The curve with equation y = e^2x ln(x – 1) has a stationary point when x = p. (i) Show that p satisfies the equation x = 1 + exp exp(2(x-1)) where exp(x) denotes ex. (ii) Verify by calculation that p lies between 2.2 and 2.6. (iii) Use an iterative formula based on the equation in part (i) to determine p correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.

1. [M1] Use correct product rule
2. [A1] Obtain correct derivative in any form dy/dx = -2e^-2x ln(x-1) + e^-2x / (x-1)
3. [A1] Equate derivative to zero and derive x = 1 + e^(e^(-2x)/(x-1)) or p = 1 + e^(e^(-2p)/(p-1)) AG
4. [M1] Calculate values of a relevant expression or pair of relevant expressions at x = 2.2 and x = 2.6
5. [A1] Complete the argument correctly with correct calculated values
6. [M1] Use the iterative process p_n+1 = 1 + exp(1 / (2(p_n - 1))) correctly at least once
7. [A1] Obtain final answer 2.42
8. [A1] Show sufficient iterations to 4 d.p. to justify 2.42 to 2 d.p., or show there is a sign change in the interval (2.415, 2.425)