight line of best fit drawn
- [M1] lny = 0.03x + 1.8
- [M1] for m = awrt 0.02 to awrt 0.03
- [M1] or c = awrt 1.7 to awrt 2.0
- [M1] or for the straight line form in terms of lnA and k: lny = lnA + kx[lne]
- [B3] A = 6 or A = 7
- [B3] Must have been found using linear points or linear equation
- [B3] B2 for A = 6 or A = 7
- [B3] or A in range: awrt 6 or awrt 7
- [B3] or B1 FT for lnA = their 1.8 or A = e^(their1.8)
- [B3] and k = 0.03 or k = 0.02
- [B3] B1 FT for k = 0.03 or 0.02 or k in range: awrt 0.02 or awrt 0.03
- [B3] Maximum of 2 marks if one or both values not rounded to 1 sf
- [B3] If B0 scored, award SC1 for A = 6 or A = 7 and SC1 for k = 0.03 or k = 0.02 found not using transformed data
- [M1] A value of x in range 33 ≤ x ≤ 37.5 nfww, isw
- [M1] for lny = 2.8 or 2.83[32...]
- [M1] OR A value of x in range 29.5 < x <33 or 37.5 ≤ x < 45 nfww
- [M1] OR STRICT FT for x = (ln17 - their lnA) / their k
- [M1] OR STRICT FT for x = (1/their k) * ln(17/their A)
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