Question 12:
12 | 1 2 1 0 2     1 0 2    
k ( c o s 1 ) d   + + or k ( c o s 1 ) d   + +
2 k k
0 0
1 2 ( ) 1 0 1 0 0    
k 2 c o s 2 2 c o s 1 + 2 k ( c o s + 1 ) + d     = + +
2 k k 2
0
12 k2  10  102
=  ( cos2+1 ) +2kk+ cos+k+  d
2 2  k   k 
0
1 k 2 1 1 0 k 2 1 0 2 2           
s i n 2 2 k k s i n k    = + + + + +
2 2 2 k 2 k
0
k 2 1 0 k 2 1 0 2      
s i n 4 k k s i n 2 2 k    = + + + + +
8 k 2 k
3 k 2 1 0 0  
2 0  = + +
2 k 2
d 3 k 2 1 0 0 2 0 0      
2 0 3 k ( 0 )   + + = − =
k d 2 k 2 k 3
1
 2 0 0  4
k =
3 | M1*
M1
M1dep*
M1
A1
M1
A1
[7] | 3.1a
1.1
1.1
1.1
1.1
2.1a
1.1 | Using 1 r2d with correct expression for r - condone
2
lack of (or incorrect) limits. Condone missing 12 only.
1
Expanding and using correct c o s 2 (1 c o s 2 )   = + -
2
need not be in an integral.
Integrating their k c o s 2 k c o s k d     + + correctly
1 2 3
to 1k sin2+k sin+k with k ,k ,k 0
2 1 2 3 1 2 3
Substitute correct limits 0 and 2 - dependent on
previous M mark. Need not see zeros from 0 limit or
other terms that would be zero when evaluated.
cao from correct integral and correct integration.
Condone ( 3 k 2 4 0 2 0 0 k 2 )  + + − from missing 1.
2
Differentiating theirA correctly which must be of the
k
form a k 2 + b + c k − 2 with a , b , c  0 . Dependent on all
previous M marks.
Dependent on all previous marks and no errors. A0 if
correct k obtained but 12 missing from area formula
(unless justified). Accept any exact equivalent.
Condone omission of (or other constant factor)
for final two marks.
PMT
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