By sketching a suitable pair of graphs on the same diagram, show that the equation ln x = 2e^(–x) has exactly one root.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2021). The mark or mark type is shown in brackets.

1. [B1] Draw correct sketch of y = ln x or y = 2e^(-x) y = ln x must extend into 1st and 4th quadrants. y = 2e^(-x) must extend into 1st and 2nd quadrants.
2. [B1] Draw correct sketch of second curve and indicate one root Point of intersection must be circled/identified or a statement such that ‘there is only one point of intersection so one root only’ or similar.
3. [M1] Consider sign of ln x – 2e^(-x), or equivalent, for 1.5 and 1.6
4. [A1] Obtain –0.04... and 0.06... or equivalents and justify conclusion
5. [B1] Replace x_(n+1) and x_n by x and apply logarithms to confirm result AG Allow if done 'in reverse' but x_n and x_(n+1) need to be seen in the final statement.
6. [M1] Use iteration process correctly at least once Need to see 3 correct values.
7. [A1] Obtain final answer 1.54 Answer required to exactly 3sf
8. [A1] Show sufficient iterations to 5sf to justify answer or show sign change in interval [1.535, 1.545]