Question 9:
--- 9(a) ---
9(a) | dV k dV 1
Obtain  or 
dt t dt kt | B1
dV
Obtain 20x3x2
dx | B1
Correct use of chain rule involving k | M1 | dV dV dx
Use =  .
dt dx dt
dV dV
Expressions for and must be seen to get M1.
dt dx
dx k
Obtain  or equivalent,
dt t  20x3x2 | A1 | If this expression is first seen with numerical values, allow A1
when their value of k is substituted back into the general
expression.
1 1 dx 20
Use t  , x and  to obtain given answer which
10 2 dt 37
must be stated
dx 20
 needed to score final A1
dt 37 | A1 | dx 1
 AG
dt 2t  20x3x2
20 k k
Need to at least see  = if
37 1  3 t
10 
10 4
20 k k
or  = if  in working for correct k.
37 1  3 t
10 
10 4
dx 20
 seen anywhere, then A0.
dt 37
5
Question | Answer | Marks | Guidance
--- 9(b) ---
9(b) | Separate variables correctly & integrate at least one side correctly | B1
Obtain terms 10x2 – x3 | B1 | May see –10x2 + x3 if negative sign moved across
or e.g. 20x2 – 2x3 if 2 moved across.
20x2 3x3
Allow  .
2 3
Obtain term lnt with ‘correct’ coefficient from their separation of
a
variables, for example alnt for .
t | B1FT | FT sign and position of 2 from their separation but B0 if error from
later manipulation.
1 1
Use t  , x to evaluate a constant or as limits in a solution
10 2
containing terms of the form x2, x3 and lnt (or ln2t) | M1 | Allow numerical and sign errors and decimals.
Allow if exponentiate before substitution, even if exponentiation
done incorrectly, allow for c or ec.
Obtain correct answer in any form, for example
lnt 19 ln0.1
10x2 x3   
2 8 2 | A1 | ln2t 19 ln0.2
10x2 x3   
2 8 2
lnt
or 10x2 x3  2.5  0.1251.15
2
Allow 1.14 to 1.16 for 1.15 and allow 2.44 to 2.46 for 2.45
2x320x2 19
Obtain answer t  1 e 4 or equivalent
10 | A1 | ISW
Need t =………
2x3 1 9
0.1 e 4 1 19
e4e2x320x2
E.g. , , .
20x22x3 19 10e20x2 10
e 4
e2x320x22.45.
Allow decimals, allow 2.44 to 2.46 for 2.45, e.g.
1
ln
A0 if e 10 present in final answer.
6
Question | Answer | Marks | Guidance