Mathematics · May / June 2025 · Paper 11 · Question 8

YMA01

As printed in the paper

The question as text

8. In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. A curve has equation y = f (x), x > 0 The point P (4, 12) lies on the curve. Given that • f′ (x) = 3 x + kx 2 where k is a constant • the equation of the tangent to the curve at P has equation y = 10x + c where c is a constant

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Show the answer and mark scheme

8(a)[4]
−
x
if they incorrectly try to simplify this) to reach at least a gathered cubic with no constant, or a
2
quadratic in x. The x may be cancelled, e.g 1.2x
(8
−
x
)
=
0.2
x
(
x −
4)
Quadratic equation in x. Do not
allow solutions which reach a cubic with nonzero constant and simply state answers from a calculator.
This is for progressing via algebra to a point at which a quadratic (possibly tacit in a cubic with no
constant) can be identified.
dM1: Attempts to solve the resulting (possibly tacit) 3 term quadratic equation, usual rules. Allow solutions
from calculator as long as a quadratic can be identified in their equation. You may need to check their
values (both must be correct).
A1: Either all three x coordinates stated as answers (the 0 must be present but may have been given earlier), or
one correct coordinate pair from (
) (
)
−
8, 153.6 , 6,14.4
−
A1: All three of (
) (
) (
)
, accept as x =…, y =… for each as long as they are paired
−
8, 153.6 , 6,14.4 , 0,0
−
correctly. The (0,0) may be stated earlier, so check the whole solution if it is not listed at the end.
2
8 (a) (i)
x
=
4,f ( )
′
x
=
10,
f ( )
′
x
=
3
x
+
k x
10
=
3 4
+
4
2
k
k
=
...
1
10
= ×+ ×
3 2
k
16
k
=
*
4
x
=
4,
y
=
12
on
y
=
10
x
+
c
12 10 4 c
=
×+
(ii)
c
= −
28
1
3
−
1
f ( )
′′
x
x
2
x
=
+
2
2
11
f (4)
′′
=
4
3
1
2
3
f( )
x
=
2
x
+
x
+
d
12
1
3
Uses
P
(4,12)
12
=
2 8
×+
×
4
+
d
d
=
...
12
3
1
28
3
f( )
x
2
x
2
x
=
+
−
12
3
Note: If work is done out of order (e.g. if an attempt to find f(x) is made in part (a) but no attempt is
made at (c)) then allow the M1A1 marks for differentiation in (b) and integration in (c) for the work in
part (a) as long is it is clear it is an attempt at the correct process. However, e.g. an attempt at integration
with a decrease in power should not be awarded the marks for (b).
(a)(i)
1
M1: Attempts to solve f (4)
′
=
10
to find k. Alternatively, substitutes x =4 and k
into f’(x) and
=
4
evaluates.
1
A1*: Shows that
with a correct intermediate line where powers/roots have been evaluated, and
k =
4
with no incorrect steps. Alternatively, correctly reaches f’(4) = 10 and conclude k
(ii)
M1: Substitutes
in
x
=
4,
y
=
12
y
=
10
x
+
c
Alternatively, substitutes x = 4, y = 12 and m = 10 into
y
−
y
=
m
(
x
−
x
)
1
1
A1: Correct value for c which may be embedded in the equation, e.g.
y
=
10
x
−
28
, as long as no
incorrect value for c is subsequently given.
(b)
M1: Differentiates with attempt to decrease power by 1 in each term. This should be to a form
1
−
2
but allow if a slip in one power has occurred before differentiating.
Ax
+
B x
1
1
3
−
1
3
−
A1ft:
x
2
+
x
Allow for
2 x
2
+
2
k x
without k replaced or follow through their k if they use a
2
2
different value than the one given. Need not be simplified.
11
A1:
f (4)
′′
=
o.e.
4
(c)
M1: Integrates with attempt to increase each power by 1. This should be to a form
px
if a slip in one power has occurred before integrating. There is no requirement for + d for this
mark. NB If they try starting from f’’(x) then ignore any spurious constant term in f’(x) for this
mark.
3
1
3
f( )
x
2
x
2
x
(
d
)
A1ft:
=
+
+
with no requirement for + d for this mark. Allow with k for this mark or
12
follow through their value of k if they use a different value from the one given. Need not be
simplified.
dM1: Substitutes
into
to find ''d''. (Must have had a constant of integration but
x
=
4,
y
=
12
y
=
f( )
x
need not be called d.) They must be using a value for k at this stage.
3
1
28
A1:
2
3
f( )
x
=
2
x
+
x
−
Allow y = ... or just the expression with no f(x).
12
3
1
9 (a) (i) Stretch parallel to the xaxis
×
or stretch parallel to the yaxis
×
2
2
0
(ii) Translate by the vector
(or translate up by 12 (units))
12
(b) (i) 12
x
2
x
−
=
(
)
12
=
2
+
1
x
12
2
−
1
(
)
x
=
×
=
12
2
−
1
*
2
+
1
2
−
1
(
)
2
Alt (i)
12
−
x
=
2
x
1
2
−
x
=
2
x
x
+
24
