Question 3:
--- 3(a) ---
3(a) | 2 6  x
 kx2dx k 1 dx1
 
0 2  6 | M1: for adding the two integrals, and
attempting to integrate, at least one
xn xn1
integral , ignore limits and
does not need to be put equal to 1. Do
not award if they add before
integrating | M1 A1
2 6
x3  x2
k k x 1
   
 3   12
0 2 | A1: correct integration, ignore limits
and does not need to be put equal to 1
8  5
k k 3 1
   
3  3 | M1: dependent on first M being
awarded, correct use of limits and
putting equal to 1.
8
This may be seen as F(2) = k and
3
using F(6) = 1 | dM1
A1cso
4k 1 | A1: cso answer given so need 4k 1
1
leading to k 
4
1
k  *
4
NB Validation – if they substitute in k = ¼ you may award the 1st three marks as per scheme. For the Final A
mark they must say “ therefore k = ¼”
(b) | 2 | B1: cao | B1
(c) | kx3
x
 kt2dt 
0 3 | x
M1: attempting to find  kt2dt
0
t2 t3
, ignore limits, may leave in
terms of k | M1
 t   t2 
 k 1 dt k t C
   
 6  12
t2
 ktk C
12
F(6) =1
1
6k3kC1  C =
4 | t
M1: attempting to find  k(1 )dt
6
tn tn1
at least one integral and
either
have + C C 0and use F(6) =1
or
have limits 2 and x and + “their
2
 kt2dt” and attempt to integrate
0
tn tn1
NB: may use any letter, need not be t
,condone use of x | M1
 0 x0

x3

0 x2
 12
F(x) 
x x2 1

  2 x6
4 48 4

 1 x6 | A1: second line correct
A1: third line correct
B1: first and fourth line correct they
may use “otherwise” instead of x < 0
or x > 6 but not instead of both | A1
A1
B1
NB: Condone use of < rather than  and vice versa
(d) | x x2 1
  0.75
4 48 4 | M1: putting their line 2 or their line 3
= 0.75 | M1 A1
x212x240
oe | A1: The correct quadratic equation –
like terms must be collected together
12 144424
x 
2 | M1d: dep on previous M1 being
awarded. A correct method for
solving a 3 term quadratic equation =
0 leading to x = …Use either the
quadratic formula or completing the
square - If they quote a correct
formula and attempt to use it, award
the method mark if there are small
errors. Where the formula is not
quoted, the method mark can be
implied from correct working with
values but is lost if there is a mistake.
If they attempt to factorise award M1
if they have
 x2 bxc  (x p)(xq),
where| pq||c|leading to x...
May be implied by a correct value for
x | dM1 A1
= 2.54 or 62 3 | A1: awrt 2.54 or 62 3 or
6 12 . If 2 values for x are given
they must eliminate the incorrect one.
Notes