Question 7:
--- 7(a) ---
7(a) | d x  1 
= k − 3 x  dx= dt oe
d t k−3x | M1
1 ( ) ( )
− ln k−3x =t +c
3 | M1
A1
1
t = 0 , x = 0  − l n k = c
3 | M1
( ) k − 3 x 3
− l n k − 3 x = 3 t − l n k  l n = − 3 t  1 − x = e − 3 t  x = . . .
k k | M1
k ( )
x = 1 − e − 3 t
3 | A1
(6)
(b) | k k
t →e−3t 0 x→  = 7  k = ...
3 3 | M1
k = 21 | A1
(2)
(c) | ( )
x = 5  7 1 − e − 3 t = 5  e − 3 t = ... | M1
1 2
 t = − l n
3 7 | dM1
Awrt 0.42 seconds | A1
(3)
(11 marks)
7(a) | x t
d x  1 
= k − 3 x  d x = d t oe
d t k − 3 x
0 0 | M1
x
 1 ( )  t
 − l n k − 3 x =  t 
3 0
0 | M1
A1
1 ( ) 1
− ln k−3x + lnk =t
3 3 | M1
( ) k k
l n k − l n k − 3 x = 3 t  l n = 3 t  = e 3 t  k = k e 3 t − 3 x e 3 t  x = ...
k − 3 x k − 3 x | M1
k ( )
x = 1 − e − 3 t
3 | A1
A1: Fully correct integration, with or without the + c for this mark.
A missing bracket should be penalised unless recovered later in the solution
M1: Attempts to find the constant of integration following any attempt at integration to an equation
involving a constant. This involves t = 0 , x = 0  c = . . . where c is non zero. For a definite integral we
would require the x = 0 to be clearly substituted to obtain a constant term.
M1: Correct log work to make x the subject. May be scored before the previous M if the constant is found
later. Allow if the constant is zero, or is missing. Condone slips on signs.
You may see versions of the following showing correct exponential work:
1 ( ) ( ) k−Ae−3t
− ln k−3x =t+cln k−3x =−3t+d k−3x=e−3t+d k−3x= Ae−3t  x=
3 3
( )
k e3t −1
A1: Correct answer or exact equivalent. E.g. x= .
3e3t
(b)
M1: Correctly identifies e − 3 t → 0 (which may be implied) and deduces the long-term limit for their
expression and sets their long-term limit to 7 and solves for k. The equation must be of the form
x= p−qe−3twith p in terms of k and q > 0. Alternatively, if they have an equation of the form
( )
k e3t −
x= they would need to solve
e3t
7 =
k
to find the value of k.

Condone work on equations of the same form where they have lost the 3 and used another positive value,
( )
k et −
usually 1. So, score for equations x= p−qe−t or x= 0 with the same conditions.
et
A1: k = 21 following a correct equation.
Correct answer following correct equation with no working scores both marks
Alt (b)
M1: Correctly identifies
d
d
x
t
= k − 3 x = 0 and substitutes in x = 7 leading to a value for k
A1: k = 21
(c)
M1: Substitutes x = 5 and their value for k into their equation from (a), then rearranges to e  3 t = . . .
This is dependent upon having a solvable equation of the form x =  p  q e − 3 t o.e such as x =
 e
e
t 3
t 3
  
Condone work on equations of the same form where they have lost the 3 and used another positive value,
et 
usually 1. So, score for equations x=pqe−t or x= 0 with the same conditions.
et
dM1: Solves an equation of the form e  3 t = k , k  0 to find a value for t. Condone work in solving
equations of the form et =k,k 0following the loss of the 3
A1: cso 0.42 (seconds)
Alt (c)
M1: Substitutes x = 5 into their solvable equation of the form  l n
(
k − 3 x
)
= t  l n k
PMT
  with their k
dM1: ...And finds a value for t
A1: cso awrt 0.42 (seconds)
Question | Scheme | Marks