Mathematics · May / June 2023 · Paper 21 · Question 4

YMA01

As printed in the paper

The question as text

4. The binomial expansion, in ascending powers of x, of 5 (3 + px) where p is a constant, can be written in the form 2 + Dx 3 ... A + Bx + Cx where A, B, C and D are constants.

Read automatically from the paper; the printed page above is exact.

Show the answer and mark scheme

4(a)[1]
−
8
+
5
+
a
=
0
a
= −
3
is M1A0
2
2
2
3
27
15
=
f
−
18
+
+
=
0
= −
3
is M1A1
a
a
2
2
2
3
=
Note the trivial case
f
3
+
=
0
= −
3
is M1A0 (not enough shown)
a
a
2
(b) Work done in (a) must be made use of in (b) to gain credit.
M1
Attempts to find a quadratic factor of
4
x
3
−
8
x
2
+
5
x
− using 2
3
x − as a factor
3
score for
2
2
±
±
...
if algebraic division is used or
x
x
score for
2
x
2
±
...
x
± if equating coefficients/inspection used.
1
3
Alternatively, they may attempt to find a quadratic factor using
as a factor so
x −
2
score for
4
x
2
±
2
x
±
...
if algebraic division is used or
score for
if equating coefficients/inspection used.
4
x
2
±
...
x
±
2
3
A1
2
+ (for 2
x − as the linear factor) or
3
2
(for
as the linear factor)
2
x
−
x
1
4
x
−
2
x
+
2
x −
2
dM1
Attempts to show their quadratic factor has no real roots. It is dependent on the previous method
mark. Factorisation attempts are M0. Accept via
an attempt at
2
−
4
for their quadratic factor so score for values correctly embedded in
b
ac
2
−
4
ac
for their quadratic
an attempt to solve their
2
x
2
−
x
+ or
1
4
x
2
−
2
x
+
2
using the quadratic formula with values
embedded or via a calculator (you must check their roots are correct for their quadratic)
2
1
an attempt to complete the square e.g.
2
±
±
...
x
4
A1*
Requires correct factorisation of the cubic, correct calculation for the quadratic, reason and
conclusion
For example, after factorising their cubic to e.g.
(2
−
3)(2
2
−
+
1)
which does not need to
x
x
x
be explicitly stated accept e.g.
3
the discriminant of
2
2
−
+ is 7
1
−<
0
so only has a root at
x =
. If correct values are
x
x
2
embedded which give 7
− then this is fine. (For
4
2
−
2
+
2
the discriminant is 28
x
x
1
7
3
2
2
−
+ =
1
0
=
±
, (2
−
3)
=
0
=
so only one real root. (The roots will be
x
x
x
i
x
x
4
4
2
the same for
4
2
−
2
+
2
=
0)
x
x
1
2
7
3
2
x
2
−
x
+ =
1
0
2
x
−
+
>
0
, hence
x =
is the only one real root (The completed
4
8
2
square form for
4
2
−
2
+
2
could be the same or a multiple of it depending on how they
x
x
have rearranged their inequality)
3
There must be some reference to the root
either by stating it or indicating the linear term has a root
x =
2
(e.g. writing “root from here” next to it), but do not accept incorrect statements such as only real root is
(2
x −
3)
. Condone “solution” for “root”
Do not allow statements such as “Math error” without interpretation of what this means.
3a(i)
r =
(8
−
2)
2
+ −−
( 3
5)
2
=
10
(
2)
2
(
5)
2
100
x
−
+
y
−
=
(ii)
4
b
Gradient between centre and P =
−
3
3
Perpendicular gradient
=
4
3
+
3
=
(
−
8)
y
x
4
3
x
−
4
y
−
36
=
0
(a)(i) Mark (i) and (ii) together
M1
Attempts to find the radius. Score for the values embedded in the expression
e.g.
(8
−
2)
2
+ −−
( 3
5)
2
or
(2
−
8)
2
+
(5
+
3)
2
. They must be attempting to find the
difference in the x coordinates and the difference in the y coordinates.
(
) 10
r =
A1
ignore any units 10
±
is A0 Do not accept 100 for this mark. isw once a correct
answer is seen.
(ii) Note that if M0 is scored in (a)(i) then this is A0 in (ii)
A1ft
