Mathematics · Oct / Nov 2023 · Paper 31 · Question 1
YMA01
As printed in the paper
The question as text
1. A curve has equation y = f (x) where 2 – 5x + e x x ∈ f (x) = x
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
1(a)[2]
(
)
f(1)
=
1 5
−+
e
= −
1.281...
<
0
(
f (2)
=
)
2
2
−×
5 2
+
e
2
=
1.389...
>
0
x is continuous over the interval [
]
As there is a change of sign and f( )
1, 2
then there is a root *
(b)(i)
x =
5 1 e
×−
1
=
awrt1.5105
2
(ii)
α=
awrt 1.7340
Notes
(a)
Attempts f(1) and f(2) with substitution seen or at least one correct to 1 d.p. rounded or
M1:
truncated and considers their signs. Note showing f(1)f(2) < 0 is a consideration of signs and is
sufficient for the “sign change” part of reasoning for the A1.
A1*: Must have
both f(1) and f(2) correct (as expressions or correct values imply these, need not be labelled)reason, which must mention continuity and state or indicate sign change in some wayconclusion, “hence root”, or accept e.g. hence “f(x)=0 between x =1 and x = 2”1(b)[3]
(i)
M1:
Attempts to find
2x using the iteration formula. Implied by sight of 1 embedded in the formula
or awrt 1.5
A1:
awrt 1.5105
(ii)
A1:
awrt 1.7340 provided M1 has been scored. Allow 1.734 with trailing zero omitted.
Note 1.7340 only may be from a graphical calculator, so scores M0A0A0. There must be evidence (i.e.
the M is scored) of an attempt at least one iteration first.
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