Question 9:
--- 9(a) ---
9(a) | y=x3 −3x+3 and y=2x3−4x2+3 x3−4x2+3x=0  | M1 | Reducing to 3-term cubic or quadratic if x cancelled.
x(x−1)(x−3)=0  | DM1 | Factorising the cubic or quadratic.
x=0, 1 and 3 {x = 0 may be seen in the working} | A1 | SC B1 for x=1, 3 only, with no M marks awarded.
3
Question | Answer | Marks | Guidance
--- 9(b) ---
9(b) | Attempt at integration of both functions. Can be before or after subtraction
of the functions or integrals | M1 | Expect integration of  (( x3−3x+3 ) − ( 2x3−4x2 +3 )) dx or
 ( −x3+4x2 −3x ) dx.
At this stage, subtraction can be done either way.
 x4 4x3 3x2  x4 3  2 4 
=− + −  or  − x2 +3x− x4 − x3+3x
 4 3 2   4 2  4 3  | A1 | OE
± covers A1 being awarded to those who subtract the ‘other’
way.
 81 108 27  1 4 3
= − + −  −  − + − ,
 4 3 2   4 3 2
or
81 27  1 3  81 108  1 4 
 − +9  −  − +3  − − + 9  −  − +3
 4 2  4 2   2 3  2 3  | DM1 | OE
63 7 27 13
Minimum required is  − − − , i.e. four fractions.
 4 4  2 6 
Correctly apply limits their 1 and 3.
Do not allow if x=0 used.
Need at least one correct substitution in every bracket.
If two integrals, need to see substitution into both.
Allow one sign error only in each expression, if brackets are not
shown.
8
=
3 | A1 | Accept if this comes from use of limits f (1)− f (3) or
 ( x3−4x2 +3x ) dx, if −8 used.
3
Only dependent on the first method mark.
Accept AWRT 2.67.
4
Question | Answer | Marks | Guidance