o.e. e.g.
or
(
x
−
2)
2
+
(
y
−
5)
2
=
"100"
(
x
−
2)
2
+
(
y
−
5)
2
=
"10 "
2
x
2
+
y
2
−
4
x
−
10
y
=
"71"
Follow through on their positive single value of r. isw once a correct equation is seen.
(b)
4
B1
Gradient between centre and P is
(may be implied by their gradient of the tangent)
−
3
M1
Attempts to find the negative reciprocal for their gradient. May be embedded within their equation
of the tangent.
M1
Attempts to find the equation of the tangent at P using a changed gradient (cannot just be a
stated gradient of the tangent which is incorrect and does not follow from any earlier work)
and (8,
−
3)
3
Score for
+
3
=
" "(
−
8)
. The coordinates (8,
−
3)
must be correctly placed.
y
x
4
If they use y
=
mx
+
c
they must proceed as far as
c =
...
A1
3
−
4
−
36
=
0
o.e. (provided all the coefficients are integers and all terms are on one side
x
y
of the equation) isw once a correct equation is seen.
Alt (b)
Alternative method using differentiation
d
y
4
−
2
x
d
y
B1
=
o.e. such as
2(
2)
2(
5)
0
or
x
−
+
y
−
=
d
2
10
x
y
−
d
x
(
)
1
d
y
(
)
1
(may be implied by 3
−
e.g.
=
100
−
(
−
2)
2
= −
(
−
2) 100
−
(
−
2)
2
4 )
y
x
2
x
x
2
d
x
Attempts to find the gradient of the tangent at P using their d
y
M1
and the given coordinate(s).
d
x
Do not be concerned with poorly differentiated equations for this mark and just look for the x (and
possibly y) coordinate(s) to be substituted into a differential equation leading to a value for the
gradient. Condone sign slips substituting in.
M1
As above in main scheme notes
A1
As above in main scheme notes
Alternative method using formula to find the equation of the tangent at a point on a circle
Using the formula
2
with centre ( , )
a b and point
( ,
)
(
x
−
a x
)(
−
a
)
+
(
y
−
b y
)(
−
b
)
=
r
x
y
1
1
1
1
B1
(
−
2)(
−
2)
+
(
−
5)(
−
5)...
(may be implied by further work)
x
x
y
y
1
1
M1
Sets equal to their radius:
(
x
−
2)(
x
−
2)
+
(
y
−
5)(
y
−
5)
=
"100"
1
1
M1
Uses the point P and sets equal to their radius: (
x
−
2)(8
−
2)
+
(
y
−
5)( 3 5)...
−−
=
"100"
A1
oe (provided all the coefficients are integers and all terms are on one side of
3
x
−
4
y
−
36
=
0
the equation)
4a
3 or 243
5
bi
(
)
(
)
=
5 3
×
4
( 405 )
=
=
10 3
×
2
3
( 90
=
3
)
B
p
p
D
p
p
1
1
B
=
18
D
=
405
=
18 90
×
3
2
=
= −
p
p
p
p
4
2
2
ii
1
135
(
)
5
3
3
C
=
C
×
×−
=
2
2
2
(a) Mark (a) and (b) together
5
px
B1
3 or 243 seen. Do not award if it is outside the bracket as
4(b)[6]
2
−
4
ac
for their quadratic
an attempt to solve their
2
x
2
−
x
+ or
1
4
x
2
−
2
x
+
2
using the quadratic formula with values
embedded or via a calculator (you must check their roots are correct for their quadratic)
2
1
an attempt to complete the square e.g.
2
±
±
...
x
4
A1*
Requires correct factorisation of the cubic, correct calculation for the quadratic, reason and
conclusion
For example, after factorising their cubic to e.g.
(2
−
3)(2
2
−
+
1)
which does not need to
x
x
x
be explicitly stated accept e.g.
3
the discriminant of
2
2
−
+ is 7
1
−<
0
so only has a root at
x =
. If correct values are
x
x
2
embedded which give 7
− then this is fine. (For
4
2
−
2
+
2
the discriminant is 28
x
x
1
7
3
2
2
−
+ =
1
0
=
±
, (2
−
3)
=
0
=
so only one real root. (The roots will be
x
x
x
i
x
x
4
4
2
the same for
4
2
−
2
+
2
=
0)
x
x
1
2
7
3
2
x
2
−
x
+ =
1
0
2
x
−
+
>
0
, hence
x =
is the only one real root (The completed
4
8
2
square form for
4
2
−
2
+
2
could be the same or a multiple of it depending on how they
x
x
have rearranged their inequality)
