Mathematics · Oct / Nov 2020 · Paper 31 · Question 3
YMA01
As printed in the paper
The question as text
3.. y O x Figure 1 Figure 1 shows a sketch of a curve with equation y = f (x) where f (x) = 2 x + 3 x > 1 x − 4 1 4
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Show the answer and mark scheme
3(a)[4]
=
(
)
d
x
4
x
−
1
1
1
1
(
)
(
)
(
)
(
)
4
x
−
1
× −
2
2
x
+
3
×
2 4
x
−
1
−
4
x
−
1
4
x
8
2
2
2
−
×
=
(
)
4
x
−
1
1
3
(
)
(
)
4
x
−
1
4
x
−
1
2
2
Turning point where d
y
3(b)[3]
x =
0
⇒
x
=
2
d
Find value of f at
x
=
2
⇒
f ( )
x
=
7
Hence range is f
7
(a)
1
1
(
)
(
)(
)
d
p
4
x
−
1
−
q
2
x
+
3
4
x
−
1
−
y
2
2
M1
Attempts the quotient rule and achieves
=
condoning slips in coefficients
(
)
d
x
4
x
−
1
1
1
Allow the denominator to appear as (
)
or even (
)
4
x −
1
2
4
x −
1
2
4
1
Alternatively attempts the product rule with (
)
and (
)
−
2
x +
3
4
x
−
1
and achieves
2
d
y
1
3
(
)
−
(
)(
)
−
=
A
4
x
−
1
2
±
B
2
x
+
3
4
x
−
1
2
d
x
(
)
1
(
)
(
)
1
4
x
−
1
× −
2
2
x
+
3
×
2 4
x
−
1
−
For a correct unsimplified d
y
d
y
2
2
=
A1
x . Look for
oe via the quotient rule
(
)
d
d
x
4
x
−
1
d
y
1
3
(
)
−
(
)(
)
−
Look for
=
2 4
x
−
1
−
2 2
x
+
3
4
x
−
1
oe via the product rule
2
2
d
x
1
Multiplies both numerator and denominator by (
4
1
)
dM1
x −
2
and simplifies to a linear numerator, or correctly takes
1
a common factor of (
)
−
4
x
−
1
out of the numerator to leave a linear factor.
2
3
If the product rule is used look for a correct attempt to factor out (
)
4
x
−
1
−
2
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