Mathematics · Oct / Nov 2025 · Paper 11 · Question 10
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As printed in the paper
The question as text
10. y l C R P Q O x Figure 4 In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. Figure 4 shows a sketch of the curve C with equation 1 3 2 y 2 x x 2 x 5 2 The line l is the normal to C at the point P where x = 0 The line l also intersects C at points Q and R as shown in Figure 4.
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10(a)[6]
d
y
d
y
3
2
2
y
=
2
x
+
1
x
−
2
x
+
5
=
6
x
+
x
−
2
so at
x =
0,
x = −
2
2
d
x
d
y
=
x
+
5
Equation of normal at P is
1
2
1
1
3
2
Q is where
y
=
2
x
+
x
−
2
x
+
5
meets
y
=
x
+
5
2
2
1
1
3
2
3
2
+
−
+
=
+
+
−
=
2
x
x
2
x
5
x
5
4
x
x
5
x
0
2
2
(
)
5
2
(
)(
)
x
4
x
+
x
−
5
=
0
x
4
x
+
5
x
−
1
=
0
x
= −
4
Notes
(a)
M1: Attempts to find the gradient of the curve at P.
It is for attempting to find the value of d
y
x when x = 0
d
1
There must be some evidence of differentiating e.g.
2
x
3
→
...
x
2
or
x
2
→
...
x
2
or
−
2
x
→−
2 or 5
→
0
followed by the substitution of x = 0 which may be implied.
Alternatively deduces that the gradient of the curve at P is −2
1
1
(
)
A1:
Equation of normal at P is
y
=
x
+ oe e.g.
5
y
−
5
=
x
−
0
2
2
1
3
2
dM1: Sets
2
x
+
x
−
2
x
+
5
=
kx
+ where k is not their gradient of the tangent at P and collects
5
2
terms to one side. Must be kx + 5 but condone a miscopy of the equation of C.
3
2
A1:
Correct cubic equation with terms collected and all on one side e.g.
4
x
+
x
−
5
x
=
1
5
3
2
equivalent e.g.
2
x
+
x
−
x
=
0.
The “= 0” may be implied by their attempt to solve.
2
2
3
2
+
+
=
ddM1: Cancels or factorises out an x term from an equation of the form
ax
bx
cx
0, , ,
and solves resulting 3TQ by any means including a calculator.
Condone factors that are inconsistent with their 3TQ so just look for the correct or correct ft
roots.
3
2
(
)(
)
+
−
=
+
−
=
=
Allow e.g.
4
x
x
5
x
0
x
4
x
5
x
1
0
x
...
2
( )(
)
−±
1
1
−
4 4
−
5
3
2
4
x
+
x
−
5
x
=
0
x
=
=
...
Do not allow e.g.
×
2 4
If roots of the 3TQ are just written down you may need to check.
They cannot just write down the roots of the cubic equation. Decimals should be correct to 2 sf
or truncated to 2sf.
Depends on the previous method mark.
5
A1:
x = −
oe only. Ignore any attempts to find the y coordinate and ignore any incorrect attempts
4
5
at finding the other x coordinate provided
x = −
is selected.
4
Note: full marks are available for those with an incorrect d
y
x that still leads to a gradient at P of −2
d
Beware in part (a) that it is possible to reach the correct answer fortuitously as follows:
d
y
=
3
+
2
−
+
=
2
+
−
y
2
x
1
x
2
x
5
6
x
x
2
2
d
x
d
y
1
2
2
x =
0
6
x
+
x
−
2
=
0
x
=
,
−
d
2
3
And then incorrectly assuming the gradient of the normal to C at P is 1
2 and that l has the
y
=
x
+
5
equation
1
2
This will generally score no marks in (a) it is entirely fortuitous.
If you are unsure if a particular response deserves credit then use Review.
10(b)[4]
d
y
1
x =
2
+
−
=
Sets
6
x
x
2
d
2
2
12
x
+
2
x
−
5
=
0
−±
2
4
+
240
=
x
24
−+
1
61
x
=
12
Notes
(b)
M1: Sets their d
y
2
equal to 1
+
−
≠
x of the form
ax
bx
2, ,
a b
0
2
d
5
2
A1:
Correct 3 term equation
2
+
−
=
or equivalent e.g.
+
−
=
12
x
2
x
5
0
6
x
x
0
2
The constant terms must be collected but the terms do not need to be all on one side so allow e.g.
2
12
x
+
2
x
=
5
1
2
+
−
=
dM1: Correct attempt to solve a 3TQ resulting from
ax
bx
2
.
2
Allow calculator solutions for solving their 3TQ and condone factors that are inconsistent with
their equation so just look for the correct or correct ft roots.
Note that only the positive solution may be seen which is acceptable.
If values are just written down you may need to check. Decimals should be correct to 2 sf or
truncated to 2sf.
−+
1
61
x
=
A1:
only.
12
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