Mathematics · Oct / Nov 2021 · Paper 11 · Question 9
YMA01
As printed in the paper
The question as text
9. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable. y y = f(x) P(9, 3) O x Figure 5 Figure 5 shows a sketch of the curve with equation y = f(x) where f(x) = x x > 0 The point P(9, 3) lies on the curve and is shown in Figure 5. On the next page there is a copy of Figure 5 called Diagram 1.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
9(a)[3]
y
=
f ( )
x
+
3
(
)
9,6
y
=
f (2 )
x
(
4.5,3
)
(
9,3
)
P
One correct sketch drawn and labelled correctly
One correct sketch drawn and labelled and with correct point
Completely correct sketches with both points
9(b)[3]
Sets
x
+
3
=
2
x
(
)
3
=
2
−
1
x
(
)
2
+
1
3
(
)
x
=
×
=
3
2
+
1
*
(
)
(
)
2
−
1
2
+
1
9(c)[3]
(
)
(
)
2
x
=
3
2
+
1
⇒
x
=
9
2
+
1
=
...
(
)
9 3
2 2
⇒
x
=
+
,
y =
3 2
+
6
(a)
Check all 3 diagrams and score the best single diagram unless the candidate
clearly indicates which one they want marked by e.g. crossing out the other(s).
M1
One correct curve drawn and labelled correctly:
For f (2x) the curve should start at O and be above and remain above f (x) and not head
back towards it significantly i.e. at least maintain the same gap.
For f (x) + 3 the curve should start on the positive yaxis and be approximately the same
shape as f (x)
A1
One correct curve drawn as above and labelled and with correct point for that curve.
The point does not have to be in the correct relative position – just look for values.
A1
Completely correct sketches with both points correct and at least one correctly labelled
– you can assume the other is the other. Allow f(2x) to cross f(x) + 3 as long as it is
beyond (9, 6) but with no other intersections for x > 0
The coordinates for the transformed P must be indicated on the sketch or if they
are away from the sketch it must be clear which curve they relate to.
For examples see below.
If you are in any doubt use review.
(b)
B1
Correct equation
x
+
3
=
2
x
2x as
2
M1
Writes
x and proceeds to collect terms in
x
Note that this may be achieved via e.g.
3
x
+
3
=
2
x
⇒
x
+
3
=
2
x
⇒+
1
=
2
x
1
1
2
Or e.g.
3
2
3
2
x
+
=
x
⇒
x
+
=
x
⇒
+
=
3
x
3
A1*
Proceeds to the given answer showing at least the steps
(
)
2
+
1
3
(
)
x
=
×
=
3
2
+
1
(
)
(
)
2
−
1
2
+
1
(
)
(
) (
)(
)
or e.g.
3
=
2
−
1
x
⇒
3
2
+
1
=
2
+
1
2
−
1
x
=
x
x
Attempts using e.g.
x +
3
=
score no marks in part (b)
2
Alternative:
B1
Correct equation
x
+
3
=
2
x
M1
x
+
3
=
2
x
⇒
x
+
6
x
+
9
=
2
x
⇒
x
−
6
x
−
9
=
0
6
36
36
±
+
x
−
6
x
−
9
=
0
⇒
x
=
2
x and attempts to solve for
Squares both sides and collects terms to obtain a 3TQ in
x
e.g. using quadratic formula
6
±
36
+
36
(
)
(
)
A1*
=
3
±
3 2
=
3
2
±
1
⇒
x
=
3
2
+
1
2
Simplifies and reaches the printed answer. If they give both answers score A0 but there
is no requirement to explain why the other answer is rejected.
(c)
M1
Attempts to square the given expression to find x. Condone a slip on the 3 (it may
α
+
β
2.
remain 3) but must result in an expression of the form
(Working need not be
shown as long as this condition is met)
(
)
A1
x =
9 3
+
2 2
oe such as
x =
27 18 2
+
B1
y =
3 2
+
6
Note that
y =
18 2
+
27
+
3
is correct but is not simplified so scores B0
Note that working for (c) must be seen in (c) i.e. do not allow working for (c) to be
credited in parts (a) and (b) unless the answers are copied into (c)
Example marking for sketches for 9(a):
M1A1A0
M0A0A0
M1A1A0
M1A1A1
M1A1A0
M0A0A0
M1A0A0
M1A1A1
M1A1A1
M1A1A0
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