3
There must be some reference to the root
either by stating it or indicating the linear term has a root
x =
2
(e.g. writing “root from here” next to it), but do not accept incorrect statements such as only real root is
(2
x −
3)
. Condone “solution” for “root”
Do not allow statements such as “Math error” without interpretation of what this means.
3a(i)
r =
(8
−
2)
2
+ −−
( 3
5)
2
=
10
(
2)
2
(
5)
2
100
x
−
+
y
−
=
(ii)
4
b
Gradient between centre and P =
−
3
3
Perpendicular gradient
=
4
3
+
3
=
(
−
8)
y
x
4
3
x
−
4
y
−
36
=
0
(a)(i) Mark (i) and (ii) together
M1
Attempts to find the radius. Score for the values embedded in the expression
e.g.
(8
−
2)
2
+ −−
( 3
5)
2
or
(2
−
8)
2
+
(5
+
3)
2
. They must be attempting to find the
difference in the x coordinates and the difference in the y coordinates.
(
) 10
r =
A1
ignore any units 10
±
is A0 Do not accept 100 for this mark. isw once a correct
answer is seen.
(ii) Note that if M0 is scored in (a)(i) then this is A0 in (ii)
A1ft
o.e. e.g.
or
(
x
−
2)
2
+
(
y
−
5)
2
=
"100"
(
x
−
2)
2
+
(
y
−
5)
2
=
"10 "
2
x
2
+
y
2
−
4
x
−
10
y
=
"71"
Follow through on their positive single value of r. isw once a correct equation is seen.
(b)
4
B1
Gradient between centre and P is
(may be implied by their gradient of the tangent)
−
3
M1
Attempts to find the negative reciprocal for their gradient. May be embedded within their equation
of the tangent.
M1
Attempts to find the equation of the tangent at P using a changed gradient (cannot just be a
stated gradient of the tangent which is incorrect and does not follow from any earlier work)
and (8,
−
3)
3
Score for
+
3
=
" "(
−
8)
. The coordinates (8,
−
3)
must be correctly placed.
y
x
4
If they use y
=
mx
+
c
they must proceed as far as
c =
...
A1
3
−
4
−
36
=
0
o.e. (provided all the coefficients are integers and all terms are on one side
x
y
of the equation) isw once a correct equation is seen.
Alt (b)
Alternative method using differentiation
d
y
4
−
2
x
d
y
B1
=
o.e. such as
2(
2)
2(
5)
0
or
x
−
+
y
−
=
d
2
10
x
y
−
d
x
(
)
1
d
y
(
)
1
(may be implied by 3
−
e.g.
=
100
−
(
−
2)
2
= −
(
−
2) 100
−
(
−
2)
2
4 )
y
x
2
x
x
2
d
x
Attempts to find the gradient of the tangent at P using their d
y
M1
and the given coordinate(s).
d
x
Do not be concerned with poorly differentiated equations for this mark and just look for the x (and
possibly y) coordinate(s) to be substituted into a differential equation leading to a value for the
gradient. Condone sign slips substituting in.
M1
As above in main scheme notes
A1
As above in main scheme notes
Alternative method using formula to find the equation of the tangent at a point on a circle
Using the formula
2
with centre ( , )
a b and point
( ,
)
(
x
−
a x
)(
−
a
)
+
(
y
−
b y
)(
−
b
)
=
r
x
y
1
1
1
1
B1
(
−
2)(
−
2)
+
(
−
5)(
−
5)...
(may be implied by further work)
x
x
y
y
1
1
M1
Sets equal to their radius:
(
x
−
2)(
x
−
2)
+
(
y
−
5)(
y
−
5)
=
"100"
1
1
M1
Uses the point P and sets equal to their radius: (
x
−
2)(8
−
2)
+
(
y
−
5)( 3 5)...
−−
=
"100"
A1
oe (provided all the coefficients are integers and all terms are on one side of
3
x
−
4
y
−
36
=
0
the equation)
4a
3 or 243
5
bi
(
)
(
)
=
5 3
×
4
( 405 )
=
=
10 3
×
2
3
( 90
=
3
)
B
p
p
D
p
p
1
1
B
=
18
D
=
405
=
18 90
×
3
2
=
= −
p
p
p
p
4
2
2
ii
1
135
(
)
5
3
3
C
=
C
×
×−
=
2
2
2
(a) Mark (a) and (b) together
5
px
B1
3 or 243 seen. Do not award if it is outside the bracket as

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