Mathematics · May / June 2023 · Paper 21 · Question 11
YMA01
As printed in the paper
The question as text
11. A sequence u1, u2, u3, ... is defined by un+1 = b – aun u1 = 3 where a and b are constants.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
11(a)[2]
u
=
b
−
3
a
2
(
)
2
u
=
b
−
a b
(
−
3 )
a
=
b
−
ab
+
3
a
3
11(b)[3]
3
+ −
3
+ −
(
−
3 )
=
153
oe
b
a
b
a b
a
3
+
(
a
+
9)
−
3
a
+
(
a
+
9)
−
a a
((
+
9)
−
3 )
a
=
153
2
a
2
−
10
a
−
132
=
0
a
2
−
5
a
−
66
=
0
*
a = −
6
11(c)[3]
(
)
b
=
9
−
6
=
3
u
= −
3 3
−
6
=
21
or
u
=
9
−
2
a
=
9
−×−=
2
6
21
2
2
(a) Check next to the question. If there is a contradiction between the answer next to the question
and the answer in the main body of the work then the answer in the main body of the work takes
precedence.
(
)
=
−
3
B1
u
b
a
oe
2
(
)
=
−
(
−
3 )
B1
u
b
a b
a
oe isw
3
(b)
M1
Attempts to add 3, their
u and their
u and sets equal to 153. Condone slips in their copying of
2
3
their expressions for
u and
u . Invisible brackets may be implied by further work.
2
3
=
+
9
May be in terms of a or b only if they substitute in
b
a
at the same time.
e.g. 3
+
+ −
9
3
+
+ −
9
(
+ −
9
3 )
=
153
a
a
a
a a
a
dM1
Attempts to substitute
b
=
a
+
9
into their equation and then attempts to collect terms to achieve a
3TQ in a (condone the omission of = 0). The three terms do not all need to be on the same side of
the equation. It is dependent on them having a
u term which would lead to an
a term. Do not
2
3
be too concerned by the mechanics of their rearrangement for this mark. It is dependent on the
previous method mark.
A1*
Achieves the given answer with no arithmetical errors seen. Condone recovery of missing
brackets, provided the recovery is before the final answer. Must see at least one intermediate stage
of working following an equation in a only, before proceeding to the given quadratic equation.
(c)
B1
a = − seen or used in (c). Ignore the positive root for this mark.
6
M1
Attempts to substitute their negative root into
b
=
a
+
9
to find a value for b and then uses these in
their expression for
u to find a value for
u
2
2
Alternatively, attempts to substitute both their negative root and
=
+ into their expression for
9
b
a
u and proceeds to find a value for
u
2
2
Ignore working relating to where they repeat the calculations to find
u using their positive root
2
May be implied by their answer (you may need to check this)
Note: For candidates who have two roots of the same sign then this mark can be awarded for using
their smaller value (i.e. most negative/least positive) and proceeding to find a value for
A1
21 only cao (if two answers are reached they must select 21 or reject the other answer)
Correct answer on its own scores full marks.
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