Question 8:
--- 8(a) ---
8(a) |  d y  C
= A x 3 + B +
d x x 3 | M1
 d y  5 43
= 4  2 5 6 x 3 − 3 0 4 − 2  2 7 x − 3 oe e.g. 1 0 2 4 x 3 − 3 0 4 −
d x x | A1A1
(3)
(b) | d y 54
= 0 at SP 1024x3−304− =01024x6 −304x3−54=0
d x x3 | M1
 5 1 2 ( x 3 ) 2 − 1 5 2 x 3 − 2 7 = 0  ( 8 x 3 + 1 ) ( 6 4 x 3 − 2 7 ) = 0  x 3 = ... | dM1
1 2 7 1 3
x 3 = − or x 3 = or x = − or x =
8 6 4 2 4 | A1
4
 1   1  2 7
y = 2 5 6 " − " − 3 0 4 " − " − 3 5 + = . . . or with x = 3
2 2 ( " − 1 " ) 2 4
2 | dM1
 1   3 
Coordinates are − , 2 4 1 and , − 1 3 4
2 4 | A1
(5)
(8 marks)
Notes:
(a)
M1: Attempts the derivative, with at least two terms of the correct form (ie xn →..xn−1 at least
twice).
A1: At least two terms correct, need not be simplified.
A1: Fully correct derivative, need not be simplified. Isw after a correct (unsimplified) answer.
d y
Note that there is no need to see the " = " in (a) just look for the differentiation.
d x
Ignore any spurious “= 0”
(b)
M1: Sets their derivative equal to zero and multiplies through by “x3” to achieve a polynomial
equation. If in doubt at least 2 terms must be multiplied.
Allow equivalent work e.g.
1 0 2 4 x 3 − 3 0 4 − 5 43 = 0  1 ( 1 0 2 4 x 6 − 3 0 4 x 3 − 5 4 ) = 0  1 0 2 4 x 6 − 3 0 4 x 3 − 5 4 = 0
x x 3
To score this mark, the derivative must have a negative power of x so allow for e.g.
dy 
 =1024x3−304−54x−1 =01024x4 −304x−54=0
dx 
dM1: Solves a 3 term quadratic in x3 by any valid means (including calculator).
dy d2y 
Must come from an attempt at =0 not =0 or ydx=0
dx dx2
Condone use of inequalities rather than “=” as long as they solve an equation.
Note that here we accept e.g. 5 1 2 ( x 3 ) 2 − 1 5 2 x 3 − 2 7 = 0  ( 8 y + 1 ) ( 6 4 y − 2 7 ) = 0  y = ...
Or even 512 ( x3)2 −152x3−27=0(8x+1)(64x−27)=0 x=...
A1: Achieves at least one correct value for x or x3. It must be clear that they are values for x3 here so
y = … is acceptable if y = x3 is seen or implied. If they have x = … then they must have cube
rooted. So e.g. ( 8 x + 1 ) ( 6 4 x − 2 7 ) = 0  x = −
1
8
unless they recover and recognise they have x3
then this is A0
dM1: Proceeds to find the y coordinate for at least one value of x. Must cube root and not e.g.
square root the solution from the quadratic first. The x must have come from a 2 or 3 term
quadratic in x3.
If no method shown, accept any value for y having found an x value but score M0 if there is
no evidence of cube rooting or clear evidence that they have substituted into something other
than the curve equation but condone if they clearly just mis-copy the equation.
Depends on having scored at least 1 of the previous M marks in (b)
A1: Both correct pairs of coordinates. Accept if given as e.g. x = −
1
2
, y =241 and x =
3
4
, y = − 1 3 4
And no other values.
Allow equivalent fractions or values for the −
1
2
and
3
4
PMT
Question | Scheme | Marks