x
−
144
=
0
(
)
−
24
±
2
4
2
−
4
×−
144
12
x
=
= −
12
±
4
+
8
=
−
12 12
±
2
2
2
(
)
0
12 12 2
12
2
1 *
x
>
x
= −
+
=
−
(
)
2
(
)
(
)
2
(ii)
x
=
12
2
−
1
=
144 2 1 2 2
+ −
=
144 3
−
2 2
(
)
(
)
y
=
12
−
x
=
12 12
−
2 1
−
=
12 2
−
2
Or common acceptable alt forms: (
)
P
432
−
288
2
,
24 12
−
2
(a)(i)
M1: Correct type of transformation identified. Score for any of
“Stretch”, x/y“scaling” or “compression”
Condone “contraction”, “enlargement”, “magnify”, “squish”, “shrink” and “elongate”. There
may be other acceptable words, use discretion.
Do not accept such as “multiply each x value by 2”
A1: Completely correct using correct terminology. The direction (x or y or horizontal or vertical), type
(must be scaling or stretch or compression) and the correct scale factor must be given. If there are
any contradictory statements, then it is A0.
Accept e.g.
Stretch or scaling or compression in the x direction with (scale) factor 1
2
Stretch or scaling in the y direction with (scale) factor 2
Note a compression in the x direction by factor 2 is A0.
(a)(ii)
M1: Correct type of transformation identified. Score for e.g.
Translate, shift
Move Up/down
Allow translation vector given as indicating translation
Do not accept such as “add 12 to each y value”.
A1: Completely correct using correct terminology. E.g. “move up by 12” is minimal acceptable answer.
The type (translation or “move” or accept “shift”) direction (up or translation vector or positive y
direction), and the correct distance must be given, but accept the translation vector to represent the
direction and distance. If there are any contradictory statements, then it is A0.
(b)(i)
M1: Sets up equation and attempts to take out a common factor of
x
dM1: Correct attempt to rationalise. Must see evidence that the numerator and denominator both
1
2
multiplied by an appropriate term e.g. seeing the denominator as just 2 1
−, ie
2
+
1
fine.
A1*: Complete proof showing all necessary steps. Condone invisible brackets when rationalising.
There may be variations where they do things in a slightly different order, but the scheme can still apply.
E.g.
(
)(
)
(
)
2
12
2
12
2
1
2
12
2
1
x
=
−
x
x
+
x
=
−
x
+
x
=
−
(
)
(
)
x
2
+
2 1
−−
2
=
12
2
−
1
scores M1 when
x
factored out
(
)
x
=
12
2
−
1
scores dM1A1 when achieving
x
=
...
Alt:
M1: Sets up equation and squares both sides, then rearranges to a quadratic in
x .
dM1: Solves the quadratic (allow if it is called x =..)
A1: Gives correct justification for the choice of roots and completes to the given answer.
(b)(ii)
(
)
M1: Attempts to square
12
2
−
1
achieving a not necessarily simplified form
a
2
+
b
(
)
2
coefficient is miscopied. E.g.
x =
12
2
−
1
=
1 4
4
×
2
−
2 2
+ can score M1.
1
(
)
A1:
x =
144 3
−
2 2
or 432
−
288
2
(
)
B1:
y =
24 12 2
−
or 12 2
−
2
Note: It is possible some will find y before x and attempt a longer method to find x. Score the M at the
point at which they attempt to square out the relevant bracket for their method.
(
k
−
1)
x
6
+
4
x
3
+
(
k
−
4)
=
0
8(b)[3]
f ( )
′′
x
x
2
x
=
+
2
2
11
f (4)
′′
=
4
3
1
2
3
8(c)[4]
f( )
x
=
2
x
+
x
+
d
12
1
3
Uses
P
(4,12)
12
=
2 8
×+
×
4
+
d
d
=
...
12
3
1
28
3
f( )
x
2
x
2
x
=
+
−
12
3
Note: If work is done out of order (e.g. if an attempt to find f(x) is made in part (a) but no attempt is
made at (c)) then allow the M1A1 marks for differentiation in (b) and integration in (c) for the work in
part (a) as long is it is clear it is an attempt at the correct process. However, e.g. an attempt at integration
with a decrease in power should not be awarded the marks for (b).
(a)(i)
1
M1: Attempts to solve f (4)
′
=
10
to find k. Alternatively, substitutes x =4 and k
into f’(x) and
=
4
evaluates.
1
A1*: Shows that
with a correct intermediate line where powers/roots have been evaluated, and
k =
4
with no incorrect steps. Alternatively, correctly reaches f’(4) = 10 and conclude k
(ii)
M1: Substitutes
in
x
=
4,
y
=
12
y
=
10
x
+
c
Alternatively, substitutes x = 4, y = 12 and m = 10 into
y
−
y
=
m
(
x
−
x
)
1
1
A1: Correct value for c which may be embedded in the equation, e.g.
y
=
10
x
−
28
, as long as no
incorrect value for c is subsequently given.
(b)
M1: Differentiates with attempt to decrease power by 1 in each term. This should be to a form
1
−
2
but allow if a slip in one power has occurred before differentiating.
