Mathematics · Oct / Nov 2020 · Paper 11 · Question 2
YMA01
As printed in the paper
The question as text
2 2. f(x) = 3 + 12x – 2x
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
2(a)[3]
b = 2
(
)
2
... ...
±
x
±
3
(
)
( )
(
)
2
f
x
=
21 2
−
x
−
3
2(b)[4]
R is (
)
0,
−
4
or “h” = 4
(
)
2
f ( )
x
−
7
=
14
−
2
x
−
3
⇒
x
=
...
or
f ( )
x
−
7
= −+
4 12
x
−
2
x
2
⇒
x
=
...
(
)
NB
x =
3
±
7
(
)
1
(
)
Area
=
×
"3
+
7 "
−
"3
−
7 "
×
"4"
2
=
4 7
Notes
(a)
B1
b = 2 implied from their expression or stated. Eg look for
−
2(
x
±
...)
2
(
)
2
Beware of expressions such as
21
+
2
x
−
3
which would be B0
M1 Sight of (
)
2
x ±
3
A1
21 2
(
3
)
2
. Condone
21 2
(
3
)
2
or they may state the values for a,b and c. If there is a
−
x
−
−
x
+ −
contradiction between what appears to be their final expression and values for a, b or c which are
incorrectly stated afterwards then treat this as isw.
(b)
R is (
)
B1
0,
−
4
or “h” = 4 (The coordinate may be seen on a diagram as a point of their triangle or
quadratic so allow just 4
− rather than the full coordinate to be indicted on the yaxis ). It may
also be implied by their working.
M1 Attempts to solve f ( )
x −
7
=
0
Score for an attempt to solve a 3TQ of the form:
2
x
2
12
x
C
0
where
3
or
A
2(
x
3)
2
0
where
A ≠
21
−
+
+
=
C ≠
−
−
=
or if they use their part (a)
" " " "(
a
b
x
" ")
c
2
0
d
" "(
b
x
" ")
c
2
0
where d
−
+
=
⇒
−
+
=
≠
a
(It must have a different y intercept to the givenf ( )
x or their f ( )
x from part (a) but the
coefficients of x and
2x must remain the same.)
Condone slips in their working. They should be solving their quadratic using either the formula
or completing the square. (usual rules for solving). Allow to be solved directly from a
calculator. If only the roots are written, then these will need to be checked.
dM1 Fully correct method for the area of their triangle. It is dependent on the previous method mark.
Score for values embedded in 1
(
) "
intercept"
where
α β are the roots of their 3TQ.
,
×
β
−
α
×
y
−
2
This should be 1
×
(2
×
...) "
×
y
−
intercept"
oe
2
Alternatively, if their quadratic has a negative yintercept they may find the area of a large right
angled triangle and subtract the area of the small rightangled triangle. Look for expressions of
the form:
1
1
×
β
×
"
y
−
intercept"
−
×
α
×
"
y
−
intercept"
2
2
They may even attempt finding all the lengths of the triangle, applying the cosine rule to find an
angle and then applying the area sine rule. (See diagrams below).
Alternatively, they may apply the shoelace method:
1 (3
(
) (
)
+
7) 0
× +
(3
−
7) ( 4)
× −
+ ×
0 0
−
0 (3
×
−
7)
+ × + −
0 0
( 4) (3
×
+
7)
=
4 7
oe
2
You may see other credit worthy methods so if in doubt send to review.
A1
4 7 Cao (Allow e.g.
=
2 28,
112
). They may work in decimals and state the correct exact
answer of 4 7 which can score full marks if it follows from a correct method.
Note
1
(
)
f ( )
x
+
7
=
0
⇒
28
−
2(
x
−
3) or 10 12
2
+
x
−
2
x
2
⇒
x
=
3
±
14
⇒ Area
=
×
2 14
×
10
2
would score B0M1M1A0
α
2 7
α
2 7
β
35°
32
−
6 7
−
4
−
4
32
+
6 7
32
+
Large triangle = 6
+
2 7
=
11.29...
Small triangle = 6
−
2 7
=
0.708...
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