Mathematics · Oct / Nov 2021 · Paper 41 · Question 3
YMA01
As printed in the paper
The question as text
3. x + x x + x + 3 ( 3 )
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
3(a)[5]
Attempts a correct identity
(
)
3
x
3
+
8
x
2
−
3
x
−
6
≡
Ax
+
B x x
(
+
3)
+
C x
(
+
3)
+
Dx
o.e.
x
0,
x
3
C
...,
D
...
Correct method for two constants e.g.
=
= −⇒
=
=
Two correct constants e.g.
C = − and
2
D =
2
Correct method to find all four constants (e.g. by comparing coefficients)
A
=
3,
B
= −
1,
C
= −
2,
D
=
2
2
2
3(b)[2]
g( )
x
=
3
x
−−
1
+
x
3
x
+
2
2
⇒
g ( )
′
x
=
3
+
−
x
2
(
)
2
x
+
3
2
2
2
2
Explains that (if x > 0)
>
so
−
>
0
and
⇒
g ( )
′
x
>
3
3(c)[1]
x
2
(
)
2
x
2
(
)
2
x
+
3
x
+
3
3 (a)
Via division
ALT
Attempts to divide
3
x
3
+
8
x
2
−
3
x
−
6
by
x
2
+
3
x
forming a linear quotient
Correct method for two constants implied by quotient of 3
x +
...
Correct quotient 3
x −
1
rem
Correct method to find all four constants (e.g. uses PF on (
)
x x +
3)
A
=
3,
B
= −
1,
C
= −
2,
D
=
2
(a)
M1: Establishes a method of starting the problem
either by attempting a correct identity
x
2
+
3
x
or attempting to divide the cubic by
to obtain a linear quotient
Condone slips in the cubic expression but the intention must be correct
dM1: Attempts a correct method to find two constants. An attempt must be made at a correct equation
For the main method this could be via substituting both
x
=
0,
x
= −⇒C and D
3
They may set up 4 simultaneous equations and solve to find two of the constants.
FYI the equations are A =3, 3A+B = 8, 3B+C+D = −3 and 3C = −6
For division it will be arriving at a linear quotient of the form 3
x ±
...
A1: Achieves any two correct constants.
Allow 3
x −as the quotient via the alt method. Condone this just written down, i.e A = 3, B = −1
1
ddM1: Attempts a full method to find all four constants.
In the main method this may be achieved by substituting both
x
=
0,
x
= − followed by comparing
3
coefficients of
3x and
2x for instance. Other methods can involve substituting other values of x.
In the alternative method it is for dividing and then setting remainder
C
D
≡
+
x x
(
+
3)
x
x
+
3
before finding values of C and D
A1: Correct values for all four constants
A
=
3,
B
= −
1,
C
= −
2,
D
=
2
Allow for these to be embedded within the identity
(b)
C
D
...
...
M1: Attempts to differentiate and achieves
+
→
+
x
+
3
x
x
2
(
3
)
2
x
+
C
D
C
D
Ax
B
A
−
−
A1ft: Differentiates
+
+
+
to
following through on their constants.
x
x
+
3
2
(
)
2
x
x
+
3
Candidates should simplify their coefficients.
You may award as above with constants A, C and D instead of the numerical values.
(c)
B1: This requires a correct part (a) and a correct (b). Look for a reason and a conclusion
2
2
Accept as a minimum
−
>
0
so g ( )
′
x
>
3
x
2
(
3
)
2
x
+
2
2
>
so g ( )
′
x
3
Or
>
2
(
)
2
x
x
+
3
2 6
(
9
)
2
2
x
+
3
3
and explains that this is 3
(
)
3
Also
+
−
=
+
+ +
ve
>
(
)
x
2
x
+
3
2
x
2
(
3
)
2
x
+
1
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