Question 9:
--- 9(a) ---
9(a) | 1 A B
 + 1 A ( 2x−1 )+Bx A=...or B=...
x ( 2x−1 ) x 2x−1 | M1
1 2 1
 −
x ( 2x−1 ) 2x−1 x | A1
(2)
(b) | d h 1  t   1  1  t 
= h ( 2 h − 1 ) c o s  d h = c o s d t
d t 5 0 1 0 h ( 2 h − 1 ) 5 0 1 0 | M1
 1   2 1 
d h = − d h = l n ( 2 h − 1 ) − l n h
h ( 2 h − 1 ) 2 h − 1 h | M1
  2 1   1  t  1  t 
− d h = c o s d t  l n ( 2 h − 1 ) − l n h = s i n ( + c )
2 h − 1 h 5 0 1 0 5 1 0 | A1ft
e.g. t = 0 , h = 2 . 5  ln ( 22.5−1 )−ln2.5=c or
2 h − 1 1  t  2 h − 1 15 sin  t1 
e.g. l n = s i n + c  = X e 0 t = 0 , h = 2 . 5  X = 1 . 6
h 5 1 0 h | M1
1  t  2 h − 1 15 sin  t1 
l n ( 2 h − 1 ) − l n h = s i n + l n 1 . 6  = 1 . 6 e 0
5 1 0 h
15  t 
sin 
 2 h − 1 .6 h e 1 0  = 1  h = ... | M1
5
h =
15  t1 
s in
1 0 − 8 e 0 | A1
(6)
(c) | ( )
t t 5 9   
s in 1 , , , ... t ... =  =  =
1 0 1 0 2 2 2 | M1
t 4 5 1 4 1 ( s )  = = | A1
(2)
Total 10
(b) Note that they may work in x throughout, but the final answer must be h = . . .
M1: Attempts to separate the variables. Do not be concerned by slips with the constants so look
 1 
for as a minimum = cos ( ...t ) o.e. May omit an integral sign on one of the sides
h ( 2h−1 )
or may be implied by their integrated expression.
M1: For
A
h 2 h
B
1
d h l n h l n ( 2 h 1 )  
 
+
−

= + −
or e.g.
" A
h
"
2
"
h
B "
1
d h l n h l n ( 2 h 1 )   
 
+
−

= + − (if they attempt some rearranging first)
Condone invisible brackets.
A1ft: Fully correct equation following through their A and B. The “+c” is not required here.
  A
h
+
2 h
B
− 1

d h =

5
1
0
c o s

1
t
0

d t  A l n h +
B
2
l n ( 2 h − 1 ) =
1
5
s i n

1
t
0

( + c ) o.e.
Condone invisible brackets.
M1: States or uses t = 0, h = 2.5 followed by c = ... is sufficient for M1. You do not need to check
their substitution/rearrangement but if there is just a value for the constant then you may need
to check they used t = 0, h = 2.5. May be implied by e.g. l n 8 − l n 5 , l n 1 . 6 , awrt 0.47
following correct integration. Note that they may have removed lns first.
M1: To score this mark they must have integrated and achieved an expression of the form
 p l n h  q l n ( 2 h − 1 ) =  r s i n

1
t
0

+ c o.e.
Rearranges using correct algebra to make h the subject including when combining logs and
when dealing with the constant. May occur before finding their constant but there must have
been a constant.
Expect to see terms in h collected on one side as part of their manipulation rather than
proceeding to the given answer once they have established the value of k. Alternatively, they
may cancel e.g.
2 h
h
− 1
→ 2 −
1
h
and rearrange to make h the subject using correct algebra.
Do not condone sign slips for this mark.
A1: Correct expression (including h = ...)
PMT
in the required form. Requires
• Correct integration
• Correct algebra
Do not penalise brackets recovered in their work or working in x rather than h until the final
answer.
(c) Note that an answer with no working seen scores M0A0
 t 
M1: Solves sin  =1 to find a value for t. Do not allow this mark to be implied by
10
t 
e.g. 5, 25, 45but allow for e.g. = t =.. Must be working in radians.
10 2
Note that attempts that do not use part (b) and just use the differential equation proceeding to
 t   t 
cos  =0 score M0A0. (Note if their k is negative condone if they solve sin  =−1)
10 10
A1: awrt 141 (but do not allow just 45) Units not required.