The variables x and y satisfy the equation y = Ae^(A-B)x, where A and B are constants. The graph of ln y against x is a straight line passing through the points (0.4, 3.6) and (2.9, 14.1), as shown in the diagram.

Mark scheme

Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.

1. [B1] State or imply equation In y = ln A + (A - B)x
2. [M1] Equate A-B to gradient of line
3. [A1] Obtain A- B = 4.2
4. [M1] Substitute appropriate values to find value of In A
5. [A1] Obtain In A = 1.92 and hence A = 6.82 and B = 2.62
6. Alternative Method 1:
7. [B1] State or imply equation In y = ln A + (A - B)x
8. [M1] Use of coordinates to obtain equation of line (ln y-3.6)/(14.1-3.6) = (x-0.4)/(2.9-0.4) condone use of y in place of In y .
9. [A1] Obtain gradient equal to 4.2
10. [M1] Substitute appropriate values to find value of In A
11. [A1] Obtain In A = 1.92 and hence A = 6.82 and B = 2.62
12. Alternative Method 2:
13. [B1] State or imply equation In y = ln A + (A - B)x
14. [M1] For one correct equation. (3.6 = ln A + 0.4(A - B))
15. [A1] For both correct equations. (14.1 = ln A + 2.9(A-B))
16. [M1 A1] For attempt at solution by elimination of In A to obtain both values. SC B1 for y = 4.2x+1.92 .