O-LevelAdditional MathematicsLogarithmic and exponential functions4037/21 · May/June 2017 · Q1011 Marks

The table shows values of the variables t and P. t: 1, 1.5, 2, 2.5 P: 4.39, 8.33, 15.8, 30.0

(i)Draw the graph of In P against t on the grid below. [Figure 10.1][2]
(ii)Use the graph to estimate the value of P when t = 2.2.[2]
(iii)Find the gradient of the graph and state the coordinates of the point where the graph meets the vertical axis.[2]
(iv)Using your answers to part (iii), show that P = abᵗ, where a and b are constants to be found.[3]
(v)Given that your equation in part (iv) is valid for values of t up to 10, find the smallest value of t, correct to 1 decimal place, for which P is at least 1000.[2]

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Mark scheme

Marking points from the official mark scheme (May/June 2017). The mark or mark type is shown in brackets.

[M1] allow lnP values to 1 dp rounded or truncated (1.5, 2.1, 2.8, 3.4)
[A1] All points within 1 square of line / must not pass through origin
[M1] e^(their y-value from graph at t=2.2)
[A1] 18 to 22.2
[B1] (0, c) with 0.1 ≤ c ≤ 0.3 (0.2) allow y = c condone c=...
[B1] m in the range 1.25 < m < 1.34 (1.28)
[M1] ln P = (their 1.28)t + their 0.2 or ln P = (ln b)t + ln a
[M1] P = e^(their 1.28)t+their 0.2 or ln b = m = their 1.28 and ln a = c = their 0.2
[A1] P = e^(their 0.2) (e^(their 1.28))t or 1.10 <= a <= 1.35 3.49 <= b <= 3.82
[M1] 1000 >= e^(their 0.2) * e^(their 1.28t) or 1000 >= their a × their b^t A correct relationship e.g. 1.3t *ln(1000) -0.2 where * is = or an inequality sign
[A1] 5.2 to 5.5 must be to 1dp

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