Ax
+
B x
1
1
3
−
1
3
−
A1ft:
x
2
+
x
Allow for
2 x
2
+
2
k x
without k replaced or follow through their k if they use a
2
2
different value than the one given. Need not be simplified.
11
A1:
f (4)
′′
=
o.e.
4
(c)
M1: Integrates with attempt to increase each power by 1. This should be to a form
px
if a slip in one power has occurred before integrating. There is no requirement for + d for this
mark. NB If they try starting from f’’(x) then ignore any spurious constant term in f’(x) for this
mark.
3
1
3
f( )
x
2
x
2
x
(
d
)
A1ft:
=
+
+
with no requirement for + d for this mark. Allow with k for this mark or
12
follow through their value of k if they use a different value from the one given. Need not be
simplified.
dM1: Substitutes
into
to find ''d''. (Must have had a constant of integration but
x
=
4,
y
=
12
y
=
f( )
x
need not be called d.) They must be using a value for k at this stage.
3
1
28
A1:
2
3
f( )
x
=
2
x
+
x
−
Allow y = ... or just the expression with no f(x).
12
3
1
9 (a) (i) Stretch parallel to the xaxis
×
or stretch parallel to the yaxis
×
2
2
0
(ii) Translate by the vector
(or translate up by 12 (units))
12
(b) (i) 12
x
2
x
−
=
(
)
12
=
2
+
1
x
12
2
−
1
(
)
x
=
×
=
12
2
−
1
*
2
+
1
2
−
1
(
)
2
Alt (i)
12
−
x
=
2
x
1
2
−
x
=
2
x
x
+
24
x
−
144
=
0
(
)
−
24
±
2
4
2
−
4
×−
144
12
x
=
= −
12
±
4
+
8
=
−
12 12
±
2
2
2
(
)
0
12 12 2
12
2
1 *
x
>
x
= −
+
=
−
(
)
2
(
)
(
)
2
(ii)
x
=
12
2
−
1
=
144 2 1 2 2
+ −
=
144 3
−
2 2
(
)
(
)
y
=
12
−
x
=
12 12
−
2 1
−
=
12 2
−
2
Or common acceptable alt forms: (
)
P
432
−
288
2
,
24 12
−
2
(a)(i)
M1: Correct type of transformation identified. Score for any of
“Stretch”, x/y“scaling” or “compression”
Condone “contraction”, “enlargement”, “magnify”, “squish”, “shrink” and “elongate”. There
may be other acceptable words, use discretion.
Do not accept such as “multiply each x value by 2”
A1: Completely correct using correct terminology. The direction (x or y or horizontal or vertical), type
(must be scaling or stretch or compression) and the correct scale factor must be given. If there are
any contradictory statements, then it is A0.
Accept e.g.
Stretch or scaling or compression in the x direction with (scale) factor 1
2
Stretch or scaling in the y direction with (scale) factor 2
Note a compression in the x direction by factor 2 is A0.
(a)(ii)
M1: Correct type of transformation identified. Score for e.g.
Translate, shift
Move Up/down
Allow translation vector given as indicating translation
Do not accept such as “add 12 to each y value”.
A1: Completely correct using correct terminology. E.g. “move up by 12” is minimal acceptable answer.
The type (translation or “move” or accept “shift”) direction (up or translation vector or positive y
direction), and the correct distance must be given, but accept the translation vector to represent the
direction and distance. If there are any contradictory statements, then it is A0.
(b)(i)
M1: Sets up equation and attempts to take out a common factor of
x
dM1: Correct attempt to rationalise. Must see evidence that the numerator and denominator both
1
2
multiplied by an appropriate term e.g. seeing the denominator as just 2 1
−, ie
2
+
1
fine.
A1*: Complete proof showing all necessary steps. Condone invisible brackets when rationalising.
There may be variations where they do things in a slightly different order, but the scheme can still apply.
E.g.
(
)(
)
(
)
2
12
2
12
2
1
2
12
2
1
x
=
−
x
x
+
x
=
−
x
+
x
=
−
(
)
(
)
x
2
+
2 1
−−
2
=
12
2
−
1
scores M1 when
x
factored out
(
)
x
=
12
2
−
1
scores dM1A1 when achieving
x
=
...
Alt:
M1: Sets up equation and squares both sides, then rearranges to a quadratic in
x .
dM1: Solves the quadratic (allow if it is called x =..)
A1: Gives correct justification for the choice of roots and completes to the given answer.
(b)(ii)
(
)
M1: Attempts to square
12
2
−
1
achieving a not necessarily simplified form
a
2
+
b
(
)
2
coefficient is miscopied. E.g.
x =
12
2
−
1
=
1 4
4
×
2
−
2 2
+ can score M1.
1
(
)
A1:
x =
144 3
−
2 2
or 432
−
288
2
(
)
B1:
y =
24 12 2
−
or 12 2
−
2
Note: It is possible some will find y before x and attempt a longer method to find x. Score the M at the
point at which they attempt to square out the relevant bracket for their method.
(
k
−
1)
x
6
+
4
x
3
+
(
k
−
4)
=
0

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