Mathematics · May / June 2024 · Paper 31 · Question 6
YMA01
As printed in the paper
The question as text
6. y l P R O x Figure 3 In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. Figure 3 shows a sketch of part of the curve with equation y x 4 7 The line l, shown in Figure 3, is the normal to the curve at the point P (8, 5)
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
6(a)[5]
y
x
2
x
2
d
x
At (
)
−
1
2
8,5 gradient of tangent is (
)
2 4 8
×−
7
=
2
5
5
(
)
Equation for l is
y
−
5
= −
x
−
8
2
2
*
y
−
10
= −
5
x
+
40
5
x
+
2
y
−
50
=
0
3
(
)
4
x
−
7
1
2
(
)
4
x
−
7
d
x
=
(see notes)
6(b)[4]
2
6
8
3
8
(
4
7
)
1
x
−
2
(
)
Complete area =
4
−
7
d
=
+
5
x
2
x
6
7
4
7
4
155
=
6
(a)
1
Attempts to differentiate to achieve (
)
−
M1:
4
±
7
or equivalent. The index does not need to be
a
x
2
processed.
d
y
Alternatively, attempts to differentiate implicitly
2
=
4
−
7
=
y
x
Ay
B
d
x
1
Achieves (
)
−
A1:
2 4
x
−
7
o.e. (which may be unsimplified but the index processed) or using implicit
2
d
y
differentiation achieves
2
y x =
4
d
1
d
y
dM1: Substitutes x = 8 into their (
4
7
)
−
a
x
±
2
or substitutes y = 5 into their
x =
. It does not need to
Ay
B
d
be evaluated. It is dependent on the first method mark but allow this mark to be scored if they have
incorrectly manipulated their derivative before substituting in, or made transcription errors
1
including if they lose the
index from the bracket.
−
2
ddM1: A full method for the equation of l. Look for
substitution of x = 8 (ory =
5
if implicit differentiation) into an attempt at a derivative
the application of the negative reciprocal rulethe use of ()
8,5 with a correct gradient for their value of m to form an equation for the normal.
5
1
−
5
= −
"
"(
−
8)
with the coordinates in the correct positions. If
= −
+ is used they must
y
x
y
x
c
2
m
c =
...
proceed as far as
It is dependent on both of the previous two method marks.
A1*: Correctly achieves 5
+
2
−
50
=
0
(in any order on the same side of the equation).
x
y
There must be a correct intermediate stage of working following their initial equation for l before
5
(
)
achieving the given answer, e.g.
y
−
5
= −
x
−
8
2
y
−
10
= −
5
x
+
40
5
x
+
2
y
2
(b)
Note that if no integration is seen then 0 marks.
3
Integrates to achieve a correct form (
)
4
x ±
7
M1:
b
2
o.e. (the index does not need to be processed)
1
u
2
3
May use substitution e.g.
=
4
±
7
d
(the integral must be correct but allow a slip
u
x
u
Qu
2
4
on the coefficient for the integrated expression)
3
Correct integration (
)
4
x −
7
2
A1:
(which may be unsimplified) with or without a constant of integration.
6
Condone poor notation e.g. if the integral sign and/or dx is still present. Index needs to be processed.
3
u
2
Using the substitution method score for e.g.
where
u
=
4
x
−
7
6
dM1: Full method to find the shaded area. Note the area of the triangle is 1
2 ×
(10
−
8) 5
×
=
8
3
(
)
Look for
b
4
x
±
7
2
+
5
o.e. (Provided this is seen then it is sufficient to proceed to the answer)
7
4
The limits must be correct for the integration
May see integration used for the triangle
8
10
5
8
5
10
1
3
(
)
(
)
2
e.g.
4
x
±
7
d
x
+
−
x
+
25
d
x
b
4
x
±
7
+ −
x
+
25
x
2
2
2
4
7
7
8
8
4
4
It is dependent on the previous method mark, but you do not need to see the limits substituted in.
If they do not use the given equation of the straight line l then this mark cannot be scored.
Condone poor notation provided the intention is clear.
25
3
u
2
If using the substitution you may see e.g.
+5 (send to review if unsure)
6
0
155
5
A1:
6 or exact equivalent e.g.
25 6 or 25.83
provided all the previous marks have been scored.
(
)
(
)
2 sin
+
45
°=
cos
−
60
°
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