The curve with equation y = 5e2x – 8x2 – 20 crosses the x-axis at only one point. This point has coordinates (p, 0).

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2018). The mark or mark type is shown in brackets.

1. M1 Rearrange at least as far as 2x = ln(...)
2. A1 Obtain x = 1/2 ln(1.6x² + 4)
3. Allow if in terms of p, need to see y equated to 0
4. AG; necessary detail needed
5. M1 Consider sign of x −1/2ln(1.6x² + 4) for 0.75 and 0.85 or equivalent
6. A1 Obtain -0.04 and 0.03 or equivalents and justify conclusion
7. Need to see substitution of numbers
8. AG; necessary detail needed, change of sign or equivalent must be mentioned
9. M1 Consider sign of 5e2x – 8x2 – 20 for 0.75 and 0.85
10. A1 Obtain -2.09... and 1.58... or equivalents and justify conclusion
11. Need to see substitution of numbers
12. AG; necessary detail needed, change of sign or equivalent must be mentioned
13. M1 Use iteration process correctly at least once
14. A1 Obtain final value 0.80956
15. A1 Show sufficient iterations to justify value or show sign change in interval (0.809555, 0.809565)
16. Starting with value such that iterations converge to correct values
17. Must be 5sf for the final answer
18. B1 Obtain first derivative 10e2x -16x
19. M1 Substitute value from iteration to find gradient, must be in the form pe2x + qx
20. A1 Obtain 37.5
21. Or greater accuracy, allow awrt 37.5 from use of x = 0.8096, 0.80955 oe