Variables P and T are known to be connected by the relationship P = Abᵀ, where A and b are constants. Values of P are found for certain values of time, T.
- (a)Show that a graph of lg P against T will be a straight line.[2]
- (b)The diagram shows the graph of lg P against T. The graph passes through (0, 6) and (14, 12). Find the values of A and b.[4]
- (c)Using the graph or otherwise, find the length of time for which P is between 100 million and 1000 million.[3]
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 2023). The mark or mark type is shown in brackets.
- [B2] lg P = lg A + T lg b oe nfww Must be seen and not from wrong working
- [B1] for lg P = lg A + T lg b isw, nfww
- [B1] and correct comparison with y = mx + c soi
- [B2] A = 10⁶ oe isw and b = 10^(3/7) oe isw
- [B2] for A = 10⁶ oe isw
- [B1] correct method which could be used to find A e.g. lgA = 6 or 12 = (3/7)×14 + lg A
- [B2] for b = 10^(3/7) oe isw
- [B1] correct method which could be used to find b e.g. lgb = (12-6)/(14-0) oe or 12 = 14lgb+6
- [M1] lg P₁ = 8 and lg P₂ = 9 soi
- [M1] leading to T₁ = 4.6 to 4.8 or T₂ = 6.8 to 7.2
- [M1] If graph not used then allow M1 for substitution of their A and their b in the exponential equation as far as 10⁸/theirA = (theirb)^T and 10⁹/theirA = (theirb)^T OR substitution of their A and their b or their lgA and their lg b in the log equation lg10⁸ = their lg A + T(their lgb) or better and lg10⁹ = their lg A + T(their lgb) or better
- [A1] Difference of correct times: T₂-T₁ where T₂ = 6.8 to 7.2 T₁ = 4.6 to 4.8 Answer in range 2.2 to 2.4 nfww
- [M2] M1 for lg10⁸ = 8 and lg10⁹ = 9 and Change in lg P / Change in T = 3/7
- [A1] Answer in range 2.2 to 2.4 nfww
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