Question 10:
10 | DR
C and C intersect when 1 2 5 3 c o s h  =
5 5
c o s h 1   = − so intersection at (  ) c o s h − 1 or
3 3
()ln3 or awrt ()1.10 | B1 | 3.1a | Correct condition for
intersection of curves leading to
a correct expression or value for
an angle at a PoI
1 5
A 5 2 ( ) 2 5 ln 3 o r 2 5 c o s h 1   = − = − or
s e c to r 2 1
2 3
awrt 27.5 | B1ft | 2.2a | For finding (±) the correct area
of the sector (or half of it) either
using 1 r2 or 1   2(5)2 d .
2 2 
1
Accept lower limit of 0.
May be embedded in a
calculation for total area. | 5 5
Here c o s h 1 , c o s h 1   = − − = − or
1 2
3 3
( =−ln3 ),( =ln3 )
1 2
i.e. may later evaluate integral…
for some k
1 2 ( 3 c o s h ) 2 d o r 1 2 ( 3 c o s h ) 2 d        
2 2 0 
1 | M1 | 3.4 | Correct use of area formula with
±their value for limits. Condone
lower limit of 0 | 5 5
Here c o s h 1 , c o s h 1   = − − = −
1 2
3 3
Limits can be seen later.
Later doubling may be seen
e e 2    + − 
c o s h 2 d d     =  =
2
1
e 2 e 2 2 d    =  + − +
4 | *M1 | 3.1a | Correct conversion of cosh2
into a form which can be
integrated. | 1
or cosh2d= 1+cosh2d
2
ln 3
e 2 e 2 2 d     + − +
ln 3 −
ln 3
1 1  
2 e e 2 2    = − − +
2 2
ln 3 −
1 1 1 1 1 1  
9 2 ln 3 9 2 ln 3 =  −  + −  −  −
2 2 9 2 9 2
8 0
4 ln 3 o r a w r t 1 3 .3 = +
9 | dep*M1 | 1.1 | Correctly integrating cosh2
and substituting their limits.
Could be embedded.
NB
9 180  9
A =   +4ln3=10+ ln3
C2 2 4 9  2 | ln 3 ln 3
1
or c o s h 2 d 1 c o s h 2 d      =  +
2
ln 3 ln 3 − −
ln 3
1 1 2 0  
s in h 2 ln 3   = + = +
2 2 9
ln 3 −
 9  41
 trapped area =25ln3−10+ ln3= ln3−10
 2  2
or awrt 12.5 (12.521...) | A1 | 1.1 | Accept unsimplified form | If B1B1M1M0M0 can score ScB1
both here and in the next line for
correct answers
12.52 / 0.5 = 25.04
so 26 tins of paint are needed | A1 | 3.2a
[7]
PMT
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