Question 9:
--- 9(a) ---
9(a) | Finds a difference between 2
terms | AO3.1a | M1 | 3e p – 5 = 5 – 3e–p (*)
3e p – 10 + 3e–p = 0
3e2p – 10ep + 3 = 0
1
e p= , 3
3
1
p = ln , ln 3
3
ALT to (*)
2(53ep)3ep 3ep
Or
2(3ep 5)3ep 3ep
Forms an equation using two
differences | AO3.1a | M1
ep
Forms a quadratic equation in | AO1.1a | M1
Obtains a correct quadratic
equation | AO1.1b | A1
ep
Obtains 2 correct solutions for
from ‘their’ quadratic
FT only applies if previous mark
has been awarded | AO1.1b | A1F
Obtains final answers in an exact
form
FT applies if previous mark has
been awarded | AO2.2a | A1F
Q | Marking Instructions | AO | Marks | Typical Solution
--- 9 (b) ---
9 (b) | Finds a ratio between two
consecutive terms
(no requirement to use a and r) | AO3.1a | M1 | Assume it is possible that
3e–q 3eq
, 5 and are three
consecutive terms of a
geometric sequence
a 3eq, ar 5, ar2 3eq
ar 5 5eq
 r 
a 3eq 3
ar2 3eq 3eq
 r 
ar 5 5
5 3eq
 259
3eq 5
This is a contradiction therefore
3e–q 3eq
, 5 and cannot form
three consecutive terms of a
geometric sequence.
Compares two ratios
(could be ratios of successive
terms, no requirement to use a
and r) | AO3.1a | M1
Identifies a contradiction | AO2.1 | R1
Draws a conclusion about k | AO2.4 | R1
Total | 10
Q | Marking Instructions | AO | Marks | Typical Solution