Mathematics · Oct / Nov 2024 · Paper 41 · Question 2
YMA01
As printed in the paper
The question as text
2. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable. The curve C1 has equation 4 2 8 x ∈ y x 10 x The curve C2 has equation 2 y 2 x 7 x ∈ Use algebra to prove by contradiction that C1 and C2 do not intersect. (4)
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Show the answer and mark scheme
16
64
1536
nx provided it is not clearly on the denominator. e.g. condone 1
Condone the position of
16 x
but not 1
1
16x . Ignore terms with higher powers of x. Do not accept
16 x
1
(b)
2
M1: Attempts to substitute
into their expansion from part (a) to achieve a fraction and
x =
3
attempts to find the reciprocal of that fraction. May be implied by their fraction. You may need
to check this on your calculator. They may attempt to find an additional term which is fine. Do
not be concerned as to whether the fraction is correct for their binomial expansion if they have
shown the substitution. Condone slips including losing one of their terms.
2
3
1
1
2
1
2
7
2
A
B
"
+
+
+
"
=
scores M1
2
16 3
64 3
1536 3
B
A
2851
1
÷
scores M1
5184
1
on its own scores M0
2
3
1
1
2
1
2
7
2
"
+
+
+
"
2
16 3
64 3
1536 3
2851 or 2333
5184
A1:
12851. Isw once a correct answer is seen. Correct answer scores M1A1.
2
Assume that
C and
C do intersect so that
x
4
+
10
x
2
+
8
=
2
x
2
−
7
1
2
(and has real solutions).
(
)(
)
x
4
+
8
x
2
+
15
=
0
x
2
+
5
x
2
+
3
=
0
x
2
=
...
or
−±
8
8
2
−×
4 15
x
4
+
8
x
2
+
15
=
0
x
2
=
2
or
(
)
2
x
4
+
8
x
2
+
15
=
0
x
2
+
4
−=
1
0
x
2
=
...
2
−
x = −
5,
3
e.g. The two values of x2 < 0 (so
x
4
+
8
x
2
+
15
=
0
has no real roots) which is not
possible so
C and
C do not intersect
1
2
Note that B0M1A1A1 is possible
B1: Sets up the contradiction by
C and
1
C / the curves intersect/meet/cross
2
o.e. (the equation must be correct)
x
4
+
10
x
2
+
8
=
2
x
2
−
7
Condone no reference to the equation having real solutions.
M1: Attempts to solve a threeterm quadratic equation in x2 by any valid (noncalculator)
method. Usual rules apply. If the roots are just stated then M0A0A0 follows.
A1:
−. Allow e.g.
−
x = −
2
5,
3
m = −
2
5,
3
Do not condone e.g.
y = −
5,
− without reverting back to
3
x =
2
...
or making valid deductions
about e.g.
2x being negative
A1: This mark can only be scored provided the previous M1A1 has been scored.
Acceptable conclusion which
2x being negative which is not possible or the equation has no real roots or
x = −−
2
3, 5
but
x …
2
0
C and
C not intersecting
1
2
x … (not just
2
0
x >>
2
0
)
Alternative considering the sign of each term / the expression
B1: As above in main scheme notes
M1: Either one of:
4
8
2
… e.g.
0
4
8
2
15 0
2
3
0
term. e.g.
x
+
x
x
+
x
+
> Allow to compare with 0 e.g.
x +
≠
2x or
4x e.g.
x …
2
0
x
2
(
x +
2
8)
>>
0
or
(
x +
2
8)
>
0
)
2
x +
2
4
…
16
or
x + …
2
4
4
Do not be concerned by > or …
A1: Correct threeterm quadratic in x2 and correct deductions that e.g.
x … and
4
0
x …
2
0
depending on their expression (condone > 0)
This mark cannot be scored if the expression they use to make their deductions is incorrect.
2x and (
)
e.g.
x
2
(
x +
2
8)
is positive (or zero) since
x +
2
8
are both greater than (or equal to)
zero is A1
e.g.
x
4
+
8
x
2
… since
0
2x and
4x are both
… is A1
0
e.g.
x + … since
2
4
4
x … is A1
2
0
e.g.
x
4
+
8
x
2
… is A0
0
A1: This mark can only be scored provided the previous M1A1 has been scored.
Acceptable conclusion which refers to
gives no solutions o.e. e.g.
cannot equal zero,
x
4
+
8
x
2
+
15
=
0
x
4
+
8
x
2
+
15
e.g.
x
4
+
8
x
2
cannot equal 15
−
e.g.
x +
2
4
cannot equal 1
−
C and
C not intersecting
1
2
x … (not just
2
0
x >>
2
0
).
Do not accept e.g. 3
−„
0
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